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The Riemann Hypothesis, explained (2016)

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Re: The Riemann Hypothesis, explained (2016)

#11

Post by Chinjut »

petters wrote: post #2

> The fourth and final term is an integral which is zero for x This does not seem correct. Overall a good article though.

I agree. This integral cannot be constantly zero in that region, as the integrand is not itself constantly zero.

Also, there's no reason to expect this integral (merely the last term of the explicit formula, the one arising from the trivial zeros of the zeta function) to count primes all in its own, so no reason it should be zero on its own below 2. That statement instead correctly describes the sum of all four terms, in the article's presentation.

Re: The Riemann Hypothesis, explained (2016)

#12

Post by kevinventullo »

petters wrote: post #2

> The fourth and final term is an integral which is zero for x This does not seem correct. Overall a good article though.

Yeah, that part is wrong. It's a continuous function of x, so if its value at x=2 is 0.1400101..., it couldn't suddenly be 0 at every x Note to other commenters: this has nothing to do with whether or not 1 is considered a prime, as the integral makes no reference to primes at all. OP was taking issue with the former claim in the quote, not the latter (and the latter, while true, appears to have nothing to say about the former; my guess is some kind of editing mistake).

Re: The Riemann Hypothesis, explained (2016)

#14

Post by petters »

mathfailure wrote: post #3
petters wrote: post #2

> The fourth and final term is an integral which is zero for x This does not seem correct. Overall a good article though.

A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers.

Of course. That was not the error.

Re: The Riemann Hypothesis, explained (2016)

#15

Post by petters »

kevinventullo wrote: post #12
petters wrote: post #2

> The fourth and final term is an integral which is zero for x This does not seem correct. Overall a good article though.

Yeah, that part is wrong. It's a continuous function of x, so if its value at x=2 is 0.1400101..., it couldn't suddenly be 0 at every x Note to other commenters: this has nothing to do with whether or not 1 is considered a prime, as the integral makes no reference to primes at all. OP was taking issue with the former claim in the quote, not the latter (and the latter, while true, appears to have nothing to say about…

Thanks. I was very brief because I was on my cell phone. Did not expect downvotes for that comment. :-)
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