Plugging $\zeta (1/2 + i)$ into Wolfram Alpha yields me some complex number, but I was under the understanding that $\zeta (1/2 + it) = 0$, for all $t$ we have yet calculated...
Is Wolfram Alpha just messing with me?
Thanks!
Asked
Active
Viewed 358 times
0
-
I believe your claim is incorrect. – Bombyx mori Mar 07 '15 at 20:42
-
All non-trivial zeros of the zeta function are on the form $z=1/2 + it$, it's not that every number on that form is a zero. – Winther Mar 07 '15 at 20:42
-
Oh! Man. That makes so much more sense. He he thanks! – Bliebervik Mar 07 '15 at 20:43
-
$\zeta(1/2+i) \sim 0.1439364271 - 0.7220997435 i$ – Math-fun Mar 07 '15 at 20:44
-
The Riemann hypothesis is that if $\zeta(z)=0$ (and $z$ is not a trivial zero) then $z=1/2+it$ for some real $t.$ This is not the same as saying that all values of $t$ give zeros of the zeta function. – coffeemath Mar 07 '15 at 20:45
2 Answers
1
No! Every $s$ such that $\zeta(s)=0$ is assumed to be of the form $1/2 + i t$ yet not every $1/2 + i t$ is a root of $\zeta$.
Recall for example that $\zeta$ is a meromorphic function and as such can only have countably many roots.
quid
- 42,135
-
1Silly me! Thank you. I need to research that to find some solutions then! – Bliebervik Mar 07 '15 at 20:44
-
There are many tables of roots. See http://mathworld.wolfram.com/RiemannZetaFunctionZeros.html for a few. – quid Mar 07 '15 at 20:47
-
@Bliebervik: It is okay. Zeta function is not easy to study. To find a root by hand is quite difficult for people not working in the field. – Bombyx mori Mar 07 '15 at 20:48
1
Wolfram alpha is correct. Note this for only some $t\in \mathbb{R}$ we have $\zeta(1/2+it)=0$ which are the non trivial zeroes.
science
- 2,900