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The \(\pi\) day: what's the product of all primes?

It is \(4\pi^2\) but don't stop reading.

Yesterday, on March 14th, we celebrated Einstein's birthday as well as the Pi Day (3/14). You could have bought Wolfram Mathematica at a discount, especially because it was the rounded \(\pi\) day, 3.1416.

In Germany and countries influenced by Germany, we often use the term "Ludolph number" for \(\pi\) – the "Archimedes constant", as the English-speaking world tends to say – because 400 years ago, the director of a fencing school and mathematics professor Ludolph van Ceulen calculated 20 digits of \(\pi\) by approximating it with the regular 1,073,741,284-gon after he decided that he wanted to have a cool tombstone (some digits were added later). Yes, the number of the sides is \(2^{30}\) so you may get the result by "halving and halving 30 times" – this method was already known to Archimedes 1700 years earlier – and the deviation of the circumference of this "billiongon" from the circle goes like "one over one billion squared". About 20 digits, indeed.

The number \(\pi\) is the most popular irrational number among the laymen. They associate various supernatural properties with it – as you could see in the thriller named Pi. Much of this hype is irrational (much like \(\pi\), pun intended) because all other irrational or transcendental numbers – and almost all real numbers are irrational and transcendental – share those properties (similar to irrationality) that the laymen are capable of describing by words (e.g. their digits are hard to remember). Some of the hype is justified because \(\pi\) really appears at nice places of mathematics.

Feynman's #1 favorite identity of mathematics was Euler's identity\[

\exp(i \pi ) + 1 = 0.

\] Because the exponentiation of purely imaginary numbers produces rotations of the complex plane, this formula is linked with the usual circumference of the circle, \(2\pi r\).

The number \(\pi\) appears in lots of mathematical identities that are not "obviously" related to the circumference of circles or surfaces or volumes of spheres. For example, the integral of the Gaussian function\[

\int_{-\infty}^{+\infty} dx\, \exp(-x^2) = \sqrt{\pi}.

\] Why is this true? Well, the circle isn't obviously present in the integral but the circle does appear in a clever calculation of the integral. Square the integral and you will get\[

\int_{-\infty}^{+\infty} dx\,dy \exp(-x^2-y^2) =\dots

\] which is rotationally symmetric. You may switch to the polar coordinates. The integral over \(\phi\) produces \(2\pi\) (that's how we cleverly got the "circle" back although it wasn't in the original integral) while the remaining factor (in which we don't forget the factor of \(r\) from the Jacobian compulsory for the polar coordinates) is the radial integral\[

\int_0^\infty dr\, r\,\exp(-r^2) = -\frac{1}{2} \left. \exp(-r^2)\right|_0^\infty = \frac 12

\] The total product is \(2\pi \times 1/2 = \pi\). The original integral must be the square root of that and it was positive, therefore \(\sqrt{\pi}\).

In an article about a certain hype about \(\pi\) and the hydrogen atom, I discussed the fact that \((-1/2)!=\sqrt{\pi}\), too. It is actually equal to the same integral we just calculated – but there are many other ways to look at it.

And this article gave you a proof of the Machin formula\[

\pi = 16\arctan \frac{1}{5} - 4 \arctan \frac{1}{239}.

\] which allows you to calculate \(\pi\) by Taylor series much more quickly than the "simpler" \(\pi = 4\arctan 1\). I could continue with interesting identities involving \(\pi\) for an hour or a few hours. You would hear about various sums that may be interpreted as Taylor series for trigonometric and related functions; or the zeta function of even positive integers which contains \(\pi^{2s}\) because one may calculate them using the norms of some functions and their Fourier series, and so on.

But let me jump to something more interesting or more controversial, namely "seemingly divergent" products and sums.

In lots of previous blog posts, I discussed the sum of positive integers,\[

1+2+3+4+5+\dots = -\frac{1}{12}.

