Metaphor? An infinite series ∑n=1∞an can be thought of as ∑n=1Nan+∑n=N+1∞an. So there is the Nth partial sum plus the "other" terms of the series, an infinite tail. These other terms I may sometimes call TN. I like to think of this maybe as some sort of animal. The partial sum is the body, and the infinite tail is ... well, the tail. The question of whether the series converges or not maybe is analogous to whether the whole weight of the animal is finite (this is a good analogy only for series whose terms are all positive -- we will deal later with series whose terms change sign). The weight will be finite exactly when the infinite tails→0 as n→∞. In fact, the first "few" terms of a series have nothing to do with convergence! You can change them, and the convergence of the series won't change at all. (If the series converges, the sum will change, but whether or not the series converges won't be changed.)

Maybe the simplest example
If all of the a_n's are 1, then the infinite series is 1+1+1+1+1+... and the partial sums are, well: S₁=1 and S₂=1+1=2 and S₃=1+1+1=3 and ... well, I hope you are convinced that S_N=N. The sequence {N} does not converge, so the infinite series 1+1+1+1+1+... diverges.

A simple and very deceptive example
My instructional strategy now is a bit dubious. I will present one of the (historically!) most deceptive examples of series. It is notorious and defies intuition. It is a very famous infinite series.
The harmonic series is the infinite series ∑_n=1^∞1/n. Sometimes we might write this series as 1+1/2+1/3+1/4+... and the "..." is supposed to indicate, hey, you know the pattern, you understand the formula, etc. There are examples of infinite series where I certainly don't instantly "see" what ... means for those series. The harmonic series occurs in many physical applications, and also arises in analyzing lots of computer algorithms.

The sequence of the individual terms, {1/n}, isn't very complicated. It was one of our first sequence examples. The important question here is to understand the partial sums. Well, S₁=1 and S₂=1+1/2=3/2 and S₃=1+1/2+1/3=11/6. This isn't helping. There is no known explicit formula for the partial sums of this series. I found decimal approximations for a bunch of partial sums. Look at the table to the right, please.

N	SN
1	1.00000
2	1.50000
3	1.83333
10	2.92897
100	5.18738
1,000	7.48547
10,000	9.78761
100,000	12.09015
1,000,000	14.37273

I emphasize that I know no simple shortcut for actually computing these darn numbers. The decimal approximations above were the result of lots of divisions and additions (an unreliable timing of the S_1,000,000 computation is about 5 seconds). Just so you know what the heck these numbers mean, what would S_1,000,001 be compared to the last table entry? Well, S_1,000,001=∑_n=1^1,000,0011/n=(∑_n=1^1,000,0001/n)+(1/1,000,001)=S_1,000,000+(1/1,000,001). Whew! Here are the actual (now 10-digit because the result wouldn't be observable in 5 digits!) approximations:
S_1,000,000=14.39272672 and S_1,000,001=14.39272772. They are different!

So what happens?
What can we conclude about the convergence of this series? The serious answer is nothing. We have weighed the body, S_{some numbers}, but we have no idea what the size of T_{these numbers} is: the "infinite tail" might be very thin (hey, less than 1/1,000,000) but it is very very long. Several ideas may occur.

Primitive idea #1 It diverges because we're adding up infinitely many numbers, and therefore things get to be too darn large.

Primitive idea #2 It converges because, although we're adding lots of numbers, the steps between the sums get smaller and smaller, so the sum can't get very large.

A simple argument
The numbers above are, ultimately, not very persuasive. But let me show you an elementary argument which will allow us to make a good decision. Please realize that this is a very clever argument! We first realized that the partial sums of the harmonic series were all positive and they were increasing: S_N<S_N+1. I will just look at some special partial sums. I wrote these lines on the board:
    S₁=1=2/2
    S₂=1+1/2=3/2
    S₄=1+1/2+1/3+1/4>1+1/2+(1/4+1/4)=1+1/2+2(1/4)=4/2
    S₈=1+1/2+1/3+1/4+1/5+1/6+1/7+1/8>1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)=1+1/2+2(1/4)+4(1/8)=5/2
Of course, in class (here is a reason to attend the lectures!) this was done sort of interactively. At one point I stopped and asked people what was the next line I should write, and I got the correct (?) answer.

We can group the partial sums when N is a power of 2. Here is the general result: S_2^k≥(k+2)/2. For example, I asked if we could give one specific partial sum which would be larger than 100. Uhhhh ... if we took k=200, then (k+2)/2=101, larger than 100. So S_2²⁰⁰ is bigger than 100.
How long would it take to add up this sort of partial sum? Well, 2²⁰⁰ is a large number. It is about 1.6·10⁶⁰. And, let's see, I would hope I could do about 10¹¹ additions and divisions in a second (that is an overestimate). And there are about 1/3 of 10⁸ seconds in a year. So there would be maybe 1/3 of 10¹⁹ additions and divisions in a year. The age of the universe is sometimes estimated to be about 20 billion years, or 2·10¹⁰ years. So ... in one "universe age" we could compute maybe 2/3·10²⁹ terms (this is a cruddy computational model, by the way). So all we would need is about ... uhhh ... 10³¹ universe ages. This sort of exercise is useful if it convinces you that computing some big partial sum is silly.

Therefore ...
Does the harmonic series converge? If it did converge, then eventually the partial sums would have to get close to a limit. But the estimate we have just seen shows that the partial sums get bigger and bigger and bigger. So the harmonic series diverges.
The discussion above is very special. We'll get a number of general tricks within a few lectures which can be used to show divergence rapidly. I just want to give you a reason for the divergence result. For the harmonic series, the infinite tail is very thin, but it is really really very long! Intuition?

Another kind of series
Let me turn to a series that behaves the way we'd like. A geometric series is one where successive terms are related by a constant ratio. In algebraic language, the text writes such a series as c+cr+cr²+cr³+...=∑_n=1^∞cr^n–1.
Actually, I just noticed that the text writes this as ∑_n=0^∞crⁿ, so the exponent looks more normal but the index of summation begins with n=0. But I'll continue with the way I actually did it in class.

An (exceptional!) explicit formula of a partial sum
The partial sum of this series is S_N=∑_n=1^Ncr^n–1. Unlike almost any other series, it is possible to find an explicit formula for this S_N. Here is the idea, which I think is usually shown in some high school (and maybe even some middle school!) math or physics courses:
    S_N=c+cr+cr²+cr³+...+cr^N
Multiply by r:
    rS_N=cr+cr²+cr³+cr⁴+...+cr^N+1
Subtract the second equation from the first:
    S_N–rS_N=c+(all the inside terms cancel!)–cr^N+1
Factor the left-hand side:
    (1–r)S_N=c–cr^N+1.

Now solve for S_N: if r is not equal to 1, S_N=[(c–cr^N+1)/(1–r)].

Convergent geometric series
Now if |r|<1, powers of r, that is, the sequence {rⁿ}, must approach 0. So in the formula for S_N, when |r|<1, I know that r^N+1→0 and S_N→c/(1–r). When |r|>1, the series diverges (the powers of r grow).
If |r|<1, then the geometric series with first term c and ratio between successive terms r, ∑_n=1^∞cr^n–1, converges, and its sum is c/(1–r).

Example #1
Here is something which I hoped that many people saw before college. The infinite repeating decimal 0.731731731... represents a rational number (a quotient of integers). What rational number does it represent?

Here the most interesting problem is recognizing the implied geometric series. Decimal notation is very clever and conceals some true subtleties. So 0.731 itself means 731·(.001) which is 731/1000. What about 0.000731? This is 731·(.000001) which is 731·(.001)·(.001). That is, 10^–3·10^–3=10^–6. Therefore 0.000731 is 731/(1000)². And similarly 0.000000731 is 731/(1000)³. So we see (maybe not so "clearly"!) that:
0.731731731...=[731/1000]+[731/(1000)²]+731/(1000)³]+....
We therefore recognize that the repeating decimal indicates an infinite series whose first term, c=731/1000, and whose constant ratio between successive terms is r=1/1000 (this is certainly less than 1). The sum is then c/(1–r)=[731/1000]/(1–[1/1000])=[731/1000]/[999/1000]=731/1000.

Digression: how maybe this is done in earlier "grades"
A teacher might say the following:
Consider Q=0.731731731... and try to figure out another way of looking at Q. Well, 1,000Q=(1,000)·(0.731731731...)=731.731731731... so then:
1,000Q=731.731731731... and subtract
  Q=0.731731731...
The result is 999Q=731, so that Q=731/999. Ain't that nice! My "excuse" for pointing out the geometric series approach is that I want to show you a use of geometric series, and also to maybe expose a bit of the structure of the decimal system, which is actually a very clever and intricate idea.

Maybe sometimes we might think of a picture of the sequence, but this really has limited use as you will see. To the right is a picture of the first 6 elements of the sequence {1/n}, sitting on the number line.

Preliminary remarks about convergence
Clearly (!) as n gets large, 1/n gets closer and closer to 0. This behavior is abbreviated, not too surprisingly, by limit notation. So here we would write lim_n→∞1/n=0. What could this mean in more precise language? Well, one implication that occurs is that as n increasing, 1/n gets close to 0. So maybe lim_n→∞a_n=L means that, as n gets large, a_n should get close to L. L is called the limit of the sequence.

Another example
This example will look initially quite silly, but thinking about it is useful. So the sequence is {(–1)ⁿ}. As a list, the elements of the sequence are (–1)¹=–1, (–1)²=1, (–1)³=–1, (–1)⁴=1, etc. This sequence has only two distinct values, and these values depend on the parity (even/oddness) of its argument, n. When n is odd, the sequence value is –1, and when n is even, the value is 1. A rudimentary picture of the sequence is shown to the right. Notice that there is something missing from this sequence -- the dynamic aspect as the sequence wiggles and hops left and right. Every element of the sequence is in the picture and the picture isn't very helpful to me at all. Does this sequence converge? Historically an answer to this question wasn't obvious. People finally decided that the answer should be no. The only numbers that are authentic candidates for the limit of this sequence are 1 and –1. If we were going to use sequences as ways of getting better and better approximations to a root of an equation, then saying that the root is maybe 1 or maybe –1 is probably not the best answer. People usually want one specific answer. So the definition of limit, even if we want only to consider it informally, needs to be stated slightly differently.

Other examples?
We looked at several additional weird examples. Let me see if I can remember them.

1. a_n=1/n if n≠209,342 and, if n=209,342, a_n=–100,000.
2. a_n=–42 if n<10¹⁰. Otherwise, a_n=1/n.

Better definition of convergence
We'll say that lim_n→∞a_n=L is true if for n large enough, a_n gets close and stays close to L.
Certainly {–1)ⁿ for certain values of n is close to 1 (heck, it is equal to 1, and I don't know how much closer it could be). But also we can always find even larger values of n (just use any large enough odd n) so that (–1)ⁿ is not at all close to 1. So the sequence {(–1)ⁿ} does not converge and does not have any limit. Another word is used: the sequence {(–1)ⁿ} diverges. The problem is that, although the sequence is sort of tame in that it never gets too large positive or too large negative, it wiggles and never sort of stabilizes.
There is a formal definition of limit in section 10.1 of the text. Later in your professional career, as you do more computations, you may likely need to work with that definition. The word "close" for example, is used to mean |a_n–L|, and usually people want to be able to control the size of |a_n–L| by selecting some number N so that when n>=N, then the size of |a_n–L| is "small" (and this is some number to be specified in practice). Maybe this is too detailed for Math 152 and a first visit to the definition. But here is the (generally accepted) definition:

Suppose {a_n} is a sequence. We write lim_n→∞a_n=L if, given any ε>0, there is a positive integer N (which may [usually does!] depend upon ε so that: if n≥N, then |a_n–L|<ε.

More examples

Algebra and sequences
Some straightforward limit facts still are true, such as:
If lim_n→∞a_n=L₁ and if lim_n→∞b_n=L₂, then lim_n→∞a_n+b_n=L₁+L₂ and lim_n→∞a_n·b_n=L₁·L₂.
Things like this are in the textbook, and I'd like to concentrate my attention on more subtle behavior.

{5^1/n}
Let's look at the sequence {5^1/n}. The first few terms, in decimal form, are these: 5., 2.236067977, 1.709975947, 1.495348781, 1.379729661, 1.307660486, 1.258498951. A picture of the first 6 of these is to the right. Maybe things here are not totally clear. In fact, one of the reasons I want to discuss this example is that things are not clear. We made some preliminary observations: certainly any root of 5 would have to be positive, in fact any root would have to be bigger than 1. And also any root would have to be less than 5. So what we know is that 1<5^1/n<5. The numbers in the sequence are "trapped" inside the interval from 1 to 5. I asked the class if just knowing this information was enough to guarantee that the sequence converged. After some thought, one student came up with the following example: {3+(–1)ⁿ}. Again depending on parity, the values of this sequence are either 2 (n odd) or 4 (n even). This sequence does not converge although its values are inside the interval from 1 to 5.

Comments on boundedness and convergence of sequences
If a sequence convergences, then the numbers in the sequence are bounded. (The reason is, essentially, that if there is convergence, all but a few of the terms are close to the limit of the sequence. Then the whole sequence is trapped near the limit and finitely many other numbers, and that can be put inside a bounded interval.)
The converse of the statement above is not true in general. I mean: If a sequence is bounded, then it may not converge. We've already seen several examples of this statement.

Here is a very brief discussion of some of the words used with logical implications.

Back to {5^1/n}
Here is one way of analyzing the sequence. If a_n=5^1/n, then ln(a_n)=ln(5^1/n)=(1/n)ln(5). But (1/n)ln(5) is just a constant multiple of the first sequence we looked at, and that sequence had limit 0. So lim_n→∞(1/n)ln(5)=0. Here is a general and quite useful fact:

If f(x) is a continuous function, and if {b_n} is a convergent sequence with lim_n→∞b_n=L, then the sequence {f(b_n)} converges, and lim_n→∞f(b_n)=f(L).

More examples

Some students noticed something like the following argument. Take n to be an integer larger than 100. (You will soon see why I selected 100: it is because it makes some computations and comparisons easier). Then

/ (50)n \     (50)(50)(50) ··· (50)
| ----- | =  ---------------------- 
\   n!  /      1·2·3·4·5···(n–1)n  

If T_n(x)=∑_j=0ⁿ[f^(j)(a)/j!](x–a)^j, then the difference between this and f(x) is
[f⁽ⁿ⁺¹⁾(c)/(n+1)!](x–a)ⁿ⁺¹ for some c between a and x.

How to "compute" sin(1/3) with error at most .00001
Here a=0 and x=1/3 and f is sine and we need to find n so that the absolute value of the error is at most .00001 (which is 1/100,000).

We know from the last lecture what the Taylor polynomials centered at 0 for sine look like. Here is the beginning of them, evaluated at x=1/3:
     (1/3)–(1/3)³/3!+(1/3)⁵/5!–(1/3)⁷/7!+(1/3)⁹/!9 ...
We would like to know when to stop. The absolute value of the error in this case is |f⁽ⁿ⁺¹⁾(c)|/(n+1)!](1/3–0)ⁿ⁺¹ for some c between 0 and 1/3. All of the derivatives of sine are +/–sine and +/–cosine, so a simple overestimate of |f⁽ⁿ⁺¹⁾(c)| for any n and any c is just 1. This makes life quite easy. Now consider that |error|≤(1/3)ⁿ⁺¹/(n+1)!. I plugged in numbers by hand and we learned that if n+1=6, then (1/3)⁶=1/729 and 6!=720. the product of these two is more than (700>²=490,000 and since this is bigger than 100,000 (remember, we are reciprocating, taking "one over") we're done if n+1=6. This means n=5. Therefore

sin(1/3)=(1/3)–(1/3)³/3!+(1/3)⁵/5! with |error|≤.00001

In fact, a silicon friend informs me that (10 digit accuracy) sin(1/3)=0.3271946968 and the sum on the right-hand side above is 0.3271947874. We actually have a different of about 0.0000001. This is pretty good for almost no work.

How to "compute" sin(3) with error at most .00001
Here a=0 and x=3 and f is sine and we need to find n so that the absolute value of the error is at most .00001 (which is 1/100,000). Maybe we could hope that sin(3)=(3)–(3)³/3!+(3)⁵/5! also, with just a small error. But now (remembering that the derivatives are all overestimated by 1 again) |error|≤3ⁿ⁺¹/(n+1)!. And, when n=5, we get 729 on top and 720 on the bottom. This looks bad. We have exponential growth on top. In fact, things will be fine.

Ideas about growth of common functions
Polynomials are sums of powers multiplied by constants. P(x)=33–4x⁷+66x^1,234 is a polynomial. Exponentials, which here I'll consider only with base >1, also grow. Here is one: Q(x)=1.0002^x. Now P(2) is about 1.95235·10³⁷³, pretty darn large. And Q(2) is just 1.004004, no sot big. BUT Q(x) ultimately gets larger, much larger, than P(x). For example, at 10⁷, P's value is about 6.6·10^8,639 and Q's value is about 1.64·10^8,677. Q is much larger. The discrepany gets incredibly bigger.

