New equation of prime-zeta? part 2

View part-1. note that by the möbius inversion we have Aa=ya1Aa=ya*1 By the dirichlet series : n=1Aa(n)ns=(n=1ya(n)ns)(n=11ns)\sum_{n=1}^\infty \dfrac{Aa(n)}{n^s}=\left(\sum_{n=1}^\infty \dfrac{ya(n)}{n^s}\right)\left(\sum_{n=1}^\infty \dfrac{1}{n^s}\right) P(s)=ζ(s)n=1ya(n)nsP(s)=\zeta(s)\sum_{n=1}^\infty \dfrac{ya(n)}{n^s} Prime zeta and zeta function being used.

If we can find n=1ya(n)ns\sum_{n=1}^\infty \dfrac{ya(n)}{n^s} then we can maybe have a new equation(or the old one) for the prime zeta. This concludes this note, in part three i will attempt the summation.

Feel free to leave your comments, and reshare if you enjoyed this.

Note by Aareyan Manzoor
3 years, 7 months ago

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I think this sum will be very hard to evaluate because it is neither additive nor multiplicative. Here's my progress so far:

The sum can be expressed as

n=1ya(n)ns=n=1μ(n)ns(pn1ps)n=1μ(n)ω(n)ns\sum _{ n=1 }^{ \infty } \frac { ya(n) }{ n^{ s } } =\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right) -\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) \omega \left( n \right) }{ n^{ s } }

For the second sum:

μω1={1 if n=pa0 otherwise\mu \omega \ast 1=\begin{cases} 1\text{ if }n=p^a \\ 0 \text{ otherwise}\end{cases}

ζ(s)n=1μ(n)ω(n)ns=a1p1pas=p1ps1\zeta (s)\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) \omega \left( n \right) }{ n^{ s } } =\sum _{ a\ge 1 }{ \sum _{ p }{ \frac { 1 }{ { p }^{ as } } } } =\sum _{ p }{ \frac { 1 }{ { p }^{ s }-1 } }

Julian Poon - 3 years, 7 months ago

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I did a similar thing! I easily worked out summation#1 but got stuck at summation#2(i got something similar but it was school and i didnt have time and focus). Let me show you what i did in summation one:

we want it to be of the form p2np^2 n such that n is square free with no prime factor=p.then, Ya(p2n)=μ(n)Ya(p^{2}n)=\mu(n). so lets write the summation as p=primep∤nμ(n)(p2n)s=p=prime1p2s(n=1μ(n)ns1pn=1μ(n)ns)=p=prime(1p2s1p2s+1)n=1μ(n)ns=1ζ(s)(P(2s)P(2s+1))\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}=\sum_{p=prime}\dfrac{1}{p^{2s}}\left(\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}-\dfrac{1}{p}\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}\right)\\ =\sum_{p=prime}\left(\dfrac{1}{p^{2s}}-\dfrac{1}{p^{2s+1}}\right) \sum_{n=1}^\infty\dfrac{\mu(n)}{n^s}\\ =\dfrac{1}{\zeta(s)} (P(2s)-P(2s+1)) We derive a beautiful result from this, i will perhaps include this in part 3.

Aareyan Manzoor - 3 years, 7 months ago

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For the first sum, s=2s=2, I am getting 0.060460284021-0.060460284021 numercially, summing from n=1n=1 to n=100n=100.

However, with your formula, I am getting 0.02506977202280.0250697720228.

Mind if you elaborate on your first step?

Edit: Shouldn't the first equation be

n=1p∤nμ(n)(p2n)s\sum_{n=1}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}

Julian Poon - 3 years, 7 months ago

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@Julian Poon It should be p2∤np^2\not\mid n as i dont want power higher then 2.