\] This result works assuming the zeta-function regularization of the superficially divergent sum; but the same result may actually be obtained by regularization procedures of many different types. This identity is "really true" in some deep sense – and it's useful e.g. for the simplest evaluation of the critical spacetime dimension in string theory, \(D=26\) for bosonic string theory or \(D=10\) for the superstring.

The sum of positive integers ends up being equal to \(\zeta(-1)\) and this Riemann zeta function may be extrapolated to all complex values of the argument – it is finite everywhere except for the pole at \(s=1\). Similarly, one may evaluate\[

1^s+2^s+3^s+\dots = \zeta(-s).

\] This sum of powers of positive integers is zero for even positive values of \(s\); it is nonzero for odd positive values of \(s\). For example, the sum of third powers is \(\zeta(-3)=+1/120\). Now, can you also evaluate the finite value of the product of all positive integers?\[

1\times 2 \times 3\times 4\times \dots = ?

\] Yes, you can. First, this product may be converted to the exponential\[

\dots = \exp (\ln 1 + \ln 2 + \ln 3+\dots) = ?

\] and the remaining task is to find the regularized "sum of logarithms of integers". The answer may be obtained by looking at the zeta function, too. Recall that\[

\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}

\] and the logarithmic prefactor may be easily produced if we differentiate with respect to \(s\):\[

\zeta'(s) = -\sum_{n=1}^{\infty} \frac{\ln n}{n^s}

\] We may eliminate the \(1/n^s\) factor by substituting \(s=0\). So to find the "sum of the logs of integers", we need to know \(-\zeta'(0)\). By the \(s\leftrightarrow 1-s\) functional equation, the behavior of \(\zeta\) around \(s=0\) is linked to the behavior around \(s=1\) where the original sum formula for \(\zeta\) is convergent. One may calculate \(\zeta'(0)\) and \(\zeta'(1)\) analytically (the latter has a divergent term) but it's straightforward and the "finite" calculations are not mysterious. So why don't we just ask Mathematica?

The symbolic and numerical answers agree:

a = 0.000000001; -(Zeta[a] + 0.5)/a
-D[Zeta[s], s] /. {s -> 0}

So yes, \(-\zeta'(0)\) we wanted to know is \(\ln \sqrt{2\pi}\) (note that the zeta function "knows" that the derivative is equal to the logarithm of something simpler) which implies that the product of all positive integers is\[

1\times 2\times 3\times \dots = \sqrt{2\pi}.

\] You could have gotten this result by an even faster but sloppier, ultra-heuristic method. This product is clearly the factorial of a number, \(\infty!\), and for such large numbers, one may use the Stirling's approximation\[

n! \approx \sqrt{2\pi n} \zav { \frac{n}{e} }^n

\] Now, the power and the polynomial prefactors are divergent for \(n\to \infty\) and hard to interpret. But these factors look "too simple" and multiplicatively, the only reasonable thing that such expressions may be equal to is one. On the other hand, the \(\sqrt{2\pi}\) prefactor is there and can't be ignored, so it's the result for \(\infty! = \sqrt{2\pi}\).

Euler has succeeded in rewriting the zeta function as a product over primes. That's why one may find the regularized value of the product over primes only, not all integers. You may look e.g. at the 2003 paper by Muñoz-Garcia and Pérez Marco; they wrote a followup five years later.

Their funny result for the product of primes – obtained by a similar conversion of the product to values of the \(\zeta\) function, its powers, derivatives, and product and ratios of such objects – reads\[

2\times 3 \times 5 \times 7 \times \dots = 4\pi^2.

\] The product of all primes is "four pi squared". It is the fourth power of the product of all positive integers, a fact that may probably be proven in seemingly independent ways. With the usage of the Euler product and the previous result (no "new" regularization is needed), one may also calculate products of \((p^s-1)\) for any value of \(s\) and the result is\[

\prod_p (p^s - 1) = \frac{(2\pi)^{2s}}{\zeta(s)}.

