# Diary for Math 152:72 lectures in fall 2001

1. (9/6/2001) Discussed area between curves and some examples. Discussed the volume of a pyramid with square base and some height. Find volumes of various solids of revolution. Used the phrase "profile curve" to denote the curve revolved. Examples included cylinders, cones, and spheres. Did mostly discs, with one shell method.

2. (9/11/2001) Motivated "average value" of a function by averaging equal spaced sample points. Then mentioned a considerable part of the course was computing integrals, using almost all things discovered one or twohundred years ago. Mentioned the idea of taking random samples of function in interval and averaging these, multiplying by length of interval -- the "Monte Carlo" method of approximating the definite integral. Then started by far the most important method of antidifferentiaton.

Integration by parts. Started with x sin x. Did x e^x, x^2 e^x, etc. (Idea of reduction formula), arctan x (struggle remembering what this function was, and what its derivative was). Went slowly, tried to motivate choosing the parts, emphasized that people should write out u and dv, etc. Emphasized need to do all homework problems in this.

3. (9/17/2001) Announced that the lecture would cover 7.2 & 7.3 & discuss how the course would be graded.

Integrated sin x and (sin x)^2 and (sin x)^3 and (sin x)^4 and (sin x)^5. Used double angle identity for cos(2x) and Pythagoras for (sin x)^2 + (cos x)^2.

Integrated tan x and (tan x)^2 and (tan x)^3 and (tan x)^4. Needed to "recall" the first (gave out sheet reviewed "everything" about 1st semester calculus). Used Pythag for 1+(tan x)^2 for second integral and "recalled" integral of (sec x)^2. Used tricky substitution for third.

Integrated sec x and (sec x)^2 and (sec x)^3. For first needed trick of multiplying "top and bottom" by sec x + tan x. For third, needed elaborate integration by parts with "solving for" desired integral.

In all of these needed double angle formula for cos and the Pythagoras identity for squared trig functions, but also needed familiarity with "easy" differentiation formulas, some of which had to be remembered or redisplayed.

Wanted to find area inside a piece of unit circle. This was pretext for integrating sqrt(1-x^2). Did this by saying we need a substitution, say ? for x, so that sqrt(1-?^2) was a known function, say, !. Then get ?^2+!^2=1, so "guess" that ? could be sin x. This was elaborate. Tried to play the same trick with 1/sqrt(5+x^2). This was not as successful.

Discussed grading of the course. 20% for quizzes (graded by Manning), textbook homework (graded by Scarangello), and workshops (graded by me). 80% for 2 exams (20% each) and final (40%).

4. (9/20/2001) Discussed the goal of finding antiderivatives. A general idea is how to find such for functions defined by formulas. Informally introduced hierarchy of such functions: polynomials, rational functions, exp, log, trig, inverse trig, etc.: even more horrible functions defined, e.g., by composition. Remarked that antidifferentiation is special for all except polynomials.

Handed out Maple's solution of antiderivative of 1/(2+x^4) - 3/(4+x^4). So a computer program can do this? So why should people do it?

Some possible answers: (0) It is in the course syllabus, (1) Is it the right answer? (i.e., if a calculator answers that sqrt(100) is .007, is that correct? -- so should output from partial fraction algorithm include, say, exp (cos(x^3))? I think not!) (2) How does it work -- so maybe we should understand a significant algorithm and also be sensitive to where maybe it might break down (important!) (3) The decomposition of rational functions given via partial fractions important for reasons other than "merely" finding antiderivatives, useful in applications in chemistry, physics, etc. (true statement, left unclarified).

Did a sequence of examples, commenting on them. First example something like (x^4-2x^2+5)/(x^2-5x+6). Top random, bottom darn easily factorable. Divided top by bottom and got quotient + remainder, the latter used to write a proper fraction of a rational function, with degree top < degree bottom> Remarked that in all examples following would omit that step -- easy to integrate polynomials.

An aim pf partial fractions to get things which can be easily integrated. Introduced sum of A/(x-2) + B/(x-3) here. Combined fractions, equated tops, found A and B by looking at magic numbers 2 and 3 for x.

Now did a fraction with the bottom (x-2)^2(x-3) as usual. Could have a quadratic polynomial with 3 free variables as coefficients above. Therefore will need 3 variables in the partial fraction answer. Looked at magic numbers 2 and 3 from usual decomposition. Motivated "usual decomposition" not at all -- it really is not easy to see where it comes from! I said, "it works" and is one form that does work. Got the third unknown by checking the x^2 coefficient in the equated tops, and remarked that there are many strategies.

Described what would happen if bottom was, say, (x-7)^5(x-4)^8(x-6)^11. Said, truthfully, that resulting system of unknowns was very computationally feasible to solve. However, I was concealing an essential computational difficulty, namely that it is VERY difficult to factor a random polynomial.

But a random polynomial does not factor into linear factors over R. So discussed how to integrate (3x-7)/(x^2+4). Then how to do more complicated things, and did another example (linear times quadratic in the bottom).

Indicated what the general case would look like, and what the answers would look like. Told them we would have a fiedl trip to meet my friend, Maple.

5. (9/24/2001) "Discussed" section 7.5. That is, I DID in front of folks problems 1, 5, 9, 15, 17. Then asked if they had any questions. I remonstrated with them about attending the recitation and about doing the homework. Onward I went, and did #8, which is integral from 0 to 1 of (1+sqrt(x))^8. I did this two ways, with expanding via binomial theorem (only a few of them had heard of this!), and then via the substitution t=1+sqrt(x).

Then I said how would you integrate sqrt(x), and did it via the substitution t=sqrt(x), and got the correct answer. I then did the integral of sqrt(1+sqrt(x)) via the substitution t=sqrt(1+sqrt(x)). Then I did the integral of sqrt(1+sqrt(1+sqrt(x))) via the subsitution t=sqrt(1+sqrt(1+sqrt(x))). It all converts to a polynomial, and the mess is in the computation of the inverse function, and then computing dx. I asked in general how people should compute indefinite integrals, in, say, 2 years.

Method (0): Ask me. I will charge $1,000 per integral. Method (1) Do it via a computer program (Maple, MATLAB, Mathematica, etc.). I then remarked that I had two versions of Maple at home. One was ~5 years old, the other ~2 years old (the latter is functionally similar to what is on Eden, the student computer system. And, hey, the older version can't integrate sqrt(1+sqrt(x)) while the newer version can, but it can't handle sqrt(1+sqrt(1+sqrt(x))) (!). I guess one needs to know a bit more. ANd, as I explained last time, it is useful to know when results are NOT valid (as in uses of partial fractions). Method (3). Buy a copy of Gradshteyn & Ryzhik: the last version is 1,000 pages, only ~$80, and is selling well (~#20,000 from the top) at amazon.com. If one needs to know lots of definite integrals, then, hey, LOOK at what has been done.
Method (4) Do things numerically (Next time!).

Then I did a few more integrals. #69, #60. I hope my repeated urgings to do homework were useful.

6. (9/27/2001) Began by stating we will discuss important topic, easily neglected by students: how to approximate the values of indefinite integrals. Wrote a table of x & 1/x on the side blackboard, where x=1.,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9,2. Describe my continuing example as the integral of 1/x from 1 to 2 which was of course ln(2)-ln(1)=ln(2)approx. .69315. I asked if students could discern (?) .69315 in the table at the side of the room. I said this was very important topic but harder to teach than the algebraic tricks we had discussed before.

Returned to the definition (essentially) and discussed the formalities of setting up the "Left-hand Rectangular Rule" to approximate the definite integral from a to b of f: this was mostly to make sure folks understood language. Divided the interval into n equal parts, with width h=(b-a)/n. Set x_j =a+jh, so x_j was the left-hand end-point of the (j+1)st subinterval (except that x_n was the right-hand endpoint of the last subinterval). Thus the approximation here was f(x_0)h+f(x_1)h+...+f(x_{n-1})h=h[f(x_0)+f(x_1)+...+f(x_{n-1})]. In the specific example set up, I wrote the arithmetic but had performed it before class. The approximation obtained was .71877, not so good.

In the case of the Trapezoid Rule the approximation is via a tilted line segment linking the points (x_j,f(x_j)) and (x_{j+1},f(x_{j+1})). Explained what a trapezoid was, and used the area formula (average of bases times the height) to get the area of the approximation. We then added these up. Students contributed to simplifying and compactifying the resulting algebraic formula. We got (h/2)[f(x_0)+2f(x_1)+2f(x_1)+...+2f(x_{n-1})+f(x_n)]: 2's on the inside nodes, and 1's on the edges. So the change from the previous result is actually not so much -- just taking away half of f(x_0) and adding on half of f(x_n). For the specific example we had, the arithmetic was carefully written out and the approximation obtained was .69332, better (interesting: some students seemed to think this was terrific).

I asked what the next step would be. I remarked that the Trapezoid Rule locally approximated by straight line segments, and then exactly integrated these line segements. The next step would be to go up a power. I looked at Ax^2+Bx+C. We determined that, due to the THREE free parameters in the formula, such a curve could be sent through three points (more freedom than lines). But how to relate the area to the parabola? This is in the book but I hope I went carefully through the development: given Ax^2+Bx+C, the area under this from -h to h is (2/3)Ah^3+2Ch. How to relate this to f(-h)=A(-h)^2+ B(-h)+C, f(0)=C, and f(h)=A(h)^2+ B(h)+C? Some thought was required, but we finally got it down to (h/3)[f(-h)+4f(0)+f(h)]. Then transported this result to the more general case, with, of course, n even, since we will need to look at pairs of subintervals. Got the familiar (h/3)[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_0)]. Emphasized thatmultiplication by 4 and 2 in binary is rather easy, and no one seemed to understand me. Oh well. In the computational example discussed, the answer (written out with all the arithmetic indicated) was .69315, quite nice.

