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Riemann zeta function is a function of complex variable $s$ that analytically continous the sum of Dirichlet series .defined as :$$\zeta(s)=\sum_{n=1}^{\infty}\displaystyle \frac{1}{n^s} $$ for when the real part is greater than $1$.

My question here is: Could the Riemann zeta function be a solution for a known differential equation?

Note: I would like if there is a paper or ref show that zeta function presented a solution for known Differential equation.

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    $\begingroup$ Note from zeta function universality (en.wikipedia.org/wiki/Zeta_function_universality) that $\zeta$ cannot obey any non-trivial autonomous differential equation (one which does not explicitly involve $s$). $\endgroup$ – Terry Tao Jan 21 '17 at 23:06
  • $\begingroup$ @TerryTao, thanks for your constructive comment , really what let me to ask this question is to know how do i seek for the Hilbert "Chaotic operator" $\endgroup$ – zeraoulia rafik Jan 22 '17 at 20:51
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    $\begingroup$ @TerryTao: May I ask what the rationale for the implication is? I ask because I just learned about a differential equation that appears to be similarly "universal" (in the sense that any continuous function can be approximated by some solution of it arbitrarily well), but nevertheless autonomous (the equation being 3x'⁴x''x''''² − 4x'⁴x'''²x'''' + 6x'³x''²x'''x'''' + 24x'²x''⁴x'''' − 12x'³x''x'''³ − 29x'²x''³x'''² + 12x''⁷ = 0). $\endgroup$ – Mehrdad Jan 23 '17 at 10:06
  • $\begingroup$ @Mehrdad , I think he meant since Riemann zeta fuinction is is "hypertranscendental" this cannot be a solution for algebraic differential equation $\endgroup$ – zeraoulia rafik Jan 23 '17 at 12:25
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    $\begingroup$ Because the zeta function is holomorphic (away from the pole at 1 of course), the approximation in zeta function universality is not just in the uniform topology, but is in fact in the smooth topology, by elliptic regularity. So if $\zeta$ obeyed any differential equation of the form $F( \zeta, \zeta', \dots, \zeta^{(k)}) = 0$ for some continuous $F$, then this equation $F( f, f', \dots, f^{(k)}) = 0$ has to be obeyed for any holomorphic $f$, and it is easy to see from power series expansion that no such non-trivial equation exists. $\endgroup$ – Terry Tao Jan 24 '17 at 0:24
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When posed properly, a long-standing open problem, but in the form you ask:

Robert A. Van Gorder, MR 3276353 Does the Riemann zeta function satisfy a differential equation?, J. Number Theory 147 (2015), 778--788.

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    $\begingroup$ From the abstract of this paper: In Hilbert's 1900 address at the International Congress of Mathematicians, it was stated that the Riemann zeta function is the solution of no algebraic ordinary differential equation on its region of analyticity. It is natural, then, to inquire as to whether $\zeta(z)$ satisfies any non-algebraic differential equation. [continued...] $\endgroup$ – Gerry Myerson Jan 21 '17 at 22:53
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    $\begingroup$ [continued...] In the present paper, an elementary proof that $\zeta(z)$ formally satisfies an infinite order linear differential equation with analytic coefficients, $T[\zeta−1]=1/(z−1)$ , is given. We also show that this infinite order differential operator $T$ may be inverted, and through inversion of $T$ we obtain a series representation for $\zeta(z)$ which coincides exactly with the Euler–MacLauren summation formula for $\zeta(z)$ . Relations to certain known results and specific values of $\zeta(z)$ are discussed. $\endgroup$ – Gerry Myerson Jan 21 '17 at 22:54
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    $\begingroup$ @GerryMyerson Thank you for the abstract which is attractive, it seems to me however that one should write Euler–MacLaurin $\endgroup$ – Duchamp Gérard H. E. Jan 22 '17 at 9:30
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    $\begingroup$ @GerryMyerson O.K. you are right, this is how it is written in the paper. It is then a disagreeable surprise to see that this passed the hurdles of author and referees :). This does not hold me back trying to have a look. $\endgroup$ – Duchamp Gérard H. E. Jan 22 '17 at 20:10
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    $\begingroup$ "When posed properly, a long-standing open problem" Igor, what precisely do you refer to? $\endgroup$ – Gil Kalai Aug 7 '17 at 12:12
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The Riemann zeta function is "hypertranscendental" in the sense shown HERE

It is not the solution $y(x)$ of a differential equation of the form $$ F(x,y,y',y'',\dots,y^{(n)})=0 $$ where $F$ is a polynomial (with constant coefficients).

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  • $\begingroup$ @ Gerald Edgar , Why you don't cited exeception in the case of the Riemann Hypothesis is true ? I meant it's will satisfy ODE or any PDE (algebraic from ) iff RH is true $\endgroup$ – zeraoulia rafik Jan 24 '17 at 21:31
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The fact that $\zeta$ satisfy no algebraic differential equation is due to its famous relation with the Gamma function which was proved by Hölder not to satisfy such an equation.

Detailed answer can be found here with five (commented and linked) references.

On the other hand, this function is linked with many other transcendental special functions like polylogarithms which satisfy Fuschian non commutative differential equations.

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Hint: This is just a note for my question , Generalised Riemann zeta function however it is hypertranscendental but it satisfies difference equation as shown in this paper theorem 5.1 and here hypertranscendental function couldn't satisfies any algebraic differential or differences equation .

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