User peter sheldrick - MathOverflow

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

2012-12-07T08:24:02Z

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

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

Answer by Peter Sheldrick for Why might André Weil have named Carl Ludwig Siegel the greatest mathematician of the 20th century?

2012-11-06T19:48:06Z

Well unfortunatley i don't have enough points on Math Overflow to make a comment. This is in response to Pete L. Clark's answer, especially the line

There is also something "organic" in the work of both Siegel and Shimura which naturally bristles a bit at the "Bourbakistic" influence of the French school(...)

at the start of the last paragraph.

Here is a nice little tidbit from the German Wikipedia http://de.wikipedia.org/wiki/Carl_Ludwig_Siegel#Siegels_Standpunkt_zur_Entwicklung_der_Mathematik

To finish here is my own translation of that:

The position of Siegel towards the development of math

Like no other mathematician of the 20th century Siegel was critical of the increasing abstraction and axiomatization of math. The bourbaki-project was for him the pinnacle of this "catastrophic development". For him the clarity of Gauss and Lagrange were still aspirations, just as the investigation of concrete mathematical objects.

Disclaimer: this is a best-effort translation of that Wikipedia bit from myself and it does not reflect my opinion.

EDIT: here is the reference for that Wikipedia bit in english Letter from Siegell to Louis J. Mordell, 3 March 1964

Comment by Peter Sheldrick

2012-12-09T05:33:17Z

To expand a bit more on the connection to polynomial interpolation: To do polynomial interpolation the Vandermonde matrix is used and one method to invert this matrix is to use the derivatives of the characteristic polynomial see [last formula in](de.wikipedia.org/wiki/…)

Comment by Peter Sheldrick

2012-12-09T04:27:47Z

... formula $B_1=-\frac{1}{2}$ but $-1\zeta(1-1)=\frac{1}{2}$. Otherwise the approximation is good for big $n$.

Comment by Peter Sheldrick

2012-12-09T04:26:34Z

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

Comment by Peter Sheldrick

2012-12-07T21:52:56Z

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 $q_n(x)=B_n(x)/n!$?

Comment by Peter Sheldrick

2012-12-07T09:06:32Z

This question is (cross-posted)[math.stackexchange.com/questions/252928/… with math.stackexchange.

Comment by Peter Sheldrick

2012-06-21T01:39:44Z

In fact for even $k$ both $c_k=0.25$ and $c_k=0.75$ work better and better (zeros of cos(2*pi*x) in [0,1]). Whereas as mentioned $0,0.5$ and $1$ for right away for odd $k$ (zeros of sin(2*pi*x)).

Comment by Peter Sheldrick

2012-06-21T01:27:44Z

That explains why $c_k=0.5$ fits well for odd $k$ - but $0$ or $1$ fit just as well. Since odd Bernoulli polynomials always have three roots: $0,0.5$ and $1$ (expect for $k=1$)...

Comment by Peter Sheldrick

2012-06-21T01:23:35Z

@Will, yes it is thanks. Unfortunate that i missed that. So if we integrate $k\cdot \int_r^1 B_k(t)\textrm{d}t$ where $r$ is a root of $B_{k+1}(x)$, we get $B_{k+1}(x)$... That's a weird property. Then a formula for the $c_k$ isn't far off - since the Bernoulli polynomials converge against $\sin(x)$ and $\cos(x)$ so the roots probably behave similarly to the roots of the taylor polynomials of those...