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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 $$B_n = n!\sum_{\lambda} \frac{\lambda^{2n}}{p'_n(\lambda)}$$ 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 '12 at 9:06

1 Answer 1

up vote 10 down vote accepted

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 '12 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 '12 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 '12 at 17:28

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