- #36
nuuskur
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The ##\sqrt{2\pi\sigma ^2}## is a normalising factor.
But why does ##\pi## appear there to start with?nuuskur said:The ##\sqrt{2\pi\sigma ^2}## is a normalising factor.
I think that this is a good explanation of that: https://davegiles.blogspot.com/2016/01/why-does-pi-appear-in-normal-density.htmlWWGD said:Another interesting connection is that between ##\pi## and the Normal Distribution. Just how does ##\pi## pop up in its density function?
You are effectively asking, whyWWGD said:But why does ##\pi## appear there to start with?
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 said:You are effectively asking, why
[tex]
\int _{\mathbb R} \exp \left (-t^2\right )\mathrm{d}t = \sqrt{\pi}
[/tex]
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.
Which one?TeethWhitener said:I'm having a hard time finding a derivation of the series in OP.
That looks right, it's sometimes called Euler's formula (no, not that one ).TeethWhitener said:$$\frac{\pi}{4}\prod_{p\equiv3\ mod\ 4}{\left(1-p^{-1}\right)}\prod_{p\equiv1\ mod\ 4}{\left(1+p^{-1}\right)}=1$$
Actually, @TeethWhitener's formula has the prime modulos reversed.pbuk said:That looks right, it's sometimes called Euler's formula (no, not that one ).
This part.TeethWhitener said: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
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.pbuk said:Which one?That looks right, it's sometimes called Euler's formula (no, not that one ).
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.TeethWhitener said: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 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.nuuskur said: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 mentioned finite sequences. A sequence is a list of numbers; a series is a sum. That's the formal mathematical usage of those terms.Janosh89 said:In #11 PeroK mentions finite series,
All finite series have a finite sum. It's only infinite series where we need to talk about convergence or divergence.Janosh89 said:I'm assuming these series converge to some value , π in the instance the OP
alluded to.
I don't know what you mean here.Janosh89 said:Infinite Series are treated in some other way [ eg 12x +1 which contains all the
primes of the form 6x +1 ] ?
Can you point to a proof of that?martinbn said: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.
That is Riemann's zeta at 2. It is in many places. Here is one of the first that come up.dextercioby said:Can you point to a proof of that?