The error overestimate is |x|ⁿ⁺¹/(n+1)!. This depends on both x and n, and you can see the effect of the dependence in the computations above when n=11 and x=.5 and x=2. What is perhaps more interesting is the following, which I certainly hope you believe by now:
If x is fixed, and n-->infinity, then |x|ⁿ⁺¹/(n+1)!-->0.
This is true because eventually n gets larger than x, and then the effect of multiplying by |x| on top compared to multiplying by stuff bigger than x on the bottom shows that the terms must approach 0. It may not be clear what the heck is going on. I promise that I will do many examples and applications. But here are three conclusions which I can write:

• The series _j=0^2j+1(-1)^jx^2j+1/(2j+1)! converges for all real numbers.
• The sum of that series is sin(x).
• If you consider the partial sum of the series, up to, say, degree N, then the absolute value of the difference between the value of that partial sum and the value of sin(x) is at most |x|^N+1/(N+1)!.

A collection of examples

Power series	Coefficients	Interval of convergence	Radius of convergence
n=0infinityxn/nn	cn=1/nn	All real numbers: (-infinity,infinity)	R=infinity
		One way to see this is to use the Root Test. Then the nth root of |xn/nn| is |x/n| and as n-->infinity, this-->0 for all x.
n=0infinityxn/n2	cn=1/n2	[-1,1]	R=1
		The Ratio Test results in considering |x|[n2/(n+1)2] and as n-->infinity, this-->1. So we get convergence for |x|<1. If x=+/-1, the series is a p-series with p=2>1, so it converges.
n=0infinityxn/n	cn=1/n	[-1,1)	R=1
		Use the Ratio Test again. The resulting ratio is |x|[n/(n+1)]. As n-->infinity, this-->1. So we get convergence for |x|<1. If x=1, divergence (Harmonic Series). If x=-1, convergence (Alternating Harmonic Series).
n=0infinityx2n/n	c2n=1/nn and codd=0	(-1,1)	R=1
		The Ratio Test. Now when x=+/-1, both endpoints give the Harmonic Series, so the power series diverges at both endpoints.
n=0infinitynnxn	cn=nn	Only {0}	R=0
		Use the Root Test.

It turns out that the qualitative behaviors displayed above are all that is possible for power series. That is, they must converge inside an interval, which can be all large as all of R or as small as the center of the power series. They may or may not, depending on the specific coefficients, converge or diverge at the end points of the interval. But no other types of behavior are possible.

Powerful general facts

These results say that power series, inside their intervals of convergence, can be treated, for the purposes of calculus, just like "big" polynomials: differentiationk, integration, etc.

"Reverse engineering" the result of the previous lecture
Last time we needed to compute a "payoff" or expectation or fair entry fee for a gambling game: _n=1^infinityn/2ⁿ. How can we think about this? I'd like to show you how I "invented" what I did last time.

Here is what I would do. I'd say to myself that this is a series which results from substituting x=1/2 into the power series _n=1^infinitynxⁿ=x+2x²+3x³+4x⁴.... but this series is the result of multiplying the series 1+2x+3x²+4x³+... by x. So what is the sum of 1+2x+3x²+4x³+...=_n=1^infinitynx^n-1? Well, nx^n-1 is the derivative of xⁿ, so the series 1+2x+3x²+4x³+... is the derivative of x+x²+x³+x⁴+... which is a geometric series with first term=x and ratio=x, so that its sum is x/(1-x).

Now go backwards! The sum of the series _n=1^infinityn/2ⁿ is the result of differentiating x/(1-x), then multiplying the result by x, and then substituting in x=1/2. And that's what we did.

And now another one ...
A slightly more involved process would get the sum of _n=1^infinityn²/2ⁿ. Here look at the power series _n=1^infinityn²xⁿ. Pull out an x to get _n=1^infinityn²x^n-1 which is the derivative of _n=1^infinitynxⁿ. But, hey, we just got the sum of that series (look at the previous paragraph). So we can get the sum of this series.

Connecting the function and the coefficients
If f(x)=_n=0^infinityc_nxⁿ=c₀+c₁x¹+c₂x²+c₃x³+c₄x⁴+... then:
Set x=0 and get f(0)=c₀.
Differentiate the previous series. The result is f´(x)=_n=0^infinitync_nx^n-1=0+c₁+2c₂x+3c₃x²+4c₄x³+... then:
Set x=0 and get f´(0)=c₁.
Differentiate the previous series. The result is f´´(x)=_n=0^infinityn(n-1)c_nx^n-2=0+0+2c₂+6c₃x+12c₄x²+... then:
Set x=0 and get f´´(0)=2c₂ so that c₂=f´´(0)/2.
Differentiate the previous series. The result is f⁽³⁾(x)=_n=0^infinityn(n-1)(n-2)c_nx^n-3=0+0+0+6c₃+24c₄x+... then:
Set x=0 and get f⁽³⁾(0)=6c₃ so that c₃=f⁽³⁾(0)/6.
By now most people in the class recognized the pattern: c_n=f⁽ⁿ⁾(0)/n!
Please note that most people like this formula and therefore accept that 0! should be 1, and that f⁽⁰⁾, the zeroth derivative of f, should be the original function. Then the formula, as stated, is certainly correct for n=0.
This says several things. First, if a function has a power series expansion, then it has exactly one power series expansion, because the coefficients are given by that formula. This means that any (valid!) way we get the power series expansion, the result must be the correct answer. Any trick, any contrivance is good.
How useful is this formula? For certain functions, it can be very useful. But, in general, if you give me a "random" function (say, a quotient or a composition) then computing high derivatives is tedious and difficult because the expressions for the derivatives begin to expand more and more ("expression swell"). So maybe for such functions the formula just given is not so useful.

Some examples
We do know one nice formula for a series and the sum of a series. That's:
a/(1-r)=a+ar+ar²+ar³+..., the geometric series.
If we can "rearrange" a function so that it fits the geometric series template, then maybe we can get a geometric series. Here are some examples.

1. f(x)=3/(x+5). Actually, 3/(x+5)={3/5}/(1+{x/5}) (dividing the top and bottom by 5) and this is the same as {3/5}/(1-[-{x/5}]). Now I hope you can see the a and the r: a={3/5} and r=-{x/5}. So
f(x)={3/5}-{3/5}{x/5}+{3/5}{x/5}²-{3/5}{x/5}³+...= _n=0^infinity(-1)ⁿ⁺¹(3/5ⁿ⁺¹)xⁿ.

2. (problem #8 in section 11.9) f(x)=x/(4x+1). Only a little bit of "rearrangement" is needed: x/(4x+1)=x/(1-{-4x}) so a=x and r=-4x.
f(x)=x-4x²+4²x³-4³x⁴+...=_n=1^infinity(-1)ⁿ⁺¹4^n-1xⁿ.

3. (problem #9 in section 11.9) f(x)=x/(9+x²). Here look at x/(9+x²)=(x/9)/(1+{x²/9})=(x/9)/(1-[-{x²/9}]) so that a=x/9 and r=-{x²/9}.
f(x)=x/9-x³/9²+x⁵/9³-x⁷/9⁴+...=_n=1^infinity(-1)ⁿ⁺¹x²ⁿ⁺¹/9ⁿ.

A real example
I began a discussion of the Fibonacci numbers, which occur almost everywhere.

Computation
How to actually compute values of functions is a serious question, and ideas concerning computation are the object of much pure mathematics and most of applied mathematics. It's generally agreed that polynomials are computable. Evaluating a polynomial involves multiplication and addition. Just to be clear, a polynomial is a function defined by a formula such as 22.45-9x²+101x⁵. It is a sum of monomials ("x to a positive integer power"), each possibly multiplied by some coefficient. The degree of a polynomial is the highest integer exponent where the coefficient is not zero. So the degree of the polynomial just written is 5.

A polynomial solution to an initial value problem
Since polynomials are easy to compute, we should try to use them in calculus as much as possible. In calculus we integrate and differentiate. Certainly we should try to solve Initial Value Problems (IVP's) for Ordinary Differential Equations (ODE's) with Initial Conditions (IC's). Here is a simple example:
      ODE: dy/dx=y
      IC: y(0)=1
Let's suppose that y=A+Bx+Cx²+Dx³+Ex⁴+Fx+... is a solution. What can be said about the coefficients A, B, C, D, E, F, ...?
So the IC, y(0)=1 says that when we insert x=0, the formula should give us the value 1. But all of the terms with x's in them are 0, so that 1=A.
What does the differential equation tell us? The formula for y tells us that y´=0+B+Cx+2Dx+3Ex²+4Fx³+... so we can compare the two "presentations". Since two polynomials will be equal exactly when their coefficients agree, we learn that:
A=B (the constant term), which (using the previous value of A) tells us that 1=B.
B=2C (the coefficient of the linear or x term), which (using the previous value of B) tells us that 1/2=C.
C=3D (the coefficient of the quadratic or x² term, which (using the previous value of C) tells us that 1/6=C.
D=4E (the coefficient of the cubic or x³ term, which (using the previous value of D) tells us that 1/24=D.
I'll bet that the alert student could answer this question:

ETC.
That is, I hope you (by now!) recognize the pattern of the coefficients, so that
y=1+1x+(1/2)x²+(1/6)x³+(1/24)x⁴+(1/120)x⁵+...
Right now I will define 0! to be 1, so that my notation will be easier to use. With the definition (or understanding!) y can be written as _n=1^infinity(1/n!)xⁿ.

