5
$\begingroup$

I found interesting results when I put in certain values for z into zeta(z). I believe my findings do not disprove the Riemann hypothesis, but I am not so sure.

When I input z = 1/3 + i into the function, I got a result with a positive real component and a negative imaginary component, which can be written as a - bi.

When I input z = 1/3 - i, I got a result with a positive real component and a positive imaginary component, which can be written as c + di.

For z = 2/5 - i/2, I got -f + gi.

For z = 2/5 + i/2, I got -h - li.

Because the Riemann Zeta Function is continuous, these results imply that there is a number in this range z for which zeta(z) has a real component of zero. In fact, there would be infinitely many of these numbers in an unbroken chain. It also implies that there is a number in this range z for which zeta(z) has a complex component of zero and there there are an infinitely many of these in a broken chain. Also, these two chains should meet with each other at some point. This seems like it would disprove the Riemann Hypothesis.

$\endgroup$
4
  • 12
    $\begingroup$ The intermediate value theorem does not hold for complex-valued functions. Example: $e^0 = 1$, $e^{ \pi i} = -1$, but $e^z$ has no zeros. $\endgroup$
    – Martin R
    Mar 18, 2021 at 20:35
  • 13
    $\begingroup$ What are all these downvotes for? Is it a sin to be wrong? $\endgroup$
    – uniquesolution
    Mar 18, 2021 at 20:38
  • 1
    $\begingroup$ I take the function $e^z$; when I input $z=1/2+ 2n\pi i \pm 1$ I get results with positive real part and plus minus imaginary part; if I input same but with an odd multiple of pi over two in the imaginary part and a little more I get results with negative real part and plus minus imaginary part; I conclude by continuity and reasoning above that $e^z$ has infinitely many zeroes which is of course absurd $\endgroup$
    – Conrad
    Mar 18, 2021 at 20:38
  • $\begingroup$ @uniquesolution, plus, OP recognized that something must be wrong and only sought clarification. $\endgroup$
    – Randall
    Mar 19, 2021 at 17:12

1 Answer 1

Reset to default
6
$\begingroup$

Yes, you can draw a curve separating $\{\frac13 + i, \frac13 - i\}$ from $\{\frac25 + \frac i2, \frac25 - \frac i2\}$ where the real component of the zeta function is $0$.

Yes, you can draw a curve separating $\{\frac13 + i, \frac25 + \frac i2\}$ from $\{\frac13 - i, \frac25 - \frac i2\}$ where the imaginanary component of the zeta function is $0$.

But you have no control over where these curves intersect; they do not have to intersect inside the trapezoid whose corners are at $\{\frac13 + i, \frac13 - i, \frac25 + \frac i2, \frac25 - \frac i2\}$, because the curves we drew may leave that trapezoid. So the zero you find may be a trivial zero, or it may be a nontrivial zero with real part $\frac12$, neither of which would disprove the Riemann hypothesis.

enter image description here

In fact, here is the picture we get if we actually draw these curves: in blue is the curve with real component $0$, and in orange is the curve with imaginary component $0$. (There's more going on outside the range $[-3,3] \times [-3,3]$, of course.) We see them intersecting once at $z=-2$, and we see them pretending to intersect at $z=1$. (But they actually don't, because the zeta function has a singularity at $1$, corresponding to the harmonic series $1 + \frac12 + \frac13 + \dots$.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .