Wikipedia:Reference desk/Archives/Mathematics/2017 December 13

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December 13

Infinite Product

Where can I find a worked example of how to convert an infinite series into an infinite product, presumably through Weierstrass factorisation? All I can find are convergence properties of these products, and proofs, but no actual demonstrations of application. Plasmic Physics (talk) 09:12, 13 December 2017 (UTC)

How about $\sum _{i=0}^{\infty }a_{i}=\ln(\prod _{i=0}^{\infty }e^{a_{i}})$?--Jasper Deng (talk) 11:29, 13 December 2017 (UTC)

There are also several given at Euler product, where the resulting product is over the primes. –Deacon Vorbis (carbon • videos) 15:43, 13 December 2017 (UTC)

The Fundamental_theorem_of_algebra basicly says that $\sum _{i=0}^{N}a_{i}x^{i}=a_{N}\prod _{r\in C}(x-r)^{m_{r}}$. If $r$ is a root then $m_{r}$ is its multiplicity, otherwise $m_{r}=0$. Bo Jacoby (talk) 16:04, 13 December 2017 (UTC).

This does not extend to infinite series and products. The Taylor series of the exponential function has no roots. and so one cannot speak of defining that function as an infinite product of monomials.--Jasper Deng (talk) 18:22, 13 December 2017 (UTC)

It does extend to some infinite series and products, but not to all infinite series and products. The (taylor series of the) sine function has roots, and one can indeed speak of defining that function as an infinite product of monomials.

$\sin(z)=z\prod _{k=1}^{\infty }(1-({z \over k\pi })^{2})$

([1]). Bo Jacoby (talk) 20:01, 13 December 2017 (UTC).

As in the OP's question, Weierstrass factorization is the relevant generalization of the fundamental theorem to all entire functions (including ones defined by series of course), using "elementary factors" instead of monomials.John Z (talk) 02:41, 14 December 2017 (UTC)

My function is: $\ z=\sum _{j=0}^{\infty }(2j+1)e^{{\epsilon _{0} \over \tau }(j^{2}+j-{1 \over 2})}$. Plasmic Physics (talk) 06:33, 14 December 2017 (UTC)

What is your independent variable, $\epsilon _{0}$ or $\tau$ or ${\epsilon _{0} \over \tau }$? Why not call the independent variable $\ z$ and write $\ f(z)=\sum _{j=0}^{\infty }(2j+1)e^{z(j^{2}+j-{1 \over 2})}$ ? Bo Jacoby (talk) 10:39, 14 December 2017 (UTC).

I beg your pardon, I thought that $\tau$ was implied. Plasmic Physics (talk) 10:48, 14 December 2017 (UTC)

I assume ε₀ is a positive real number, say the permittivity of free space? In that case your series diverges whenever the real part of τ is positive. Are you hoping to find an analytic continuation of the part where it converges, similar to what is done with the Riemann zeta function? --Trovatore (talk) 11:08, 14 December 2017 (UTC)

You need to find the complex zeroes of $\ f({\epsilon _{0} \over \tau })$. You need to use analytical continuation as the sum is divergent for ${\epsilon _{0} \over \tau }\geq 0$. Bo Jacoby (talk) 11:04, 14 December 2017 (UTC).

You can rewrite you expression as

$\ z=e^{-{\frac {3\epsilon _{0}}{4\tau }}}\sum _{n=1,3,..}^{\infty }n\exp \left({{\frac {\epsilon _{0}}{4\tau }}n^{2}}\right)$

and then try to find its sum using a standard method when $\epsilon _{0}<0$. Ruslik_Zero 20:59, 14 December 2017 (UTC)

Apologies - I made an mistake. The actual function is: $z=\sum _{j=0}^{\infty }(2j+1)e^{-{\epsilon _{0} \over \tau }(j^{2}+j)}$, which rearranges to $z=2e^{\epsilon _{0} \over 4\tau }\sum _{j=0}^{\infty }(j+{1 \over 2})e^{-{\epsilon _{0} \over \tau }(j+{1 \over 2})^{2}}$. It should converge for $\tau >0$, $\epsilon _{0}$ is a positive constant. Plasmic Physics (talk) 05:57, 15 December 2017 (UTC)

Let $w=e^{-{\epsilon _{0} \over \tau }}$ to get the simpler sum $\sum _{j=0}^{\infty }(j+{1 \over 2})w^{(j+{1 \over 2})^{2}}$. The exponential function has disappeared from the problem. Bo Jacoby (talk) 09:35, 15 December 2017 (UTC).