We learned a while ago that solutions for Initial Value Problems should be unique. So we see clearly that
e^x=_n=1^infinity(1/n!)xⁿ.
This is certainly not clear. I sort of weaseled (?) a reason why the infinite sum should be considered a solution of the IVP. So maybe the infinite sum "is" such a solution. But there is no reason to suspect that it has to represent e^x. In fact, everything I've done is actually correct, and the implied conclusions (that the series converges, and that the sum of the infinite series is e^x, and that computing partial sums can be efficiently done and provides a great way to compute values of the exponential function) are all totally correct.

Definition of power series
A power series centered at a is a series of the form
_n=0^infinityc_n(x-a)ⁿ.
Here the numbers c_n are the coefficients of the series, and the whole thing looks like some sort of infinitely long polynomial.

An example
Here is a power series centered at a=0:
_n=0^infinityxⁿ/(3n+1)
For which x's does this series converge? Well, we just return to the thinking mode of the previous lecture, with a_n=xⁿ/(3n+1), and we try to think of a strategy to investigate convergence. The prominence of xⁿ in the formula for a_n makes me want to use either the Ratio Test or the Root Test. I always feel more at home (?) with the Ratio Test (personal preference) so I'll try that.

           |x|n+1
 |an+1|   --------
          3(n+1)+1          |x|n+1(3n+1)          (3n+1)
------ = -------------- = --------------- = |x|· -------
            |x|n           {3(n+1)+1}|x|n         (3n+3)
 |an|      ------
            3n+1

What happens if |x|=1? We need to use other methods. There are two x's which must be considered.

To the right is a picture (yes, I work diligently to "translate" almost any information into pictures, so just ignore if you are not a picture person!) of the real line. I have tried to indicate how the line is divided into two collections of points. The red points (with D) are where this series diverges, and the green points (with C) are where this series converges. The behavior at the boundary of the intervals is sort of difficult to draw (!) but I hope the use of ] and [ and ( and ) help a bit.

Another example ...
Here is a power series centered at a=4:
_n=0^infinity(x-4)ⁿ/(3n+1)
For which x's will this series converge?
My hope was that students would make their own mental connections between the work needed to detect where this series converges and the work we just did. For example:

The chunk of text on the left should change to the chunk of text on the right.

... Well, we just return to the thinking mode of the previous lecture, with an=xn/(3n+1), and we try to think of a strategy to investigate convergence. The prominence of xn in the formula for an makes me want to use either the Ratio Test or the Root Test. I always feel more at home (?) with the Ratio Test (personal preference) so I'll try that. |x|n+1 |an+1| -------- 3(n+1)+1 |x|n+1(3n+1) (3n+1) ------ = -------------- = --------------- = |x|· ------- |x|n {3(n+1)+1}|x|n (3n+3) |an| ------ 3n+1 Now we need to have n-->infinity. The result is |x| ...

... Well, we just return to the thinking mode of the previous lecture, with an=(x-4)n/(3n+1), and we try to think of a strategy to investigate convergence. The prominence of (x-4)n in the formula for an makes me want to use either the Ratio Test or the Root Test. I always feel more at home (?) with the Ratio Test (personal preference) so I'll try that. |x-4|n+1 |an+1| -------- 3(n+1)+1 |x-4|n+1(3n+1) (3n+1) ------ = -------------- = --------------- = |x-4|· ------- |x-4|n {3(n+1)+1}|x-4|n (3n+3) |an| ------ 3n+1 Now we need to have n-->infinity. The result is |x-4| ...

The logic should go totally in parallel, with x changing to x-4. Another way of seeing what's going on is to consider the series _n=0^infinity(x-4)ⁿ/(3n+1) and to ask, as I tried to do, what's the simplest (?) value of x to worry about? I claimed that this value of x is 4 because x=4 makes all of the (x-4)^{some power} equal to 0. So I hope this these considerations make the following conclusion believable.

The series converges if |x-4|<1, which is the same as -1<x-4<1, which is the same as 3<x<5. The series diverges if |x-4|>1, which is the same as x<3 or x>5. The series diverges for x=5 and converges for x=3.

And maybe another ...
How about _n=0^infinity(x+20)ⁿ/(3n+1)? I hoped that students would "see" the answer, which is shown graphically to the left. With more definiteness, I declare that this series converges if |x+20|<1, which is the same as -1<x+20<1, which is the same as -21<x<-19. The series diverges if |x+20|>1, which is the same as x<-21 or x>-19. The series diverges for x=-19 and converges for x=-21.

Actually, the coefficients know it all
The information about how a series _n=0^infinityc_n(x-a)ⁿ converges is really contained in its collection of coefficients, the c_n's. The "center" x=a just moves it around. Therefore here (and in almost all of the applications I know!) I will almost always consider series whose center is 0, _n=0^infinityc_nxⁿ, and not even worry about the x=a case. If we need to be concerned with a=500 or a=-sqrt(2), well, then, I will do the computations around x=0 and then just translate the results to make things work. The example is (more or less!) generally correct
I'll try to illustrate with some general reasoning here why people like power series. I'll give an argument to show that convergence questions follow some straightforward logic. Suppose that I have a power series _n=0^infinityc_nxⁿ which converges at x₁, shown on the real line to the right. What can I say about convergence at x₂, also shown to the right. All that I know is that x₂ is closer to the origin than x₁. And I do not tell you anything more about the coefficients of the series. What follows is a very important and also slightly tricky bit of reasoning.

I know _n=0^infinityc_nx₁ⁿ converges. Then (look at the first entry in the left-hand column of the table which began the previous lecture!) c_nx₁ⁿ-->0 because the individual terms in any series which converges must get smaller. But this means that there is some (maybe big!) number M so that |c_nx₁ⁿ|<M (if there were not such an M, some collection of the terms of the series would grow and grow, and they couldn't approach 0).

I want to know if the series converges at x₂. I'll consider absolute convergence. Maybe that's a stricter condition, but it is easier to work with. Look:
_n=0^infinity|c_n|·|x₂|ⁿ=_n=0^infinity|c_n| |x₁|ⁿ(|x₁|/|x₂|)ⁿ<_n=0^infinityM(|x₁|/|x₂|)ⁿ.
The first equality occurs because |x₂|=|x₁|·(|x₂|/|x₁|) and the second inequality occurs because we're overestimating by the M on each term. Please realize that these sorts of estimates are commonly used in real computations! But the series we ended up with is a geometric series with ratio equal to |x₁|/|x₂| and this is less than 1 and therefore it must converge. The Comparison Test then implies that the series we began with, the power series at x₂, must also converge. The only relationship we needed between x₁ and x₂ to make this work is that the distance to 0 of x₂ is less than the distance to 0 of x₁. The x₂ can be on the other side of the origin (because only absolute values appear in the discussion!). If you'd like to put the idea in terms of disease or contagion or something, then all of the x₂'s closer to the origin than x₁ "catch" convergence from x₁. Maybe the picture to the right sort of illustrates the situation. (The only peculiarity is that the "opposite" point, -x₁, may not have convergence, but that's because a geometric series with ratio=1 does not converge, so the argument presented here does not allow any conclusion.)

Back to the example
In the case of the example just presented, _n=0^infinityxⁿ/(3n+1), the interval of convergence is -1<=x<1, with 0, the center of the power series, sitting in the middle of the interval.

What sort of function is the sum?
It turns out that the function which is sum of the power series is very nice inside the interval of convergence: it is continuous (no breaks or jumps) and it is differentiable (very smooth). But before this is discussed in detail, I wanted to show people ...

How to gamble
What I discussed is presented here.

Convergence of series

Facts to know

If an converges, then an-->0. So if the limit is not 0 or if the limit does not exist, the series must diverge. Series with positive terms Comparison Test; Integral Test; Limit Comparison Test. Alternating series If three conditions are true, then the series converges. If some of the conditions are not true, generally there's no conclusion. Absolute convergence implies convergence. Ratio Test Consider an, and try to compute limn-->infinity|an+1|/|an|=L. If the limit exists, then: L<1 implies absolute convergence; L>1 implies divergence; L=1 gives no information. Root Test Consider an, and try to compute limn-->infinity|an|1/n=L. If the limit exists, then: L<1 implies absolute convergence; L>1 implies divergence; L=1 gives no information.