I discussed error estimates. First, for the "Rectangle Rule": I wanted to relate int from 0 to h of f(x) with f(0)h, one part of the Rect Rule approximation. This was difficult to motivate, and I probalby didn't do it too well. So I took the integral of f from 0 to h and integrated by parts, with u=f(x) and dv=dx. But one of the other parts was a bit surprising. du is f'(x)dx, certainly, but v=x-h (allowed, its x-derivative is still 1). I think I lost a number of students here -- too much symbolic stuff. Anyway, if one follows through, the resulting equation has the integral on one side and the rectangle on the other, and there is an error term: here it is int from 0 to h of f'(x) multiplied by (x-h). Then I estimated that by pulling out the max of |f'(x)| and integrating (just finding the area!) h-x (need absolute value) from 0 to h. So we got Max of |f'| multiplied by h^2/2. Then add up errors to get worst estimate. There are n of these pieces, and h is (b-a)/n so result is Max of |f'| multiplied by (b-a)^2/n. Clearly the important ingredient here is the n, because as n gets big, the error goes to 0.

I did not derive the error formulas for the Trap Rule which is M_2(b-a)^3/(12n^2) or for Simpson's Rule which is M_4(b-a)^5/(180n^4). Here M_j is the max of the absolute value of the jth derivative of f on the inteval from a to b. I told them, correctly, that that one way to get these is by integrating by parts with "weird" choices for parts, twice for Trap and four times for Simpson. You earn an extra power for Simpson because it turns out the formula is accurate for cubic polynomials -- not at all clear. I mentioned that the important contrasts were the "Rules" were, respectively, O(1/n) (that's "big-Oh"), O(1/n^2), and O(1/n^4). Some students had heard of big-Oh notation, but many had not. Even those who had were not familiar with the real power of the different rates of convergence.

I applied this to the integrl of 1/x on the interval from 1 to 2. Then I asked how many subdivisions Simpson's Rule would require to approximate cos(x^2) to accuracy < 1/10,000 on the interval from 0 to 1. I found the fourth derviative, and BRUTALLY overestimated this mess, and answered the question.

I told them that there would be an exam two weeks from today, with more information to follow in e-mail.

7. (10/1/2001) Went back to question of how to make numerical integration fast. One can improve the schemes (going past Simpson) in order to reduce n, but things get more complicated. One does want reduced n in order to decrease errors in arithmetic, due to floating point computation. But another source of computational effort is the number of "function calls", the number of times one needs to evaluate f. Can one reduce this?

Went back to the integral of a quadratic polynomial over the interval, say, from -1 to 1. For Simpson's Rule, one uses three function evaluations. Is there a way to do integration of quadratic functions opn this interval exactly with only two function evaluations? This is a conceptually new thing. Rephrasing: can I find a, b, c, and d so that the integral of f(x) from -1 to 1 equations af(c)+bf(d) when f is the function 1 and the function x and the function x^2. Quite a lot of conversation followed, and the associated system of equations was deduced. I mentioned that all we were after was one solution, if such existed (and since there were three equations in four unknowns, one could hope for solutions, although this also seemed difficult for people).

A solution was found: a=1, b=1, c=-1/sqrt(3), and d=1/sqrt(3). We talked more about why this was not much described in decades ago calc courses, and one reason is that it is hard to do by hand, since the sample points are no longer equally spaced. But in fact such things are used in, say, the TI-82, along with another wrinkle (concentrate on where the function "wiggles"!).

Then improper integrals, defined to be those where the original definition of integral fails, due to defects in the domain or the range. I tried the integral from 1 to infinity of 1/x and 1/x^2. The first diverged and the second converged, and I emphasized that it was impossible to just glance at the graphs and see this. I then asked if something like 1/(x^2+354) integrated from 1 to infinity converged. The answer was "yes" after some discussion. I stated the comparison theorem for such improper integrals.

I tried to indicate that imporpoer integrals arose in chemistry and physics because of difference of scale: from the point of view of one electron, a finite area flat plate external to it might as well be infinite in extant. The same occurs when studying big molecules -- the difference in scale makes finite integrals so close to infinite improper integrals that one might as well compute the imporper integrals. And that's what people do.

I computed the integral of exp(-x) from 0 to infinity. Then I computed the integral of x*exp(-x) from 0 to infinity, mentioning that the comparison result would not apply. Then I computed the integral of x^2*exp(-x) from 0 to infinity. Then I indicated how to compute the integral of x^3*exp(-x), etc. In fact, it turns out that the integral of x^n*exp(-x) is n!, and this is accepted as the correct generalization of the factorial function. So, for example, (1/2)! "is" the integral from 0 to infinity of sqrt(x)*exp(-x). The other kind of improper integral will be discussed next time.

8. (10/5/2001) I began by giving students 8 to 10 minutes to do a test question from the very last time I taught second semester calculus. The question asked students to sketch part of a region in the plane (to the right of y=1 and between y=4exp(-3x) and y=4exp(-2x). The integral involved is, of course, improper. I went over the problem.

Then I began study of improper integrals with "problems" in the range-- that is, the integral of f from a to b with, say, f(x)-->infinity as x-->a or b. So I discussee the integral of 1/sqrt(x) from 0 to 1, and the integral of 1/x^2 from 0 to 1. I remarked again that one could not tell from looking at the graph whether the integrals converges or diverged.

In fact, in the specific case of 1/x^2, there is actually an interesting physical interpretation. electrons, for example, repel each other with a force proportional to the inverse of the distance between them squared. The fact that the integral from 0 to 1 diverges leads one to believe that the amount of work done to bring two electrons next to one another is infinite, so "it can't be done" but since the integral from, say, 1 to infinity is finite, the total amount of work needed is finite, and certainly quantities like escape velocity can be calculated.

I then sketch y=ln(x). The integral from 1 to infinity certainly diverges (compare with just a positive constant!). What about the integral from 0 to 1? First I needed to find the antiderviative of ln(x), and then we needed to analyze the limit as A-->0+ of Aln(A)-A. The first part of this is difficult to understand: A-->0 but ln(A)-->-infinity. The limit can be analyzed with l'Hopital's rule, as Aln(A)=ln(A)/(1/A), so the result is now like infinity/infinity. I applied L'H and got the answer. This took a while.

I then picked up a copy of Hamming's "Coding and information Theory", used sometimes as an upper-level text in an electrical engineering course, and read off from there the definition of the Entropy Function: f(p)=pln(1/p), where p is here a probability (so naturally between 0 and 1) and f(p) is "information". I asked what the graph of this function looked like. There was some delay while a calculator was found. I asked whether f naturally had a "value" at 0. A discussion resulted. Then again the limiting value of f at 0 was gotten with L'H's rule. I further asked students to find
1. What could be called f'(0) (that is, what is the limit of f'(p) as p-->0+?)? This needs another use of L'H's rule.
2. What is the total area under the "bump" of the entropy function? (This is yet another use of integration by parts, followed by yet another use of L'H's rule.)

I then sketched a derivation of the formula for arc length. This was uninspired, and followed the text. I did not have time to do interesting examples, which I hope I will next class.

I handed out another "test question" from an old exam: what is the area enclosed between sec(x) and tan(x) when x is between 0 and Pi/2 (after sketching the region)? Yet another improper integral. I told students to e-mail me the answer and I would check it for them.

9. (10/8/2001) I sketched out the schedule for the rest of the week (Tuesday, the workshop will be an old exam, Wednesday I will have extra office hours, Thursday the exam, Friday a Maple "lab" and return of the exam).

Examples of the arc length formula, derived last time (integral from a to b of sqrt of 1+(dy/dx)^2. First, some simple test cases: the length of a straight line segment, which is a straightforward computation, and then the length of a circular arc. The arc I choose to explore was the quarter arc of x^2+y^2=1 in the first quadrant.

Here y=sqrt(1-x^2). The derivative needs the chain rule, and then amazingly the formula sqrt of 1+(dy/dx)^2 turns out to be (after computation) just 1 over sqrt(1-x^2), and its antiderivative is arcsin. When 0 and 1 are inserted for a and b, the answer is Pi/2 as predicted. Of course, some amazing coincidence took place to get something nice to antidifferentiate. And, in fact, the integral is more subtle than appears because the integral from 0 to 1 of 1 over sqrt of 1-x^2 is actually an improper integral, and should therefore be handled somewhat carefully!

Then I looked at the problems assigned in the text. I wrote all of the functions listed inn section 8.1's problems: 5, 7, 8, 11, 36. The functions looked weird. Then I did problem 6, again a function with a "weird" formula. Of course the weird formula is exactly chosen so that the arc length computation gives something that has a simple antiderivative.