Ok, so what then? Plasmic Physics (talk) 11:21, 15 December 2017 (UTC)

Then solve the equation $\sum _{j=0}^{\infty }(j+{1 \over 2})x^{(j+{1 \over 2})^{2}}=0$. Sorry, I don't know how. Bo Jacoby (talk) 12:11, 15 December 2017 (UTC).

Do you mean find the general solution, or just evaluate the zeroes? $\lim _{k\to 0^{+}}z=0$.

The equation is $f(x)=0$ where $f(x)=\sum _{j=0}^{\infty }(2j+1)x^{j^{2}+j}$. The truncated equation $1+3x^{2}+5x^{6}+7x^{12}=0$ has the 12 complex solutions (x₁, . . . , x₁₂) = (–0.54122i, 0.54122i, –0.5983–0.616984i, –0.5983+0.616984i, 0.5983–0.616984i, 0.5983+0.616984i, –0.898509–0.371691i, –0.898509+0.371691i, 0.898509–0.371691i, 0.898509+0.371691i, –i, i) and the product representation is $f(x)\approx 7\prod _{j=1}^{12}(x-x_{j})$. Bo Jacoby (talk) 17:41, 15 December 2017 (UTC).

In general, one cannot approximate the roots of an analytic function using roots of its Taylor polynomials, so this will not suffice for an exact infinite product representation.--Jasper Deng (talk) 18:38, 15 December 2017 (UTC)

May be not in general, but in this case it works fine! Bo Jacoby (talk) 23:03, 15 December 2017 (UTC).

All you did was demonstrate that the fundamental theorem of algebra holds for Taylor polynomials (like any polynomials). Whether the sequence of roots converges to that of the function remains to be shown.--Jasper Deng (talk) 00:26, 16 December 2017 (UTC)

Criticism without improvement is not helpful. Show it yourself! (Note that small roots, such as |x₁|<0.6, are unaffected by terms of degree ≥20) Bo Jacoby (talk) 08:22, 16 December 2017 (UTC).

This problem seemed familiar to me. I eventually remembered that it is rotational partition function. There is certainly rich literature discussing its properties like this. Ruslik_Zero 17:57, 15 December 2017 (UTC)

I think that the question is meaningless as the function $z(\epsilon _{0}/kT)$ is not an entire function and can not be factorized: it has infinite number of poles located at $\epsilon _{0}/kT=2\pi im,\;m\in Z$. Ruslik_Zero 08:35, 16 December 2017 (UTC)

I'm not quite sure what a pole is, but from the introduction in our article on the matter, a pole exists where the limit of a function tends to infinity as the variable tends to some constant. If this is true of all poles, then this function has at most only one pole for tau approaching zero from the left. Plasmic Physics (talk) 10:39, 16 December 2017 (UTC)

See pole (complex analysis). Weierstrass factorization won't work in the presence of them, even if you only intend real values for the variables here.--Jasper Deng (talk) 10:47, 16 December 2017 (UTC)

Yes, that's the article. What if I use absolute tau instead, then I'll have no poles? Plasmic Physics (talk) 10:58, 16 December 2017 (UTC)

I'm not sure what you mean, but an entire function must have a globally convergent Taylor series. In this case, we can then rule out the function being entire because the Taylor series clearly does not have an infinite radius of convergence.--Jasper Deng (talk) 11:15, 16 December 2017 (UTC)

Ok, I see. Well, if I substitute τ with |τ| in the function, then, the series should be absolutely convergent. Plasmic Physics (talk) 11:43, 16 December 2017 (UTC)

Then it fails to be differentiable as a function of |τ| and hence can't be entire. This is not to say you can't find an infinite product representation - just not one (directly) using the Weierstrass factorization theorem.--Jasper Deng (talk) 20:19, 16 December 2017 (UTC)

OK, with some manipulation, I've found the taylor series for the function of |τ|, in terms of $(1-\tau ^{2})$, which should be entire differentiable. It yields a triple sum, which provided some entertaining mental stimulation to find. The triple sum is as follows:

$z\left(|\tau |\right)=2\sum _{j=0}^{\infty }\sum _{k=0}^{\infty }\sum _{l=0}^{\infty }{\left({k \over 2}+l-1\right)! \over {k!l!\left({k \over 2}-1\right)!}}\left(j+{1 \over 2}\right)\left(\epsilon _{0}\left({1 \over 4}-\left(j+{1 \over 2}\right)^{2}\right)\right)^{k}\left(1-\tau ^{2}\right)^{l}$.

