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I have pored over the internet and can't seem to find any proof of a particular property of Bernoulli Polynomials. The property is:

$P'_{n}(x) = nP_{n-1}(x)$

I have not found anything conclusive in any of my textbooks that lay out a clear proof or anything on the internet. Any help or resources as to what the proof might be would be greatly appreciated!

Somos
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djohnson
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  • This is well-known and comes from the explicit formula: https://en.wikipedia.org/wiki/Bernoulli_polynomials#Explicit_formula . Also there one finds an integral formula https://en.wikipedia.org/wiki/Bernoulli_polynomials#Integrals which is pretty much the same as the derivative formula. – Angina Seng Nov 27 '18 at 21:36
  • Of course, any proof would require a definition of Bernoulli polynomials. Which one do you want to use? – Somos Nov 27 '18 at 21:40
  • See Appell Sequence – Eleven-Eleven Nov 27 '18 at 21:45

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From the definition of bernoulli polynomials, we have

$$B_n(x)=\sum_{k=0}^n\binom{n}{k}B_{n-k}x^k$$

where $B_k$ are the traditional Bernoulli numbers. Differentiate both sides to get

\begin{eqnarray*}B'_n(x)&=&\frac{d}{dx}\sum_{k=0}^n\binom{n}{k}B_{n-k}x^k\\&=&\sum_{k=1}^n\binom{n}{k}kB_{n-k}x^{k-1}\\&=&\sum_{k=0}^{n-1}\binom{n}{k+1}(k+1)B_{n-k+1}x^k\\&=&n\sum_{k=0}^{n-1}\binom{n-1}{k}B_{n-1-k}x^k\\&=&nB_{n-1}(x)\end{eqnarray*}

Where we used the combinatorial identity

$$(k+1)\binom{n}{k+1}=n\binom{n-1}{k}$$

Eleven-Eleven
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