This is, by the way, 2.704813829. Well, you can see it in the accompanying table. This method of "computing" or approximating e is actually very very slow. The first several million digits of e are online, if you need them.

Then I worked on building new functions. If F(x)=f(x)+g(x), and the derivatives of f and g exist, what can one predict about the existence and value of the derivative of F?
Since F(x)=f(x)+g(x) we know that F(x+h)=f(x+h)+g(x+h), and the difference quotient for F can be written this way:
(F(x+h)-F(x))/h=[(f(x+h)+g(x+h))-(f(x)+g(x))]/h=[f(x+h)-f(x)+g(x+h)-g(x)]/h=[f(x+h)-f(x)]/h+[g(x+h)-g(x)]/h.
And now let h-->0, and we see that the derivative of the sum is the sum of the derivatives.

Now I did a hard problem from the textbook (I think #50 of section 3.1), to show that we have gotten already to some level of achievement. The problem asks us to find the equations of the lines tangent to the parabola y=x²+x which also go through the point (2,-3). Note that although the problem statement does not request it, I would almost always begin the solution by making a sketch.

If P=(x,y) is the point of tangency on the parabola, we can solve the problem by realizing that the slope of the tangent line at P, m_TAN, can be written in two different ways. First, since the tangent line goes through P and (2,-3), its slope is (y-(-3))/(x-2), which is (x²+x+3)/(x-2). But m_TAN is also f´(x) if f(x)=x²+x. So m_TAN=2x+1. Therefore 2x+1=(x²+x+3)/(x-2), and (2x+1)(x-2)=x²+x+3. Then 2x²-3x-2=x²+x+3 so that moving everything to one side, we get x²-4x-5=0. Since this is a problem in a textbook, the quadratic factors into (x-5)(x+1)=0. If x=5, then y=5²+5=30 and the derivative is 2(5)+1=11. So the tangent line is y-30=11(x-5). We can make a cheap check: does this line go through (2,-3)? Well, -3-30=-33 and 11(2-5)=-33, so the answer is "Yes." The point of tangency is (5,30) which explains why we can't see it in the picture. You can find the equation of the other line yourself.

Now we began to discuss what is called the product rule. The statement of the product rule begins "The derivative of the product is ..." There is an expectation of simplicity and symmetry here, which should be eliminated as soon as possible. Consider x² which is also, of course, x·x. The derivative of x is 1, and 1·1=1, but the derivative of x² is 2x, so the product of the derivatives is not the formula we want.

If F(x)=f(x)·g(x), then F(x+h)=f(x+h)·g(x+h), so that (F(x+h)-F(x))/h=(f(x+h)·g(x+h)-f(x)·g(x))/h. Now the game is to somehow write this fraction in terms of the difference quotient of f and the difference quotient of g. Here the picture may help. It tries to show a sort of decomposition of f(x+h)·g(x+h)-f(x)·g(x). The suggestion is that f(x+h)·g(x+h)-f(x)·g(x)=(f(x+h)-f(x))·g(x)+f(x)·(g(x+h)-g(x))+(f(x+h)-f(x))·(g(x+h)-g(x)). If we now divide by h and let h-->0, then:
[(f(x+h)-f(x))·g(x)]/h-->f´(x)·g(x) and [f(x)·(g(x+h)-g(x))]/h-->f(x)·g´(x).
The blue rectangle in the corner is a curiosity. It is algebraically (f(x+h)-f(x))·(g(x+h)-g(x)) (then divided by h) In the [REMARKABLE] equation I mentioned last time: F(x+h)=F(x)+F´(x)h+Err·h the blue rectangle belongs to the Error term. So what happens is this:
[(f(x+h)-f(x))·(g(x+h)-g(x))]/h=[(f(x+h)-f(x))/h]·(g(x+h)-g(x)). The first term -->f´(x) but the second term: g(x+h)-->g(x as h-->0 since g is continuous (because differentiable functions are continuous). Therefore the blue rectangle divided by h -->f´(x)·0 which is 0: it contributes nothing to the limit. This is a bit elaborate, but I'd like to be honest when I can be (!?). So now we know that the derivative of f(x)·g(x) is f´(x)·g(x)+f(x)·g(x). This is called the product rule or, sometimes, the Leibniz rule, memorializing one of the inventors of calculus.

Examples: If f(x)=x and g(x)=x, the product rule gives us 1·x+x·1=2x, the correct answer. We can also differentiate something like e^x·x¹⁷⁸. And we can differentiate 37·x²³ with f(x)=37 (a constant function) and g(x)=x²³. The product rule predicts that the derivative is 0·x³⁷+37·23x²². Frequently people use this special case of the product rule without thinking about it: the derivative of a constant times a function is a constant times the derivative of the function.

One example
I'm going to look at g(x)-f(x). This is sqrt(x²+1)-x. I want to investigate
lim_x-->infinity sqrt(x²+1)-x.
"Clearly" (always use that word when you can't explain where the idea comes from) multiply this by a fraction which has the same thing on the top and the bottom (so the fraction is a fancy way of writing "1".) Here the following fraction was suggested:

sqrt(x2+1)+x
-------------
sqrt(x2+1)+x

                            (sqrt(x2+1)-x)(sqrt(x2+1)+x)
g(x)-f(x) = sqrt(x2+1)-x = ------------------------------- =
                                  sqrt(x2+1)+x)

   (x2+1)-x2               1
----------------- = ---------------
   sqrt(x2+1)+x)     sqrt(x2+1)+x

Another example
Now let me look at h(x)-f(x). This is sqrt(x²+x)-x. I'd like to consider
lim_x-->infinity sqrt(x²+x)-x.
Now the appropriate conjugate is (sqrt(x²+x)+x).

                            (sqrt(x2+x)-x)(sqrt(x2+x)+x)
h(x)-f(x) = sqrt(x2+1)-x = ------------------------------- =
                                  sqrt(x2+x)+x)

   (x2+x)-x2               x
--------------- = ---------------
  sqrt(x2+x)+x)    sqrt(x2+x)+x

Now look: sqrt(x²+x)=sqrt(x²[1+{1/x}])=sqrt(x²)sqrt(1+{1/x}). We've seen sqrt(x²) before, and what is it? If x is positive, then sqrt(x²)=x. (If x were negative, sqrt(x² is -x.) Here we go:

      x               x                              x                    1
-------------- = ------------------------- = ------------------- = -------------------
 sqrt(x2+x)+x     sqrt(x2)sqrt(1+{1/x})+x     x[sqrt(1+{1/x})+1]    [sqrt(1+{1/x})+1]

Numerical evidence?
I haven't talked enough about numerical evidence for limits in class, mostly because I am scared of using a calculator or computer in front of people. But I certainly use such things on my own. So here is some numerical information.

	x=10	x=102	x=103	x=104	x=105
sqrt(x2+1)-x	0.0498756211	0.0049998750	0.0004999998	0.0000499999	0.0000049999
sqrt(x2+x)-x	0.4880884817	0.4987562112	0.4998750624	0.4999875006	0.4999987500

I needed 25 digit accuracy to get the last few entries in the first row to 10 digit accuracy. Computations with small numbers which are almost equal can be imprecise. But the numbers should help you accept the previous results which were obtained with algebraic manipulation.

