Problem # Student	Link to the student's solution; some further remarks
12.3: 3 Mr. Klumb	Here is Mr. Klumb's witty response to the question. He showed that the function was not odd and also was not even by examining integrals of the function. Another solution is the following: f(2)=22+2=6 and f(-1)=(-1)2+(-1)=0. If f were odd, then f(-2) should be -f(2) but it is not. If f were even, then f(-2) should be f(2) but it is not. So f is neither odd nor even.
12.3: 5 Mr. Klami	Here Mr. Klami verifies that e|x| is even. To check that a function is even or is odd one must supply evidence that the necessary equations (f(-x)=f(x) for f even, and f(-x)=-f(x) for x odd) are correct for all of the implied values of x. Since, for example, the collection of x's is frequently all positive x's, it isn't feasible to check these computations by listing equations verifying agreement at each x. (I can't write an infinite number of equations, anyway!) Therefore people usually use algebra (in this case, relying on |x|=|-x|) or use geometry (a sketch of e|x| which is symmetric with respect to the y-axis. Mr. Klami actually provides both sorts of evidence.
12.3: 7 Mr. Novak	Here Mr. Novak checks that a function is odd. Again, he supplies both graphical and algebraic evidence.
12.3: 13 Mr. Shah	Here is the first (in this set of problems) computation of a Fourier series. In this case it is a Fourier cosine series for x on the interval [0,Pi]: the even extension of x is used. Here is a picture on [-Pi,Pi] of the function and its partial sum up to the cos(3x) term.
12.3: 19 Mr. Shaw	Here is an odd function on [-Pi,Pi] and its corresponding Fourier series. Since the function is odd, all of the cosine coefficients are 0, and the series is a sine series. Here's a picture of the function and the partial sum of its sine series up to sin(10x). I think the jump and overshoot behaviors are apparent.
12.3: 23 Ms. Horn	Here is a Fourier cosine series of an even function, and below is a picture of the function and its extension to [-Pi,Pi] together with the terms of its cosine series up to cos(4x).
12.3: 29 Mr. Pierre-Louis	Here are some elaborate computations of Fourier sine and cosine expansions of the same function. What a lot of work! Here is a picture of the function analyzed by Mr. Pierre-Louis and (on the left) the terms of the Fourier sine series up to sin(4x) and (on the right) the terms of the Fourier sine series up to cos(4x). Both the odd and even extensions of this function are continuous everywhere, so there is no jump phenomenon. The corner does disappear in the graphs of these partial sums, however.
12.3: 35 Mr. Sosa	I'm sorry but I don't seem to have the student solution for this problem. Maple tells me that the zeroth cosine coefficient is 4Pi2/3, and the nth cosine coefficient is 2 2 4 (2 n Pi sin(Pi n) cos(Pi n) - sin(Pi n) cos(Pi n) 2 / 3 + 2 Pi n cos(Pi n) - Pi n) / (Pi n ) / and the nth sine coefficient is 2 2 2 2 2 2 4 (-2 n Pi cos(Pi n) + n Pi + cos(Pi n) - 1 / 3 + 2 Pi n sin(Pi n) cos(Pi n)) / (Pi n ) / The n3's on the bottom occur because of two integrations by parts. Here's a picture of the function and a partial sum of the Fourier series up to sin(10x) and cos(10x). You can see the Gibbs phenomenon, I hope, for the jump located at 0"="2Pi (the quotes are because I don't really believe the numbers are equal, but Fourier series think that these two numbers are the same.

Parseval
The homework problems in 12.2 (such as #15) should have convinced you that there are many weird equations which Fourier series can verify. There's one that I should tell you about because it has physical meaning and you should know it. It is called Parseval's Theorem or Parseval's equation, or, maybe, just Parseval. Suppse f(x) is a 2Pi periodic function. Then we know
f(x)=a₀/2+SUM_N=1^infinity(a_ncos(nx)+b_nsin(nx))
If you try to compute _-Pi^Pi(f(x))²dx and use the equality, then interesting things occur. Because of orthogonality, when we "expand" the sum and square it and then integrate, all of the "cross terms" (say terms like sin(17x)cos(53x)) will integrate to 0. Also, because of the normalization coefficients, we get the following

Parseval's Equation -PiPif(x))2dx=a02/2+SUMn=1infinity(an2+bn2)

Again, in many applications, _-Pi^Pi(f(x))²dx has a physical meaning. For example, in many simple systems, this might be the energy of a signal or a vibration. The a_n²'s and b_n²'s could then be the amount of energy in various harmonics. In many physical situations, a few low harmonics have most of the energy.

