I Is Riemann's Zeta at 2 Related to Pi through Prime Numbers?

[sysprog]

I am fascinated by sequences of prime numbers among integers, twin prime occurrence, intervals between Mersenne primes and related numbers. I sense or intuit a relation to π and the above series utilizing prime numbers to approximate π from studying related numeric series that converge on trigonometric identities.

Pi is ratio of circle's circumference to diameter. Trigonometric functions defined on unit circle contain π. I intuit or perhaps remember an old text that describes a relation to prime number sequences yet cannot put my finger on it. Perhaps numeric sequences approximating transcendental functions and those approximating transcendental numbers such as π resemble each other such that I am conflating series. This connection has bothered me since this thread began. Thanks.

We know that the uncountably infinite is of greater magnitude than the countably infinite, and we know that there are fewer algebraic irrationals than transcendentals, but we have not found a simple way to prove an arbitrary designatable real or imaginary number to be transcendental.

Regarding the special case of the transcendental number $\pi$, Euler showed that $\frac π 4 = \frac 3 4 \cdot \frac 5 4 \cdot \frac 7 8 \cdot \frac {11} {12} \cdot \frac {13} {12} \cdots$ with the numerators being the odd primes (and the denominators being the nearest thereto multiples of 4) $\dots$

Using e.g. the Sieve of Eratosthenes to find the odd primes to obtain that series to produce digits of $\pi$ is not as fast as using the more rapidly converging Gauss-Legendre Algorithm, but that algorithm is a core-hog (i.e. requires much memory) $-$ as you know, CPU time vs memory is a frequently-occurring trade-off.

 

[Klystron]

I had deleted my post, fearing I was off-topic but @sysprog ties it in nicely.

I am fascinated by sequences of prime numbers among integers, twin prime occurrence, intervals between Mersenne primes and related numbers. I sense or intuit a relation to π and the above series utilizing prime numbers to approximate π from studying related numeric series that converge on trigonometric identities.

Pi is ratio of circle's circumference to diameter. Trigonometric functions defined on unit circle contain π. I intuit or perhaps remember an old text that describes a relation to prime number sequences yet cannot put my finger on it. Perhaps numeric sequences approximating transcendental functions and those approximating transcendental numbers such as π resemble each other such that I am conflating series. This connection has bothered me since this thread began. Thanks.

 

[pbuk]

Pi Day was a couple of weeks ago so this conversation is a bit late really. For those that missed the fun, Chris Caldwell's Prime Pages are a great year-round resource e.g. here's a nice big prime.

 

[WWGD]

IIRC, the series described in the OP comes from a Fourier series.

And this year's $\sqrt 10$ square root of 10 day; 3.1622 was a much better approximation than any $\pi$ day
Edit

You use the Fourier series for f(x)=x in $[ \pi, \pi)$, when combined with Parseval's identity, comes down to:

$\Sigma_{n=1}^{\infty} \frac {2(-1)^{n+1}}{n} sin(nx)$

$\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)^2 dx = \Sigma _{n=1}^{\infty} \frac {4}{n^2}$; still for f(x)=x on

$[-\pi, \pi)$

 

[WWGD]

Another interesting connection is that between $\pi$ and the Normal Distribution. Just how does $\pi$ pop up in its density function?

 

[nuuskur]

The $\sqrt{2\pi\sigma ^2}$ is a normalising factor.

 

[WWGD]

The $\sqrt{2\pi\sigma ^2}$ is a normalising factor.

But why does $\pi$ appear there to start with?

 

[sysprog]

Another interesting connection is that between $\pi$ and the Normal Distribution. Just how does $\pi$ pop up in its density function?

I think that this is a good explanation of that: https://davegiles.blogspot.com/2016/01/why-does-pi-appear-in-normal-density.html

 

[nuuskur]

But why does $\pi$ appear there to start with?

You are effectively asking, why
<br /> \int _{\mathbb R} \exp \left (-t^2\right )\mathrm{d}t = \sqrt{\pi}<br />
is true. It is proved. If you start looking for reasons why this equality holds and relate it to $\pi$ somehow, you are endangered by confirmation bias.

Similarly, there is no call for esoterics between $\pi$ and the prime numbers. Just because some equality holds that contains said quantities does not imply there has to be some "deep" connection "intertwining" them. The fancy lingo is impressive, isn't it?

