Brett Nortj

Riemann Hypothesis.

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{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots }

This would be [infinity] * [2n] = [3] * [2] = [6] * [x]

Then, [5] * [2] = [10] * [x]

Then, [7] * [2] = [14] * [x]

So, the  answer is double prime times by [x].

Or,

[N] * [2] *[X].

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1 hour ago, Brett Nortj said:

{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots }

This would be [infinity] * [2n] = [3] * [2] = [6] * [x]

Then, [5] * [2] = [10] * [x]

Then, [7] * [2] = [14] * [x]

So, the  answer is double prime times by [x].

Or,

[N] * [2] *[X].

OK, I'll bite what is x?

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Just now, dimreepr said:

OK, I'll bite what is x?

[X] is just an expression for the symbols combined, a new symbol - no rule says we cannot add or use our own symbols.

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2 hours ago, dimreepr said:

Trivial or non-trivial?

Why don't we walk on the wild side? Let's make it a process of elimination?

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8 minutes ago, Brett Nortj said:

Why don't we walk on the wild side? Let's make it a process of elimination?

Why not, which version of zero do you prefer?

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3 minutes ago, dimreepr said:

Why not, which version of zero do you prefer?

I think [X] would be the symbol of the log, of course...

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3 minutes ago, Brett Nortj said:

I think [X] would be the symbol of the log, of course...

The wrong version, never mind please try again... :o

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6 minutes ago, dimreepr said:

The wrong version, never mind please try again... :o

[X] would be the indent on the log?

Look, I have always struggled with logarithms, but I know I have something going here.

Why don't you take a shot? Nothing to lose.

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23 hours ago, Brett Nortj said:

[X] would be the indent on the log?

Look, I have always struggled with logarithms, but I know I have something going here.

Why don't you take a shot? Nothing to lose.

You struggle with logarithms but want to solve the Riemann hypothesis?

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20 minutes ago, wtf said:

You struggle with logarithms but want to solve the Riemann hypothesis?

He said he wants to talk about it, not necessarily solve it, wtf.

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2 minutes ago, taeto said:

He said he wants to talk about it, not necessarily solve it, wtf.

X does neither.

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Posted (edited)
24 minutes ago, dimreepr said:

X does neither.

It depends on the X. Maybe it was a very brilliant X. I admit I did not bother to read far enough into the OP to discover what it means. I trust that you investigated and determined that the X is actually kind of stupid. Note taken.

Maybe he just has some questions about logarithms or RH.

I will attempt to elucidate. The "natural logarithm" function \( \ln, \) or nowadays mostly written \( \log,\) can be conveniently defined by \[ \ln(x) := \int_1^x \frac{1}{t} dt \] for any positive real number \(x.\)

Then \( \ln(1) = 0\) is obvious by definition, and using standard substitution of variables in the Riemann integral, you can easily deduce basic properties, such as  \( \ln(ab) = \ln(a) + \ln(b) \) for all \( a,b > 0, \) as well as \( \ln(a^b) = b \ln(a), \) for all \(a > 0\) and all \(b.\) 

To formulate the Riemann Hypothesis in terms of logarithms it also helps to know about primes. I will wait for the OP to indicate whether this is the intention before I go on to explain about primes.

 

Edited by taeto

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42 minutes ago, taeto said:

It depends on the X. Maybe it was a very brilliant X. I admit I did not bother to read far enough into the OP to discover what it means. I trust that you investigated and determined that the X is actually kind of stupid. Note taken.

Maybe he just has some questions about logarithms or RH.

I will attempt to elucidate. The "natural logarithm" function ln,  or nowadays mostly written log, can be conveniently defined by 

ln(x):=x11tdt

for any positive real number x.

 

Then ln(1)=0 is obvious by definition, and using standard substitution of variables in the Riemann integral, you can easily deduce basic properties, such as  ln(ab)=ln(a)+ln(b) for all a,b>0, as well as ln(ab)=bln(a), for all a>0 and all b.  

To formulate the Riemann Hypothesis in terms of logarithms it also helps to know about primes. I will wait for the OP to indicate whether this is the intention before I go on to explain about primes.