I will try to have extra office hours Wednesday afternoon and early evening. Please study for the exam. Also please begin reading chapter 13.

I began the study of what the text calls Classical Equations and Boundary-Value Problems. So we need a new diary!.

Vocabulary of the day
acolyte 1. [Relig] a person assisting a priest in a service or procession. 2. an assistant; a beginner.	electrolyte 1. a substance which conducts electricity when molten or in solution, esp. in an electric cell or battery. 2. a solution of this.

We had more fun with Fourier series. I reviewed the formulas for Fourier coefficients. I wrote also how these were used to assemble the Fourier series for a function. I noted that if a function f were periodic with period 2Pi, then any interval of length 2Pi will be good for computing the Fourier coefficients. So, for example, if I wanted a₁₄, what I wrote last time is (1/Pi)₀^2Pif(x)cos(14x) dx and the textbook has (1/Pi)_-Pi^Pif(x)cos(14x) dx. But if for some peculiar reason you wanted to use (1/Pi)₆₆₈^668+2Pif(x)cos(14x) dx you would get the same answer. Of course, for this, you should realize that the function f(x) must be periodic with period equal to 2Pi.

Now what should we expect about the Fourier series of f(x)? I really tried to think about the levels of information engineering students should know about Fourier series.

Primary level
(What you really need to know)
On average, if you look at a "high" partial sum of the Fourier series for f, then random samples of the values of this partial sum will be close to the values of f(x). A precise statement is that the mean square error will --> 0 as more terms are taken of the partial sums.

Secondary level
(What you should know for Math 421)
The sum of whole Fourier series for a function, f, will be f(x) if f is continuous at x. If f has a jump discontinuity at x, then the sum of the whole Fourier series will be (f(x^-)+f(x⁺))/2, the average of the left and right hand limits of f. Notice, though, that from the point of view of Fourier series, 0 and 2Pi are the same, so the left side of 0 is the left side of 2Pi, and the right side of 0 is the right side of 2Pi.
Comment This property really isn't just for 421, but may also be useful in applications: I may be exaggerating about my classifications!

Tertiary level
(What Fourier series enthusiasts might know)
The Gibbs phenomenon: if f has a jump discontinuity at x, then the partial sums exhibit over and undershoots very near the jump, and the bumps are opposite direction of the jump.
Notice, please that the sum of the whole Fourier series does not does not have this behavior. Its behavior was described above. I remarked that I did know of some real-world applications where this Gibbs phenomenon was important, but I didn't know very many.

Example 1
Suppose f(x)=5sin(x)-2cos(3x)+8cos(17x). What is the Fourier series of f(x)? This is a very cute problem. The Fourier series of f(x) is ...5sin(x)-2cos(3x)+8cos(17x). It is its own Fourier series. Why is that? Any other sine/cosine coefficient would be gotten by integrating (the a_n or b_n formulas). But the other sine/cosine functions are all orthogonal to these. So, for example, a₁₇ is gotten by multiplying f(x) by cos(17x) and integrating from -Pi to Pi. Hey: by orthogonality this is 0. What about a₁₇? Well, by orthogonality you only need to "worry" about (1/Pi)_-Pi^Pi8(cos(17x))²dx, and (we discussed this at great length!) this is just 8. The darn 1/Pi in the original formula is included (orthonormalization!) to make the coefficient come out correctly.