The following is as uncharitable as I can be at this hour.

Spoiler

Take any conditionally convergent series. Then there exists a rearrangement that converges to $\pi$. Now find a foundational discovery in the fabric of space time continuum that makes this happen (or at least insist that there exists one).

 

[WWGD]

You are effectively asking, why
<br /> \int _{\mathbb R} \exp \left (-t^2\right )\mathrm{d}t = \sqrt{\pi}<br />
is true. It is proved. If you start looking for reasons why this equality holds and relate it to $\pi$ somehow, you are endangered by confirmation bias.

Similarly, there is no call for esoterics between $\pi$ and the prime numbers. Just because some equality holds that contains said quantities does not imply there has to be some "deep" connection "intertwining" them. The fancy lingo is impressive, isn't it?

Take any conditionally convergent series. Then there exists a rearrangement that converges to $\pi$. Now find a foundational discovery in the fabric of space time continuum that makes this happen.

Well, it's not clear whether there is a connection or not. Naively, you'd expect to find it in connection with settings related to circles. But there are coincidences as well.

 

[nuuskur]

Indeed, the connection to circles is an example of potential confirmation bias. That in theory $\pi$ appears in circles and geometry does not imply that $\pi$ is exclusively about circles. Of course, it would be nice if there was some pretty (and consistent) explanation -- maybe there is! But we can't assume a priori there exists one.

 

[TeethWhitener]

I'm not sure it's confirmation bias. Finding the integral requires squaring it and transforming to polar coordinates.
$$\left(\int_{-\infty}^{\infty}{e^{-x^2}\ dx}\right)^2=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^2+y^2\right)}\ dx\ dy=\int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2}r\ dr \ d\theta$$
To be absolutely rigorous, you have to treat the improper integrals correctly, since the Cartesian integral is over a square and the polar integral is over a circle. In practice, this involves comparing the integral over the square with the integrals over its incircle and circumcircle, and using the squeeze theorem in the limit that the size of the square goes to infinity. So circles are in fact involved directly in the proof.

 

[TeethWhitener]

I'm having a hard time finding a derivation of the series in OP. I would think it would converge to $\frac{4}{\pi}$ instead of $\frac{2}{\pi}$. Maybe someone can spot my math error.

I started with the Leibniz formula:
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots$$
Multiplying both sides by $\frac{1}{3}$ and subtracting the result from the original series gives the series with all the multiples of 3 missing:
$$\left(1-\frac{1}{3}\right)\frac{\pi}{4}=1+\frac{1}{5}-\frac{1}{7}-\frac{1}{11}+\cdots$$
Repeating the process with $-\frac{1}{5}$ gives the series with all the remaining multiples of 5 missing:
$$\left(1+\frac{1}{5}\right)\left(1-\frac{1}{3}\right)\frac{\pi}{4}=1-\frac{1}{7}-\frac{1}{11}+\cdots$$
Continuing ad infinitum gives:
$$\frac{\pi}{4}\prod_{p\equiv3\ mod\ 4}{\left(1-p^{-1}\right)}\prod_{p\equiv1\ mod\ 4}{\left(1+p^{-1}\right)}=1$$
and multiplying both sides by $\frac{4}{\pi}$ would give me the series, except there's a factor of 2 that I've picked up somewhere.

Any thoughts?
Edit: a few sign errors

 

[TeethWhitener]

BTW, if the derivation of the series does in fact ultimately come from the Leibniz formula, that itself is related to circles through the arctangent function, in that
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots$$
is simply the Taylor series expansion of $\arctan(1)$.

 

[pbuk]

I'm having a hard time finding a derivation of the series in OP.

Which one?

$$\frac{\pi}{4}\prod_{p\equiv3\ mod\ 4}{\left(1-p^{-1}\right)}\prod_{p\equiv1\ mod\ 4}{\left(1+p^{-1}\right)}=1$$

That looks right, it's sometimes called Euler's formula (no, not that one ).

 

[DrClaude]

That looks right, it's sometimes called Euler's formula (no, not that one ).

Actually, @TeethWhitener's formula has the prime modulos reversed.

