Answer
Remember that the functions cos(5x) and sin(2x) and sin(33x) are all orthogonal on the interval [-Pi,Pi] for the "dot product" coming from the formula _-Pi^Pi(function₁)(function₂) dx. Also, the square of a sine or cosine on the interval has integral pi. Therefore the value of _-Pi^Pi(f(x))² dx is (pi)[3²+9²+8²].

Answer
I know that sin(w) is [e^iw-e^-iw]/[2i]. Let me concentrate temporarily on e^iw. If I plug in nx for w and divide by 2ⁿ and sum from 1 to infinity I will get SUM_n=1^infinity[e^i nx]/[2ⁿ]. But e^i nx is also (e^ix)ⁿ so that what is inside the sum is [e^ix/2]ⁿ. Therefore the sum is a geometric series with first term e^ix/2 and ratio between successive terms e^ix/2.

The sum of a geometric series is a/(1-r) if a is the first term and r is the ration between successive terms. So here the sum would be [e^ix/2}/[1-e^ix/2] and this is e^ix/[2-e^ix]. The sum of the other terms, with e^-ix, turns out to be e^-ix/[2-e^-ix].

We need 1/(2i) multiplied by the difference (the complex formula of sine has a minus in it!) of e^ix/[2-e^ix] and e^-ix/[2-e^-ix]. This difference is a fraction, which should be combined in the standard way for fractions.
The top of the fraction is e^ix(2-e^-ix)- e^-ix(2-e^ix)= 2e^ix-1 -2e^-ix+1= 2e^ix-2e^-ix.
The bottom of the fraction is [(2-e^-ix)(2-e^-ix)]=4-2e^ix-2e^-ix+1 =5-2e^ix-2e^-ix.

I "recognize" the bottom as 5-4cos(x) since cos(x) is (1/2)[e^ix+e^-ix].
The top (put in the division by 2i) is [2e^ix-2e^-ix]/2i which is 2sin(x).

a) Find the Fourier coefficients of f(x).

b) Suppose g(x) is the sum of the first 100 terms of the Fourier series of f(x) (both sine and cosine). Draw a graph of both y=g(x) and y=f(x) on the interval [0,3]. Draw a graph of both y=g(x) and y=f(x) on the interval (3,Pi). Label the graphs.

c) Suppose now h(x) is the sum of the whole Fourier series of f(x). Graph f(x) and h(x) on the interval [-Pi,Pi]. Label the graphs.

Answer
a) Here we go. f(x) is 0 for x<0 and is (1/Pi)x otherwise. The integrals only matter on [0,Pi].

The Fourier sine coefficients are (1/Pi)_-Pi^Pif(x)sin(nx) dx=(1/Pi)₀^Pi(1/Pi)x sin(nx) dx=(1/Pi²₀^Pix sin(nx) dx.
We integrate x sin(nx) by parts with u=x and dv=sin(nx)dx. I won't go through the details. The result is -(x/n) cos(nx)+(1/n²)sin(nx). We need then to |₀^Pi. The result of that is -(Pi/n)cos(nPi) because sin(nPi) is 0. And cos(nPi) is (-1)ⁿ, so that (putting it all together) the n^th Fourier sine coefficient, b_n, is (1/Pi n)(-1)ⁿ⁺¹.

For n=0, we know that a₀ is just the total area, which is Pi/2. The other Fourier cosine coefficients are (1/Pi²₀^Pix cos(nx) dx. The result of this integration turns out to be [(-1)ⁿ-1]/(Pi²n²).

b) Here are the graphs of f(x) and g(x). I checked with Maple. f(x) and g(x) are virtually indentical on this scale on [0,3]. On [3,Pi) we see the Gibbs overshoot phenomenon, and the fact that g(x) tries to be periodic, and "exits" the interval on the right at the halfway point between 1 and 0.

On [0,3]	On [3,Pi)

c) The whole Fourier series is equal to the original function except that the Fourier series attempts to compromise at -Pi and Pi, and there it inserts a value halfway between 0 and 1.

a) f(x)=-1 if x=<0 and 2 if x>0.

b) f(x)=5.

c) f(x)=21+2sin(5x)+8cos(2x).

d) f(x)=SUM_n=1⁸c_nsin(nx); c_n=1/n.

e) f(x)=-4+SUM_n=1⁶(c_nsin(nx)+7cos(nx)) ; c_n=(-1)ⁿ.

Answer
Part a) is the only part which needs any computation. That's because the f(x)'s in b), c), d) and e) are already Fourier series: they are their own answers. (All the other Fourier coefficients are 0 because of orthogonality!) I still need to do a), though.

