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We know that the Fourier Series $$s(x)=\sum_{k\neq0}\frac{1}{k}\exp\left(2\pi ik x\right)$$ corresponds to the sawtooth function, $s(x)=\left\{x\right\} -\frac{1}{2}$. Suppose that $\left(\frac{k}{d}\right)$ is the Jacobi Symbol, which is zero when $(d,k)>1$.

My question is: What is the function $f_d(x)$ corresponding the the Fourier Series $$f_d(x)=\sum_{k\neq0}\left(\frac{k}{d}\right)\frac{1}{k}\exp\left(2\pi ikx\right),$$ and how can we find this function?

Thank you,

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up vote 8 down vote accepted

For any primitive character $\chi$ mod $N$, the finite sum $\sum_{a\;mod\;N} \;\chi(a)\,s(x+{a\over N})$ produces your sum, multiplied by the corresponding Gauss sum.

Edit: in response to comment/question, I don't think there is a simpler expression for the sum over $a$ mod $N$, exactly because of the assumption that the character has conductor $N$. In the opposite case(s), e.g., that the character $\chi$ is trivial, indeed there is cancellation. But with the primitivity assumption, the riff is that $\sum_a \chi(a)\,e^{2\pi i na/N} = \overline{\chi}(n)\,\sum_a \chi(a)\,e^{2\pi i a/N}$ for $n$ prime to $N$. In fact, since for (odd!) quadratic characters this essentially gives the class number (by computing $L(1,\chi)$, any serious simplification would have to give striking new information on class numbers, which is a non-trivial issue (see Siegel, Heegner, Stark, Goldfeld, et alia).

Further edit: yes, indeed, setting $x=0$ in the Fourier expansion literally gives (up to some constants and normalizations) $L(1,\chi)$ for odd $\chi$. We must believe that the Fourier series converges pointwise away from the discontinuities at integers, and it does: the usual Dirichlet kernel argument proves convergence for finitely-piecewise continuous functions at points where the function is once-continuously-differentiable.

Edit: @J.M., about whether the quadratic symbol should be a Jacobi symbol or Kronecker, ... the real point is that $k\rightarrow (k/d)_2$ should be a primitive character mod $N$. So, no, the "d" need not be prime (which is the earlier, "Jacobi" sense), but can be composite (the later, "Kronecker" sense).

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Is there a more concise way to rewrite this finite sum? For $s(x)$ as an example, we have $$\sum_{a\text{ mod } b} s\left(x+\frac{a}{b}\right)=s(xb).$$ Can something similar be done when characters are involved? – Eric Naslund Dec 18 '11 at 18:31
Ahhh thank you for the very useful edit, I think this is exactly what I am looking for. I was wondering can you elaborate a bit more? How is this sum $\sum_{a\text{ mod }N} \chi(a)s\left(x+\frac{a}{N}\right)$ related to $L(1,\chi)$? Can we make things explicit? I know that $\frac{-i\pi}{N G(\chi)}\sum_{a\text{ mod }N} a\chi (a)=L(1,\chi)$ for $\chi$ quadratic (or something close to this), is that used here? – Eric Naslund Dec 18 '11 at 18:53
Paul, do your results still carry over if we replace "Jacobi" with "Kronecker" in Eric's question, or are there modifications that have to be done? – J. M. Dec 19 '11 at 0:25

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