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At this link: http://www.phy.ohio.edu/~phillips/Mathmethods/Notes/Chapter8.pdf

The author writes that one can prove from the generating function of Legendre polynomials that $P_{2n}$ are all even and $P_{2n+1}$ are odd, by generalizing the argument for finding $P_n(-1)$.

I tried doing that, but can't quite see how the generalization works.

Of course we should start from considering $(1+2xt+t^2)^{-1/2} = \sum_{n=0}^\infty P_n(-x)t^n$. The argument given for $P_n(-1)$ involved simplifying to a form with a recognizable Taylor expansion.

I would guess that we should write this in something like the form

$$ (1\pm z^2)^{-1/2} $$

If we choose $(1-z^2)^{-1/2}$ this has expansion $\sum\binom{-1/2}{n}1^{-1/2-n}(-z^2)^n$. Looks promising, so we try to wrangle the given generating function into that form. It is

$$(1-x^2+x^2+2xt + t^2)^{-1/2} = ([1-x^2]-[-(x+t)^2])^{-1/2} $$

$$ = (1-x^2)^{-1/2}\left(1-\left[\frac{-(x+t)^2}{1-x^2}\right]\right)^{-1/2} $$

$$ = (1-x^2)^{-1/2}\left(1-\left[-\left(\frac{x+t}{\sqrt{1-x^2}}\right)^2\right]\right)^{-1/2} $$


So (1) this just feels wrong. It feels way too over-complicated and drifting off course. (2) I'm not sure what to do with that factor of $(1-x^2)^{-1/2}$. (3) Even if I could get all the i's dotted and t's crossed, this looks like both $x$ and $t$ will, when expanded as a Maclauren series, be under the exponent of $n$. So then what do you do, when you should only have $t^n$ and then a coefficient function of $x$?

I feel that this simply must not be the way to go, but don't see an alternative.

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  • $\begingroup$ Hint: By considering $[(1-2xt+t^2)(1+2xt+t^2)]^{-1/2}$, prove that $\sum_{k=0}^n P_{n-k}(x)P_k(-x)$ is even (in $x$). Now use $P_0=1$ and induct. $\endgroup$ Sep 13, 2023 at 3:10

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We consider according to the referenced paper the bivariate function \begin{align*} g(t,x)=\sum_{n=0}^{\infty}P_n(x)t^n=\frac{1}{\sqrt{1-2xt+t^2}}\tag{1} \end{align*} Here we use $g(t,x)$ to show that even indexed Legendre polynomials $P_{2n}$ are even.

We obtain from (1) \begin{align*} \color{blue}{g(t,x)+g(-t,x)} &=\sum_{n=0}^{\infty}P_n(x)t^n+\sum_{n=0}^{\infty}P_n(x)(-t)^n\\ &\,\,\color{blue}{=2\sum_{n=0}^{\infty}P_{2n}(x)t^{2n}}\tag{2} \end{align*} It follows from (1) and (2) \begin{align*} \color{blue}{2\sum_{n=0}^{\infty}P_{2n}(x)t^n}\tag{3} &=g(t,x)+g(-t,x)\\ &=\frac{1}{\sqrt{1-2xt+t^2}}+\frac{1}{\sqrt{1+2xt+t^2}}\\ &=g(-t,-x)+g(t,-x)\\ &\,\,\color{blue}{=2\sum_{n=0}^{\infty}P_{2n}(-x)t^{2n}}\tag{4} \end{align*} and the claim that $P_{2n}$ are even follows by comparing coefficients of $t$ in (3) and (4).

A similar calculation could be done to show that odd indexed Legendre polynomials $P_{2n+1}$ are odd.

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  • $\begingroup$ Small question: Shouldn't (2) have a factor of 2 on it? I assume the logic is that the odd powers of $t$ cancel, so shouldn't the even powers combine? $\endgroup$
    – Addem
    Sep 14, 2023 at 15:05
  • $\begingroup$ @Addem: Yes, many thanks - mistake corrected. $\endgroup$ Sep 14, 2023 at 15:11

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