Title: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 03:31:42 PM
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.
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.
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 03:39:55 PM
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: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.
Title: Re: about Riemann's hipothesys
Post by: cKleinhuis on July 15, 2012, 03:45:45 PM
??? attachment size is now limited to 256 k sorry, but try jpg ?!?!?
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 04:01:41 PM
I'm trying to limit 256kb as JPG
Title: Re: about Riemann's hipothesys
Post by: Sockratease on July 15, 2012, 04:02:27 PM
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. :police:
NEXT TIME YOU SAY THAT I WILL DELETE THE POST IMMEDIATELY!! :siren: :siren:
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 04:04:43 PM
The trouble image was vanished
The site don't accept BMP images
Title: Re: about Riemann's hipothesys
Post by: cKleinhuis on July 15, 2012, 04:06:59 PM
bmp is not allowed because bmp has no compression inherent, and thus consuming too much space...
Title: Re: about Riemann's hipothesys
Post by: cKleinhuis on July 15, 2012, 04:07:29 PM
back to topic ... @sock, please calm down ;)
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 04:10:13 PM
bmp is not allowed because bmp has no compression inherent, and thus consuming too much space...
Now I stood new posting rulesTitle: Re: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 04:23:58 PM
If a<0 the graphic can be render because the sum going to infinity
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 04:25:42 PM
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:If a<0 the graphic can be render because the sum going to infinity
1+2+3+4+5+... = n(n-1)/2
1+sqrt(2)+sqrt(3)+ ... = 1+1.41421+1.73205+ ... + = NaN (to infinity)
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on July 15, 2012, 04:31:31 PM
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.
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.
Title: similarity between Rieman function and irrational squares root
Post by: hgjf2 on August 12, 2012, 02:38:51 PM
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
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
Title: Re: similarity between Rieman function and irrational squares root
Post by: hgjf2 on August 12, 2012, 02:42:46 PM
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 equationLet 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
x^2=100x-100 were x give 50+-sqrt(2400) = 50+-20*sqrt(6) . Normally sqrt(6) is irrational.
Title: Re: similarity between Rieman function and irrational squares root
Post by: hgjf2 on August 12, 2012, 02:50:33 PM
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.x^2=100x-100 were x give 50+-sqrt(2400) = 50+-20*sqrt(6) . Normally sqrt(6) is irrational.
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.
Title: Re: about Riemann's hipothesys
Post by: jehovajah on September 15, 2012, 09:51:19 AM
So, why is this important to know?
The Riemann Zeta function and hypothesis was used to develop a strange branching fractal, if I remember correctly. It was referenced in this forum some years ago now. How would your insight affect general fractal production, zooming and surface determination, even colour cycling?
A lot of questions I know,mbut I am in no rush . Take your time to develop your insight.
The Riemann Zeta function and hypothesis was used to develop a strange branching fractal, if I remember correctly. It was referenced in this forum some years ago now. How would your insight affect general fractal production, zooming and surface determination, even colour cycling?
A lot of questions I know,mbut I am in no rush . Take your time to develop your insight.
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on June 25, 2016, 12:15:02 PM
A new hint founded for to solve this the Riemann's hypothesis: I founded few formulas on sites: the relationship between the numbers of Bernoulii and the function Zeta : Z(2n)=[(-1)^(n+1)]*{[(2pi)^(2n)]/2*(2n)!}*B(2n) and Z(-n)=-[-B(n+1)]/(n+1) and the hint formula:
Z(1-s)=2[(2pi)^(-s)]*[cos(s*pi/2)]*(s-1)!*Z(s) the "holly grail".
Z(z) is the Zeta function. Z(z)=1+1/(2^z)+1/(3^z)+1/(4^z)+... when z=a+bi.
B(n) are the numbers of Bernoulli: 1;-0,5;0,1(6);0;-0,0(3);0;0,0(238095);0;-0,0(3);0;0,0(75);0;-0,2(531135);0;1,1(6);0;-7,09216...;0;54,97118...;0;
-529,1(24);0;6192,12319...;0;-86580,2(531135);0;1425517,1(6);0;-27298231,07...;0;601580873,9;0;-1511631576.7...;0;... etc.