But how does factorial figure into this? Let me consider a big exponential, say R(x)=543^x, and this with S(x)=x!. Now R(10) is about 2.228·10²⁷ while S(10), as we saw last time, is only the miserably small (by comparison!) 3,628,800, seven digits compared to 27 digits. Now let's try, say (not at random -- I just had some fun discovering the crossing point) x=1472. Then the exponential is about 4.22·10^4,025 and the factorial is about 7.28·10^4,025. The factorial has pulled ahead, and you should see that each time we increase x by 1, the exponential will increase by "only" a factor of 543 while that factorial increasing by a steadily increasing (!!) factor, beginning for this x with 1473. Wow!

If you tell me that big numbers don't mean much, well then I will respond in several ways. First, in real-life biological/chemical/physical applications, you don't know the widely varied time scales and rate constants which occur. Look at the logistic curves again. Also, if we are interested in limits as x→∞, well, there are lots and lots and lots of numbers "up there" as x grows, and 1,473 is not very large from that point of view.

What I have written is an effort to support this hierarchy ("a system in which grades or classes of status or authority are ranked one above the other"):

As x→∞, Polynomial growth is (eventually) slower than Exponential growth which in turn is (eventually!) slower than Factorial growth.

Back to sin(3). Well, we considered, I think 3ⁿ⁺¹/(n+1)!. I believe we chopped it up multiplicatively this way:
    3ⁿ⁺¹/(n+1)!=(3⁷/7!)({other powers of 3}/{product of numbers all at least 8})
The reason we did this was that we knew in the second fraction, everything would always be less than 1/2. What do I mean? Well, look at 3⁴/(8·9·10·11)= (3/8)(3/9)(3/10)(3/11). Each term is at most 3/8 and that's less than 1/2. The product of 4 of them is less than (1/2)⁴=1/16 which is less than 1/10. So each 4 new "n"'s gains us at least one decimal place. Therefore to get 3ⁿ⁺¹/(n+1)! less than 1/10⁵ we will need at least five groups of 4 n's (weird way to write this!). But also we need to deal with what we started with, which is 3⁷/7!. I was advised that was about .4 (it is, in fact, more like .433928...) so I bet n+1=20 will work. That means I could take the Taylor polynomial of degree 19 for sine centered at 0 and evaluate it at 3, and get the desired accuracy.

So my answer, now written in summation form, is
    sin(3)=∑_j=0⁹(–1)^j3^2j+1/(2j+1)! with accuracy of less than .00001.

The strange things in the summation are all how people write the sine Taylor polynomials in more compact form. The (–1)^j flips the sign, and since we start at j=0, we start with a +1. The (2j+1) in both the exponent and the factorial give us the desired odd integers, and the top bound in the sum, j=9, means that we go up to 2(9)+1=19.

And now the numbers: sin(3) is 0.4112 00080 and T₁₉(3) is 0.4112 00078. In fact, we get accuracy to .00000 0001, a heck of a lot better than we aimed for. This is not an error -- I did not request the best possible Taylor approximation. I just wanted, as happens very frequently in practice, some approximation which will work.

The exponential values
I don't have time to write the diary entries, but we described how to compute e^1/3 and e³. These weren't too much more difficult than the sine computations. In fact, the Taylor polynomials involved are simpler. The estimations are slightly different.

A different game Suppose f(x)=x^1/3 and a=8. Then f(8)=2. I wanted to find T₃(x)=f(8)+f´(8)(x–8) +[f´´(8)/2](x–8)² +[f´´´(8)/3!](x–8)³.

We computed derivatives and plugged in. A silicon friend told me that the answer is 2+1/12*(x–8)–(1/288)*(x–8)^2+(5/20736)*(x–8)^3.

Remainder or error estimate in the textbook
There are many ways of estimating the error between the function f(x) and the Taylor polynomial T_n(x). I think I've actually only ever used one estimate, which is given in your book on page 506 and is called the ERROR BOUND. The text also calls the error the remainder. Here it is:

Suppose we want to consider f(x), maybe hard to compute, and the nth degree Taylor polynomial centered at a, Tn(x), which we hope is easier to compute. Then |Tn(x)–f(x)|≤K[|x–a|n+1/(n+1)!] where K is some overestimate of |f(n+1)(x)| between x and a.

I looked at this error estimate for f(x)=x^1/3, n=3, a=8, and x in the interval [6,8]. I chose an interval centered at 8 so that |x–8|≤2. To the right is a picture containing graphs of both y=x^1/3 and y=T₃(x) as x goes from 0 to 16.

Look at the curves. Near x=8, the curves look very, very close. A bit farther away, the curves separate. Indeed, the best way I can see which curve is which is that x^1/3 goes through (0,0) so that must be the green curve.

I hope you can see that knowing the polynomial and using it instead of x^1/3 might be advantageous, maybe, maybe (the idea is that polynomials are very easy to compute and other functions might not be).

To the right is another picture. Look at it very carefully please. It is also a graph of both y=x^1/3 and y=T₃(x) as x goes from 6 to 10. Both curves are displayed. But I can only see one curve. Maybe if I look really really really closely, maybe I can see some dots of two colors but I am not sure. How close are these curves?

Well, let's try to use the error estimate in the textbook. That error estimate is quoted above. So an overestimate of the absolute value of the difference between T₃(x) and f(x) for x in [6,10] is
|T_n(x)–f(x)|≤K[|x–a|ⁿ⁺¹/(n+1)!] where K is some overestimate of |f⁽ⁿ⁺¹⁾(x)| between x and a.
How can we use this? Well, here f(x)=x^1/3 and n=3 and a=8 and the x's involved are between 6 and 10. So we need to understand K[|x–a|ⁿ⁺¹/(n+1)!]. This becomes K[|x–8|⁴/4!]. Of course 4! is 24. What about |x–8|? This is the distance from x to 8. If x is in [6,10], the largest distance to 8 is 2, so |x–8| is always ≤2. Good: now we know that the error is at most K[16/24] (because 2⁴ is 16). What about K? To compute K we need f⁽⁴⁾(x). Well, here are f(x) and a few of its derivatives:
    f(x)=x^1/3;     f´(x)=(1/3)x^–2/3;     f´´(x)=f⁽²⁾(x)=–(2/9)x^–5/3;     f⁽³⁾(x)=(10/27)x^–8/3;     f⁽⁴⁾(x)=–(80/81)x^–11/3.
I am interested in the absolute value (the size) of the error, so I will forget the minus sign. Therefore I need to know how large (80/81)x^–11/3 can get on [6,10]. Because of the negative sign in the exponent, this is a decreasing function of x, and its largest value is at the left-hand endpoint. So the biggest this can be is (80/81)6^–11/3. O.k.: I gave up -- we used a calculator. This number is about (rounding up!) .0014, and I'll take that as my value for K.

Therefore the error is at most (.0014)(16/24) and this is about .00092. Now look carefully at the picture. Do you think you can see a difference in height of less than .001? The difference between two of the "hashmarks" on the vertical scale is .05, and the graphs of the functions are separated by one-fiftieth (that's 1/50) of that difference. I don't think that the pixels in the picture are small enough to show this!

The picture to the right shown here is on a different interval, from 6 to 6.25 for x. The vertical scale is also different. Each vertical hashmark here represents a height difference of .001, and now, in the part of this interval which is farthest away from 8, near 6, I hope you can indeed see two curves, separated by just a little bit. The error estimate actually gives useful quantitative information.

We can use the polynomial instead of x^1/3 in the interval [6,10] and loose little information. Here are two more pictures (yes, I like pictures, maybe too much). The left picture is just T₃ (it is a cubic polynomial with no critical points [like x³+x). The right picture has both T₃ in blue again, and a broad "curve" in green of x^1/3, with the latter just shown on [6,8].

What are sequences and what are they used for?
A sequence is a sort of ordered list of numbers (precise definition below).
Much computation is done sequentially. For example better and better approximations to roots of equations are gotten by Newton's method. This scheme creates numbers which (ideally!) get closer and closer to the root desired, even though there may be no exact algebraic formula for this root. Definite integrals (except in classrooms!) can only rarely be computed exactly. Instead, there are usually collections of approximations which are computed, and this collection of approximations is used to give a "value" for the definite integral. In this part of the course (really this is the last major part of the course!) we will study sequences and accompanying ideas.

Formal definition of a sequence
A sequence is a real-valued function whose domain is the positive integers. I will begin with what I hope are some rather simple examples but the later examples will be complicated enough for everyone.

Example and notation
Well, one example of a sequence is a sequence whose n^th term or value or element (all these words are used!) is 1/n. So this is the function f(x)=1/x when x is a positive integer. The usual notation for this sequence is {1/n}. Yeah, a different notation is used -- usually the sequence "function" doesn't get shown. The formula is enclosed by the braces { and }. The letters that are used inside the braces are usually n and m and p and q. If needed, we might refer to the formula with a subscript. So, for example, we might write a_n=1/n. Therefore, a₅=1/5 and a₁₇=1/17. Also a_n+1=1/(n+1) and a_5m=1/(5m). Notice especially the last two equations which might look strange. The stuff "down" in the subscript is the argument to the function which defines the sequence. Please try not to get confused. For example, if a_n=1/n, then a₄=1/4, but a₄₊₁=a₅=1/5 and a₄+1=(1/4)+1=5/4.
People also frequently "define" functions by listing their first few members. So this sequence might look like 1, 1/2, 1/3, 1/4, 1/5, ... and the "..." is supposed to indicate that the reader should now recognize the pattern. To me this is another use of the strange mathematical word "clearly". Clearly sometimes people will recognize the pattern, but also clearly many times there will be difficulty.

I will use some workshop time tomorrow continuing this topic and catching up with the syllabus.

I wrote this on the board before the class began. It is a result I'll use several times in the initial discussion. Here:

Suppose we have a differentiable function which is 0 at two distinct points. Then the derivative of the function will be 0 at least once between these two values.

Important and useful
Much of the remainder of the semester will be devoted to material which is extremely useful to everyone involved with engineering and science. Today is likely to be your first exposure to the ideas which are generally used to compute values of functions. It is almost certain that these ideas are at the center of how the calculators and computers you use compute (and graph) most functions. You should know something about these ideas.

When I taught this last year I jumped right into an example, and didn't show people any of the structure. The examples, even the easiest, can be intricate. I don't feel I was too successful here last year, so I want to try a different approach today. Some of what I say will not be in the textbook. I will begin with a totally unmotivated question. The question will look somewhat weird.

What's K?
Suppose f(x) is a differentiable function, a and b are numbers with a≠b, and we consider the equation:
    f(b)=f(a)+f´(a)(b–a)+[f´´(a)/2](b–a)²+K(b–a)³.
Then there is some number K which makes this equation correct. Why? Because since a≠b, (b–a)³ isn't 0, and I could solve for K in this whole messy equation. So I want to investigate what K is, and get another way of writing K. To do this I will use the Mean Value Theorem repeatedly in the form quoted above.

Consider the function
    G(x)=f(x)–(f(a)+f´(a)(x–a)+[f´´(a)/2](x–a)²+K(x–a)³).
What do I know about G(x)? Well, I know that G(a)=f(a)–(f(a)+f´(a)(a–a)+[f´´(a)/2](a–a)²+K(a–a)³) so that G(a)=0. I also know that G(b)=f(b)–(f(b)+f´(a)(b–a)+[f´´(a)/2](b–a)²+K(b–a)³). This is also 0 because K was chosen so that this is true. So G(a)=0 and G(b)=0. Therefore (MVT/RT above) there is some number in between a and b, I'll call it c₁, so that G´(c₁)=0.

Now let's compute the derivative of G(x). There are many letters around. I am differentiating with respect to x. With this in mind, I see:
    G´(x)=f´(x)–(0+f´(a)1+[f´´(a)/2]2(x–a)+3K(x–a)²).
Now what? Well, G´(a)=f´(a)–(0+f´(a)1+[f´´(a)/2]2(a–a)+3K(a–a)²) is 0 because of the a–a's and because the f´(a)'s cancel. We also know that G´(c₁)=0. Now MVT/RT applied to the function G´(x) tells us that its derivative is 0 somewhere between a and c₁. That is, there is c₂ between a and c₁ so that G´(c₂)=0.

Now let's compute the derivative of G´(x), again being careful.
    G´´(x)=f´´(x)–(0+0+[f´´(a)/2]21+3·2K(x–a)).
Of course we consider G´´(a) which is f´´(a)–(0+0+[f´´(a)/2]21+3·2K(a–a)) and this is 0. Also G´(c₂)=0. So MVT/RT again applies to tell us that there is c₃ between a and c₂ with G´´´(c₃)=0. Wow.

The derivative of G´´(x):
    G´´´(x)=f´´´(x)–(0+0+0+(3·2·1)(K)1).
Now we know when x=c₃ this is 0. So we know that f´´´(c₃)–(3·2·1)K is actually equal to 0. We can solve this for K, and get K=[f´´´(c₃)/(3·3·1)]. c₃ is between a and c₂ and c₂ is between a and c₁ and c₁ is between a and b so that c₃ is itself between a and b.

What's going on?
Here is what we know: If
    f(b)=f(a)+f´(a)(b–a)+[f´´(a)/2](b–a)²+K(b–a)³.
then there is a number c between a and b so that K=[f´´´(c)/(3·2·1)] (three of the ´ in this). This turns out to be the beginning of a marvelous and successful computational strategy.
So what's your problem with all this? You might claim to not understand what the heck is going on, and, more particularly, why any moderately sane person might want to go through these algebraic contortions. Yes I totally agree with you. But it turns out (sit here, look at the fireworks (?) that follow) this is extremely useful. There have been centuries (!!) of thought involved in preparing and using all this stuff -- it is really clever.

Taylor polynomials
This is copied from page 502 of your text. The Taylor polynomial of degree n for the function f(x) centered at a (wow, what a collection of words!) is
T_n(x)=f(a)+f´(a)(x–a)+[f´´(a)/2](x–a)²+[f⁽³⁾(a)/3!](x–a)³+...+[f⁽ⁿ⁾(a)/n!](x–a)ⁿ
There are a whole bunch of things to discuss. Let's see. First, if you've never seen it before, the appearance of the excitement mark, !. This is called a factorial. The value of the factorial of a positive integer is the product of the integer together with all of the integers less than it down to 1: n!=n(n–1)(n–2)(n–3)···(3)(2)(1). Here is a very brief table of factorials:

n	1	2	3	4	5	6	7	8	9	10
n!	1	2	6	24	120	720	5,040	40,320	362,880	3,628,800

The major thing you should notice right now is that the factorials grow very big very quickly. That's computationally important. I also should mention that most people define 0! to be 1. That's so certain formulas are easier to write (really). (It turns out to be possible to define factorials of other numbers. For example, in 251, you can define and compute (1/2)! -- wait for that.)

More notation is in such things as f⁽⁴⁾(a). This means the fourth derivative of f evaluated at a. So f´´(a) can also be written as f⁽²⁾(a), and even just f´(a) is f⁽¹⁾(a). Again, in order to make writing certain formulas easier, most people think that f⁽⁰⁾(a), the zeroth derivative of f evaluated at a (so no derivatives are done!), should just be f(a).

If all of this notation is clear, then here's another, very compact way to write the Taylor polynomial.
T_n(x)=∑_j=0ⁿ[f^(j)(a)/j!](x–a)^j.
I hope you can see where the zero factorial and zeroth derivative make this much easier to write.

Example 1
Let's get T₈(x) for sin(x) and a=0. So we need derivatives, and here they are:

Derivative #	Function	Value at a=0
0	sin(x)	0
1	cos(x)	1
2	–sin(x)	0
3	–cos(x)	–1
4	sin(x)	0
5	cos(x)	1
6	–sin(x)	0
7	–cos(x)	–1
8	sin(x)	0

The numbers in the last column, together with the factorials, are the coefficients which build the Taylor polynomial. I emphasized with all of these examples that noticing patterns is the way to go. In the case of sine, the derivatives repeated every four. I am not cheating by choosing something excessively simple. Almost every function I know that arises in modeling physical and geometrical situations has patterns in its derivatives, and part of the fun (?) is finding these patterns. Well, now I can write T₈(x):
T₈(x)=[0/1](x–0)⁰+[1/1](x–0)¹+[0/2](x–0)²+[–1/6](x–0)³+[0/24](x–0)⁴+[1/120](x–0)⁵+
   +[0/720](x–0)⁶+[–1/5,040](x–0)⁷+[0/40,320](x–0)⁸
Only a nitwit or a very pedantic math instructor would write it that way. Most people would drop the 0 terms, change (x–0) to x, make a few other notational simplifications,and get
T₈(x)=x–[1/6]x³+[1/120]x⁵+[1/5,040]x⁷
Then there are some questions.
What is T₄(x)? It must be x–[1/6]x³.
What is T₇(x)? It must be =x–[1/6]x³+[1/120]x⁵+[1/5,040]x⁷, in this case just the same as T₈(x).
What is T₁₀(x)? I bet it is the same as T₉(x), and this would be T₁₀(x)=x–[1/6]x³+[1/120]x⁵+[1/5,040]x⁷–[1/362,880]x⁹.