If you prefer graphical evidence then maybe the picture to the right can help. It shows graphs of f(x)=x and g(x)=sqrt(x²+1) h(x)=sqrt(x²+x) on the interval [0,10]. Although 10 is not very large, I think the asymptotic relationships between the functions are already apparent.

Short cuts are good but ...
So I know lim_x-->infinityx=infinity and lim_x-->infinitysqrt(x²+1)=infinity, but lim_x-->infinitysqrt(x²+1)-x=0.
And I also know lim_x-->infinityx=infinity and lim_x-->infinitysqrt(x²+x)=infinity, but lim_x-->infinitysqrt(x²+x)-x=1/2.
     Please notice that 1/2 and 0 are not equal.
Therefore for this kind of limit, simple algebraic manipulations are not guaranteed to give valid results. We all like computational shortcuts, but limits involving infinity can be difficult to manipulate. Such limits should be treated more as descriptions of geometric situations rather that anything else.

Vocabulary?
On a well-known web page (concerning the Fungi of Australia, the word conjugate is defined to mean "copulation, especially isogamic copulation". I think I am too scared to look up "isogamic" since it might mean something illegal.

How to evaluate limits
As was remarked in the previous lecture, if the limiting function were defined with familiar formulas, I'd first try to "plug in" if there is any chance this strategy can be applied. If this simplest method can't be used, I'd try to "massage" the function (algebraically) and get to some algebraically equivalent restatement where plugging in can be used.

The derivative
Here is the most important single use of limit in Math 151.
Suppose f(x) is a function. Then we write f is differentiable at x=a if lim_h-->0[f(a+h)-f(a)]/h exists. If the limit exists, it is called the derivative of f at a and the notation f´(a) is used for the value of the limit.
Comments There are other notations for the derivative (almost every chunk of applied science and engineering has its favorite notation!). We'll see some of them. Please notice that the quotient involved in the definition of the derivative is exactly of the form which prevents direct "plugging in". If we insert h=0 in [f(a+h)-f(a)]/h we get [f(a+0)-f(a)]/0 and that's 0/0, a meaningless arithmetic expression.

One nice example
Suppose f(x)=1/x². We already looked at this function in the lecture on September 13 (whose diary entry is not done, which I regret). Let's look at [f(a+h)-f(a)]/h:

                        1         1             
                     -------  - ----- 
  1/(a+h)2-1/a2       (a+h)2      a2
----------------  = ------------------
        h                 h

    1         1          a2-(a+h)2     
 -------  - -----      ------------- 
 (a+h)2      a2          (a+h)2a2          a2-(a+h)2
------------------- = --------------- = -------------- 
       h                   h              h(a+h)2a2

  a2-(a+h)2       a2-[a2+2ah+h2]      -2ah-h2        h(-2a-h)      -2a-h 
------------- = ---------------- =  ----------- = ------------ = --------
  h(a+h)2a2        h(a+h)2a2          h(a+h)2a2     h(a+h)2a2    (a+h)2a2

Comments There are many opportunities to make algebraic errors in what's done above. One amazing thing is that we will develop very easy ways to compute derivatives for almost all familiar functions combined in interesting ways (including algebraic combinations and composition and "inversing"). Most of next week will be devoted to stating and understanding these results. Although you should (and will, darn it!) practice these rules (algorithms!), learning them is not the peak of the course. There are very nice programs which can compute such derivatives. A major purpose of the course is to understand why derivatives are interesting to people. We need to know how to use them. And maybe machines are not yet up to that level of cognition.

Not a nice example
Consider f(x)=|x|. I would like to see if this f is differentiable at a=0. Therefore I must consider:

f(0+h)-f(0)    |0+h|-|0|    |h|
----------- = ---------- = -----
    h              h         h

h<0 so we are considering h-->0-	h>0 so we are considering h-->0+
Here h is negative, and |h|=-h. Therefore |h|/h is -h/h and this is -1: there's no appearance of h in this result. I think that the limit as h-->0- of |h|/h must be -1.	Here h is positive, and |h|=h. Therefore |h|/h is h/h and this is +1: there's no appearance of h in this result. I think that the limit as h-->0+ of |h|/h must be +1.

But the limits from the two sides (+/-) don't agree. Therefore the limit lim_h-->0[f(0+h)-f(0)]/h does not exist.
We conclude that this function is not differentiable at x=0.

Interpretations
Here are a collection of simple interpretations of what we are discussing. Other interpretations will be shown in you in virtually every technical course you take from now on.

Math 151	Geometry in the plane	Simple physical motion
We're given some function f, and want to understand how it "changes".	The object studied is the graph of f, which is the collection of points in the plane with coordinates (x,y) which satisfy y=f(x).	Here we study rectilinear motion, where the position of a point on the coordinate line is given by f(x) (usually the variable is called t, not x, because this would help people remember that f(x) is the position at time "x").
The change in f over an interval from a to a+h is just f(a+h)-f(a).	f(a+h)-f(a) denotes the difference in the heights of the function at the x-values a and a+h.	f(a+h)-f(a) is the difference in position at the two times indicated. This is also called displacement. In general, this is not the distance that the point travels between those two times, because the moving point could wiggle back and forth.
[f(a+h)-f(a)]/h is called the average rate of change of f over the interval.	[f(a+h)-f(a)]/h is the slope of the secant line through the points (a,f(a)) and (a+h,f(a+h)).	[f(a+h)-f(a)]/h is called the average velocity of the point over the time interval.
If, as h-->0, the average rate of change of f approaches a limit, this limit is called the derivative of f at a and written f´(a).	If, as h-->0, the slope of the secant line approaches a limit, this limit is called the slope of the tangent line to y=f(x) at x=a.	If, as h-->0, the average velocity of the point over the time interval approaches a limit, this limit is called the instantaneous velocity of the point at the time x=a. The limit in this case is frequently written ds/dt (difference in s, a Latin abbreviation for distance, divided by a difference in t, meaning time.

