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I was reading about Bernoulli polynomials in this article: and I saw this property:

$$B_n(1−x) = (−1)^nB_n(x)$$

But the only proof the article gives is by induction. I was interested about the intuition and reason of this property. Like, how it was discovered, or why this is useful.

And about the definition of a bernoulli polynomial: $$(a)B_0(x) = 1;$$ $$(b)B_n'(x) =B_{n−1}(x);$$ $$(c)\int_0^1 B_n(x)dx= 0 \tag{for n>1}$$

How they became with this definition? I mean, how they felt a necessity of defining these rules to something they would call a "Bernoulli polynomial". I'm not satisfied with induction proofs, I want to know about the intuition.

Thanks by any kind of help!

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the new link is so http:/… – Peter Sheldrick May 10 '14 at 3:54
@PeterSheldrick thanks :) – Lucas Zanella May 10 '14 at 4:55
up vote 0 down vote accepted

Essentially, it relies on the fact that the derivative of a nice odd functions is even and derivative of nice even functions is odd. The Bernoulli polynomials are odd and even alternatingly about $x=\dfrac12$ because of the above reason, and to start of with $B_0(x) = 1$ is an even function about $x=\dfrac12$. Hence, $B_1(x)$ is odd about $x=\dfrac12$, $B_2(x)$ is even about $x= \dfrac12$ and in general $B_n(x)$ is odd or even about $x= \dfrac12$ depending on whether $n$ is odd or even.

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what do you mean by "odd about x=1/2"? Thanks – Lucas Zanella Apr 11 '13 at 0:07
@LucasZanella Essentially $f(1/2-x) = -f(x-1/2)$ – user17762 Apr 11 '13 at 0:10

Let us go in order of your questions:

  1. Derivation of the first equation.

Here is proof without using induction. The Bernoulli polynomials could be found by expanding the generating function as follows:

\begin{equation} G(x,z) = \frac{z e^{z x}}{e^z -1 } = \sum_{i=0}^{\infty} B_i(x) \frac{z^i}{i!} \end{equation}

We verify that $G(1-x,-z)= G(x,z)$ and by matching coefficients the result we want.

\begin{equation} G(1-x, -z) = \frac{(-z) e^{-z(1-x)}}{e^{-z} -1} = \frac{z e^{z x}}{e^z-1}= G(x,z). \end{equation}

Now choose the $n$ power of $z$, and expand $G(1-x,-z)$. This shows that \begin{equation} (-1)^n B_n(1-x) = B_n(x) \end{equation} or, what is the same: \begin{equation} B_n(1-x) = (-1)^n B_n(x) \end{equation}

  1. That is not the only definition. I suggest you to look in the the Wikipedia site for other definitions.

The numbers were born when Bernoulli was trying to find the solution to the problem

\begin{equation} \sigma_k(n) = \sum_{i=1}^{n-1} i^k \end{equation}

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