 

Okay, seems to be right to me.

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Posted (edited)

Okay then: primes. 

Formally a natural number \(p\) is called "prime" if \(p > 1\) and it is true that if you take any integer numbers \(a\) and \(b,\) and \(p\) divides the product \(ab\), then \(p\) divides \(a\) or \(b.\)

The first primes are \(2,3,5,7,11,13,\) and then there are some more larger ones as well.

The "primorial" \(p\#\) of a prime \(p\) is the product \(2\times 3 \times 5 \times 7 \times 11 \times \cdots \times p\) of all the primes \(2,3,5,7,11,...,p.\)

The "Euler totient function" \(\varphi\) is a more complicated function, but all you have to worry about is its value at primorial \(p\#,\) which is simply the product \( \varphi(p\#) = 1 \times 2 \times 4 \times 6 \times 10 \times \cdots \times (p-1)\) of all numbers less by 1 than a prime up to \(p.\)

This is enough that you have to know, before you can meaningfully consider the Riemann Hypothesis. 

The Riemann Hypothesis is logically equivalent to the following statement:

\[ \frac{p\#}{\varphi(p\#) \ln \ln p\#} > e^\gamma, \]

for all primes \(p,\) where \(\gamma\) is known as the Euler-Mascheroni constant, with \[ \gamma := \int_1^\infty (\frac{1}{\lfloor x \rfloor} - \frac{1}{x}) dx \approx 0.577\ldots, \] where the value of the floor function \( \lfloor x \rfloor \) is the largest integer that is not greater than \(x.\) 

Reference: J. Nicolas, Petites valeurs de la fonction d’Euler, J. Number Theory 17, 3, 1983, pp. 375–388.

 

Edited by taeto

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38 minutes ago, taeto said:

Okay then: primes. 

Formally a natural number p is called "prime" if p>1 and it is true that if you take any integer numbers a and b, and p divides the product ab , then p divides a or b.

The first primes are 2,3,5,7,11,13, and then there are some more larger ones as well.

The "primorial" p# of a prime p is the product 2×3×5×7×11××p of all the primes 2,3,5,7,11,...,p.

The "Euler totient function" φ is a more complicated function, but all you have to worry about is its value at primorial p#, which is simply the product φ(p#)=1×2×4×6×10××(p1) of all numbers less by 1 than a prime up to p.

This is enough that you have to know, before you can meaningfully consider the Riemann Hypothesis. 

The Riemann Hypothesis is logically equivalent to the following statement:p#φ(p#)lnlnp#>eγ,

for all primes p, where γ is known as the Euler-Mascheroni constant, with

γ:=1(1x1x)dx0.577,

where the value of the floor function x is the largest integer that is not greater than x.  

 

Reference: J. Nicolas, Petites valeurs de la fonction d’Euler, J. Number Theory 17, 3, 1983, pp. 375–388.

 

Well, it was I that wrote the script on primes, namely; [n] = {[n] + [n] +1}.

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1 minute ago, Brett Nortj said:

Well, it was I that wrote the script on primes, namely; [n] = {[n] + [n] +1}.

I could not figure out what \( [n] \) means. So maybe you are way ahead of me.

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On 8/3/2018 at 12:24 PM, Brett Nortj said:

{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots }

This would be [infinity] * [2n] = [3] * [2] = [6] * [x]

Then, [5] * [2] = [10] * [x]

Then, [7] * [2] = [14] * [x]

So, the  answer is double prime times by [x].

Or,

[N] * [2] *[X].

I prefer this version
There was a young man from Nepal
Who had one triangular ball
The square of its weight
Times his pecker, times eight
Is his number, so give him a call

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Posted (edited)
3 minutes ago, John Cuthber said:

I prefer this version

 

You obviously played for a different rugby club from me.

 

:)

Edited by studiot

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27 minutes ago, John Cuthber said:

I prefer this version
There was a young man from Nepal
Who had one triangular ball
The square of its weight
Times his pecker, times eight
Is his number, so give him a call

Definite +1 for an on-topic Limerick, good work! 

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