I will usually abbreviate the word "instantaneous" while discussing velocity, because I will almost never again refer to average velocity.

O.k.: a tangent line
Well, I made a mistake in class. I will not repeat the mistake here. Here I will use a different function. I will try hard not to make another mistake.

Suppose f(x) is the function defined by the formula sqrt(17+x³). Then it turns out that this function is differentiable (if you are "naive" this is not at all obvious and would be quite difficult to verify directly from the definition!) and f´(x)=[3x²]/[2sqrt(17+x³)]. Please: we will very soon see how this formula gets computed! Suppose I ask for an equation of the line tangent to the graph of this f(x) when x=2? We need some information.
A point on the line Well, (2,f(2)) is a point on the line, and f(2)=sqrt(17+2³)=sqrt(17+8)=sqrt(25)=5.
Slope of the line Well, f´(2) is supposed to be the slope of the tangent line, and this is f´(2)=[3·2²]/[2sqrt(17+2³)]=3·4/[2·5]=6/5.
Therefore an equation of the line tangent to the graph of this f(x) when x=2 is (y-5)=[6/5](x-2). I am lazy and I will not "simplify".

Some numerical "work"
Very very very few of you will need to write the equations of tangent lines after getting through a first calculus course. But you will almost certainly need to consider many, many derivatives and make judgements based on these derivatives. What the heck is going on? Well, let us think about the example above. Since f´(2)=6/5, I know that
lim_h-->0[f(2+h)-f(2)]/h=6/5.
If I omit the "lim_h-->0" phrasing, then the equality is no longer ture. But the limit definition implies that what is true is a statement with an error term, and this error term is small when h is small. That is:
[f(2+h)-f(2)]/h=6/5+{ERROR [small when h is small]}. I can multiply by h and then add f(2) and get the following equation:
f(2+h)=f(2)+(6/5)h+{ERROR [small when h is small]}h

This equation is really why people study derivatives. The important qualitative aspect is that the Error is multiplied by h, and when h is small the Error is small. Products of "smalls" are even smaller, and the effect on the output of changing the input to f by h for small h is almost entirely determined by the multiplier, (6/5)h. Here are some numbers.

Input value x Output value f(x)=sqrt(17+x3)

x=2 f(x) is exactly 5. x=2+h f(2+h)=f(2)+(6/5)h+{ERROR [small when h is small]}h x=2.1 so h=.1 f(x) is about 5.124548 and f(2)+(6/5)(.1) equals 5.12 x=2.01 so h=.01 f(x) is about 5.012045 and f(2)+(6/5)(.01) equals 5.012 x=1.9 so h=-.1 f(x) is about 4.884567 and f(2)+(6/5)(-.1) equals 4.88 x=1.99 so h=-.01 f(x) is about 4.988045 and f(2)+(6/5)(.01) equals 4.988

Reinterpreting the definition
The function f is differentiable at a if for any very small {perturbation|kick|change} to the input, h, the output will be approximately f(a) (the old output) plus a constant multiplying the perturbation. That constant is the derivative of f at a.

The strategies available now for computing limits are:

1. "Plug in": if the function is given by a formula involving standard functions and you know the darn thing is continuous.
2. Transform what's given using standard algebraic manipulations into a form where you can "plug in".
3. There will be some other strategies, later.

Intermediate Value Theorem, restated
Suppose that the function f is defined and continuous on the interval [a,b]. Then the equation f(x)=y has at least one solution for every y which is between f(a) and f(b).

QotD, version 2 (fill in the blanks)
The QotD last time was not suitable. Here, maybe is something I should have asked, an alternative QotD for the last lecture, with a fill-in-the-blanks format.
Suppose f(x)=x³+cos(7x²+5)+sin(3x⁴-8)+2. Then f(-2) is  # 1  because (-2)³=-8 and the remainder of the formula describing the function f(x) is at most  # 2  at x=-2. And f(+2) is  # 3  because 2³=8 and the most negative the remainder of the formula describing the function f(x) can be at x=2 is  # 4 . Since f(x) is  # 5  in the interval [-2,2] and the signs of f(x) at the endpoints differ, f(x) must have  # 6  root in [-2,2].

So the QotD would have been to fill in the blanks. Here we go:

1. negative
2. 4: the "outputs" of sine and cosine are between +1 and -1, and therefore cos(7x²+5)+sin(3x⁴-8)+2 must be somewhere in the interval [-4,4]. The "stuff" inside the sine and cosine terms is mostly irrelevant here.
3. positive
4. -4: the reasoning is the same as for #2.
5. continuous
6. at least one

Horizontal and vertical asymptotes: geometric and algebraic vocabulary
The limit idea was fairly successful, and people over years decided to use it to cover more and more situations. Sometimes the extensions were not as simple as the original setting. One extension that is useful describes certain geometric behavior of graphs called vertical and horizontal asymptotes.

Near x=3
To the right is the graph of a function, f(x), whose domain includes all x except for 3. The "arrows" at the top of the curve are supposed to indicate that as x gets closer and closer to 3 (on either the right or the left side) then f(x) gets large. But there's more, really. It doesn't just "get" large in some uncontrolled way. The function is supposed to

• Get as large as "you" want So if any number is named (10? 100? 100,000? 10¹⁰?), the function gets larger than that number close enough to 3.
• Stay large Not only does it get large, but eventually it stays large. For example, the graph is supposed to indicate that there is some (perhaps small) positive number, epsilon, so that if 0<|x-3|<epsilon, then f(x)>10¹⁰. This is, for me, difficult to imagine.

Examples
1/(x-3)? 1/|x-2|?? 1/(x-3)²??? 1/(x-3)^{EVEN POS INT}???

A rational function
x²/[(x+2)³(x+4)⁶]

Horizontal asymptotic behavior
Be a bit careful ...

sin(x)/x
Not damned oscillation, just damped oscillation

x/sqrt(x²+3)

Review of the definition of limit

Discussion, criticism, comments about the definition of limit

Framework of ideas supporting the definition of limit

Algebra

Order (1)

Wiggling

Squeezing the wiggling

Order (2)

Plugging in

An important official word: continuity

The Garden State Parkway, from Cape May to Montvale and my friend Francine ...
The Garden State Parkway runs most of the length of New Jersey. Mile 0 is at Cape May, while the other end, mile 172, seems to be close to Montvale. Suppose that my friend Francine leaves Cape May at 7 AM one morning, and drives north on the Garden State Parkway. Further, suppose she arrives at mile 172, the northern end, at, say, 10 AM. Must Francine at some time be at mile 135 (fairly near Busch campus)? The parkway seal here was "borrowed" from a State of New Jersey webpage.

