I – 6 Spezial Themen -Special Themes-

I- 6- 1- Zeta function

https://complex-pictures.com/i-5-potenzieren-mit-wurzel-1-i/When you use and think about complex numbers, you come to Bernhard Riemann (1826-1866) and his basic works on the zeta-function. We begin with a phase portrait of it (s.Lit.(1) p. 202, 214) , see Wipikedia ‚Riemann‘ Zetafunktion‘ and we work on the following question: is it of any use to operate with –to the power of -i- to get a clearer presentation of the trivial and non trivial zeros? It is our opinion that the ‚blue‘ spots in fig.1,3,5,7 represent such zeros.

Even if you add up the definition zeta-sum to 4000, you receive only a poor picture of the function. Zeros at 14.2 , 21.1, 24.9, 30.3, 33.1 and 37.7 could roughly be seen but they are numerically not precise and show additional false ‚zeros'(Lit(1),p202,214). To calculate the following portraits we used matlab (R), other computer languages like Wolfram Alfa (R) do these calculations too. 40.000 calls of the zeta commands affords a relatively long calculation time (> 4 – 5 Min).

Wer sich mit komplexen Zahlen beschäftigt, kommt an Bernhard Riemann (1826-1866) und seinen Arbeiten zu der Zeta-Funktion nicht vorbei. Beginnen wir mit einem Phasen-Portrait der Zeta-Funktion (vgl. Lit(1) S.202 u.,214 ; und Wipikedia ‚Riemann Zetafunktion“ ) und der Frage, ob das Potenzieren mit i eine übersichtliche Darstellung der trivialen und nichttrivialen Nullstellen erlaubt. Unserer Meinung nach stellen die ‚blauen‘ Punkte in Abb.: 1,3,5,7 Nullstellen dar.

Figure 1: f(x) =( Σ 1/(n^z) ) ^ i , n=1 to 4000 .

Selbst bei der Summe bis 4000 in Abb.: 1 ergibt sich nur eine sehr unvollkommene Annäherung an die Funktion (Lit(1) S.215) . Nullstellen um 14,2 , 21,1, 24,9, 30,3 , 33,1 und 37,7 mit x ca. 0.65 deuten sich zwar deutlich an, sie sind allerdings numerisch ungenau und es gibt zahlreiche Falschsignale. Die Funktionsberechnungen der folgenden Phasenpotraits wurden mit Matlab (R) vorgenommen; auch andere Programmsysteme z.B. Wolfram Alpha, berechnen die zeta-Funktion für den Zahlenbereich komplexer Zahlen. Für 40.000 Berechnungen der Funktion wird die Computer-Belastung jedoch erheblich ( >4-5 Min).

Fig. 2 : Zeta ( ζ ) f = zeta(200*200 Matrix) .
Fig. 3: Zeta (200*200 matrix) ^ i .
Fig. 4: Zeta (200*90 pixel) trivial zeros, -2, -4, -6 ..
Fig. 5: see Fig.: 4 , power i , trivial zeros , -2, -4, -6 …
Fig.: 6 , 7 Zeta function tests, left : g(z) ^2 with g(z) = f(z) ^i and right : f(z ) ^ i . Fig. 6 and 7 schow a clear visualisation of the zeros .

Fig. 8: Zeta zeros from 0 to 30i .
Fig. 9: Zeta zeros 30i to 80i .

Fig. 10: Zeta zeros 0 to 27*i , x = 0.1 to 0,9 , at x=0.5 the zero line is touched

I- 6- 2 Gamma function, ( Γ ) . Lit(2) Handbuch der Theorie der Gammafunktion , 1906 , Dr. Niels Nielson, Kopenhagen .

