Power series for rational function

Question

Before your read my problem: i am thankful for every help i can get but i need a closed form (formal power series) and not any other representation of my function.

I am working on a problem involving generating functions. I derived a function for my counting problem of the form:

$$ f(x) = \frac{4x^2 + 4x}{1 - 3x - 3x^2}. $$

Now I want to find a closed formal power series for this function in the form:

$$ \sum_{n=0}^\infty a_n x^n. $$

First, I concluded that I had to use polynomial division because the degree of $P(x)/Q(x)$ are the same. After performing polynomial division, I found that:

$$ f(x) = -\frac{4}{3} + \frac{4}{3} \cdot \frac{1}{1 - 3x - 3x^2}. $$

Here, the constant $ -\frac{4}{3} $ and the $ \frac{4}{3}$ multiplied with the series have "no effect" on how I calculate the formal power series, as I can apply them later to the closed-form solution. So I consider:

$$ f(x) = -\frac{4}{3} + \frac{4}{3} \cdot \left( \sum_{n=0}^\infty a_n x^n \right). $$

To proceed, I want to find the decomposition of $ \frac{1}{1 - 3x - 3x^2} $. I start with partial fraction decomposition:

$$ \frac{1}{(x - x_1)(x - x_2)} = \frac{A}{x - x_1} + \frac{B}{x - x_2}, $$

where $ x_1 $ and $ x_2 $ are the roots of $ 1 - 3x - 3x^2 $, calculated as:

$$ x_1 = \frac{3 + \sqrt{21}}{-6}, \quad x_2 = \frac{3 - \sqrt{21}}{-6}. $$

From the decomposition, we write:

$$ 1 = A(x - x_2) + B(x - x_1), $$

which expands to:

$$ 1 = x(A + B) - Ax_2 - Bx_1. $$

By equating coefficients, we find:

$$ A + B = 0 \quad \text{so} \quad A = -B, $$

and:

$$ -Ax_2 - Bx_1 = 1 \quad \text{then plug in A = -B we get:} \quad B(x_2 - x_1) = 1. $$

Substituting the values of $ x_1 $ and $ x_2 $, we calculate:

$$ B = \sqrt{\frac{3}{7}}, \quad A = -\sqrt{\frac{3}{7}}. $$

Thus, the decomposition becomes:

$$ \frac{1}{(x - x_1)(x - x_2)} = \frac{-\sqrt{\frac{3}{7}}}{x - x_1} + \frac{\sqrt{\frac{3}{7}}}{x - x_2}. $$

Next, consider the term $ \frac{-\sqrt{\frac{3}{7}}}{x - x_1} $. This can be rewritten as:

$$ \frac{-\sqrt{\frac{3}{7}}}{x - x_1} = \frac{-\sqrt{\frac{3}{7}}}{-x_1 \left( 1 - \frac{x}{x_1} \right)} = \frac{\sqrt{\frac{3}{7}}}{x_1} \cdot \frac{1}{1 - \frac{x}{x_1}}, $$

which is:

$$ \frac{\sqrt{\frac{3}{7}}}{x_1} \cdot \sum_{n=0}^\infty \left( \frac{x}{x_1} \right)^n. $$

Similarly, for $ \frac{\sqrt{\frac{3}{7}}}{x - x_2} $, we have:

$$ \frac{\sqrt{\frac{3}{7}}}{x - x_2} = \frac{\sqrt{\frac{3}{7}}}{-x_2} \cdot \sum_{n=0}^\infty \left( \frac{x}{x_2} \right)^n. $$

Adding these terms together gives:

$$ g(x) = \frac{\sqrt{\frac{3}{7}}}{x_1} \cdot \sum_{n=0}^\infty \left( \frac{x}{x_1} \right)^n + \frac{\sqrt{\frac{3}{7}}}{-x_2} \cdot \sum_{n=0}^\infty \left( \frac{x}{x_2} \right)^n. $$

Substituting the values of $x_1$ and $x_2$ and simplifying, we get:

$$ g(x) = \frac{3}{14} \cdot (7 + \sqrt{21}) \cdot \sum_{n=0}^\infty \left( \frac{x}{6(-3 + \sqrt{21})} \right)^n + \frac{3}{14} \cdot (7 - \sqrt{21}) \cdot \sum_{n=0}^\infty \left( \frac{x}{6(-3 - \sqrt{21})} \right)^n. $$

Combining this with the earlier expression, we write the final form as:

$$ f(x) = -\frac{4}{3} + \frac{4}{3} \cdot g(x), $$

which expands to:

$$ f(x) = -\frac{4}{3} + \frac{4}{3} \cdot \left[ \frac{3}{14} \cdot (7 + \sqrt{21}) \cdot \sum_{n=0}^\infty \left( \frac{x}{6(-3 + \sqrt{21})} \right)^n + \frac{3}{14} \cdot (7 - \sqrt{21}) \cdot \sum_{n=0}^\infty \left( \frac{x}{6(-3 - \sqrt{21})} \right)^n \right]. $$

However, when I compare this result to the Taylor expansion of the original function:

$$ f(x) = 4x + 16x^2 + 60x^3 + 228x^4 + 864x^5, $$

the coefficients $(4, 16, 60, 228, 864, \dots)$, which represent the possible combinations in my counting problem, do not match the coefficients obtained from my final expression. For n=0 i need the expression to be equal to 1.