Let me explain why T₉(.5) and sin(.5) agree to so many decimal places. Well, we need a generalization of the first computation we did. Here is the result, a version of what's called Taylor's Theorem:

If T_n(x)=∑_j=0ⁿ[f^(j)(a)/j!](x–a)^j, then the difference between this and f(x) is
[f⁽ⁿ⁺¹⁾(c)/(n+1)!](x–a)ⁿ⁺¹ for some c between a and x.

Example 2
We get T₈(x) for f(x)=e^x and a=0.

Derivative #	Function	Value at a=0
....blah.......blah.......blah.......blah.......blah..........blah...
any n you want!	ex	1
....blah.......blah.......blah.......blah.......blah..........blah...

T₇(x)=1+x+[x²/2]+[x³/6]+[x⁴/24]+[x⁵/120]+[x⁶/720]+[x⁷/5,040]
It's supposed to be easy: this is a technique people actually use.

How good is T₇(–.4) as an approximation to e^–.4? Here the remainder is     [f⁽ⁿ⁺¹⁾(c)/(n+1)!](x–a)ⁿ⁺¹ when a=0, x=–.4, and n=7 becomes e^c[1/(8!)](–.4)⁸. Now c is between –.4 and 0 and since e^x is increasing, the biggest value occurs at c=0, where e⁰=1. (It is supposed to be easy, otherwise people would not use it!) So the error is at most [1/(8!)](–.4)⁸. This is less than .00000002. The "true value" of e^–.4 is 0.670320046035639 the value of T₇(–.4) is 0.670320030476191. I had a machine do these computations, of course. The reason I put quotes around "true value" is that the machine used Taylor polynomials to do the computation, of course.

Modeling this story to solve the problem
Declaring that the quantity FROG is directly proportional to the quantity TOAD is language meaning that FROG=KTOAD for some positive constant K. So if T represents the temperature of the hot object, then Newton's Law of Cooling implies that RATE OF CHANGE OF T=K(difference between T and 30). With the usual notation, realizing that "rate of change" is a derivative, and using T in ^oC and t for time measured in minutes since the object was brought into the room, Newton's Law of Cooling becomes dT/dt=K(T–30) with some unknown constant, K. What about the other information in the story? I've collected the information in a possible graph of the temperature of the object to the right. Our guess for the solution curve (the description of the temperature as time changes) is in magenta. It should tend to 30 as t→∞.

Solving the model equation
dT/dt=K(T–30) is separable, so we get (I'm beginning to skip steps) ∫[1/(T–30)]dT=∫K dt so that ln(T–30)=Kt+C. We have two constants, K and C, to be determined. But we know two "chunks" of information about the temperature, (10,80) and (20,40). Therefore (plugging in):
(10,80) gives us ln(50)=K(10)+C and (20,40) gives us ln(10)=K(20)+C.
We have two linear equations in two unknowns. We can solve them. Use your favorite method, or:
Double the first equation and subtract the second to get 2ln(50)–ln(10)=C. I will simplify using log properties, so C=2ln(50)–ln(10)=ln({50}²)–ln(10)=ln(2500)–ln(10)=ln(250).
Let's put this value of C in the first equation. We have ln(50)=K(10)+ln(250) so 10K=ln(50)–ln(250)=ln(50/250)=ln(1/5) and K=(1/10)ln(1/5).

The solution and the solution curve
The implicit form of the equation is therefore ln(T–30)=(1/10)ln(1/5)t+ln(250). Let's exponentiate, and use exponentiation properties and log properties:
T–30=e^{(1/10)ln(1/5)t+ln(250)}=e^{(1/10)ln(1/5)t}e^ln(250)=e^{ln[(1/5)(1/10)t]}(250)=250(1/5)^(1/10)t, and, finally, T=250(1/5)^(1/10)t+30 or maybe 250(5^–(1/10)t)+30. Whew! Please note that on an exam I would not require or expect so much "simplification" but when you go out and work on such problems, people usually like the answers written in such "simple" ways. Since T(t)=250(1/5)^(1/10)t+30, the initial temperature, T(0), was 250+30=280. The temperature after one hour is T(60) (60 minutes=1 hour) and that is 250(1/5)⁶+30 which is 30.016 (no, I did not compute this!). Also please note that as t→∞, since (1/5) is a positive number less than 1, 250(1/5)^(1/10)t→0 so that T(t)→30 as we had guessed.
To the right is a graph of T(t)=250(1/5)^(1/10)t+30 in magenta with dashed black lines at 80, 40, and 30. The 80 and 40 lines cross at 10 and 20, respectively, and the curve overlays the 30 line as t grows. The curve seems to cross the t=0 line at T=280.

Bacteria again ...
The differential equation dy/dx=Ky with K constant (so the solutions can be written as y=Ce^Kx) is widely used for modeling drug levels in blood and radiocarbon dating (K negative) and for modeling growth of bacteria, yeast, fungus, etc. when K is positive. x is usually time. How valid is this? Well, if we had E. coli which would "divide every twenty minutes" we should be 50 feet under E. coli even if the darn things started only a few weeks ago. In fact, for such growth there are limits to nourishment and other factors which retard growth as the population increases. Sometimes such things are bundled into the model as the "Carrying Capacity" of the system. So people have considered many differential equations which model growth. Let me discuss a very simple case of a widely accepted model equation.

A better model: the logistic equation
Consider the differential equation dy/dx=y(2–y). If y is close to 0 but positive, then the right-hand side seems to be about "2-ish" multiplying y. So this is very much like dy/dx=2y, exponential growth. If y is close to 2 but less than 2, then dy/dx is (really small number) multiplying y, so y will grow very slowly. Therefore 2 serves as the Carrying Capacity of this model. I will try to solve a specific initial value problem for this equation, which is an example of the Logistic Differential Equation. The numbers are selected to make the algebra as easy as possible.

The initial value problem will consist of the differential equation dy/dx=y(2–y) and the initial condition y(0)=1.

1. Separate and integrate: ∫[1/{y(2–y)}]dy=∫dx.
2. The left-hand side needs partial fractions: [1/{y(2–y)}]=(z/y)+(w/{2–y}) so (tops only) 1=z(2–y)+wy. y=0 gives z=1/2 and y=2 gives w=1/2. Now integrate.
3. (1/2)ln(y)–(1/2)ln(2–y)=x+C where the minus sign before one of the 1/2's comes from the antiderivative of 1/(2–y). If x=0, y is supposed to be 1. Then (1/2)ln(1)+(1/2)ln(1)=0+C so C=0 ("select the numbers to make everything as easy as possible").
4. The solution implicitly is therefore (1/2)ln(y)–(1/2)ln(2–y)=x but people like explicit forms. So multiply by 2 and use the ln property to get ln(y/{2–y})=2x. Exponentiate so y/{2–y}=e^2x. We need still to solve for y. Multiply by 2–y and y=e^2x(2–y)=2e^2x–e^2xy and y+e^2xy=2e^2x and y(1+e^2x)=2e^2x. Huh. So clearly ...

But what happens if ...
What if we wanted to look at a different, even just slightly more complicated model, say dy/dx=y(y–1)(y–2) etc. Then separating etc. is possible. But trying to convert the implicit description of the solution to an explicit description, one that can be analyzed more easily, is essentially impossible unless we are very lucky.

A different kind of reasoning
There is a different way to study such equations, one where geometric reasoning is used instead of lots of algebraic computations. When this sort of reasoning applies, getting asymptotic information is usually much easier than one would think possible. The tool we will consider for this way of studying differential equations is called the direction field in your textbook (in others it is called the slope field). I will use this tool to get an approximate idea of the shape and behavior of what are called solution curves or integral curves of the differential equation. These are the graphs of particular solutions of the differential equation.

Direction fields
Let's look at the differential equation y´=x²–x–4y². I can't "solve" this but let me tell you just a small amount of information about the solution near the point (3,1). A solution curve passing through the point (3,1) will have its slope determined by the differential equation. I mean that y´ when x=3 and y=1 will be the value of x²–x–4y² when x=3 and y=1. This is 3²–3–4·1²=9–3–4=2. So the tangent line to the curve at (3,1) will have slope 2. Well, let's think about it. I'll draw a thickened-up line segment of slope 2 at (3,1) and consider some curves going through that point.

Curve #1 I've drawn a little piece of a curve, tangent to the chunk of line (it is formally called a line element, I think). I think this curve, which is increasing, might possibly be a solution to this differential equation near this point. A possibility
Curve #2 Here is another curve going through (3,1). I believe that this could also be the solution curve. O.k.: it is concave down, and I surely don't know what the concavity of the solution curve is at (3,1) without more study. But in my casual way, I have two possible candidates for the solution curve. Certainly it is also increasing near (3,1) because the slope at (3,1), which is 2, is positive. A possibility
Curve #3 Now this candidate is really a different sort of curve. This curve does pass through (3,1) but it is decreasing, and its slope is certainly not 2 at (3,1). This is a losing candidate (!) and certainly can't be a little piece of a solution curve. Not possible!

I don't think that this "reasoning" is profound. I am merely asserting that I can eliminate and (tentatively) accept certain candidate curves as good candidates for solution curves. Now I will extend the reasoning, by drawing lots of line segments at lots of points and see what my brain's visual processing power can tell me about the curves. In fact, (and this is somewhat amazing to me) with some practice, most people can "see" the curves quite clearly. So here are some examples. The direction fields were drawn by computer, but the curves were (badly) drawn by me. In class, I was so proud of our boys and girls who did not quarrel about the crayons!

dy/dx=(1/20)x2
This is the direction field	This is the direction field with some solution curves.
Discussion I had Maple draw a bunch of direction fields. I admit that the instructions took a bit of practice, but once I understood them it wasn't too difficult to produce the pictures, and I honestly tell you that there was a great deal less computation than if I had requested any approximations to solution curves numerically. There are 12·12=144 line elements in each direction field picture. The rather strange numbers which appear in the differential equations (for example, the (1/20) in this equation) were chosen so that the tilts of the direction field elements would be easier to see in these pictures. This differential equation can be solved easily and its solutions are (1/20)(1/3)x3+C. But I want to look at the curves and not the formulas. Here to the left is the direction field without any decoration. To the right is my attempt to draw by "hand" curves which, whenever they touch a line element, have the line elements as segments of tangent lines. I think you can see that the solution curves are increasing, and that the concavity is down on the left and up on the right. That's all I want from this example.
dy/dx=(1/10)xy
This is the direction field	This is the direction field with some solution curves.
Discussion O.k., algebraically this differential equation is still something we can handle fairly easily. This is a separable equation, and we can solve it fairly easily. Let me take the 144 pieces of the direction field and try to draw some curves which, when they touch any part of the direction field, will have that tiny line segment looking like it is tangent to the curve. To me these words more make things harder to understand. Draw some curves! The curves I "drew" are concave up when they are above the x-axis, and they are concave down when they are below the x-axis. I feel that I understand how initial conditions would "evolve" (?) forward and backward. People frequent thing that x is a variable representing time. So an initial condition located in the upper halfplane, above the x-axis, evolves forward (the future) and gets very very large. Also, in the past, it came from something very very large. That's all I want here: just approximate qualitative information about solutions. There is one special solution curve (not drawn here) which is the x-axis, with y=0. You can check that y=0 is a solution of this differential equation. The y values don't change. Such a solution is called an equilibrium solution.
dy/dx=(1/30)(y+2)(y–2)
This is the direction field	This is the direction field with some solution curves.
Discussion I can still solve this algebraically, I think (I haven't tried). This is a variant on the logistic equation which has a nicely worked out solution above. Here the numbers are a bit more random, though. The direction field tells me more in a short amount of time than working algebraically would tell me in a long time. Look at the picture of the direction field, and then look at the solution curved indicated. There is a great deal going on in the picture of the curves, and let me try to pick out some features. First, there are some horizontal lines which are solutions. How can I detect them? If y=Constant is a solution to some complicated differential equation dy/dx=F(x,y), then because the horizontal line always has dy/dx=0, it had better be true the the complicated function F(x,y) is 0 when we plug "Constant" in for y. This may seem complicated to you, but here look at the right-hand side: (1/30)(y+2)(y–2). When is this equal to 0? Without being silly, I think I can write that clearly this occurs for y=–2 and y=+2. The horizontal lines defined by these questions are then solutions to the differential equation dy/dx=(1/30)(y+2)(y–2) because both sides are 0. These solutions don't change with "time" if we think of x as time. Here is some vocabulary which is used when discussing these solutions: they are called equilibrium solutions or steady-state solutions. Please also note that the two equilibrium solutions divide the plane into three chunks. A solution curve which starts in one chunk can't go into a different one. It would have to cross one of the equilibrium curves, and, at that crossing point, there would be two solution curves, which would contradict the Existence and Uniqueness Theorem. For in that theorem is the assertion that each point has at least a little piece of exactly one, only one, solution curve. Now I want to discuss other (non-equilibrium) solutions for this differential equation. Consider first solutions near the equilibrium solution y=–2. If we take initial conditions (x0,y0) with y0 not equal to –2 but near –2, something interesting happens. The curves above y=–2 get moved down as we go futureward (x increases). They get moved down towards –2, and as close to –2 as you want (yes, a limit indeed). And if y0 is below or less than –2, the curves below y=–2 move up towards –2. More precisely, if we take an initial condition (x0,y0) with y0 close to –2, the resulting solution y=f(x) has the property that limx→∞f(x)=2. Nearby solution curves converge to the equilibrium solution y=–2. This situation has its own name: y=–2 is called a stable equilibrium. There are many physical situations (mechanical, chemical, etc.: many!) where learning about stable equilibriums (equilibria?) is very important. In real life, determining an exact measurement (y=–2) may be very difficult. It is far better to know that, hey, starting something off near –2 will have the same long-term effect as starting it at –2. This is useful and neat. Let's look more closely at the other equilibrium solution, y=2, and its neighboring solution curves. Suppose we start near but above y=2. If you look at the picture, you'll see that the direction field pushes (?) the curve away from y=–2 (these curves go to ∞). Similarly, if we start near but below y=2, the curves get pushed away from y=2. The lower curves go towards y=–2, which right now is not interesting: I want to concentrate are things near y=2. So if this differential equation described a physical situation, I would now know that if my initial condition (x0,y0) had y0 exactly at 2, then on the solution curve, I would have y0 always equation to 2 as x0 increased (the future). But if I didn't get y0 exactly equal to 2, I have no ability to predict the future. Maybe y will tend to ∞ or maybe to –2. Here's the vocabulary, which shouldn't be surprising: y=2 is called an unstable equilibrium.
Physical situation Stable equilibrium Think of a ball at the bottom of a bowl. It will just sit there, position not changing. If you jiggle the bowl a bit, then the ball will move, but eventually (if you stop jiggling) it will go towards the bottom again. This "bottom" situation is a stable equilibrium. Unstable equilibrium Turn the bowl over and put the ball on top of the bowl. You may conceive of a situation where you have placed the ball so carefully on the top that is it totally balanced and still, motionless. Ideally the situation would go on forever -- equilibrium. But now jiggle the ball in any way, with a very small jiggle, in any direction. The ball will roll off. This is unstable equilibrium.
dy/dx=(1/30)y2(y–2)
This is the direction field	This is the direction field with some solution curves.
Discussion O.k.: I definitely would not want to try to "solve" this equation algebraically and get explicit solutions. But the direction field approach combined with simple reasoning allows me to find equilbrium solutions and even to detect which are stable and which are unstable, and this can be very useful in practice. I diagnose equilibrium solutions by considering (1/30)y2(y–2)=0. The roots are 0 and 2. So y=0 and y=2 are the only equilibirum solutions for this differential equation. Yes, I admit it: this example is very carefully chosen so that the method works, but the ideas are sufficiently nice so that the method will work even for examples which aren't so "pretty". What about stable and unstable? Again, look at the direction field, and then examine the solution curves which are drawn. Both of the equilibrium solutions are unstable. Nearby solution curves are not "sucked into" the equilibrium solution. Yes, it is true that curves above y=0 get pulled towards y=0, but to be an equilibrium solution, both sides (up and down) must have solution curves which are attracted to the equilibrium, and this does not occur here. Again, both of these equilibrium solutions are unstable. Mr. Priestner commented that the equilibrium solution y=2 sort is is stable when "time" (x) runs backwards, that is, as x→-∞. My comment was that certainly it was true that y=2 attracts solutions, but the word "stable" is usually reserved only for solutions which are attracting as x→+∞, so y=2, in that standard sense, is not stable.
dy/dx=(1/20)(x+y2)
This is the direction field	This is the direction field with some solution curves.
Discussion O.k., the last example for this method is really nearly random. The right-hand side, (1/20)(x+y2), is a low-degree polynomial, but I don't know how to solve the equation (see the remark below if you are curious, though!). There are no equilibrium solutions because there is no constant C which makes (1/20)(x+C2) equal to 0 for all values of x. So here's a complicated situation. But still, consider the direction field, which is easy to have drawn, needing very little computational effort. And consider the curves I drew. I bet that the following occurs: if y=f(x) is a solution to this differential equation, then there will be exactly one critical point on the solution curve, and this critical point will be an absolute minimum. To the right of the solution curve, the function will increase. In fact, it will explode in finite time, that is, there will be some Q so that limx→Q–f(x)=∞, but I don't think the picture necessarily tells that. What about to the left of the solution curve? I bet that somewhere the concavity of the solution changes, and it becomes concave down, and that the curve sort of tapers off to a sort of flat situation. Everything predicted here is actually correct. It would be rather difficult (but possible, I think, but I wouldn't want to do it!) to confirm all this algebraically. I think the direction field idea is worth using as a tool when analyzing differential equations, especially when some quick asymptotic information is wanted.

