1. ## Taylor series

Find the Taylor series of the given function at a = 0

$
f(x)=\frac{1}{(1-x)^2}
$

2. I got to the point where I found that

$
f^{(n)}(a) = (n + 1)!
$

But I am not sure how to proceed from here...

3. Originally Posted by hasanbalkan
Find the Taylor series of the given function at a = 0

$
f(x)=\frac{1}{(1-x)^2}
$

It's been a while since I've done these, but let's give it a try.

$f(x)=\frac{1}{(1-x)^2}$

So now we need to start taking some derivatives, and see what kind of pattern shows up.

$f'(x) = \frac{2}{(1-x)^3}$

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, $f^{n}(x) = \frac{(-1)^{n+1}(n+1)!}{(1-x)^{n+2}}$

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...

$f(x) = \sum_{n=0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^n$

Can you take it from here?

4. Originally Posted by hasanbalkan
I got to the point where I found that

$
f^{(n)}(a) = (n + 1)!
$

But I am not sure how to proceed from here...
I think the sign will alternate for the derivatives. When you plug this in to Taylor's formula, the (n+1)! and n! can reduce nicely.

Once you do that, write out the first few terms and see what pattern you can see.

5. Originally Posted by hasanbalkan
Find the Taylor series of the given function at a = 0

$
f(x)=\frac{1}{(1-x)^2}
$

Do you how to find a Taylor series?

You need to find derivatives of the function for the following formula.

6. Thank you for the quick response! Now I know what I need to do.

7. Originally Posted by hasanbalkan
Find the Taylor series of the given function at a = 0

$
f(x)=\frac{1}{(1-x)^2}
$

Use the fact that the maclaurin series for $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

Now notice that $\frac{D[\frac{1}{1-x}]}{dx}=\frac{1}{(1-x)^2}$

Now that we have shown this relationship between $\frac{1}{1-x}$ and $\frac{1}{(1-x)^2}$ we can make the logical next step

Since the derivative of $\frac{1}{1-x}$ is $\frac{1}{(1-x)^2}$

and we know that $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

THen it is logical to assume that the power series for $\frac{1}{(1-x)^2}$ is the derivative of $\sum_{n=0}^{\infty}x^n$

Which is $\sum_{n=1}^{\infty}nx^{n-1}$

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 $(-1,1)$ because that is the radisu of convergence of $\sum_{n=0}^{\infty}x^{n}$

But now we must check -1 and 1

To do this we imput the two values...if we imput 1 we get

$\sum_{n=1}^{\infty}n1^{n-1}=\sum_{n=1}^{\infty}n$

and using the n-th term test we see that $\lim_{n\to\infty}n=\infty$
so this new series diverges at x=1

Now we must check when x=-1
Maknig the series $\sum_{n=1}^{\infty}n(-1)^{n-1}$

But we dont even have to check the Alternating series test because as we showed $\lim_{n\to\infty}a_n=\lim_{n\to\infty}n=\infty$
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 $\sum_{n=1}^{\infty}nx^{n-1}$ and is convergent on $(-1,1)$

Just for fun and to check $\frac{1}{(1-.5)^2}=4$

and $\sum_{n=1}^{\infty}n(.5)^{n-1}=4$

haha I know that was much more than was needed but I hoped it would be helpful in learning how to find these!

Mathstud