Algebraic simplifications

Question

Given the following line of maths:

$$(1-x)(1+x)(1+x^2)(1+x^3)...(1+x^{2^n})$$

we can simplify it:

$$(1-x^2)(1+x^2)...(1+x^{2^n})$$

and so on to obtain:

$$(1-x)(1+x)(1+x^2)(1+x^3)...(1+x^{2^n}) = (1-x^{2^n})(1+x^{2^n})=(1-x^{2^n2})=(1-x^{2^{n+1}})$$

Therefore we can say that:

$$\frac{1-x^{2^{n+1}}}{1-x}=\prod_{i=0}^{n}(1+x^{2^i})$$

My question is how to show that:

$$\ln{\left( \frac{1}{1-x}\right)} = \sum_{i=0}^{\infty}\ln{(1+x^{2^i}})$$

I do not follow where we "lose" the: $\frac{-x^{2^{n+1}}}{1-x}$ part, and how we suddendly get the infinite series, instead of up to $n$.

Answer 1

My question is how to show that:

$$\ln{\left( \frac{1}{1-x}\right)} = \sum_{i=0}^{\infty}\ln{(1+x^{2^i}})$$

The above expression follows from the identity:

$$\frac{1-x^{2^{n+1}}}{1-x}=\prod_{i=0}^{n}(1+x^{2^i})$$

Only when $|x|<1$ or in other words, $-1<x<1$.

So assuming that this condition holds, from the identity you have already proved, we have that

$$\frac{1-x^{2^{n+1}}}{1-x}=\prod_{i=0}^{n}(1+x^{2^i})$$

As $n\to \infty$, $2^{n+1} \to \infty$ and hence $x^{2^{n+1}} \to 0$. Thus we get,

$$\frac{1}{1-x}=\prod_{i=0}^{\infty}(1+x^{2^i})$$

And then taking logarithm on both sides, we get

$$\ln{\left( \frac{1}{1-x}\right)} = \sum_{i=0}^{\infty}\ln{(1+x^{2^i}})$$

Hope this helps.

Answer 2

HINT:

For $|z|<1,$ $$\lim_{n\to\infty}z^n=0$$

and $$\dfrac{1-x^m}{1-x}=1+x+x^2+\cdots+x^{m-1}$$

and for $-1\le u<1,$ $$-\ln(1-u)=\sum_{r=1}^\infty\dfrac{u^r}r$$