It is a well known fact that the geometric series $$1+x+x^2+x^3+\ldots$$ has the following form $$\frac{1}{1-x}$$ Another possible representation is $$\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$ This comes from the identity $$1+x+x^2+x^3+\ldots+x^{2^{k}}=\frac{1-x^{2^{k}+1}}{1-x}$$ now taking the numerator of the rhs we have $$1-x^{2^{k}+1}=\left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)=\left(1-x^{2^{k}-1}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$ proceeding this way we eventually get $$\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)\ldots\left(1+x^{2^{k}-2}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$ Taking the limit for the geometric series $$\sum_{k=0}^{\infty}x^{k}=\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$ Now taking the zeta function $$\zeta(z)=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}+\frac{1}{p^{2z}}+\frac{1}{p^{3z}}+\ldots\right)$$ we can express it as $$\zeta(z)=\prod_{k=0}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$

Now considere for

$$G(z)=\prod_{k=1}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$ note that now $k\geq 1$ and that $G(z)$ converges absolutely for $z>\frac{1}{2}$

Can we say that, after analytic continuation, that

$$H(z)=\sum_{k=0}^{^\infty}\frac{|\mu(k)|}{k^{z}}=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}\right)$$ has exactly the same zeros as $\zeta(z)$?

  • $\begingroup$ How can you say that G(z) converges for $z > \frac{1}{2}$ ? $\endgroup$ – Roupam Ghosh Mar 3 '11 at 7:44
  • $\begingroup$ @Roupa: because $w=2z,4z,8z,\ldots$. $\endgroup$ – Neves Mar 3 '11 at 7:54
  • $\begingroup$ what is $w$? And how do you say that the expression converges for $\Re(z) > \frac{1}{2}$? Can you please elaborate? $\endgroup$ – Roupam Ghosh Mar 3 '11 at 7:57
  • $\begingroup$ @roupam, all the powers are multiples of a power of 2. $\endgroup$ – Neves Mar 4 '11 at 22:08

Hint: $$ \sum \frac{|\mu(k)|}{k^z} = \frac{\zeta(z)}{\zeta(2z)} $$ for $\Re(z) > 1$

Check out http://en.wikipedia.org/wiki/Dirichlet_series It might help you out.

  • $\begingroup$ where exactly do you think the answer is? $\endgroup$ – Neves Mar 4 '11 at 22:10
  • $\begingroup$ @Neves: You can factor $\sum \mu(n)^2 n^{-s}$ as the Euler product $\prod (1+p^{-s})$. To get the identity Roupam mentions, note that $1+x = (1-x^2)/(1-x)$ and then use the product of the original Riemann zeta function. Assuming RH, all zeros are therefore on $\zeta$'s critical line. $\endgroup$ – anon Jul 31 '11 at 18:45

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