The Riemann hypothesis in various settings

So Terence Tao has posted a blog regarding the Riemann Hypothesis. He notes that his blog is one that makes “no new progress” on the hypothesis, but later refers to the entry as one in which he is “simply arranging existing facts together”.
And THAT, ladies and gentlemen, would be the perfect way for someone as amazing as Tao to provide a subtle, suave introduction to a long series of posts, the culmination of which could be an actual proof to Riemann’s actual hypothesis.
Has the already-brilliant Terry Tao solved the Riemann Hypothesis? Stay tuned to his quadrant of internet space to find out….

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[Note: the content of this post is standard number theoretic material that can be found in many textbooks (I am relying principally here on Iwaniec and Kowalski); I am not claiming any new progress on any version of the Riemann hypothesis here, but am simply arranging existing facts together.]

The Riemann hypothesis is arguably the most important and famous unsolved problem in number theory. It is usually phrased in terms of the Riemann zeta function $latex {\zeta}&fg=000000$, defined by

$latex \displaystyle \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}&fg=000000$

for $latex {\hbox{Re}(s)>1}&fg=000000$ and extended meromorphically to other values of $latex {s}&fg=000000$, and asserts that the only zeroes of $latex {\zeta}&fg=000000$ in the critical strip $latex {\{ s: 0 \leq \hbox{Re}(s) \leq 1 \}}&fg=000000$ lie on the critical line $latex {\{ s: \hbox{Re}(s)=\frac{1}{2} \}}&fg=000000$.

One of the main reasons that the Riemann hypothesis is so important to number theory is that the zeroes of…

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One thought on “The Riemann hypothesis in various settings

  1. Near the beginning of his 1859 paper Riemann incorrectly assumes that the complex variable s = (1/2) + ti is a zeta power. Riemann fails to recognise that an expression containing an imaginary number such as (1/2) +ti cannot be a power unless the base is a log base such as e. The best known example of this is Cotes’s formula cosu + isinu equals e^(iu), where it is not possible for e to be replaced by other values.

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