Then I went further. I asked people what _-Pi^Pi (f(x))²dx was. Well, look at [5sin(x)-2cos(3x)+8cos(17x)]². "Expand" this square. Some terms are 0 very rapidly (the "cross terms") because of orthogonalization. What we have left is _-Pi^Pi5²(sin(x))²+(-2)²(cos(3x))²+8²(cos(17x))²dx. This is just Pi(5²+2²+8²)=93Pi (I hope, if I added correctly).
You certainly can do this computation by integrating everything in sight. This would be a lot of work, and I think you might not compute things correctly.

Example 2
This example is more computationally intricate, especially when done "by hand". f(x) is defined initially on the interval [0,Pi]. It is piecewise linear, and is the sort of function we encountered in our study of Laplace transform methods. The points (0,0) and (Pi/2,1) and (Pi,1) are on the graph, which suggests that f(x) be (2/Pi)x in the interval 0<x<Pi/2 and f(x)=1 for Pi/2<x<Pi. I asked for the Fourier series of f(x). Objections were raised immediately, because this function wasn't defined on [-Pi,Pi]. We decided to use the extension suggested by Mr. Novak, mainly, just let f be 0 for negative x's. Well, but remember this is only a recipe for -Pi to Pi, and f(x) is extended periodically elsewhere. I then actually computed, by hand (with a little help from the devil's machine owned by Mr. Klumb, the Fourier coefficients. This involved integrating by parts. I hope that students can integrate by parts.

I just asked my friend (?) Maple to do the same computation. The results were:

The cosine coefficients, a(n):
  
                                          Pi n
                   sin(Pi n) Pi n + 2 cos(----) - 2
                                           2
                   --------------------------------
                                    2
                                Pi n
The sine coefficients, b(n):

                                            Pi n
                    -cos(Pi n) Pi n + 2 sin(----)
                                             2
                    -----------------------------
                                    2
                                Pi n

a(0);

                                  3
                                 ---
                                  8

        2 cos(x)   (Pi + 2) sin(x)   cos(2 x)       sin(2 x)
  3/8 - -------- + --------------- - -------- - 1/2 --------
            2              2             2             Pi
          Pi             Pi            Pi

               cos(3 x)       (3 Pi - 2) sin(3 x)
         - 2/9 -------- + 1/9 -------------------
                   2                    2
                 Pi                   Pi

And here is a picture of the Novak extension together with the 10th partial sum of its Fourier series. You can see that the Fourier series is trying to get close to the Novak extension. On mostg of the horizontal line segments and on the tilted line, the partial sum of the Fourier series is wiggling above and below, causing the visible alternation in colors (yes, I regret the color choices, but I am too busy to try to fix them!). At the endpoints, though, the partial sum wants to have the same value at -Pi and Pi. So the value the partial sum takes is 1/2, the appropriate average of 0 and 1. Also, the Gibbs phenomenon is visible, if you care about it.

Graph of the Novak extension	The sum of the whole Fourier series of the Novak extension of f

The even extension

There are several standard ways of extending a function defined on [0,Pi]. One is the even extension, which asks for a function so that f(-x)=f(x). To get the graph, just flip what you are given across the y-axis. There are some interesting consequences. One is that all of the Fourier sine coefficients are 0. Why? Look at         bn=(1/Pi)-PiPiF(x)sin(nx)dx When we change x to -x, the integrand, F(x)sin(nx) changes to F(-x)sin(-nx) which is the same as -F(x)sin(nx). Since we're looking at an interval balanced around 0 (from -Pi to Pi) the contribution at x of F(x)sin(nx) is exactly balanced out by -F(x)sin(nx) at -x. So all of the bn's are 0.

I had Maple compute the third partial sum of the Fourier series for the even extension of f. Here it is:

                    4 cos(x)   2 cos(2 x)       cos(3 x)
              3/4 - -------- - ---------- - 4/9 --------
                       Pi          Pi              Pi

I also had Maple graph the even extension and some partial sums. The approximation is really good. On the left is the even extension and just the first three terms (the constand and cos(x) and cos(2x) terms): already quite close. The picture on the right shows the even extension and the terms up to cos(10x). In this picture, two distinct graphs can hardly be seen since they are so close. Here the sum of the whole Fourier series will exactly be equal to the function -- there are no jumps.

