16ciii. (1 Viewer)

Axio · Aug 22, 2015

For this proof, can I do what I did from lines 3 to 4?

$\sum_{n=0}^{\infty }\frac{x^{2^{n}}}{1-x^{2^{n+1}}}\\=\lim_{N\rightarrow \infty }\frac{1-x^{2^{N+1}}-1+x}{(1-x)(1-x^{2^{N+1}})}\\=\lim_{N\rightarrow \infty }\frac{x(1-x^{2^{N+1}-1})}{(1-x)(1-x^{2^{N+1}})}\\=\frac{x(1-x^\infty )}{(1-x)(1-x^\infty )}\\=\frac{x}{1-x}$

Drsoccerball · Aug 22, 2015

I think you got to mention that the reason you can cancel them out so easily is because theyre equally infinity.

Drsoccerball · Aug 22, 2015

Hold on theyre not equally infinity i wouldve said that as the exponential approaches infinity the +1 -1 becomes meaningless

porcupinetree · Aug 22, 2015

How are you making the transition from line 1 to 2?

braintic · Aug 22, 2015

Is there supposed to be a restriction on x in this question?

Axio · Aug 22, 2015

porcupinetree said:
How are you making the transition from line 1 to 2?

I used a previous part of the question.

braintic said:
Is there supposed to be a restriction on x in this question?

-1 < x < 1

kawaiipotato · Aug 22, 2015

I think it'd be better to say since -1 < x < 1 then x^(infinity) approaches zero

Axio · Aug 22, 2015

kawaiipotato said:
I think it'd be better to say since -1 < x < 1 then x^(infinity) approaches zero

Yeah that sounds like a good addition.

InteGrand · Aug 22, 2015

Axio said:
For this proof, can I do what I did from lines 3 to 4?

$\sum_{n=0}^{\infty }\frac{x^{2^{n}}}{1-x^{2^{n+1}}}\\=\lim_{N\rightarrow \infty }\frac{1-x^{2^{N+1}}-1+x}{(1-x)(1-x^{2^{N+1}})}\\=\lim_{N\rightarrow \infty }\frac{x(1-x^{2^{N+1}-1})}{(1-x)(1-x^{2^{N+1}})}\\=\frac{x(1-x^\infty )}{(1-x)(1-x^\infty )}\\=\frac{x}{1-x}$

$Avoid writing $x^{\infty}.$

16ciii. (1 Viewer)

Axio

=o

Drsoccerball

Well-Known Member

Drsoccerball

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porcupinetree

not actually a porcupine

braintic

Well-Known Member

Axio

=o

kawaiipotato

Well-Known Member

Axio

=o

InteGrand

Well-Known Member

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