I figured this out last night
[math]\displaystyle \left (1-x \right )^{-1}= \prod_{n=0}^{ \infty} \left (1+x^{2^n} \right )[/math]
for |x| < 1 , amirite?
This must be a well-known expansion, ainnit?!
-1/12
>>7955380
Look up the binomial theorem
>>7955380
If you multiply it out, you'll get every possible combination of powers of 2 added together. This is exactly ever number, because you can represent every number in base 2. So it is not too surprising that this is equal to the geometric series.
>>7955380
Heres proof
>>7955410
>here's examples
Ftfy
>>7955394
wat
>>7955380
[math]\displaystyle (1-x)\prod_{n=0}^N (1+x^{2^n}) = 1-x^{2^{N+1}}[/math] (quick induction)
Letting N go to infinity, you get the identity
>>7955423
Since you were too fucking lazy to google it:
https://en.m.wikipedia.org/wiki/Binomial_theorem
You just have a specific example of the theorem imho
>>7955380
left side is just
[eqn] 1 + x + x^2 + x^3 + x^4 + \cdots [/eqn]
right side is
[eqn](1+x) (1+x^2) (1+x^4)\cdots [/eqn]
which gives powers of x equal to all the sums of powers of 2:
[eqn]1+2+4+8 +16+32+\cdots[/eqn]
Which obviously gives you each natural number once (you can see this by just treating it as binary and noting that every binary number corresponds to every natural number).
So its a neat, trivial identity.
>>7955429
Mobile users should just fuck off.
>>7955427
niiice, thx
>>7955429
>specific example of the theorem
no
>>7955437
>(you can see this by just treating it as binary and noting that every binary number corresponds to every natural number).
I do not see this
>>7955685
32 + 16 + 8 + 4 + 2 + 1
you need each combination to correspond to a natural number
lets take 32 + 16 + 2, in binary that is
1*32 + 1*16 + 0*8 + 0*4 + 1*2 + 0*1 = 110010