The odd extension
Now with f defined on [0,Pi] we ask that f(-x)=-f(x). Flip the graph over the y-axis, and then over the x-axis. Now because         a_n=(1/Pi)_-Pi^PiF(x)cos(nx)dx
and F(-x)cos(-nx)=-F(x)cos(nx) using the oddness of this extension, we see that all of the a_n's are 0. Here's the beginning of this Fourier series:

        2 (Pi + 2) sin(x)   sin(2 x)       (3 Pi - 2) sin(3 x)
        ----------------- - -------- + 2/9 -------------------
                 2             Pi                    2
               Pi                                  Pi

Here's a Maple graph of the odd extension of f(x) together with the sum of the first 10 terms of the Fourier sine series (up to and including the sin(10x) term). The series gets quite close on the tilted line segment, and attempts to be near the two horizontal segments. Of course, there is, in effect, a jump discontinuity at -Pi and Pi. From the Fourier point of view, the odd extension is repeated every 2Pi. So at, for example, x=Pi, the function has a left limit of 1 and a right limit of -1, so the series hops from 1 to -1. To me the Gibbs bumps are showing up.

Now what does theory predict here?

Graph of the even extension	The sum of the whole Fourier series of the even extension of f

The sum of the Fourier sine series of f (that is, the Fourier series of the odd extension of f) is equal to the original function except at the ends, where it averages the left and right behavior.

Volunteers
The syllabus for the course contains the following entry:

12.3

Fourier Sine and Cosine Series

3, 5, 7, 13, 19, 23, 29, 35

Since the exam will cover this section, I requested volunteers to do the problems in claas on Thursday. The following students either couldn't evade my attention, or, rarely, actually volunteered to put a problem on the board at the beginning of class Thursday. I thank them for their efforts in advance. Mr. Lin could not volunteer since he was in Texas. After class, he mysteriously precipitated (chem engineers do that).

Problem #	Student
12.3: 3	Mr. Klumb
12.3: 5	Mr. Klami
12.3: 7	Mr. Novak
12.3: 13	Mr. Shah
12.3: 19	Mr. Shaw
12.3: 23	Ms. Horn
12.3: 29	Mr. Pierre-Louis
12.3: 35	Mr. Sosa

QotD I quickly sketched something on the board, and asked people to sketch the even and odd extensions of it.

Original random function's graph	Even extension	Odd extension

Complaint Department
Ms. Tozour complained that the qotd was not "well-posed"(badly stated question). She stated (see the ?'s in the third graph) that the value(s?) of the odd extension at 0 was not clear. I agree with her.

Orthogonality
This is in the text, but following what we did last time, you should know:

• ₀^2Pisin(nx)sin(mx)dx=0 when n and m are different integers.
• ₀^2Picos(nx)cos(mx)dx=0 when n and m are different integers.
• ₀^2Pisin(nx)cos(mx)dx=0 when n and m are integers.

To continue the analogy I started last time, I need to orthonormalize these functions. Since sin²+cos²=1, and the wiggles are the same over an interval of length 2Pi, I know that
₀^2Pisin(nx)²+cos(nx)²dx=2Pi and each of ₀^2Pisin(nx)²dx and ₀^2Picos(nx)²dx are the same, so
if n is an integer >0, ₀^2Pisin(nx)²dx=Pi and ₀^2Picos(nx)²dx=Pi.
Of course, for the silly case (n=0), ₀^2Picos(0x)²dx=2Pi. (We don't care about the 0 function, so from now on I'll drop any mention of stuff corresponding to sin(nx) when n=0.)

The Fourier series of a function
If F(x) is a function defined on the interval [0,2Pi], define
        a_n=(1/Pi)₀^2PiF(x)cos(nx)dx
        b_n=(1/Pi)₀^2PiF(x)sin(nx)dx
The Fourier series of F(x) is the infinite series of functions
        a₀/2+SUM_N=1^infinity(a_ncos(nx)+b_nsin(nx))
where
        a_n=(1/Pi)₀^2PiF(x)cos(nx) dx for n integer, n>=0
        b_n=(1/Pi)₀^2PiF(x)sin(nx) dx for n integer, n>0

Weird things to note
Well, these are weird but they are what's usual in the subject. Notice that the a₀ term is divided by 2. That's because the normality constant for cos(0x) is 2Pi, not Pi. And also notice that the rest of the normalizing constants come off the formulas for the coefficients. In many standard linear algebra contexts, the darn formulas have the normalizations (those silly square roots) somehow in both the vectors and the coefficients of the vectors. Maybe what is done with Fourier series is more sensible.

