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Show that $$\Sigma_{r=0}^{n}(2r+1)P_r(x)=P_{n+1}'(x)+P_n'(x).$$

I've started with the relation $$P_l'(x)=\Sigma_{r=0}^{\frac{1}{2}(l-1)}(2l-4r-1)P_{l-2r-1}(x).$$ I tried adding the sums for $P_n'(x)$ and $P_{n+1}'(x)$ to show that this is equal to the first expression, but that didn't get me anywhere. Any ideas?

J. Hunt
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  • That can be proved also by exploiting the fact that Legendre polynomials give an orthogonal base of $(-1,1)$ with respect to the standard inner product. It is enough to compute $$\int_{-1}^{1} P_m'(x)P_r(x),dx$$ for $m\in{n,n+1}$ and $r\in{0,\ldots,n}$. – Jack D'Aurizio Apr 26 '17 at 13:52

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Even simpler: Bonnet's recursion formula gives

$$ (2n+1) P_n(x) = P_{n+1}'(x)-P_{n-1}'(x) $$ hence the given sum is a telescopic sum.

Jack D'Aurizio
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