User:Sure Beae: Difference between revisions

My website is https://foil.town/

I’m interested in maths, FreeBSD/HardenedBSD, cybersecurity, networking, and biology. Specifically, discrete calculus, and its relations/intersections with Riemann-Liouville fractional calculus, analytic number theory, polypi (offset polygamma) function (see A generalized polygamma function by Oliver Espinosa and Victor H. Moll for a related function with nice analytic properties), the generalized Bernoulli numbers, and so on.

Here is a graph based on the YouTube video How to Extend the Sum of Any* Function

https://www.desmos.com/calculator/xjzu4jqghl

Here is the exact version that isn’t using the difference between two divergent summations

https://www.desmos.com/calculator/y2i9qmoqjn

Here is a graph of the famous Faulhaber’s formula, which is misnamed as Faulhaber never wrote or interacted with it directly.

https://www.desmos.com/calculator/ywv6hjiq0l

Hasse’s Newton-like approx

https://www.desmos.com/calculator/eyz2xo1fa8

Polygamma/Polypi

https://www.desmos.com/calculator/tdbchepkst

https://www.desmos.com/calculator/60yjlcu1ko

A fun game I would play when I was in school, doing complex mappings using the zeros of 3D trig-based functions, here’s 1/z as example

https://www.desmos.com/3d/weknvqvovz

The easiest way to contact me on other platforms is ‘sure’ on IRC or @bord:chat.foil.town on matrix.

Notes to myself

We’ll start off with extending ∑ k = 1 x 1 k p {\displaystyle \sum _{k=1}^{x}{\frac {1}{k^{p}}}} , the generalized harmonic number, into a function. First, we need to recognize two things:

1 x p + 1 = 1 Π ( p ) ∫ 0 ∞ t p e − t x d t , for  x > 0  and  ℜ ( p ) > − 1 {\displaystyle {\frac {1}{x^{p+1}}}={\frac {1}{\Pi \left(p\right)}}\int _{0}^{\infty }t^{p}e^{-tx}dt,\quad {\text{for }}x>0{\text{ and }}\Re (p)>-1}

and that ∑ k = 1 x e − k t {\displaystyle \sum _{k=1}^{x}e^{-kt}} can be extended by using ∇ − 1 {\displaystyle \nabla ^{-1}} . We recognize the geometric series and write:

∇ − 1 e − x t = e − t − e − ( x + 1 ) t 1 − e − t {\displaystyle \nabla ^{-1}e^{-xt}={\frac {e^{-t}-e^{-\left(x+1\right)t}}{1-e^{-t}}}}

Now we’re ready for the manipulation:

1 x p = 1 Π ( p − 1 ) ∫ 0 ∞ t p − 1 e − t x d t {\displaystyle {\frac {1}{x^{p}}}={\frac {1}{\Pi \left(p-1\right)}}\int _{0}^{\infty }t^{p-1}e^{-tx}dt}

take the antidifference of both sides:

∇ − 1 1 x p = 1 Π ( p − 1 ) ∫ 0 ∞ t p − 1 e − t − e − ( x + 1 ) t 1 − e − t d t {\displaystyle \nabla ^{-1}{\frac {1}{x^{p}}}={\frac {1}{\Pi \left(p-1\right)}}\int _{0}^{\infty }t^{p-1}{\frac {e^{-t}-e^{-\left(x+1\right)t}}{1-e^{-t}}}dt}

Our extension of the generalized harmonic number here is valid for ℜ ( p ) > 0 {\displaystyle \Re (p)>0} . From here, you can take the limit as x {\displaystyle x} approaches infinity, and obtain the integral for ζ ( p ) {\displaystyle \zeta (p)} . However, we can see that the limit as x approaches infinity in the indefinite sum and resulting integral for ζ ( p ) {\displaystyle \zeta (p)} do not converge for p < 1 {\displaystyle p<1} . There isn’t a way to logically directly extend this limit as x approaches infinity other than using analytic continuation as Bernhard Riemann did in Ueber die Anzahl der Primzahlen unter einergegebenen Grösse.

