I read the generating functionology, where author handles $$b_k(x) = {x \over 1-kx} b_{k-1}(x) = {x ^k \over (1-x)(1-2)(1-3x) \cdots (1-kx)}$$ since $b_0(x) = 1.$ I see that if denominator $(1-kx)$ would be 1 in the recurrence, I would have simpler recurrence, $b_k = x b_{k-1} = x^k b_0$. I understand the logic. You unroll $b_{k-1}$ in terms of $b_{k-2}$ and so on: $b_k = x b_{k-1} = x (x b_{k-2}) = x (x(xb_{k-3})) = x^k b_{k-k}$. But I fail to expand $b_k = x b_{k-1}$ into $x^k$ using the generating functions.
Say you have $a_{n+1} = 3 a_n$. You may spot that $a_n = 3^n a_0$ using the logic above or work out $$a(x) = \sum_{n=0} {a_n x^n}$$ so that $$\sum_{n=0} a_{n+1} x^n = {\sum_{n=0} {a_{n+1} x^{n+1}} ± a_0 \over x} = {(a(x) - a_0) \over x}.$$ Now, equation $a_{n+1} = 3 a_n$ is rewritten as $${a(x) - a_0 \over x} = 3 a(x)$$ and $$a(x) = \frac{a_0}{1-3x} = \sum_{n=0} {(a_0 3^n) x^n} = \sum_{n=0} {a_n x^n} $$ so that $a_n = a_0 3^n$ in the series. Now, I have $x$ instead of 3 in $a_{n+1} = x a_n$. I may write down the solution $a_n = a_0 3^n$ right away. But, if I try to solve it the same way, I get the $a(x) = a_0 / (1-x^2)$ instead of $a(x) = a_0 / (1-3x)$ and $$a(x) = {a_0 \over 1-x^2} = {a_0 \over 1-x} \cdot {a_0 \over 1+x} = a_0\sum_{n=0} {1^n x^n} + a_0\sum_{n=0} {(-1)^nx^n}$$ so that $a_{n+1} = x a_n$ has solution $a_n = a_0 (1+(-1)^n)$ instead of $a_n = a_0 x^n.$ What is the mistake? Why does Wilf not fall into it?
Isn't it too localized? Can I reword the problem so that it is interesting for general audience?
Let's say we want $a_{n+1}(x) = x a_n(x)$. Then $$ a(x) = \frac{a_0(x)}{1-x^2} $$ is correct.
But if $$ \sum_{n=0}^\infty a_n(x) x^n = \sum_{n=0}^\infty u_n x^n, $$ for all $x$, it does not follow that $a_n(x)=u_n$