Fig. 11: Complex gamma function ( Γ ) f = sqrt(2pi) *( z ^(z – 0.5) ) *exp( -z ) = sqrt(2pi) *exp((z-0.5)*log(z)) *exp(-z); approximation with Stirling-formula, 3800*3800 pixel .
Fig. 12: (like Fig. 11 with two correction terms – * (1+1/(12*z) + 1/(288*z^2 ) . Mostly the pictures have ca. 4 *10^6 pixel .
Fig. 12a: enlarged by factor 25

f = derivative of gamma function, two correction terms

Fig. 13: f = sqrt(2pi) *(exp(z)*(z^(z-3/2)*(z-1/2) + z^(z-1/2)*log(z))*(1/(12z) – 1/(288*z^2) +1) – z^(z-1/2) *exp(-z) * (1/(12*z) – 1/(288*z^2)+1 -z^(z(-1/2) * exp-z)*(1/(12*z^2) – 1/(144*z^3)) ). – derivative formula produced by matlab (R) symbolic toolbox-. (Here it is the derivative built ‚by hand‘ without correction terms:

Γ ‚ =sqrt(2pi) *(exp((z-1/2)*log(z))*(log(z)+(z-1/2)/z )*exp(-z)) – (exp(-z))*(exp((z-1/2)*log(z)))

Fig. 14 : Γ- function (Product representation , Euler) f = ( 1/z ) * Π ((1 + 1/n)^z) /(1 + z/n) , n=1:4000

The phase portrait of the Euler/Mascheroni/Weierstrass definition f = exp(-0.577216*z) / z * Π exp(z/n) * (1+ z/n )^-1 is as assumed totally identical, but the computation times are much longer.

  • Fig. I- 6- 15 Γ ‚/ Γ f = c+ Π (1/ (z + n) – 1/( 1 +n) ) , n= 0 : 2000, c=0 .
In Fig. 15: we see the negative digamma function, Γ ‚/ Γ., ca 4 mio. pixel .

In Wikipedia we found some phase portraits of the zeta- and gamma function as well as of their derivatives , they came from Wiki -Common. Unfortunately no detailed information of the used formulas for these pictures were given. The authors called themselves ‚ Rumil ‚ or ‚ Empetrisor ‚ with no given way of contacting them .

  • Phi-function f ^i , rotated
  • Phi-function or Digamma-function f ^i ; n=0:2000

Fig. I- 6- 16 / 16a Γ ‚/ Γ Phi-function or Digamma – function to the power of i , f = – Σ ( 1 / (z + n) – 1/ (1 + n)), n=0:2000, left rotated.

  • f = log(gammafunction two correction terms)
  • f = log(gammaf. two correction terms ) ^i .
Fig.17 /17a: f = log( Γ ) and f = log( Γ ) ^i
  • Fig. 18 ( 1/ ( Γ(z) ) ^i , Γ(z) – Eulerproduct, n=1:2000
  • Fig. 19 1 / Γ(z) , Γ(z) – Euler (fig.14) , n=1:2000

We cannot detect any severe differences between gamma and 1 over gamma in this projection.

Fig. 20: f = Γ(z)‘ 1. Derivative of gamma function ; Γ(z) ‚/ Γ(z) = (here called Phi – function multiplied by gamma function , results in Γ(z)‘ ; see Fig . 15 and 14 , 2.200*2.200 pixel . This picture is also shown in Wekipedia ‚Gamma-Functioen‘ (author ‚Empetrisor‘). It reveals that the derivative of the Stirling approximation in Fig.13 yields only a poor picture of the function fig. 20 .
  • Fig f 2. deriv gamma function

In summary both groups of functions of section I- 6- mostly do not really contribute to attractive pictures to this website. The zeros of the zeta -function are clearly visible but not always to the wanted extent of focus. We feel that only Fig. 16 and 17 are prominent examples to see the zeros more clearly.

  • Fig . 21 f = 2. deriv. gamma function
  • Fig. 22 function as fig 21 ^ i
The author is of the opinion that the 2. and third derivative of Γ has not yet been shown in the literature. Calculation Second derivative Γ “ = (Γ ‚/Γ) ‚ * Γ + (Γ ‚/Γ) * Γ ‚ .
  • Fig. 23 3rd derivative of gamma function
The 3rd derivative calculates as following : Γ “‘ = (Γ ‚/Γ) “ * Γ + 2* (Γ ‚/ Γ) ‚ * Γ ‚ + (Γ ‚/ Γ) * Γ “ . With the obtainable derivatives of the poly gamma functions it is straight forward to calculate all derivatives of the gamma function.