We discussed various curves which could represent the position of Francine on the parkway in terms of miles from the start of the parkway at time t, in terms of hours elapsed from 7 AM. I tried to show that our everyday intuition lead to the graph being increasing (as you travel from left to right, the points on the graph go up). The graph can have level spots, where Francine pulls over for a rest stop. Legally Francine isn't supposed to drive backwards, though.

If we believe that motion is continuous (so Francine does not have a Star Trek transporter or other device) then the graph of Francine's position goes from (7 AM, 0 miles) to (10 AM, 172 miles) and therefore the graph must have on it at least one point with coordinate description (*,135). All of this, by the way, rests on some complicated assumptions, some of them philosophical (why should motion be continuous?). Today, though, I believe that motion is continuous, and therefore at sometime Francine must be at Mile 135. By the way, I will retain this information for later, when we analyze the rate of change of position (velocity) so that we can see whether Francine deserves a speeding ticket.

The Intermediate Value Theorem
Suppose that the function f is defined and continuous on the interval [a,b]. Then the equation f(x)=y has at least one solution for every y which is between f(a) and f(b).

In mathematics, the word theorem is applied to results that are deduced from basic principles, and usually the term is used for more important conclusions in the subject. In this case, the Intermediate Value Theorem follows from basic principles governing the real numbers. A particular basic principle which is used in the proof of the theorem is the "least upper bound" property of the real numbers. This essentially declares that there are "no holes" in the real numbers. A precise statement is fairly delicate, and this property essentially shows that the reals and the rationals are distinct. Several upper-level math courses spend quite a bit of time exploring the statement. You can read about it in Wikipedia but I do add that detailed knowledge of such foundational material is not needed for success in Math 151 (or, for that matter, for successful careers in almost all of science and engineering!).

The square root of 2
If we were desperate to compute sqrt(2) (that is, really, desperate to approximate sqrt(2)), for example, we could look at f(x)=x²-2 on the interval [0,2]. This f(x) is certainly continuous. (We already observed that we could "plug in" values to evaluate limits for polynomials. I know that f(0)=-2<0 and f(2)=+2>0. Therefore according to the Intermediate Value Theorem there will be at least one x inside the interval [0,2] so that f(x)=0: x²-2=0. This is a positive number whose square is 2, which we call sqrt(2). Now we have "trapped" sqrt(2) inside the interval [0,2]. If we compute f(1)=1²-2=-1, we know that a root must be inside [1,2] since the signs of f(x) at the two endpoints differ. We can continue this "game", each time halving the interval, and chosing a half-subinterval so that the signs of f(x) differ on the endpoints. The graph of f(x) on the first few subintervals is shown below.

Interval: [0,2]	Interval: [1,2]	Interval: [1,1.5]	Interval: [1.25,1.5]	Interval: [1.375,1.5]

The bisection method for root-finding
I discussed this too rapidly and then asked the QotD about it. I regret the hurry. In particular, I mentioned the word "algorithm". Let me give some further information about this word in the form of quotes from The Art of Computer Programming by D. E. Knuth:

The modern meaning for algorithm is quite similar to that of recipe, process, method, technique, procedure, routine, except that the word "algorithm" connotes something just a little different. Besides merely being a finite set of rules which gives a sequence of operations for solving a specific type of problem, an algorithm has five important features:

1. Finiteness An algorithm must always terminate after a finite number of steps.
2. Definiteness Each step of an algorithm must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case.
3. Input An algorithm has zero or more inputs, i.e., quantities which are given to it initially before the algorithm begins. These inputs are taken from specified sets of objects.
4. Output An algorithm has one or more outputs, i.e., quantities which have a specified relation to the inputs.
5. Effectiveness An algorithm is also generally expected to be effective. This means that all of the operations to be performed in the algorithm must be sufficiently basic that they can in principle be done exactly and in a finite length of time \e

Let us try to compare the concept of an algorithm with that of a cookbook recipe: A recipe presumably has the qualities of finiteness (although it is said that a watched pot never boils), input (eggs, flour, etc.) and output (TV dinner, etc.) but notoriously lacks definiteness. There are frequently cases in which the definiteness is missing, e.g., "Add a dash of salt." A "dash" is defined as "less than 1/8 teaspoon"; salt is perhaps well enough defined; but where should the salt be added (on top, side, etc.)?
... a computer programmer can learn much by studying a good recipe book

Discussion of the bisection algorithm	Specification of the bisection algorithm
The bisection algorithm is one of the simplest and neatest algorithms. It approximates roots very nicely. The entry conditions are a continuous function defined on an interval which includes the interval's endpoints. Also the sign of the function at the endpoints differs: the function's value is positive at one endpoint and negative at the other. Then, according to the Intermediate Value Theorem, there must be at least one root (f(x)=0) inside the interval. Another entry condition is a positive number, here called epsilon, which serves as the error tolerance for the root.	Entry conditions A continuous function f(x) defined on an interval [a,b], with f(a)·f(b)<0; a positive tolerance epsilon for the error.
Here we check if the interval we're looking at already fulfills the error tolerance condition. In later steps, we will be altering the interval, and shrinking it.	Exit condition If b-a<epsilon, report the interval [a,b] as the answer.
Compute the value of f(x) at the middle of the interval.	Computation Let c=(1/2)(a+b). Compute f(c).
Here's the heart (?) of the algorithm. If f(a) and f(c) have different signs, then the root desired is in [a,c], the left half of the interval. We then change the interval (shrinking it) and see if the length of the interval is small enough. If the signs of f(a) and f(c) are not different, the root we're looking for (whose presence is guaranteed by the Intermediate Value Theorem!) is in the right half of the interval. So we redefine [a,b] as the right half interval and check if the exit condition is satisfied.	Decision If f(a)·f(c)<0, then change b to c, and return to check the Exit condition. Otherwise, change a to c, and return to check the Exit condition.

Workshop work ...
Workshops are important, as I've said previously. And they will be graded (by the recitation instructor and by me) equally for both mathematical content and technical exposition. Please try to do both, and please ask for advice about writeups from either of us.

Textbook homework assignments
The textbook homework assignments to be handed in on a Thursday should be available each week on Monday. I hope this will help you schedule your work.