You tell me how the Fourier series of a function relates to the function

I gave the class a handout. I wanted, in observation and discussion with students, to discover relationships (some subtle) between a function and its Fourier series (or, rather, since one can't add up all of any real infinite sum, the partial sums of the Fourier series): more heuristic stuff.
I would also like to have the Maple commands shown there available for you to copy, if you have the time and desire to experiment with them. Here they are:

• g:=n->(1/Pi)*int(F(x)*sin(n*x),x=0..2*Pi); Creates the sine coefficients.
• h:=n->(1/Pi)*int(F(x)*cos(n*x),x=0..2*Pi); Creates the cosine coefficients.
• Q:=N->h(0)/2+sum(h(n)*cos(n*x)+g(n)*sin(n*x),n=1..N); Adds things up and provides a linear combination of trig functions which is the N^th partial sum of the Fourier series of F.
• plot({F(x),Q(3)},x=0..2*Pi,thickness=3,scaling=constrained); Plots the function F(x) together with the third partial sum of its Fourier series.

>Q(3);
            2
        2 Pi
        ----- + 2/5 cos(x) - 2/5 Pi sin(x) + 1/10 cos(2 x)
         15

                - 1/5 Pi sin(2 x) + 2/45 cos(3 x) - 2/15 Pi sin(3 x)

This F(x) and the 3rd partial sum of its Fourier series	This F(x) and the 10th partial sum of its Fourier series	This F(x) and the 20th partial sum of its Fourier series

The graphs of the Q(n)'s (the partial sums) get closer to the graph of F(x) as n increases.

What does closer mean? This turns out to be a rather difficult question, both theoretically and in practice.

The pictures should show some of the difficulty. For example, you may want a function to be small on [a.b]. A very strict interpretation might be to have the values, f(x), very close to 0 for all x. But suppose you were really modelling some process which you expected to sample, somehow "randomly", on the interval, a few times (10 or 100 or ...). Maybe you would be happy enough controlling the average distance to 0. So things are complicated.

In the pictures of our function F(x) and various partial sums, inside the interval the partial sums are getting close to the values of the function. At the end points (0 and 2Pi) they aren't getting close ... what the heck. Also, if you look really closely at the graphs, you can see tiny bumps near the "ends" which represent some complicated phenomena. Well, one thing at a time.

What the Fourier series sees...
We get the Fourier coefficients by integrating the product of a sine or cosine on [0,2Pi] (the solid green curve) by our function F(x) (the solid magenta [?]) curve). One point of view is that everything goes on inside the shaded box. But the trig function goes on forever, and it is periodic with period 2Pi. To the trig function, our F(x) might as well be "extended" with period 2Pi to the left and to the right forever. Notice that the trig function will try at, say, 0, to approximate the values from both the left and right of the extended F(x). This extended F(x) has a jump discontinuity at 0, and the trig function, in trying its approximation, settles on being halfway between the ends of the jump. This is the collection of black dots in the picture at half the height of F(x) at x=2Pi.

The partial sums of the Fourier series try very hard to get close to F(x). If F is continuous at x, then they will converge to F(x). If F has a jump discontinuity at x, then they will converge to the average (really!) of the left and right hand limits of F at x (the middle of the jump).

Gibbs: the overshoot
J. Willard Gibbs received the first U.S. doctorate in engineering in 1863. He saw that at a jump discontinuity, there is always an overshoot of about 9% in the Fourier series. On the top side, the overshoot is above, and on the bottom side, below. These bumps get narrower and closer to the jump, but they never disappear!