Sequence facts     rn-->0 if |r|<1.     a1/n-->1 if a>0.     n1/n-->1.     (1+{1/n})n-->e. Series facts Geometric series 0infinityarn=1/{1-r} if |r|<1. Diverges if |r|>=1 and a is not 0. p-series 1infinity1/np converges if p>1 and diverges if p<=1.

Remarks

What was done
I believe we discussed
11.6: 6 (look also at 5), 8, 12, 23, 27 (look also at 25) and 11.7: 1, 6, 13, 15, ...
We tried to determine if these series converge absolutely, converge conditionally, or diverge.

The QotD was 11.7: 17: does ₁^infinity(-1)ⁿ2^1/n converge or diverge?

A possible vocabulary word: retrograde, meaning "move backward in an orbit" (under certain circumstances, planets being observed seem to move backwards in their orbits) and then also "move in a direction contrary to the usual one".

Comparison
We've been dealing with results which really depend on the terms of the series being positive. For example, the Comparision Test. Here's a version:

Hypothesis Suppose 0<=a_j<=b_j.
Conclusion If _j=1^infinityb_j converges, then _j=1^infinitya_j converges.
If _j=1^infinitya_j diverges, then _j=1^infinityb_j diverges.

Another comparison result
Please look in the text for some good examples using the Limit Comparison Test which can be useful. There are numerous textbook questions which you can try. Here is a statement:

Hypothesis Suppose a_j and b_j are both always positive, and that the lim_j-->infinitya_j/b_j exists and is positive.
Conclusion Either both _j=1^infinitya_j and _j=1^infinityb_j converge or
both _j=1^infinitya_j and _j=1^infinityb_j diverge.

Series with terms of varying signs
So now we will complicate things a bit, and look at series whose signs vary. Let me start really easily but things will get more intricate rapidly. (Varying stop signs)

1-1+1-1+1-...
This is just about the simplest example I could show. We got a formula for the j^th term. We need the sign to alternate, and that will be given by (-1)^{something here}. The sign will alternate if the "something here" is either j or j+1. The first term will be +1 and the second term will be -1 if we use j+1. So an explicit formula is a_j=(-1)^j+1. Next I asked about convergences of the series _j=1^infinity(-1)^j+1. For this we must consider the sequence of partial sums.
      s₁=1; s₂=1-1=0; s₃=1-1+1=1; s₄=0, etc.
It isn't too difficult to see that s_{even integer}=0 and s_odd integer=1. The partial sums flip back and forth. This is exactly the kind of behavior we did not get when we considered series with all positive terms. There the partial sums just traveled "up", to the right. Well, this particular infinite series does not converge, since the partial sums do not approach a unique limit.
      _j=1^infinity(-1)^j+1 diverges.

2-(1{1/2})-(1{1/3})+(1{1/4})-(1{1/5})+...
This is a more complicated series. I suggested that we try to "guess" a formula by first getting a formula for the sign, and then a formula for the absolute value. In this case, the sign is surely given by (-1)^j+1, just as before. The magnitude (these are 1-dimensional vectors, after all!) or absolute value is 1+{1/j} (some students suggested {j+1}/j, which is also a good formula. So putting these together, a_j=(-1)^j(1+{1/j}). And now we looked at the {con|di}vergence of _j=1^infinity(-1)^j+1(1+{1/j}).

The partial sums are more complicated and more interesting.
      s₁=2; s₂=2-(1{1/2})-{1/2}=.5; s₃=2-(1{1/2})+{1{1/3})=11/6=1.8333; s₄=2-(1{1/2})+(1{1/3})-(1{1/4})={7/12}=.58333
This is where I stopped computing in class, but, golly, I have a friend who could compute s₁₇ either exactly ({4218925/2450448}) or approximately (1.72169). This is nearly silly. Richard Hamming, one of the twentieth century's greatest applied mathematicians, remarked that

From s₁ to s₂, we move left since the second term in the series is negative. From s₂ to s₃ we move right, because the third term in the series is positive. But notice that we don't get to s₁. because the jump right has magnitude 1{1/3} and this is less than 1{1/2}the magnitude of the previous jump left.

I hope you are willing to believe that what's described persists in general.

Does this series converge? Students in both classes had interestingly varied opinions about this, and I will admit that I tried to encourage discussion. But the collection of partial sums does not approach a unique limit. Why? Well, the distance between any odd partial sum and any even partial sum will be at least 1, since the magnitude of the j^th term is 1{1/j}, which is certainly >1. The partial sums can't get close.
      _j=1^infinity(-1)^j+1(1+{1/j}) diverges.

1-1/2+1/3-1/4+1/5-...
Here a_j has sign (-1)^j+1 again, and the absolute value or magnitude is 1/j. Does _j=1^infinity(-1)^j+1(1/j) converge? The partial sums are more complicated and more interesting.
      s₁=1; s₂=1-(1/2)-1/2=.5; s₃=1-(1/2)+{1/3)=5/6=.8333; s₄=1-(1/2)+(1/3)-(1/4)=7/12=.58333
Here's a picture of these partial sums. Things are a bit more crowded (that's good for convergence!) than in the previous picture.

The previous three qualitative properties still hold. Since the signs alternate, the partial sums wiggle left and right. Since the absolute values decrease, the odd sums are less than the even sums, and all of the even sums are less than all of the odd sums. But now the distance between the odd and even sums-->0 since the magnitude of the terms is 1/j, and this-->0. So here is a rather subtle phenomenon:
      _j=1^infinity(-1)^j+1(1/j) converges.

The theorem on alternating series (Alternating Series Test)
The following is the major result of section 11.5 of the text.

Hypotheses Suppose that

The terms of a series alternate in sign (+ to - to + etc.).

The absolute value or magnitude of the terms decreases.

The limit of the absolute values of these terms is 0.
Conclusion The series converges.

Finally, here is some "experimental evidence" which might help you believe that the alternating harmonic series converges.

Some partial sums of the alternating harmonic series
s10=0.6456349206 s100=0.6456349206 s1,000=0.6881721793 s10,000=0.6926474305 s100,000=0.6930971829 s1,000,000=0.6931466807

In fact, as several people guessed, the sum of the alternating harmonic series is ln(2). But the convergence is actually incredibly slow. The one millionth partial sum (which took almost 8 seconds for a moderately fast PC to compute) only has 5 accurate decimal digits. This is not the best and fastest way to compute things!

But what if ...
The sign distribution of terms in an infinite series could be more complicated. I suggested that we consider something like

             7cos(36j7-2j2)+2sin(55j+88)
j=1infinity ----------------------------
                         2j

Please notice that with a few modifications, the corresponding question can be answered very easily. Look at:

 
             7|cos(36j7-2j2)|+2|sin(55j+88)|
j=1infinity ----------------------------------
                           2j

Proof via manipulative
There was a spectacular demonstration in class! It wa inspired by thinking about old-fashioned folding carpenter's rulers. If we have an infinite series _j=1^infinitya_j, we could consider the associated series _j=1^infinity|a_j|, where we have stripped off the signed of the terms, and are just adding up the magnitudes. This is sort of like an unfolded carpenter's rule, stretched out as long as possible. It may happen that the series of absolute values, a series of positive terms, may converge. So when "the ruler" is stretched out as long as possible, it has finite length. Well, if we then fold up the ruler, so some segments point left (negative) and some point right (positive) then the resulting length is also positive.

The picture to the left as an attempt to show this statement and to duplicate the physical effect of what I displayed in class. The top has the segments stretched out as far as possible. The next picture shows some of the segments rotating, aimed backwards (negatively). The last segment shows in reg segments which are negative and in green the other segments, oriented postively. I hope this makes sense, and justifies the following:
If _j=1^infinity|a_j| converges, then _j=1^infinitya_j must converge also (and, actually, |_j=1^infinitya_j|<=_j=1^infinity|a_j|).

Proof via algebra
There is a verification of these statements in the textbook, using algebra, on p.741 in section 11.6, if you would like to read it. Sigh.

And conversely?
Notice that the converse of the assertion about absolute values and series may not be correct. That is, a series may converge, and the series of absolute values of its terms may not. The simplest example, already verified, used the alternating harmonic series, divergent, and the harmonic series, convergent.

Vocabulary
A series _j=1^infinitya_j which has _j=1^infinity|a_j| converging is called absolutely convergent. Then the correct implication above is:

If a series is absolutely convergent, then it is a convergent series.

