Given a non-negative integer we define to be a solution* to:

.

Show that:

.

*)Yes, I realize what I said is notwell-definedbut it makes no difference. Pickanysolution.

Printable View

- Jul 25th 2007, 06:28 AMThePerfectHackerProblem 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. - Jul 27th 2007, 02:57 PMmathisfun1
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).

- Jul 27th 2007, 04:49 PMtopsquark
- Jul 28th 2007, 05:54 PMThePerfectHacker
There is not need to use the Rodrigues formula, nor any identities that exist among Legendre polynomials.

- Jul 29th 2007, 04:17 AMtopsquark
- Jul 29th 2007, 01:30 PMThePerfectHacker
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,

Multiply by to get,

And integrate,

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,

Since and they are integers we have,

.

Q.E.D. - Jul 29th 2007, 02:35 PMThePerfectHacker
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,

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. - Aug 2nd 2007, 09:26 AMRebesquesQuote:

The trick is to write the differencial equation as (in Sturm-Louiville form)

Since, by identifying a Sturm-Liouville operator here, the are eigenfunctions and thus orthogonal for the inner product - that is, ; qed! - Aug 2nd 2007, 10:24 AMThePerfectHacker
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? - Aug 2nd 2007, 11:23 AMRebesquesQuote:

Change "eigenvectors" to "eigenfunctions" to fix your post

Quote:

this is not a boundary value problem. Thus, Sturm-Louiville Theory does not apply

- Aug 2nd 2007, 06:09 PMmathisfun1
- Aug 2nd 2007, 06:15 PMThePerfectHacker
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.