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proof that $y_1$ or $y_2$ are not a polynomial for any $n$

$$ y_1(x)=1-\frac{n(n+1)}{2!}x^2+\frac{(n-2)n(n+1)(n+3)}{4!}x^4-+\cdots$$

$$ y_2(x)=x-\frac{(n-1)(n+2)}{3!}x^3+\frac{(n-3)(n-1)(n+2)(n+4)}{5!}x^5-+\cdots$$

also can be seen as

$$ y_1(x)=1+\sum_{s=1}^{\infty}(-1)^s \frac{n(n-2)\cdots(n-2s+2)\cdot(n+1)(n+3)\cdots(n+2s-1)}{(2s)!}x^{2s}$$

$$ y_2(x)=x+\sum_{s=1}^{\infty}(-1)^s \frac{(n-1)(n-3)\cdots(n-2s+1)\cdot(n+2)(n+3)\cdots(n+2s)}{(2s+1)!}x^{2s+1}$$

the linear combination of $y_1$ and $y_2$ are solution of the legendre equation.

I've been unable to demonstrate this

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1 Answer 1

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Hint: Let $a_{2s}=(-1)^sn(n-2).\ldots(n-(2s-2))\dfrac{(n+2s-1)!}{n!(2s)!}$ and $a_{2s+1}=(-1)^s(n-1).\ldots(n-(2s-1))\dfrac{(n+2s)!}{n!(2s+1)!}$ so that $$y_1(x)=1+\sum_{s=1}^{+\infty}a_{2s}x^{2s}$$ and $$y_2(x)=x+\sum_{s=1}^{+\infty}a_{2s+1}x^{2s+1}$$ If $n=2s'$ (even), then for all $s$, $a_{2s+1}\neq 0$ ($ (n-1).\ldots(n-(2s-1))\neq 0$) hence $y_2$ is not polynomial.
If $n=2s'+1$ (odd) then for all $s $, $a_{2s}\neq 0$ ($n(n-2).\ldots(n-(2s-2))\neq 0$) hence $y_1$ is not polynomial.

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