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Legendre hints from Bell

for some hints about Legendre Polynomials.

Question Page 597, Problem 5:

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

Answer: When you do a Laplace transform wrt t, the x floats along like when you do d/dt(x*t).

Additional question: I get to this point after taking the Laplace Transform:

d/dx*W(x,s) + s/x*W(x,s) = 1/s^2.

Is this correct?

Answer: Yes, that's what I got. Check out lecture 23 around the 28 minute mark to see what to do next.

Message from Bell: When you solve that ODE as a first order linear ODE in the x variable, you should get an integrating factor of x^s. Remember that the arbitrary constant that you get in the solution might depend on s. So I like to write C(s) for the arbitrary constant. You will need to show that C(s) must be zero by showing that


This follows from one of the boundary conditions given in the problem.

Page 209 Problem 5: I've done this problem 5 times over, (yes, again after reading Prof. Bell's notes) and I keep getting 2/3 out of the integral rather than the necessary 0. Whether it's cos^3 or x^3, I don't see any way to get anything other than Pm*Pn/3*(-1 - 1) = 2/3 * PnPm rather than the 0 we're looking for. Or do I just not understand orthogonality??

New Question: How is everyone else "showing orthogonality?" Are you evaluating all the available (11) polynomials, doing just a few and making a statement, or have you figured out how to integrate (11) itself?

From Bell: My hint about p. 209, problem 5 was to make a change of variables,

x = cos theta.

After that, you'll reduce the orthogonality of P_n(cos theta) with respect to the weight function Sin(theta) on [0,pi] to the orthogonality of the Legendre Polynomials on [-1,1]. The orthogonality of the Legendre Polynomials on the interval [-1,1] is a given fact, so you shouldn't be computing actual integrals to verify the orthogonality for the first few Legendre Polynomials listed on p. 180.

Question Page 209 Problem 7:

Where does the $ \pi $ come from in this solution?

Answer: When you do the positive lambda case, you get A = 0 and let B = 1 => Sin(5*mu) = 0. If mu = m*pi/5, this equation is true. I let B=1 because we cannot have both A and B = 0.

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)

p. 216, #s 1 and 3:

I am using the hints given, but I'm still not sure I'm doing this correctly. For example, for #1, I've calculated c4 as c4 = [((2*4)+1)/2] * integral from -1 to 1 of (7x^4-6x^2)(P_4(x)) dx, where P_4= (1/8)(35x^4-30x^2+3), and then I would do something similar for C3, C2, C1, and C0. But if I find C4 this way, I'll get an answer where there's an x^9 term, and I don't see how to get an answer in terms for P_4 and P_1, like there is in the back of the book.

Answer: These are definite integrals carried out from -1 to 1, so you should get a numerical value for the solution, which will be the coefficient cm of the mth legendre polynomial. For instance, for m=0, you'll have c_0 =(2*0+1)/2 * integral from -1 to 1 of (7x^4-6x^2)*P_0(x)dx, where P_0(x)=1. The solution of that integral is -3/5, which matches the value for the coefficient of P_0(x) in the back of the book. --Jkusnick 08:47, 25 October 2010 (UTC)

p. 209 #7 & #17:

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

Answer: The theorem says the eigenfunctions are orthogonal. However, to VERIFY that, you'll have to compute the integrals.

p. 209, #17


You can solve for lambda in a way similar to the way Prof. Bell did it at the beginning of class on 10/20/10. From the original problem you get: r^2+8r+(16+lamda)=0. So r = sqrt(-lambda)-4. As usual, the lambda >0 case gives you non-zero solutions. Since our root is complex of the form -4 plus/minus i*sqrt(lambda), where mu = sqrt(lambda). We can find lambda that allow non-zero solutions like in 7 by using the general form of:

y = exp(-4*x)*(A*cos(mu*x)+B*sin(mu*x)). Plug in boundary conditions and Ta-Da.

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