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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?

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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

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

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Thank you very much, Mhenni! – diimension Nov 7 '12 at 21:34
@diimension:You are welcome. – Mhenni Benghorbal Nov 8 '12 at 2:29

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