As my final test case for the arc length formula, I tried to find the arc length of a parabolic arc, y=x^2 from 0 to 1. This resulted in trying to antidifferentiate sqrt(1+4x^2) which I could do with a secant substitution, changing it to the integral of (sec theta)^3, which we did several lectures ago. With the help of a student, I resurrected the formula for the antiderivative of sec^3, and managed to (sort of) finish the problem. I remarked that the next step "up" in complexity (x^3) yielded an integrand, sqrt(1+9x^4), which could NOT be computed in terms of "familiar functions" by anyone in the world.

The secret is that almost all arc length intgrals which arise in practice must be approximated numerically.

Then I started ordinary differential equations. I looked at dy/dx=x^2, and it was claimed that the "solutions" were (1/3)x^3+C, for any constant C. I sketched these curves. I looked at dy/dx=y, where students had a very, very hard time finding solutions other than e^x. We finally decided that (const)e^x were solutions, where const could be any fixed number (including 0 and negative numbers). Again I sketched the solution curves.

I discussed a more difficult model, the vibrating spring. Some of the students knew of F=ma, Newton's second (?) law, and also some knew, perhaps in a garbled form, Hooke's "law", an experimentally verifiable fact that the restoring force in a spring varies directly with the displacement of the mass from equilibrium. Then I got a second order differential equation, y"=-(k/m)y, and described some solutions, sin(sqrt(k/m)t) and cos(sqrt(k/m)t). Are these all of the solutions (i.e., maybe with this setup, you could start a spring vibrating and the mass could sail off to Alpha Centuri? Well, no, but that there are only "vibrations" as solutions to the equation is not obvious.

These are all examples of differential equations. I tried to tell the CS people in class that CS problems could be studied analogously. Here the "variable" would be the input size of the problem and the functions to be studied would be computation time or storage time. And as one increased, say, the size of the input, the algorithm would change (maybe it is described recursively, so maybe that makes it easier to see) and that the resulting setup looks very much like a differential equation. Differential equations describing elementary physics have been around for >300 years, and people have been systematically studying algorithms for maybe ~30 years. Learn about differential equations to have some "paradigm" for the study of algorithms.

I defined differential equation and solution of a differential equation.

10. (10/15/2001) I discussed and defined the terms: differential equation, solution of a differential equation, initial condition, intiial value problem, direction field, equilibrium solution, stable and unstable equilibrium solutions, and Euler's method (this is sections 9.1 and 9.2).

These were illustrated with some examples. I played with y'=x+y and sketched the direction field. We got some pictures. Then I used Euler's method to approximate y(3) for the initial condition y(1)=2. I remarked that Euler's method converged very slowly.

I tried to "solve" y'=y(1-y)(2-y) and I couldn't. I asked for some solutions and was given the solutions y=constant where constant is 0 or 1 or 2 (the "equilibirum" solutions). I looked at the direction field for this equation and was able to deduce much more about the solutions, especially their qualitative behavior. We could even find stable and unstable equibrium solutions.

I stated the following: given a differential equation y'=stuff involving x and y, and given an initial condition, then there is a unique solution, and Euler's method will converge to the solution. (but very very slowly in practice, so other methods are used).

Problems due Tuesday: 7.4:45, 7.8: 30, 9.1: 4,6, 9.2: 2,12

Began to look at separable equations.

11. (10/18/2001) I discussed problem 2 of section 9.2 for a few minutes.

I started separable equations (section 9.3) first by finding dy/dx when x^2+y^2=1 and mentioning that the chain rule. Then I did the "general" case: F(y)=G(x) so that (applying d/dx) we get F'(y)dy/dx=G'(x) so that dy/dx=G'(x) (times) 1/F'(y). A separable euqation is one we can reverse this process on. So it looks like: dy/dx=product of some function of x alone multiplied by some function of y alone. I asked which of the following were separable equations:

1) dy/dx=x+y 2) dy/dx=y 3) dy/dx=e^y sin(x) 4) dy/dx=x/(x+y) 5) dy/dx=e^(x+y).

There was some controversy. Then I solved #3 with the initial condition y(Pi/2)=0. The integrals were fairly easy, and I addressed the issue of how many constants were needed in the integration (at first sight, two, but really it works out to one). And I got the specific solution, and I was even able to "solve for y". I checked that the answer was correct by finding dy/dx and then seeing that it was actually e^y sin(x).

I noted that one step of this was frequently not computationally feasible, that is, the solving for y. I tried the separable equation dy/dx=(x^2+1)/(y^2+1) subject to the initial condition y(-1)=2. This merely (?) involved integrating some polynomials. I then got the specific solution by substituting in the initial conditions. The solution was y^3/3+y=x^3/3+x+6, I think. I remarked that I couldn't "solve" for y very easily. Also even if I could, it isn't clear how useful the result would be.

I got Maple to graph the solution, using the command "implicitplot". The solution seemed to be asymptotic to y=x when x and y were large positive and negative -- this in fact agrees with the algebra because the dominant terms on both sides are x^3 and y^3. Also the curve doesn't have any critical points (e.g. no local max or min) and this agrees with the dy/dx formula since it can never be 0.

I looked at y'=y^2 and y'=y and y'=sqrt(y) with the initial condition y(0)=1. I asked students to sketch the direction field and give some qualitatively correct solution curve through this initial condition. They did, and the curves all sailed up from (0,1) as time (the independent variable) --> infinity.

They all looked the same, but then since these are all rather simple separable differential equations I was able to solve them explicitly and note that the domain of the y^2 solution "ended" at 1 (!!!). So this equation had finite time blowup while the others didn't. There is no apparent reason that such happens since y^2 looks really really nice. If this all represents a physical theory then there is significant difference between the solutions, one which is not discernible by loose use of direction fields.

I "solved" y'=ky as in the text to get y=Aexp(kt) and then did problem #2 of section 9.4. I advertised this as such a well-known use of differential equations that most people just quote the answer rather than go through the rigmarole.

Problems due Tuesday: 7.2:14, 7.5:20, 9.3:10,14, 9.4:9,12

12. (10/22/2001) I went over #34, 9.3, a mixing problem. This took quite a while.

Then I began the second major topic of the semester, infinite sequences and series. I began by remarking that chapter 11 begins rather abstractly, and that I would try to motivate the material.

How can one compute the values of various functions (trig functions, exp and log, etc.)? The only functions one really knows how to compute are polynomial functions (one definition: a function which can be written as a finite sum of constant multiples of non-negative integer powers of x). So how could one compute values of exp? Well, begin by characterizing exp as the unique solution of y'=y with y(0)=1. Then assume (outrageously!) that y can be written as a polynomial: y=A+Bx+Cx^2+Dx^4+.. By comparing y' with y and equating coefficients of the same powers of x, we see that A=B, and B=2C and C=4D and D=5E, etc. Since y(0)=1, A must be 1. Then we can evaluate the coefficients so that the solution we get is 1+x+(1/2)x^2+(1/6)x^3+(1/24)x^4+... Here the general term is (1/n!)x^n.

Of course this expression is NOT a polynomial. "Nevertheless" we try to get an approximate value of, say, exp(.5), by inserting .5 for x in this, we get an infinite sum. Suppose I want an approximation for exp(.5) by looking at 1+.5+(/5)^2/2+(.5)^3/6+(.5)^4/24+(.5)^5/120. The error is the terms left out. We can't just evaluate this, since it would involve an infinite number of additions. I tried to show how one could overestimate this with a simpler expression.

I "reviewed" geometric series, purely formally. That is, if S=a+ar+ar^2+ar^3+... then rS=ar+ar^2+ar^3+... so S-rS=a and S=a/(1-r). Then I asked if the sum of (.5)^6/6!+(.5)^7/!7+(.5)^8/8!+... was a geometric series. I was told not, and we thought about that for a while. Then I wrote the geometric series (.5)^6/6! with ratio .5/7 and we realized that this was an overestimate of the "error" and the sum of this overestimate was (.5)^6/6!/(1-(.5)/7), a very small number (computed by students).

This whole scheme would be systematized in the coming weeks. I finally spent the last few minutes of the meeting discussing the following gambling game: flip a fair coin repeatedly, and if the first head occurs on the nth toss, I pay you n^2 dollars. It took some time to get understanding of this, and then I asked what was the expectation of this game -- the average winnings. It seems to be 1/2+2^2/4+3^2/8+4^2/16+... I sasked what a fair entry fee was. We discussed this.

13. (10/25/2001) I discussed definitions and results in the text.

Definition of sequence. Examples started with sequences with simple explicit formulas such as sqrt(n) and (-1)^n. I went on to sequences defined by Newton's method, recursively. When applied to F(x)=x^2-a (searching for sqrt(a)) the "rule" looks like: x_1=initial guess, and successive guesses defined by x_{n+1}=(x_n+a/x_{n}). It is not at all clear what an explicit formula for such approximations such look like. I also defined the Fibonacci numbers and noted that we later would derive an explicit formula for them.

I attempted a definition of convergent sequence, and tried to see why, say, (-1)^n did NOT converge. But 1/n did converge, and its limit was 0. I gave the text's definition of convergent sequence.

I analyzed r^n first for r=1/2. I took logs and looked at the limit of n ln(.5), which -->-infinity as n-->infinity. Then exponentiating we saw that .5^n--> 0 as n-->infinity. More generally, we saw that r^n-->0 as n-->infinity for any r in the interval [0,1).