The real thing Much computation is needed to produce an "explicit" solution to dy/dx=(1/20)(x+y2). To the right is a picture of the solution curve which goes through (0,0). The function defining the curve can't be written in terms of the standard functions you know. The rather simple-looking picture is the result of a bunch of Airy functions combined in very strange ways. I am not inventing all this! The situation is very complicated. Here is the solution in detail, written by a silicon friend of mine: 2/3 2/3 50 x (2/3) 50 x (1/3) (1/6) AiryAi(1, - -------) 150 + 5 AiryBi(1, - -------) 20 3 100 100 1/5 --------------------------------------------------------------------- 2/3 2/3 50 x (2/3) 50 x (1/6) AiryAi(- -------) 3 + AiryBi(- -------) 3 100 100 I don't understand this complicated formula at all. But I do understand the direction field and the solution curves.

A possible QotD
As usual, there wasn't enough time. But if there had been, I would have asked the following question. Find the solution of y´=(e^x/3)/y which satisfies y(0)=2. Be sure to write the equation as y=some function of x. Here is a solution which you should look at after you try the problem yourself.

What's a differential equation?
A differential equation is an equation relating an unknown function and one or more derivatives of the unknown function with some other functions.

Examples
y´=x² Done in 151 (Rutgers Calc 1) ; to be reviewed here.
y´=5y Done in 151 (Rutgers Calc 1) ; to be reviewed here.
y´=y² We'll discuss this here.
y´=xy We'll discuss this here.
y´=x+y To be discussed in your differential equations course.
y´´=–y To be discussed in your differential equations course. This equation governs simple harmonic motion (the movement of an ideal vibrating spring).
y´=x²+y² This can't be "solved" in terms of standard functions.

Order of a differential equation
This is the highest numbered derivative which occurs in the differential equation. In these two lectures we will look only at first order equations. All of the examples above are first order, except one (the simple harmonic motion equation) which is second order. Please note that the differential equations course which most of you will take will indeed study equations of higher order, and that these do occur. Some standard vibrating beam equations studied in mechanical engineering are fourth order, and many of the equations in physics and chemistry are second order.

Story #1
Probably we have all been told that bacteria (usually) reproduce by, say, binary fission. This is more or less correct, and more or less the fact means that the rate of increase of bacteria at any time is directly proportional to the number of bacteria at that time. So twice as many bacteria "now" means that twice as many bacteria are being born now. This is certainly dreadfully simplified, but this approximation works in many circumstances. I wondered, when I first heard this fact, why, if, say, E. coli doubles rather rapidly, shouldn't the world be covered very soon by a layer of E. coli which is 40 feet thick? In fact,

A single cell of the bacterium E. coli would, under ideal circumstances, divide every twenty minutes.
(From Michael Crichton (1969) The Andromeda Strain, Dell, N.Y. p.247)

Story #2
We start with an 800 gallon tank of pure water. It is being filled with a fluid at 50 gallons per minute, and these 50 gallons contain 5 lbs of salt. At the same time, 50 gallons per minute of the solution in the tank is being drained. How much salt is in the tank at any time? How much salt would you expect to be in the tank after a long time?

Let's construct a differential equation which models the salt in the tank. We'll call S(t) the number of pounds of salt in the tank at time t. How much salt is being added? Well, 5 pounds per minute. How much salt is being taken away? This is more subtle, and we had some discussion of our assumptions during class. The simplest analysis, which we will do here, is to assume that the tank contents are mixed well: it is homogeneous. The situation with a large real tank might not match this, of course. But, actually, real containers and tanks sometimes have mixing devices installed to try to match this assumption. Well, if there are S(t) pounds of salt in the tank at time t, and if the tank holds 800 gallons, and if 50 gallons are taken out, then the proportion of 50/800 of the salt is taken out: [1/(16)]S(t). Now we put things together.

dS/dt, the rate of change of the salt, is +5–[1/(16)]S(t). The differential equation is dS/dt=5–[1/(16)]S. We also shouldn't forget that we start with no salt at all in the tank: S(0)=0.
Prediction? What should happen over the long term to the amount of salt in the tank? It starts out at 0, and then increases ... to what? Well, a guess is that the amount of salt in the tank should increase to 80 pounds, which is the same as the salt concentration (1 pound per 10 gallons) incoming. We will see how to solve the differential equation and check this prediction. The solution is discussed here.

Solution of a differential equation
A solution of a differential equation is a function which, when it and all of its relevant derivatives are inserted into the differential equation, makes the equation true for all values of the domain variable. I know this may seem long-winded, but I hope the discussion and examples which follow will shown why such elaboration is necessary.

An example: y´=x²
Well, we know how to solve y´=x²: just integrate. We did this repeatedly last semester (and even this semester). The solutions are y=(1/3)x³+C, where C is any constant. There are infinitely many solutions. A few of them are shown to the right.
The blue curve has C=1: y=(1/3)x³+1. It is the solution curve which goes through (0,1).
The red curve has C=3: y=(1/3)x³+3. It is the solution curve which goes through (0,3).
The green curve has C=–2: y=(1/3)x³–2. It is the solution curve which goes through (0,–2).
The solution curves are just vertical translates, up and down, of each other. They are all the same shape, have the same domain, etc. This situation is rather straightforward, as you will see.

General solution; particular solution
There is some special vocabulary used. The differential equation y´=x² has the general solution f(x)=(1/3)x³+C. When C has a specific value, then the function is called a particular solution. So f(x)=(1/3)x³–2 is a particular solution, and it is the only particular solution which passes through (0,–2). The specification (0,–2) is called an initial condition. That comes from the physical situation where x represents time, and we think that the y-value corresponding to the given x-value represents a certain starting place. Sometimes people write y(0)=–2 as the initial condition. That can confuse me. The combination of an initial condition and a differential equation is called an initial value problem.

An example: y´=y²
Now let's change and consider y´=y². I guess that the general solution is f(x)=1/(C–x). (I'll show you how to guess it also, very soon!) How could you check that my suggestion for a solution actually is a solution? Well, if f(x)=1/(C–x) then, since f(x)=(C–x)^–1, we know that f´(x)=(–1)(C–x)^–2(–1). The first –1 comes from the power, and the second –1 comes from the Chain Rule, so they cancel. But (C–x)^–2, the derivative of the function, is actually the square of (C–x)^–1, the original function. We have now verified that f(x)=1/(C–x) does solve y´=y². Now let's look at some particular solutions.
The blue curve has C=1: y=1/(1–x). It is the solution curve which goes through (0,1). It has domain (–∞,1), and is increasing and concave up.
The red curve has C=1/3: y=1/({1/3}–x). It is the solution curve which goes through (0,3). It has domain (–∞,1/3), and is increasing and concave up.
The green curve has C=–1/2: y=1/(–{1/2}–x). It is the solution curve which goes through (0,–2). It has domain (–1/2,∞), and is increasing and concave down.
These solution curves are not just vertical translates of each other. Their domains are different (yeah, this matters in real life) and the solution curves have different shapes. The particular solutions "blow up" at different numbers. And this is still a fairly simple differential equation.

A big theorem and a joke
When I was young, so much younger than today, I was told the following BIG THEOREM about differential equations.

THEOREM Suppose we have some function of two variables, F(x,y), and we are interested in the differential equation y´=F(x,y), and a solution going through the point (x₀,y₀). Then there always is a solution, and there is exactly one solution.

After I learned about this theorem, I thought that all this worry about differential equations was totally silly -- the theorem tells you everything. This is false. In practice, the theorem doesn't tell you how to compute or approximate solutions efficiently. It doesn't tell you what the domains of the solutions are (this is important in applications). It doesn't tell you the asymptotic behavior of the solutions (how much salt there is after a long time). These questions are very important, and they are the questions which need to be answered in practice. All this makes me think of my favorite math joke.

JOKE Several people are in a hot-air balloon, trying to land over a fog-shrouded countryside at the end of a long day. The balloon dips down low and they see the ground faintly. Spotting a person, one of them calls down: "Where are we?" Some minutes later the wind is carrying them away and they hear faintly, "You're in a balloon!" One person in the balloon gondola says thoughtfully to the other, "It's so nice to get help from a mathematician." The other says, "How do you know that was a mathematician?" The first replies, "There are three reasons: it took a long time to get the answer, it was totally correct, and, finally, it was absolutely useless."

Separable equations
A separable first order differential equation is one which can be written in the following way: dy/dx=F(y)G(x). The right-hand side is a product of some function in y multiplied by some function in x. This is the definition.
I'll describe a procedure which leads, in many cases, to a solution.

1. Separate: put all the y stuff on one side and put all the x-stuff on the other side. The equation becomes dy/F(y)=G(x)dx.
2. Integrate:∫dy/F(y)=∫G(x)dx. Of course, the practicality of this (at least "by hand") depends on the specific functions F and G.
3. Solve for y as a function of x. Again, this may or may not be practical. Examples will help you to understand.

Some examples

• Consider y´=y² which is dy/dx=y². Here F(y)=y² and G(x)=1. Well, the first step gives dy/y²=dx. Now integrate both sides (remember that 1/y²=y^–2, please). We get –1/y=x+C. This can be solved for y fairly easily, and so y=1/(C–x). (Yes, if you are very alert you would have noticed that I changed –C to C, and everyone does that, because I can stuff the minus sign into the unknown constant C.) So at least I've shown you where my guess above comes from. I've forgotten what I did, but I think it was something like: let's find a solution of y´=x²y which goes through the point (1,2). So if dy/dx=x²y then dy/y=x²dx and then ∫dy/y=∫x²dx so that ln(y)=(1/3)x³+C. If we want this to "go through" (1,2) then ln(2)=(1/3)13+C so that C=ln(2)–(1/3). And the particular solution is ln(y)=(1/3)x³+ln(2)–(1/3). Most people, especially in applications, prefer an explicit solution, so an attempt to solve for y is made. I can do this here (the situation is simple) but in general things may be too complicated. The equation ln(y)=(1/3)x³+ln(2)–(1/3) can be exponentiated to give y=e^{(1/3)x3+ln(2)–(1/3)}. Although for me on an exam this would be fine, I would not want to mislead you. Again, especially in applications, appropriate simplification is very useful. Here this is what would be done:
  e^{(1/3)x3+ln(2)–(1/3)}=e^(1/3)x3e^ln(2)e^–(1/3)=2e^{(1/3)[x3–1]}. So the particular solution of y´=x²y satisfying the initial condition y(1)=2 is f(x)=2e^{(1/3)[x3–1]}. Part of a graph of y=f(x) is shown to the right. The function increases very quickly when x is positive.
• Here is an important counterexample: y´=x²+y² is a differential equation which is not separable. This is not totally obvious. Please let me try to convince you.

  If x²+y²=F(y)G(x) is true, I will "explore" the possibilities using values of x and y. For example, if x=0 and y=0, we see that 0=F(0)G(0). Therefore either F(0)=0 or G(0)=0 (or both, of course). What if G(0)=0? Then try x=0 and y=1. The equation becomes 0²+1²=F(1)G(0), and this is 1=F(1)G(0). If G(0)=0, this is impossible! If the alternative holds, that is, F(0)=0, just insert x=1 and y=0 to get a contradiction.

  To the right is a picture of the solution curve to y´=x²+y² which goes through (1,2). The function defining the curve can't be written in terms of the standard functions you know. Quite a bit of computation is needed to produce the rather simple-looking picture (the darn picture needs a dozen Bessel functions combined in very strange ways).

Back to the salt tank ...
The differential equation dS/dt=5–[1/(16)]S is separable. Maybe this is not totally obvious. Look:
5–[1/(16)]S=(5–[1/(16)]S)(1), and 5–[1/(16)]S is a function of S alone, and 1 is a function of t alone.

Let's separate and solve. So dS/{5–[1/(16)]S}=dt, and the right-hand side integrates to t+C. The left-hand side is maybe a bit more intricate. You could substitute: w=5–[1/(16)]S so dw=–[1/(16)]dS and dS=–16dw. The result is –16ln(w)=–16ln(5–[1/(16)]S). I generally guess, get it wrong, and need to guess again. So after integrating we have –16ln(5–[1/(16)]S)=t+C. The initial condition here, a result of the tank originally being filled with pure water, is S(0)=0. So we can get C:
–16ln(5–[1/(16)]S)=t+C become –16ln(5–[1/(16)]0)=0+C and C is –16ln(5).

The solution is –16ln(5–[1/(16)]S)=t–16ln(5). Most people prefer a more explicit formulation, so we solve for S as a function of t.
Divide by –16: ln(5–[1/(16)]S)=–[1/(16)]t+ln(5).
Exponentiate: 5–[1/(16)]S=e^{–[1/(16)]t+ln(5)}.
Some algebra on the right: e^{–[1/(16)]t+ln(5)}=e^–[1/(16)]te^ln(5)=5e^–[1/(16)]t.
Now get S: 5–[1/(16)]S=5e^–[1/(16)]t becomes –[1/(16)]S=–5+5e^–[1/(16)]t which turns into S=80–80e^–[1/(16)]t.

What does the solution look like?
If we believe S(t)=80–80e^–[1/(16)]t, let's check the initial condition:
S(0)=80–80e^–[1/(16)]0=80–80·1=0. Good!
How about the long-range asymptotic behavior? That is, what happens when t gets very large (t→∞)?
If t→∞, then –[1/(16)]t→–∞, so e^–[1/(16)]t→0. The combination 80–80e^–[1/(16)]t must therefore →80, which is what we expected. Let's see what the S(t) curve looks like. But here, unlike in class, I will try to explain the final picture.

Here is 80e[1/(16)t for t between –50 and 0. This is part of an exponential growth curve, and it starts small and increase up to 80, the value at 0. It is concave up.	Now I've flipped the curve across the vertical axis. This is 80e–[1/(16)t for t between 0 and 50. The curve is still concave up, but it is decreasing: since the constant is negative, this is exponential decay.	I flipped the previous curve across the horizontal axis. This is a graph of –80e–[1/(16)t for t between 0 and 50. It is concave down and increasing, from 80 to near 0. We'll get the real picture of S(t) by translating this up 80.

To the right is a graph of S(t)=80–80e^–[1/(16)t for t between 0 and 100. The dashed red horizontal line is at height 80, the asymptotic level of the salt in the tank. You can see that the salt starts at 0, the initial condition, and then increases and sort of curves underneath the line at height 80. The curve is concave down. The difference between 80 and S(t) becomes rather small as t grows.

QotD
If we had the time, I would have asked you to find the solution of y´=y²x³ which goes through (2,1). Be sure to write the solution as y=some function of x.

The idea of polar coordinates You have found a treasure map supposedly giving directions to the burial spot of a chest full of gold, jewels, mortgages, etc., stolen by the Dread Pirate Penelope. The information you have is that A buried treasure is located 30 feet from the old dead tree, in a NorthNorthWest direction. So there you are, on the island. Perhaps the Old dead tree is still visible. You could mentally draw a circle 30 feet in radius around the Old Dead Tree. Then you find the North direction. π/4=45o to the West is NW (Northwest) and then NNW (Northnorthwest) is π/8 towards North (anyway, you decide on the direction). Where that direction intersects the circle is probably where to dig, unless Penelope is tricky, etc.

The whole idea of locating a point in a 2-dimensional setting using distance from a fixed point and angle with respect to a fixed direction is called polar coordinates.

"Standard issue" polar coordinates Fix a point (usually called "the center" or sometimes "the pole" and in most common situations, the origin of the xy-coordinate system). Also fix a direction -- if needed this might be called "the initial ray". Almost always this is the positive x-axis in an xy-coordinate system. Then locate another point in the plane by giving its distance from the center (called r) and by drawing the line segment between the center and the point you are locating. Measure the angle between that and the initial ray (note: counterclockwise is a positive angle!): this is called θ. Then r and θ are the polar coordinates of the point.

An example and the problem with polar coordinates Well, make the standard choices for "the pole" and "the initial ray". Let's get polar coordinates (the values of r and θ) for the point whose rectangular coordinates are x=sqrt(3) and y=1. Of course this is not a random point (sigh). So we consider the picture, and decide that the hypotenuse (r) should be 2 units long, and the acute angle (θ) should be π/6. Fine. But suppose that the point (sqrt(3),1) is operating in a sort of dynamic way. Maybe it is the end of a robot arm, or something, and suppose that the arm is swinging around the pole, its angle increasing. It might be true that we somehow are computing various angles, and since the arm is moving continuously (still no teleporting robot arms!) the angles which are θ's should change continuously. If the arm swings completely around the pole, and comes back to the same geometric location, it would make more sense to report its polar coordinates as r=2 and θ=13π/6 (which is better understood as 2π+π/6).

