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07-25-2007, 08:28 AM
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| | Problem 32 Given a non-negative integer  we define  to be a solution* to:  .
Show that:  .
*)Yes, I realize what I said is not well-defined but it makes no difference. Pick any solution.
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07-27-2007, 04:57 PM
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| | 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). | 
07-27-2007, 06:49 PM
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Originally Posted by mathisfun1 | But Rodrigues' formula is the simplest way to go with it!
-Dan
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07-28-2007, 07:54 PM
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| | There is not need to use the Rodrigues formula, nor any identities that exist among Legendre polynomials.
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Last edited by ThePerfectHacker; 07-29-2007 at 03:31 PM.
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07-29-2007, 06:17 AM
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Originally Posted by ThePerfectHacker 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
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07-29-2007, 03:30 PM
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| | 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),
Thus, ![[(1-x^2)P_n'(x)]'+n(n+1)P_n(x)=0 [(1-x^2)P_n'(x)]'+n(n+1)P_n(x)=0](http://www.mathhelpforum.com/math-help/latex2/img/09749436fbc23be7c427fad817f4b1f3-1.gif)
Multiply by  to get, ![[(1-x^2)P_n'(x)]'P_m(x)+n(n+1)P_n(x)P_m(x)=0 [(1-x^2)P_n'(x)]'P_m(x)+n(n+1)P_n(x)P_m(x)=0](http://www.mathhelpforum.com/math-help/latex2/img/63811478048d2c0aec8d34ce219a0093-1.gif)
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 \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](http://www.mathhelpforum.com/math-help/latex2/img/9cd83d38656e2d701e77def453efbab3-1.gif)
Use integration by part on the first integral, 
The first term vanishes, 
Doing the similar steps with  instead we have, 
Now subtract these two equations to get, ![[n(n+1)-m(m+1)]\int_{-1}^1 P_n(x)P_m(x) dx = 0 [n(n+1)-m(m+1)]\int_{-1}^1 P_n(x)P_m(x) dx = 0](http://www.mathhelpforum.com/math-help/latex2/img/b314287c0786955d1417f59c43b9efa2-1.gif)
Since  and they are integers we have,  .
Q.E.D.
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07-29-2007, 04:35 PM
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| | Here is a way to get the Rodrigues formula since it was mentioned.
Let  .
We see that,
Take the derivative  times to get, 
Take the derivative one last time, ![\left[(1-x^2)\frac{dz}{dx}\right]'+n(n+1)z^n=0 \left[(1-x^2)\frac{dz}{dx}\right]'+n(n+1)z^n=0](http://www.mathhelpforum.com/math-help/latex2/img/8bc0844ad4cdd77b75dc3b65744cac3d-1.gif)
Hence, 
Is a polynomial which solves the equation.
But we write it as, 
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.
We begin by noting that, 
Where  is any peicewise smooth simple closed curve containing  .
This is immediately true by Cauchy's Integral Formula.
Let  define  for  . That is a circle centered at  with radius  .
Evaluating the above integral with this parametrization (and doing some simplification, not shown) we get Laplace's formula.
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08-02-2007, 11:26 AM
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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  are eigenfunctions and thus orthogonal for the  inner product - that is,  ; qed!
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Last edited by Rebesques; 08-02-2007 at 01:24 PM.
Reason: eigenfunctions, more precisely
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08-02-2007, 12:24 PM
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Originally Posted by Rebesques I claim this is cheating!
Since, by identifying a Sturm-Liouville operator here, the  are eigenvectors and thus orthogonal for the  inner product - that is,  ; 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?
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08-02-2007, 01:23 PM
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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.
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