Then we looked at 10^(1/n). Again by taking logs and limits and then exponentiating back we found that 10^(1/n)-->1 as n-->infinity. In fact, we saw that if A is positive, then A^(1/n)-->1 as n-->infinity. Part of the discussion involved acknowledging that A^(1/n) was decreasing for A>1 and increasing for 0 We looked at n^(1/n), and again took logs and looked at limits, this time with the help of L'Hospital's rule. So we saw that this sequence had a limit and the limit was 1.

Sequences and limits and algebra: sums of convergent sequences are convergent, and the limits are the sums of the limits. The saem with products, etc. I said that a more complete statement was in the textbook.

Sequences and limits and inequalities: if a_n<=b_n and a_n converges to L and b_n converges to M, then L<=M. We talked about whether just knowing that a_n converged implies that b_n would have to (it wouldn't).

A squeeze result: if a_n<=b_n<=c_n, and if a_n and c_n both converge with the SAME limits, then b_n would have to converge, and would share the common limit. I applied this to considering w_n=(3^n+4^n)^(1/n). This was difficult. Here I bounded w_n below by (4^n)^(1/n)=4 and above by (4^n+4^n)^(1/n)=2^(1/n)4 so the common limit was 4. This was not obvious. After much further discussion (including the asymptotic behavior of (3^n+4^n+5^n)^(1/n)!) we arrived at the following result: if A and B are positive, then the sequence (A^n+B^n)^(1/n) converges and its limit is the MAXIMUM of A and B (the largest of the two, or, if the two are equal, the common value).

I tried to echo the definition of infinite series which is in the book. I defined partial sums and sequence of partial sums and sum of an infinite series.

Problems due Tuesday: 9.3:16,17; 11.1:36,59,61; 11.2:10.

We discussed sequences and series again and did a homework problem. In this course we will have convergent and divergent improper integrals, convergent and divergent improper sequences, and convergent and divergent improper series: complicated terminology. I defined again what a sequence was, what a series was, and what the sequence of partial sums of a series was. This last concept allowed the definition of a convergent series and the sum of a convergent series (as the limit of the sequence of partial sums). We saw:

• If a series converges, then the sequence of individual terms must also converge, and the limit of this sequence is 0.
• The converse of the previous statement is in general false: that is, there are series whose nth term --> 0 but which do not converge.
The example I gave was 1+ ten terms of size 1/(10)+ 100 terms of size 1/(100)+ etc.

For a while I would discuss series with nonnegative terms. For these series, the sequence of partial sums was increasing, and there were two mutually exclusive possibilities: EITHER the partial sums had some upper bound, and the series converged, OR the partial sums were unbounded.

Then I discussed the series sum (from 1 to infinity) of 1/(n^2) and sum (from 1 to infinity) of 1/sqrt(n). I remarked that so far we knew how to handle geometric series, and reminded people about such series. I asked if the series sum (from 1 to infinity) of 1/(n^2) was a geometric series. After some thought (What's a? What's r?) we seemed to convince ourselves that this series was not a geometric series.

Then I compared the partial sum 1+1/4+1/9+1/16 with an integral. By writing certain "boxes" correctly, we cleverly saw that this was less than 1 + the integral of 1/x^2 from 1 to 4. This integral I could compute, and we saw that the partial sum was less than 1+the integral from 1 to 4 of 1/x^2, and this was an overestimate which was easy to compute. Then I asked about the partial sum s_{700}, and we agreed that an overestimate was 1+the integral from 1 to 700 of 1/x^2: this was (if I remember correctly) 2-1/700. More generally we saw that just 2 alone was an overestimate of any partial sum of this series. Therefore the sequence of partial sums was bounded above, and the series converged.

The series sum_{n=1}^infinity 1/sqrt{n} was also not a geometric series. Here it turns out that the series diverged. For a partial sum one uses a comparison with the integral from 1 to T of 1/sqrt{x} dx, where T was some positive integer. This was 2sqrt{T}-2, and it --> infinity as T --> infinity. Since the partial sums are greater than this integral, the partial sums are unbounded.

I stated the Integral Test, as in the textbook. I then applied the Integral Test to analyze the convergence of sum_{n=1}^infinity 1/n, called the Harmonic Series. We saw that the series diverged since int_1^T=ln(T)-->infinity as T-->infinity. But I then asked: can we find a specific partial sum which is bigger than 100? I remarked that this is not a task to assign (directly!) to a computer for reasons which would become clear. I underestimated this partial sum by the integral from 1 to T+1 of 1/x, and therefore in order to be sure that the partial sum was > 100 I needed ln(T+1)>100, or T>e100-1. Since e^2 is approximately 9, it turns out that this means T should be about 1045 (the exponenet is halved). This is one heck of a lot of terms!

Further, we know that sum_{n=1}^infinity 1/n^2 converged. What partial sum will be within .01 of the true value? For this I split up the infinite series into s_N (a partial sum) + t_N, an "infinite tail", the sum from N+1 to infinity of 1/sqrt{n}. I overestimated this infinite tail by the integral from N to infinity of 1/x^2 which was 1/N. So s_{100} will be within the true value of the sum of the whole infinite series. In fact, Maple tells me that s_{100} is 1.634983903 while I just happen to know that the "true value" (to 10 decimal places) is 1.644934068. We are correct.

First I wrote a review of what we've done so far: for sequences with positive terms, bounded partial sums and covergence coincide, and unbounded partial sums and divergence coincide. Also, I rewrote the integral test, and remarked that it could be used to estimate infinite tails and also to get underestimates for partial sums. And I wrote that p-series (sum as n runs from 1 to infinity of 1/n^p) converges if p>1 and diverges otherwise.

I tried to give some reason for what we are doing now. That is, suppose one is trying to analyze some physical phenomenon, or maybe some computer algorithm. One gets experimental evidence and builds a mathematical model. Frequently the model involves some differential equation or maybe (especially in the case of the algorithm) what's called a difference equation. Then how to "solve" the equation? One can try for a polynomial solution, because polynomials can be easily computed (just multiplications and additions). But what happens? The "polynomial" solution turns out to have infinite degree -- to be a a power series. Then how can values of this solution be computed? The techniques which we are studying are used to give good approximations to these values, which are in turn used to go back and study the model, refining and changing it if necessary.

All this is too abstract. I then asked if sum_{n=1}^infinity 1/n^3 converged. I was told that it did (p-series, p=3>1). I asked what partial sum s_N=sum_{n=1}^N 1/n^3 was within .0001 of the "true value". I broke the series up into a partial sum, s_N, + an "infinite tail", t_N=sum_{n=N+1}^infinity 1/n^3 which I over estimated using the integral from N to infinity of 1/x^3. When I got that (1/2N^2) I asked what value of N would make this < .0001 and got one such N: 80 (I think). Maple reports that the partial sum here is 1.201979 while the "true value" is 1.20205: looks close, doesn't it?

I then asked the same questions for the series sum_{n=1}^infinity 1/(n^3+sqrt{n}). Here the question of convergence was settled with the following logic: since a bigger series converged, its partial sums are bounded, and therefore the partial sums of the smaller series must be bounded and THEREFORE (sigh!) the smaller series must converge. This comparison technique is used quite a lot. Again, the same partial sum will be within .0001 of the true sum.

I then asked the same questions of the series sum_{n=1}^infinity 1/(2^n + sqrt{n}). Here I said that each term satisfied the following inequality: 1/(2^n + sqrt{n})< 1/sqrt{n} and since 1/sqrt{n} diverged (p-series, p=1/2<1) can we assert that the smaller series must diverge? The answer is no, actually. So I asked how to determine that the series with general term 1/(2^n + sqrt{n}) converged? Well, look at the fastest growing part of the denominator (if that's what the bottom is called!): so compare this series with 1/2^n which is a convergent geometric series (a=1/2, and r=1/2 so |r|<1). Now how about a partial sum which is within .0001 of the true sum? Here the infinite tail of the overestimating series is sum_{N+1}^infinity 1/2^n, which is again a geometric series, with a=1/2^{N+!} and r=1/2. The sum is 1/2^N so to make this less than .0001 takes N=14 I think.

We then looked at a problem in the book for a while: for which p does the sum_{n=1}^infinity ln(n)/n^p converge? First we analyzed the case p=1. This can be done with the integral test, since ln(x)/x^2 has a neat antiderivative (look at the substitution ln(x)=u). It can also be done by comparison with the divergent harmonic series whose n'th term is 1/n since ln(n)>1 for n at least 3. Now ln(n)/n^p is also larger than ln(n)/n^1 for p<1, so, again, the series diverges for p less than or equal to 1. For p>1 the situation is more complicated. I suggested looking at an example. We considered the series whose nth term was ln(n)/n^1.0007. I rewrote this as 1/n^1.003 multiplied by ln(n)/n^.0004. This was motivated by the workshop problems involving rate of growth. Which of the functions (ln(x) and x^.0004) grows faster as x-->infinity? Here a graph of x^.0004 reveals very litte on the interval [1,5], say -- it essentially looks like a horizontal line at height 1. But l'Hospital's rule applied to the quotient ln(x)/x^.0004 actually shows that the limit is 0. Therefore the series with nth term ln(n)/n^1.0007 actually grows smaller than a series with nth term 1/n^1.0003, a p-series with p=1.0003>1, so it therefore converges. In fact, the series with nth term ln(n)/n^p converges if p>1.

I wrote the comparison test and the limit comparison tests, as in the text.

