Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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

share|cite|improve this question
$\int^{1}_{0}b_{r}(x)dx=0$ does hold for all $r=0..\infty$ ? – k1next Dec 5 '12 at 9:15
Try to put your $b_r$ into an exponential generating function and then analyze how this looks as a function of $x$, i.e. when we are keeping $y$ fixed. It will turn out that it factors as an exponential and a constant (i.e. if we allow $y$ to vary again this means a product of an exponential and a fucntion only of y). To determine the constant/function of $y$ perhaps look at which conditions on $b_r$ still remain to be used. – newguy Dec 5 '12 at 9:21 – user27126 Dec 5 '12 at 9:24
I mean for all $r=1..\infty$ – k1next Dec 5 '12 at 9:26
up vote 1 down vote accepted

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 $$ f_y(x) = \frac{y\exp(xy)}{\exp y- 1}. $$

share|cite|improve this answer
I think we need to some conditions to integrate and take derivatives termwise. – Ash GX Dec 5 '12 at 9:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.