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Does any one know, how to compute any of those two things?

  1. The relationship between Legendre polynomials and Shifted Legendre Polynomials.

  2. $\displaystyle\int_{-1}^1P_n^2(x)dx=\dfrac{2}{(2n+1)}$ for $n\geq0$.

I tried to use Bonnet's equation:

$(2n-1)xP_{n-1}(x)=nP_n(x)+(n-1)P_{n-2}(x)$

but I couldn't move. Thanks :)

Edit: The second problem refers to regular Legendre Polynomials.

Community
  • 1
Adam
  • 470

2 Answers2

3

I will answer your question about determining the value of $\int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula $$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$

$\newcommand{\partial}[1]{\left[#1\right]}$ $\newcommand{\bracket}[1]{\left(#1\right)}$ \begin{equation} \begin{split} I&=\int_{-1}^1 P_n(x)^2dx \\ &=\frac{1}{2^{2n}n!^2}\int_{-1}^1 [(x^2-1)^n]^{(n)} \cdot [(x^2-1)^n]^{(n)}dx \\ &=\frac{1}{2^{2n}n!^2} \bracket{\partial{[(x^2-1)^n]^{(n)}\cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-\int_{-1}^1 [(x^2-1)^n]^{(n+1)}\cdot [(x^2-1)^n]^{(n-1)}dx} \\ &=\frac{1}{2^{2n}n!^2} \bracket{0-\int_{-1}^1 [(x^2-1)^n]^{(n+1)}\cdot [(x^2-1)^n]^{(n-1)}dx} \\ &=... \\ &=\frac{(-1)^n}{2^{2n}n!^2} \int_{-1}^1 [(x^2-1)^n]^{(2n)}\cdot [(x^2-1)^n]^{(0)}dx \\ &=\frac{(-1)^n}{2^{2n}n!^2} \cdot (2n)! \cdot \int_{-1}^1 (x^2-1)^ndx \end{split} \end{equation}

Now

\begin{equation} \begin{split} \int_{-1}^1 (x^2-1)^ndx&=\int_{-1}^1 (x+1)^n(x-1)^ndx \\ &=\bracket{\partial{\frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - \int_{-1}^1 \frac{(x+1)^{n+1}}{n+1}\cdot n(x-1)^{n-1}dx} \\ &=\bracket{0 - \int_{-1}^1 \frac{(x+1)^{n+1}}{n+1}\cdot n(x-1)^{n-1}dx} \\ &=... \\ &=(-1)^n\int_{-1}^1 \frac{n!\cdot (x+1)^{2n}}{(2n)!}\cdot n!(x-1)^0dx \\ &=(-1)^n\frac{n!^2}{(2n)!}\int_{-1}^1 (x+1)^{2n}dx \\ &=(-1)^n\frac{n!^2}{(2n)!}\partial{\frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1} \\ &=(-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1} \end{split} \end{equation}

So finally we get our desired value:

\begin{equation} \begin{split} I&=\int_{-1}^1 P_n(x)^2dx \\ &=\frac{(-1)^n}{2^{2n}n!^2} \cdot (2n)! \cdot (-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1} \\ &=\frac{2}{2n+1} \end{split} \end{equation}

Q.E.D.

Maestro13
  • 1,960
1

Hints:

Part 1:

See Shifted Legendre Polynomials.

I am not exactly sure what you intend to do for part 1., since it is not clear from your question. Maybe you can clarify.

Look at the DLMF and what do you notice about the Legnedre versus SHifted Legendre. So, if you can prove one of them, do you see an approach to deriving the other?

Part 2:

Try evaluating the integral using Rodrigues' Formula.

Amzoti
  • 56,093
  • Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:) – Adam Mar 19 '13 at 20:03
  • See update - no more comments for a while – Amzoti Mar 19 '13 at 20:22
  • In the part 2 I'm supposed (forced) to use Bonnet's equation. – Adam Mar 19 '13 at 20:43
  • Then, you might be interested in the answers here: http://math.stackexchange.com/questions/205932/problems-regarding-integrals-involving-legendre-polynomials – Amzoti Mar 19 '13 at 21:07
  • How did this go unvoted until now? +1 – amWhy Apr 17 '13 at 00:54
  • Don't I know it! – amWhy Apr 17 '13 at 00:58
  • @amWhy: Yes, I have seen many of your nice answers get no or very low votes- but I'm working on that! :-) – Amzoti Apr 17 '13 at 01:00