Slope of a tangent line
Let's look at the curve y=1/x² near x=3. The point (3,1/9) is on the curve, since 1/3²=1/9. We can try to find the equation of a line tangent to the curve at (3,1/9). Since we know a point this line goes through, we will be able to find an equation for the line if we know a slope. I'll call the slope of the tangent line m_tan. The traditional calculus way to find this slope is to approximate it with m_sec, the slope of a secant line. This will be a line which goes through the point (3,1/9) and (3+h,1/(3+h)²) when h is small. We can get this slope by writing it as the difference in the second coordinates divided by the diffeence in the first coordinates:

   1      1
 ----- - ---
 (3+h)2 - 9
--------------
   (3+h)-3

   1      1            9-(3+h)2 
 ----- - ---         -----------      
 (3+h)2 - 9            9(3+h)2            9-(3+h)2  
--------------  =  --------------  =  ---------------
   (3+h)-3              h                 h9(3+h)2    

 9-(3+h)2        9-(9+6h+h2)        -6h-h2
-----------  =  -------------  =  -----------
 h9(3+h)2         h9(3+h)2          h9(3+h)2

  -6h-h2        -6-h
---------  =  ---------
 h9(3+h)2      9(3+h)2 

Is m_tan=-2/27 "reasonable"? Well, if we look at the graph we can see that the y values on the curve are getting smaller as the x values increase (near x=3). So the tangent line should tilt down, which makes the minus sign of the answer more agreeable. The magnitude of the slope, 2/27, is small, and, in fact, if you really look at the curve, the tilt is quite small. Some graphs are shown below. Of course 0 is not in the domain of 1/x².

y=1/x2; x between -5 and 5	y=1/x2; x between 2 and 4	y=1/x2; x between 2.9 and 3.1

The graphs need to be looked at a bit carefully. The vertical interval in the first graph is [0,20]. In the second graph it is [0,.5], and in the third graph it is [0,.25]. This graph is locally linearizable near x=3 in the following sense: if we "zoom in" on it near the point (3,1/9) a sufficient number of times, the graph will look as near to a straight line as we would like. The line that results will got through (3,1/9), and will have slope -2/27.

Possibly the slope of another tangent line
Suppose now that we define f(x) in a piecewise fashion:

      11 -2x if x<3  
f(x)=
      x2-4 if x≥3

 f(3+h)-f(3)      (3+h)2-4-5        9+6h+h2-9
------------  =  ------------  =  -----------  =  6+h
      h                h               h

m_sec if h<0:

 f(3+h)-f(3)       11-2(3+h)-5        11-6-2h-5 
-------------  =  --------------  =  ------------  =  -2
      h                 h                 h

There does not seem to be one value of m_tan here. So I guess there may not be a unique satisfactory tangent line. Some pictures may help.

y=1/x2; x between -2 and 6	y=1/x2; x between 2 and 4	y=1/x2; x between 2.9 and 3.1

Again look closely at these pictures, which all have differing horizontal and vertical scales. In this case, there's no amount of "zooming in" which will make the graph look like a straight line segment. This graph is not locally linerizable. The behavior is qualitatively very different from the previous example.

Limits
The relationship between the approximating m_sec and m_tan needs to be investigated more precisely. There's a name to the process: LIMIT. Limits are fundamental to precise statements about asymptotic relationships of all kinds.

Exactness and reality
"Real" functions are not very exact. A chemical engineer might have (I'm simplifying hugely) some process which produces the correct kind of plastic if, say, a certain amount of benzene is used in the mix. (Hey: benzene, with chemical formula X₃Y₇Z₁₁ (nah), was suggested by a student. So a certain percent of benzene, say 3%, may produce a plastic which has the desired amount of, say, translucence. But in reality, even very precise measurements may not create an input with exactly 3% benzene. Maybe some days we get 3.5, or 4.2. And maybe the desired output measurement is not necessary -- we can deal with some error in the output. Real measurements imply that we need to contemplate error, not as a moral defect (at least in this case), but as part of our mathematical model which we must deal with.

Output tolerance as controlled by input tolerance
So maybe we need to understand the following idea: we want a certain output tolerance: that is what we can "live with" in the material created. Is there an input tolerance describing inputs in an interval around the ideal input which makes the corresponding outputs close enough to the desired output? This is too darn abstract. Let me consider a numerical example, with a very simple function.

x² near x=3

If the input to f(x)=x2 is 3, the output will be 9. Suppose we are willing to live with a +/- error of 1 in the output. That is, we can tolerate |f(x)-9|<1. Is there some simple specification of an interval of inputs near 3 which will guarantee this? In this case, almost everyone thinks that the simplest kind of input specification would be an interval of x's centered at 3 (your feelings may differ here, but this is what is usually done). So I ask if there is some convenient number ("CN") so that if |x-3|<CN then |x2-9|<9. Most people don't want an exact, precise, most perfect (?) value of CN. It may be difficult to get something like that. They are willing to settle for something convenient. In this case, the graph to the right shows the needed output tolerance using horizontal red lines.

The next graph, shown here to the right, adds part of the vertical lines corresponding to x=2.9 and x=3.1. The part of f(x)=x2 shown which is between the vertical blue lines is "clearly" forced to be between the red horizontal lines. This geometry reflects the following algebraic statement:      If |x-3|<.1 then |x2-9|<1. So an input tolerance of .1 around 3 will guarantee that the output tolerance of 1 around 9 is true.

Change the output tolerance to .2
What if we changed the desired output tolerance to .2? That is, we want to get |x²-9|<.2 by controlling the size of |x-3|.

Here's the picture of f(x)=x2 with the two horizontal red lines indicating the output tolerance needed.
Now here is the "old" input tolerance lines which we just used. Please note that this input tolerance won't work: there is part of the graph of x2 between 2.9 and 3.1 which is not between y=9-.2 and y=9+.2.1.
Here is a display showing a satisfactory input tolerance. I used .1 as the input tolerance (yes, I experimented a bit). I repeat that in reality people rarely care about the "best possible" input tolerances. They usually just want to get some number that works. To me this graph doesn't show very much. So let's try a different scale.
I hope that you can see here a graphical "verification" of the algebraic implication:      If |x-3|<.01 then |x2-9|<.2 An input tolerance of .01 satisfies the output tolerance of .2.

x² near x=10
Now let me change the game a bit. Suppose I want to consider the same function, f(x)=x², but now investigate what happens near 10. Since f(10)=100, the "ideal" input is 10 to get the "ideal" output of 100.

