text: # Positivity of $L(1,\chi)$ for real Dirichlet's character
  
  Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.
  
  In my answer to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples of the real characters modulo 3 and 4. The proof goes as follows: one takes $$f(x)=\sum_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum_{j=1}^{m-1}\chi(j)x^j$$ and uses Abel's theorem to write $$L(1,\chi)=\int_0^1f(x)dx;$$ since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.
  
  Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of the polynomial $$g_\chi(x)=\sum_{n=1}^{m-1}\chi(n)x^n$$ on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works perfectly further; for example, $$g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5$$ or $$g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11.$$ Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is always holds for small values $m\le30$.
  
  Question. Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$, is it true that $g_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed in the literature?
  
  -
  
  These are called Fekete polynomials, and you can find out a great deal about them here. Unfortunately they tend to have lots of real zeros in $(0,1)$ when $m$ is large.
  
  -
  Thanks, David! Never heard of them... Feteke or Fekete? And is $L(1,\chi)>0$ for some $m$? – Wadim Zudilin May 26 '10 at 1:00
  fekete, sorry! I'm not sure I understand your second question. – David Hansen May 26 '10 at 1:01
  Oh, no, the inequality $L(1,\chi)>0$ for $\chi$ real goes back to Dirichlet, and is an immediate consequence of his famous class number formula. – David Hansen May 26 '10 at 1:46
  Wadim and David: there's an easier reason why L(1,chi) can't be negative for real nontrivial chi: the function L(s,chi) for s > 0 is obviously real-valued. As s --> infty it tends to 1, which is positive, and we know L(s,chi) is nonzero for s > 1 by the Euler product. Therefore by continuity L(s,chi) > 0 for s > 1, hence by taking a limit from the right L(1,chi) is definitely not negative. It's not 0 either (harder!), so L(1,chi) > 0. – KConrad May 26 '10 at 2:22
  It is not the same, but Chowla then Rosser chased around vaguely similar ideas (weighted sums) to show that $L(s,\chi)>0$ for real $s>0$ for various real $\chi$. Chowla: matwbn.icm.edu.pl/ksiazki/aa/aa1/aa119.pdf Rosser: ams.org/journals/bull/1949-55-10/S0002-9904-1949-09306-0/… – Junkie May 26 '10 at 7:17
  
  As David pointed out, it is known that for large m these polynomials have many zeroes. This is very unfortunate for the following reason: If you assume $g_{\chi}$ is non-negative then it follows by Mellin inversion that the L-function $L(s,\chi)$ can not have a Siegel zero (or, more generally, any zero on the positive real axis).
  
  -
  Thanks, Marc! The heuristics is already indicated in David's reference, but only for the Legendre symbol. I acknowledge your contribution by upvoting. – Wadim Zudilin May 26 '10 at 1:54
explanation: The document discusses the positivity of the value of the Dirichlet L-function \( L(1, \chi) \) for a real nonprincipal Dirichlet character \(\chi\) modulo \(m\). It introduces a method to show that \( L(1, \chi) > 0 \) by considering the polynomial \( g_\chi(x) = \sum_{n=1}^{m-1} \chi(n) x^n \) and its positivity on the interval \((0, 1)\). The document also mentions that this method works for small values of \(m\) but fails for larger \(m\) due to the presence of real zeros of the polynomial. The text includes a question about the general positivity of \( g_\chi(x) \) and whether the method is discussed in the literature.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Positivity of $L(1,\chi)$ for real Dirichlet's character
  
  Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.
  
  In my answer to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples of the real characters modulo 3 and 4. The proof goes as follows: one takes $$f(x)=\sum_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum_{j=1}^{m-1}\chi(j)x^j$$ and uses Abel's theorem to write $$L(1,\chi)=\int_0^1f(x)dx;$$ since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.
  
  Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of the polynomial $$g_\chi(x)=\sum_{n=1}^{m-1}\chi(n)x^n$$ on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works perfectly further; for example, $$g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5$$ or $$g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11.$$ Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is always holds for small values $m\le30$.
  
  Question. Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$, is it true that $g_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed in the literature?
  
  -
  
  These are called Fekete polynomials, and you can find out a great deal about them here. Unfortunately they tend to have lots of real zeros in $(0,1)$ when $m$ is large.
  
  -
  Thanks, David! Never heard of them... Feteke or Fekete? And is $L(1,\chi)>0$ for some $m$? – Wadim Zudilin May 26 '10 at 1:00
  fekete, sorry! I'm not sure I understand your second question. – David Hansen May 26 '10 at 1:01
  Oh, no, the inequality $L(1,\chi)>0$ for $\chi$ real goes back to Dirichlet, and is an immediate consequence of his famous class number formula. – David Hansen May 26 '10 at 1:46
  Wadim and David: there's an easier reason why L(1,chi) can't be negative for real nontrivial chi: the function L(s,chi) for s > 0 is obviously real-valued. As s --> infty it tends to 1, which is positive, and we know L(s,chi) is nonzero for s > 1 by the Euler product. Therefore by continuity L(s,chi) > 0 for s > 1, hence by taking a limit from the right L(1,chi) is definitely not negative. It's not 0 either (harder!), so L(1,chi) > 0. – KConrad May 26 '10 at 2:22
  It is not the same, but Chowla then Rosser chased around vaguely similar ideas (weighted sums) to show that $L(s,\chi)>0$ for real $s>0$ for various real $\chi$. Chowla: matwbn.icm.edu.pl/ksiazki/aa/aa1/aa119.pdf Rosser: ams.org/journals/bull/1949-55-10/S0002-9904-1949-09306-0/… – Junkie May 26 '10 at 7:17
  
  As David pointed out, it is known that for large m these polynomials have many zeroes. This is very unfortunate for the following reason: If you assume $g_{\chi}$ is non-negative then it follows by Mellin inversion that the L-function $L(s,\chi)$ can not have a Siegel zero (or, more generally, any zero on the positive real axis).
  
  -
  Thanks, Marc! The heuristics is already indicated in David's reference, but only for the Legendre symbol. I acknowledge your contribution by upvoting. – Wadim Zudilin May 26 '10 at 1:54
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document discusses the positivity of the value of the Dirichlet L-function \( L(1, \chi) \) for a real nonprincipal Dirichlet character \(\chi\) modulo \(m\). It introduces a method to show that \( L(1, \chi) > 0 \) by considering the polynomial \( g_\chi(x) = \sum_{n=1}^{m-1} \chi(n) x^n \) and its positivity on the interval \((0, 1)\). The document also mentions that this method works for small values of \(m\) but fails for larger \(m\) due to the presence of real zeros of the polynomial. The text includes a question about the general positivity of \( g_\chi(x) \) and whether the method is discussed in the literature.
  
  Label: question-and-answer