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Edit: correcting the polynomial division result from $ f(x) = \frac{4}{3} - \frac{4}{3} \cdot \frac{1}{1 - 3x - 3x^2}. $ to $ f(x) = -\frac{4}{3} + \frac{4}{3} \cdot \frac{1}{1 - 3x - 3x^2}. $

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My question is: What have I done wrong? Am I using an incorrect approach, or do I have a misunderstanding about some concept?

Answer 1

It seems there is a small mistake when doing the partial fraction decomposition. We have \begin{align*} \frac{1}{1-3x-3x^2}&=\left(-\frac{1}{3}\right)\frac{1}{x^2+x-\frac{1}{3}}\\ &=-\frac{1}{3}\left(\frac{A}{x-x_1}+\frac{B}{x-x_2}\right)\tag{1} \end{align*} Since \begin{align*} -\frac{1}{3}A&=-\frac{1}{3}\left(-\sqrt{-\frac{3}{7}}\right)=\frac{1}{\sqrt{21}}\\ -\frac{1}{3}B&=\frac{1}{3}A=-\frac{1}{\sqrt{21}} \end{align*} we get from (1) \begin{align*} \color{blue}{\frac{1}{1-3x-3x^2}=\frac{1}{\sqrt{21}}\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)}\tag{2} \end{align*} from where you can continue in the way you have described.

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[2024-11-18] With respect to OPs comment some more details. We can continue from (2) with \begin{align*} \frac{1}{\sqrt{21}}&\left(\frac{1}{x-x_1}-\frac{1}{x-x_2}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\cdot\frac{1}{1-\frac{x}{x_1}}+\frac{1}{x_2}\cdot\frac{1}{1-\frac{x}{x_2}}\right)\\ &=\frac{1}{\sqrt{21}}\left(-\frac{1}{x_1}\sum_{n=0}^{\infty}\left(\frac{x}{x_1}\right)^n +\frac{1}{x_2}\sum_{n=0}^{\infty}\left(\frac{x}{x_2}\right)^n\right)\\ &=\frac{1}{\sqrt{21}}\sum_{n=0}^{\infty}\left(\frac{1}{{x_2}^{n+1}} -\frac{1}{{x_1}^{n+1}}\right)x^n\tag{3} \end{align*} We know from OP that \begin{align*} x_1=-\frac{1}{6}\left(3+\sqrt{21}\right)\qquad\qquad x_2=-\frac{1}{6}\left(3-\sqrt{21}\right) \end{align*} from which we derive by also rationalizing the denominator \begin{align*} \frac{1}{x_1}&=-\frac{6}{3+\sqrt{21}}=\frac{3-\sqrt{21}}{2}\\ \frac{1}{x_2}&=-\frac{6}{3-\sqrt{21}}=\frac{3+\sqrt{21}}{2}\\ \end{align*} Putting this into (3) we finally obtain \begin{align*} {\color{blue}{f(x)}}&{\color{blue}{=\frac{4x^2-4x}{1-3x-3x^2}}}\\ &=-\frac{4}{3}+\frac{4}{3}\cdot\frac{1}{1-3x-3x^2}\\ &\color{blue}{=-\frac{4}{3}+\frac{4}{3\sqrt{21}} \sum_{n=0}^{\infty}\left(\left(\frac{3-\sqrt{21}}{2}\right)^{n+1}-\left(\frac{3+\sqrt{21}}{2}\right)^{n+1}\right)x^n}\\ &\color{blue}{=4x+ 16x^2 + 60x^3 + 228x^4 + 864x^5+\cdots} \end{align*} in accordance with the claim. The last line was calculated with some help of Wolfram Alpha.

Answer 2

The denominator $1-3x-3x^2$ is invertible in the power series ring since it has a constant term $1$ which is a unit. Hence all you need to do is to find a power series $\sum_{n=0}^\infty c_n x^n$ which is its inverse, and multiply this with the numerator $4x+4x^2$ to get the resulting expression as a closed formal power series. The coefficients $c_n$ can be determined by multiplying this power series with the denominator ($1-3x-3x^2$) and setting the product equal to $1$ (for example, $c_0 =1$, $c_1 - 3c_0 = 0;\ {\rm hence} \ c_1 = 3$, $c_2 -3c_1 -3c_0 = 0$; hence $c_2 = 12$; in general, $c_n -3c_{n-1} - 3c_{n-2} =0$ for $n \geq 2$, so the coefficients $c_n$ may be determined recursively).