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When we consider the function $f(x)=1+x+x^2+x^3+...+x^n$ where $n$ tends to infinity, we can rewrite this as $$f(x)=1+x(1+x+x^2+x^3+...)=1+x(f(x))\qquad (1)$$ After some algebraic manipulations, we arrive at $$f(x)=\frac{1}{1-x}\qquad (2)$$ This expression is used to assign a "sum" to certain divergent series. If we take another look at (1), however, we can also write $$f(x)=1+x(1+xf(x))$$ from which we can deduce that $f(x)=\frac{1+x}{1-x^2}$. Of course, we can produce infinitely many functions that satisfy $f(x)$ in this manner, but I only see (2) in the literature. Is there a good reason for this phenomenon? Thanks,


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up vote 8 down vote accepted

$1-x^2=(1-x)(1+x)$ so you did not find a new solution.

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Aha! Thank you user9325! I don't want to insult you with your good answer, but I feel a tad stupid now. I will accept it as soon as the minimum time to wait is over. – Max Muller May 5 '11 at 13:11
It's not stupid to ask "stupid" question. There is one aspect I did not address: Your functional equation is wrong for $x=1$ which is not so surprising as the series is divergent there but it shows that this kind of functional equation demands some subtleness in general. – Phira May 5 '11 at 13:17
All right, what kind of "subtleness" do you mean? Is it perhaps related to Tauberian theorems? – Max Muller May 5 '11 at 13:22
No, I just meant that a functional equation is not unique without thinking about the domain of the function. – Phira May 5 '11 at 13:28

Notice that

$$f(x) = \frac{1+x}{1-x^2} = \frac{1+x}{(1+x)(1-x)} = \frac{1}{1-x}$$

so you end up with the same function.

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And you can cancel $1+x$ because the series does not converge for $x=-1$. – lhf May 5 '11 at 13:11
Interestingly, you can just plug $x=-1$ into both sides of the formula, and get $1-1+1-1+1-1+\cdots = 1/2$. – Chris Taylor May 5 '11 at 13:14
Indeed! It is called the Cesaro Summation of Grandi's Series ;) See – Max Muller May 5 '11 at 13:19
that's Cesàro summation. – lhf May 5 '11 at 13:19

Let $f(x) = \sum x^k$ and let $g(x) = 1/(1-x)$. Then where $f$ makes sense (namely for $|x| < 1$) the functions $f$ and $g$ agree. This answers your last question "Is there a good reason for this phenomenon?"

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