Also the function Zeta is good helpfull for who is curious how made B(z) when z<-C\{N} where N is the lot of the natural numbers.
Also B(-1)=(pi^2)/6=1,6449340668... and B(-2)=2,40405... and B(-3)=(pi^4)/30=3.2469697011... etc.
If the Riemann's hypothesis "say" that null values for Z(z) are placed at the lines {z=a+bi|a=0} and {z=a+bi|b=1/2} at this formula
Z(1-s)=2[(2pi)^(-s)]*[cos(s*pi/2)]*(s-1)!*Z(s) if s=1/2+it then 1-s=1/2-it and 1/2=1-1/2 for any t<-R.
If I will solve the Riemann's hypothesis but and will obtain the copyright for this future big theorem, then I will publish and here at this topic this theorem.
Z(1-s)=2[(2pi)^(-s)]*[cos(s*pi/2)]*(s-1)!*Z(s) the "holly grail".
Z(z) is the Zeta function. Z(z)=1+1/(2^z)+1/(3^z)+1/(4^z)+... when z=a+bi.
B(n) are the numbers of Bernoulli: 1;-0,5;0,1(6);0;-0,0(3);0;0,0(238095);0;-0,0(3);0;0,0(75);0;-0,2(531135);0;1,1(6);0;-7,09216...;0;54,97118...;0;
-529,1(24);0;6192,12319...;0;-86580,2(531135);0;1425517,1(6);0;-27298231,07...;0;601580873,9;0;-1511631576.7...;0;... etc.
Also the function Zeta is good helpfull for who is curious how made B(z) when z<-C\{N} where N is the lot of the natural numbers.
Also B(-1)=(pi^2)/6=1,6449340668... and B(-2)=2,40405... and B(-3)=(pi^4)/30=3.2469697011... etc.
If the Riemann's hypothesis "say" that null values for Z(z) are placed at the lines {z=a+bi|a=0} and {z=a+bi|b=1/2} at this formula
Z(1-s)=2[(2pi)^(-s)]*[cos(s*pi/2)]*(s-1)!*Z(s) if s=1/2+it then 1-s=1/2-it and 1/2=1-1/2 for any t<-R.
If I will solve the Riemann's hypothesis but and will obtain the copyright for this future big theorem, then I will publish and here at this topic this theorem.
Title: Re: about Riemann's hipothesys
Post by: TheRedshiftRider on June 25, 2016, 12:44:10 PM
Thanks for adding new information, it is nice to see people are still trying to solve it.
But please note that only one post at the time is enough if only one of the posts adds something. If there is something to add to the post afterwards or you think you've forgotten something please just edit your post instead of posting a new one.
The second post has been removed because it did not add anything. If you have questions about this please send me a personal message.
Edit: Well, actually you should contact Sockratease. He was a little faster. Maybe a bit too fast?
But please note that only one post at the time is enough if only one of the posts adds something. If there is something to add to the post afterwards or you think you've forgotten something please just edit your post instead of posting a new one.
The second post has been removed because it did not add anything. If you have questions about this please send me a personal message.
Edit: Well, actually you should contact Sockratease. He was a little faster. Maybe a bit too fast?
Title: Re: about Riemann's hipothesys
Post by: TheRedshiftRider on July 14, 2016, 06:08:12 PM
Shouldn't you contact a mathematical institute about this? It could help to the whole process of solving this theorem.
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on July 16, 2016, 08:52:11 AM
I was tried to contact one math institute, but sadly in Romania don't exist functionally math institute. The self research institute in sciences will be only this at Magurele whick studying lasers.
Also not only Riemann hypothesis awaiting to be theorem, that I'm discovered a new theorem about the holomorphy in R^n with n>2 whick awaits to been published and presented proper, also I contacted the math society at University of Bucharest , named "Gazeta Matematica" , but if I want to gain copyright for my discoveries, I must be carefull that those discoveries to not been takes by otherwise, also I search an editure for to can publishing those how demanding the office of trademarks, because only an editure will representing me in instance.
The university "Spiru-Haret" where I graduated the master and the faculty of maths, don't have math society like "Gazeta Matematica"
I still working at presentation, and ensuring that theories don't have lags.
