Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The norm of the Legendre polynomial, where $P_k(t) = \frac{(2k)!}{2^k(k!)^2}q_k(t), k = 0,1,2,....$, is based on the following $L^2$ inner product:

Note: $q_k(t) = t^k - \sum\limits_{j=0}^{k-1} \frac {\langle t^k,q_j \rangle}{||q_j||^2}q_j(t)$ for $k= 1,2,...$

  • First, prove that $||R_{k,k}||^2 = (-1)^k(2k)! \int^{1}_{-1} (t^2 -1)^kdt$ by a repeated integration by parts, where $R_{j,k}(t) = \frac {d^j}{dt^j}(t^2 -1)^k$ which is a polynomial of degree $2k -j$. In particular, the Rodrigues formula claims that $P_k(t)$ is a multiple of $R_{k,k}(t)$.

  • Second, prove that $\int^{1}_{-1} (t^2 -1)^kdt = (-1)^k \frac{2^{2k+1}(k!)^2}{(2k+1)!}$ by using the change of variables $t = cos\theta$ in the integral.

  • Finally, use Rodrigues' formula to complete the proof and also use Rodrigues' formula to prove $P_k(1) = 1$.

How will I be able to prove this?

share|cite|improve this question
up vote 1 down vote accepted

Recalling the Beta function

$$ \mathrm{B}(x,y)=2\int^{\pi/2}_{0} \sin^{2x-1}(\theta) \cos^{2y-1}(\theta) d\theta = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \,. $$

We will prove the result, $$\int^{1}_{-1} (t^2 -1)^kdt = (-1)^k \frac{2^{2k+1}(k!)^2}{(2k+1)!}\,.$$

$$ \int^{1}_{-1} (t^2 -1)^k dt= 2\int^{1}_{0} (t^2 -1)^k dt = 2(-1)^k\int^{\pi/2}_{0} \cos^{2k+1}(\theta) dt = (-1)^k\frac{\Gamma(k+1)\Gamma(\frac{1}{2})}{\Gamma(k+\frac{3}{2})}$$

$$ = (-1)^k\frac{\sqrt{\pi}\,\Gamma(k+1)}{\Gamma(k+\frac{3}{2})}\,. $$

Now, to relate the two answers, just use the identity

$$k!=\Gamma(k+1),\quad (2k+1)! = \Gamma(2k+2)= \frac{1}{\sqrt{\pi}}2^{2k+1}\Gamma(k+1)\Gamma(k+\frac{3}{2}) $$

share|cite|improve this answer
    
Thank you very much, Mhenni! – diimension Nov 7 '12 at 21:34
1  
@diimension:You are welcome. – Mhenni Benghorbal Nov 8 '12 at 2:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.