# Bernoulli Polynomials

I am having a problem with this question. Can someone help me please.

We are defining a sequence of polynomials such that:

$P_0(x)=1; P_n'(x)=nP_{n-1}(x) \mbox{ and} \int_{0}^1P_n(x)dx=0$

I need to prove, by induction, that $P_n(x)$ is a polynomial in $x$ of degree $n$, the term of highest degree being $x^n$.

Suppose $P_{n+1}$ has leading term $ax^r$ for some $a$ and $r$. The defining equation and the induction hypothesis tell you what about $a$ and $r$? –  Gerry Myerson Nov 8 '12 at 2:17
Recall that $\displaystyle \int x^n dx = \dfrac{x^{n+1}}{n+1}$. Hence, if $P_n(x)$ is a polynomial of degree $n$, then it is of the form $$P_n(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$ Since $P_{n+1}'(x) = (n+1) P_n(x)$, we have that $$P_{n+1}(x) = \int_{0}^x (n+1) P_n(y) dy + c$$ Hence, $$P_{n+1}(x) = \int_{0}^x (n+1) \left(a_n y^n + a_{n-1} y^{n-1} + \cdots + a_1 y + a_0\right) dy + c\\ = a_n x^{n+1} + a_{n-1} \left(\dfrac{n+1}n\right) x^n + \cdots + a_{1} \left(\dfrac{n+1}2\right) x^2 + a_{0} \left(\dfrac{n+1}1\right) x + c$$ Now finish it off with induction.