You are wrong because $|\tau |^{2}\neq \tau ^{2}$ for complex numbers. Ruslik_Zero 14:34, 17 December 2017 (UTC)

What does it matter? Does the tripple sum not converge absolutely for complex values of tau? Plasmic Physics (talk) 04:31, 18 December 2017 (UTC)

It doesn't describe the same complex-valued function as substituting in |τ| for τ.--Jasper Deng (talk) 06:00, 18 December 2017 (UTC)

(←) I'm not sure that I understand as the original function (corrected) is not complex-valued. Plasmic Physics (talk) 09:56, 18 December 2017 (UTC)

@Plasmic Physics: Its restriction to the reals isn't, but the Weierstrass factorization theorem works on complex analytic functions, so you first need to extend your real-valued function to a complex-valued one. As it turns out, the Taylor series you gave above (in the triple sum) is not that of $z(|\tau |)$.--Jasper Deng (talk) 10:20, 18 December 2017 (UTC)

Perhaps, I only made a mistake in calling it a function of |τ|. All I actually did was convert the original function into an even function, so that the sign of τ is inconsequential, and to result in the diappearance of the pole at τ=0. Plasmic Physics (talk) 11:05, 18 December 2017 (UTC)

Actually your Taylor series can't converge to the original function, since it implies that there is a root at $\tau =\pm 1$ when it's evident from your original series that that can't be the case (since the partial sums at $\tau =1$ converge to a positive number). I believe what you might have here is a non-analytic smooth function at the origin.--Jasper Deng (talk) 17:53, 18 December 2017 (UTC)

Yes, and you can not get rid of the pole at $\tau \to 0$, which is of infinite order. Ruslik_Zero 18:41, 18 December 2017 (UTC)

The limit of the taylor series is zero as tau approaches 0, not infinity. It should be, because the original (corrected) function has the same limit, which should remain if the plot is reflect about the origin. Plasmic Physics (talk) 19:33, 18 December 2017 (UTC)

It actually doesn't exist at all. If you approach 0 along the negative real axis you get infinity.--Jasper Deng (talk) 01:21, 19 December 2017 (UTC)

I know that it does not converge to the original function. it is only intended to be equivalent for τ>0. Check your maths, it only implies a root at $\tau =\pm 1$ if you take the 1st partial sum. Plasmic Physics (talk) 19:33, 18 December 2017 (UTC)

Oh, it seems I forgot that zero to the zeroth is defined to be 1. But given that the Taylor series defines an entire function and the even version of your function is not entire, it can't be right.--Jasper Deng (talk) 01:28, 19 December 2017 (UTC)

Again, only the original function has a limit of infinity as tau approaches zero from below. As tau approaches zero from above, the orignal function reaches the limit of zero, ergo the even function has the same limit of zero as tau approaches zero from either side. Plasmic Physics (talk) 05:02, 19 December 2017 (UTC)

In the equation $\sum _{j=0}^{\infty }(2j+1)x^{j^{2}+j}=0$ only even exponents appear. So the computation can be simplified to $y=x^{2}$ and $\sum _{j=0}^{\infty }(2j+1)y^{j^{2}+j \over 2}=0$. The evaluation of the series is simplified by using the Padé approximant. Bo Jacoby (talk) 14:48, 17 December 2017 (UTC).

I'm not looking for an approximation, that is the whole aim of the exercise. Plasmic Physics (talk) 04:31, 18 December 2017 (UTC)

The solution is the limiting value of the approximations. Bo Jacoby (talk) 05:06, 18 December 2017 (UTC).

You're not answering his question, which asks for an exact infinite product representation.--Jasper Deng (talk) 06:00, 18 December 2017 (UTC)

I know, sir. Neither are you. Work with me, rather than against me. Write $f(x)=\lim _{n,m\rightarrow \infty }[m/n]_{f}(x)$ and factorize the Padé approximant $[m/n]_{f}(x)={a_{m}\prod _{i=1}^{m}(x-r_{i}) \over b_{n}\prod _{j=1}^{n}(x-s_{j})}$. Bo Jacoby (talk) 10:19, 18 December 2017 (UTC).

This won't work for the same reason why your earlier method using roots of Taylor polynomials won't: there's no guarantee about the behavior of the roots of the numerator and the denominator as you increase m.--Jasper Deng (talk) 10:53, 18 December 2017 (UTC)

No guarantee does not imply that i doesn't work. It probably works fine. Try being constructive. Bo Jacoby (talk) 14:48, 18 December 2017 (UTC).