 

[nuuskur]

I started with the Leibniz formula:
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots$$
Multiplying both sides by $\frac{1}{3}$ and subtracting the result from the original

This part.
Oop, DrClaude beat me to it.

 

[TeethWhitener]

Which one?That looks right, it's sometimes called Euler's formula (no, not that one ).

I figured the derivation might look a lot like Euler's derivation of the Riemann zeta function, but I'm not a mathematician, so apologies for lack of rigor.

 

[WWGD]

I'm not sure it's confirmation bias. Finding the integral requires squaring it and transforming to polar coordinates.
$$\left(\int_{-\infty}^{\infty}{e^{-x^2}\ dx}\right)^2=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^2+y^2\right)}\ dx\ dy=\int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2}r\ dr \ d\theta$$
To be absolutely rigorous, you have to treat the improper integrals correctly, since the Cartesian integral is over a square and the polar integral is over a circle. In practice, this involves comparing the integral over the square with the integrals over its incircle and circumcircle, and using the squeeze theorem in the limit that the size of the square goes to infinity. So circles are in fact involved directly in the proof.

I don't know if it really actually requires it. There may be other ways of finding the integral ( though none I am aware of) that doesn't use polar coordinates. But, once you've chosen the method you describe, then, yes.

 

[WWGD]

Indeed, the connection to circles is an example of potential confirmation bias. That in theory $\pi$ appears in circles and geometry does not imply that $\pi$ is exclusively about circles. Of course, it would be nice if there was some pretty (and consistent) explanation -- maybe there is! But we can't assume a priori there exists one.

I agree, but not fully convinced by your argument. If $\pi$ had been derived by manipulating a conditionally-convergent series, then yes, you have a point, but in this case, AFAIK, it had not come about that way.

 

[Jarvis323]

Another fascinating relationship is to do with points generated by repeated multiplication and modulus plotted around a circle.



These are sequences generated based on number theoretic phenomena that is related to primality and divisibility. And they form the same patterns you get if you roll circles around other circles. Which also happen to be the shapes of magnetic fields in microphones, and a basis for forming Fibonacci spirals, and the shape of the Mandelbrot fractal.

 

[Janosh89]

In #11 PeroK mentions finite series,

as someone who has rudimentary maths & is recovering from a brain injury ,
I'm assuming these series converge to some value , π in the instance the OP
alluded to.

Infinite Series are treated in some other way [ eg 12x +1 which contains all the
primes of the form 6x +1 ] ?

 

[PeroK]

In #11 PeroK mentions finite series,

I mentioned finite sequences. A sequence is a list of numbers; a series is a sum. That's the formal mathematical usage of those terms.

I'm assuming these series converge to some value , π in the instance the OP
alluded to.

All finite series have a finite sum. It's only infinite series where we need to talk about convergence or divergence.

Infinite Series are treated in some other way [ eg 12x +1 which contains all the
primes of the form 6x +1 ] ?

I don't know what you mean here.

 

[Janosh89]

Please take down my post if it does not contribute to the thread

 

[martinbn]

I might repeat what is already said but one has of course

$$\prod_{p\; prime}\frac1{1-p^{-2}} = \frac{\pi^2}{6}$$

which also proves that there are infinitely many prime numbers. Otherwise the left side is a rational number.

 

[dextercioby]

I might repeat what is already said but one has of course

$$\prod_{p\; prime}\frac1{1-p^{-2}} = \frac{\pi^2}{6}$$

which also proves that there are infinitely many prime numbers. Otherwise the left side is a rational number.

Can you point to a proof of that?

 

[martinbn]

Can you point to a proof of that?

That is Riemann's zeta at 2. It is in many places. Here is one of the first that come up.

https://www.math.cmu.edu/~bwsulliv/MathGradTalkZeta2.pdf

 

[dromarand]

$\prod_{p \ prime}{} \frac{p^{2}+1}{p^{2}-1} = \frac{5}{2}$
$\prod_{p \ prime}{} \frac{p^{4}+1}{p^{4}-1} = \frac{7}{6}$

 

[WWGD]

Can you point to a proof of that?

You can also show this using Fourier Series for $x^2$ in $[-\pi, \pi ]$
https://math.stackexchange.com/ques...m-n-1-infty-1-n2-using-fourier-series#2706955