Another example
Consider _j=1^infinity{sin(5j+8)}³⁷/j⁵. I don't know very much about {sin(5j+8)}³⁷ except that, for any j, this is a number inside the interval [-1,1]. Therefore _j=1^infinity|{sin(5j+8)}³⁷/j⁵| has terms which are all smaller than _j=1^infinity1/j⁵ (a p-series with p=5>1, so convergent)> The comparison test asserts that _j=1^infinity|(sin(5j+8)³⁷/j⁵| converges, and therefore _j=1^infinity{sin(5j+8)}³⁷/j⁵ itself must be a convergent series.

Given a series, take absolute values
The result just stated is a very powerful and easily used method. If you "give" me a series with random signs, the first thing I will do is strip off or discrad the signs and try to decide if the series of positive (non-negative, really) terms converges.

Many of the classical "tests" are about how the series resembles the geometric series. So we think about how a_j might look like ar^j for j large, and make conclusions based on this. Let's look at an example.

An example from the text
Here is problem #3 from section 11.6. It asks if this series is absolutely convergent, conditionally convergent, or divergent:
      _n=0^infinity(-10)ⁿ/n!.
The text uses n as the index of summation here, which shouldn't be too distressing. The text starts the sum from n=0 rather than n=1, but, again, this shouldn't be too bad, since {con|di}vergence depends on the infinite tail, and the initial terms don't affect that at all. More subtly, if you inspect the formula for the term in the series, the text uses 0!, which I don't think has been previously mentioned. 0! means 1, and the reason for that understand will be given later.
The answer It turns out that this series does converge (absolutely!) and, actually, its sum is e¹⁰. This is not obvious, and, again, the reasoning will be given later. Forget this premature statement of the conclusion and try to look at the sum, first using a "primitive" approach.

Let's strip off the signs. Then |a_n|=(10)ⁿ/n!. For example, |a₇|=10⁷/7!=10,000,000/5040=1984 (approximately). This is large. But what if we consider how |a₂₀| and |a₂₁| are related? Then we need to investigate how (10)²⁰/20! changes into (10)²¹/21!. The top multiplies by 10, of course, but the bottom changes by a factor of 21. The change from |a₂₀| to |a₂₁| is a multiplication by 10/21. And further thought should show that the next change involves multiplication by 10/22, and then 10/23, etc. All of the changes after n=20 use multiplication by numbers less than 1/2. Certainly the series converges (compare with a geometric series whose ratio is 1/2<1). There is a way to "mechanize" these comparisions and it is called the Ratio Test. The Ratio Test
Suppose we are given an infinite series with n^th term a_n and suppose that lim_n-->infinity|a_n+1/a_n| exists. If this limit is less than 1, then the original series converges absolutely and therefore must converge. If the limit is greater than 1, the original series diverges. If the limit happens to be equal to 1, the Ratio Test does not supply any information about convergence or divergence.

Return to the example from the text
Let's use the Ratio Test on the series _n=0^infinity(-10)ⁿ/n!. So we need to consider a limit. Here it is:

              |(-10)n+1|
              |--------| 
              | (n+1)! |                 10n+1n!                 10
limn-->infinity----------- =limn-->infinity---------=limn-->infinity-----= 0
               |(-10)n|                 (n+1)!10n                n+1
               |------| 
               |  n!  |

QotD
Simplify:

    23(n+1)
  ---------
  ((n+1)!)3
-------------
     23n
   ------
    (n!)3

Series with positive terms
Today we will consider series whose terms are positive or, at worst, non-negative. (In the next lecture we'll discuss what happens if we allow different signs, but for now, only +'s.) What can we say in general about series with positive (or even just non-negative) terms? Well, the sequence of partial sums is increasing, since we're just adding more and more non-negative terms. What can happen? One thing is that the sequence of partial sums can tend to infinity (hey, this is what happens to the silly infinite series _j=1^infinity1: the sequence of partial sums is unbounded and the series diverges. Another thing that can happen to a positive series is that the sequence of partial sums can tend to a limit (a non-negative finite limit). This happens, for example, with positive geometric series with ratio less than 1. Then the sequence of partial sums is bounded and the series converges. This is a consequence of the fact that "Bounded monotonic sequences converge".

This theoretical alternative is everything.

• For a series with positive terms, bounded partial sums means the same thing as convergence.
• For a series with positive terms, unbounded partial sums means the same thing as divergence.

Comparing series with non-negative terms
Suppose we have two infinite series, _j=1^infinitya_j and _j=1^infinityb_j. We also will assume that both of the series have non-negative terms. This means that all of the a_jj's and all of the b_j's are >=0.

Hypothesis Suppose we know that the terms of one series are always smaller than the terms of the other series. So, specifically, suppose we know that for all j, a_j<=b_j.
Conclusion Suppose s_n=_j=1ⁿa_j of the series with smaller terms, and t_n=_j=1ⁿb_j is the n^th partial sum of the series with larger terms. Then s_n<=t_n.

Comparing the series and inheriting {con|di}vergence

• Suppose the bigger series, _j=1^infinityb_j, converges. Then the smaller series, _j=1^infinitya_j, must also converge.
  Why? The bigger series has bounded partial sums. The smaller series has partial sums which are less than those of the bigger series, so the smaller series also has bounded partial sums. Therefore it converges.
• Suppose the smaller series, _j=1^infinitya_j, diverges. Then the bigger series, _j=1^infinityb_j, must also diverge.
  Why? The smaller series has unbounded partial sums, which means we can make them as large as we like (the partial sums-->infinity). The partial sums of the larger series are bigger than those partial sums, so we can make them also larger than any specified number. Therefore, since unbounded partial sums imply divergence, the larger series must also diverge.

Easy examples
Making the bottoms bigger in fractions shrinks the value. So the fraction 1/(56+7j+2^j) is always less than 1/2^j. I know that _j=1^infinity1/2^j converges, and therefore the series _j=1^infinity(56+7j+2^j) must also converge.

Making the tops larger makes the value larger. Since I know that _j=1^infinity1/j, the harmonic series, diverges, I must know that _j=1^infinity[3+sin(j)]/j also diverges. This is because 3+sin(j)>=2 always.

The prototypes
The examples you need to know are these:
      Geometric series a+ar+ar²+ar³+ar⁴+... converges if |r|<1. When |r|>=1 (and a is not zero!) then the series diverges.
      p-series 1+1/2^p+1/3^p+1/4^p+1/5^p+1/6^p+... converges if p>1 and diverges if p<=1.
You should be able to do most of the problems in section 11.4 with these in mind. You should practice. I want to discuss some useful but more complicated applications.

Bounds on tails with integrals
Look at _j=1^infinity1/(arctan(j)+j³). This is a horrible series. I don't know much about arctan(j) when j is a positive integer except that the values are between 0 and Pi/2. Compare this series with _j=1^infinity1/j³ whose individual terms are each larger than the original series. The integral test tells me that this series converges since ₁^infinity(1/x³)dx is finite. (Its value is 1/2: please check this!).

So I know that _j=1^infinity1/(arctan(j)+j³) converges. Well, that's fine but suppose I really need to know what the sum is? For example, suppose I want to know the sum to 3 decimal places (error less than 0.001). Here is a strategy. I write the infinite sum as a sum of two pieces, a partial sum and an infinite tail.
      _j=1^infinity1/(arctan(j)+j³)=_j=1ⁿ1/(arctan(j)+j³)+_j=n+1^infinity1/(arctan(j)+j³)=s_n+the n^th "infinite tail"
The infinite tail is _j=n+1^infinity1/(arctan(j)+j³). But each piece of that sum is less than _j=n+1^infinity1/j³. Look again at the picture here. So we know:
      0<=_j=n+1^infinity1/(arctan(j)+j³)<_j=n+1^infinity1/j³<_n^infinity(1/x³)dx.
Here's a computation of the improper integral:
_n^infinity(1/x³)dx= lim_{A-->infinityn}^A(1/x³)dx= lim_A-->infinity-1/(2x²)]_n^A= lim_A-->infinity-1/(2A²)+1/(2n²)= 1/(2n²).
So the error between the true sum and the n^th partial sum is less than 1/(2n²). How can I get this less than 0.001=1/1,000? Let's take n to be, say, 25. Then 2(25)²=1,250. That works.

Therefore the sum of the series _j=1^infinity1/(arctan(j)+j³) is the same as _j=1²⁵1/(arctan(j)+j³) to 3 decimal digit accuracy. The following Maple command and its answer took less than a hundredth of a second:

add(evalf(1/(arctan(j)+j^3)),j=1..25);
                                 0.7440955743

Bounds on tails with geometric series
Let's try a different sort of series. I'll write some partial sums:

1
     2.5
1 + ------
      1! 
     2.5     (2.5)2 
1 + ------ + ------
      1!       2!
     2.5     (2.5)2   (2.5)3 
1 + ------ + ------ + ------
      1!       2!       3! 
     2.5     (2.5)2   (2.5)3   (2.5)4 
1 + ------ + ------ + ------ + ------
      1!       2!       3!       4!
     2.5     (2.5)2   (2.5)3   (2.5)4   (2.5)5 
1 + ------ + ------ + ------ + ------ + ------
      1!       2!       3!       4!       5!