Problems due Tuesday: 11.3:22,23,34;11.4:32,36;11.5:28

I analyzed the alternating harmonic series, sum_{n=1}^infinity (-1)^{n+1}/n. We saw that the even partial sums moved "down" (decreased) and the odd partial sums moved "up" (increased). The even partial sums were bounded below by each of the odd partial sums, and the odd partial sums were bounded above by each of the even partial sums. Therefore the odd sums converged and so did the even sums (calling on results from sequences discussed a while ago). Since the difference between the odd and even partial sums were essentially the n^th term of the series and that -->0, we saw that the limits of the odd and even partial sums were identical and the series converged. I contrasted this result with the fact that the harmonic series itself (with all positive signs) did not converge.

I stated the Alternating Series Test as in the text. I gave another example of how this test applies. I also mentioned (and exemplified) that the nth partial sum of a series satisfying the hypotheses of the Alternating Series Test was within a_{n+1} of the true sum. So it is very easy to get an estimate of the error when using a partial sum in place of the full sum.

Then I copied the text's discussion of the fact that if sum |a_n| converges, then sum a_n must also converge. We discussed the text's discussion (!) and I tried to insure that people understood it. I gave a siumple example (something like the sum_{n=1}^infinity (cos(5n^3-33n^{3.6}+ etc)/n^5) of a complicated-looking series whose terms can be handled easily by realizing that values of cosine are between -1 and +1, so |(cos(5n^3-33n^{3.6}+ etc)/n^5)| is at most 1/n^5 and that series converges. Therefore by comparison, the original series with absolute value signs must converge, and by our result just verified, if the series with absolute value signs converged, then it must without absolute value signs. We noted that the converse to the result just used is generally false, and an example was the Alternating Harmonic Series.

I went on as in the text. I defined absolutely convergent series and conditionally convergent series. I translated our previous result as asserting that any absolutely convergent series must converge. I tried to give a metophor making sense almost surely only to me comparing series to folding and unfolding rulers with hinges. If the total length of the ruler totally unfolded is finite, then the lenght of the rule folded up any way must be finite.

The text then discusses the Ratio and Root Tests. I went on to discuss the first of these. First we reviewed again what a geometric series was. I asked how to identify a geometric series, and we discussed that for a while. Then I asked if the series sum_{n=1}^infinity (10^n)/n! converged. The first three terms seemed to be increasing. But what happens after, say, the 20^th term? The ratio between successive terms seems to always be LESS THAN 1/2. Then (this series has positive terms!) by comparison with a geoemtric series with ratio 1/2, this series converges. I then stated the Ratio Test carefully: if lim_{n-->infinity}|a_{n+1}/a_n| exists and equals L, then the series converges absolutely if L<1 and diverges if L>1. We applied the Ratio Test to the series sum_{n=1}^infinity (10^n)/n!, and had to be careful with the transition from compound fraction to simple fraction, and then with cancellation, using (n+1)!=(n+1)n!. Here we got L=0 so the series converged.

I then asked: For which x does sum_{n=0}^infinity ((-1)nx2n)/(2^2n(n!)2) converge? For those students who were still conscious (!) I explained that the sum of this series is important in analyzing the vibrations of a circular drumhead. We did this using the Ratio Test, and had to make sure that the algebra was correct. The answer: the series converges for all values of x.

I asked that students be sure to look at 11.5 and 11.6 before the workshop class tomorrow.

I began by discussing the Ratio and Root Tests. I tried to carefully apply these tests to Problem 5d) of the past workshop: sum_{n=1}^infinity {8/n^{1/5}}x^n. I used the Ratio Test as instructed, and got the results indicated in the answers: absolute convergence (and so convergence) if |x|<1, divergence if |x|>1. A separate investigation was needed for the other x's: for x=+1, p-series, p=1/5<1, divergence; alternating series test for for x=-1, convergence.

I introduced a heuristic idea for background on both the Ratio and Root Tests. Here's an appropriate definition for heuristic:

adj.
1. allowing or assisting to discover.
2. [Computing] proceeding to a solution by trial and error.

If a_n is approximately arn, that is like a geometric series, then the quotient a_{n+1}/a+n is approximately like (arn+1)/(arn)=r, and the approximation should get better as n-->infinity. This gives a background for the Ratio Test. For the Root Test, if we assume that a_n is approximately arn and take nth roots, then some algebra shows that (a_n)1/n is approximately a1/nr, and we know

Limit fact 1 lim_{n-->infinity}a1/n=1 if a>0.

so we get the Root Test, similar to the Ratio Test but with (|a_n|)^(1/n) replacing |a_{n+1}/a_n|. I applied this to the example above, with a_n=8/n^{1/5}x^n. Here we needed to do some algebraic juggling and also needed

Limit fact 2 lim_{n-->infinity}n1/n=1.

We got results about convergence using the Root Test which were (comfortingly!) the same as what we got using the Ratio Test.

I analyzed the series sum_{n=1}^infinity n^n x^n, first with the Root Test and then with the Ratio Test. The Root Test was definitely easier, and the algebra gave n |x|, which (tricky!)-->infinity if x is NOT 0 when n-->infinity, but which has limit 0 when x=0. So the series converges ONLY when x=0. Then I tried the Ratio Test. Here some possibly tricky algebra is needed to get the same answer. Also needed is the following limit:

Limit fact 3 lim_{n-->infinity}(1+(1/n))1/n=e.

With this limit, the same result was gotten as with the Root Test, although with quite a bit more work. I then verified the limit, by taking ln of (1+(1/n))1/n and using l'Hospital's rule.

In general we have been studying Power Series. I defined a power series centered at c" it is sum_{n=0}^infinity a_n (x-c)^n, a sort of infinite analog of a polynomial. Almost all of our examples would have c=0.

Guiding questions:

1. For which x does such a series converge? (Remark: such series always converge for x=0. Is there anything useful and general one can say about the x's for which the series converge?)
2. What are the properties of the sum of such a series? (Still to be understood!)
3. Why should we look at this? What are the rewards, and what are the reasons?
I asked what could happen if the series converged when x=R, some number>0. What could one say, first of all, about the individual terms of sum_{n=0}^infinity a_n R^n? Since the series converged, the terms must -->0. But if the terms -->0, then all of the terms must be bounded, that is there must be an M so that |a_n R^n| less than or equal to M. Then I compared |a_n r^n| with |a_n R^n|, and, in fact, saw that |a_n r^n| less than or equal to |a_n R^N||r/R|^n less than or equal to M |r/R|^n and sum_{n=0}^infinity M|r/R|^n is a convergent geometric series, so it must converge.

We concluded that if the series converged for |x|=R then it must converge for x's with |x|interval of convergence and must have a radius of convergence. Here are examples:

• a_n=n^n. This was analyzed earlier in this class. The resulting power series converges only for x=0. The interval of convergence is 0 and the radius of convergence is 0.
• a_n=8/n^{1/5}. This was analyzed earlier in this class. The resulting power series converged for x in the interval [-1,1) (this is interval notation and means that -1 is included while 1 is not included). So [-1,1) is the interval of convergence and 1 is the radius of convergence.
• a_n=1/(n^n), leading to a power series not previously considered, namely sum_{n=1}^infinity x^n/(n^n). The Root Test easily shows (take n^{th} roots and get x/n which -->0 as n-->infinity for ALL x) that the interval of convergence is (-infinity,infinity) here and the radius of convergence is infinity.
Generally the radius of convergence is half the length of the interval of convergence -- it is the distance from the center to the boundary of the interval of convergence. This is the answer to question 1 above.

We saw even more from the comparison with geometric series. If we wanted to analyze how to compute the series for |x|=r with rinfinite tail. The infinite tail is than less than sum_{n=N+1}^infinity M (r/R)^n, which is an infinite series with first term M(r/R)^{N+1} and ratio r/R which is less than 1. The sum is M(r/R)^{N+1}/(1-(r/R)) and by choosing N large enough we can make this as small as we want ("less than epsilon" in math language). But what about the other part? The partial sum sum_{n=0}^N a_n x^n is a polynomial in x. It is a continuous function of x. In formally this mean if we change x's by a little bit, the values of the polynomial won't change by much. (Or looked at via a graph, the graph won't jump or break anywhere.) So the power series is a sum of this continuous polynomial + a very small error. Some more thought leads to the following fact:

The sum of a power series inside its interval of convergence must be a continuous function.

This will be a partial answer to question #2 above, but much more and much nicer things are actually true and will be discussed later.

Students should prepare to hand in these problems next Tuesday: 11.6:25,32;11.7:13,35;11.8:13,30.

I finished the theory of power series (as much as I would do):

• The power series sum_{n=0}^infinity a_n x^n converges in an interval centered around 0. This is called the interval of convergence. The distance from 0 to the edge of the interval is called the radius of convergence. The series converges absolutely inside the interval of convergence.
• If f(x)= the sum of the series, then f(x) is continuous inside the interval of convergence. Also, f'(x) will be the sum of the series sum_{n=0}^infinity n a_n x^{n-1} and the integral of f will be the sum of the series sum_{n=0}^infinity (a_n)/n x^{n+1}. This is called "term-by-term" integration and differentiation.
• By evaluating f at 0 we see that f(0)=a_0. Differentiating and evaluating f' at 0 we see that f'(0)=a_1. Repeating this we get f^{(n)}(0)=n!a_n or, if we want it "the other way", a_n=f^{(n)}(0). So if we know the sum, we know the series. There can be only one power series equal to a function.
Two applications: I worked through how to get a formula for the Fibonacci numbers. See [Postscript]   |   [PDF] for this material.

