about Riemann's hipothesys

[hgjf2]

The Rieman's Hypothesys believery wrong.
Let Z[z] = 1^(-z)+2^(-z)+3^(-z)+4^(-z)+...+n^(-z), where z<-C a complex number z=a+bi.
Z[z] is Zeta Riemann function.
I looked that Z[2+bi] is unperiodic because Z[z] = exp(0)+exp(-z*ln 2) + exp(-z*ln 3) + exp(-z*ln 4) + ... just a infinity sum a exponentials put into logarithmic order.
If b<-R then, exp(-z*ln k) has a period by 2TT/ln(k) where TT=3,1415926535.. the length of circle divided by the diameter.
Than Z[2+bi] has period 2TT*f(ln(2),ln(3),ln(4),...,ln(n)) where f(a1,a2,a3,a4,...an) is the lest some multiply.
How ln(2) is irrational numer f(1, ln(2)) = infinity = 1/0.
Z[2+bi] has unperiodic.
Just is very complex for searching values z for Z[z]=0 =0+0i.

[hgjf2]

The Rieman's Hypothesys believery wrong.
Let Z[z] = 1^(-z)+2^(-z)+3^(-z)+4^(-z)+...+n^(-z), where z<-C a complex number z=a+bi.
Z[z] is Zeta Riemann function.
I looked that Z[2+bi] is unperiodic because Z[z] = exp(0)+exp(-z*ln 2) + exp(-z*ln 3) + exp(-z*ln 4) + ... just a infinity sum a exponentials put into logarithmic order.
If b<-R then, exp(-z*ln k) has a period by 2TT/ln(k) where TT=3,1415926535.. the length of circle divided by the diameter.
Than Z[2+bi] has period 2TT*f(ln(2),ln(3),ln(4),...,ln(n)) where f(a1,a2,a3,a4,...an) is the lest some multiply.
How ln(2) is irrational numer f(1, ln(2)) = infinity = 1/0.
Z[2+bi] has unperiodic.
Just is very complex for searching values z for Z[z]=0 =0+0i.

The graphs for Z[2+bi] as real part (just    [ cos(0) + cos(-z*ln2) + cos(-z*ln3) + cos(-z*ln4)]   ) rendered by VISUAL BASIC for ACCESS are:
 

[cKleinhuis]

attachment size is now limited to 256 k sorry, but try jpg ?!?!?

[hgjf2]

I'm trying to limit 256kb as JPG

[Sockratease]

I'm sorry . Yet I can't upload the images because another users has posted too many videos

HOW MANY TIMES DO YOU NEED TO BE TOLD THAT NOBODY CAN UPLOAD VIDEOS HERE?

LAST WARNING.  

NEXT TIME YOU SAY THAT I WILL DELETE THE POST IMMEDIATELY!!    

[hgjf2]

The trouble image was vanished
The site don't accept BMP images

[cKleinhuis]

bmp is not allowed because bmp has no compression inherent, and thus consuming too much space...

[cKleinhuis]

back to topic ... @sock, please calm down

[hgjf2]

bmp is not allowed because bmp has no compression inherent, and thus consuming too much space...

Now I stood new posting rules

[hgjf2]

The graphic Z[a+bi] rendered as a complex function is similar with a unperiodic watering vortex.
If a<0 the graphic can be render because the sum going to infinity

[hgjf2]

The graphic Z[a+bi] rendered as a complex function is similar with a unperiodic watering vortex.
If a<0 the graphic can be render because the sum going to infinity

Logical:
1+2+3+4+5+... = n(n-1)/2
1+sqrt(2)+sqrt(3)+ ... = 1+1.41421+1.73205+ ... + = NaN (to infinity)

[hgjf2]

The function Z[a+bi] where a>1.5 as example is inside a complex number circle with center in 0+0i and with ray m=1+2^(-1.5)+3^(-1.5)+4^(-1.5)+...
The graphic is a route of a tick tock clock whick have a infinity articulation and have rotations with the different speeds and period 2TT/ln(k).
Normally the complex track is a unperiodical curl.
The curl can touch the center 0+0i.
 

[hgjf2]

I looked that the irrational number from square root can be infinity sum with number like Catalan:
Let a series Catalan: 1;1;2;5;14;42;132;429;...;
Let's put into infinity sum: 1+1/100+2/10000+5/1000000+14/100000000

[hgjf2]

I looked that the irrational number from square root can be infinity sum with number like Catalan:
Let a series Catalan: 1;1;2;5;14;42;132;429;...;
Let's put into infinity sum: 1+1/100+2/10000+5/1000000+14/100000000

Here obtain 1,010205144337 what is irrational number because the square of this number give 1,0205144337 result of equation
x^2=100x-100 were x give 50+-sqrt(2400) = 50+-20*sqrt(6) . Normally sqrt(6) is irrational.
 

[hgjf2]

Here obtain 1,010205144337 what is irrational number because the square of this number give 1,0205144337 result of equation
x^2=100x-100 were x give 50+-sqrt(2400) = 50+-20*sqrt(6) . Normally sqrt(6) is irrational.

If look the order of numbers, the order seem as Rieman complex set like Z[2.5+ni] or Z[3+ni] defined n<-R.
Yet any, the final digit on Catalan number set are cvasiperiodic order: 1;1;2;5;4;2;2;9;0;2;6;6;2;0;0;5;0;0;0;0;0;0;0;0;1;...
Exact as components of sum of logarithms from Zeta function.