User peter sheldrick - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:49:12Z http://mathoverflow.net/feeds/user/24400 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115691/bernoulli-number-formula-involving-roots-of-taylor-polynomial-of-exp-1 Bernoulli number formula involving roots of taylor polynomial of $\exp-1$ Peter Sheldrick 2012-12-07T08:24:02Z 2012-12-08T06:12:45Z <p>Please prove, give more symbolic or numeric support (counterexample!?), simplify or drop me a reference (or some vague hunch).</p> <p>We have </p> <p>$$B_n = n!\sum_{\lambda} \frac{\lambda^{2n}}{p'_n(\lambda)}$$ 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.</p> <p>The polynomial $p_n$ is defined as follows. </p> <p>$$p_n(x):=x^{n+2}t_n(1/x)$$</p> <p>where</p> <p>$$t_n(x) = \sum_{k=0}^{n+2} \frac{x^k}{k!} -1$$</p> <p>is the truncated taylor polynomial of $\exp-1$ to power $n+2$.</p> <p>Then $p_n$ is the <a href="http://en.wikipedia.org/wiki/Reciprocal_polynomial" rel="nofollow">reciprocal polynomial</a> 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).</p> <p>Examples (already tested for $n=0..18$ with a symbolic solver and for $n=0..62$ numerically):</p> <p>$n=0$</p> <p>$$t_0(x)=\frac{x^2}{2}+x$$</p> <p>$$p_0(x)=x^2t(1/x)=x^2(\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{2}+x$$</p> <p>root of $p_0$ is $-1/2$.</p> <p>$$p_0'(x)=1$$</p> <p>$$B_0=0!\cdot \frac{(-1/2)^{2\cdot 0}}{1}=1$$</p> <p>$n=1$</p> <p>$$t_1(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x=\frac{x^3}{6}+\frac{x^2}{2}+x$$ $$p_1(x)=x^3t_1(1/x)=x^3(\frac{1}{6x^3}+\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{6}+\frac{x}{2}+x^2$$</p> <p>Roots of $p_1$ are $${\lambda}_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$</p> <p>We have $$\lambda_{1,2}^2=-\frac{1}{24}\frac{-}{+}\frac{\sqrt{15}i}{24}$$</p> <p>$$p_1'(x)=2x+\frac{1}{2}$$so $$\frac{1}{p_1'(\lambda_{1,2})}=\frac{-}{+}\frac{2 \sqrt{15}i}{5}$$ $$\frac{\lambda_{1,2}^2}{p_{1}'(\lambda_{1,2})}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt{15}i}{60}$$</p> <p>$$B_1=1!(\frac{\lambda_{1}^2}{p'_{1}(\lambda_1)}+\frac{\lambda_{2}^2}{p'_{2}(\lambda_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$</p> http://mathoverflow.net/questions/111519/why-might-andre-weil-have-named-carl-ludwig-siegel-the-greatest-mathematician-of/111674#111674 Answer by Peter Sheldrick for Why might André Weil have named Carl Ludwig Siegel the greatest mathematician of the 20th century? Peter Sheldrick 2012-11-06T19:48:06Z 2012-11-06T21:52:52Z <p>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 </p> <p><em>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(...)</em> </p> <p>at the start of the last paragraph. </p> <p>Here is a nice little tidbit from the <em>German</em> Wikipedia <a href="http://de.wikipedia.org/wiki/Carl_Ludwig_Siegel#Siegels_Standpunkt_zur_Entwicklung_der_Mathematik" rel="nofollow">http://de.wikipedia.org/wiki/Carl_Ludwig_Siegel#Siegels_Standpunkt_zur_Entwicklung_der_Mathematik</a></p> <p>To finish here is my own translation of that:</p> <p><strong>The position of Siegel towards the development of math</strong></p> <p>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.</p> <p>Disclaimer: this is a best-effort translation of that Wikipedia bit from myself and it does not reflect my opinion.</p> <p>EDIT: here is the reference for that Wikipedia bit in english <a href="https://plus.google.com/107976469380197792992/posts/LVTEUKSHeAR" rel="nofollow">Letter from Siegell to Louis J. Mordell, 3 March 1964</a></p> http://mathoverflow.net/questions/115691/bernoulli-number-formula-involving-roots-of-taylor-polynomial-of-exp-1/115741#115741 Comment by Peter Sheldrick Peter Sheldrick 2012-12-09T05:33:17Z 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](<a href="http://de.wikipedia.org/wiki/Vandermonde-Matrix#Weitere_Eigenschaften" rel="nofollow">de.wikipedia.org/wiki/&hellip;</a>) http://mathoverflow.net/questions/115691/bernoulli-number-formula-involving-roots-of-taylor-polynomial-of-exp-1/115741#115741 Comment by Peter Sheldrick Peter Sheldrick 2012-12-09T04:27:47Z 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 $n$. http://mathoverflow.net/questions/115691/bernoulli-number-formula-involving-roots-of-taylor-polynomial-of-exp-1/115741#115741 Comment by Peter Sheldrick Peter Sheldrick 2012-12-09T04:26:34Z 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 $m &lt;&lt; n$ (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 http://mathoverflow.net/questions/115691/bernoulli-number-formula-involving-roots-of-taylor-polynomial-of-exp-1/115741#115741 Comment by Peter Sheldrick Peter Sheldrick 2012-12-07T21:52:56Z 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 $q_n(x)=B_n(x)/n!$? http://mathoverflow.net/questions/115691/bernoulli-number-formula-involving-roots-of-taylor-polynomial-of-exp-1 Comment by Peter Sheldrick Peter Sheldrick 2012-12-07T09:06:32Z 2012-12-07T09:06:32Z This question is (cross-posted)[<a href="http://math.stackexchange.com/questions/252928/bernoulli-number-formula-involving-roots-of-taylor-polynomial-of-exp-1]" rel="nofollow" title="bernoulli number formula involving roots of taylor polynomial of exp 1%5d">math.stackexchange.com/questions/252928/&hellip;</a> with math.stackexchange. http://mathoverflow.net/questions/100178/generate-bernoulli-polynomials-in-one-integration Comment by Peter Sheldrick Peter Sheldrick 2012-06-21T01:39:44Z 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)). http://mathoverflow.net/questions/100178/generate-bernoulli-polynomials-in-one-integration Comment by Peter Sheldrick Peter Sheldrick 2012-06-21T01:27:44Z 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 $k=1$)... http://mathoverflow.net/questions/100178/generate-bernoulli-polynomials-in-one-integration Comment by Peter Sheldrick Peter Sheldrick 2012-06-21T01:23:35Z 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...