Wikipedia:Reference desk/Archives/Mathematics/2017 December 20

= December 20 =

Average sequences
Consider the (infinite) sequence a1, a2, a3, a4, ..., where the first two terms are integers and each term after is the average (arithmetic mean) of the two preceding terms. For which values of a1 and a2 is every term of the sequence an integer? GeoffreyT2000 (talk) 03:00, 20 December 2017 (UTC)
 * a1 = a2 and no other. $$|a_{i+2} - a_{i+1}| = 2^{-i}|a_2-a_1|$$.  The only way to keep that an integer for all $$i$$ is to have $$|a_2-a_1| = 0$$.--108.52.27.203 (talk) 03:22, 20 December 2017 (UTC)

Infinite Product(2)
I need to find out how to convert an infinite series to an infinite product. The original formula is as follows, taking $$\epsilon_0$$ to be a constant:


 * $$z=2e^\frac{\epsilon_0}{4\tau}\sum_{j=0}^\infty\frac{j+\frac{1}{2}}{e^{\frac{\epsilon_0}{\tau}\left(j+\frac{1}{2}\right)^2}}$$

In part one of this querry, I concluded that a this can be achieved for a function which is equivalent for $$\tau\ge0$$, through Weiestrass factorisation. This equivalent function was derived by making the original function even so as to create a function which is entire, without a pole at zero.

Initially, the new formula was presented as a triple sum, but has since been simplified to a double sum, which is presented below:


 * $$z=\sqrt{2\pi}e^\frac{\epsilon_0}{4}\sum_{j=0}^\infty\sum_{k=0}^\infty\frac{\left(j+\frac{1}{2}\right)I_{\frac{1}{2}-k}\left(1\right)}{k!2^{k}e^{\epsilon_0\left(j+\frac{1}{2}\right)^2}}\left(\tau^2-1\right)^k,\quad\tau\ge0$$

How to proceed now that an entire function has been obtained? Plasmic Physics (talk) 08:27, 20 December 2017 (UTC)
 * Note that an infinite sum of entire functions is not necessarily entire. For example
 * $$\sum_{k=0}^\infty x^k = (1-x)^{-1}$$.
 * Bo Jacoby (talk) 19:27, 20 December 2017 (UTC).
 * Noted. Plasmic Physics (talk) 21:05, 20 December 2017 (UTC)
 * It was already noted above that the second formula is wrong as it has a zero at $$\tau=1$$ whereas the original function does not. Ruslik_ Zero 19:35, 20 December 2017 (UTC)
 * That is not correct. Plot the partial z(1) using the second formula to prove it. WolframAlpha is quite useful for this. Plasmic Physics (talk) 20:21, 20 December 2017 (UTC)
 * Let $$x=e^{-\frac{2\epsilon_0}{\tau}}$$. Then
 * $$z = \sum_{j=0}^\infty (2j+1)x^ {\frac{j(j+1)}2} = 1+ 3x+ 5x^3+ 7x^6+ 9x^{10}+ 11x^{15}+13x^{21}+ \cdots$$
 * This series is simpler than the double sum above. Bo Jacoby (talk) 01:36, 22 December 2017 (UTC).
 * In what way is this helpfull? Plasmic Physics (talk) 03:33, 22 December 2017 (UTC)

Your area of interest is $$\frac{\epsilon_0}{\tau}>0$$. So $$0<x<1$$. The series converges for complex $$x$$ where $$|x|<1$$. Choose a number $$N$$ such that the polynomial $$z_N(x) = \sum_{j=0}^N (2j+1)x^ {\frac{j(j+1)}2} $$ of degree $$D=\frac {N(N+1)}2$$ is a sufficiently good approximation to the power series $$z_\infty(x) = \sum_{j=0}^\infty (2j+1)x^ {\frac{j(j+1)}2} $$ within your area of interest. The polynomial $$z_N(x)$$ has the $$D$$ complex roots $$r_1, \cdots, r_D$$, and $$z_\infty(x)\approx z_N(x) = (2N+1)\prod_{j=1}^D (x-r_j)$$. This is a finite product. To obtain an infinite product, let $$N\rarr\infty$$. Bo Jacoby (talk) 12:03, 22 December 2017 (UTC).
 * I am not sure that this limit exists. For example, $$\exp(x)$$ does not have zeros but any N-th order Taylor polynomial of it has N roots. Ruslik_ Zero 20:36, 22 December 2017 (UTC)
 * The product representation of $$e^x$$ is $$\lim_{N\rarr\infty} \left(1+\frac x N \right)^N$$. Bo Jacoby (talk) 20:58, 22 December 2017 (UTC).
 * It is true but what I said still holds. Ruslik_ Zero 20:07, 23 December 2017 (UTC)

I should have emphasized that the roots depend on N: $$z_N(x) = (2N+1)\prod_{j=1}^D (x-r_{N,j})$$. The limit $$z_\infty(x)=\lim_{N\rarr\infty} z_N(x)$$ certainly exists for $$|x|<1$$. For example $$z_7(x)=15\prod_{j=1}^{28}(x-r_{7,j})$$ where $$r_{7,1}=-0.292929$$, $$r_{7,2}=-0.0262713+0.709453i$$, $$r_{7,3}=-0.0262713-0.709453i$$, $$r_{7,4}=0.407881+0.745992i$$, $$r_{7,5}=0.407881-0.745992i$$, $$r_{7,6}=-0.909933+0.102748i$$, $$r_{7,7}=-0.909933-0.102748i$$, $$r_{7,8}=0.642708+0.659149i$$, $$r_{7,9}=0.642708-0.659149i$$, $$r_{7,10}=-0.66325+0.652759i$$, $$r_{7,11}=-0.66325-0.652759i$$, $$r_{7,12}=-0.305051+0.900312i$$, $$r_{7,13}=-0.305051-0.900312i$$, $$r_{7,14}=0.801119+0.536062i$$, $$r_{7,15}=0.801119-0.536062i$$, $$r_{7,16}=0.910997+0.377235i$$, $$r_{7,17}=0.910997-0.377235i$$, $$r_{7,18}=0.977577+0.194432i$$, $$r_{7,19}=0.977577-0.194432i$$, $$r_{7,20}=-0.878192+0.472676i$$, $$r_{7,21}=-0.878192-0.472676i$$, $$r_{7,22}=-1$$, $$r_{7,23}=i$$, $$r_{7,24}=-i$$, $$r_{7,25}=0.317872+1.00891i$$, $$r_{7,26}=0.317872-1.00891i$$, $$r_{7,27}=-0.62899+0.853767i$$, $$r_{7,28}=-0.62899-0.853767i$$.

Bo Jacoby (talk) 08:15, 24 December 2017 (UTC).