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Suppose that $p$ is a real polynomial of degree $n$. Prove that for $|x|<1$, $$\sum\limits_{m=0}^\infty{p(m)x^m}=h((1-x)^{-1})$$ for some real polynomial $h$ of degree $n+1$ without the constant term.

I have started trying by taking a specific expression of the polynomial $p$, but I have completely lost. I understand that this result generalizes the following identity $$(1-x)^{-1}=\sum\limits_{m=0}^\infty{x^m}$$ for $|x|<1$ but couldn't anymore rather thinking about it. Help is highly appreciated.

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  • $\begingroup$ @ Michael Hardy: edited! $\endgroup$
    – user149418
    May 30 '15 at 18:15
  • $\begingroup$ Consider the derivatives of $(1-x)^{-1}$ and the derivatives of the power series. $\endgroup$
    – jgon
    May 30 '15 at 18:21
  • $\begingroup$ @ jgon Yes, this can be done. I did this for a polynomial of degree 2 or 3. But this seems to be more complicated for polynomials of arbitrary degree. Is there any other way of solving it? $\endgroup$
    – user149418
    May 30 '15 at 18:58
  • $\begingroup$ you shouldn't need to find it explicitly if you can show the left hand side's derivatives form a basis for the space $\endgroup$
    – jgon
    May 30 '15 at 19:07
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Consider the polynomials $$P_0:=1,P_1:=X, P_2:=X(X-1),\dots,P_l:=X(X-1)\dots (X-l+1).$$ Then we may express $p$ as $\sum_{j=0}^nc_jP_j$ for some coefficients $c_j$. Since for $|x|\lt 1$, we have $$\sum_{m=0}^{+\infty}P_j(m)x^m=\frac{d^j}{dx^j}\left(\frac 1{1-x} \right) $$ and the later can be expressed as a constant times $(1-x)^{-j-1}$, we get the wanted polynomial $h$ by summing over $j$.

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    $\begingroup$ Isn't $P_k(x)=(x+1)\dotsm (x+k)$ for $k\ge 1$ ? $\endgroup$
    – user149418
    May 31 '15 at 8:25
  • $\begingroup$ No, the minus signs are important here otherwise the displayed equality would not hold. $\endgroup$ May 31 '15 at 8:48
  • $\begingroup$ Nice answer !!! $\endgroup$
    – mick
    May 24 at 15:24

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