Question

find the maclaurin series for the function: (x(5x+1))/(1-x)^3

Answer 1

Given \[ f(x) = \frac{x(5x+1)}{(1-x)^3} = \frac{(5x^2+x)}{(1-x)^3}\]

We could use the definition of the MacClaurin series and find f(0), f'(0), f''(0), f'''(0), etc and plug it in, but that would be very tedious. You are more than welcome to do so, but we can also look at the power function:

\[ f(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n \]

If we take the derivative twice, we can get a function that starts to look like our function

\[ \frac{d}{dx}f(x) = \frac{1}{(1-x)^2} =\sum_{n=0}^{\infty}nx^{n-1} \]
the second derivative gives:

\[ \frac{d^2}{dx^2}f(x) = \frac{2}{(1-x)^3} =\sum_{n=0}^{\infty}n(n-1)x^{n-2} \]
so:

\[ \frac{1}{(1-x)^3} =\frac{1}{2}\sum_{n=0}^{\infty}n(n-1)x^{n-2} \]

Can you see how we would manipulate the above series to make it look like the function we are looking for? (There is one more step and some simplification)

Answer 2

yes but what about the numerator?

Answer 3

So, try and multipy each side of our equation (both the rational function and the series) by our desired numerator, the resulting series should be the answer

Answer 4

im lost there now

Answer 5

Okay, so lets multiple each side, we get the Left hand side to be:

\[ \frac{5x^2+x}{(1-x)^3} = \frac{(5x^2+x)}{2}\sum_{n=0}^{\infty} n^2-n(x^{n-2})\]
Note that the series is depedent on n, so we can bring (5x^2+x) inside:

\[ \frac{5x^2+x}{(1-x)^3} = \frac{(1)}{2}\sum_{n=0}^{\infty} n^2-n(5x^2+x)(x^{n-2})\]
Distribute the x^(n-2) and simplify using laws of exponents
\[ \frac{5x^2+x}{(1-x)^3} = \frac{(1)}{2}\sum_{n=0}^{\infty} n^2-n(5x^2(x^{n-2})+x(x^{n-2}))\]

\[ \frac{5x^2+x}{(1-x)^3} = \frac{1}{2}\sum_{n=0}^{\infty} (n^2-n)(5x^{n}+(x^{n-1}))\]

Should be the solution

Answer 6

Then you can also break it into two series:

\[ \frac{1}{2}\sum_{n=0} (n^2-n)5x^n+\frac{1}{2}\sum_{n=0} (n^2-n)x^{n-1}\]
\[ \frac{5}{2}\sum_{n=0} (n^2-n)x^n+\frac{1}{2}\sum_{n=0} (n^2-n)x^{n-1}\]

Answer 7

ok great thank you!