The sum of the terms up to and including the term with (2.5)5/5! is 11.6705. The sum of the terms up to and including the term with (2.5)10/10! is 12.1817. The sum of the terms up to and including the term with (2.5)15/15! is 12.1824. The sum of the terms up to and including the term with (2.5)20/20! is 12.1824.

I bet this series converges, and that its sum is 12.18 (two decimal place accuracy). I'll verify this by looking at the tail _j=11^infinitya_j. Here a_j=(2.5)^j/j! The first term is a₁₁ which is 0.000597. I computed this -- the value is not "obvious" from the formula. This is not too interesting, because I need information about all of the terms, and there are infinitely many. Let me investigate how the terms are related to one another in the following way:

            (2.5)j+1
            ------
 aj+1       (j+1)!      (2.5)j+1j!     (2.5)j!     2.5
------ =  ---------- = ----------- = --------- = -----
  aj        (2.5)j     (2.5)j(j+1)!   (j+1)j!     j+1
            ------ 
              j!

If j is at least 11, then 2.5/(j+1) will be at most 2.5/12 and this is less than 0.209. Therefore we know that if j is at least 11, then a_j+1/a_j<0.209, so that a_j+1<a_j(0.209). We can replace each term by 0.209 multiplied by the term before it, and this is an overestimate.

Now what? Look at the tail more closely:
_j=11^infinitya_j=a₁₁+a₁₂+a₁₃+a₁₄+a₁₅+a₁₆+a₁₇+...<
a₁₁+a₁₁(0.209)+a₁₂(0.209)+a₁₃(0.209)+a₁₄(0.209)+a₁₅(0.209)+a₁₆(.209)+...<
a₁₁+a₁₁(0.209)+a₁₁(0.209)²+a₁₂(0.209)²+a₁₃(0.209)²+a₁₄(0.209)²+a₁₅(0.209)²+...<
a₁₁+a₁₁(0.209)+a₁₁(0.209)²+a₁₁(0.209)³+a₁₂(0.209)³+a₁₃(0.209)³+a₁₄(0.209)³+...<
a₁₁+a₁₆(0.209)+a₁₁(0.209)²+a₁₁(0.209)³+a₁₁(0.209)⁴+a₁₂(0.209)⁴+a₁₃(0.209)⁴+...<
a₁₁+a₁₁(0.209)+a₁₁(0.209)²+a₁₁(0.209)³+a₁₁(0.209)⁴+a₁₁(0.209)⁵+a₁₂(0.209)⁵+...<
a₁₁+a₁₁(0.209)+a₁₁(0.209)²+a₁₁(0.209)³+a₁₁(0.209)⁴+a₁₁(0.209)⁵+a₁₁(0.209)⁶+...<
ETC.
Here what I mean by "etc." is that we can keep pushing the subscripts "down" until they reach 11, each time "paying" by multiplying by another power of 0.209. And these are successive overestimates. So we finally end up with a geometric series whose first term is a₁₁=0.000597, with ratio=0.209. This series has sum equal to 0.000597/(1-0.209) which is about 0.00076. Hey! This number is less than .001, and I bet that the true sum of the series therefore has leading decimal expansion 12.181 or 12.182.

Another example
Here we go:

1
     9.8
1 + ------
      1! 
     9.8     (9.8)2 
1 + ------ + ------
      1!       2!
     9.8     (9.8)2   (9.8)3 
1 + ------ + ------ + ------
      1!       2!       3! 
     9.8     (9.8)2   (9.8)3   (9.8)4 
1 + ------ + ------ + ------ + ------
      1!       2!       3!       4!
     9.8     (9.8)2   (9.8)3   (9.8)4   (9.8)5 
1 + ------ + ------ + ------ + ------ + ------
      1!       2!       3!       4!       5!

The sum of the terms up to and including the term with (9.8)10/10! is 10965.326. The sum of the terms up to and including the term with (9.8)20/20! is 18011.195. The sum of the terms up to and including the term with (9.8)30/30! is 18033.744. The sum of the terms up to and including the term with (9.8)40/40! is 18033.744.

I bet this series converges, and that its sum is 18033.74 (two decimal place accuracy). I'll verify this by looking at the tail _j=31^infinitya_j. Here a_j=(9.8)^j/j! The first term is a₃₁ which is 0.000650. I computed this -- the value is not "obvious" from the formula. This is not too interesting, because I need information about all of the terms, and there are infinitely many. Let me investigate how the terms are related to one another in the following way:

            (9.8)j+1
            ------
 aj+1       (j+1)!      (9.8)j+1j!     (9.8)j!     9.8
------ =  ---------- = ----------- = --------- = -----
  aj        (9.8)j     (9.8)j(j+1)!   (j+1)j!     j+1
            ------ 
              j!

If j is at least 31, then 9.8/(j+1) will be at most 9.8/32 and this is 0.306 (exactly! wow!). Therefore we know that if j is at least 31, then a_j+1/a_j<0.306, so that a_j+1<a_j(0.306). We can replace each term by 0.306 multiplied by the term before it, and this is an overestimate.

Now what? Look at the tail more closely:
_j=31^infinitya_j=a₃₁+a₃₂+a₃₃+a₃₄+a₃₅+a₃₆+a₃₇+...<
a₃₁+a₃₁(0.306)+a₃₂(0.306)+a₃₃(0.306)+a₃₄(0.306)+a₃₅(0.306)+a₃₆(.3)+...<
a₃₁+a₃₁(0.306)+a₃₁(0.306)²+a₃₂(0.306)²+a₃₃(0.306)²+a₃₄(0.306)²+a₃₅(0.306)²+...<
a₃₁+a₃₁(0.306)+a₃₁(0.306)²+a₃₁(0.306)³+a₃₂(0.306)³+a₃₃(0.306)³+a₃₄(0.306)³+...<
a₃₁+a₃₆(0.306)+a₃₁(0.306)²+a₃₁(0.306)³+a₃₁(0.306)⁴+a₃₂(0.306)⁴+a₃₃(0.306)⁴+...<
a₃₁+a₃₁(0.306)+a₃₁(0.306)²+a₃₁(0.306)³+a₃₁(0.306)⁴+a₃₁(0.306)⁵+a₃₂(0.306)⁵+...<
a₃₁+a₃₁(0.306)+a₃₁(0.306)²+a₃₁(0.306)³+a₃₁(0.306)⁴+a₃₁(0.306)⁵+a₃₁(0.306)⁶+...<
ETC.
Here what I mean by "etc." is that we can keep pushing the subscripts "down" until they reach 31, each time "paying" by multiplying by another power of 0.306. And these are successive overestimates. So we finally end up with a geometric series whose first term is a₃₁=0.000650, with ratio=0.306. This series has sum equal to 0.000650/(1-0.306) which is about 0.000936. Hey! This number is less than .001, and I bet that the true sum of the series therefore has leading decimal expansion 18033.744 or 18033.745.

Random (?) facts
e2.5 is approximately 12.18249396 e9.8 is approximately 18033.74493

Not at all random, and this will be explained later.

QotD
The series _j=1^infinity1/(j⁵+3^j) converges. I told students that _j=1¹⁰1/(j⁵+3^j)=0.27951. Then I asked that students find an error estimate. So some sort of overestimate of _j=11^infinity1/(j⁵+3^j) is needed. Several strategies can be used to get valid answers. Here are two strategies.

Compare to a p-series and then estimate with an integral
Certainly for any positive integer j, 1/(j⁵+3^j)<1/(j⁵). Therefore _j=11^infinity1/(j⁵+3^j)<_j=11^infinity1/(j⁵). This sum is, in turn, less than the improper integral ₁₀^infinity(1/x⁵)dx. Computation of the integral:
₁₀^infinity(1/x⁵)dx=lim_{A-->infinity10}^A(1/x⁵)dx=lim_A-->infinity-1/(4x⁴)₁₀^A=lim_A-->infinity-1/(4A²)+1/(4·10⁴)=1/(4·10⁴). This is a fine answer just as it is!

Compare to a geometric series and then get the sum of the geometric series
Certainly for any positive integer j, 1/(j⁵+3^j)<1/3^j. Therefore _j=11^infinity1/(j⁵+3^j)<_j=11^infinity1/3^j. But this infinite series is a geometric series whose first term is 1/3¹¹ and whose ratio is 1/3. The sum of this series is {1/3¹¹}/(1-{1/3}). This is a fine answer just as it is!