My next example was U(x-{Pi/2}), the Heaviside step or jump at Pi/2. This function is 0 to the left of Pi/2 and is 1 to the right of Pi/2. In Maple, the following formula describes the function: F:=x->piecewise(x<Pi/2,0,1);
Here are the pictures for this function.

This F(x) and the 3rd partial sum of its Fourier series	This F(x) and the 10th partial sum of its Fourier series	This F(x) and the 20th partial sum of its Fourier series

I hope you see that the partial sums detect two jump discontinuities, one at Pi/2, certainly, but another one at 0=2Pi (well, they are the same numbers to sine and cosine) as well!

HOMEWORK due Tuesday, November 9.
Read sections 12.1, 12.2, and 12.3 of the text. Hand in:
12.1: 7, 15
12.2: 1, 5, 9, 15, 17

#2
Why I wouldn't want to be an engineer
This is an advisory note to both chemical and mechanical engineers. When mistakes are made, the results can be serious. When an academic mathematician makes a mistake, the grade can be computed again. Here are links to newspaper articles about the breaking of a 460,000 gallon tank full of 50% sodium hydroxide solution this past weekend in nearby New Jersey. Maybe there is no routine engineering, just routine engineers!
The New York Times       The Home News Tribune
I think a study of this incident would probably be an excellent senior project!

#3
I want to begin the next part of the course. I attempted to convince students that Fourier series would be an extended metaphor (is this good or bad educationally?), analogous to the material we just covered about symmetric matrices.So we looked at the symmetric matrix C=

( 0 -1 1)
(-1  1 2)
( 1  2 1)

( 2/sqrt(6) -1/sqrt(3)    0     )
(-1/sqrt(6)  1/sqrt(3) 1/sqrt(2))
(-1/sqrt(6)  1/sqrt(3) 1/sqrt(2))

Of course, P^tCP is the matrix D=

(6 0 0)
(0 3 0)
(0 0 2)

Then I asked the following weird question, which has a wonderful answer. If I take a "random" vector in R³ then it is possible to write the vector as a linear combination of the w_j's (j=1,2,3). So if Q=(11,33,-7), then
Q=(some #₁)w₁+(some #₂)w₂+(some #₃)w₃
In general it might be irritating (possible, but irritating) to find the coeffients. Here it turns out the computation is rather easy. If Q=(some #₁)w₁+(some #₂)w₂+(some #₃)w₃ then take the dot product of Q with w₁, say. The dot product distributes over the linear combination. Yes, you should know what this all means after 4 weeks of linear algebra and lots and lots of vector manipulation in a bunch of different courses:
Q·w₁=((some #₁)w₁+(some #₂)w₂+(some #₃)w₃)·w₁. If you distribute the dot product, two terms become 0 and one term is just 1. So Q·w₁=(some #₁. In our specific case, with Q=(11,33,-7) and w₁=(2/sqrt(6),-1/sqrt(6),-1/sqrt(6)) the dot product is exactly (22-33-1(-7))/sqrt(6)=-4/sqrt(6).

Many computer systems are optimized to take dot products, so this is really useful.

Now the analogy
Things will get complicated here. For the remainder of this course, the object is to study certain "classical" (a century old) ways of getting solutions to the partial differential equations which are supposed to model things like heat transfer or diffusion (the same equation: H(x,t) is a function of 1-dimensional position) or string vibration (double derivative in t is the double derivative in x: X(x,t) is the height at position x at time t of a vibrating string) or plate vibration (double derivative in t is the sum of the double derivative in both x and y where the deflection is a function of position [x and y] and time, t).

It turns out that these partial differential operators are all linear, and analyzing there solution can be accomplished by looking for eigenvectors. In all cases, one needs functions whose second derivative is a multiple of the original function (so the "eigen" characterization will be valid). So we throw out such things as x. For the most part, we also won't consider, say, e^47t, because this function gets big as you differentiate it.

Historically the functions which Mr. Fourier used were sin(nx) and cos(nx) where n is an integer. These functions will turn out to be eigenfunctions for essentailly all of the partial differential equations we will consider.

