Find the Taylor series of the given function at a = 0
$\displaystyle
f(x)=\frac{1}{(1-x)^2}
$
Your help is very appreciated!
It's been a while since I've done these, but let's give it a try.
$\displaystyle f(x)=\frac{1}{(1-x)^2}$
So now we need to start taking some derivatives, and see what kind of pattern shows up.
$\displaystyle f'(x) = \frac{2}{(1-x)^3}$
I can see a pattern now just from this derivative that a few things are going to happen. The sign is going to alternate each term, the power in the denominator will increase by 1, and the numerator will have a coefficient of (n+1)!, where n is the nth derivative.
So generally, $\displaystyle f^{n}(x) = \frac{(-1)^{n+1}(n+1)!}{(1-x)^{n+2}}$
I think that looks good, but I apologize if I messed up. Also what's nice is since you are doing this for a=0, the denominator will always be 1.
As for the rest of the Taylor formula...
$\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^n$
Can you take it from here?
Use the fact that the maclaurin series for $\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$
Now notice that $\displaystyle \frac{D[\frac{1}{1-x}]}{dx}=\frac{1}{(1-x)^2}$
Now that we have shown this relationship between $\displaystyle \frac{1}{1-x}$ and $\displaystyle \frac{1}{(1-x)^2}$ we can make the logical next step
Since the derivative of $\displaystyle \frac{1}{1-x}$ is $\displaystyle \frac{1}{(1-x)^2}$
and we know that $\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$
THen it is logical to assume that the power series for $\displaystyle \frac{1}{(1-x)^2}$ is the derivative of $\displaystyle \sum_{n=0}^{\infty}x^n$
Which is $\displaystyle \sum_{n=1}^{\infty}nx^{n-1}$
But now that we have doen this we must also recall that taking the derivative of a series can change its endpoint behavior...so therefore we know this series converges on $\displaystyle (-1,1)$ because that is the radisu of convergence of $\displaystyle \sum_{n=0}^{\infty}x^{n}$
But now we must check -1 and 1
To do this we imput the two values...if we imput 1 we get
$\displaystyle \sum_{n=1}^{\infty}n1^{n-1}=\sum_{n=1}^{\infty}n$
and using the n-th term test we see that $\displaystyle \lim_{n\to\infty}n=\infty$
so this new series diverges at x=1
Now we must check when x=-1
Maknig the series $\displaystyle \sum_{n=1}^{\infty}n(-1)^{n-1}$
But we dont even have to check the Alternating series test because as we showed $\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}n=\infty$
So the second prerequisite of the alternating series test has already failed so this series is divergent at x=-1
So we no know that the series is $\displaystyle \sum_{n=1}^{\infty}nx^{n-1}$ and is convergent on $\displaystyle (-1,1)$
Just for fun and to check $\displaystyle \frac{1}{(1-.5)^2}=4$
and $\displaystyle \sum_{n=1}^{\infty}n(.5)^{n-1}=4$
haha I know that was much more than was needed but I hoped it would be helpful in learning how to find these!
Mathstud