A week ago, I derived pic related, then extended it to
[math] \displaystyle (1-a)^{-1}= \prod_{n=0}^{ \infty}(1+a^{3^n}+a^{3^n2})[/math]
then to
[math] \displaystyle (1-a)^{-1}= \prod_{n=0}^{ \infty}(1+a^{4^n}+a^{4^n2}+a^{4^n3})[/math]
then generalised it to
[math] \displaystyle (1-a)^{-1}= \prod_{n=0}^{ \infty} \sum_{j=1}^{k}a^{k^n(k-j)}[/math]
for [math]|a|<1<k \epsilon \mathbb{N}[/math]
amirite?
No.
used [math]a[/math] instead of [math]x[/math], sorry for fckup
Yes you are right. This is essentially a demonstration of the fact that every natural number has a unique representation in any natural number base k, k ≥ 2.
>>7973378
thx Anon, I have no idea of any application, can you suggest? I was thinking maybe modular analysis?
Can you show/link to the derivation?
I'm pretty loaded and don't see it trivially. The most general case would be nice, if possible.
>>7973394
It's e.g. on StackExchange. For small numbers, you can check what happens
[math] (1+x) (1+x^2) (1+x^4) [/math]
[math] = (1+x^2) (1+x^4) + x (1+x^2) (1+x^4) [/math]
[math] = (1+x^4) + x^2 (1+x^4) + (x+x^{2+1}) (1+x^4) [/math]
[math] = (1+x^4) + (x^2 + x^{4+2}) + (x (1+x^4) +x^{2+1} (1+x^4)) [/math]
[math] = (1+x^4) + (x^2 + x^{4+2}) + (x+x^{4+1} + x^{2+1} + x^{4+2+1}) [/math]
[math] = 1 +x+x^2+x^3+x^4+x^5+x^6+x^7 [/math]
The point is that at n=3 numbers, each number in the scheme 2^n (so 1 and 2 and 4) never added to itself, give you
1
2
2+1
4
4+1
4+2
4+2+1
similarly, for n=2 and the scheme 3^n & 2·3^n (so 1 & 2 and 3 & 6 and 9 & 18)
1
2
3 (not 1+2, which is excluded as they come in one sum)
3+1
3+2
6
6+1
6+2
9 (not 3+6, which is excluded as they come in one sum)
9+1
9+2
9+3
9+3+1
9+3+2
9+6
well and so on.
You take a grid like 1,2,4,8,16,... or 1,3,9,27,... and provide stuff so that everything up the the next number can be reached.
Smells a lot like ideals in ring theory.
>>7973361
Here's something I came up with, a method of translating any infinite product to an infinite sum:
[math] \prod_{k=0}^\infty b_k = \prod_{k=0}^{M-1}b_k + \sum_{n=M}^\infty(b_n-1)\,\prod_{k=0}^{n-1}b_k [/math]
E.g.
[math] (1-x)^{-1} = 1 + \sum_{n=1}^\infty x^{2^n} \, \prod_{k=0}^{k-1} (1-x^{2^k}) [/math]
[math] (x+1) x^2+(x+1) \left(x^2+1\right) x^4+(x+1) \left(x^2+1\right) \left(x^4+1\right) x^8+x+1 + ... [/math]
>>7973394
This is what I did:
[math](1-x)^{-1}=1+x+x^2+x^3+x^4 \cdots [/math]
[math]=(1+x)(1+x^2+x^4+x^6+ \cdots )[/math]
[math]=(1+x)(1+x^2)(1+x^4+x^8+x^{12}+ \cdots )[/math]
[math]=(1+x)(1+x^2)(1+x^4)(1+x^8+x^{16}+x^{24}+ \cdots )[/math]
[math]=(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16}+x^{32}+ \cdots )[/math]
and so on by induction.
>>7973411
>a method of translating any infinite product to an infinite sum
I'ma hafta check this out tomorrow, saved screenshot, this looks really interesting, thx Anon.
>>7973438
I tracked my derivation here
https://axiomsofchoice.org/infinite_product_of_complex_numbers
>>7973361
Well op. I've proved your first statement.
First let us define the function [math]F_m(x)[/math] that denotes your product.
[eqn]
F_m(x) = \prod_{n=0}^\infty \sum_{k=0}^{m-1} x^{k m^n}
[/eqn]
We can derive the following recurrence
[eqn]
\begin{align}
F_m(x^m) &= \prod_{n=0}^\infty \sum_{k=0}^{m-1} x^{k m^{n+1}} \\
&= \prod_{n=1}^\infty \sum_{k=0}^{m-1} x^{k m^n} \\
F_m(x^m) \sum_{k=0}^{m-1} x^k &= \prod_{n=0}^\infty \sum_{k=0}^{m-1} x^{k m^n} \\
F_m(x^m) \frac{x^m-1}{x-1} &= F_m(x)
\end{align}
[/eqn]
After applying this n times we get (induction left to the reader)
[eqn]
F_m(x) = F_m(x^{m^n}) \prod_{k=1}^n \frac{x^{m^k}-1}{x^{m^{k-1}}-1}
[/eqn]
You might note that is a telescoping product! Formally we can use linear difference operators to determine the partial product. (Though you could just as easily distribute, shift indices,cancel,etc.)
[eqn]
\begin{align}
\log(F_m(x)) &= \log(F_m(x^{m^n})) + \sum_{k=1}^n \left( \log(x^{m^k}-1) - \log(x^{m^{k-1}}-1) \right) \\
&= \log(F_m(x^{m^n})) + \sum_{k=1}^n \nabla \log(x^{m^k}-1) \\
&= \log(F_m(x^{m^n})) + \log(x^{m^n}-1) - \log(x^{m^0}-1) \\
F_m(x) &= F_m(x^{m^n}) \frac{x^{m^n}-1}{x-1} \\
\end{align}
[/eqn]
All that's left to do is take the limit as n goes to infinity, given that m is positive and -1<x<1.
[eqn]
\begin{align}
F_m(x) &= \lim_{n \to \infty} F_m(x^{m^n}) \frac{x^{m^n}-1}{x-1} \\
&= F_m(\lim_{n \to \infty} x^{m^n}) \lim_{n \to \infty} \frac{x^{m^n}-1}{x-1} \\
&= F_m(0) \frac{-1}{x-1} \\
&= \frac{1}{1-x} \\
\end{align}
[/eqn]
>>7973793
Nice work! Here is the first step of the induction:
[math] \displaystyle F_m(x^{m^2})= \prod_{n=0}^ \infty \sum_{k=0}^{m-1} x^{km^{n+2}}= \prod_{n=2}^ \infty \sum_{k=0}^{m-1}x^{km}[/math]
so
[math] \displaystyle F_m(x^{m^2})=F_m(x^m) \div \sum_{k=0}^{m-1}x^{km}[/math]
where
[math] \displaystyle \sum_{k=0}^{m-1}x^{km}=\frac{1-x^{m^2}}{1-x^m}[/math]
therefore
[math] \displaystyle F_m(x^m)= \frac{1-x^{m^2}}{1-x^m}F_m(x^{m^2})= \prod_{k=1}^{2} \frac{1-x^{m^k}}{1-x^{m^{k-1}}}F_m(x^{m^2})[/math]
and further induction gives your result above.