Some valid polar coordinates for the point whose rectangular coordinates are x=sqrt(3) and y=1:
r=2 and θ=π/6 r=2 and θ=13π/6 r=2 and θ=25π/6 ETC.
But the "robot arm" could also swing around backwards, so other possible polar coordinates for the same geometric point include
r=2 and θ=–11π/6 r=2 and θ=–23π/6 ETC.
Generally, r=2 and θ=π/6+2π(multiplied by any integer): the integer could be 0 or positive or negative.

The irritation ("It's not a bug, it's a feature ...") is that there are further "reasonable" polar coordinate pairs for the same point! For example, go around to π/6+π. If you position your robot arm there, and then tell the arm to move backwards 2 units, the arm will be positioned at (sqrt(3),1). Sigh. So here are some more polar coordinates for the same point:
r=–2 and θ=7π/6 and r=–2 and θ=–19π/6 and r=–2 and θ=–31π/6 ETC.
but we are not done yet, because there are also (going backwards in the angle and the length)
r=–2 and θ=–5π/6 and r=–2 and θ=–17π/6 and r=–2 and θ=–29π/6 ETC.
Generally, r=–2 and θ=7π/6+2π(multiplied by any integer): the integer could be 0 or positive or negative.

Common restrictions on polar coordinates and the problems they have
This is irritating. Any point in the plane has infinitely many valid "polar coordinate addresses". In simple applications, people frequently try to reduce the difficulty. Much of the time, we expect r>0 always. And maybe we also make θ more calm. The restriction 0≤θ<2π is used, except when it isn't, so the restriction –π<θ≤π is used in other circumstances. I am not trying to be even more incomprehensible than usual. I am merely reporting what different people do. As we will see, this is all very nice, except that there are natural circumstances, both in physical modeling (the robot arm I mentioned and sometimes in things like understanding the mathematical modeling of big molecules) and in the mathematical treatment, where it makes sense sometimes to ignore the artificial restrictions, even if this makes life more difficult. You'll see a few of these circumstances.

Conversion formulas
If you consider the picture to the right, I hope that you can fairly easily "read off" how to go from r and θ to x and y:
    x=r cos(θ)
    y=r sin(θ)

Going from x and y to r is easy enough: r=sqrt(x²+y²). If we divide the y equation above by the x equation, the r's drop out and we get y/x=tan(θ) so that θ=arctan(y/x). Please note that there are infinitely many valid r and θ pairs for every point, so this method will only give you one such pair! Be careful in real applications, please.

Specifying regions in the plane in polar fashion
It is useful to try to get used to thinking in polar fashion, because then you will be able to see problems (usually physical or geometric problems with lots of central symmetry) where this coordinate system can be used to really simplify computations. So here are some simple examples of regions which can be easily specified with polar inequalities.

1<r<3 r=1 and r=3 represent two circles centered at 0, one of radius 1 and one of radius 3. The inequality, which has no restriction on θ, means that every point between these circles is in the region specified. This sort of region is frequently called an annulus.
π/4<θ<π Here there is no restriction on r. I will assume (!!) the standard restriction that r should not be negative. Then a point is in this region if its angle with respect to the positive x-axis is more than π/4 and less than π (the angle π specifies the negative x-axis!!!). I guess I would call this region a wedge. I also "guess" that 0 is in this region. The wedge goes out radially "to infinity" in all of the legal θ directions.
0<r<1 and π/2<θ<5&pi/4 Here are combined restrictions on both polar coordinates. I hope you can "see" the geometric result. I could describe this region in rectangular coordinates, certainly, but the needed equations and inequalities would be much more complicated to me. Maybe this sort of thing will give you some appreciation of the usefulness of polar coordinates.

We will study the equations and graphs of some polar curves next time, and we will do a bit of calculus (arc length and area). That will conclude the course lectures.

A first polar curve (a weird one!)
We tried to sketch as well as possible all points in the plane which have r=θ using any (!) determination of polar coordinates (no restrictions on r and θ). The result is weird until you get used to it (as usual!).

To the right is Maple's graph of r=θ when –12≤θ≤12. Think hard about what the graph would look like for θ small positive (in the first quadrant) and θ small negative (although the "robot arm" is pointing in the fourth quadrant, the corresponding r value is negative, so the curve appears in the second quadrant, near the negative x-axis). The curve is smooth with no corners or cusps, and has infinitely many self-intersections on the y-axis.

Most people understand the θ>0 part of the curve but the part with θ<0 was more difficult, because we must remember to go backwards when r is negative (this happens because the curve is r=θ). The picture shown is something which might help people to understand.

I don't think negative r's will occur very often in practice (but they do -- see a few of the curves below). This curve is an example of an Archimedean spiral (although usually r>0 in the description of that spiral).

Today I'll do some calculus in polar coordinates with the object being (as I mentioned in class) to get two formulas which you are supposed to know. I'll first sketch some "classical" polar coordinate curves, and then I'll do some computations with arc length and area. Here are some examples with numbers maybe a bit nicer than what was done in class.

A collection of examples

r=3+sin(θ) Let's consider r=3+sin(θ). Since the values of sine are all between –1 and 1, r will be between 2 and 4. Any points on this curve will have distance to the origin between 2 and 4 (the green and red circles on the accompanying graph). When θ=0 (the positive x-axis) r is 3. As θ increases in a counterclockwise fashion, the value of r increases to 4 in the first quadrant. In the second quadrant, r decreases from 4 to 3. In the third quadrant, corresponding the sine's behavior (decrease from 0 to –1) r decreases from 3 to 2. In all of this {in|de}crease discussion, the geometric effect is that the distance to the origin changes. We're in a situation where the central orientation is what matters, not up or down or left or right. Finally, in the fourth quadrant r increases from 2 to 3, and since sine is periodic with period 2Π, the curve joins its earlier points. The picture to the right shows the curve in black. I'd describe the curve as a slightly flattened circle. The flattening is barely apparent to the eye, but if you examine the numbers, the up/down diameter of the curve is 6, and the left/right diameter is 6.4.
Converting to rectangular coordinates A naive person might think, "Well, I could convert the equation r=3+sin(θ) to rectangular coordinates and maybe understand it better." Except under rare circumstances (I'll show you one below), the converted equation is very irritating and difficult to understand. For example, Let's start with r=3+sin(θ) and multiply by r. The result is r2=3r+r·sin(θ). I multiplied by r so that I would get some stuff I'd recognize from the polar/rectangular conversion equations. r2 is x2+y2 and r·sin(θ) is y. So I have x2+y2=3r+y, or x2+y2–y=3r. I would rather avoid square roots so I will square this, and get (x2+y2–y)2=9r2=9(x2+y2). This is a polynomial equation in x and y of highest degree 4, defining this curve implicitly. The equation doesn't give me much insight.
r=2+sin(θ) Now consider r=2+sin(θ). Again, the values of sine are all between –1 and 1, so r will be between 1 and 3. Any points on this curve will have distance to the origin between 1 and 3. We can begin (?) the curve at θ=0 when r=2, and spin around counterclockwise. The distance to the origin increases to r=3 at θ=Π/2 (the positive y-axis). The distance to the origin decreases back to r=2 when θ=Π (the negative x-axis). The curve gets closest to the origin when θ=3Π/2 (the negative y-axis) when r=1. Finally, r increases (as θ increases in the counterclockwise fashion) to r=3 again when θ=2Π. Here the "deviation" from circularity in the curve is certainly visible. The bottom seems especially dented.
r=1+sin(θ) We decrease the constant a bit more, and look at r=1+sin(θ). The values of sine are all between –1 and 1, so r will be between 0 and 2. The (red) inner circle has shrunk to a point. This curve will be inside a circle of radius 2 centered at the origin. We begin our sweep of the curve at 0, when r is 1. Then r increases to 2, and the curve goes through the point (0,2). In the θ interval from Π/2 to Π, sin(θ) decreases from 1 to 0, and the curves moves closer to the origin as r decreases from 2 to 1. Something rather interesting now happens as θ travels from Π to 3Π/2 and then from 3Π/2 to 2Π. The rectangular graph of 1+sine, shown here, decreases down to 0 and then increases to +1. The polar graph dips to 0 and then goes back up to 1. The dip to 0 in polar form is geometrically a sharp point! I used "!" here because I don't believe this behavior is easily anticipated. The technical name for the behavior when r=3Π/2 is cusp. This curve is called a cardioid from the Latin for "heart" because if it is turned upside down, and if you squint a bit, maybe it sort of looks like the symbolic representation of a heart. Maybe.
r=1/2+sin(θ) Let's consider r=1/2+sin(θ). The values of sine are all between –1 and 1, so r will be between –1/2 and 3/2. The (red) inner circle actually had "radius" –1/2, and it consists, of course, of points whose distance to the pole, (0,0), is 1/2. When θ is 0, r is 1/2. In the first two quadrants, 1/2+sin(θ) increases from 1/2 to 3/2 and then backs down to 1/2. In the second two quadrants, when θ is between Π and 2Π, more interesting things happen. The rectangular graph on the interval [0,2Π] of sine moved up by 1/2 shows that this function is 0 at two values, and is negative between two values. The values are where 1/2+sin(θ)=0 or sin(θ)=–1/2. The values of θ satisfying that equation in the interval of interest are Π+Π/6 and 2Π–Π/6. The curves goes down to 0 distance from the origin at Π+Π/6, and then r is negative until 2Π–Π/6. The natural continuation of the curve does allow negative r's, and the curve moves "behind" the pole, making a little loop inside the big loop. Finally, at 2Π–Π/6, the values of r become positive, and the curve links up to the start of the big loop. This curve is called a limaçon ( got thing wrong in class -- I'm sorry). The blue lines are lines with θ=Π+Π/2 and θ=2Π–Π/6. These lines, for the θ values which cross the pole, are actually tangent to the curve at the crossing points.
r=0+sin(θ) Let's try a last curve in this family, with the constant equal to 0. What does r=sin(θ) look like? A graph is shown to the right. There are several interesting features of this graph. First, this is a polar curve which does have a nice rectangular (xy) description. If we multiply r=sin(θ) by r, we get r2=r·sin(θ), so that x2+y2=y. This is x2+y2–y=0 or, completing the square, x2+y2–2(1/2)y+(1/2)2–(1/2)2=0 so that (x–0)2+(y–1/2)2=(1/2)2. This is a circle of radius 1/2 and center (0,1/2), exactly as it looks. The moving "picture" of this curve is quite different. Between 0 and π it spins once around the circle but then from π to 2π it goes around the circle another time! So this is really somehow two circles, even though it looks like only one geometrically. This is hard to understand.

More information about these curves is available here.

Area of polar curves
The area of the region enclosed by the rays r=α and r=β and the polar curve r=r(θ) (r is some function of θ) is the (initially) silly-looking integral {1/2}∫_θ=α^θ=β(r(θ))²dθ.

I tried to motivate this formula by considering areas of circular sections of radius R and central angle ψ. After considering examples, we first "guessed" that the area would be directly proportional to R² and that the constant of proportionality was ψ/2 (this only works so easily if we use radian measure for angles). This observation can then be applied to find the area of the region by slicing the region using many narrowly placed "rays" coming from the origin so that we had regions with central angle dθ and radius r(θ). This is discussed in the textbook on p. 649. I mentioned that we could apply this to find the area inside a limaçon.

Length of polar curves
The formula is ∫_θ=α^θ=βsqrt(r²+(dr/dθ)²)dθ. This formula is gotten from the parametric curve formula on p.652 of the textbook. Please look at the examples in the textbook.

There is no class time for more examples of polar curves next time but what follows may be useful.

"Sketching" roses Here are dynamic pictures of two roses. The first is r=cos(3θ). It is "covered" or traced twice and has 3 "petals". The second is r=cos(4θ). It is only covered once, and it has 8 petals! Wow, polar coordinates can be annoying!   r=cos(3) This is a three-leafed rose. Please note that the graph shows one sweep, as θ goes from 0 to 2Π. The rectangular graph, shown here, has three pairs of ups and downs. The polar trace covers the leaves twice. The six up-and-downs of cos(3θ) (magically?) reduce to retracings of half of the loops. I hope I made this evident. I introduced some deliberate distortion in the second tracing. (!) Without the distortion, the second tracing could not be seen at all. The imaginary "point" travels over the computer screen's pixels and colors them (from white to black in this case). The second trip could not be noticed since the pixels had already been flipped to black. So I put in a small perturbation so that the second layer of travel could be seen.   r=cos(4θ) This "rose" has 8 leaves or petals, and the dynamic way it is traced is weird and wonderful to me. The rectangular graph. to the right, shows four bumps up and four bumps down. There are no retracings of already colored points, so that the wiggles up and down of cos(4θ) all result in 8 leaves.     Area inside one petal of r=cos(3θ) Well, cos(3θ) "first" (going from 0 to 2π) is 0 when 3θ=π/2. So we get half a petal by integrating from 0 to π/6. The formula is ∫αβ(1/2)r2dθ for area in polar coordinates (see the discussion on p.649 of the textbook), so this becomes (for the whole petal, we need to double): 2·(1/2)∫0π/6 cos(3θ)2dθ. This can be computed using a trig identity.   Another kind of spiral I didn't talk about Exponentials and snails, darn it! Curves of the form r=a ebθ are spirals of a different kind than what we've drawn (for example, different from the spiral drawn for the QotD last time). All of these spirals have a strange and wonderful geometric property. If a ray is drawn from the pole (the origin) then the angle the ray makes with the tangent line of the spiral at any intersection is the same. I attempted to illustrate this with the first picture to the right. It turns out that this silly geometric property has natural consequences in terms of the energy efficiency of its construction. A snail shell when considered transversal (perpendicular) to the axis of symmetry usually is one of these curves and different species has different constants, b. You can read a wikipedia article about these spirals, which have a number of different names (of course!). To the right is a machine-drawn picture of r=e.25θ as θ goes from –Π to 5Π. The .25 was put in to make the exponential not grow too fast so the picture would be tolerable.

A bug drawing out a thread ...
Thread is wound around the unit circle centered at the origin. A bug starts at (1,0) and is attached to an end of the thread. The bug attempts to "escape" from the circle traveling counterclockwise. The bug moves at unit speed.

I would like to find an expression for the coordinates of the bug at time t. Look at the diagram. The triangle ABC is a right triangle, and the acute angle at the origin has radian measure t. The hypotenuse has length 1, and therefore the "legs" are cos(t) (horizontal leg, AB) and sin(t) (vertical leg, BC). Since the line segment CE is the bug pulling away (!) from the circle, the line segment CE is tangent to the circle at C. But lines tangent to a circle are perpendicular to radial lines. So the angle ECA is a right angle. That means the angle ECD also has radian measure t. But the hypotenuse of the triangle ECD has length t (yes, t appears as both angle measure and length measure!) so that the length of DE is t sin(t) and the length of CD is t cos(t).

The coordinates of E can be gotten from the coordinates of C and the lengths of CD and DE. The x-coordinates add (look at the picture) and the y-coordinates are subtracted (look at the picture). Therefore the bug's path is given by x(t)=cos(t)+t sin(t) and y(t)=sin(t)–t cos(t).

Radian measure is used in several ways here. The angle BAC is the same as the arc length from (1,0) to C along the circle and this is the same as the length of the line segment EC.

t between 0 and 1	t between 0 and 10 Note that the scale is changed!

Finally to the right is an animated picture of the bug moving. Maybe you can understand this picture better: maybe (!!). This curve is more typical of parametric curves. I don't know any easy way to "eliminate" mention of the parameter. This seems to be an authentically (!) complicated parametric curve, similar to many curves which arise in physical and geometric problems. It has an official name. It is called the evolute of the circle.

Parametric curves
We begin our rather abbreviated study of parametric curves. These curves are a clever way of displaying a great deal of information. Here both x and y are functions of a parameter. The parameter in your text is almost always called t. The simplest physical interpretation is that the equations describe the location of a point at time t, and therefore the equations describe the motion of a point as time changes. I hope examples will make this more clear. The t here is usually described for beginners as time, but in applications things can get a great deal more complicated. Parametric curves could be used to display lots of information. For example, some steels contain chromium. Maybe the properties of the steel such as ductility (a real word: "The ability to permit change of shape without fracture.") and density, might depend on the percentage of chromium. So the t could be that and the x and y could be measurements of some physical properties of the steel. Here x=f(t) and y=g(t), as in the text. Now a series of examples.

Example 1
Suppose x(t)=cos(t) and y(t)=sin(t). I hope that you recognize almost immediately that x and y must satisfy the equation x²+y²=1, the standard unit circle, radius 1, center (0,0). But that's not all the information in the equations.