Remarks about the QotD solutions

1. Why should the improper integral have its lower bound equal to 10 rather than 11? A number of students used 11 as the lower bound. The aim is to overestimate the infinite tail _j=11^infinity1/(j⁵+3^j) by something easier to compute. The most important word segment here is over in "overestimate". With this in mind, please look at the picture to the right. It should convince you that if you want to overestimate _j=11^infinity(1/j⁵) by something like _?^infinity(1/x⁵)dx that the correct choice for ? is 10.
2. Which error estimate is "better"? Except for rather rare circumstances, in real life getting any useful estimate is good enough. In this case, the first error estimate, 1/(4·10⁴), is 0.000025, and the second error estimate, {1/3¹¹}/(1-{1/3}), is 0.00000847. I guess I would generally think that 1/3^j will give something smaller than 1/j⁵, since exponentials grow faster than polynomials. But both estimates show that the number given for the tenth partial sum, 0.27951, is accurate to at least 4 decimal places.

Again, a series is a formal sum, a₁+a₂+a₃+... Your text writes this using sigmas, _j=1^infinitya_j.

One comment Please realize that the j doesn't really matter. It is what is called a dummy variable. The j is a logical placeholder. The sum _k=1^infinitya_k represents the same thing, and the sum _q=1^infinitya_q represents the same thing, and so, I suppose, the sum which follows represents the same thing: _TOAD=1^infinitya_TOAD (why can't we use words like TOAD as index variables?).

Associated with each series are two different sequences. One sequence is the sequence of terms of the series: {a_j}. Another sequence is the sequence of partial sums of the series: this sequence looks like this: a₁, a₁+a₂, a₁+a₂+a₃, a₁+a₂+a₃+a₄. So the 400^th partial sum would be _j=1⁴⁰⁰a_j.

A specific example: the harmonic series
The harmonic series 1+1/2+1/3+1/4+1/5+...=_j=1^infinity1/j has these two series associated with it:

The sequence of individual terms
The first five terms: 1, 1/2, 1/3, 1/4, 1/5, .... I hope that the asymptotic behavior of this sequence whose j^th term is 1/j as j gets large is obvious: the sequence of individual terms -->0.

The sequence of partial sums
The first five terms: 1, 1+1/2=3/2 (1.5), 1+1/2+1/3=11/6 (about 1.8333), 1+1/2+1/3+1/4=25/12 (about 2.0833), 1+1/2+1/3+1/4+1/5=137/60 (about 2.2833). The asymptotic behavior of this sequence is certainly not obvious or clear or ... We used some quite tricky reasoning last time to see that this sequence of partial sums is not bounded. They get really really large. I'll try to show this again later in today's lecture using a different approach which is maybe more systematic.

A series converges if ...
A series converges if the sequence of partial sums converges. The limit of the sequence of partial sums is frequently called the sum of the series.

It can be difficult to decide if a series converges (heck, the harmonic series does not, even though the terms go to 0). Even if you know a series converges, it can be difficult to approximate the sum of the series. This is interesting, because most of the computations that are commonly done using calculators and computers (such as sin(.567) or e^3.45) consist of numerical approximations to sums of infinite series. So what we are doing is interesting and useful.

The infinite tail
An infinite series _j=1^infinitya_j can be thought of as _j=1ⁿa_j+_j=n+1^infinitya_j. So there is the n^th partial sum plus the "other" terms of the series. 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 weight of the animal is finite (this is a good analogy only with 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-->infinity. In fact, the first "few" terms of a series have nothing to do with convergence!

For example, we already know that the harminic series 1+1/2+1/3+1/4+1/5+... does not converge. Then the series 56+37.8+409+1/4+1/5+1/6+... (all other terms are the terms in the harmonic series, unchanged) also does not converge. The sequence of partial sums has some change (1-->56, 1/2-->37.8, 1/3-->409) but the fact that the partial sums eventually do not approach a one fixed finite number is still correct.

In a few minutes (geometric series) we will see that the series 1/2+1/4+1/8+1/16+... (here the n^th term is 1/2ⁿ) does converge, and its sum is 1. If I change the first term to 15 and the third term to 88, the series still will converge. Its sum will be 1+(15-1)+(88-1/8) (the changes are made to the old sum).

So really when we study convergence we are looking at infinite tails.

Geometric series
One kind of series has nice, simple formulas for partial sums and for sums, and that's geometric series. A geometric series consists of some sort of starting term, and each further term is formed by multiplying the previous term by some constant factor. So:

1. A series
   5+5/7+5/7²+5/7³+5/7⁴+... is a geometric series. The starting term is 5, and the constant factor, the multiplier, is 1/7. The use of "..." here is traditional. It is supposed to be telling the reader that, "Hey, I've given you enough evidence, enough of the pattern, you should now be able to figure out what the general formula of the series is." Sometimes the ... implied pattern is clear, and sometimes, darn it!, it is not.
   
2. Another series
   3-3/19+3/(19)²-3/(19)³+3/(19)⁴-3/(19)⁴+3/(19)⁵... This is also a geometric series. Here the first term is 3, and the multiplier, usually called the ratio, between successive terms is -1/19.
   
3. And another
   Look carefully, please, at 7+2·7/9+3·7/9²+4·7/9³+5·7/9⁴+6·7/9⁵+... and try to tell if this is a geometric series. In fact, looking at any three consecutive terms is enough to disqualify this series. For example, we start with 7 and the next term is 2·7/9. These terms dictate that the ratio is 2/9. But the term after 2·7/9 is 3·7/9² and these two terms indicate that the ratio must be (3/2)/9. Now 2/9 and 3/18 are not the same. The ratio between successive terms must be constant in a geometric series, so this series is not a geometric series.

Formula for the nth partial sum of a geometric series
a-arn sn= --------      1-r

This is a neat formula which is sometimes very useful. Please realize that an explicit formula for the partial sum of a series is very, very rare. It should not be expected.
We saw last time (look in section 11.1, page 707) that if |r|<1, then rⁿ-->0 as n-->infinity. Therefore:

Formula for the sum of a geometric series when |r|<1
a s = -------      1-r

Use #1 of geometric series
Here is something which many people are supposed to see (!) 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, a=731/1000, and whose constant ratio between successive terms is r=1/1000. The sum is then a/(1-r)=[731/1000]/(1-[1/1000])=[731/1000]/[999/1000]=731/1000.

Digression: how maybe this is done in earlier "grades"
A teaching might say the following:
Let's consider the number 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! (Thanks to Ms. Panova for pointing out this approach.) 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 intellectual construction.

Use #2 of geometric series
A square of side length 5 has another square whose side length is half of that, placed outside but so that corners and an edge coincide. Another square whose side length is half of that, placed outside of both squares but so that corners and an edge coincide. And ...
My language is perhaps not too precise. A sort of picture of this object (just the first 6 squares) is shown to the right. The object is an example of a fractal. General information about fractals is here and a source which is very accessible is here

The question What is the total area of all of the squares?

The first square has area 5·5=52.	The second square has area (5/2)·(5/2)=52/4.	The third square has area (5/2/2)·(5/2/2)=(5/22)·(5/22)=52/42

The pattern may convince you that the total area is the sum of
5²+5²/4+5²/4²+...
This is a geometric series whose first term is a=5², and the constant ratio between successive terms is r=1/4. The sum is then a/(1-r)=5²/(1-[1/4])=100/3. This is the total area.

I remarked to students that questions about some other geometric quantities can easily be asked. For example,
The question What is the total perimeter of all of the squares?

The first square has perimeter 4·5=20.	The second square has perimeter 4·(5/2)=20/2.	The third square has perimeter 4·(5/2/2)=4·(5/22)=20/22

The pattern may convince you that the total perimeter is the sum of
20+20/2+20/2²+...
This is a geometric series whose first term is a=20, and the constant ratio between successive terms is r=1/2. The sum is then a/(1-r)=20/(1-[1/2])=40. This is the total perimeter.

And here's another question. We could think about the region inside all of the squares as one part of the plane. Then the boundary between this region and the remainder of the plane could be analyzed. It has a border which is one relatively long horizontal line segment on the bottom, and one vertical line segment on the left. The region is shown to the right. I hope you "see" that it sort of fades off in a complicated way to the right. There's lots of border turnings on the right.

The question What is the total length of the border of this region?

One way to analyze this is to take the total perimeter of the squares, 40, which we got above, and to subtract the interior border lengths. Look at these pictures.

The first inner wall has length 5/2.	The second inner wall has length (5/2)/2=5/22.	The third inner wall has length (5/22)/2=5/23.

The pattern may convince you that the total length of the inner walls is the sum of
(5/2)+5/2²+5/2³+...
This is a geometric series whose first term is a=(5/2), and the constant ratio between successive terms is r=1/2. The sum is then a/(1-r)=(5/2)/(1-[1/2])=5. This is the total length of the inner walls, so the perimeter of the region is 40-5=35. Whew! I think that's enough for this.

Use #3 of geometric series
Bruno and Igor have a loaf of bread. Bruno eats half the loaf and passes what remains to Igor. Igor eats half of what he is given and passes what remains to Bruno. Brundo eats half of what he is given and passes what remains to Igor. Igor eats half of what he is given and passes what remains to Bruno...

