# How do you use differentiation to find a power series representation for 1/(6+x)^2?

Apr 10, 2015

Basically:
1. Integrate to get a form of $\frac{1}{1 - x}$.
2. Modify the equation to achieve getting precisely $\frac{1}{1 - X}$ where X is some variant of x, whether it's $\frac{x}{6}$ (here), $- \frac{y}{2}$, $\frac{\theta}{\pi}$, etc.
3. Write out the power series with $x$ = some substituted value, like $\frac{x}{6}$, here.
4. Reverse what you did to re-acquire the original function. i.e. re-multiply by what you divided by (-1 and 1/6, here), then re-differentiate. Whatever returns your original function.

Notice how the power series $\frac{1}{1 - x}$ can be written as the power series:
$1 + x + {x}^{2} + {x}^{3} + \ldots$
$= \sum {x}^{n}$

Similarly, use -x instead of x. Every odd power is negative, and every even power is positive by virtue of squaring to some order of magnitude (e.g. ${\left({x}^{2}\right)}^{n}$).

$= \sum {\left(- 1\right)}^{n} {x}^{n} = 1 - x + {x}^{2} - {x}^{3} + {x}^{4} - \ldots$

Integrate $\frac{1}{6 + x} ^ 2$ to get $- \frac{1}{6 + x}$. Divide by -(1/6) to get $\frac{1}{1 + \frac{x}{6}}$.

x/6 is your new x. Plug it in, use this alternating series from a few lines up, factor in the 1/6 to get back to $\frac{1}{6 + x}$, and incorporate the negative to get back to $- \frac{1}{6 + x}$.

$\implies - \left(\frac{1}{6}\right) \left[{\left(- 1\right)}^{0} {\left(\frac{x}{6}\right)}^{0} + {\left(- 1\right)}^{1} {\left(\frac{x}{6}\right)}^{1} + {\left(- 1\right)}^{2} {\left(\frac{x}{6}\right)}^{2} + {\left(- 1\right)}^{3} {\left(\frac{x}{6}\right)}^{3} + \ldots\right]$
$\implies - \frac{1}{6} + \frac{x}{36} - {x}^{2} / 216 + {x}^{3} / 1296 - \ldots$

Then, re-differentiate the result to get back to $\frac{1}{6 + x} ^ 2$.

$\implies \frac{1}{36} - \frac{x}{108} + \frac{{x}^{2}}{432} - \ldots$