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

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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 $\tau \geq 0$, 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:

$z={\sqrt {2\pi }}e^{\frac {\epsilon _{0}}{4}}\sum _{j=0}^{\infty }\sum _{k=0}^{\infty }{\frac {\left(j+{\frac {1}{2}}\right)I_{{\frac {1}{2}}-k}\left(1\right)}{k!2^{k}e^{\epsilon _{0}\left(j+{\frac {1}{2}}\right)^{2}}}}\left(\tau ^{2}-1\right)^{k},\quad \tau \geq 0$

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 $N\rightarrow \infty$. 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 $\lim _{N\rightarrow \infty }\left(1+{\frac {x}{N}}\right)^{N}$. 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 $z_{\infty }(x)=\lim _{N\rightarrow \infty }z_{N}(x)$ 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).