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I was reading about Bernoulli polynomials in this article: http://ocw.mit.edu/courses/mathematics/18-100c-analysis-i-spring-2006/projects/silva.pdf 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 ocw.mak.ac.ug so http:/ocw.mak.ac.ug/courses/mathematics/… –  Peter Sheldrick May 10 at 3:54
    
@PeterSheldrick thanks :) –  Lucas Zanella May 10 at 4:55
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1 Answer 1

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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
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