Suppose I want the outputs to be within 1 of the ideal output. That is, I want |x2-9|<1. The horizontal lines shown indicate this restriction. Warning! There is a huge horizontal/vertical distortion in this graph. The horizontal scale ranges from 9.5 to 10.5, a length of 1. The vertical scale goes from about 90 to about 110, a length of 20. The "true" picture is a long and thin vertical rectangle, with a 1-to-20 aspect ratio.
The vertical lines shown are x=9.9 and x=10.1, which would be the geometric side of the input restriction |x-10|<.1, the input tolerance we used in "If |x-3|<.1 then |x2-9|<1" The same input tolerance won't work about different "ideal inputs". If you really believe in the graph, this isn't too surprising since the graph is much more tilted around x=10 than it was around x=3.
The vertical segments shown are parts of the lines x=9.97 and x=10.03. I hope this is pictorial evidence that the following implication is correct:      If |x-10|<.03 then |x2-100|<1 An input tolerance of .03 satisfies the output tolerance of 1 near the ideal input/output pair (10,100).

Complicated? Yes, it is.
These pictures should convince you that the limit business when addressed "officially" is indeed complicated. In your courses and careers, there may rarely be times when you'll need to cope with these implications. But I believe you should have some idea of what's going on.
Graphs don't really work too well for complicated functions or irritating input/output pairs. People have therefore developed extremely intricate strategies relating the inequalities needed. Serious investigation of this algebra is not the aim of the course but the equipment is there when/if you need it.

The official definition
The statement lim_x-->af(x)=L means:

Given any (positive) output tolerance, there is a (positive) input tolerance so that
If 0<|x-a|<the input tolerance, then |f(x)-L|<the output tolerance.

The piecewise linear function

The inverse(s)

Example

Some trig functions and their inverses

Some exponential functions and their inverses

Dwarf apple trees

Growth rate

Ultimate height

Chapter 2 and its fictions

Slope of a tangent line

QotD

The diary
I mentioned the diary and remarked that it served several purposes. First, even very enthusiastic students may sometimes miss a class. The diary can help them find out what was covered. Another reason for the diary's existence is so that I can fix errors which I will certainly make, so while the diary will be a representation of what went on in class it won't be a totally accurate one: improvements may occur! Also, even if I don't make a mistake in the lecture, sometimes I may not explain things optimally, so the diary may give students another chance to understand the presentation. I really don't think that a diary entirely takes over the teaching/learning aspect of the lectures. There is still a good amount to be said about the interaction between the instructor and the student, provided that students are willing to ask questions and (sigh!) the instructor is really willing to listen and answer. I will try to do the latter (also with errors, of course, since I plead the defects of humanity!).

QotD
Most days I will ask students who attend the lecture a question or two, and receive their answers in writing. Students will get full credit for any answer. This "Question of the Day" has several purposes. First and most minor, it is a way of taking attendance. More important is that it serves as an additional method of communicating between the lecturer and the audience. If I ask some question which I think is "simple" and, hey, almost no one answers the question correctly, well ... something's wrong. I will "grade" the question very roughly, only retaining for records whether or not an answer was made. For the student, this gives an opportunity to see if some content in the lecture has been learned satisfactorily, and, if not, maybe that also signals that something's wrong. I'll try to get the graded QotD's back to you in recitation.

Study groups
I asked students in the class to tell me their e-mail addresses and likely majors, and, with appropriate permission, this information is now displayed on a web page. Now at the beginning of this course, there may be little that is new to students. I caution you, though, that even if you have seen much of this material before, the course will move swiftly. It is a good idea to form study groups meeting at appropriate times and durations to go over homework, etc. I hope that the student list will help people get in touch with each other.

Calculus [Advertisement]
Calculus originated as a way of describing certain methods used in understanding and solving problems originating in geometry and physics. The language of the subject shows this ancestry. Calculus has been very successful. Calculus also "happens" to be suitable to describe and study many problems in biology and economics. Any problems which involve rates of change (how soluble is substance A in substance B as the concentration of substance C varies?) or accumulation of changes (what is the present value under certain interest rate assumptions of a stream of mortgage payments to be made monthly over the next 20 years?) is a problem which likely can be described using calculus. Frequently this description will suggest certain ways of solving the problems using the methods of calculus. That's why the subject is required in so many majors.

Functions
The word function is used in a technical sense in calculus, and is one of the most important vocabulary words. It is the logical setting for how things are transformed to other things. In the case of Math 151, the "things" are numbers. So functions will change numbers to numbers. I decided to begin with what I hoped was a known example described verbally.

Hooke's law
Think of a spring with a rest or "equilibrium" position. Then Hooke's Law states that the deviation from the rest position is in direct proportion to the impressed force. There are many assumptions involved in Hooke's Law, but to a considerable extent it is a widely valid shorthand for a great variety of real phenomena. If I further add something like: the spring gets stretched 10 inches by a weight of 300 pounds, then I can formulate an equation relating the impressed force and the deviation from the equilibrium position:
D(x)=(1/30)x
Here D(x) is the deviation in inches from the equilibrium position (with positive meaning the spring is forced to be longer) and x is the impressed force in poinds (I'm thinking maybe of a thick mechanical spring or something like that). x is called the independent variable and D(x) is called the dependent variable. Notice that sigh matters. If x is negative, then we could be compressing the spring, and the end of the spring is higher than the equilibrium position in our model.
Notice that as in all mathematical models, you need to be alert a bit about how "correct" this is. Uhhh ... if we stretch a rubber band with a force of six trillion tons, I don't necessarily think it will end up having a length of one lightyear.

Functions
One metaphor I will use (and probably overuse!) in this course is the idea of a function as a machine with an input and a unique output associated to each input. The collection of all valid inputs to the machine, those inputs which don't cause the machine to break, is called the domain. The collection of all of the outputs for these valid inputs is called the range.

Functions via formulas
Functions can be described simply with formulas. Actually, most of the functions we will study in this course will have such descriptions. Warning: students interested in obviously experimental disciplines (engineering, physics, chem, etc.) will mostly encounter functions which are described by data sets, and must work with them using the techniques of this course which will mostly be presented using functions defined by formulas. This can be ... irritating.
The domain of a function defined by a formula is usually assumed to be what's called the natural domain: this is the collection of all inputs of real numbers which make sense in the formula. But there will be exceptions (see below!).

• Polynomials
  Polynomials are sums of constant multiplies of monomials. So on example of such a function is f(x)=53x⁴⁰-9.7x²+88. Polynomial formulas only involve multiplication and addition, and therefore the natural domain of a polynomial is all real numbers.

Equal-sized squares are cut from the corners of an 8 inch by 11 inch rectangular piece of paper. The flaps in the resulting piece of paper are folder up. Write a formula for the resulting volume which the paper object encloses, using the edge length of the squares as the independent variable. What is a useful domain for this formula?