Then I discussed playing games and fare entry fees. Suppose we have a "spinner", that is, a circular piece of material with an arrow pivoted to swing around the center. Half of the area is red, one-third is green, and a sixth is blue. Then the chance that the spinner hits red is 1/2, hits green is 1/3, and hits blue is 1/6.

Suppose now that a player wins $10 if the spinner hits red, loses$18 if it hits green, and wins $24 if it hits blue. What is the average winning? It is (1/2)(10)-(1/3)(18)+(1/6)(24)=3. So a fair "entry fee" to play this game would be$3: that is, in the long run, neither the contestant nor the proprietor of the game would win/lose money if such an entry fee were charged.

Now a more complicated game: consider a "fair coin" with a head and a tail. Toss the coin until it lands heads up. How many different ways are there for the game to play? It could land H (heads), 1/2 the time. It could land tails then heads, TH, 1/4 of the time. It could land TTH, 1/8 of the time. Generally, it could have the sequence TTT(n-1 times)TTTH, which is n tosses, 1/(2^n) of the time. There are infinitely many different ways for this game to end.

Suppose the proprietor of the game will pay the contestant n^2 dollars if the game ends after n tosses. What is the average amount that could be won in the game? Analogous to what was done above in the simpler game with a finite number (just 3) outcomes, here the average winning amount would be sum_{n=1}^infinity (n^2)/(2^n)?

Does this sum converge? That is, can one expect on average to win a certain finite amount in this game? We used the ratio test to show that it did indeed converge. But what is the sum, exactly?

Here we need to think a bit. The sum of the series sum_{n=1}^infinity x^n is x/(1-x), the sum of a geometric series. The derivative of both sides gives the series sum_{n=1}^infinity nx^{n-1} on the series side, and on the other side (using the quotient rule) 1/(1-x)^2. If we multiply both sides by x, we'll get sum_{n=1}^infinity nx^n on the series side and x/(1-x)^2 on the function side. So if we let x=1/2, the series will be sum_{n=1}^infinity n/2^n and the function value will be (1/2)/(1-1/2)^2=2. So the sum of this series is 2.

To get n^2 to appear on top we need to differentiate again. That is, take sum_{n=1}^infinity nx^n =x/(1-x)^2 and differentiate. The series side becomes sum_{n=1}^infinity n^2 x^{n-1} and the function side (again using the quotient rule) is (1+x)/(1-x)^3. If we multiply by x, the series becomes sum_{n=1}^infinity n^2 x^n and the function side becomes (x+x^2)/(1-x)^3. If we substitute x=1/2, the series side is sum_{n=1}^infinity n^2/2^n (finally what we want!) and the function side is (1/2+(1/2)^2)/(1-(1/2))^3 which is 6. So \$6 is a "fair entry fee" to pay the proprietor of this game. I don't think the sum is at all obvious. What I've done is find what is called the expectation of this game.

I announced that the next exam would be on the Thusday after Thanksgiving, which is Thursday, November 29. It will cover differential equations and the material of chapter 11. An old exam will be gone over in the workshop session on Tuesday, November 27. A draft of a formula sheet will be given out earlier.

We returned to power series, and used the general name, Taylor series, to indicate that if f is the sum of a series, sum_{n=0}^infinity a_n (x-c)^n, then a_n=f^{(n)}(c)/n!. The partial sum sum_{n=0}^N a_n (x-c)^n is called the N^{th} Taylor polynomial. The difference between the partial sum and the true value of the function is called the remainder, R_N(x) (this is what I previously called the "infinite tail", and is sum_{n=N+1}^infinity a_n (x-c)^n).

The game is to usefully estimate R_N(x) and use the idea that when the series converges (R_N(x)-->0) then the series can be manipulated (inside its radius of convergence) to yield other series which converge. The manipulations can either be algebraic (add, multiply, etc.) or from calculus (differentiate, integrate). So how to estimate R_N(x)? The techniques we've already learned to estimate "infinite tails" can all be applied here, but they can be supplemented by a powerful new idea.

Taylor's Theorem: f and T_N and R_N as before. Then |R_N(x)| is less than or equal to (M_{N+1} |x-c|^{N+1}}/(N+1)!. Here M_{N+1} is an overestimate of the absolute value of the (N+1)^{st} derivative of f on the interval whose endpoints are c and x.
This is a very complicated statement. The estimate for |R_N(x)| actually has 4 different algebraic variables: x and c and N and M_{N+1}.
The reason why the theorem is true is not straightforward. The text has one method. I prefer to use a technique involving integration by parts, a method similar to what I indicated was used for estimating the error in various integration approximation techniques. In this, int_c^x f'(t)dt=f(x)-f(c), and the parts are u=f'(t), dv=dt, and du=f'(t)dt and v=the non-obvious t-c. I didn't pursue this, but it really does work.

I indicated that I would try to show a lot of simple examples explaining why Taylor's Theorem is useful both as a computational mechanism and as a theoretical tool.

First looked at exp(x)=e^x. Here c=0 (almost always c=0, actually!). Then all the derivatives of exp are exp, and f^{(n)}(0)=exp(0)=1, so the Taylor series is \sum_{n=0}^infinity(1/n!)x^n, a very very simple series. I applied the Ratio Test to show that the radius of convergence was infinity (because the limit was always 0, and didn't depend on x, and 0 is less than 1). But how can one use the series to compute values of exp?

How can exp(-.3) be computed with an accuracy of .001? So we can try to write exp(-.3)=T_N(-.3)+R_N(-.3), and select N (if we can) with |R_N(-.3)| less than or equal to .001. If we can do that, then T_N(-.3) is a finite sum and can be directly computed. So here R_N is what? . Here M_{N+1} is an overestimate of the absolute value of the (N+1)^{st} derivative of f on the interval whose endpoints are c and x.
This is a very complicated statement. The estimate for |R_N(x)| actually has 4 different algebraic variables: x and c and N and M_{N+1}.
To see that the theorem is true is not straightforward. The text has one method. I prefer to use a technique involving integration by parts, a method similar to what I indicated was used for estimating the error in various integration approximation techniques. In this, int_c^x f'(t)dt=f(x)-f(c), and the parts are u=f'(t), dv=dt, and du=f'(t)dt and v=the non-obvious t-c. I didn't pursue this, but it really does work.

I indicated that I would try to show a lot of simple examples explaining why Taylor's Theorem is useful both as a computational mechanism and as a theoretical tool.

First looked at exp(x)=e^x. Here c=0 (almost always c=0, actually!). Then all the derivatives of exp are exp, and f^{(n)}(0)=exp(0)=1, so the Taylor series is \sum_{n=0}^infinity(1/n!)x^n, a very very simple series. I applied the Ratio Test to show that the radius of convergence was infinity (because the limit was always 0, and didn't depend on x, and 0 is less than 1). But how can one use the series to compute values of exp?

How can exp(-.3) be computed with an accuracy of .001? So we can try to write exp(-.3)=T_N(-.3)+R_N(-.3), and select N (if we can) with |R_N(-.3)| less than or equal to .001. If we can do that, then T_N(-.3) is a finite sum and can be directly computed. So here R_N is what? (M_{N+1} |x-c|^{N+1}}/(N+1)!. Here c=0 and x=-.3 and what should M be? M is a max of the (N+1)^{st} derivative of exp (just exp) on [-.3,0]. But exp is increasing so a max is just exp(0), which is 1, There's NO N in the answer. This simplifies things a great deal. So what we need is an N so that |-.3|^{N+1}/(N+1)!<.001 and think about this: the TOP-->0 and the BOTTOM-->infinity, so we should expect some N for which this is actually <.001, so I wrote a list of N!'s, which seemed to grow very quickly. We found out that N=6 actually works, and so T_6(-.3) will be within .001 of exp(-.3).

Then I worked on exp(2). Here the situation was more complicated. x=2 and c=0 and what is M_{N+1}? It is an overestimate of exp on [0,2]. This is e^2. But we want to compute e^2? Isn't it silly to need to know e^2 in order to compute e^2? Ahhhh ... but the key here is OVERESTIMATE: loose information about e^2 can be converted to better and better estimates. For example, e^2 is certainly less than 10 (since e<3, e^2<3^2=9<10). So we know that |R_N(2)| can be overestimated by 10(2^{N+1})/(N+1)!. Now both the top (2^{N+1}) and bottom ({N+1}!) go to infinity. The key here is that factorials grow much faster than "geometric" growth. We saw this explicitly because I wrote a list of powers of 2 and factorials for N going from 1 to 10. In this case we needed to go to N=8 to get T_8(2) within .01 of the true value of e^2.

Note that life is actually a bit complicated here. One gets, even for a fairly simple function (and exp is very simple in this context!) that the N's can be different for different x's, and for different error tolerances.

