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President Trump Donald John TrumpOmar fires back at Trump over rally remarks: ‘This is my country’ Pelosi: Trump hurrying to fill SCOTUS seat so he can repeal ObamaCare Trump mocks Biden appearance, mask use ahead of first debate MORE on Saturday suggested that he wasn’t wrong to claim that up to 3 million people voted illegally in the 2016 election — even though he later walked back that number.
« Of the 11,879 people who requested an ‘oh by the way, I’m not wrong’ retweet, only one replied, » Trump tweeted in response to his claim that former President Obama had wiretapped his phones in Trump Tower.
Of the 11,879 people who requested an « oh by the way, I’m not wrong » retweet, only one replied.
….most people just clicked like. This was also in a day the Fake News Media insisted I was calling the first ever NATIONAL VOTE for President Election “an epic landslide.” But not only was it not an « epic landslide, » it was not a landslide. — Donald J. Trump (@realDonaldTrump) March 4, 2018
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The president’s tweet came after a Friday court filing by his lawyer, Rudy Giuliani, that said his campaign had received numerous messages that demanded « an apology and an explanation for the now discredited ‘wiretap’ charge. »
The filing claimed the messages began « within minutes » of the president’s tweet about former Obama administration officials wiretapping his phones.
The messages were cited in the court filing as a proof-of-claim against the Obama
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Are all The Riemann Zeta values given as $\zeta(2n)$ equivalent?
So I have come across this idea:
All (rational) The Riemann Zeta values can be written as $\zeta(2n)$.
Any idea how to prove this?
A:
Yes, all the $\zeta(2n)$ are equivalent. There’s just an indexing problem. Perhaps you have come across the standard indexing of the $\zeta(s)$, and so are wondering if all the the $\zeta(2n)$ are the same thing: no, they are.
If you’re coming from the usual « We can define the $\zeta(n)$ as real numbers, and then we can define a complex $\zeta(s)$ by analytic continuation », then the idea is you can do that, and at some point the real and imaginary parts of $\zeta(s)$ will agree. However, the complex $\zeta(s)$ is defined in $\mathbb{C}$, so the real and imaginary parts will generally disagree, giving rise to the real numbers with an even index, and the complex numbers with an odd index.
For each real $s$ not equal to $0$ or $1$, there is a unique $\zeta(s)$ such that $\zeta(2n)=\zeta(s)$ for all integers $n$. You can calculate
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