Answer to a): if n>0, a_n=(1/Pi)_-Pi^Pif(x)cos(nx) dx. We'll compute the integral by splitting it into two parts because of the piecewise definition of f(x):
_-Pi⁰(-1)cos(nx) dx=[(-1)/n](sin(n·0)-sin(-n·Pi)]=0.
₀^Pi2cos(nx) dx=[2/n](sin(n·Pi)-sin(n·0)]=0.
So a_n=0 when n>0 because sine at integer multiples of Pi is 0.

Now a₀ is (1/Pi)(the integral of f(x) from -Pi to Pi), which is 1. Why are all the other a_n's equal to 0? There is a simple reason. We can write f(x)=(1/2)+(another function)? The "another function" here is -3/2 on [-Pi,0) and 3/2 on (0,Pi]. So "another function" is odd and its cosine Fourier coefficients are all 0. Therefore f(x)'s cosine Fourier coefficients, except for the constant (n=0) one, are all 0.

The sine coefficients:
Here n>0. b_n=(1/Pi)_-Pi^Pif(x)sin(nx) dx. Again, the integral is more easily computed by splitting it into two parts because of the piecewise definition of f(x):
_-Pi⁰(-1)sin(nx) dx=-[(-1)/n](cos(n·0)-cos(-n·Pi)].
₀^Pi2sin(nx) dx=-[2/n](cos(n·Pi)-cos(n·0)]=0.

The total integral is (3/n)[1-cos(n·Pi)] because the antiderivative of sine is -cosine, and because cos(0)=1 and cos(-w)=cos(w). Now cos(n·Pi)=(-1)ⁿ, so that b_n=(3/n)[1+(-1)ⁿ⁺¹].

The Fourier series of f(x) is therefore (1/2)+SUM_n=1^infinity(3/n)[1+(-1)ⁿ⁺¹]sin(nx).

b) Suppose f(x)=cos(x⁵)+sin(x²) for x in [-Pi,Pi]. Which coefficients of the Fourier series of f(x) must be 0?

Answer
a) f(x) is an odd function. Indeed, f(-x)=-x+(-x)³=-x-x³=-(x+x³)=-f(x). Therefore all the a_n's must be 0 (the cosine coefficients).

b) g(x) is an even function. Indeed, g(-x)=cos((-x)⁵)+sin((-x)²)=cos(-x⁵)+sin(x²)=cos(x⁵)+sin(x²)=g(x). Therefore all the n_n's must be 0 (the sine coefficients).

a) If F(x) is the odd extension of f(x) to [-Pi,Pi], write a formula or formulas for F(x). Which terms must be 0 in the Fourier series of F(x)?

b) If G(x) is the even extension of f(x) to [-Pi,Pi], write a formula or formulas for G(x). Which terms must be 0 in the Fourier series of G(x)?

Answer
a) F(x) is x+x⁴ for x>0, and F(x) is x-x⁴ for x<0. Then we know that F(-x)=-F(x) so F(x) is odd. For an odd function, the cosine coefficients are 0: what we generally call the a_n's.

b) G(x) is x+x⁴ for x>0, and G(x) is -x+x⁴ for x<0. Then we know that G(-x)=G(x) so G(x) is even. For an even function, the sine coefficients are 0: what we generally call the b_n's.

Answer
The function F(x) is the Fourier sine expansion of f(x). On the domain of f, that is, for x in [0,Pi], we have F(x)=f(x). Therefore, since 3 is in [0,Pi], then F(3)=f(3)=2e^-12.

For the negative values of x, the sine series converges to the odd extension of f(x). That is, F(x)=-f(-x) for x<0. Therefore F(x)=-2e^{4x} for x<0 so that F(-2)=-2e^-8.

Answer
y_tt=9y_xx for x [0,25]; y(0,t)=y(25,t)=0; y(x,0)=4sin([2Pix]/25); y_t(x,0)=0.

Find y(x,t).

Answer
Here is an answer:
y(x,t)=5sin(2x)cos(2t)+8sin(6x)cos(6t)
I got this by recalling the separation of variables pairings we used when we solved the wave equation.

What is the speed of wave propagation along the string? What is the initial displacement of the string at the point x=20? What is the initial velocity of the string at the point x=50? At what point of the string is the initial velocity the largest?

Answer
The speed of wave propagation along the string is sqrt{50}. The initial displacement of the string at the point x=20 is 20²(100-20)=32,000. The initial velocity of the string at the point x=50 is (1/3)(100-50)=50/3. The maximum of the initial velocity is at the point x=25. The easiest way to see this is to plot y_t(x,0), or we could use calculus: in [0,25] the function is increasing and in [25,100] it is decreasing and the function is continuous.

Suppose we also know ₀^Pig(x)sin(nx) dx=1/n³ for all positive integers, n. Find y(x,t).

Answer
Here we are considering the wave equation with initial displacement equal to 0. Then the solution is y(x,t)=SUM_n=1^infinityc_nsin{nx)sin(nt) with c_n=2/[nPi]₀^Pig(x)sin(nx) dx=2/[n⁴Pi]

Answer
Assume that y(x,t)=X(x)T(t). Then substitute this in the equation and get
X(x)T'(t)=12X(x)T(t)-5X'(x)T(t)+7X''(x)T(t).

