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Given, a series of polynomials $\{{b_r(x)}\}^{\infty}_{r=0}$ on $[0,1]$, such that $b_{0}=1\;\;,b^{'}_{r}=rb_{r-1}(x)\;(r\ge 1)\;,\int^{1}_{0}b_{r}(x)dx=0\;(r\ge 1)$ How can we prove

$\sum^{\infty}_{r=0}b_{r}(x)y^r/r!=\frac{y\mathrm{e}^{xy}}{\mathrm{e}^{y}-1}$

And is it possible to directly calculate (rigidly) this sum without knowing the result?

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    I mean for all $r=1..\infty$2012-12-05

1 Answers 1

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Let $y \in \mathbb R$ and define $f_y\colon [0,1] \to \mathbb R$ by $ f_y(x) := \sum_{r=0}^\infty b_r(x)\frac{y^r}{r!} $ We have, taking derivatives, that for $x \in [0,1]$: \begin{align*} f_y'(x) &= \sum_{r=0}^\infty b_r'(x) \frac{y^r}{r!}\\ &= \sum_{r=1}^\infty rb_{r-1}(x) \frac{y^r}{r!}\\ &= y\cdot \sum_{r=0}^\infty b_r(x) \frac{y^r}{r!}\\ &= y \cdot f_y(x) \end{align*} Hence $f_y(x) = \exp(xy)f_y(0)$. Integrating, we have by uniform convergence \begin{align*} \int_0^1 f_y(x)\, dx &= \sum_{k=0}^\infty \int_0^1 b_r(x)\, dx \cdot \frac{y^r}{r!}\\ &= 1. \end{align*} On the other hand \begin{align*} \int_0^1 f_y(x)\, dx &= \int_0^1 f_y(0)\exp(xy)\, dx\\ &= f_y(0) \cdot \left.\frac{\exp(xy)}y\right|_{x=0}^1\\ &= f_y(0) \cdot \frac{\exp y -1}y \end{align*} So $ 1 = f_y(0) \cdot \frac{\exp y - 1}y \iff f_y(0) = \frac y{\exp y -1} $ This gives $ f_y(x) = \frac{y\exp(xy)}{\exp y- 1}. $

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    I think we need to some conditions to integrate and take derivatives termwise.2012-12-05