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How to calculate $L'(1,\chi)/L(1,\chi)$?

asked 2012-05-08 10:14:26 -0500

anonymous user

Anonymous

Question as in title, where $L(s,\chi)$ is the Dirichlet $L$-function associated with the nontrivial character modulo $3$. Please provide complete SAGE code. Thank you in advance.

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Maybe you could start by showing us how much you have already?

niles gravatar imageniles ( 2012-05-08 10:47:22 -0500 )edit

Have you had a look at the [L-function tutorial](http://wiki.sagemath.org/days33/lfunction/tutorial) from Sage days 33? In particular, it has some fairly recent comments on the development status. I have no idea whether Sage yet implements, for example, the insights in Ihara, Murty & Shimura's paper from circa 2007, [On the Logarithmic Derivative of Dirichlet L-Functions at s=1](http://www.kurims.kyoto-u.ac.jp/~kenkyubu/emeritus/ihara/Publications-and-Recent-Preprints/RecentArticles/pdf-files/IMS.main.pdf).

bgins gravatar imagebgins ( 2012-05-08 10:55:22 -0500 )edit

Actually I managed to calculate the particular example "by hand", using that $L(1,\chi)=\sum\chi(n)/n$ and $L'(1,\chi)=-\sum\chi(n)\log(n)/n$. Of course it would be nicer to use some built-in function for that purpose. Note that I am a beginner at SAGE. Also, I looked at http://wiki.sagemath.org/days33/lfunction/tutorial, but the command "LSeries" did not work at http://www.sagenb.org/ (upon calling "L=LSeries(DirichletGroup(3).0)" I get the error message "name 'LSeries' is not defined").

Anonymous gravatar imageAnonymous ( 2012-05-08 11:28:29 -0500 )edit

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answered 2017-03-13 15:38:08 -0500

dan_fulea gravatar image

We can try to compute using bare hands. The following is thus not useful if the question wants more than this particular case. I cannot see how sage supports better this now.

First of all, we can compute $\displaystyle L(\chi,1)=\sum_{n\ge 0}\left( \frac 1{3n+1} - \frac 1 {3n+2}\right)=\sum_{n\ge 0}\frac 1{(3n+1)(3n+2)}$ exactly, for instance:

sage: sum( 1/(3*n+1)/(3*n+2), n, 0, oo )                           
1/9*pi*sqrt(3)

and the result is connected to "polylogarithmic computations".

Comment: We may start with $1+x^3 +x^6+\dots =1/(1-x^3)$ and integrate twice, first with $x$ from $0$ to $y$, then with $y$ from $0$ to $1$. Fubini shows we can forget about polylogarithms, since $$ L(\chi, 1) = \int_0^1 dy\int_0^ydx\; \frac 1{1-x^3} = \int_0^1dx\int_x^1dy\; \frac 1{1-x^3} =\int_0^1\frac {dx}{1+x+x^2}=\frac 1{\sqrt 3}\pi\ .$$

The derivative is more complex. Pari/GP gave

? sum( n=0,10000000, -log(3*n+1)/(3*n+1)+log(3*n+2)/(3*n+2) )
%6 = 0.2226631782653383756620209560

but suminf( n=0,-log(3*n+1)/(3*n+1)+log(3*n+2)/(3*n+2) ) was testing my patience.

At any rate, the rest can be estimated by rewriting the sum as a sum over $$ \frac 1{(3n+1)(3n+2)} \left[\ \ln \left(1+\frac 1{3n+1}\right)^{3n+2} -\ln(3n+2) \ \right] $$ or over $$ \frac 1{(3n+1)(3n+2)} \left[\ \ln \left(1+\frac 1{3n+1}\right)^{3n+1} -\ln(3n+1) \ \right]\ . $$

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Asked: 2012-05-08 10:14:26 -0500

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Last updated: Mar 13