Now divide both sides by X(x)T(t):
[T'(t)/T(t)]=12-5[X'(x}/X(x)]+7[X''(x)/X(x)].

The left hand side does not depend on x. The right had side does not depend on t. They are equal to each other. Therefore, both the left hand side and the right hand side are constants. We set the constant to be -, and obtain two ordinary differential equations:
T'(t)=-T(t)
-5X'(x)+7X''(x)=-X(x)
and that's the desired pair of ODE's.

y_tt=9y_xx for x in R; y(x,0)=x(2-x) for x in [0,2] and y=0 otherwise; y_t(x,0)=0 for x in R.

a) Find y(x,t).

b) Draw the solution for t=5 and t=10 (two graphs).

c) How long will it take before an observer located at point x=27 receives a signal?

Answer
a) The D'Alembert form of the solution is: y(x,t)=(1/2)(f(x-3t)+f(x+3t)) where f(x) is the initial displacement profile. The first part is a wave traveling right and the second part is a wave traveling left. Since the initial velocity is 0, we don't need the "g" terms.

But what is f(x)? It is actually not the simple algebraic x(2-x), but (since we're going to substitute x-/+3t in for x) a function which is equal to x(2-x) in [0,2] but is 0 otherwise. Here is a way of writing this, using the Heaviside function from Laplace transforms:
the initial profile is x(2-x)H(x)H(2-x).

Now I can write y(x,t). It is (1/2){[x-3t](2-[x-3t])H([x-3t])H(2-[x-3t])+[x+3t](2-[x+3t])H([x+3t])H(2-[x+3t])}.

b)

t=5	t=10

The graphs are two bumps half the height of the original profile. One is traveling left and the other, right. The speed is 3 distance units per time unit. The original bump is centered at x=1. For t=5, the bumps are centered at -14 and 16. For t=10, they are centered at -29 and 31. I have drawn the horizontal and vertical axes with different units.

c) The right edge of the bump moving right will reach x=27 at time (27-2)/3=25/3.

a) Sketch a graph of u(x,1/100) and a graph of u(x,100).

b) Explain why this initial temperature distribution can't result from an earlier temperature distribution.

Answers
a)

t=.01	t=100

b) The initial temperature distribution shown is not smooth: more precisely, it is not differentiable at x=2 and x=2.5 and x=3. Temperature distributions which are the result of the time "evolution" of earlier temperature distributions must be smooth. Therefore this initial temperature distribution is not the "evolution" of an earlier temperature distribution.

a) Write the initial and boundary value problem for u(x,t), the temperature of the bar.

b) Write the solution, u(x,t).

c) What is the limiting temperature distribution as t-->infinity?

Answer
a) The boundary value problem is this: u_t=ku_xx for some constant k (the heat equation) and u_x(0,t)=0 and u_x(Pi,t)=0 for t>=0 (the boundary conditions, representing insulated ends) and u(0,x)=f(x) for x in [0,Pi] (the initial condition).

b) u(x,t)=[c₀/2]+SUM_c=1^infinityc_ncos(nx)e^-kn2t where:
If n>0, c_n=(2/Pi)₀^Pif(x)cos(nx) dx= (2/Pi)₀^Pi/22cos(nx) dx+(2/Pi)_Pi/2^Pi4cos(nx) dx=[4/Pi]sin([nPi]/2). sin([nPi]/2) is 1 if n=4k+1 and is -1 is n=4k+3. Otherwise it is 0. We could invent some sort of "formula" but I think the value of c_n is now clear.
For n=0, then c₀=(2/Pi)₀^Pif(x) dx=6.

c) The limiting temperature distribution is the constant 3 (that's c₀/2, or the average of the initial temperature, f(x).) All of the exponentials with negative constants force the other terms to 0 very rapidly.

Answer
In a bar with insulated ends, we saw (using both a mathematical treatment and thinking about the physical situation) that the limiting temperature distribution as t-->infinity is the average value of the initial temperature distribution.
a) The average value is (0+2+0+5+2·6)/(10)=(19)/(10).
b) The average value is (1/10)₀¹⁰x+2x² dx=(1/10)[(1/2)x²+(2/3)x³]|₀¹⁰=(215)/3. Remember that the (1/10) comes from the average value over the interval [0,10].

Answer

t=0	t=.01	t=100

Comments
The first picture is a parabola, with vertex at (3,9) (hey: the derivative of 6x-x² is 6-2x etc.). The next picture shows the initial heat distribution "relaxing" a bit towards equilibrium, and the next picture shows the steady state temperature distribution for these boundary conditions. u(x,100) would like very much like this, I believe. Check me with Maple or Matlab if you wish.