\] For example, the product of \((p-1)\) over primes happens to be zero because \(\zeta(1)\) has the pole. Similarly,\[

\prod_p (p^2 - 1) = 48\pi^2.

\] In fact, for all even positive values of \(s\), the product of \((p^s-1)\) over primes is, just like \(\zeta(s)\) itself, a rational multiple of \(\pi^s\).

I believe that much of mathematics would become more natural – and some nice insights (perhaps even unknown ones) would become more easily accessible – if the zeta function (and/or similar, general and good enough) regularization were implicitly used everywhere. For example, I believe that it makes sense to say that\[

\dots (t-2)(t-1)t(t+1)(t+2)\dots = C \sin \pi t

\] where the constant \(C\) may be determined by some consistency checks and will probably include the factor of \(\zeta(-1)=-1/12\). The sign may be undetermined. Or is some exponential factor missing in my result? Similarly, the product\[

t(t+1)(t+2)\dots = C' \Gamma(-t).

\] Just don't be afraid to write many of these products – which are singular without regularization – because the finite value associated with them by all "good enough regularization schemes" is uniquely determined. I do think that such regularized values could be useful for accelerating knowledge and proofs – in number theory, the research into the Riemann Hypothesis, physics, and beyond.

Bonus: Ludolph's calculation

Many people click at Ludolph's tombstone above so let me say a few words about the calculation. Draw a regular \(n\)-gon inside a unit circle where \(n=2^{30}\). You may see that the circumference of this polygon is a bit shorter than \(2\pi\). The circumference is \(2^{30}\) times the side, and the side is \(2\times \sin(\pi/2^{30})\). So the circumference is \(2^{31} \sin (\pi/2^{30})\).

A funny thing is that \(\sin(\pi/2^{30})\) may be calculated by halving the angles many times. You iterate the formula \(\sin(2y)=1-2\sin^2 y\) i.e.\[

\sin(x/2) = \sqrt{\frac{1-\sin x}{2}}

\] and because you know how to calculate the square roots numerically and because you know that \(\sin(\pi/4)=\sqrt{1/2}\), you may get an accurate result after some numerical calculation that is doable without a computer. Numerically, \(2^{31} \sin (\pi/2^{30})\) is about \(2\pi - 8.96455\times 10^{-18}\).

You may also consider the \(n\)-gon for \(n=2^{30}\) that contains the circle inside. This one has a longer circumference than \(2\pi\) and this circumference is also calculable. It's \(\cos(\pi/2^{30})\) times longer than the previous one – and this cosine is also numerically calculable because \(\cos^2 x = 1-\sin^2 x \) – i.e. \(2^{31}\tan(\pi/2^{30})\). This ends up being equal to \(2\pi+1.7929105\times 10^{-17}\) or so.

If you were clever enough (and Ludolph himself wasn't, I think), you could figure out that a much better approximation for \(2\pi\) is the weighted average – taking the sine (underestimate) with the weight \(2/3\) and the tangent (overestimate) with the weight \(1/3\). If you do so, the weighted average will be \(2\pi+2.3\times 10^{-35}\) or so – 35 digits of \(2\pi\) will be correct (exactly enough for Ludolph's modernized tombstone).

There surely exist more schoolkid-friendly ways to show that the optimal weights are \(2/3\) and \(1/3\) but the simplest proof for adults, I think, is to notice that \(\sin z= z-z^3/6+\dots\) and \(\tan z = z+z^3/3+\dots\) while we would really like to include "just" the term \(z\) to get \(\pi\). The error \(-z^3/6\) is twice as small as the error of \(+z^3/3\), so if the sine gets twice as high a weight than the tangent, the errors from the \(z^3\) terms (around \(10^{-18}\) in our calculation of \(\pi\)) cancel and only the \(z^5\) terms survive, giving the error \(10^{-35}\) to \(\pi\).

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