NTIC Connecting to Zeros

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Section 25.6 Connecting to Zeros

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Subsection 25.6.1 Where are the zeros?

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Our next goal is to see how this connection

$\begin{equation*} \log(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx \end{equation*}$

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relates to the zeros of the

$\zeta$ function (and hence the Riemann Hypothesis).

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L = lcalc.zeros_in_interval(10,100,0.1)
[l[0] for l in L]

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We see all the zeros for

$\sigma=1/2$ between

$0$ and

$100\text{;}$ there are 29 of them.

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We will connect to

$\zeta$ by means of a very powerful analogy, the one which Euler used to prove

$\zeta(2)=\frac{\pi^2}{6}$ (see the end of Subsection 20.4.2) and which, correctly done, does yield the right answer.

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Begin the analogy by recalling basic algebra. The Fundamental Theorem of Algebra states that every polynomial factors over the complex numbers. For instance,

$\begin{equation*} f(x)=5x^3-5x=5(x-0)(x-1)(x+1)\text{.} \end{equation*}$

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If we take the logarithm of such a factorization, we can say things like

$\begin{equation*} \log(f(x))=\log(5)+\log(x-0)+\log(x-1)+\log(x+1) \end{equation*}$

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Then if it turned out that

$\log(f(x))$ was useful to us for some other reason

$R\text{,}$ it would be reasonable to say that we can get information about the otherwise-mysterious

$R$ from adding up information about the zeros of

$f$ (and the constant

$5$ ), because of the addition of

$\log(x-r)$ for all the roots

$r\text{.}$

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You can't really do this with arbitrary functions, of course. Disappointingly,

$\zeta$ is definitely a function where this doesn't work, mostly because

$\zeta(1)$ diverges so badly, no matter how you define the complex version of

$\zeta\text{.}$

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But it so happens that

$\zeta$ is very close to a function you can analyze this way,

$(s-1)\zeta(s)\text{.}$ Applying the logarithm factoring idea to

$(s-1)\zeta(s)$ (and doing lots of relatively hard complex integrals, or some other formal business with difficult convergence considerations) allows us to essentially invert the equation

$\begin{equation*} \log(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx \end{equation*}$

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to the even more surprising formula

$\begin{equation} J(x)=Li(x)-\sum_{\rho}Li(x^\rho)-\log(2)+\int_x^\infty\frac{dt}{t(t^2-1)\log(t)}\label{men-j-zeros}\tag{25.6.1} \end{equation}$

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Subsection 25.6.2 Analyzing the connection

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It is hard to overestimate the importance of the formula (25.6.1). Each piece comes from something inside

$\zeta$ itself, inverted in this special way.

First, $Li(x)$ comes from the fact that we needed $(s-1)\zeta(s)$ to apply this inversion, not just $\zeta(s)\text{.}$ In fact, this particular inversion can be seen by integrating, as it's true that

$\begin{equation*} s\int_1^\infty Li(x)x^{-s-1}dx=-\log(s-1) \end{equation*}$

so one can see that $s-1$ and $Li$ seem to correspond.
Second, each $Li(x^\rho)$ comes from each of the zeros of $\zeta$ on the line $\sigma=1/2$ in the complex plane. This is the part which most closely corresponds to the factoring.
The constant term $\log(2)$ comes from the constant when you do the factoring, similarly to the $5$ in the example above using $f(x)=5x^3-5x\text{.}$
Finally, the integral in (25.6.1) comes from the zeros of $\zeta$ at $-2n$ we mentioned just before the statement of 25.3.7.

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To give you a sense of how complicated (25.6.1) really is, here is a plot of just one small piece of it.

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This is the plot of

$Li(20^{1/2+it})$ up through the first zero of

$\zeta$ above the real axis. It's beautiful, but also forbidding. After all, if takes that much twisting and turning to get to

$Li$ of the first zero, what is in store if we have to add up over all infinitely many of them to calculate

$J(20)\text{?}$

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So at the very least, it would be helpful to know where all of those mysterious zeros live! This is why the Riemann Hypothesis is so important; it pins them down quite dramatically.