# Legendre proofs

• Feb 24th 2008, 12:53 PM
Legendre proofs
I have two Legendre questions, can anyone help me with either of them?
Suppose $\displaystyle 0 \le k < n$. Let p be a polynomial. Show that

$\displaystyle (\frac{d}{dx})^k ((x^2 -1)^n p(x)) = (x^2 -1)^{n-k} q(x)$

for some polynomial q.

Taking p(x) = 1 this implies

$\displaystyle (\frac{d}{dx})^k ((x^2 -1)^n) = (x^2 -1) q(x)$

for some polynomial q.

and the other one is...

Use integration by parts to show that

$\displaystyle (L_n , x^k) = 0$

for all $\displaystyle 0 \le k < n$. Deduce that $\displaystyle (L_0, L_1, ... ,L_n)$ is an orthogonal basis of $\displaystyle P_n$.

$\displaystyle L_n$ may be $\displaystyle L_n(x) = \frac{(2n)!}{{2^n}(n!)^2}x^n +...$ or it may be $\displaystyle L_n(x) = \frac{1}{2^n n!}(\frac{d}{dx})^n ((x^2 - 1)^n)$... Im not sure really which is why im stuck! Both are used in my handout
• Feb 24th 2008, 02:32 PM
mr fantastic
Quote:

I have two Legendre questions, can anyone help me with either of them?
Suppose $\displaystyle 0 \le k < n$. Let p be a polynomial. Show that

$\displaystyle (\frac{d}{dx})^k ((x^2 -1)^n p(x)) = (x^2 -1)^{n-k} q(x)$

for some polynomial q.

[snip]

Are you allowed to use proof by induction? Or do you have to derive (ha ha) the result?
• Feb 24th 2008, 02:40 PM
i thought about that as they had the p(x) = 1 example so thought that would be the base case. I suppose you could yeah, it doesnt specify that you have to derive it.
• Feb 25th 2008, 02:07 AM
mr fantastic
Quote:

I have two Legendre questions, can anyone help me with either of them?
Suppose $\displaystyle 0 \le k < n$. Let p be a polynomial. Show that

$\displaystyle (\frac{d}{dx})^k ((x^2 -1)^n p(x)) = (x^2 -1)^{n-k} q(x)$

for some polynomial q.

Taking p(x) = 1 this implies

$\displaystyle (\frac{d}{dx})^k ((x^2 -1)^n) = (x^2 -1) q(x)$

for some polynomial q.

and the other one is...

Use integration by parts to show that

$\displaystyle (L_n , x^k) = 0$ Mr F says: The integration is over the interval (-1, 1).

for all $\displaystyle 0 \le k < n$. Deduce that $\displaystyle (L_0, L_1, ... ,L_n)$ is an orthogonal basis of $\displaystyle P_n$. Mr F says: On the interval (-1, 1).

$\displaystyle L_n$ may be $\displaystyle L_n(x) = \frac{(2n)!}{{2^n}(n!)^2}x^n +...$ or it may be $\displaystyle L_n(x) = \frac{1}{2^n n!}(\frac{d}{dx})^n ((x^2 - 1)^n)$... Im not sure really which is why im stuck! Both are used in my handout

Mr F says: They are equivalent expressions. The former is the series solution to Legendre's Equation. The latter is Rodrigues' Formula.

..
• Feb 25th 2008, 02:26 AM
right so what im integrating is;

$\displaystyle \int_{-1}^{1} \frac{(2n)!}{2^n(n!)^2} x^n x^k dx$

can the $\displaystyle \frac{(2n)!}{2^n(n!)^2}$ be taken outside the integral as a constant? My guess is no which means im nowhere new. I already knew the integral was between 1 and -1, I just dont know how to integrate that. Do I use integration by parts? Substitution?
• Feb 25th 2008, 02:34 AM
CaptainBlack
Quote:

right so what im integrating is;

$\displaystyle \int_{-1}^{1} \frac{(2n)!}{2^n(n!)^2} x^n x^k dx$

can the $\displaystyle \frac{(2n)!}{2^n(n!)^2}$ be taken outside the integral as a constant? My guess is no which means im nowhere new. I already knew the integral was between 1 and -1, I just dont know how to integrate that. Do I use integration by parts? Substitution?

I'm not sure what you are doing or if it is right but:

$\displaystyle \int_{-1}^{1} \frac{(2n)!}{2^n(n!)^2} x^n x^k dx= \frac{(2n)!}{2^n(n!)^2}\int_{-1}^{1} x^{n+k} dx$

RonL
• Feb 25th 2008, 02:38 AM