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I'm given $$ a_n = \sum_{i=2}^{n-2} a_ia_{n-i} \quad (n\geq 3), a_0 = a_1=a_2 = 1 $$ and I need to find the functional equation for the generating function satisfying the above equality. I obtained the correct answer, but I don't understand why some of the steps are allowed.

$$ \begin{align} f(x) &= \sum_{n=0}^{\infty} a_nx^n\\ &=\sum_{n=0}^{\infty} \sum_{i=2}^{n-2} a_i a_{n-i} x^n\\ &=\sum_{n=0}^{\infty} \sum_{i=2}^{n-2}a_ix^i a_{n-i}x^{n-i}\\ &=\sum_{n=2}^{\infty} \sum_{i=2}^{n-2} a_ix^i a_{n-i}x^{n-i} \text{ (since it doesn't make sense for $n$ to start at $0$})\tag1\\ &=\sum_{n=2}^{\infty} \sum_{i=2}^{\infty} a_ix^i a_{n-i}x^{n-i} \text{ (since terms past $n-2$ don't exist)} \tag2\\ &=\sum_{n=2}^{\infty} \sum_{i=2}^{\infty} a_ix^i a_{n}x^{n} \text{ (not sure)} \tag3 \end{align} $$ Therefore, $$ f(x) -1-x-x^2 = (f(x) -1 -x)^2. $$ So the equations $(1),(2)$, and $(3)$, I'm not sure why those steps are allowed (or if they're even correct). Could someone please help?

TheRealFakeNews
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1 Answers1

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For this problem, I would go in the other direction:

$\begin{align} f^2(x) &= (\sum_{n=0}^{\infty} a_n x^n) (\sum_{m=0}^{\infty} a_m x^m)\\ &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n x^n a_m x^m\\ &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n a_m x^{n+m}\\ &= \sum_{i=0}^{\infty} \sum_{j=0}^{i}a_j a_{i-j} x^{i}\\ &= \sum_{i=0}^{\infty} x^i \sum_{j=0}^{i}a_j a_{i-j} \\ &= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i \sum_{j=0}^{i}a_j a_{i-j} \\ &= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i (2a_0 a_i + 2 a_1 a_{i-1} +\sum_{j=2}^{i-2}a_j a_{i-j} )\\ &= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i (2a_0 a_i + 2 a_1 a_{i-1} +a_i )\\ &= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)\sum_{i=2}^{\infty} x^i a_i + 2 a_1\sum_{i=2}^{\infty} x^i a_{i-1}\\ &= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)\sum_{i=2}^{\infty} x^i a_i + 2 a_1 x\sum_{i=2}^{\infty} x^{i-1} a_{i-1}\\ &= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)(f(x)-a_0-x a_1)+ 2 a_1 x\sum_{i=1}^{\infty} x^{i} a_{i}\\ &= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)(f(x)-a_0-x a_1)+ 2 a_1 x (f(x)-a_0)\\ &= a_0^2+ 2 a_0 a_1 x + f(x)(2 a_1 x+2 a_0+1)-(2 a_0+1)(a_0x+ a_1)- 2 a_0 a_1 x\\ &= a_0^2 + f(x)(2 a_1 x+2 a_0+1)-(2 a_0+1)(a_0x+ a_1)\\ &= 1 + f(x)(2 x+3)-3(x+1)\\ &= f(x)(2 x+3)-3x-2\\ \end{align} $

This gives a quadratic equation for $f(x)$ (as expected, from the recursion for $a_n$):

$f^2 - (2x+3)f + 3x+2 = 0$.

Solving, the discriminant is $(2x+3)^2 - 4(3x+2) = 4x^2+12x+9-12x-8 =4x^2+1 $, so the solutions are $f(x) = \frac{2x+3 \pm \sqrt{4x^2+1}}{2} $. Since $f(0) = 1$, we have to take the minus sign, so $f(x) = \frac{2x+3 - \sqrt{4x^2+1}}{2} $.

There is a significant chance that there is some error here (since I did it off the top of my head), but the basic idea is that this type of recurrence always leads to a quadratic equation in the generating function.

Note that the coefficients of all the odd powers of $x$ (from $x^3$ on) are zero, which agrees with Glenn O's comment about $a_3$.

Also, I do not know whether or not it is a coincidence that the discriminant does not have an $x$ term - if it did, expanding the square root would be a lot harder.

marty cohen
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