Problems with the Geometric Series Factored

Question

Consider the geometric series generated by 1/(1-x).If I were to factor the series out I would get $$\frac{1}{1-x}= (1+x)(1+x^2)(1+x^4)(1+x^8)...(1+x^{2^n}) ...$$Suppose I were to take the natural logarithm of both sides I would get the following: $$-\ln(1-x)= \ln(1+x)+\ln(1+x^2)+\ln(1+x^4)+...\ln(1+x^{2^n})+...$$ Let $x=-1$; I would get $$-\ln(2)=\ln(0)+\ln(2)+\ln(2)...$$

But this should be wrong because the ln(0) is infinity. What's going on?

Answer 1

Note that the identity you wrote is incorrect. It is the case that $$ (1+x)(1+x^2)(1+x^4)\dotsb(1+x^{2^n}) =\frac{1-x^2}{1-x}\times\frac{1-x^4}{1-x^2}\times\dotsb\times\frac{1-x^{2^{n+1}}}{1-x^{2^n}}=\frac{1-x^{2^{n+1}}}{1-x}\tag{1} $$ so $$ \sum_{i=1}^n\log(1+x^{2^i})=\log(1-x^{2^{n+1}})-\log(1-x);\quad (-1<x<1) $$ where the restriction is necessary so that the log laws hold.

Answer 2

You do not have to go too far to get a surprize out of $$\frac{1}{1-x}= (1+x)(1+x^2)(1+x^4)(1+x^8)...(1+x^{2^n})$$

Let $x=-1$ and you get $$ 1/2 =0 $$

The problem is with the interval of convergence of $$ \frac {1}{1-x} = 1+x+x^2+... $$

Answer 3

The problem arises for all $n$. For $n=0$. You get $\,(1-x^2)/(1-x) = (1+x\,)$ which is true for all $\,x\neq 1.\,$ If $\,x=-1\,$ then you have $\,0=0.\,$ Now $\,\log 0\,$ is undefined but equation is true. The next case for $n=1$ you get $\,(1-x^4)/(1-x) =(1+x)(1+x^2)\,$ and now if $\,x=-1\,$ you have $0=0$ and you still have $\log 0$ problem. In general $\, (1-x^{2^{n+1}})/(1-x) = (1+x)(1+x^2)\cdots(1+x^{2^n})\,$ and for $\,x=-1\,$ you still have $\log 0$ problems. In the limit, $\,x^{2^{n+1}} \to 0\,$ but only if $\,|x|<1.$

An interpretation of the equation $\, 1 + x + x^2 + \dots + x^{2^{n+1}-1} = (1+x)(1+x^2) \cdots (1+x^{2^n}) \,$ is that every integer $\, 0 <= k < 2^{n+1} \,$ has a unique binary representation as a sum of powers of $2$ using $\, 2^0, 2^1, 2^2, \dots, 2^n. \,$ If $\, x \neq 1 \,$ then the left side of the equation equals $\, (1-x^{2^{n+1}})/(1-x). \,$

Answer 4

The proof of the infinite product relies on identities of the form

$$(1+x)(1+x^2)(1+x^4)=1+x+x^2+x^3+x^4+x^5+x^6+x^7.$$

For $x=-1$,

$$(1-1)(1+1)(1+1)=1-1+1-1+1-1+1-1$$ still holds but nothing allows you to equate it to

$$\frac1{1-(-1)}$$ "in the limit".

For the same reason that

$$1+x+x^2+x^3+x^4+x^5+x^6+x^7+\cdots=\frac1{1-x}$$ is only valid for $|x|<1$.