# How is $3 + 4\cos \theta + \cos 2\theta \geq 0$ related to $\zeta(s)^3|\zeta(s + it)^4\zeta(s + 2it)| \geq 1$?

The inequality

$$\zeta(s)^3 | \zeta(s + it)^4 \zeta(s + 2it)| \ge 1$$

follows from

$$3 + 4 \cos(\theta) + \cos(2 \theta) \ge 0.$$

How is that done? What is the relationship between zeta and the trigonometry?

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grab any book with a proof of the prime number theorem ($\zeta(1+it)\neq0$) e.g. lang's algebraic number theory. (that's where this comes up isnt it?) – yoyo Feb 27 '11 at 22:17

Just in case, everyone recall the notation $s=\sigma+it$ to denote complex numbers.

The inequality you are looking for is

$$|\zeta(\sigma)^3 \zeta(\sigma + it)^4 \zeta(\sigma + 2it)| \geq 1.$$

for $\sigma>1$. From this we can prove the Prime Number Theorem (by using a contour integral and some limits switching) because it shows that the zeta function has no zeros on the line $\sigma=1$. To see why, count the zeros of the numerator and the zeros of the denominator when $\sigma\rightarrow 1$. We see that a zero at $1+it$ would force a pole to exist at $1+2it$, but that is impossible since the only pole of $\zeta(s)$ is at $s=1$.

So why does it follow from the trigonometric identity

$$3 + 4 \cos(\theta) + \cos(2 \theta) \ge 0?$$

First take logarithms. Then the above is equivalent to showing that $$3\log|\zeta(\sigma)|+4\log|\zeta(\sigma+it)|+\log|\zeta(\sigma+2it)|\geq 0.$$

Recall that $\log|z|=\Re(\log z)$ and $\Re(z)+\Re(w)=\Re(w+z)$ where $\Re(z)$ denotes the real part of $z$. Hence we need to show that

$$\Re \left(3\log\zeta(\sigma)+4\log\zeta(\sigma+it)+\log\zeta(\sigma+2it)\right) \geq 0.$$

Now since $$\log(\zeta(s))=\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n}n^{-s}$$ when $\sigma>1$, where $\Lambda(n)$ is the Von Mangoldt Lambda Function, grouping the sums changes the left hand side into

$$\Re \left(\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n}\left( 3n^\sigma+4n^{\sigma+it}+n^{\sigma+2it} \right) \right).$$

Since everything is sufficiently convergent, we can bring $\Re$ inside the summation to get $$\left(\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n}\Re\left( 3n^\sigma+4n^{\sigma+it}+n^{\sigma+2it} \right) \right)$$

Now, notice $\Re(n^{x+iy})= n^{x}\cos(y\log n)$, so that the above becomes

$$\left(\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n} n^{-\sigma}\left(3+4\cos(t\log n)+\cos(2t\log n) \right) \right)$$

Lastly, the term above must be greater than or equal to zero since every term in the summation is non-negative. Thus we conclude

$$\Re \left(3\log\zeta(\sigma)+4\log\zeta(\sigma+it)+\log\zeta(\sigma+2it)\right) \geq 0$$ and hence

$$|\zeta(\sigma)^3 \zeta(\sigma + it)^4 \zeta(\sigma + 2it)| \geq 1$$ as desired.

Hope that helps,

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The inequality you state does not seem to make too much sense.

Anyway, you can find a related discussion in the book Riemann's Zeta Function, by H.M. Edwards on pages 79/80.