Then I tried to compute sin(2). For this I found the Taylor series of sine. This involved taking derivatives of sine, and finding that these derivatives repeated every fourth time (sine --> cosine --> -sine --> -cosine --> sine). So the values at 0 (necessary for getting the Taylor series) were 0, 1, 0, -1, etc. I wrote the Taylor series for sine centered at 0: sum_{n=0}^infinity (-1)^n x^{2n+1}/{2n+1}!. To compute sin(2), I needed again to estimate (M_{N+1}|x-c|^{N+1}}/(N+1)!. Here how should we estimate M? We need to know a useful max of the absolute value of some derivative of sine. But, all of these functions (there are exactly 4 of them!) can be estimated by 1 (amazing, and easy!). So if x=2 and c=0 we need 2^N/(N+!)! < .01. This can be done with, I think, N=7. We then discussed what T_7(2) for sine. It is actually the first 4 terms of the series: sum_{n=0}^4 (-1)^n 2^{2n+1}/{2n+1}!, since these terms go up to DEGREE 7. This is a case where the number of terms and the degree of the terms do NOT coincide.

I also mentioned the series for cosine: : sum_{n=0}^infinity (-1)^n x^{2n}/{2n}!

I gave out a draft review sheet for the second exam, and a schedule for the last part of the course.

I reviewed Taylor's series, the Taylor polynomial, and Taylor's Theorem, with attention to the remainder estimate. The simplest functions to apply this "machine" to are exp and sine and cosine because these have simple families of derivatives. I reviewed much of what we did last time, giving numerical evidence for the truth of our estimate. For example, in the first problem we did then, the Taylor polynomial T_6(-.3) is approximately .7408182625 while the "true value" of exp(-.3) was reported by Maple to be .74089182207. This is quite a bit more accurate than the pessimistic .001 error given by Taylor's Theorem.

I remarked that the graph of sin x and T_7(x) (which is x-(1/6)x^3+(1/120)x^5-(1/5040)x^7) when looked at on the interval [0,2] were actually very close, certainly within .01, and that therefore for many practical purposes the functions could be exchanged with one another.

One application of such ideas could be to the evaluation of difficult integrals. The example I chose to look at was the integral from 0 to .3 of cos(x^2). This function actually arises in a number of applications, such as optics. How could we compute int_0^.3 cos(x^2)dx to an accuracy of .001? One technique might be to try to find an antiderivative of cos(x^2). This turns out to be impossible (that is, finding an antiderivative in terms of known functions). It isn't easy to see that getting such an antiderivative is impossible. Another method might be to use some numerical techniques to approximate the integral, such as the Trapezoid Rule or Simpson's Rule. But there is also a third approach, which I carried out in the lecture.

The Taylor series of cos(x) is sum_{n=0}^infinity (-1)^n/(2n)! x^{2n}. I also know that the Remainder is going to be something like M|x|^{a power}/{that power}! and that the M here will be at most 1, because the derivatives of cosine are +/- sine and +/- cosine, and these are all bounded in absolute value by 1. In fact, I know that the Remainder will -->0 because factorial growth in the bottom will overcome power growth in the top. So the series will actually equal cosine for all x's. Then cos(x^2) is gotten by substituting x^2 in the series for cosine, so we get sum_{n=0}^infinity (-1)^n/(2n)! (x^2)^{2n} which is sum_{n=0}^infinity (-1)^n/(2n)! x^{4n} because repeated exponentials multiply. If we want the integral of cos(x^2), we just need to antidifferentiate the series sum_{n=0}^infinity (-1)^n/(2n)! x^{4n}, treating it like an infinite polynomial. So the result will be sum_{n=0}^infinity (-1)^n/(2n)! (1/(4n+1))x^{4n+1}. This is then an indefinite integral, and we need to evaluate it at .3 and subtract off its value at 0. The value at 0 is 0, since all the terms have powers of x in them. The value at .3 is sum_{n=0}^infinity (-1)^n/(2n)! (1/(4n+1))(.3)^{4n+1}. This is actually an alternating series, and its value can be approximated by a partial sum with accuracy at least that of the first omitted term. But the third term (with n=2) is actually less than .00001, so the integral is just the sum of two fractions, the one with n=0 and with n=1. I did this computation both in summation notation and by writing out the first few terms, in order to give people practice in understanding "sigmas".

Then I went through problem 6 of the previous workshop, making sure that the power series for arctan was discussed. I went through all parts of the problem (a & b & c & d & e & f). The total number of students present was the same as the number of parts in the problem.

I found a power series expansion for ln(1+x) by taking the derivative, 1/(1+x), setting it equal to a sum of a geometric series, sum_{n=1}^infinity (-1)^n x^n, and integrating each term to get sum_{n=1}^infinity (-1)^n/(n+1) x^(n+1). The radius of convergence of the series resulting is 1, since the original geometric series converges for |x|<1 and integration (and differentiation) don't change the radius of convergence. I mentioned that ln(x) can't have a nice Taylor series because it has a vertical asymptote at x=0, so alternatively a Maclaurin series for ln x centered at x=1 is given by rewriting the series just obtained: ln x= sum_{n=1}^infinity (-1)^n/(n+1) (x-1)^(n+1).

I attempted to use the series just derived to show how ln(1.5) could be computed to accuracy .00001 (1/100,000). Here I used the fact that sum_{n=1}^infinity (-1)^n/(n+1) (.5)^(n+1) was an alternating series. We found that n=10 would make (.5)^(n+1) certainly less than 1/(100,000) so that the partial sum of the first 19 terms would be close enough. Here we needed results on alternating series and their partial sums.

I remarked that the series for exp and sin and cos were widely used for computation but the series for arctan and ln were not, becuase they converged relatively slowly. The first three series converged rapidly because factorials grow very fast.

Then I asked what x^{1/3} "looked like". This was very fuzzily stated. To compute x^{1/3} one can use various buttons on a calculator. For example, 8^{1/3} is 2. The computation is probably done using some variant of Newton's method. That is, a starting "guess" x_0 is gotten, and then x_{n+1} is some interesting (?) rational combination of x_n, a recursively defined sequence. This will tell you rather rapidly what x^{1/3} is like. But in some sense it doesn't give you a feeling for the function. It is an algorithm, but not with much information directly. For example, we know that if x is close to 8 then x^{1/3} should be close to 2 (this is because x^{1/3} is continuous). Maybe not even that is totally clear from the Newton's method iteration.

I asked what the first Taylor polynomial of x^{1/3} centered at x=8 was. We found out that it was T_1(x)=2+(1/12)(x-8) (this is f(8)+f'(8)(x-8) when f(x)=x^{1/3}). I asked people to graph T_1 and f in the window 6 We looked at T_2(x). I carried out the computations and the error analysis. Here T_2(x)=2+(1/12)(x-8)-(1/(288))(x-8)^2. Again I asked for a graph of f versus T_2. Here it was very difficult to tell the graphs apat locally, even though T_2's graph was a parabola opening down with max around 20. Locally they are very close. What govern's the error for x in [8,10]? I looked at (M_3/6!)|x-8|^3. M_3 again was gotten by taking the absolute value of f^(3) (x)=(1/3)(-5/3)(-8/3)x^{-11/3} and noticing that this function is decreasing and therefore dominated by its value at x=8 which is 5/(3456). So the error here is (5/(3456))(1/6)(2^3), less than .002. So for x in the interval [8,10] x^{1/3}, a "mysterious" function, is just like T_2, a quadratic polynomial, up to an error of less than .002. I also remarked that the fact the x^{1/3} is close to 2 for x close to 8 is just saying that T_0 closely approximates f when x is close to 8.

I chose this example so that the numbers would work up relatively nicely when done by hand, but the general philosophy works with little change in much more complicated situations. So one can frequently "locally" replace a function by a polynomial.

I asked for questions about the material of chapter 11.

I did two problems from the text: #37 and #41 of section 11.10.

Then I began parametric curves. I first studied the "unit circle" (x=cos t, y=sin t) and discussed how this compared with x^2+y^2=1. There is more "dynamic" information in the parametric curve description, but there is also more difficulty: two equations rather than one.

I sketched (x=cos t,y=cos t) and (x=t^2,y=t^2) and (x=t^3,y=t^3): all these are pieces of the line y=x: I tried to describe the additional dynamic information in the parametric definitions.

I showed some parametric curves from book Normal Accidents by Charles Perrow: pictures of ship collision tracks.

Then I tried to sketch the parametric curves (x=2 cos t, y = 3 sin t) and (x=1+sin t, y= cos t -3) which I got from another textbook. The first turned out to be an ellipse whose equation is (x/2)^2+(y/3)^2=1 and the second turned out to be a circle of radius 1 centered at (1,-3). These curves intersect, but do they actually describe the motion of particles which have collisions? Well, one intersection is a collision and the other is not (this is generally not too easy to see!).

I tried to describe the parametric curves for a cycloid. This is done with a slightly different explanation in section 10.1 of the text. This took a while.

Then I described my "favorite" parametric curve: (x=1/(1+t^2),y=t^3-t).

Next time: calculus and parametric curves. These problems are due Tuesday: 10.1: 12,18; 11.11: 11,14; 11.12: 5,28.

I started again to discuss parametric curves, beginning by sketching again my favorite, x=t^3-t, y=1/(1+t^2). I wanted to do calculus with parametric curves. The object was to get formulas for the slope of the tangent line to a parametric curve and for the length of a parametric curve.