Also not only Riemann hypothesis awaiting to be theorem, that I'm discovered a new theorem about the holomorphy in R^n with n>2 whick awaits to been published and presented proper, also I contacted the math society at University of Bucharest , named "Gazeta Matematica" , but if I want to gain copyright for my discoveries, I must be carefull that those discoveries to not been takes by otherwise, also I search an editure for to can publishing those how demanding the office of trademarks, because only an editure will representing me in instance.
The university "Spiru-Haret" where I graduated the master and the faculty of maths, don't have math society like "Gazeta Matematica"
I still working at presentation, and ensuring that theories don't have lags.
Title: Re: about Riemann's hipothesys
Post by: zebastian on July 16, 2016, 03:57:34 PM
hi hgjf2,
when you got your paper ready you can upload it to arxiv.org to claim your copyright and make it public.
I think you are getting fast feedback for dealing with the riemann zeta function. :) Be sure your theory is bullet proof.
Has someone else of you had contact with arxiv.org as a publication platform?
when you got your paper ready you can upload it to arxiv.org to claim your copyright and make it public.
I think you are getting fast feedback for dealing with the riemann zeta function. :) Be sure your theory is bullet proof.
Has someone else of you had contact with arxiv.org as a publication platform?
Title: Re: about Riemann's hipothesys
Post by: hermann on October 02, 2016, 01:47:43 PM
Did you know, that one can post LaTex formulars here in fractal forums?
 = \sum_{n=1}^\infty {\frac {1} {n^s}} = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} + \ldots)
 = \prod_{p \in \mathbb{P}} {\left( 1-p^{-s} \right)}^{-1} = \prod_{p \in \mathbb{P}} {\frac{1}{1-\frac{1}{p^{s}}}} = \frac{1}{(1-\frac{1}{2^{s}}) (1-\frac{1}{3^{s}})(1-\frac{1}{5^{s}}) (1-\frac{1}{7^{s}}) \ldots})
 = 2^{s}\pi^{s-1}\sin \left(\frac{\pi s}{2} \right) \Gamma(1-s)\zeta(1-s))
} \Gamma\left(\frac{s}{2} \right)\zeta(s)=\pi^{-\left(\frac{1-s}{2} \right)}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s))
May be this will help you to make expressions more readable.
Hermann
May be this will help you to make expressions more readable.
Hermann
Title: Re: about Riemann's hipothesys
Post by: hgjf2 on October 08, 2016, 08:17:47 AM
Thanks!
:) :cantor_dance:
:) :cantor_dance:
Title: The One Million Dollar Question
Post by: hermann on October 10, 2016, 04:33:13 PM
I you realy have a proofs for Riemann's Hypothesis you will be a rich man!
If you have 10% for me it will help me for an early retirement.
Hermann
https://www.youtube.com/watch?v=rGo2hsoJSbo (https://www.youtube.com/watch?v=rGo2hsoJSbo)
P.S There is a misspelling in the title it should be Riemann Hypothesis
If you have 10% for me it will help me for an early retirement.
Hermann
https://www.youtube.com/watch?v=rGo2hsoJSbo (https://www.youtube.com/watch?v=rGo2hsoJSbo)
P.S There is a misspelling in the title it should be Riemann Hypothesis
Title: Re: about Riemann's hipothesys
Post by: TheRedshiftRider on October 10, 2016, 06:36:10 PM
I you realy have a proofs for Riemann's Hypothesis you will be a rich man!
If you have 10% for me it will help me for an early retirement.
That would be nice. Maybe the forum itself could use some of it as well I suppose, to expand a little. :D If you have 10% for me it will help me for an early retirement.
Edit: thank you for sharing that video. I really needed a refresher.
Title: Riemann Hypothesis
Post by: hermann on October 20, 2016, 09:37:31 PM
Another nice video on Riemann's Hypothesis.
https://www.youtube.com/watch?v=VTveQ1ndH1c (https://www.youtube.com/watch?v=VTveQ1ndH1c)
https://www.youtube.com/watch?v=VTveQ1ndH1c (https://www.youtube.com/watch?v=VTveQ1ndH1c)