Difference between revisions of "HWK9MA527Fall2010" - Rhea

Revision as of 09:38, 23 October 2010

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Question Page 597, Problem 5:

What do we do with the x in the first term of this problem?

I don't understand what to do with this problem, either. Again, it would be fantastic if the book gave us harder examples.

Here are some hints about the problems on Legendre Polynomials.

See p. 180 for a list of the first few Legendre Polynomials.

The even numbered Legendre Polynomials only involve even powers of x, so they are even functions.

The odd numbered Legendre Polynomials only involve odd powers of x, so they are odd functions.

The Legendre Polynomials are orthogonal on the interval [-1,1].

p. 209, 5. asks you to show that

$ P_n(\cos\theta) $

are orthogonal on [0,pi] with respect to the weight function

$ \sin\theta, $

i.e., to show that

$ \int_0^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta\ d\theta=0 \qquad\text{if }n\ne m. $

The key here is to use the change of variables

$ x=\cos\theta $

and convert the integral to one in x over the interval [-1,1], where you can use the orthogonality of the Legendre Polynomials.

p. 216, problems 1 and 3 ask you to expand a given function in terms of Legendre Polynomials. Here, you will use the fact that if Q(x) is a polynomial of degree N, then

$ Q(x)=\sum_{n=0}^N c_nP_n(x) $

where the coefficients c_n are computed via orthogonality:

$ \int_{-1}^1 Q(x)P_m(x)\ dx=c_m\int_{-1}^1 P_m(x)P_m(x)\ dx. $

You will need to use the fact given on page 212 that

$ \int_{-1}^1 P_m(x)^2\ dx=\frac{2}{2m+1}. $

p. 216, 5. asks you to show that if f(x) is even, then all the odd coefficients in its Legendre expansion must vanish, i.e., that

$ \int_{-1}^1 f(x)P_n(x)\ dx=0 $

if n is odd. Recall that if n is odd, P_n is odd. An even times an odd is a ... etc.

Question: Page 209, Problem 17: For the given equation, shouldn't p=1, q=16, r=1? These values differ from the textbook's values.

Answer: If that were the case, then the equation would be

[py']' + (q+ lambda r) y =

[1 y']' + (16 + lambda) y =

y" + (16 + lambda) y = 0

and it ain't. You need to use problem 6 in the same section to get p,q, and r.

Question: Why isn't q=pg=16*exp(8x)?

Answer: Here is the idea of problem 6. We have the equation

$ y'' + 8 y' + (\lambda + 16)y=0. $

Multiply that equation by p(x). You get

$ py'' + 8p y' + (\lambda p+ 16p)y=0. $

If this were in Sturm-Liouville form, it would look like

$ [py']'+ (q + \lambda r) y = $

$ py'' + p'y' + (q+ \lambda r) y = 0. $

By comparing those two, we see that we need

$ p'=8p $

and q=16p and r=p. Solving the ODE for p yields

$ p(x)=e^{8x}. $

(We can take the arbitrary constant in the solution to be a convenient value because we just want one p(x) that has this property.)

Finally, we get

$ p(x)=e^{8x},\quad q(x)=16e^{8x},\quad\text{and }r(x)=e^{8x}. $

Hmmm. I see what you mean. I think the answer in the back of the book is wrong.--Steve Bell 12:07, 23 October 2010 (UTC)

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p. 209, #17

I don't understand how to solve for lamba. The examples in the book are WAAAAY easier! Help! Thank you!

p. 209 #7 & #17:

To verify orthogonality, do you just use Theorem , or do you have to do the integral?