Finding the power series representation of $\frac{x^2+1}{8x-1}$

Question

So, I'm honestly at a loss of what I am doing wrong here. I know that the $x^2+1$ needs to be factored out of the summation, but apparently, this was not the answer to the first text box. Also, I'm not sure how to find the rate in this particular geometric series. I have tried to factor out $-1$ to make the rate $(8x)$, but I am not sure if this is even the right approach. Where did I go wrong in this problem?

Answer 1

\begin{align*} & \frac{x^2+1}{8x-1} \\ & \\ & = \frac{\color{red}{x^2 - \frac{1}{8}x} + \frac{1}{8}x + 1}{\color{red}{8x-1}} = \color{red}{\frac{x}{8}} + \frac{\frac{1}{8}x + 1}{8x-1} \\ & \\ & = \frac{x}{8} + \frac{\color{blue}{\frac{1}{8}x - \frac{1}{64}} + \frac{65}{64}}{\color{blue}{8x-1}} = \frac{x}{8} + \color{blue}{\frac{1}{64}} + \frac{65}{64}\frac{1}{8x-1} \\ & \\ & = \frac{x}{8} + \frac{1}{64} - \frac{65}{64}\frac{1}{1-8x} = \frac{x}{8} + \frac{1}{64} - \frac{65}{64}(1-8x)^{-1}, \\ & \\ \end{align*}

and you can expand $(1-8x)^{-1} = (1+(-8x))^{-1}$ using the Binomial expansion, or Geometric series, as Sil pointed out in the comments.

If you're not comfortable with the method I used above, you can instead use polynomial long division:

$8x-1 \ \overline{)x^2 + 0x + 1}$

although my method is in fact polynomial long division in disguise.

$$$$

It looks like you went the quickest route and skipped the fraction/polynomial long division steps entirely:

$\frac{x^2+1}{8x-1} = \frac{-x^2-1}{1-8x} = (-x^2-1) \left(1 + (8x) + (8x)^2 + ...\right) = \ ...$

which is correct, but it isn't in the form required in your worksheet.

Answer 2

Using the method as suggested in the comments and by Adam Rubinson:

You just have $(x^2+1) \left( \frac{-1}{1 - 8x} \right)$. Now expand $\left( \frac{-1}{1 - 8x} \right)$ using geometric series, where $a = -1$, $r = 8x$ to get:

$$x^2 \left(- 1-8x-64x^2 - \cdots \right) + \left(- 1-8x-64x^2 - \cdots \right)$$ $$= -1-8x+(-1-64)x^2+(-8-512)x^3$$

Notice that for the $x^n$ term, the coefficient of the second summation is $8^2$ times that of the coefficient in the first summation, because the first summation is shifted two places to the right (which means it is a factor of $8^2$ behind).

Hence, after $-1-8x$, the general term is $(-8^{n-2} - 8^n)x^n$, or $-8^{n-2} (1+8^2)x^n = -1 \cdot 8^{n-2} \cdot 65x^n$ starting from $n = 2$. In summation notation, this is just:

$$f(x) = -1-8x - 65 \sum_{k=2}^\infty 8^{k-2} \cdot x^k, |x| < \frac{1}{8}.$$

and factoring the $8^{-2}$ outside the summation gives the nicer form:

$$f(x) = -1-8x - \frac{65}{64} \sum_{k=2}^\infty (8x)^k, |x| < \frac{1}{8}.$$

which is equivalent to the form stated in the comments.