Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > MHF Lounge > Problem of the Week
Closed Thread
 
Thread Tools Display Modes
  #1  
Old July 25th, 2007, 08:28 AM
ThePerfectHacker's Avatar
Global Moderator

 
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,751 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default Problem 32

Given a non-negative integer n we define P_n(x) to be a solution* to:
(1-x^2)y''-2xy'+n(n+1)y=0 \mbox{ on }(-1,1).

Show that:
\int_{-1}^1P_n(x)P_m(x) dx = 0 \mbox{ for }n\not = m.


*)Yes, I realize what I said is not well-defined but it makes no difference. Pick any solution.
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Advertisement
 
  #2  
Old July 27th, 2007, 04:57 PM
Newbie
 
Join Date: Jun 2007
Posts: 18
Country:
Thanks: 0
Thanked 1 Time in 1 Post
mathisfun1 is on a distinguished road
Default

The given equation is the Legendre equation and its solutions are the Legendre orthogonal polynomials (Legendre polynomials - Wikipedia, the free encyclopedia). I'm not sure how to show the orthogonality of the solutions from scratch though (i.e. w/o Rodrigues's formula).
  #3  
Old July 27th, 2007, 06:49 PM
topsquark's Avatar
Generous Contributor
 
Join Date: Jan 2006
Location: Angelica, NY
Posts: 7,605
Country:
Thanks: 643
Thanked 2,305 Times in 2,093 Posts
topsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond repute
Default

Quote:
Originally Posted by mathisfun1 View Post
The given equation is the Legendre equation and its solutions are the Legendre orthogonal polynomials (Legendre polynomials - Wikipedia, the free encyclopedia). I'm not sure how to show the orthogonality of the solutions from scratch though (i.e. w/o Rodrigues's formula).
But Rodrigues' formula is the simplest way to go with it!

-Dan
__________________
Got a Physics question? Come on over to
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert
  #4  
Old July 28th, 2007, 07:54 PM
ThePerfectHacker's Avatar
Global Moderator

 
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,751 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default

There is not need to use the Rodrigues formula, nor any identities that exist among Legendre polynomials.
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

Last edited by ThePerfectHacker; July 29th, 2007 at 03:31 PM.
  #5  
Old July 29th, 2007, 06:17 AM
topsquark's Avatar
Generous Contributor
 
Join Date: Jan 2006
Location: Angelica, NY
Posts: 7,605
Country:
Thanks: 643
Thanked 2,305 Times in 2,093 Posts
topsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond reputetopsquark has a reputation beyond repute
Default

Quote:
Originally Posted by ThePerfectHacker View Post
There is not need to use the Rodrigues formula, nor any identities tha exist among Legendre polynomials.
Well, I suppose you could plug the differential equation into the integral and get a diffeo-integral equation and work from there, but I've never seen it done.

-Dan
__________________
Got a Physics question? Come on over to
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert
  #6  
Old July 29th, 2007, 03:30 PM
ThePerfectHacker's Avatar
Global Moderator

 
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,751 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default

As mentioned this problem can be solved with Legendre Polynomials and their identities, but there is no need. There are many nice identities among the Legendre Polynomials and not a single one is used here.

The trick is to write the differencial equation as (in Sturm-Louiville form),
[(1-x^2)y']'+n(n+1)y=0

Thus,
[(1-x^2)P_n'(x)]'+n(n+1)P_n(x)=0
Multiply by P_m(x) to get,
[(1-x^2)P_n'(x)]'P_m(x)+n(n+1)P_n(x)P_m(x)=0
And integrate,
\int_{-1}^1 [(1-x^2)P_n'(x)]'P_m(x) dx + n(n+1)\int_{-1}^1 P_n(x)P_m(x)dx=0
Use integration by part on the first integral,
(1-x^2)P_n'(x)P_m(x)\big|_{-1}^1 - \int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+n(n+1)\int_{-1}^1P_n(x)P_m(x)dx=0
The first term vanishes,
- \int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+n(n+1)\int_{-1}^1P_n(x)P_m(x)dx=0
Doing the similar steps with P_m(x) instead we have,
-\int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+m(m+1)\int_{-1}^1P_n(x)P_m(x)dx=0
Now subtract these two equations to get,
[n(n+1)-m(m+1)]\int_{-1}^1 P_n(x)P_m(x) dx = 0
Since n\not = m and they are integers we have,
\int_{-1}P_n(x)P_m(x) dx = 0.
Q.E.D.
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  #7  
Old July 29th, 2007, 04:35 PM
ThePerfectHacker's Avatar
Global Moderator

 
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,751 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default

Here is a way to get the Rodrigues formula since it was mentioned.

Let z=(x^2-1)^n.