If x=f(t) and y=g(t), then f(t+delta t) is approximately f(t)+f'(t)delta t +M (delta t)^2 (this is from Taylor's Theorem). Also g(t+delta t) is approximately g(t)+g'(t)delta t +M (delta t)^2. The reason for the different "M" is that these are different functions to which Taylor's Theorem is being applied. Then we analyze the slope of the secant line, (delta y)/(delta x): this is (g(t)+g'(t)delta t +M -g(t))/(f(t)+f'(t)delta t +M (delta t)^2) and we see that the limit as delta t --> 0 must be g'(t)/f'(t), and this must be the slope of the tangent line, dy/dx. The formula in the book, developed in a different way using the Chain Rule, is also nicely stated in classical Leibniz notation as (dy/dt)/(dx/dt).

I applied this to find the equation of a line tangent to my favorite curve, x=t^3-t, y=1/(1+t^2), when t=2. I found a point on the curve and I found the slope of the tangent line, and then found the line. It sloped slightly down, just as the picture suggested. Then I found the angle between the parts of the curve at the "self-intersection", t=+/-1. This involved some work with geometry and with finding the slopes of the appropriate tangent lines. The best we could do is with some numerical approximations to the angles.

I asked people what the curve x=sin t+2 cos t, y-2 cos t - sin t, looked like. We decided it was bounded because sine and cosine are always between -1 and +1, so the points of the curve must be in the box -3 One way to answer this question is to just look at the graph and estimate the box, which would be, say, x approximately 2.2. Another way would be to try to use calculus. If dx/dt=0 then maybe we have a vertical tangent to the curve, which might give one side of the bounding box. So we looked for where dx/dt=0. This meant cos t-2sin t=0 or t=arctan(1/2) which is approximately .4636. When substituted into the equation for x, we get x approximately 2.23, just like the graph!

I remarked that for "well=behaved" parametric curves, vertical tangents occur when dx/dt=0, and horizontal tangents, when dy/dt=0.

Then I briefly discussed the arc length between (f(t),g(t)) and (f(t+delta t),g(t+delta t)). With the approximations f(t+delta t)=f(t)+f'(t)(delta t) and g(t+delta t)=g(t)+g'(t)(delta t) we saw that this distance (using the standard distance formula) was approximately sqrt(f'(t)^2+g'(t)^2) delta t. We can "add these up" and take a limit. The length of a parametric curve from t=START to t=END is the integral_START^END sqrt(f'(t)^2+g'(t)^2) dt.

How useful is this formula? Maybe not too. As a first example, I tried to find the length of the loop in my favorite parametric curve. I ended up with int_{-1}^{1} sqrt((3t^2-1)^2+(-2t/(1+t^2)^2)) dt, which is a very intractible integral. I would not expect to be able to use the Fundamental Theorem of Calculus to "do" this integral. That is, I would not expect that I could find an antiderivative in terms of familiar functions. I'd expect to need to compute this integral numerically, approximately.

The integrand in the arc length formula for parametric curves is horrible if one wants to compute the length exactly. It is difficult to find many examples. Here are a few, and section 10.3 has others. I first looked at x=cos t and y=sin t. for t=0 to 2Pi. This is, of course, the unit circle. I found the length of this curve and it was easy because the integrand turned out to be exactly 1!

I looked at the curve x=(1-t^2)/(1+t^2) and y=2t/(1+t^2) for t=0 to t=1. This looks initially very forbidding. But direct computation of sqrt(f'(t)^2+g'(t)^2) yields (miraculously?) 2/(1+t^2). This integrated from 0 to 1 is exactly Pi/2. Remarkable! (In fact, 10.3's problem section has a number of such "accidents" which take a great deal of dexterity to create!) What "is" this curve? It is a curve going from (1,0) (when t=0) to (0,1) (when t=1) and has length Pi/2. Sigh. It is one-quarter of the unit circle! This is a famous "rational parameterization" of the unit circle: a description using rational functions, quotients of polynomials.

These problems are due Tuesday: 10.1: 9,10; 10.2: 3,8; 10>3: 4,7.

The final exam is Monday, 12/17/2001, 4-7 PM, in Pharmacy 115, on Busch Campus.

We talked a bit about parametric curves, and then began polar coordinates. I "wrote" an excerpt from a really lousy adventure novel.

 Go to the old dead tree at dawn. Then at sunrise, walk ten paces in a north-north-east direction from the tree. Dig there for treasure.
We tried to understand this, and decided that it represented locating a point in the plane with reference to a fixed origin, "the pole", and a fixed direction, the polar axis. Ther origin was the dead tree and the fixed direction was sunrise. Thus what is specified is the distance from the tree (r=10) and the angle from the sun's direction, assumed east (theta=Pi/2+Pi/8=(5Pi/8). Of course in the purported text I quoted from it would be forgotten that the sunrise's direction changed with the season, or that there was no treasure but a booby prize, or the steps were taken by a giant or a dwarf or ... many things.

I introduced polar coordinates and got the equations relating (x,y) to (r,theta). I looked at the point (3,0) (in rectangular coordinates) and saw that there were infinitely many pairs of polar coordinates which described the position of this point (I gave examples). I remarked that in many applications and computer programs, there were restrictions on r (usually r is non-negative) or on theta (say, theta must be in the interval [0,2Pi) or in the interval (-Pi,Pi] or something) in an effort to restore uniqueness to the polar coordinate representations of a point's location. No such restrictions were suggested by the text used in this course. I sketched the regions in the plane specified by the inequalities 2 Polar coordinates are useful for looking at things with circular symmetry. For example, the equation r=3 is much easier to contemplate than the equation x^2+y^2=9.

What about other polar curves? We spent some time sketching the curve r=2+cos theta. This curve was inside the circle r=3 and outside the circle r=1. I tried to sketch what it might looklike. Even with this rather simple curve we got into difficulties. It looked "smooth" but when theta was 0 or Pi, it wasn't clear whether the curve has a "point", that is, a cusp.

Therefore I studied tangent lines to polar curves and we derived a formula in rectangular coordinates for the slope of a tangent line to a polar curve r=f(theta). Here I noted that polar curves were a special type of parametric curve. So, since x=r cos theta, we know that x=f(theta)cos theta and y=f(theta)sin theta. Then using the ideas previously derived, namely that dy/dx= (dy/d theta)/(dx/d theta), I got the formula in the book for dy/dx in polar coordinates.

I used this to find the equation of a line tangent to y=2+cos theta when theta was Pi/6. This was ugly and difficult, but at least we got the correct sign (a negative one!) for the slope. Then we looked at what happened when theta=Pi. It turned out that dy/d theta was not 0 and dx/d theta =0. For this we inferred that the curve had a vertical tangent at that point. Iwas able to refine my picture a bit. There was no cusp (a cusp is what happens at the origin in the curve y=x^{3/2}). I also used this to try to get the "top" of the curve: where dy/dx=0. We needed to solve an equation with cosine and sine approximately to see where this was, and we were able to with the help of calculators.

I wasn't finished with the curve r=2+cos theta yet. I asked if the point r=-3 theta=Pi was on this curve. Discussion followed. The answer was yes although the equation was NOT satisfied by that pair of polar coordinates, but it was satisfied by another pair of polar coordinates describing the same point (r=3, theta=0). This is a difficulty with polar coordinates.

I wasn't finished with the curve r=2+cos theta yet. I asked how this curve could be describe in rectangular coordinates. We multiplied by r and got r^2=2r+r cos theta which was x^2+y^2=2+/-sqrt{x^2+y^2}+x. Then if we wanted a polynomial equation, we wrote (x^2+y^2-x)^3=4(x^2+y^2). This is quite unwieldy, and could only be graphed with some effort, while the polar representation could be graphed much more easily.

We looked at the curve r=(1/2)+cos theta. This had a new difficulty, in that in part of the theta range the curve had negative r. We traced out this curve, including moving "backwards" for theta in that range.

Next time students will present some polar coordinate area problems.

Take a spool of red thread which is 150 yards long. How much heart-shaped area can be enclosed by this thread (as a Valentine's Day present)? For a heart-shaped area I suggested as a model a cardioid, r=H(1-cos theta) with a parameter H to be specified later.

We used various formulas in section 10.5. The arc length of a polar curve r=f(theta) between theta=alpha and theta=beta is int_alpha^beta sqrt{f(theta)^2+f'(theta)^2 d theta. So I tried to find the integral where theta=0 and theta=2Pi and f(theta)=H(1-cos theta). Here we needed to use double- (actually, half-) angle formulas and be fairly careful. The length turned out to be 8H. So for our curve, H would be 150/8.

The area enclosed by r=f(theta) and theta=alpha and theta=beta is the int_alpha^beta (1/2) f(theta)^2 d theta. Here the integral turned out to be 3Pi(150/8)^2, approximately 1600 square yards. In this integral and the one above, we needed 1=cos^2+sin^2 and cos(2theta)=cos^2-sin^2.

I tried to answer some question about area in polar coordinates. For example, consider the curve r=(1/2)+cos theta. This has loop inside a loop. We tried to figure out how to get the area of the big loop and the little loop. For the little loop (between theta = 2Pi/3 and theta =4Pi/3) r is negative, but the r^2 is still positive. We tried to figure out how to get the two loops. We then looked at the spiral r=theta, and figured out what limits to put to get various areas enclosed by the spiral.

Students put 4 problems from section 10.5 on the board.

I mentioned when the final would be given, and said that I would be available in Hill 542 from 10 AM to Noon to answer questions, and I would certainly try to answer questions by e-mail.