Timeline for Scaled Riemann zeta function with no zero in the critical strip
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2021 at 4:47 | vote | accept | Vincent Granville | ||
| Apr 27, 2021 at 14:03 | comment | added | Vincent Granville | @Conrad: more changes are needed. More about this later today. | |
| Apr 27, 2021 at 13:59 | comment | added | Conrad | why do you think that will make the product converge? the eta function doesn't have a product in the critical strip either for the same reasons as zeta as the arguments behave wildly | |
| Apr 27, 2021 at 13:58 | comment | added | Wojowu | Introducing one factor won't impact convergence of the product. | |
| Apr 27, 2021 at 12:53 | comment | added | Wojowu | The equivalent of argument in the series expansion is the imaginary part of the summands, and this is something you cannot adjust by multiplying by real numbers. | |
| Apr 27, 2021 at 12:44 | comment | added | Wojowu | I think in turn you missed the point of my answer. By taking absolute values you are ignoring the argument of the terms, and the argument is what prevents the product from converging. | |
| Apr 27, 2021 at 12:32 | comment | added | Vincent Granville | In my previous question, I was dealing with a product equivalent (w.r.t convergence) to a series whose $k$-th term is $\sim \cos(t\log p_k)/p_k^\sigma$. It could never converge, agreed. Here by rescaling I fixed the issue: the $k$-th term is equivalent to $\sim \cos(t\log p_k)/p_k^{2\sigma}$. And since $\sigma>1/2$, meaning $2\sigma>1$, it actually always converge. That was the whole purpose of this new post (fixing the problem in the previous one) and I think you missed it. | |
| Apr 27, 2021 at 11:59 | history | answered | Wojowu | CC BY-SA 4.0 |