Can we write $B_n(x)-B_n(0)=nx^{n-1}$?

Question

Bernouli polynomials satisfies the relation $B_n(x+1)-B_n(x)=nx^{n-1}$.

Can we write $B_n(x)-B_n(0)=nx^{n-1}$ or something like $B_{n+1}(x)-B_n(0)$ to be equal to $nx^{n-1}$?

I mean I want to change the subscripts to be $B_{n+1}$ and $B_n$ instead of keeping both subscripts to be $B_n$.

No, $B_n(x)$ is certainly not a constant plus $nx^{n-1}$. – darij grinberg Oct 24 '19 at 20:39 — darij grinberg, Oct 24 '19 at 20:39

Cardboard Box · Answer 1 · 2019-10-24T19:01:00.740

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Well, $B_2(x) = x^2 - x - 1/6$, so $B_2(x) - B_2(0) = x^2 - x \neq 2x$.

Maybe you're looking for something like this: $$\sum_{i=0}^x i^n = \frac{B_{n+1}(x+1) - B_{n+1}(0)}{n+1}.$$

edited Oct 24 '19 at 19:01

answered Oct 24 '19 at 18:41

Cardboard Box

Can you check the updated question? – MAS Oct 24 '19 at 18:52
Sure, I'll take a look – Cardboard Box Oct 24 '19 at 18:54
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Do you really intend to ask about something of the form f(x) - g(0)? If you take a polynomial with many nonzero coefficients, like $B_{n+1}(x)$ and subtract a constant, it will still have many nonzero coefficients. – Cardboard Box Oct 24 '19 at 18:57
I've updated my answer to include something that may be similar to what you're looking for. – Cardboard Box Oct 24 '19 at 19:02
Thanks but I was looking something like $B_m(x)-B_n(0)=f(x^n)$, where $m, \ n$ are different positive integers – MAS Oct 24 '19 at 19:02
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Maybe there is something like that, but I don't know of it. Good luck! – Cardboard Box Oct 24 '19 at 19:03
ok, many thanks for your help – MAS Oct 24 '19 at 19:04

1 Answers1