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The function $\Gamma(s)$ does not have zeros, but $\Gamma(s)\pm \Gamma(1-s)$ does.

Ignoring the real solutions for now and assuming $s \in \mathbb{C}$ then:

$\Gamma(s)-\Gamma(1-s)$ yields zeros at:

$\frac12 \pm 2.70269111740240387016556585336 i$ $\frac12 \pm 5.05334476784919736779735104686 i$ $\frac12 \pm 6.82188969510663531320292827393 i$ $\frac12 \pm 8.37303293891455628139008877004 i$ $\frac12 \pm 9.79770751746885191388078483695 i$ $\frac12 \pm 11.1361746342106720656243966380 i$ $\frac12 \pm 12.4106273718343980402685363665 i$

$\dots$

and

$\Gamma(s)+\Gamma(1-s)$ gives zeros at:

$\frac12 \pm 4.01094805906156869492043027819 i$ $\frac12 \pm 5.97476992595365858561703252235 i$ $\frac12 \pm 7.61704024553573658642606787126 i$ $\frac12 \pm 9.09805003388841581320246381948 i$ $\frac12 \pm 10.4760650707765536619292369200 i$ $\frac12 \pm 11.7804020877663106830617193188 i$ $\frac12 \pm 13.0283749883477570386353012761 i$

$\dots$

By multiplication, both functions can be combined into: $\Gamma(s)^2 - \Gamma(1-s)^2$

After playing with the domain of $s$ and inspecting the associated 3D output charts, I now dare to conjecture that all 'complex' zeros of this function must have a real part of $\frac12$.

Has this been proven? If not, appreciate any thoughts on possible approaches.

Thanks!

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  • $\begingroup$ Should be an easy consequence of Euler's reflection formula, I'd guess. $\Gamma(z) \Gamma(1-z) = \pi / \sin(\pi z)$. $\endgroup$ – Marty Feb 23 '12 at 20:59
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    $\begingroup$ @Marty: It does not seem so easy. $\endgroup$ – GH from MO Feb 23 '12 at 21:27
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    $\begingroup$ I guess something like this works: apply complex Stirling approximation to $|\Gamma(s)| = |\Gamma(1-s)|$ to show no non-real zeros with real part outside $[0,1]$, then contour integrate to count zeros of $\Gamma(s) \pm \Gamma(1-s)$ in a rectangle $[0,1]+i[-T,T]$, and compare with the number of roots of real part $1/2$ that can again be estimated by Stirling. $\endgroup$ – Noam D. Elkies Feb 23 '12 at 23:52
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    $\begingroup$ humbee (mathoverflow.net/users/21930/humbee) posted in an answer (now moved to this comment): A. Does this mean that Gamma(s) +- Gamma(1-s) has an Euler product? B. How does the existence of an Euler product affects the location of the on the critical strip? $\endgroup$ – Anton Geraschenko Mar 7 '12 at 19:48
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    $\begingroup$ How are my approximations of zeros with 5000 digits precision explained? Checked with precision 100 in sage, pari and maple. $\endgroup$ – joro Apr 9 '12 at 9:07
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In the first part, we show that there are no zeros for $z = s + i t$ with $|t| \ge 4$ .

Let $\psi(z):= \Gamma'(z)/\Gamma(z)$ be the digamma function. If $z = s + i t$, then $$\frac{d}{ds} |\Gamma(z)|^2 = \frac{d}{ds} \Gamma(z) \Gamma(\overline{z}) = |\Gamma(z)|^2 \left(\psi(z) + \psi(\overline{z})\right).$$ (Both $\Gamma(z)$ and $\psi(z)$ are real for real $z$, and so satisfy the Schwartz reflection principle.) The product formula for the Gamma function implies that there is an identity $$\psi(z) = - \ \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{z + n} \right) = 1 - \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n + 1} - \frac{1}{z + n} \right),$$ and hence $$\psi(z) + \psi(\overline{z}) = 2(1 - \gamma) + \sum_{n=1}^{\infty} \left(\frac{2}{n + 1} - \frac{1}{z + n} - \frac{1}{\overline{z} + n} \right).$$ Suppose that $z = s + i t$, and that $s \in [0,1]$. Then $$ \frac{2}{n + 1} - \frac{1}{s + i t + n} - \frac{1}{s - i t + n} = \frac{2(s^2 + t^2 + n s - s - n)}{(1+n)(n^2 + 2 n s + s^2 + t^2)} \ge \frac{-2}{(n^2 + t^2)}.$$ (The last inequality comes from ignoring all the positive terms in the numerator, and then setting $s = 0$ in the denominator.) It follows that $$\psi(z) + \psi(\overline{z}) \ge 2(1 - \gamma) - \sum_{n=1}^{\infty} \frac{2}{n^2 + t^2},$$ which is positive for $t$ big enough, e.g. $|t| \ge 4$. On the other hand, $$\psi(z + 1) + \psi(\overline{z} + 1) = \psi(z) + \psi(\overline{z}) + \frac{1}{z} + \frac{1}{\overline{z}} = \psi(z) + \psi(\overline{z}) + \frac{2s}{|z|^2}.$$ In particular, if $\psi(z) + \psi(\overline{z})$ is positive for $s \in [0,1]$ for some particular $t$, it is positive for all $s$ and that particular $t$. It follows that, if $|t| > 4$, that $|\Gamma(s + it)|^2$ is increasing as a function of $s$. In particular, if $|t| > 4$, then any equality $$|\Gamma(s + i t)| = |\Gamma(1 - (s + i t))| = |\Gamma(1 - s + i t)|$$ implies that $s = 1/2$.