The question How much bread (the total amount) does Bruno eat?

As I mentioned in class, I know people who can somehow "solve" these problems by inspection, that is, they read or listen to the problem, and ZAP!!! the answer is clear. (The same is true for the geometric problems mentioned in Use #2.) I am not one of these "zap" people -- some of the students seem to be! I would probably solve the problem by computing the amount of bread Bruno and Igor eat, for at least a few rounds. I would try to discover the pattern, and then I'd use the observed pattern.

Round #	Bruno eats	Igor eats
1	1/2	1/4
2	1/8	1/16
3	1/32	1/64

I filled out this table dynamically in class, with explanations being given as I did it. For example, I remarked that after Bruno ate half the loaf, Igor would receive the other half loaf. Igor would eat half of that, which is 1/4 loaf, and pass 1/4 loaf to Bruno. Bruno would eat half of a 1/4, which is 1/8 loaf, and pass the remaining 1/8 loaf to Igor, etc. It seems apparent ("clear") that Bruno eats 1/2+1/8+1/32+..., a quantity which we recognize as a geometric series. The first term, a, is 1/2, and the constant ratio between successive terms is 1/4. Therefore Bruno must eat a/(1-r)=(1/2)/(1-[1/4])=2/3 of the loaf. Poor Igor will eat 1-2/3=1/3 of the loaf.

It is easy to change this problem. You could imagine the named people eating different quantities, or you could imagine there being more people, etc. Sums like this do arise in real applications, and I hope that you will be able to recognize them and cope with them.

Integrals and sums
Section 11.3 contains a full presentation of what I'll discuss here. I want to study the harmonic series again, and this time really convince you that it diverges. After that I want to study another series, convince you it converges, and then actually get a fairly good decimal approximation for its sum.

The harmonic series again
Let's look at 1+1/2+1/3+1/4+1/5+... but accompany this with a picture. In fact, let me begin with a specific partial sum, the fifth partial sum: s₅= 1+1/2+1/3+1/4+1/5 (look, there are no ...'s!). I've sketched part of the graph of y=1/x to the right. I only sketched the graph on an interval beginning a bit less than 1 and ending a bit more than 6. The function f(x)=1/x is strictly decreasing. If I draw the boxes shown (this is the Riemann sum approximation to the area under the curve from x=1 to x=6 using a partition equally spaced with 5 subintervals and sample points at the left-hand endpoints, it just happens (NO! This is arranged very precisely.) that the Riemann sum approximation is equal to the fifth partial sum: the boxes all have widths equal to 1 and they have heights 1, 1/2, 1/3, 1/4, and 1/5. Therefore ₁⁶1/x dx<s₅. But I "know" the integral. It is ln(x)]₁⁶=ln(6) since ln(1)=0. Therefore ln(5)<s₅.

If you are curious, ln(6) is approximately 1.791 and s₅ is approximately 2.28333. So the arithmetic reinforces the picture.

Now I would like you to imagine a similar picture for y=1/x, where the interval goes from slightly to the left of 1 up to slightly to the right of n+1. Again, think of boxes over the graph, indicating a Riemann sum with left-hand endpoints at the integers, a bunch of boxes with width equal to 1. The function is still strinctly decreasing, so the tops of the boxes are above the graph. The sum of the areas is s_n=_j=1ⁿ1/j, the n^th partial sum of the harmonic series. Therefore s_n>₁ⁿ⁺¹1/x dx=ln(n+1). But as n-->infinity, ln(n+1)-->infinity, and since s_n is bigger than ln(n+1), s_n-->infinity also, and the harmonic series diverges.

I asked in class how big n should be for the n^th partial sum of the harmonic series to be larger than, say, 200. I can be sure that s_n>200 if I know that ln(n+1)>200 since s_n>ln(n+1). But ln(n+1)>200 happens when (exponentiate both sides of the inequality) n+1>e²⁰⁰. How big is e²⁰⁰? It is about 7·10⁸⁶. If you could do 10¹⁰ additions each second, then (there are 31,556,926 seconds in a year) to compute this partial sum would take about 2·10⁶⁹ years. One estimate for the age of the universe is 13.7 billion years, or 1.37·10¹⁰ years. You'd need a lot of universes. I think it is amazing we can estimate such a partial sum.

A convergent series
Let's look at 1+1/2⁴+1/3⁴+1/4⁴+1/5⁴+.... It will turn out that this series converges. Let me first convince you of this with a geometric argument whose computations are based in calculus. So let me look at the graph of y=1/x⁴ on an interval from slightly to the left of 1 to slightly to the right of 6. Now let me consider Riemann sums with right-hand endpoints. Again each of the widths is 1 and the heights come from f(x)=1/x⁴ at x=2, 3, 4, 5, and 6. What we have (again, f(x) is strictly decreasing) is ₁⁶1/x⁴dx>1/2⁴+1/3⁴+1/4⁴+1/5⁴+1/6⁴. Again, let's compute the integral:
₁⁶1/x⁴dx=-1/(3x³)]₁⁶=-1/(3·6³)+1/(3·1³)=0.33179.
The sum 1/2⁴+1/3⁴+1/4⁴+1/5⁴+1/6⁴ is 0.08112. This isn't exactly s₆ for this series, since we left out the first term, which is 1. In fact, though,
s₆<1+₁⁶1/x⁴dx

A (possible) complaint!?

This time we put the boxes under the curve. Before we put the boxes on top of the curve. How do we know where to put the boxes? And before we got a simple relationship between the integral and the partial sum, and now we've got to add the first term to the integral to get information connecting the integral and the partial sum. What's going on? If you are just exploring, sometimes you will need to look at the improper integral needed and see if it converges or diverges. You'll then need to adjust your logic to the situation. You can make mistakes, but these mistakes can be fixed. The situation really is very forgiving. To overestimate an infinite tail with an integral, put the boxes underneath the curve. To underestimate a partial sum with an integral, put the boxes on top of the curve.

Now imagine the situation from 1 to n+1 with the graph of f(x)=1/x⁴. Similar consideration of boxes under the graph leads to
s_n<1+₁ⁿ⁺¹1/x⁴dx
Now ₁ⁿ⁺¹1/x⁴dx=-1/(3x³)]₁ⁿ⁺¹=-1/(3(n+1)³+1/3
As n-->infinity, this-->1/3. Therefore all of the s_n's are bounded above by 1+(1/3). Since the s_n's form an increasing sequence (hey, everything we're adding is positive!) we know ("Bounded monotonic sequences converge") that the sequence of partial sums {s_n} converges, and therefore the infinite series converges!

Yeah, but what is the sum?
It happens that with more advanced methods the sum _j=1^infinity1/j⁴ can be computed exactly. You won't believe me but the value is Pi⁴/90, approximately 1.0823! The sums of most series can't be computed exactly in terms of "standard" constants (for example, if I changed the 4 to a 3 the series will still converge, and no one knows the exact value of the sum). Let me show you how to estimate an infinite tail.

The partial sum _j=1ⁿ1/j⁴ has an infinite tail beginning with the term 1/(n+1)⁴ and going on and on from there. If you consider the picture to the right, the infinite tail is overestimated by the improper integral _n^infinity1/x⁴dx=(I will omit the details!)=1/(3n³).
The infinite tail is the error between the partial sum and the true sum. If I want the partial sum to be accurate to 3 decimal places, we just need the infinite tail to be less than .001. I should find an n so that 1/(3n³) be less than 1/1,000. This will certainly happen when n=3. Therefore the true value of the sum will be approximated to 3 decimal places by s₁₀. My silicon friend can compute this easily, and the value is 1.082.

The integral test and p-series
The textbook discusses the integral test in section 11.3. Here is a version of the integral test.

The integral test
Suppose f(x) is a positive decreasing function, defined for x>=1. Then the series j=1infinityf(j) converges exactly when the improper integral 1infinityf(x) dx converges.

The logic behind the integral test is used to underestimate partial sums and to overestimate infinite tails, as previously discussed. The best-known application of the Integral Test is the following fact:

p-series
Consider the infinite series j=1infinity1/jp=1+1/2p+1/3p+1/4p+1/5p+... where p is a constant. This series converges if p>1 and diverges otherwise.

When p=1 this is the harmonic series which diverges. We analyzed the series for p=4 and showed that it converged.

Big ideas
Sequence

Convergence and limit

Increasing

Decreasing

Bounded

Monotonic & bounded implies ...

Series

The sequence of partial sums ...

The harmonic series
The partial sums 1., 1.500000000, 1.833333333, 2.083333333, 2.283333333 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 are adding up lots of numbers, the steps between them get smaller and smaller and smaller, and so the sum can't get very large.

So what does happen?