We see that,
(x^2-1)\frac{dz}{dx} - 2nx z = 0

Take the derivative n times to get,
(1-x^2)\frac{dz}{dx}+n(n+1)z^{(n-1)}=0
Take the derivative one last time,
\left[(1-x^2)\frac{dz}{dx}\right]'+n(n+1)z^n=0
Hence,
\frac{d^{n}(x^2-1)^n}{dx^n}
Is a polynomial which solves the equation.
But we write it as,
\frac{1}{2^nn!}\cdot \frac{d^n(x^2-1)^n}{dx^n}
Because if we solve the Legendre equation with a power series we find that the leading coefficient of the polynomial starts with that factor. So to make Rodrigues formula agree with an infinite series solution we make the leading term equal to that factor.
-------------------------------
Warning Complex Analysis Ahead

Here is a fabulous identity due to Laplace.
P_n(x) = \frac{1}{\pi}\int_0^{\pi} (x+\sqrt{x^2-1}\cos \phi)^n d\phi

We begin by noting that,
P_n(x) = \frac{1}{2\pi i}\oint_{\gamma} \frac{1}{2^n}\frac{(z^2-1)^n}{(z-x)^{n+1}}dz
Where \gamma is any peicewise smooth simple closed curve containing x.
This is immediately true by Cauchy's Integral Formula.

Let x>1 define \gamma = x+\sqrt{x^2-1}e^{i\phi} for 0\leq \phi \leq 2\pi. That is a circle centered at x with radius \sqrt{x^2-1}.

Evaluating the above integral with this parametrization (and doing some simplification, not shown) we get Laplace's formula.
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  #8  
Old August 2nd, 2007, 11:26 AM
Rebesques's Avatar
Senior Member
 
Join Date: Jul 2005
Location: At my house.
Posts: 396
Thanks: 30
Thanked 49 Times in 44 Posts
Rebesques will become famous soon enough
Send a message via ICQ to Rebesques Send a message via AIM to Rebesques Send a message via MSN to Rebesques Send a message via Yahoo to Rebesques
Default

Quote:
The trick is to write the differencial equation as (in Sturm-Louiville form)
I claim this is cheating!

Since, by identifying a Sturm-Liouville operator here, the P_n(x) are eigenfunctions and thus orthogonal for the L^2 inner product - that is, \int_{-1}P_n(x)P_m(x) dx = 0; qed!
__________________
Never leave home without some Latex in your back pocket.

Last edited by Rebesques; August 2nd, 2007 at 01:24 PM. Reason: eigenfunctions, more precisely
  #9  
Old August 2nd, 2007, 12:24 PM
ThePerfectHacker's Avatar
Global Moderator

 
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,751 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default

Quote:
Originally Posted by Rebesques View Post
I claim this is cheating!

Since, by identifying a Sturm-Liouville operator here, the P_n(x) are eigenvectors and thus orthogonal for the L^2 inner product - that is, \int_{-1}P_n(x)P_m(x) dx = 0; qed!
Yes, that is cheating. It takes all the fun away from the problem if it can be done.*

(Change "eigenvectors" to "eigenfunctions" to fix your post.)


*)I say "if it can be done" because this is not a boundary value problem. Thus, Sturm-Louiville Theory does not apply. Are you sure?
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  #10  
Old August 2nd, 2007, 01:23 PM
Rebesques's Avatar
Senior Member
 
Join Date: Jul 2005
Location: At my house.
Posts: 396
Thanks: 30
Thanked 49 Times in 44 Posts
Rebesques will become famous soon enough
Send a message via ICQ to Rebesques Send a message via AIM to Rebesques Send a message via MSN to Rebesques Send a message via Yahoo to Rebesques
Default

Quote:
Change "eigenvectors" to "eigenfunctions" to fix your post
thnx!


Quote:
this is not a boundary value problem. Thus, Sturm-Louiville Theory does not apply
Doesn't have to be, Sturm-Louiville operators are really flexible about this.
__________________
Never leave home without some Latex in your back pocket.
  #11  
Old August 2nd, 2007, 08:09 PM
Newbie
 
Join Date: Jun 2007
Posts: 18
Country:
Thanks: 0
Thanked 1 Time in 1 Post
mathisfun1 is on a distinguished road
Default

Quote:
Originally Posted by ThePerfectHacker View Post

The trick is to write the differencial equation as (in Sturm-Louiville form),
[(1-x^2)y']'+n(n+1)y=0

So did you change your signature just for this problem :P?
  #12  
Old August 2nd, 2007, 08:15 PM
ThePerfectHacker's Avatar
Global Moderator

 
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,751 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default

Quote:
Originally Posted by mathisfun1 View Post
So did you change your signature just for this problem :P?
You mean my Sturm-Louiville signature? No, it is just I sometimes post equations that I love. And sometimes that which I just learned. So for instance, the Sturm-Louiville signature was posted when I was doing boundary value problem from my Partial Differencial Equations book.
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Closed Thread
Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 01:08 AM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.