Aareyan Manzoor - 3 years, 7 months ago

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@Aareyan Manzoor I am getting n=1p∤nμ(n)(p2n)s=P(2s)1ζ(s)n=1pnμ(n)nsp2s\sum_{n=1}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s} = P(2s)\frac{1}{\zeta(s)}-\sum_{n=1}\sum_{p\mid n} \dfrac{\mu(n)}{n^sp^{2s}}

Julian Poon - 3 years, 7 months ago

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@Julian Poon Yes i am getting similar except it should be p∤np\not \mid n

Aareyan Manzoor - 3 years, 7 months ago

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@Aareyan Manzoor The frist equation I used ∤\not\mid while the second one I used \mid. Assuming my way of calculating numerically is correct, I am getting 0.0604826745959-0.0604826745959, which is close to the original numerical calculation I made.

Julian Poon - 3 years, 7 months ago

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@Julian Poon Now speaking logically, what do you think? We remember that it has one prime factor squared. So it is p^2times square free number. We the can say n=p^2*k. Mobius dissapears over non square frees so we can just sum over all nin muliples of p.

Aareyan Manzoor - 3 years, 7 months ago

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@Aareyan Manzoor Well my point is that your equation:

p=primep∤nμ(n)(p2n)s\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}

Should be

n=1p∤nμ(n)(p2n)s\sum_{n=1}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}

Julian Poon - 3 years, 7 months ago

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@Julian Poon I think they are the same thing. I am summing over primes you are summing over natural numbers but ebery prime divdes n numbers

Aareyan Manzoor - 3 years, 7 months ago

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@Aareyan Manzoor So are you saying it should be

n=primep∤nμ(n)(p2n)s\sum_{n=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}

?

Or else, what domain are you summing nn over?

Julian Poon - 3 years, 7 months ago

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@Julian Poon let me make it better p=primen is NOT a multiple of pμ(n)(p2n)s\sum_{p=prime} \sum_{\text{n is NOT a multiple of p}} \dfrac{\mu(n)}{(p^2n)^s}

Aareyan Manzoor - 3 years, 7 months ago

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@Aareyan Manzoor Ah opps... sorry...

I am getting a different answer though'

p=primep∤nμ(n)(p2n)s=p=prime1p2s(n=1μ(n)nsn=1μ(pn)(pn)s)=p=prime1p2s(n=1μ(n)ns+1psn=1μ(n)(n)s)\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}=\sum_{p=prime}\dfrac{1}{p^{2s}}\left(\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}-\sum_{n=1}^\infty \dfrac{\mu(pn)}{(pn)^s}\right)=\sum_{p=prime}\dfrac{1}{p^{2s}}\left(\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}+\frac{1}{p^s}\sum_{n=1}^\infty \dfrac{\mu(n)}{(n)^s}\right)

=1ζ(s)(P(2s)+P(3s))=\frac{1}{\zeta(s)} \left(P(2s)+P(3s)\right)

Also, since

n=1μ(n)ns(pn1ps)=p=primep∤nμ(n)(p2n)s\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right)=-\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}

n=1μ(n)ns(pn1ps)=1ζ(s)(P(3s)+P(2s))\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right)=-\frac{1}{\zeta(s)} \left(P(3s)+P(2s)\right)

Though I am sure there is problems with this working too since it doesn't match numerically

Julian Poon - 3 years, 7 months ago

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@Julian Poon in the step where you μ(pn)=μ(n)\mu(pn)=-\mu(n) you assumed p,n are coprime. try p=2, n=6 we get μ(26)=0μ(6)\mu(2*6)=0\neq -\mu(6)

Aareyan Manzoor - 3 years, 7 months ago

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@Aareyan Manzoor Oh yeah

p is primep∤nμ(n)(p2n)s=p is prime1p2s(pip,pi is prime(11pis))=p is prime1p2s(pi is prime(11pis)11ps)\sum_{p\text{ is prime}}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}=\sum_{p\text{ is prime}}\dfrac{1}{p^{2s}}\left(\prod_{p_i \neq p,p_i \text{ is prime}}\left(1-\frac{1}{p_i^s}\right)\right)=\sum_{p\text{ is prime}}\dfrac{1}{p^{2s}}\left(\frac{\prod_{p_i\text{ is prime}}\left(1-\frac{1}{p_i^s}\right)}{1-\frac{1}{p^s}}\right)