There will be a quiz tomorrow. It will cover improper integrals and differential equations. Exponential decay
dy/dt=ry y=Ae^rt (here r is negative!) radioactive tracers in blood radiocarbon dating http://njpaleo.com/articles/article14.html

Exponential growth dy/dt=ry y=Ae^rt (here r is positive!)

QotD Bacteria

Exponential growth with a "carrying capacity" Consider the differential equation dy/dt=y(2-y). If y is close to 0 but positive, then ... If y is close to 2 but less than 2, then ... Example of the Logistic differential equation, numbers selected to make everything easy. Let me try to solve this equation with the initial condition y(0)=1. Separate and integrate. The result (after use of partial fractions!) is -(1/2)ln(2-y)+(1/2)ln(y)=x+C Here C=0 (wow!). And then ln(y/[2-y])=2x, so y/[2-y]=e^2x and then y=(e^2x)[2-y]=2e^2x-e^2xy, so (1+e^2x)y=2e^2x and, finally, y=[2e^2x]/[1+e^2x]. This may be in a form that "real people" can understand.
A picture of this curve is shown to the right. For large x, For x negative,

Direction fields, a useful qualitative tool

dy/dx=(1/20)x2 Discussion
This is the direction field	This is the direction field with some solution curves.
dy/dx=(1/10)xy Discussion
This is the direction field	This is the direction field with some solution curves.
dy/dx=(1/30)(y+2)(y-2) Discussion
This is the direction field	This is the direction field with some solution curves.
dy/dx=(1/30)y2(y-2) Discussion
This is the direction field	This is the direction field with some solution curves.
dy/dx=(1/20)(x+y2) Discussion
This is the direction field	This is the direction field with some solution curves.

The 20% of the students who were not in class can come to my office before or after class to pick up your exam. They also can come to my office during an office hour to pick up your exam. Finally, they can come to my office at any other mutually convenient time to pick up your exam. (I did spend more than 12 hours this weekend grading the exams, in order to get them back to you in a timely manner.)

Differential equations

Differential equations, more of a definition

Sample scenarios
First story Salt dissolves in water. Put more salt in the water. It will dissolve, although perhaps less rapidly. Add more and more salt. Eventually the solution becomes saturated, and the salt just drops to the bottom (precipitates?). Differential equations can model the salt/water interaction.
A second story 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)

What is a solution of a differential equation?

Solving an example

What is going on?

Separable differential equations and another example

Initial conditions and initial value problems

A solution

A BIG THEOREM

Is it useful?

One example
dy/dx=e^(x2)

Another example
dy/dx=2x with the initial condition (0,0): when x=0, then y=0.

A last example (today!)
dy/dx=xy² with the initial condition y(0)=1.

Two silly (?) formulas
The object of this lecture is to tell you about two formulas, one for arc length (section 8.1) and one for surface area (section 8.2). I called the formulas silly because of their limited usefulness, at least limited in the sense that "hand computation" is not very practical. Both arc length and surface area will be revisited in calc 3, where much better perspectives can be given for both.

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 delta x). "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 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).

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. The semicircle I considered in class was 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

      x2     52-x2+x2      52
1 + ----- = --------- = ------
    52-x2      52-x2     52-x2

This should look slightly familiar. The trig substitution x=5sin() makes this integral into 5d=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)=Pi/2 and arcsin(-1)=-Pi/2) this works out to 5Pi, 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 here are some of the formulas:

(2/3)(x²-1)^3/2
[x³/6]+[1/(2x)]
[x⁵/6]+[1/{10x³}]
[x²/2]-[ln(x)/2]
ln([e^x+1]/[e^x-1])

These are all very weird and ludicrous formulas. Let me try one, and maybe you will see some reason for the structure of the formulas.

Doing a book problem
Let's compute the length of f(x)=[x³/6]+[1/(2x)] as x goes from 1 to 3. We will need sqrt(1+[f´(x)]²). So:
f´(x)=[x²/2]-[1/(2x²)]
(f´(x)}²)=[x²/2]²+2[x²/2](-[1/(2x²)])+[1/(2x²)]²=[x⁴/4]-1/2+[1/(4x⁴)]
You should examine this computation with a bit of suspicion. The "middle" term has lots of coincidences. (The original formula of the function is designed for so these "coincidences" happen!) 2's cancel and x²'s cancel. Now look:
1+(f´(x)}²)=1+[x⁴/4]-1/2+[1/(4x⁴)]=[x⁴/4]+1/2+[1/(4x⁴)]
The darn original -1/2 has somehow changed to +1/2. And the -1/2 came from 2[x²/2](-[1/(2x²)]) so the +1/2 could be replaced by 2[x²/2](+[1/(2x²)]). A "miracle" has occurred! (Ehhh ... not very much of a miracle. It is really a designed algebraic event.) Therefore, let us replace the +1/2 by the stuff suggested and then see:
1+(f´(x)}²)= [x⁴/4]+1/2+[1/(4x⁴)]=[x⁴/4]+2[x²/2](+[1/(2x²)])+[1/(4x⁴)]=the square of [x²/2]+[1/(2x²)]
Therefore, sqrt(1+[f´(x)]²) becomes (take the square root of the thing that is squared): [x²/2]+[1/(2x²)].
To find the arc length, integrate this from x=1 to x=3. The antiderivative is (relatively!) easy, and the result is [x³/3]-[1/(2x)]]₁³ etc. etc. (I'm really not too interested in the actual numbers here!)

A parabola
Let me find the arc length of y=x² from x=0 to x=1. Here the arc length formula, _a^bsqrt(1+[f´(x)]²)dx becomes (since f(x)=x² and f´(x)=2x) ₀¹sqrt(1+4x²)dx. I can compute this, sigh, using a trig substitution. I could "try" 2x=tan() etc., etc. Indeed, I am advised by a friend that the antiderivative is

                           2 1/2
                 x (1 + 4 x )                           2 1/2
                 --------------- + 1/4 ln(2 x + (1 + 4 x )   )
                        2

I decided to save on pictures. The picture shown here will work just as well for the example below. They look sort of the same.

Cubic curve
Let me find the arc length of y=x³ from x=0 to x=1. Here the arc length formula, _a^bsqrt(1+[f´(x)]²)dx becomes (since f(x)=x³ and f´(x)=3x²) ₀¹sqrt(1+9x⁴)dx. Now for the Hot News: there is no antiderivative of this function in terms of the standard functions. This actually can be proved. Therefore, I can't go any further "by hand". If I actually wanted to know the length, I would need to approximate the definite integral using some sort of numerical technique.

"Truth"
The truth for arc length is that, more or less, the computability of the arc length integral using FTC 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 lieing 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 not be more than an approximate rectangle. I will try later to address these sorts of slight (?) distortions.) So a piece of the surface area is 2Pi 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 img src="gifstuff/is11.gif" width=8>_a^b2Pi f(x)sqrt(1+[f´(x)])dx. Notice that sqrt(1+[f´(x)])dx (this uses what we had for ds).

Sphere
Here is a result from a long time ago: the surface area of a sphere of radius R is 4Pi 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 4Pi(5²).

We need to compute _a^b2Pi f(x)sqrt(1+[f´(x)])dx. Notice that sqrt(1+[f´(x)])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(2Pi)5dx which does indeed work out to 100Pi as it should.

Parabola
We can try to find the surface area which happens when y=x² from x=0 to x=1 is revolved around the x-axis. So the formula _a^b2Pi f(x) sqrt(1+[f´(x)]²)dx becomes ₀¹2Pi x² sqrt(1+4x²)dx and, oh my goodness! I can find an antiderivative of this. It is 2Pi multiplied by the following:

                 2 3/2             2 1/2
       x (1 + 4 x )      x (1 + 4 x )                            2 1/2
       --------------- - --------------- - 1/64 ln(2 x + (1 + 4 x )   )
             16                32

Cubic curve
We can try to find the surface area which happens when y=x³ from x=0 to x=1 is revolved around the x-axis. So the formula ₀¹2Pi f(x) sqrt(1+[f´(x)]²)dx becomes _a^b2Pi x³ sqrt(1+9x⁴)dx and, oh my goodness! I can find an antiderivative of this. If u=1+9x⁴ then du=36x³dx so (1/36)du=x³ and we have (2Pi/36) u^1/2du=(2Pi/36)(2/3)u^3/2+C= (2Pi/36)(2/3)(1+9x⁴)^3/2+C. Then the surface area is (2Pi/36)(2/3)(1+9x⁴)^3/2]0¹ and won't bother to finish.

More "truth"
x² and x³ are two of only a few simple powers of x 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.

QotD
I asked people to write an integral for the arc length of something like f(x)=5x⁷+2x as x goes from 1 to 3: just the integral, and do nothing with it.

Maintained by greenfie@math.rutgers.edu and last modified 2/20/2007.