• Rational functions
  These are quotients of polynomials, for example f(x)=(x⁴+6)/(x²-5). Since we are now "allowed" to divide, we'd better avoid dividing by 0. Here we need to find out where x²-5=0 which is x=+/-sqrt(5). So the domain is all other x's: the intervals (-infinity,-sqrt(5)), (-sqrt(5),sqrt(5)), and (sqrt(5),infinity). Things can get complicated, especially because if you give a "random" polynomial, finding the roots exactly may not be possible!
• Algebraic functions
  Here we allow roots to occur, so we can get things like f(x)=sqrt(x²-5). Since even powers can't be negative, we'd better have inputs to square roots (and other even roots) be non-negative. So here we need x²-5≥0, which means that the domain needs to be (-infinity,-sqrt(5)] and [sqrt(5),infinity).
• Exponentials and logs
  We'll deal more with these in the future, but exponentials are always positive, so the domains of logs have restrictions to only positive numbers. These functions are needed to model many sorts of growth problems.
• Trig and inverse trig functions
  So we need to consider sine and cosine and tangent (domain restricted to not allow division by 0) in order to model periodic phenomena. The inverse trig functions have various kinds of domain restrictions.

Piecewise defined functions
Here is a silly example:

      x2 if x<1
f(x)=
      x+2 if x≥1

Part of the graph of this function is shown to the right. Please note the "empty circle" on the parabolic arc, because of the restriction x<1 (x is not allowed to be equal to 1). There's a "filled circle" on the other piece, a half line.

A more elaborate "real" example
Now here is a somewhat more complicated example:

f(x)= 
     .1x if 0≤x≤15,100
     1,510+.15(x-15100) if 15,100<x≤61,300
     8,440+.25(x-61300) if 61300 <x≤123,700
     24,040+.28(x-123700) if 123,700<x≤188,450 
     42,170+.33(x-188450) if 188450<x≤336,550
     91,043+.35(x-336,550) if 336,550<x

Graph of a function

The graph of a function is the collection of all points in the plane with coordinates (x,f(x)) if x is in the domain of the function. To the right is the graph of a function. A collection of points in the plane is the graph of a function if every vertical line which touches the graph intersects it exactly once. This is called the Vertical Line Test, and corresponds to the fact that we want exactly one output determined by every "legal" input. If we have the graph, the domain is all x's on the x-axis for which the vertical line through that x hits the graph. Similarly the range is all y's on the y-axis for which the horizontal line through that y hits the graph. Sometimes if you only have the graph and no other information, precise descriptions of the domain and range can be difficult.

An inverse function Consider the function defined by f(x)=2x3+7. Part of the graph of this function is shown to the right. It was produced by my silicon friend, Maple. (Available on eden, just type maple). The graph is a bit deceptive. I asked that the function be shown for x's in the interval from -3 to -3. The program automatically "autoscales" and squeezes so that the image is in a square. This can be confusing at times if you don't realize it. In this case, the vertical range seems to be between -45 and 60.

The part of the graph I want to emphasize here is the qualitative aspect: certainly every vertical line hits the graph once (the Vertical Line Test) but so does every horizontal line. That is, every output corresponds to a unique input. For this function, this uniqueness claim can be verified algebraicaly quite easily: if w is an output, we can write a formula for the unique x it came from. w=2x3+7 --> w-7=2x3 --> (w-7)/2=x3 -->[(w-7)/2]1/3=x A function which has unique outputs for each input is called 1-to-1. Such a function has an inverse. The function g(w)=[(w-7)/2]1/3 is inverse to f. It undoes f, so that f(g(w))=w and g(f(x))=x. The graph of g is shown to the right. It is gotten algebraically by interchanging the two coordinates of each point. Geometrically the graph of the inverse function is gotten by picking up the original graph of f and "flipping it over" the main diagonal, y=x.

A piecewise linear function to play with
I defined a function by sketching its graph, as shown to the right. I declared that the function, f(x), was piecewise linear. One part was a half infinite line (a ray) passing through (-3,0) and ending at (0,2). Then there was a line segment from (0,2) to (3,-1), followed by a line segment from (3,-1) to (5,4). The domain of this function was (-infinity,5] and its range was (-infinity,4]. My real intention was to use this function to study the problems involved in specifying inverses, but ...

QotD
Suppose H(x)=f(x²) and K(x)=f(x)². I asked students to tell me the domain and range of H and K. Below, by the way, are graphs of f and H and K, as Maple "sees" them. Notice again that vertical and horizontal scales differ.

Graph of part of f	Graph of all of H	Graph of part of K

QotD answers
Let's see: H(x)=f(x²). The inputs to f are (-infinity,5]. But squaring makes things non-negative, so the only numbers I can push into x so x² feeds into f are numbers which won't square to bigger than 5. Therefore I should put in only numbers between -sqrt(5) and +sqrt(5). So the domain will be [-sqrt(5),+sqrt(5)]. The range will be the outputs of f when the inputs of f are in the interval [0,5], and those outputs are [-1,4].
K(x)=f(x)². Now K has the same domain as f: (-infinity,5] because the domain of squaring is every possible number -- there are no restrictions to worry about. The outputs, however, are squared. The collection of outputs of f(x) are (-infinity,16]. When these are squared we will get every non-negative number.
Here is more detail. For x>0, the outputs of f(x)² go from 0 to 16. On the left, though, the outputs of f itself go down to -infinity, so squaring means the results are all non-negative numbers. The range of K is [0,infinity).

Numbers

• Positive integers
  We can start with 0 (the "additive identity": anything plus 0 is that anything) and 1 (the "mulitplicative identity": anything times 1 is that anything). 1's added to 1's create the positive integers.
• Integers
  If subtraction is considered, then we need "additive inverses". The additive inverses of the positive integers are negative integers. These together with 0 are all integers.
• Rational numbers
  Now multiplicative inverses force the consideration of reciprocals of integers (except for 0, because that would lead to real algebraic difficulties). Further multiplication makes the collection of rational numbers.
• All the others (these are most of them!)
  Rational numbers have finite or repeating decimal expansions. In this course we will use various sequential approximation processes, and we will need to consider numbers with all sorts of decimal expansions, not just repeating expansions. These weird and "chaotic" numbers are not rational: they are irrational. In various ways which can be made quite precise, there are many more irrational numbers than there are rational numbers: a typical number is irrational. Any physical experiment or process when measured gives only an approximation to some sort of "true" answer, and this sequence of approximations, like the decimal expansion, is likely to be an approximation of an irrational number.

