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

We have

$$B_n = n!\sum_r n!\sum_{\lambda} \frac{r^{2n}}{p'_n(r)}$$ frac{\lambda^{2n}}{p'_n(\lambda)}$$ where $r$ \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 $p_n(x):=x^{n+2}t_n(1/x)$ - like 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 $$r_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt ${\lambda}_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$

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

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

$$B_1=1!(\frac{r_{1}^2}{p'_{1}(r_1)}+\frac{r_{2}^2}{p'_{2}(r_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$$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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Please prove, give more symbolic or numeric support (counterexample!?), simplify or drop me a reference (or some vague hunch).

We have

$$B_n = n!\sum_r \frac{r^{2n}}{p_n'(r)}$$ frac{r^{2n}}{p'_n(r)}$$ where $r$ 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.

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

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

Then $p_n(x):=x^{n+2}t_n(1/x)$ - like the reciprocal polynomial 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 $$r_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$

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

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

$$B_1=1!(\frac{r_{1}^2}{p'_{1}(r_1)}+\frac{r_{2}^2}{p'_{2}(r_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$
show/hide this revision's text 2 edited tags; edited body

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

We have

$$B_n = n!\sum_r \frac{r^{2n}}{p_n'(r)}$$ where $r$ 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.

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

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

Then $p_n(x):=x^{n+2}t_n(1/x)$ - like the reciprocal polynomial 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 $$r_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$

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

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

$$B_1=1!(\frac{r_{1}^2}{p'_{1}(r_1)}+\frac{r_{2}^2}{p'_{2}(r_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$
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