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I am trying to understand Bernoulli polynomials, and so I came across this abomination in the article: $$ B_n(x) = \frac{D}{e^D-1} x^n $$ where $D$ is the differentiation operator and the fraction is "expanded out as a formal power series"

Since we have $$\frac{t}{e^t-1} = \sum_{n=0}^\infty B_n \frac{t^n}{n!}$$ Does this imply that I should interpret the fraction as $$ \frac{D}{e^D-1} = \sum_{n=0}^\infty B_n \frac{D^n}{n!} $$

gist076923
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  • Yes, see (holomorphic) functional calculus – reuns Dec 22 '22 at 19:55
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    Yes. This sort of "abomination" is common for many purposes. – anon Dec 22 '22 at 19:56
  • First time I saw the fraction of doom was a few months ago, and now today I finally understand it. Thanks all – gist076923 Dec 22 '22 at 20:03
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    This abomination is also used to "derive" the Euler-Maclaurin summation formula. See for example Spivak - Calculus 3rd edition, problems in the chapter on complex power series. – Tob Ernack Dec 22 '22 at 20:30
  • I just learned about Euler-Maclaurin the other day, and I was wondering how that was proved. Now I will look into that. It's all coming full circle – gist076923 Dec 22 '22 at 20:33

2 Answers2

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I'm so silly $$ \begin{align*} &\frac{D}{e^D-1}x^n \\ &= \sum_{k=0}^\infty B_k \frac{D^k(x^n)}{k!} \\ &= \sum_{k=0}^n B_k \frac{(n)_k}{k!} x^{n-k} \\ &= \sum_{k=0}^n \binom{n}{k} B_k x^{n-k} \\ &= \sum_{k=0}^n \binom{n}{k} B_{n-k} x^k \\ &= B_n(x) \end{align*} $$

gist076923
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We are looking for solutions to

$$B_n(x+1) - B_n(x) = n x^{n-1}$$

that is

$$(e^D -1) B_n(x) = D x^n$$

So take

$$B_n(x) = \frac{D}{e^D-1} x^n$$

Note: $D$ is the derivative. $B(x+1) = e^D B\,(x)$ from Taylor's formula.

$\bf{Added:}$

The Euler polynomials, satisfying

$$\frac{1}{2}( E_n(x+1) + E_n(x))= x^n$$

are given by

$$E_n(x) = \frac{2}{e^D+1} x^n$$

They are also useful in calculating alternating sums

$$\sum_{k=0}^{m-1} (-1)^k (x+k)^n$$

orangeskid
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  • That is a neat proof. How do you justify the division of $e^D-1$? – gist076923 Dec 22 '22 at 22:17
  • The division $\frac{D}{e^D-1}$ is formal, but works when RHS is a polynomial. There is something similar with Euler polynomials. Learned this from a basic book, probably Norlund has it. – orangeskid Dec 22 '22 at 22:37