The point (x(t),y(t)) is on the unit circle. At "time t" (when the parameter is that specific value) the point has traveled a length of t on the unit circle's curve. The t value is also equal to the radian angular measurement of the arc. This is uniform circular motion. The point, as t goes from –∞ to +∞, travels endlessly around the circle, at unit speed, in a positive, counterclockwise direction.

Example 2
Here is a sequence of (looks easy!) examples which I hope showed students that there is important dynamic (kinetic?) information in the parametric curve equations which should not be ignored.

x(t)=t and y(t)=t. Surely the "particle" travels on the main diagonal line y=x. The travel (remember, the domain unless otherwise limited, is all values of t which makes sense to the defining functions) is from lower left (third quadrant) to upper right, (first quadrant). Also I hope that you see the particle moves at uniform speed. This is uniform linear motion.
x(t)=t3 and y(t)=t3. The path of this point is also on the main diagonal line y=x. But the motion of this point, while in the same direction as the first example (from third to first quadrant), is very different. The difference is in the "clock". Try a few values of t. Between t=0 and t=1 we travel from (0,0) to (1,1). Between t=1 and t=2, we travel from (1,1) to (8,8). Just roughly, the distance changes from sqrt(2) to sqrt(98). That's a big change. The particle, near –∞, travels fast. Then as t goes around 0, it is slower, and, as t gets very large positive, the particle moves faster.
Now let's consider the motion described by x(t)=t2 and y(t)=t2. Again, we can "eliminate the parameter". That rather grim phrase (if you read murder mysteries!) is what the text uses to describe getting rid of t by manipulating the two equations. Here we just realize that x=y. Much of the information about the motion of this point on the main diagonal is lost if we go to just y=x. The t2 means that both x and y must be non-negative. When t is large negative, the point is way up high in the first quadrant, and traveling towards the origin rather rapidly. It begins to slow, and then "stops" (!) but only "instantaneously" (!!) at (0,0) (when t=0). It turns around (how the heck does a particle "turn around"?) and begins to retrace its path, up towards the open end (??) of the first quadrant. As it travels, its speed increases. So this is really quite complicated motion, and very different from the first two examples.
The final example in this series is x(t)=cos(t), and y(t)=cos(t). Again, the path traveled by the point is on the main diagonal line y=x. But now the dynoamics are extremely different. Because cosine oscillates endlessly between –1 and +1, the motion of the point whose position is described by these equations is on the line y=x and only between (–1,–1) and (1,1). It moves back and forth between these points, completing a round trip in every time interval of length 2Pi. This is very different motion from the other examples.

Example 3

I considered x(t)=5cos(t)and y(t)=5sin(t). This is a slight variant of the very first example. Since x(t)/5=cos(t) and y(t)/5=sin(t), I "know" that the particle's path must be on (x/5)2+(y/5)2=1, so x2+y2=52, a circle whose center is (0,0) and radius is 5. The fact that the path lies on the circle whose equation is given is fairly clear. But I was asked a slightly different question in class: why does the point travel in a counterclockwise direction? When you learn about velocity vectors, this may also be "fairly" clear, but right now, if you plug in t=0, the point (x(t),y(t)) is (5,0), and if you plug in π/2, the point is (0,5). Therefore I feel that travel proceeds in a counterclockwise direction, from (5,0) to (0,5) and so on.

What if x(t)=5cos(t) and y(t)=3sin(t)? Then the trig identity gives (after elimination of the parameter) (x/5)2+(y/3)2=1 as the possible path of the point. This is an ellipse centered at (0,0). Its wider part (total length 10) is on the horizontal axis, as shown. The vertical extant of the ellipse is 6. So I attempted to create a moving image of this curve. I hope it is helpful to you. The parameterization is by central angle. Speed varies. (I will discuss this more below.)

My favorite parametric curve This is x(t)=1/(1+t2) and y(t)=t3–t. I would like to analyze this curve a bit differently from the other examples by separately considering the horizontal and vertical components. The horizontal control Here we consider x as a function of t. The function has even degree and is therefore symmetric: if (t,x) is on the curve, so it (–t,x). Actually, the function is relatively simple. Consider positive t. As t increases, x decreases, and since 1+t2→∞ x→0. So we get a picture as shown below. Since this represents the horizontal part of the motion described by this parametric curve, the result is this in the (x,y) plane: the point for large negative t starts close to the x-axis. Then as t increases, it slowly moves right. At its largest it is 1 unit to the right of the vertical axis. Then it slowly moves back towards the vertical axis again. The vertical control In this case, I'm considering the y-component of the moving point. I could think of this as the vertical control of some sort of scanning machine. The function has odd degree and is antisymmetric (if (t,y) is on the curve, then (–t,–y) is on the curve). It is a cubic polynomial with three real roots, at 1, –1, and 0. The graph is shown to the right. The curve in the (t,y) plane starts very low left, then up, then down, then up, way way up on the right. But this is a description of the vertical motion. Now what could we "see" in the (x,y) plane? The point starts way way down (this is large negative y!), moves up, bounces down, and then finally moves way way up. Combining these two motions can be difficult to do but with practice you will find it easier. A part of the combined motion, the whole parametric curve, is shown to the right. Indeed, it is a curve with a self-intersection. I don't know how to describe this curve accurately and efficiently without the parametric "apparatus". The self-intersection occurs when t=1 and t=–1 (that's where x=0 and y<1, as shown in the picture). The point at which this occurs is (0,1/2).

Calculus?
Finally, very late in the lecture, I attempted some calculus. Here's what I said.

Suppose we want to analyze what happens when the parameter changes just a little bit, from t to t+Δt. Well, the point starts at (f(t),g(t)). What can we say happens at t+Δt? Well, f(t+Δt)≈f(t)+f´(t)Δt. Why is this true? You can think of this either 151 style (calc 1) as linear approximation, or you will see a more sophisticated 152 approach in a few weeks. Similarly for g(t) we know g(t+Δt)≈g(t)+g´(t)Δt. Therefore the point in the interval [t,t+Δt] moves from (f(t),g(t)) to (approximately!) (f(t)+f´(t)Δt,g(t)+g´(t)Δt). What is the slope of the line segment connecting these points?

Slope
Take the difference in second coordinates divided by the difference in the first coordinates. The result (there is a lot of cancellation) is g´(t)/f´(t). If this were an xy curve, this would be noted as dy/dx, the slope of the tangent line. In fact, people usually remember the result in the following way:

dy   dy/dt
-- = -----
dx   dx/dt

A tangent line to the evolute Let me find a tangent line to the evolute of a circle when t=3Π/4. For this, x(t)=cos(t)+t sin(t) and y(t)=sin(t)–t cos(t). When t=3Π/4, x=–[1/sqrt(2)]+3Π/sqrt(2) (approximately 0.95897) and y=[1/sqrt(2)]+3Π/sqrt(2) (approximately 2.37319). So we know a point the line goes through. How about the slope? dy/dx=(dy/dt)/(dx/dt)=[cos(t)–cos(t)+t sin(t)]/[–sin(t)+sin(t)+t cos(t)]=tan(t) (amazing that it is so relatively simple). At t=3Π/4, this is –1. So the line is y–[1/sqrt(2)]+3Π/sqrt(2)=(–1)(x–{–[1/sqrt(2)]+3Π/sqrt(2)}). The line and the curve are shown to the right.

Speed
Since Distance=Rate·Time, and in the time interval [t,t+Δt] we move from (f(t),g(t)) to (approximately!) (f(t)+f´(t)Δt,g(t)+g´(t)Δt), we can get the speed (the Rate) by taking the distance between these points and dividing by Δt. There is more cancellation here, and the result is Speed=sqrt(f´(t)²+g´(t)²) or

         ___________
        /(dx)2 (dy)2   
Speed= / (--)+ (--) 
      /  (dt)  (dt) 

Length of a curve
Well, suppose we move along a parametric curve given by x=f(t) and y=g(t) from t=START to t=END. If we believe that the speed is sqrt(f´(t)²+g´(t)²), then we know that this speed can vary. In a short time interval (a time interval which is dt long!) the distance traveled is Speed·Time, or sqrt(f´(t)²+g´(t)²)dt. We can add up all these distances from t=START to t=END using the integral idea. So the distance traveled along the curve from t=START to t=END will be given by ∫_t=START^t=ENDsqrt(f´(t)²+g´(t)²)dt. (We are integrating the magnitude of the velocity vector). Is this a reasonable formula?

Almost no speed functions have nice, neat, simple antiderivatives. In the real world, you'll need to use numerical approximation. However, Math 152 is not the real world.

A textbook problem Section 11.3, problems 3 through 15, are all "Find the length of the path over the given interval" with some rather silly-looking functions specified. Accidentally (exactly not accidentally, actually!) all of the problems can be computed exactly with antiderivatives and values of standard functions. There is a fair amount of ingenuity involved in constructing such examples. I urged students to practice with several of them. Here is problem #7:     Find the length of the path described by (3t2,4t3), 1≤t≤4. The solution Here x=3t2 and y=4t3. We will compute the speed and then attempt to "integrate" (actually using FTC, so we'll need to find the antiderivative). Now dx/dt=6t and dy/dt=12t2. Therefore the speed is sqrt(6t)2+(12t2)2). This is sqrt(36t2+144t4). We'll need to integrate this, and maybe I will "simplify" first. Indeed, since we consider t in an interval where t≥0, sqrt(t2)=t (otherwise we would need to worry about |t| or –t etc.). But 36t2+144t4=36t2(1+4t2) so that the square root is 6t·sqrt(1+4t2). Therefore the distance traveled along the curve is an integral, ∫t=1t=46t·sqrt(1+4t2)dt. Several students immediately suggested various substitutions. Here is one which does the job efficiently. So:     If u=1+4t2, du=8t dt, so (1/8)du=dt.     ∫6t·sqrt(1+4t2)dt=(6/8)∫sqrt(u)du.     Now (6/8)∫sqrt(u)du=(6/8)(2/3)u3/2+C=(1/2)u+C=(1/2)(1+4t2)3/2+C.     So ∫t=1t=46t·sqrt(1+4t2)dt=(1/2)(1+4t2)3/2|14=(1/2)(65)3/2–(1/2)(5)3/2. Tangent lines I'll find the tangent lines at the self-intersection point of my favorite curve. The point involved is (1/2,0) and the two values of t are +1 and –1. Since I have the point, the other information I need to find the tangent lines is the slope. Well, slope is dy/dx=(dy/dt)/(dx/dt). Since x=1/(1+t2), dx/dt=(–2t)/(1+t2)2. I only fouled this up two or three times in class, but when t=–1, dx/dt=1/2, and when t=+1, dx/dt=–1/2. Since y=t3–t, dy/dt=3t2–1. When t=–1 or when t=+1, dy/dt=3–1=2. Therefore when t=–1, dy/dx=(2)/(1/2)=4. The line goes through (1/2,0), so an equation for it is y=4(x–1/2). Therefore when t=1, dy/dx=(2)/(–1/2)=–4. The line goes through (1/2,0), so an equation for it is y=–4(x–1/2). I had Maple graph the parametric curve and the two lines just found (in green and blue). I then asked what the angle between the lines (the angle which encloses the x-axis). The angle is 151.9 degrees. Hey, please remember that the slope of a line is the tangent of the angle that the line makes with the positive x-axis. Here the angle between the line y=4(x–1/2) and the positive x-axis has angle equal to arctan(4), approximately 1.326 radians. Double this is about 151.9 degrees. Hey: we have the machines. Please use them. Uniform speed? We developed a formula for speed. It was sqrt(f´(t)2+g´(t)2) where x=f(t) and y=g(t). For the circle x=5cos(t) and y=5sin(t), previously considered, we know dx/dt=–5sin(t) and dy/dt=5cos(t), so that (dx/dt)2+(dy/dt)2=25((sin(t))2+(cos(t))2)=25·1=25, so the speed, which is the square root, is always 5. So this is uniform circular motion: the word "uniform" here means that the speed is constant. (Notice, oh physics people, that I am not saying the velocity is constant and I am not saying that the acceleration is 0. Indeed, both of those statements are false. The direction of the velocity is changing, so the acceleration is not 0. We need to look at these quantities as vectors, and this will be done in Math 251.) Nonuniform speed Then I tried to analyze an ellipse we had parameterized earlier. We have x(t)=5cos(t) and y(t)=3sin(t) so dx/dt=–5sin(t) and dy/dt=3cos(t), and the speed is the square root of 25(sin(t))2+9(cos(t))2= 16(sin(t))2+9(sin(t))2+9(cos(t))2= 16(sin(t))2+9 since sin2+cos2 is 1. Sigh. The speed at time t is sqrt(16(sin(t))2+9). When t=0, the particle whose motion we are describing is at (5,0) and the speed is sqrt(9)=3 since sin(0)=0. When t=Π/2, the particle is at (0,3) and the speed is sqrt(25)=5 since sin(&pi/2)=1. So certainly the particle is not moving at the same speed. Indeed, a graph of the speed, sqrt(16(sin(t))2+9), is shown to the right. Now this does resemble what I know about the motion of a planet in orbit. When it is far away from the center of the orbit, the planet will have large potential energy and relatively small kinetic energy (it will move slowly). This is near t=0 and t=Π. When it is close to the center of the orbit, the planet moves faster, and the kinetic energy is larger, while the potential energy, measured by the work needed to move closer/farther from the center, decreases. The total energy is conserved, but the way the total is divided between kinetic and potential energy varies. Maybe if you look again at the moving picture, you will "see" both of these phenomena. They are not too clear to me visually. Back to my favorite curve The length of the loop in my favorite curve can be gotten by computing the integral of the speed from –1 to +1 (the two self-intersection times). So this is (I'll use the formulas we already have) ∫–11sqrt((–2t/(1+t2)2)2+(3t2–1)2)dt. Maple takes more than half a second (quite a lot of time!) to acknowledge that it can't find an antiderivative, so we can't use FTC. In less than a tenth of a second, the approximate value 1.971944073 is reported.

The philosophy behind the definite integral and its use
Maybe the formulas are not totally silly. Both of them are illustrations of how definite integrals can be used to compute various quantities. The procedure (which we have already used in various area and volume situations, and also with work) represents an attempt to compute "something" complicated:

1. Break up the complicated quantity into little pieces.
2. Approximate the little pieces by something simple.
3. Add up the little pieces, and take a limit.

Arc length
We're given a function, f(x), defined on the interval [a,b]. The quantity to be computed is the length of the graph, the curve y=f(x). This is called arc length. Here is the idea.
Break up [a,b] into many little subintervals, whose length we will call dx (or Δx). "Above" each little subinterval is a little piece of the curve. The usual name for a little piece of curve is ds. If you magnify the little piece, as shown, well, the result is almost a right triangle. The curve length is still somewhat curvy, but, well, maybe I can approximate it by a straight line segment. The resulting picture is just about a right triangle. dy is the change in y (the function) when the input variable, x, Pythagoras then declares that (ds)² should be the same (really, approximately the same!) as (dx)²+(dy)². Therefore ds=sqrt{(dx)²+(dy)²). Let's rewrite what's inside the square root:
(dx)²+(dy)²=(dx)²(1+{dy/dx}²).
So sqrt(=(dx)²(1+{dy/dx}²))=dx·sqrt(1+{f´(x)}²). ("Factoring out" the dx is strange to me, and so is recognizing the quotient dy/dx as the derivative.)

Now we should add up these pieces and take limits. In this context, this is all done by writing a definite integral. So the arc length formula is ∫_a^bsqrt(1+[f´(x)]²)dx. This is the official formula. Let's see how well it works with some examples.

Line segment
Maybe the simplest curve is a straight line segment. Let me "find" the length of the line segment joining (1,1) and (4,3). This should be the same as the distance from (1,1) to (4,3), which is (square root of the sum of the squares!) sqrt(13). Let's find this number using the calculus formula above.

We need a formula for the line segment. The slope will be (3–1)/(4–1) which 2/3. So f(x)=(2/3)x+something. What will the "something" be? Since the line should pass through (1,1), when we put x=1, the result should be 1. Therefore (2/3)(1)+something=1, so something is 1/3. The formula is f(x)=(2/3)x+(1/3). The derivative is f´(x)=(2/3). Now the arc length is ∫_a^bsqrt(1+[f´(x)]²)dx which is ∫₁⁴sqrt(1+[2/3]²)dx. The integrand is a constant, so the result is sqrt(1+[2/3]²)x|₁⁴=sqrt(1+[2/3]²)4–sqrt(1+[2/3]²)1=sqrt(1+[2/3]²)3. This is the same as sqrt(13).