The second part is a continuation of the argument above, which completes the argument. (merged from a different answer.)

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros outside the box $z = s + it$ with $|t| \le 4$ and $|s| \le 15$ are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so. If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral $$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$ around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero $\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and $1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four conjugates of the root found by joro. A similar argument applies for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

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    $\begingroup$ There are some results in the literature that prove this monotonicity property for fairly small t but I don't recall offhand how small. $\endgroup$ – Matt Young Feb 24 '12 at 3:44
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    $\begingroup$ In fact the monotonicity property holds for $|t|>5/4$ but fails for $|t|\leq 1$. See math.ca/10.4153/CMB-2010-107-8 $\endgroup$ – GH from MO Feb 25 '12 at 0:59
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This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so. If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral $$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$ around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero $\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and $1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four conjugates of the root found by joro. A similar argument applies for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

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    $\begingroup$ Wonderful. You can simplify and strengthen the proof by using a generalized Rouché's theorem. This tells us that $\Gamma(z)+\theta\cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$ when $|\Gamma(1-z)|<|\Gamma(z)|$ holds on the boundary. This is equivalent to $\pi/|\sin(\pi z)|<|\Gamma(z)|^2$, hence it suffices to have $\pi<|\Gamma(z)|^2$ on $\partial C_n$. It seems that the last inequality holds for $n\geq 5$. $\endgroup$ – GH from MO Feb 25 '12 at 0:15
  • $\begingroup$ Very impressive, although I honestly have to say that fully understanding the proof is beyond my math skills. Still got the goosebumps from reading it though :-) The proof does induce two follow up questions: 1) could the function $\Gamma(s)^2 - \Gamma(1-s)^2$ be uniquely represented by an infinite product involving its 'complex' zeros (via Weierstrass factorization)? 2) is there a function for locating the zeros (similar to $Z(t)$ for the Riemann non trivial zeros)? Thanks. $\endgroup$ – Agno Feb 25 '12 at 0:33
  • $\begingroup$ @Agno: Rouché's theorem is contained in basic textbooks, and this is all you need (actually a slight generalization of it). Using this you can shorten the above proof to a few lines (e.g. no integrals), see my comment above. $\endgroup$ – GH from MO Feb 25 '12 at 0:39
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    $\begingroup$ @GH: Rouché? Touché! $\endgroup$ – user631 Feb 25 '12 at 3:19
  • $\begingroup$ @Agno: the logarithmic derivative is the tool to count zeros and it is always available. $\endgroup$ – Marc Palm Mar 6 '12 at 18:30
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I would like to expand on Guild of Pepperers's answer by noting that the zeros are essentially uniformly spaced and may easily be approximated to a high degree of accuracy. Using Stirling approximation, I obtained the formula $$ \Gamma\left(\frac12+it\right) = \sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}\exp\left(-\frac\pi2|t|+i(t\log|t|-t+\varepsilon(t))\right), $$ valid for real $t$, where the error $\varepsilon(t)$ is an odd, bounded, real-valued function asymptotically equal to $\frac{1}{24t}$. (Indeed, $\varepsilon(t)$ has asymptotic and convergent expansions coming from the asymptotic and convergent versions of Stirling approximation, respectively.) We then have, for $s = \frac12+it$ on the critical line, $$ \Gamma(s)+\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\cos\left(t\log|t|-t+\varepsilon(t)\right), $$ $$ \Gamma(s)-\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\sin\left(t\log|t|-t+\varepsilon(t)\right). $$ One may show by means fair or foul that $t\log|t|-t+\varepsilon(t)$ is monotonically increasing for $|t|\geq1.05$, is bounded between $-0.96$ and $0.96$ for $|t|<1.05$, and is only zero when t = 0. Therefore, the zeros of $\Gamma(s)+\Gamma(1-s)$ on the critical line occur, with multiplicity one, very near those $t$ for which $t\log|t|-t$ is an odd integer multiple of $\frac{\pi}{2}$, and similarly for $\Gamma(s)-\Gamma(1-s)$ and the even integer multiples of $\frac{\pi}{2}$.

It's interesting that the number of zeros up to a given height $T$ is of the same order of magnitude, $T \log(T)$, as for the Riemann zeta function, but that these zeros have (essentially) uniform spacings rather than GUE spacings.

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Just to point out there are very good approximation to complex zeros off your line of $$ \Gamma(s)-\Gamma(1-s) \qquad(1)$$

At $\rho \approx -1.69711183621729718874218687438 - 0.305228379993226071272967719419 i$ (1) appears to vanish while $\Gamma(\rho) \approx 1.4648039 + 0.3642699441 i$

Root finding with better precision converges to $\rho$ while (1) still appear to vanish in both sage and gp/pari (modulo bugs).