The vectors in our setup will be functions. The dot product we will use will be the following: if f and g are functions on the interval [a,b], then the "dot product" will be _a^bf(x)g(x) dx. You may well object that this is too weird. Well, at least algebraically this behaves like the inner product of vectors in Rⁿ (f·g=g·f and f·(g₁+g₂)=f·g₁+f·g₂ etc.).
What might be a bit more useful is to tell you the distance between f and g which this inner product defines: this distance is (_a^bf(x)-g(x) dx)^1/2. This is just about the same as the root mean square error between f and g (usually people might want to divide by b-a). This quantity should measure the average error between the functions f and g. If it is small, then, on average, the graphs of the functions f and g should be close.

Then just as in the Rⁿ case we will try to write a function f(x) as a sum, a very big sum:
SUM_n=0^infinity(coefficients)sin(nx)+(other coefficients)cos(nx).
This sort of sum is called a Fourier series.

Caution: the problem with convergence
I believe that the principle object of this course is to teach engineering students methods which they can use to model and predict practical problems. What I've just written is a real difficulty, a conflict between my own "trade", and the course objective. I know that the free and unrestricted use of infinite series almost inevitably leads to problems and sometimes even errors. I will show you a very simple example below. I remark that I will try to keep from stating false results in this course, and try to help students from making mistakes related to convergnece. But all I can do is caution you that, generally, if you "push" known methods to handle new or unusual situations, you may run into problems with convergence, and you may make errors. This has happened repeatedly historically.

The simple example
For various reasons in such applications as probability, one may take a matrix and compute its row sums and its column sums:

           Row sums 
   (horizontal sums of the entries)
( 3 -7  8)    4
(10  2 11)   23
(-5  6 -8)   -7
  8  1 11 
Column sums
(vertical sums of the entries)

  4
 23
 -7
---
 20 column sum of the row sums

 8  1 11 | 20 row sum of the column sums

  FOREVER ------>
F( 1 -1  0  0  0  0  0  0 .....
O( 0  1 -1  0  0  0  0  0 .....
R( 0  0  1 -1  0  0  0  0 .....
E( 0  0  0  1 -1  0  0  0 .....
V( 0  0  0  0  1 -1  0  0 .....
E( 0  0  0  0  0  1 -1  0 .....
R( 0  0  0  0  0  0  1 -1 .....
|( 0  0  0  0  0  0  0  1 .....
|( .  .  .  .  .  .  .  . ......
V

0 and 1 are not equal

I then restated

Euler tells me ...
eit=cos(t)+i sin(t) and cos(t)=[eit+e-it]/2 and sin(t)=[eit-e-it]/(2i)

and used this to compute ₀^2Pisin(17x)cos(5x) dx. There are several ways to do this integral. You can integrate by parts (you must do this twice, and be somewhat careful). You can use certain trig identities (this is the way the textbook does it). Or you can use Euler:
sin(17x)=[e^17ix-e^-17ix]/(2i) and cos(5x)=[e^5ix+e^-5ix]/2 so that sin(17x)cos(5x)=(1/4i)(e^22ix-e^-12ix+e^12ix-e^-22ix. This may seem involved, but all I'm doing is manipulating exponents in a standard fashion.

Easy integrals?
Suppose A is a non-zero integer. Then ₀^2Pie^iAxdx={1/iA}e^iAx]₀^2Pi. Now if x=0, e^iA·0=1. If x=2Pi, e^iAx=e^iA(2Pi)=cos(2Pi A)+i sin(2Pi A). The cosine term is 1 (cosine at an even multiple of Pi is 1) and the sine term is 0 (sine at any multiple of Pi is 0). Therefore the integral is 0.

Now use this result with A=22 and A=-12 and A=12 and A=-22 to see that ₀^2Pisin(17x)cos(5x) dx=0. The functions sin(17x) and cos(5x) are orthogonal.

The QotD was to compute ₀^2Pisin(7x)cos(7x) dx. I suggested using the same (complex) methods. Mr. Pierre-Louis instead successfully used a trig identity.

HOMEWORK
The next exam will be given during the regular class period on Thursday, November 18, 2004. Please begin to read sections 12.1, 12.2, and 12.3 of the text.

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