Circle Maybe the next curve to look at is a circle, but we need the graph of a function so let's try to find the arc length of a semicircle. Let's look at the upper semicircle, radius 5, center at (0,0). For this curve, f(x)=sqrt(52–x2). Now I need sqrt(1+[f´(x)]2). So: f´(x)=(1/2)(52–x2)–1/22x using the Chain Rule. The 2's cancel, and we need to square the derivative, so: (f´(x))2=(52–x2)–1x2 but this is the same as x2 ----- 52–x2 to which we must add 1: x2 52–x2+x2 52 1 + ----- = --------- = ------ 52–x2 52–x2 52–x2 Finally we supposed to take the square root of this result, so that the integral we need to compute is ∫–55 5/sqrt(52–x2)dx. This should look slightly familiar. The trig substitution x=5sin(θ) makes this integral into ∫5θ dθ=5arcsin(x/5)+C. I am skipping the details because I've done many of these integrals already. Now evaluate the definite integral: 5arcsin(x/5)|–55=5arcsin(1)–5arcsin(–1), and (since I know arcsin(1)=Π/2 and arcsin(–1)=–Π/2) this works out to 5Π, which is indeed half the circumference of a circle of radius 5.

Problems in the book
These two curves work out fairly well. But let's look at section 8.1, and some of the problems there. The problems mostly have the form, "Find the length of the graph of the function defined by the following formula" and I think the instructions should be modified to read "the following absurd formula." Here are some of the formulas from there:
(1/12)x³+x^–1 (problem #3)
(x/4)⁴+(1/{2x²}) (problem #4)
x^3/2 (problem #7)
(1/3)x^3/2–x^1/2 (problem #8)
(1/4)x²–(1/2)ln(x) (problem #9)
ln(cos(x)) (problem #10)
{e^x+e^–x}/2 (problem #18)

Why didn't the book ask something simpler, instead of functions defined by such bizarre formulas? Let's see why. I will answer problem 1 of section 8.1, which asks for the arclength of y=x⁴ between x=2 and x=6. The problem actually asks only for the definite integral and adds but do not evaluate.

We consider ∫_a^bsqrt(1+[f´(x)]²)dx. Here a=2 and b=6, and since f(x)=x⁴, f´(x)=4x³. The answer to problem 1 is therefore ∫₂⁶sqrt(1+16x⁶)dx. What about evaluation? In the sense most often used in calculus courses, this integral can't be evaluated. That is, there is no antiderivative of sqrt(1+16x⁶) which can be written in terms of standard functions. This isn't because we're ignorant, but because it is impossible to do this. If you wanted to compute this arclength, you would need to use one of the numerical techniques.

The secret to the problems in the textbook which were quoted above is that all of the bizarre functions were selected so that sqrt(1+[f´(x)]²) becomes something which it is possible to integrate (in the sense of "find an antiderivative and use FTC"). I did problem #18. Here it is.

Section 8.1, problem #18
Let's find the arc length of f(x)=(e^x+e^–x)/2 from x=–10 to x=10. Now f´(x)=(e^x–e^–x)/2. Now let's square.

          (ex)2  –2  +(e–x)2  
(f´(x))2= ------------------
                  4

               (ex)2  –2	   +(e–x)2   4 + (ex)2  –2  +(e–x)2  
1+(f´(x))2= 1+ ------------------ =  ---------------------
                        4                     4

These functions and these graphs
The functions quoted from the problems in section 8.1 are mostly rather silly. This one, (e^x+e^–x)/2, is not. Some pictures:

ex Exponential growth	e–x Exponential decay	(ex+e–x)/2 The average of the two

The curve shown in the third box is called a catenary, and it is the curve a "uniformly loaded chain" describes. The function (e^x+e^–x)/2 is called the hyperbolic cosine and is usually abbreviated cosh(x) (pronounced kosh ecks). Its derivative is (naturally!) the hyperbolic sine, (e^x–e^–x)/2, and is abbreviated sinh(x) (pronounced cinch ecks). Really. We used (and proved!) the hyperbolic function identity 1+(sinh(x))²=(cosh(x))².

"Truth"
The truth for arc length is that, more or less, the computability of the arc length integral using FTC and familiar functions is impossible almost all of the time! Therefore, from the elementary, student point of view, maybe this is all a waste of time. But, really, it isn't. As soon as you give me a definite integral and want to approximate the values, there are all sorts of strategies. So what's important is that arc length can be computed by a definite integral, and what's important for you to try to understand is the philosophy of going from the vague idea of arc length to the integral formula for the arc length. And that philosophy will now be displayed again as we get an integral formula for a certain type of surface area.

Surface area Suppose we are again given a function y=f(x) defined on an interval [a,b]. I would like to "compute" (the quotes are because we will get a definite integral formula which will share the benefits and defects of the previous result) the surface area which results when the graph of y=f(x) is revolved around the x-axis. We will get our formula using the same philosophical approach. We can chop up [a,b] into many little pieces, each having length, say, dx. Then (the picture!) the little piece of arc length laying over dx, which we called ds, will be revolved around the x-axis. This gets us a sort of ribbon. What is the area of that ribbon? We won't be able to compute it exactly, but maybe we can approximate the area of the ribbon nicely. Well, we can take the magic scissors (hey: I was able to draw the darn scissors almost correctly this time!) and cut the ribbon and then, sort of, almost, lay it out flat. The result will sort of, almost, be a rectangle. What are the dimensions of this rectangle? One side is the length of the piece of arc, ds. The other side is the circumference of a circle whose radius is f(x), the height of that part of the curve away from the x-axis. (The reason for the repeated "sort of, almost" is that this is actually a distortion of the true value - the ribbon really would not lie flat, and the ribbon really would just approximately be a rectangle. I will try later to address these sorts of slight (?) distortions.) So a piece of the surface area is 2Π f(x) ds. We use a definite integral to get the total surface area and add everything up. The result for the area when the curve is revolved around the x-axis is ∫ab2Π f(x)sqrt(1+[f´(x)]2)dx. Notice that sqrt(1+[f´(x)]2)dx (this uses what we had for ds). Sphere Here is a result from a long time ago (thousands of years!): the surface area of a sphere of radius R is 4Π R2. (This is the area of four "great circles" of the sphere, circles made by intersecting a plane with the center of the sphere.) I would like to verify this result using the surface area formula. I'll use the same semicircle as before: f(x)=sqrt(52–x2), with a=–5 and b=5. Please note that revolving this semicircle around the x-axis gets the area of the whole sphere of radius 5, so that the answer should be 4Π(52). We need to compute ∫ab2Π f(x)sqrt(1+[f´(x)]2)dx. Notice that sqrt(1+[f´(x)]2)dx is what we called ds before, and we did compute ds in a previous example. We saw that ds was equal to 5/sqrt(52–x2)dx. But f(x)=sqrt(52–x2) so, wow! (yeah, wow) there is cancellation and the arclength becomes ∫–55(2Π)5dx which does indeed work out to 100Π as it should. More "truth" There are very few simple powers of x (x2 and x3 and maybe x1) which give me integrands in the surface area formula that I can find antiderivatives of. (That's a horrible sentence!) If I want to compute surface areas for almost any "random" function defined by a formula, I'll need to use numerical approximations.

We actually ended about 4 minutes early. I owe you.

A quote from von Neumann
John von Neumann (1903-1957) was a mathematician who was raised in Hungary and spent most of his career in the United States. He worked in many areas of pure and applied mathematics. His ideas were influential in quantum mechanics, the development of nuclear weapons, game theory, and the theory and construction of digital computers.

In mathematics you don't understand things. You just get used to them.

Another kind of improper integral
The integrals we've looked at are called improper because their domains are infinite. But there is another collection of integrals which are also labeled improper because something goes wrong in their ranges: the function to be integrated becomes infinite. Here are some simple examples.

Toy example #1
Let's look at ∫₀¹(1/x³)dx. Here f(x)=1/x³ and surely as x→0⁺, f(x)→+∞. Here is the official way to analyze this improper integral. Suppose that w is a small positive number. Compute ∫_w¹(1/x³)dx. We can do this with FTC by finding an antiderivative. The integral is –1/(2x²)]_w¹= –1/(2·1²)–[–1/(2·1²)]=(1/2w²)–(1/2). Now as w→0⁺, certainly this→+∞. Therefore the improper integral ∫₀¹(1/x³)dx diverges. Sometimes people declare that the integral "diverges to infinity".

The picture accompanying this is, again, useless. It helps me to organize the computation, but does not show enough quantitative information to help decide convergence or divergence.

Toy example #2
Let's look at ∫₀¹(1/sqrt(x))dx. This does converge and its value is 2. Why?
Suppose that w is a small positive number. Compute ∫_w¹(1/sqrt(x))dx. We can do this with FTC by finding an antiderivative. The integral is 2sqrt(x)]_w¹= 2sqrt(1)–2sqrt(w). Now as w→0⁺, certainly this→2.
Therefore we say that the improper integral ∫₀¹(sqrt(x))dx converges and its value is 2.

I tried to change things a bit after we considered these examples. So what about ∫₀¹((1+e^x)/x³)dx? Does this converge or diverge? We can compare this to Toy Example #1. Indeed, I know that for all x in the unit interval (actually, inside the unit interval, since x≠0) 1+e^x>1, so that (1+e^x)/x³>1/x³. So the "area" under y=(1+e^x)/x³ is larger than the area under y=1/x³. But the latter area, by the analysis in Toy Example #1, is infinite, so the larger area must also be infinite. So this is good: without much work, we can conclude that the improper integral ∫₀¹((1+e^x)/x³)dx diverges. Of course you can complain to me that I arranged this example. I will certainly agree with that, but we need to walk before we can run. And we need to run before we fall down and get hurt a lot. (No, I don't think that last sentence is part of efficient, warm educationese.)

How about ∫₀¹(1/[sqrt(x)+x³])dx? Here we know that 0≤>(1/[sqrt(x)+x³])≤1/sqrt(x). But ∫₀¹(sqrt(x))dx does converge -- it has finite area. So a smaller set, the one designated by ∫₀¹(1/[sqrt(x)+x³])dx, will also converge.

Both of these computations are examples of a comparison technique, a nice indirect way to conclude that certain improper integrals converge. Here is one statement of this comparison technique.

Comparison for Improper Integrals
Suppose I know that 0≤f(x)≤g(x) for all x with 0<x≤A.
If ∫₀^Ag(x)dx converges, then ∫₀^Af(x)dx converges. The value of this integral will be at most the value of the first integral.
If ∫₀^Af(x)dx diverges, then ∫₀^Ag(x)dx diverges.
(Convergence is "inherited" downwards, and divergence is inherited upwards. In general, the other implications (the converses) are not true.

A Maple dialog
Here is a record of a Maple session I just created.

> int(1/x^3,x=0..1);
                                   infinity
> int(1/sqrt(x),x=0..1);

                                       2
> int((1+exp(x))/x^3,x=0..1);
                                   infinity

> int(1/(sqrt(x)+exp(x)),x=0..1);
                                1
                               /
                              |         1
                              |   ------------- dx
                              |    1/2
                             /    x    + exp(x)
                               0
> int(1/(sqrt(x)+exp(x)),x);          
                               /
                              |        1
                              |  ------------- dx
                              |   1/2
                             /   x    + exp(x)
> int(1/sqrt(x),x=0..(.001));    
                                 0.06324555320
> int(1/(sqrt(x)+exp(x)),x=(.001)..1);
                                 0.4636551521

The last two Maple instructions tell me that ∫₀¹1/(sqrt(x)+e^x)dx is 0.46 with .06 as an overestimate of the error. You may think this is bizarre, but this strategy works and it is fairly simple.

Here is one part of the session I didn't discuss in class. I did this immediately before class, and I was a bit startled by the result. Look at it, please. > int(1/(x^3+sqrt(x)),x=0..1); / 1/2 | 5 - 1 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/20 |8 arctan(----------------) 5 (10 - 2 5 ) + 10 ln(2) (10 + 2 5 ) (10 - 2 5 ) | 1/2 1/2 \ (10 + 2 5 ) 1/2 1/2 5 + 1 1/2 1/2 1/2 5 + 5 1/2 1/2 1/2 + 8 arctan(----------------) 5 (10 + 2 5 ) - 12 arctan(----------------) 5 (10 + 2 5 ) 1/2 1/2 1/2 1/2 (10 - 2 5 ) (10 - 2 5 ) 1/2 5 - 3 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 + 4 arctan(----------------) 5 (10 + 2 5 ) + ln(5 + 3) 5 (10 - 2 5 ) (10 + 2 5 ) 1/2 1/2 (10 - 2 5 ) 1/2 1/2 -5 + 5 1/2 1/2 1/2 5 + 3 1/2 1/2 - 12 arctan(----------------) 5 (10 - 2 5 ) + 20 arctan(----------------) (10 - 2 5 ) 1/2 1/2 1/2 1/2 (10 + 2 5 ) (10 + 2 5 ) 1/2 1/2 5 + 5 1/2 1/2 -5 + 5 1/2 1/2 + 20 arctan(----------------) (10 + 2 5 ) - 20 arctan(----------------) (10 - 2 5 ) 1/2 1/2 1/2 1/2 (10 - 2 5 ) (10 + 2 5 ) 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 - 2 5 (10 - 2 5 ) (10 + 2 5 ) ln(5 - 5 ) + 2 ln(5 + 5 ) 5 (10 - 2 5 ) (10 + 2 5 ) 1/2 5 - 3 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 - 20 arctan(----------------) (10 + 2 5 ) - 5 (10 - 2 5 ) (10 + 2 5 ) ln(-5 + 3) 1/2 1/2 (10 - 2 5 ) 1/2 5 + 3 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 + 4 arctan(----------------) 5 (10 - 2 5 ) - ln(-5 + 3) (10 - 2 5 ) (10 + 2 5 ) 1/2 1/2 (10 + 2 5 ) \ 1/2 1/2 1/2 1/2 1/2| / 1/2 1/2 1/2 1/2 - ln(5 + 3) (10 - 2 5 ) (10 + 2 5 ) | / ((10 - 2 5 ) (10 + 2 5 ) ) | / / This took quite a lot of time (about 1.5 seconds) and memory (almost 70 megabytes). I think that Maple computed this with a rationalizing substitution. I was surprised that an explicit answer was given. The numerical value of this (extensive?) answer is (approximately!) 1.776627144. I know the original integral converges by comparing it with the 1/sqrt(x) integral, and also I know that the value of the integral will be less than 2. And 1.77662... is certainly less than 2.

L'H
I wrote a version of L'Hôpital's Rule on the board, since I would need it twice in the next few minutes. I copied it from the book (page 272).

L'H Suppose f(x) and g(x) are differentiable functions inside an interval containing a and that f(a)=g(a)=0. Also assume g (x) is not 0 for x near but not equal to a. Then

     f(x)          f´(x)
lim ------ =  lim ------ 
x→a  g(x)     x→a  g´(x)

if the limit on the right-hand side exists. This result is also valid if both of the limits of f(x) and g(x) as x→a are +∞ or –∞.

It is very important that there is a quotient in the algebraic form. Also, the symbolic quotients 0/0 or ∞/∞ are sometimes called indeterminate forms.

I applied this to two examples: lim_x→2(x²–3x+2)/(x²–4) and something like lim_x→1(e^x–x)/x².

lim_x→2(x²–3x+2)/(x²–4)
Please notice that both the top and the bottom are 0 when x=2, since 2²–3·2+2=4–6+2=0 and 2²–4=0. Therefore we can use L'H, which is the beginning calculus student's dream of the product rule. The resulting quotient is (2x–3)/(2x) and the limit of that as x→2 is (4–3)/(2·2)=1/4.
There is another, more elementary way, to do this, since the top and bottom are simple and can be factored:
   (x²–3x+2)/(x²–4)=[(x–2)(x–1)]/[(x–2)(x+2)]=(x–1)/(x+2)
and certainly the limit of the last quotient can be obtained by "plugging in" (more politely, "instantiating x at 2") since the functions involved are continuous. The resulting limiting value is again 1/4.

lim_x→1(e^x–x)/x²
Here we take the derivative of the top and bottom and a quotient is gotten: (e^x–1)/(2x), and we could indeed take more derivatives and get (e^x/2. Then plugging in to all three quotients gets (e–1)/1²=e–1 and (e–1)/2 and e/2. Of course, the use of L'H is wrong. The original quotient does not satisfy the eligibility criterion for L'H. There is no "indeterminaste form". This error is easy to make since L'H is such a tempting technique. Please: always check for eligibility first.

Sometimes some algebraic manipulation is needed to apply L'H (this will be true in what follows). And sometimes L'H needs to be used more than once (if the original indeterminate form is, after the differentiation step, again an indeterminate form, etc.).

x·ln(x)
I asked people for the graph of y=x·ln(x) on the interval [0,1]. Some people were willing to use their graphing calculators. A graph much like what is shown to the left was produced.

The graph seems to be a slightly assymetrical bump below the x-axis, hanging from (0,0) and (1,0). The "bottom" is at, about, –1/3 (near x=1/3, actually). We actually found the exact "bottom" by looking at the derivative of x·ln(x): it was at (1/e,–1/e). If you "ask" your computer or calculator for the value of x·ln(x) at x=0, however, there will be some sort of complaint rather than a number. The machines have been advised that 0 is not in the domain of ln. But the graph certainly seems to indicate that (0,0) is there (wherever there is!). What's going on?

