How do you find a power series representation for f(x)=1/(1+x)^3?

Sep 24, 2015

Oh... I remember this one. Yeah, expect a long answer with this one.

I got:

$\frac{1}{2} {\sum}_{n = 0}^{N} \left(n + 1\right) \left(n + 2\right) {\left(- x\right)}^{n} = \frac{1}{1 + x} ^ 3$

$= 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + \ldots$

for some arbitrarily large finite $N$.

So, what you would do is start from the one series you should have been taught, which was:

$\setminus m a t h b f \left({\sum}_{n = 0}^{N} \left(x \text{'")^n = 1/(1-x"'}\right)\right)$

where $x '$ can be any variable and $N$ is arbitrarily large and finite. Notice how $1 + x = 1 - \left(- x\right)$, meaning that $x ' = - x$ for this instance.

What you could do is figure out how to go from the original function to the function whose power series you DO know.

$\textcolor{h i g h l i g h t}{\int} \frac{1}{1 + x} ^ 3 \mathrm{dx} = \frac{{\left(1 + x\right)}^{-} 2}{-} 2 = - \frac{1}{2 {\left(1 + x\right)}^{2}}$

$\frac{1}{2} \textcolor{h i g h l i g h t}{\int} - \frac{1}{1 + x} ^ 2 \mathrm{dx} = \frac{1}{2} \cdot \frac{1}{1 + x}$

Then, to achieve the form $\frac{1}{1 - x '}$:

$\frac{1}{2 \left(1 + x\right)} \cdot \textcolor{h i g h l i g h t}{2} = \frac{1}{1 - \left(- x\right)}$

Therefore, what you have just done is:

1. Integrated the original function
2. Integrated the resultant function
3. Multiplied by 2

Note that you will have to reverse these steps, so keep that in mind.

Now, we can proceed by writing out the power series for $\frac{1}{1 - \left(- x\right)}$:

$= {\sum}_{n = 0}^{N} {\left(- x\right)}^{n} = 1 - x + {x}^{2} - {x}^{3} + \ldots$

Next, we operate on the series to achieve the original function again.

To get the original function back, we have to undo what we did to the original function that allowed us to get to the series which we know the answer to, because operating on a power series mirrors operating on the function that is represented by said power series.

Therefore, we have to reverse the steps we performed earlier, but now on the power series rather than the function. To do this:

1. Divide by 2
2. Take the first derivative of the resultant function
3. Take the first derivative of the resultant function to get back to the original function

Then, we should have achieved the power series for $\frac{1}{1 + x} ^ 3$. We get:

1. Divide by 2

$\textcolor{h i g h l i g h t}{\frac{1}{2}} \cdot \left[1 - x + {x}^{2} - {x}^{3} + {x}^{4} - {x}^{5}\right]$

$= \frac{1}{2} {\sum}_{n = 0}^{N} {\left(- x\right)}^{n}$

$= \frac{1}{2} - \frac{x}{2} + {x}^{2} / 2 - {x}^{3} / 2 + {x}^{4} / 2 - {x}^{5} / 2 + \ldots$

2. Take the first derivative of the resultant function (remember that the index increases by $1$ now)

$= \textcolor{h i g h l i g h t}{\frac{d}{\mathrm{dx}}} \left[{\sum}_{n = 0}^{N} {\left(- x\right)}^{n} / 2\right]$

$= \textcolor{h i g h l i g h t}{\frac{d}{\mathrm{dx}}} \left[\frac{1}{2} - \frac{x}{2} + {x}^{2} / 2 - {x}^{3} / 2 + {x}^{4} / 2 - {x}^{5} / 2 + \ldots\right]$

$= \cancel{0} - \frac{1}{2} + x - \frac{3}{2} {x}^{2} + 2 {x}^{3} - \frac{5}{2} {x}^{4} + \ldots$

3. Take the first derivative of the resultant function to get back to the original function (remember that the index increases by $1$ again, since the first term disappears)

$= \textcolor{h i g h l i g h t}{\frac{d}{\mathrm{dx}}} \left[{\sum}_{n = 1}^{N} \frac{- n {\left(- x\right)}^{n - 1}}{2}\right]$

$= \textcolor{h i g h l i g h t}{\frac{d}{\mathrm{dx}}} \left[0 - \frac{1}{2} + x - \frac{3}{2} {x}^{2} + 2 {x}^{3} - \frac{5}{2} {x}^{4} + \ldots\right]$

$= \cancel{0 - 0} + 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + \ldots$

And there you have it!

$\textcolor{b l u e}{\frac{1}{1 + x} ^ 3}$

$= \textcolor{h i g h l i g h t}{\frac{1}{2} \frac{d}{\mathrm{dx}} \frac{d}{\mathrm{dx}}} \left[\frac{1}{1 - \left(- x\right)}\right]$

$= \textcolor{h i g h l i g h t}{\frac{d}{\mathrm{dx}}} \left[{\sum}_{n = 1}^{N} \frac{- n {\left(- x\right)}^{n - 1}}{2}\right]$

$= \frac{1}{2} {\sum}_{n = 2}^{N} n \left(n - 1\right) {\left(- x\right)}^{n - 2}$

Finally, if we shift the index so that $n$ starts at $0$, then the terms with $n$ become $n + 2$, $n + 1$, and ${\left(- x\right)}^{n}$, respectively.

$= \textcolor{b l u e}{\frac{1}{2} {\sum}_{n = 0}^{N} \left(n + 2\right) \left(n + 1\right) {\left(- x\right)}^{n}}$

$= \textcolor{b l u e}{1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + \ldots}$

with $| x | < 1$.