Let $F_0, F_1, F_2, ...$ be the Fibonacci numbers and let $f$ be the function defined $$f(x) = \frac{x}{1-x(1+x)}$$


The function $f$ is called the "generating function of the sequence $F_0, F_1, F_2, ...$ it is a "formal" algebraic expression in the sense that it behaves in a natural algebraic way but the $x$ never takes a numerical value, the powers of it merely acting as 'place-markers' in the power series$^1$. Bearing that in mind, we see that $$f(x) = F_0 + F_1x + F_2x^2 + F_3x^3 + ....$$ $$xf(x) = F_0x + F_1x^2 + F_2x^3 + ....$$ $$x^2f(x) = F_0x^2 + F_1x^3 + ....$$ But we know that $F_2 = F_1 + F_0, F_3 = F_2 + F_1$, etc., and so subtracting the second and third lines from the first gives us $f(x)(1 - x(1 +x)) = f(x) - xf(x) - x^2f(x) = **F_0 + (F_1 - F_0)x** = x$

Please understand me it. I highlight what the most confuesed me.

$^1$ So what is x? f(x) is function or not? I'm so confused.


1 Answer 1


This is a subtle point. The idea is that when you were manipulating the power series for $f(x)$ you never said anything about whether or not the series converged -- in other words, you were formally manipulating the series. A formal power series is just a power series used without talking about convergence.

To answer your question, $f(x)$ is not a function. In the (paraphrased) words of Herbert Wilf, a formal power series is a "clothesline on which we hang up a sequence of numbers for display". We use $f(x)$ to keep track of the Fibonacci numbers; that's all. In the same vein, you could use the series $$ g(x) = 0! + 1!\cdot x + 2!\cdot x^2 + \cdots + n!\cdot x^n + \cdots $$ to keep track of the factorial numbers even though the series converges only for $x=0$.

  • $\begingroup$ " to keep track of the factorial numbers even though the series converges only for x=0" Why is it important to series would be converged? $\endgroup$
    – user180834
    Dec 18, 2014 at 17:26

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