L'H
Really what the graph indicates is a suggested value for lim_x→0⁺x·ln(x). The guess for this is 0. To verify this guess (in a math course!) we will use L'Hôpital's Rule. As x→0⁺, we have a 0 multiplied by (–) ∞. L'Hôpital's Rule works on quotients, so we need to rearrange things algebraically. Here we go:
lim_x→0⁺x·ln(x)=rearranging=lim_x→0⁺[ln(x)/{1/x}]=using L'H=lim_x→0⁺[{1/x}/{–1/x²}]=rearranging back=lim_x→0⁺–x=0 so the limit is verified.

The integral
How much "area" is included in the bump above? More precisely, what is the definite integral of x·ln(x) from 0 to 1. The bump is below the x-axis so this integral should be negative. It sort of resembles a triangle with base [0,1] and altitude about 1/3. So a guess is that the area should be about –1/6. An actual graph of the function and of this approximating triangle is shown to the right. The function is concave down so it bulges beneath the triangle. And, in fact, the "point" of the triangle is above the graph. So there is more (absolute value!) area in the bump than in the triangle.

Let's compute. The area is ∫₀¹x·ln(x)dx. This is officially an improper integral (it really is difficult to evaluate the integrand at x=0!). So first I'll compute ∫_Q¹x·ln(x)dx for Q small positive, and then let Q→0⁺. To use FTC we need an antiderivative of x·ln(x). Integration by Parts again works. If u=ln(x), then dv=x dx, and du=(1/x)dx and v=(1/2)x². So:
∫x·ln(x)dx=(1/2)x²ln(x)–∫(1/2)x²(1/x)dx=(1/2)x²ln(x)–(1/4)x²+C
The definite integral is (1/2)x²ln(x)–(1/4)x²|_Q¹=((1/2)Q²ln(Q)–(1/4)Q²)–((1/2)1²ln(1)–(1/4)1²).

L'H
I know that lim_Q→0⁺–(1/4)Q²=0 because Q² is continuous and I can just "plug in". What about lim_Q→0⁺(1/2)Q²ln(Q)? Here we (officially!) need L'Hôpital's Rule again. So:
lim_Q→0⁺(1/2)Q²ln(Q)=rearranging=lim_Q→0⁺(1/2)[ln(Q)/{1/Q²}=using L'H=lim_Q→0⁺(1/2)[1/Q}/{–2/Q³}=rearranging back=lim_Q→0⁺(–1/4)Q²=0.

If you now put everything together (and don't lose track of the minus signs!) you can see:
∫₀¹x·ln(x)dx converges and its value is –1/4. The –1/4 is certainly consistent with the estimate we made earlier, backed up by the red and green graph.

A real application ...
This strange function with its somewhat strange graph is related to a function used in applications. The function used in applications is actually a bit more complicated (sigh). Here is how to think about it.

A graph of y=x·ln(x) on [0,1]. The function is concave up and below the x-axis. The area is –1/4, as just computed.
Flip the curve over the x-axis. So this is a graph of y=–x·ln(x) on [0,1]. The function is concave down and above the x-axis. The area is +1/4, because now the region we're considering is above the x-axis.
This is a more complicated flip. Replace x by 1–x. This is a flip which exchanges left and right because of the +/– change in the x multiplier. In fact, the left and right are exchanged, and the y-axis and x=1 are interchanged. So this is a graph of y=–(1–x)·ln(1–x) on [0,1]. The function is concave up and above the x-axis. The area is 1/4.
Now add up the two previous functions. Here is a graph of y=–x·ln(x)–(1–x)·ln(1–x) on [0,1]. Yes, it is weird. This function officially has two strange behaviors, at both 0 and 1. The function is called the binary symmetric entropy function and it is used to study the amount of information flowing through a "channel" (you could think of a channel as a wire, and the information as 0's and 1's -- bits). The entropy function helps to analyze what happens in complicated situations where there may be interference (noise). This function has one bump, and its total area is 1/2.

Comment You can't actually predict that the result of adding the two bumps would be just one symmetric bump. In fact, that result may not occur. Here is a graph of x[ln(x)]2 on [0,1]. Because of the square, the function is positive, and the bump is more pronounced. And this is a graph of x[ln(x)]2+ (1–x)[ln(1–x)]2. So two distinct bumps can show up.

Return of the exam
Here is the first exam in a somewhat more compact form, with one version of the answers also available. A discussion of the grading of the exam and the results is here. Let me make some comments about how this exam did and did not resemble a standard 152 exam.

• Problem 1 Standard, except for the use of the parameter, C. The problem with C resembles slightly one of the homework or workshop problems (the one with C–x²). This is, without C, a nice 152 exam problem.
• Problem 2 Standard problem.
• Problem 3 This is an easy computation but a difficult idea to explain. (The computation is so easy that several people essentially seemed to refuse to do it!) This is a possible 152 problem, but slightly more conceptual than usual (although I did give a rapidly vibrating (sin(Nx))³ as n→∞ as a 152 exam problem).
• Problem 4 A difficult problem -- some people not in the class were interested that I asked this question. Almost half the students got it entirely correct and many others got most credit. I was impressed.
• Problem 5, 6, and 7 Routine 152 exam problems.
• Problem 8 Mostly routine, but, of course, the estimate is somewhat irritating. I would give this problem in a standard 152 course, but not expect great performance.

Die Mathematiker sind eine Art Franzosen: Redet man zu ihnen, so übersetzen sie es in ihre Sprache, und dann ist es alsobald ganz etwas anders.

As translated by a native speaker:

Mathematicians are some sort of Frenchmen: if you talk to them, they translate it into their language, and then it is soon something completely different.

This was to alert you that when you meet and "use" mathematicians, you should realize that they will try to find out what the rules of the (logical) game are and then, maybe, change and break them. Sometimes this is just for fun (!) and can be sort of useless. But sometimes breaking the rules can be in response to needs for additional mathematical tools. We consider that kind of rule-breaking today.

Improper integrals: what are they?

So far we have discussed computing standard definite integrals. Although certainly a definite integral like ∫abf(x)dx can represent many different ideas, the most familiar instantiation is as an area (assuming that f(x)>0 for x in [a,b]) bounded by the x-axis, x=a, x=b, and the graph of y=f(x). In most of the computations we've done, the function f has been rather "nice" -- differentiable, mostly, and only a few times has it had some discontinuities. In fact, there are many applications where this simplicity is made more complicated because the applications themselves demand a "stronger" kind of integral.

For example, we might consider a situation where, say, b gets larger and larger and larger. Where, say, b→∞. I will give a real physical example of this at the next lecture. This integral will have a domain which is an infinite interval. It could be [a,∞) or (–∞,b] or even (–∞,∞). Such integrals also occur very frequently in statistics (and therefore they "infiltrate" almost all experimental sciences!). A completely naive interpretation of the area would declare that since the length of the base is infinite, the total area must somehow be forced to be infinite. More subtly, the height in certain cases will decrease fast enough so that the total integral can be thought of as finite.

There may also be a sort of defect (?) in the range of f. For simplicity, consider the situation where, although f is "nice" for x>a, as x→a⁺, f(x) gets larger and larger and larger: in fact, we might need to try to "integrate" or compute the value of ∫_a^bf(x)dx even if f tends to ∞ as x→a.
Again, a first look could convince you that such "areas" need to be infinite, also, because the height is infinite. But, actually, the way f grows as x→a⁺ is what matters. It is possible to imagine that the growth of f is so controlled that the total approximating areas don't tend to ∞. And maybe, in such situations, we should be able to compute the integral.

I will first consider the "defect in the domain" case.

Toy example #1
Look at y=1/x². The integral ∫₁^B(1/x²)dx (I'm using B as an abbreviation for BIG) can be computed directly:
∫₁^B(1/x²)dx=–1/x|₁^B=1–(1/B) (be careful of the signs!)
Now if B→∞, certainly 1–(1/B)→1. Then we will declare that the improper integral ∫₁^∞(1/x²)dx converges and its value is 1.

Toy example #2
Look at y=1/x. Consider the analogous integral ∫₁^B(1/x)dx (again think of B as a BIG number). We compute it:
∫₁^B(1/x)dx=ln(x)|₁^B=ln(B)–ln(1)=ln(B)–0
I wanted also to consider here the behavior as B→∞, but some students seemed to be confused (this is confusing!). What does happen to ln(B) when B gets large? If you only have a loose idea of the graph in your head, well the log curve might not look like it is increasing too fast. Well, it actually is not increasing very fast, but it is increasing. Look: ln(10) is about 2.3, so ln(10²)=2ln(10) is about 4.6 (that's 2·2.3), and ln(10³)=3ln(10) is about 6.9 (that's 3·2.3), etc. Here etc. means I can get values of ln as large as I want by taking ln's of large enough powers of 10 (hey, ln of 10^5,000 is bigger than ... 10,000: so there!). So the values of ∫₁^B(1/x)dx do not approach a specific number as B→∞. Therefore we say: the improper integral ∫₁^∞(1/x)dx diverges.

The distinction between converges (approaches a specific finite limit) and diverges (does not approach a specific finite limit) is the one that is important in applications and that motivates the distinct use of the two words.

Geometric constrast
I love pictures. I like computation, but I can barely tolerate (!) "algebra". But I introduced the actual definition of {con|di}vergent improper integrals with some algebra, and didn't draw any pictures. Why? Well, because pictures, while useful, don't help too much. Here are qualitative pictures of the two graphs.

Well, they are actually different graphs, sorta. But I wanted to emphasize, through sketching them not too carefully (and on different axes, without scales!) these curves both "start at" (1,1), and as they go right, are always positive, decreasing, concave up, with 0 as limits. My eyes, at least, can't tell that one of them (on the left?) has finite total area, and one of them (to the right!) has infinite total area. The difference is quite surprising.

The exponential probability distribution
Many real-life phenomena are described using something called the exponential probability distribution. For example, the probable life-time of a lightbulb could be described with it. See here for more information.

I will just discuss lightbulbs now. Here is a first attempt to be precise. If the probability that a light bulb will fail in t minutes is proportional to e^–Ct then the lifespan of the bulb is said to have an exponential probability distribution. More specifically, if t=0 is NOW, and t₁<t₂ are later times, then then the probability of lightbulb failure between the times t₁ and t₂ is proportional to ∫_t1^t₂e^–Ctdt.
That is, that portion of the area to the right which is shaded blue represents the chance of a lightbulb burning out during that particular time interval.

What is the total probability from NOW to FOREVER?
Every (real!) lightbulb is going to fail some time. Probabilities of an event range from 0 to 1, where an event which is certain will have probability 1. Since any lightbulb will fail between t=0 (NOW) and t=∞ (FOREVER), we should consider the (improper) integral ∫₀^∞e^–Ctdt. Let me analyze this carefully.
Suppose A is a large positive number. Let's compute ∫₀^Ae^–Ctdt=–(1/C)(e^–Ct)]₀^A=–(1/C)e^–CA+(1/C)e⁰=(1/C)–(1/C)e^–CA.
The antiderivative has a –(1/C) factor because when e^–Ct is differentiated, the Chain Rule produces a multiplicative –C, and the –(1/C) cancels this. Now what happens to (1/C)–(1/C)e^–CA as A gets large, A→∞? e^–CA (with a negative sign, with A and C positive) must go to 0 (this is exponential decay). Therefore the improper integral ∫₀^∞e^–Ctdt converges, and its value is 1/C. But, wait: since every lightbulb fails, shouldn't this be 1? Yes, surely. Let's fix this up. The key is "proportional to". We should multiply the function e^–Ct by a constant so that the improper integral will turn out to be 1. The computation we just did shows that the constant should be C. So the probability distribution is actually Ce^–Ct.

The {expectation|mean|average} of an exponential probability distribution
Let me try to discuss something a little bit harder. First, some background with (maybe) some easier ideas. We could imagine a population of, say, bugs. Maybe there are three types of bugs in my bug collection and I know the following information:

My bug collection
Bug type	Observed lifespan	Proportion in my collection
Orange spotted bug	20 days	30%
Blue striped bug	50 days	50%
Red plaid bug	80 days	20%

What is the average lifespan of this bunch of bugs? Well, it isn't the average of the 3 lifespans (the sum of 20, 50, and 80 divided by 3 -- a rather naive computation) since that doesn't take into account the varying proportions of the bug types. If you think about it, the average in this case is 20·(.3)+50·(.5)+80·(.25): a sort of weighted sum. It is a sum of lifetimes multiplied by the proportion of the population. You should convince yourself that this is the correct number.
What about lightbulbs? What proportion of a lightbulb population will "die" at time t? Well, that proportion is about Ce^–Ctdt. The appropriate weighted sum in this case multiplies the proportion by t and adds it all up with an integral: ∫₀^∞tCe^–Ctdt. If we compute this integral, we maybe can get some idea of when an average lightbulb dies. This quantity is called the mean or the expectation.

We can compute ∫tCe^–Ctdt using integration by parts. If u=t then dv=Ce^–Ctdt and du=dt and v=–e^–Ct. Therefore uv–∫v du is –te^–Ct–∫–e^–Ctdt=–te^–Ct–(1/C)e^–Ct. The definite integral from t=0 to t=A is –te^–Ct–(1/C)e^–Ct]₀^A=–Ae^–CA–(1/C)e^–CA–{–(1/C)e⁰}.
What happens at A→∞? Well, e^–CA certainly goes to 0 (radioactive decay!). But the term Ae^–CA has a sort of conflict. Although the exponential decays, certainly A→∞? Which factor "wins"? Exponential decay is faster than any degree of polynomial growth, actually, so the limit is 0. You certainly can see this with L'Hopital's rule:
lim_t→∞te^–Ct=lim_t→∞t/[e^Ct]=lim_t→∞1/[Cexp(Ct)]=0.
So the integral from 0 to A, as A→∞, approaches a limit, which is –{–(1/C)e⁰}=(1/C). This is the average lifespan of a lightbulb. Incidentally, if we want to check that this is a valid model, we can look at sample lifespans, and this can be used to identify the value of the parameter C. The average lifespan written on a package I just examined is 750 hours. Other technical words which refer to average lifespan are mean and expectation.

Other things ...
People who study statistics are interested in more details about life and lightbulbs than are described here. For example, they may want to know how dispersed the lifespans are around the mean. That is, do all of the lightbulbs tend to die out right around 1/C, or is there a considerable amount of variability? Various numbers measure this, including variance and standard deviation. They all need computation of ∫₀^∞t²Ce^–Ctdt, a different improper integral.

Here is an extended example of how to use improper integrals in a physical setting. I've used this example in our 152 instantiations, but don't have time this semester. Escape velocity When I was very young, I read a science fiction novel by Robert Heinlein which stated "... the escape velocity from the Earth is 7 miles per second ..." and now I would like to sort of verify this using only well-known (?) facts and some big ideas of physics. The radius of the earth Well, a sketch of the (continental) United States is shown to the right. There are 4 time zones. The U.S. is about 3000+ (maybe 3200?) miles wide. Therefore one time zone is about 1000 miles wide (I think the Pacific time zone actually slops a bit into the ocean), and since there are 24 time zones around the world, the circumference of the world is ... uhh ... about 25,000 miles. Or so. And therefore the radius of the earth is that divided by 2Pi, and therefore the radius of the earth is about 4,000 miles. Newton and gravitation Two masses attract each other with a force whose magnitude is proportional to the product of the masses and inversely proportional to the square of the distance between them. Therefore, if I have a mass, m, and if the Earth has mass M, the magnitude of the force of gravity is GmM/r2. G is a constant. Work lifting up Suppose we want to lift a mass m from the surface of the earth to a distance R, where R is very large. Then the work done is force multiplied by distance. The force needed to act against gravity certainly changes with distance. So I will compute the work with calculus. If x is some distance between 4,000 and R, then the force is GmM/x2. If the distance is a little bit, say, dx, then the work dW needed is GmM/x2 dx. The total work, W, is ∫x=4,000RGmM/x2 dx which I can compute readily as –GmM/x]4,000R=GmM({1/4,000}–1/R). I think I did the minus signs correctly. Notice that as R→∞, this work →GmM/4,000: this is the most work you can do, to get anywhere in the universe (assuming the universe is empty except for the mass m and the earth, of course). Kinetic energy How much kinetic energy would we need to supply to the mass m so that it would equal the potential energy the mass would have if it were lifted out to anywhere in the universe? Well, kinetic energy is (1/2)mv2 and that potential energy we already computed is GmM/4,000. So (1/2)mv2=GmM/4,000, and thus v2=2GM/4,000. But what is GM? But F=ma On the earth, a, the acceleration of gravity, is 32 ft/(sec)2. Yes, this is an archaic system of measurement, but that's part of the fun. But also F=GmM/(4,000)2. So GmM/(4,000)2=32m. Therefore GM=(4,000)2·32. And the answer is ... v2=2GM/4,000=2[(4,000)2·32]/4,000=8,000·32= (256,000)/(5,280)=(approximately)50. The 5,280 came from converting feet to miles. Therefore v, the escape velocity, is about 7mps. I think this computation is so silly that it is cool.

Maintained by greenfie@math.rutgers.edu and last modified 10/13/2009.