=p is prime1p2s(1ζ(s)111ps)=1ζ(s)p1p2s(1ps)=\sum_{p\text{ is prime}}\dfrac{1}{p^{2s}}\left(\frac{1}{\zeta(s)}\frac{1}{1-\frac{1}{p^s}}\right)=\frac{1}{\zeta(s)}\sum_{p}\frac{1}{p^{2s}(1-p^{-s})}

=1ζ(s)(p1ps1P(s))=\frac{1}{\zeta(s)}\left(\sum_p\frac{1}{p^s-1}-P(s)\right)

It matches numerically. This equation is disappointing... It gives neither insight into the original problem nor on the evaluation of the sum p1ps1\sum_p\frac{1}{p^s-1}

Julian Poon - 3 years, 7 months ago

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@Julian Poon Oh well..... damn it. Thanks for your help julian poon, evaluating this series has been fun.

Aareyan Manzoor - 3 years, 7 months ago

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Extra, might be useful (derived from conversation below):

n=1μ(n)ns(pn1ps)=n=1μ(n)ns(pn1p2s)P(2s)ζ(s)\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right)=\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ 2s } } \right)-\frac{P(2s)}{\zeta(s)}

Julian Poon - 3 years, 7 months ago

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More extra, in case my numerical calculation is off: I used desmos to calculate pn1ps\sum _{ p|n }^{ } \frac { 1 }{ p^{ s } }

pn1ps=j=1nPrime(j)js(gcd(n,j)1.1gcd(n,j)1.12gcd(n,j)2.2+1)\sum _{ p|n }^{ } \frac { 1 }{ p^{ s } }=\sum _{j=1}^n\frac{P_{rime}\left(j\right)}{j^s}\left(\frac{\left|\gcd \left(n,j\right)-1.1-\left|\gcd \left(n,j\right)-1.1\right|\right|}{2\gcd \left(n,j\right)-2.2}+1\right)

Prime(x)=G(P(x))P_{rime}\left(x\right)=G\left(P\left(x\right)\right)

P(x)=(k=1floor(x)gcd(x,k))(2x)P\left(x\right)=\left(\sum _{k=1}^{\operatorname{floor}\left(x\right)}\gcd \left(x,k\right)\right)-\left(2x\right)

G(x)=(x+0.9x+0.92x1.8+x1.1x1.12x+1.121)G\left(x\right)=\left(\frac{\left|x+0.9-\left|x+0.9\right|\right|}{-2x-1.8}+\frac{\left|-x-1.1-\left|-x-1.1\right|\right|}{2x+1.1\cdot 2}-1\right)

μ(n)=k=1floor(x)gcd(k,floor(x))1.1gcd(k,floor(x))1.12gcd(k,floor(x))+2.2cos(2πkfloor(x))\mu{(n)}=\sum _{k=1}^{\operatorname{floor}\left(x\right)}\frac{\left|\gcd \left(k,\operatorname{floor}\left(x\right)\right)-1.1-\left|\gcd \left(k,\operatorname{floor}\left(x\right)\right)-1.1\right|\right|}{-2\gcd \left(k,\operatorname{floor}\left(x\right)\right)+2.2}\cos \left(\frac{2\pi k}{\operatorname{floor}\left(x\right)}\right)

Essentially, Prime(x)P_{rime}(x) outputs 1 if x is prime, and outputs 0 otherwise.

Julian Poon - 3 years, 7 months ago

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I can't help but notice that the function's name is the letters in your name. :P. gtg, got to do hw.

Julian Poon - 3 years, 7 months ago

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I have verified the above numerically with the method shown below.

Julian Poon - 3 years, 7 months ago

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