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== Homework 9 Collaboration Area ==
 
== Homework 9 Collaboration Area ==
  
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Here are some hints about the problems on Legendre Polynomials.
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Click on
  
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See p. 180 for a list of the first few Legendre Polynomials.
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[[LegendreMA527Fall2010 |Legendre hints from Bell]]
  
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The even numbered Legendre Polynomials only involve even powers of  x,
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for some hints about Legendre Polynomials.
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so they are even functions.
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The odd numbered Legendre Polynomials only involve odd powers of  x,
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Question Page 597, Problem 5:
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so they are odd functions.
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The Legendre Polynomials are orthogonal on the interval [-1,1].
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What do we do with the x in the first term of this problem?
  
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p. 209, 5. asks you to show that
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Answer: When you do a Laplace transform wrt t, the x floats along like when you do d/dt(x*t).
  
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<math>P_n(\cos\theta)</math>
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Additional question:  I get to this point after taking the Laplace Transform:
  
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are orthogonal on [0,pi] with respect to the weight function
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d/dx*W(x,s) + s/x*W(x,s) = 1/s^2.
  
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<math>\sin\theta,</math>
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Is this correct?
  
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i.e., to show that
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Answer: Yes, that's what I got.  Check out lecture 23 around the 28 minute mark to see what to do next.
  
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<math>\int_0^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta\ d\theta=0
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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
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\qquad\text{if }n\ne m.</math>
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to write  C(s) for the arbitrary constant.  You will need to show that  C(s) must be zero by showing that
  
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The key here is to use the change of variables
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W(0,s)=0.
  
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<math>x=\cos\theta</math>
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This follows from one of the boundary conditions given in the problem.
  
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and convert the integral to one in  x  over the interval [-1,1], where you can use the orthogonality of the Legendre Polynomials.
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Page 209 Problem 5:
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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??
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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?
  
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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
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From Bell:  My hint about p. 209, problem 5 was to make a change of variables,
  
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<math>Q(x)=\sum_{n=0}^N c_nP_n(x)</math>
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x = cos theta.
  
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where the coefficients c_n  are computed via orthogonality:
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After that, you'll reduce the orthogonality of P_n(cos theta) with respect
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to the weight function Sin(theta) on [0,pi] to the orthogonality of the
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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.
  
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<math>\int_{-1}^1 Q(x)P_m(x)\ dx=c_m\int_{-1}^1 P_m(x)P_m(x)\ dx.
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Question Page 209 Problem 7:
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</math>
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You will need to use the fact given on page 212 that
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Where does the <math>\pi</math> come from in this solution?
  
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<math>\int_{-1}^1 P_m(x)^2\ dx=\frac{2}{2m+1}.</math>
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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.
  
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p. 216, 5. asks you to show that if  f(x)  is even, then all the odd
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Question: Page 209, Problem 17:
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coefficients in its Legendre expansion must vanish, i.e., that
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<math>\int_{-1}^1 f(x)P_n(x)\ dx=0</math>
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if  n  is odd.  Recall that if n is odd,  P_n  is odd.  An even times an odd is a ... etc.
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Page 209, Question 17:
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For the given equation, shouldn't p=1, q=16, r=1?  These values differ from the textbook's values.
 
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
 
Answer:  If that were the case, then the equation would be
  
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[py']' + (q+ lambda r) y = 0
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[py']' + (q+ lambda r) y =
  
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[1 y']' + (16 + lambda) y = 0
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[1 y']' + (16 + lambda) y =
  
 
y" + (16 + lambda) y = 0
 
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.
 
and it ain't.  You need to use problem  6  in the same section to get p,q, and r.
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Question: Why isn't q=pg=16*exp(8x)?
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Answer:  Here is the idea of problem 6.  We have the equation
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<math>y'' + 8 y' + (\lambda + 16)y=0.</math>
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Multiply that equation by p(x).  You get
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<math>py'' + 8p y' + (\lambda p+ 16p)y=0.</math>
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If this were in Sturm-Liouville form, it would look like
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<math>[py']'+ (q + \lambda r) y = </math>
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<math>py'' + p'y' + (q+ \lambda r) y = 0.</math>
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By comparing those two, we see that we need
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<math>p'=8p</math>
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and q=16p  and r=p.  Solving the ODE for p yields
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<math>p(x)=e^{8x}.</math>
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(We can take the arbitrary constant in the solution
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to be a convenient value because we just want one
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p(x) that has this property.)
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Finally, we get
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<math>p(x)=e^{8x},\quad q(x)=16e^{8x},\quad\text{and }r(x)=e^{8x}.</math>
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Hmmm.  I see what you mean.  I think the answer in the back of the book is wrong.--[[User:Bell|Steve Bell]] 12:07, 23 October 2010 (UTC)
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p. 216, #s 1 and 3:
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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.
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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. --[[User:Jkusnick|Jkusnick]] 08:47, 25 October 2010 (UTC)
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p. 209 #7 & #17:
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To verify orthogonality, do you just use the Theorem, or do you have to do the integral?
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Answer:  The theorem says the eigenfunctions are orthogonal.  However, to VERIFY that, you'll have to compute the integrals.
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p. 209, #17
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Answer:
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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:
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y = exp(-4*x)*(A*cos(mu*x)+B*sin(mu*x)).  Plug in boundary conditions and Ta-Da.
  
 
[[2010 MA 527 Bell|Back to the MA 527 start page]]  
 
[[2010 MA 527 Bell|Back to the MA 527 start page]]  

Latest revision as of 11:00, 27 October 2010

Homework 9 Collaboration Area

Click on

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

W(0,s)=0.

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

Answer:

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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