## Recommended Posts This would be [infinity] * [2n] =  *  =  * [x]

Then,  *  =  * [x]

Then,  *  =  * [x]

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

Or,

[N] *  *[X].

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1 hour ago, Brett Nortj said: This would be [infinity] * [2n] =  *  =  * [x]

Then,  *  =  * [x]

Then,  *  =  * [x]

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

Or,

[N] *  *[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... ##### Share on other sites
6 minutes ago, dimreepr said:

The wrong version, never mind please try again... [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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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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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: This would be [infinity] * [2n] =  *  =  * [x]

Then,  *  =  * [x]

Then,  *  =  * [x]

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

Or,

[N] *  *[X].

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

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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
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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