We can, however, extend the indefinite sum directly. ∇ − 1 x p {\displaystyle \nabla ^{-1}x^{p}} can be extended to an entire and holomorphic function only using the limit of Müller’s method and Newton series. Then, we can take the partial derivative of ∇ − 1 x p {\displaystyle \nabla ^{-1}x^{p}} with respect to x {\displaystyle x} evaluated at x = 0 {\displaystyle x=0} and obtain a function that extends the Bernoulli numbers, is entire, holomorphic, and doesn’t require using analytic continuation. The location of the trivial zeros changes to odd integers 3 and greater (given by Faulhauber’s formula producing polynomials which have an even multiplicity root at x=0 for those values of p rather than the sine term in the reflection formula), whilst the location of the nontrivial zeros and critical strip remain in the same location.

Π ( s 2 ) ( 2 τ ) s 2 B ( 1 − s ) = Π ( 1 − s 2 ) ( 2 τ ) 1 − s 2 B ( s ) {\displaystyle \Pi \left({\frac {s}{2}}\right)\left({\frac {2}{\tau }}\right)^{\frac {s}{2}}B\left(1-s\right)=\Pi \left({\frac {1-s}{2}}\right)\left({\frac {2}{\tau }}\right)^{\frac {1-s}{2}}B\left(s\right)}

https://www.desmos.com/calculator/ppmuwcuhaw

Reflection formulas

B ( s ) = Π ( s 2 ) Π ( 1 − s 2 ) ( 2 τ ) s − 1 2 B ( 1 − s ) {\displaystyle B\left(s\right)={\frac {\Pi \left({\frac {s}{2}}\right)}{\Pi \left({\frac {1-s}{2}}\right)}}\left({\frac {2}{\tau }}\right)^{s-{\frac {1}{2}}}B\left(1-s\right)}

I prefer

B ( s ) = Π ( s 2 ) Π ( 1 − s 2 ) ( τ 2 ) 1 2 − s B ( 1 − s ) {\displaystyle B\left(s\right)={\frac {\Pi \left({\frac {s}{2}}\right)}{\Pi \left({\frac {1-s}{2}}\right)}}\left({\frac {\tau }{2}}\right)^{{\frac {1}{2}}-s}B\left(1-s\right)}

B ( s ) = 2 τ − s Π ( s ) cos ⁡ ( τ s 4 ) B ( 1 − s ) 1 − s {\displaystyle B(s)=2\tau ^{-s}\Pi \left(s\right)\cos \left({\frac {\tau s}{4}}\right){\frac {B\left(1-s\right)}{1-s}}}

Do consider here that ( − i ) s + i s = 2 cos ⁡ ( τ s 4 ) {\displaystyle \left(-i\right)^{s}+i^{s}=2\cos \left({\frac {\tau s}{4}}\right)} .

B ( s ) = τ − s Π ( s ) ( ( − i ) s + i s ) B ( 1 − s ) 1 − s {\displaystyle B\left(s\right)=\tau ^{-s}\Pi \left(s\right)\left(\left(-i\right)^{s}+i^{s}\right){\frac {B\left(1-s\right)}{1-s}}}

Read

https://www.claymath.org/wp-content/uploads/2023/04/Wilkins-translation.pdf

if you ever forget how you came up with this. Everything lines up quite nicely. Also, remember the exact values of the Riemann Zeta function Euler found with is just an application of this reflection formula.

Euler Product

∂ ∂ x ( ∇ − 1 x s ) | x = 0 = { − s ∏ p 1 1 − p s − 1 for  s < 0 − 2 τ − s Π ( s ) cos ⁡ ( τ s 4 ) ∏ p

1 1 − p − s for  s > 1 {\displaystyle {\frac {\partial }{\partial x}}\left(\nabla ^{-1}x^{s}\right){\bigg |}_{x=0}={\begin{cases}-s\prod _{p}{\frac {1}{1-p^{s-1}}}&{\text{for }}s<0\\-2\tau ^{-s}\Pi \left(s\right)\cos \left({\frac {\tau s}{4}}\right)\prod _{p}^{}{\frac {1}{1-p^{-s}}}&{\text{for }}s>1\end{cases}}}

https://arxiv.org/pdf/2009.06743