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## Bernoulli number formula involving roots of taylor polynomial of $\exp-1$

Please prove, give more symbolic or numeric support (counterexample!?), simplify or drop me a reference (or some vague hunch).

We have

$$B_n = n!\sum_{\lambda} \frac{\lambda^{2n}}{p'_n(\lambda)}$$ where $\lambda$ ranges over the roots of the polynomial $p_n$ and $p_n'$ is the derivative of $p_n$. $B_n$ is the $n$-th bernoulli number.

The polynomial $p_n$ is defined as follows.

$$p_n(x):=x^{n+2}t_n(1/x)$$

where

$$t_n(x) = \sum_{k=0}^{n+2} \frac{x^k}{k!} -1$$

is the truncated taylor polynomial of $\exp-1$ to power $n+2$.

Then $p_n$ is the reciprocal polynomial of $t_n$ just without conjugation (though it probably makes no difference if there is conjugation or not - since $p$ is a polynomial with real coefficients and the roots come in conjugate pairs).

Examples (already tested for $n=0..18$ with a symbolic solver and for $n=0..62$ numerically):

$n=0$

$$t_0(x)=\frac{x^2}{2}+x$$

$$p_0(x)=x^2t(1/x)=x^2(\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{2}+x$$

root of $p_0$ is $-1/2$.

$$p_0'(x)=1$$

$$B_0=0!\cdot \frac{(-1/2)^{2\cdot 0}}{1}=1$$

$n=1$

$$t_1(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x=\frac{x^3}{6}+\frac{x^2}{2}+x$$ $$p_1(x)=x^3t_1(1/x)=x^3(\frac{1}{6x^3}+\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{6}+\frac{x}{2}+x^2$$

Roots of $p_1$ are $${\lambda}_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$

We have $$\lambda_{1,2}^2=-\frac{1}{24}\frac{-}{+}\frac{\sqrt{15}i}{24}$$

$$p_1'(x)=2x+\frac{1}{2}$$so $$\frac{1}{p_1'(\lambda_{1,2})}=\frac{-}{+}\frac{2 \sqrt{15}i}{5}$$ $$\frac{\lambda_{1,2}^2}{p_{1}'(\lambda_{1,2})}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt{15}i}{60}$$

$$B_1=1!(\frac{\lambda_{1}^2}{p'_{1}(\lambda_1)}+\frac{\lambda_{2}^2}{p'_{2}(\lambda_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$

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This question is (cross-posted)[math.stackexchange.com/questions/252928/… with math.stackexchange. – Peter Sheldrick Dec 7 at 9:06

Start by the definition of the Bernoulli numbers via their generating function: we have $$\big(e^x-1\big)\sum_{k=0}^\infty B _ k\frac{x^k}{k!}=x\ .$$ Therefore, writing with your notation $t _ n(x)= \sum_{k=1}^{n+2}\frac{x^k}{k!}$,

$$t _ n (x) \sum _ {k=0}^{n } B _ k \frac {x ^ k}{k!} =x + x^{n+1}S _ n(x) + \frac {B _ n}{n!(n+2)!}x ^ {2n+2}\ ,$$ for some polynomial $S _ n (x)$ of degree not larger than $n$. Passing to the reciprocal polynomials we obtain an equation in form of a division by $p_n$ with remainder:

$$x^{2n+1}=p _ n (x) q _ n(x) + r _ n(x)$$

for a polynomial $q _ n(x)$ of degree $n$ ( precisely, it's $q _ n(x):= B_{n }(x)/n!$) and a polynomial $r _ n(x)$ with $\mathrm{deg\ } r \le n+1 < \mathrm{deg\ } p_n$ and $r _ n (0) = - \frac{B _ n}{n!(n+2)!}$ .

Finally, let's write the partial fraction decomposition of the (proper) fraction $\frac{r _n(x)}{p_n(x)}$. We have, summing over the $n+2$ (simple) roots of $p _ n(x)$, by a general elementary identity

$$\frac{r_n(x)}{p _ n (x)}=\sum _ \lambda \frac{r _ n(\lambda )}{p' _ n (\lambda ) }\frac{1}{x-\lambda }\ .$$

Computing the identity at $x=0$ we finally find your identity:

$$B _n =n!\sum _ \lambda \frac{ \lambda^ {2n} }{p' _ n (\lambda ) } .$$

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Hi, thanks for your work! This doesn't effect the proof, but if $q_n(x)=B_n(x)$ then $q_n(0)=B_n(0)=B_n$. It is $p(0)=1/(n+2)!$ so $p_n(0)q_n(0)+r_n(0)$=$B_n/(n+2)!-B_n/(n!(n+2)!)\neq 0$. So is $q_n(x)=B_n(x)/n!$? – Peter Sheldrick Dec 7 at 21:52
Yes! Thank you for checking. Indeed it's $q_n(x):=\sum_{k=0}^n \frac{B_k}{k!}x^{n-k}=B_n(x)/n!$. – Pietro Majer Dec 7 at 22:35
Note that, if we consider, more generally, truncations of the series of degree $n+2$ and $m$, we can evaluate as well sums over the roots of $p _ n$ of the form $$\sum_ { \lambda } \frac{ \lambda^ {r} }{p' _ n ( \lambda ) }\\ .$$ These vanish for small values of $r$. For $r$ between $n$ and $2n$, one gets a similar expression --some $B _ m/m!$. However, for even larger $r$ one gets less simple expressions (linear cominations of $B _ m$'s). – Pietro Majer Dec 8 at 17:28
We have $$B_m = m!\sum_{\lambda} \frac{\lambda^{m}\lambda^{n}}{p'_n(\lambda)}$$ for $0\leq m \leq n$. Is that what your comment is going towards? I also tested this for real $m$ in the interval $[2,n]$ and then it approximates $-m\zeta(1-m)$ for $m << n$ (if you replace $m!$ with $\Gamma(m+1)$). One angle for me to prove the orginal formula was to derive it from a straight up interpolation problem for $-m\zeta(1-m)$, but i didn't have much success in that direction yet (and now that there is your proof that is maybe unnecessary). Around $m=1$ the approximation fails completely since in this fo – Peter Sheldrick Dec 9 at 4:26
... formula $B_1=-\frac{1}{2}$ but $-1\zeta(1-1)=\frac{1}{2}$. Otherwise the approximation is good for big $n$. – Peter Sheldrick Dec 9 at 4:27
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