Geometry
The real numbers are usually thought of as corresponding to a specific geometric object, the real line. I usually think of this line as horizontal with 0 sitting "in the middle". 1 is to the right of 0. And this geometric picture brings up the idea of order. In addition to the algebraic structure, there is order: a<b. To me this means (in my picture of the line) that a is to the left of b. Negative numbers are to the left of 0 and positive numbers, including all the positive integers, are to the right of 0. Here are some interesting aspects to note.

The decimal expansion of a real number provides an "address" to locate the number geometrically. For example, 23.4680472... means we first move right from 0 by 23 steps, each 1 long. Then look between 23 and 24, and divide that length in tenths. The number we're looking for resides (?) in the 4^th subinterval. Now divide that subinterval into tenths. The number is in the 6^th subinterval of that collection. Etc. The "Etc." of course conceals an approximation process which will locate the real number with the specified decimal expansion.

Some real numbers have two valid decimal expansions. This doesn't bother me, since, say, a house on a corner of a grid system of streets could also possibly have two valid addresses (if it is on the corner of Cedar Avenue and Third Street, the house could have the address 12 Cedar Avenue and also 42 Third Street). So, for example, the real number with decimal address 23.45999... (with the 9's repeating) also have the decimal address 23.46, a terminating expansion (there are a string of 0's which are usually not written.

Distance including a discussion of | |
The distance between two points is a non-negative real number whose size expresses how far apart the numbers are. This will be important when we study approximation schemes. We'd like to know that the approximation gets "close" to the correct answer, and the closeness will be measure by the distance. Algebraically, if the points correspond to the real numbers a and b, the distance between them is |a-b| and this is the same as |b-a|, so that distance has some symmtery. But I just used absolute value, and here is the piecewise definition of absolute value:
|x|=x if x≥0 and |x|=-x if x<0
Therefore absolute value is always a non-negative number. The absolute value of a number is 0 only when the number itself is 0. And absolute value of a product is the product of the absolute values (this actually is not totally obvious, and needs a bit of thought, I believe).

Intervals
Suppose we want to discover what numbers x are closer to 9 than a distance of 4. Algebraically this requirement translates to |x-9|<4. We can sort of "unroll" the inequality. The absolute value will be less than 4 if the number itself is both less than 4 and greater than -4. The two inequalities can be compactly written as follows:
-4<x-9<4 which implies
5<x<13
This is an interval and an interval which does not contain either endpoint is called an open interval. The notation for this interval is (5,13). Intervals which contain both endpoints are called closed. An examples of such an interval is [-4,6], which means the numbers x satisfying -4≤x≤6. There are also half-open intervals, unbounded intervals (with notation using + or - infinity), etc. Please see the textbook.

Warning!
If you wanted to "solve" (better: understand!) the inequality |x-9|>4 you can't just "unroll" it to -4<x-9>4. This inequality has no solutions. There is no number which is simultaneously less than -4 and greater than 4. You can't write this so compactly and using such implications represents an invalid (wrong!) method of solution.

A valid method of solution would involve separately solving the inequalities:
-4<x-9 orx-9>4 which gives 5<x or x>13. This is actually, therefore, two intervals:
(-infinity,5) and (13,infinity). So the inequality |x-9|>4 has a solution set which is two intervals.

The plane
The conventional way to describe the plane algebraically is to drop down two lines perpendicular to each other: coordinate axes. A point in the plane will then be described by an ordered pair of real numbers. The first coordinate will usually be called the x-coordinate and the second, the y-coordinate. This pair describes ordered distances from the horizontal (the x-axis) and vertical (the y-axis) lines. Please see the text for more about this.

The embarrassment of all this, especially with "new" students, is that (3,8) could describe both a point in R², the plane, and could also describe an open interval (with [missing!] endpoints 3 and 8). The context is supposed to help, but still the notational confusion is possible, and this is lousy.

Distance in the plane, R²
Well, I went through the usual diagram to try to motivate the algebraic definition of distance in R². Look, please, at the diagram to the right. In the plane, points correspond to ordered pairs of numbers. So a point p might correspond to an ordered pair, (x₁,y₁), and q might correspond to (x₂,y₂). Then the point (x₁,y₂) is the vertex of a right triangle whose hypotenuse is a line segment connecting p and q. One leg of the right triangle is on a line where all the first coordinates are x₁, and the length of that leg is given by the one dimensional formula, |y₁-y₂|. The other leg is on the line where all the second coordinates are y₂, and the length of that leg is |x₁-x₂|. Then by Pythagoras, the hypotenuse has length sqrt(|x₁-x₂|²+|y₁-y₂|²). And usually the absolute values are discarded since we are squaring the quantities. Therefore we officially define:
dist(p,q)=sqrt((x₁-x₂)²+(y₁-y₂)²) if p has coordinates (x₁,y₁) and q has coordinates (x₂,y₂).

Example
The distance between between (3,-2) and (6,4) is sqrt([3-6]²+[4-(-2)]²)=sqrt(3²+2²)=sqrt(13).

A circle and its algebraic description
Suppose we wanted to describe algebraically the collection of all points which have distance sqrt(13) from (6,4). This is, of course, a circle of radius sqrt(13) and center (6,4). Well, if (x,y) is such a point, then sqrt([x-6]²+[y-4]²)=sqrt(13). This does describe the circle. Of course, if you must, all sorts of algebraic things could be done to the equation. But I am the laziest person in the room, and therefore ...

A (non-vertical) line and its algebraic description
Suppose we wanted an algebraic description of points with coordinates (x,y) which lie on the straight line which goes through (4,3) and (8,13). If (x,y) is such a point, then look at the picture: two right triangles indicated are similar, so the corresponding sides have the same ratios:

13-y   13-3
---- = ----
 8-x    8-4

QotD
I gave two points and asked for an equation for the line containing the two points. Most people were able to do this.

Maple
Much intricate algebraic computation can now be done by computer programs. A course like Math 151 can therefore comcentrate more on understanding rather than elaborate computational skills. On the other hand, knowing what results should look like (if you calculator declares that 56.007 is the sum of 1,345 and -14,891, do you believe it?) is important, so a secure command of various algebraic algorithms is still important. At Rutgers, the program Maple is widely available since there is a university license for it. Students may also get copies under certain circumstances. Basic information about the program is here and here are some local help pages.

Read the text!
Please read the first few sections of chapter 1, as on the Math 151 syllabus.

Maintained by greenfie@math.rutgers.edu and last modified 9/7/2006.