Checked to precision $5000$ digits and (1) still appears to vanish.

Here is $\rho$ with $100$ digits of precision:

-1.697111836217297188742186874382163077146364585981726518217373889827452772242797069678994954785699956 - 0.3052283799932260712729677194188512919331197338088909477524842921187943642970297308885952936796125572*I

... for $ \Gamma(s)+\Gamma(1-s)$ approximation of zero appears $\rho \approx -0.60940537628997711023 - 0.82913081575572747216 i$ checked to $5000$ digits of precision.

With 100 digits:

 -0.6094053762899771102337308158313839002012166649163876907688596366808893391382113824494098816671945331 - 0.8291308157557274721587141536678087800797120641344787653174391388417832472543392187032283839972409848*I

Edit In comments juan suggested using x-ray to investigate the zeros.

The primary reference for x-ray I found is X-Ray of Riemann zeta-function, J. Arias-de-Reyna

AFAICT x-ray are the plots of Re(f(s))=0 and Im(f(s))=0. The zeros are the intersection.

The x-ray and juan's comments suggest the above quadruples of zeros are indeed zeros off $\frac12$ and possibly there are no more complex zeros zeros off the line.

Here is the x-ray of $ \Gamma(s)-\Gamma(1-s)$. Blue is $\Re(\Gamma(s)-\Gamma(1-s))=0$ and red is $\Im(\Gamma(s)-\Gamma(1-s))=0$.

(source)


(source)

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  • $\begingroup$ @Joro, you are the "Master of the Counter Example" :-) As you pointed out, the zeros of $\Gamma(s) \pm \Gamma(1-s)$ are very small, so precision of the calculation can be an issue, however with 5000 digits accuracy, your two counter examples could also easily fall in the category: (...)this reduces the claim to z=s+it with |s|≤15 and |t|≤4 where the claim can be checked directly(...). If your counter examples are correct, then my only escape is to restrict the claim to the critical strip only (similar to the $\zeta(s)$ and $\zeta^{(k)}(s)$ equivalents). $\endgroup$ – Agno Apr 9 '12 at 11:04
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Since we have

$\Gamma(1/2+it)=\sqrt{\pi/\cosh(\pi t)}\exp[i(2 \vartheta(t)+t \log(2\pi)+\arctan(\tanh(\pi t/2)))]$

where $\vartheta(t)$ is the Riemann Siegel function. The zeros on the critical line have ordinates the zeros of

the cosine or sine of the real function

$2 \vartheta(t)+t \log(2\pi)+\arctan(\tanh(\pi t/2))$

But there are real zeros, for example there is one at $s = 4.0260426340124070065475\dots$

X-ray:

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    $\begingroup$ juan, how do you explain the approximations of zeros in my answer with precision 100 checked in pari, sage and maple? Have I misunderstood the question? $\endgroup$ – joro Apr 9 '12 at 9:26
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    $\begingroup$ With mpmath I check also this zero. I think now that my parenthesis "(with some effort)" contains an error. :-/ What it is clear is that the real zeros are those of the function given. But the behavior of this function for t complex is really not simple. I will try to do an X-ray of this function. Later we will try to post it. If I know how to do it. $\endgroup$ – juan Apr 9 '12 at 19:39
  • $\begingroup$ Apparently I can not post here a plot. The x-ray gives little doubt that the zeros of $\Gamma(s)-\Gamma(1-s)$ are the ones with real part $1/2$ that appeared computed in this question. The complex at $-1.69-0.30 i$ its complex conjugate the symmetrical of this with respect the critical line $2.69+0.30i$ and its complex conjugate. And then the real zeros one at $0.5$. The others real zeros can be obtained best from a real plot of the function. In the x-ray this zeros, that are very near the poles at $4$, $5$, $\dots$, can not be seen since they are contained in very short lines. $\endgroup$ – juan Apr 10 '12 at 16:11
  • $\begingroup$ Of course the real zeros are symmetric with respect to 0.5 so that there are zeros near $-3$, $-4$, $\dots$ $\endgroup$ – juan Apr 10 '12 at 16:24
  • $\begingroup$ juan, you can post images (possibly unless you don't have enough reputation). The format is HTML, i.e. write <img src="server/file.gif"> in your answer. The x-rays i found are at drememi.ludost.net/gamma1.png and drememi.ludost.net/gamma2.png $\endgroup$ – joro Apr 11 '12 at 5:19
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Given that $\Gamma(s)$ and $\Gamma(1-s)$ are complex conjugates when $\Re(s)=1/2$, it is not surprising that $\Gamma(s)+\theta\Gamma(1-s)$ has an infinitude of zeros on the line $\Re(s)=1/2$, as long as $|\theta|=1$. The monotonicity argument given in the first answer then shows that there are no other zeros with $0<\Re(s)<1$. With the possible exception when the imaginary part of $s$ is small, the zeros for two different $\theta$ should interlace (if $\theta$ goes around the unit circle once, a zero is carried to an adjacent zero).

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