# Taylor series

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• Apr 28th 2008, 07:31 AM
hasanbalkan
Taylor series
Find the Taylor series of the given function at a = 0

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

Your help is very appreciated!
• Apr 28th 2008, 07:44 AM
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...
• Apr 28th 2008, 07:49 AM
Jameson
Quote:

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

$
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.

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

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?
• Apr 28th 2008, 07:53 AM
Jameson
Quote:

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.
• Apr 28th 2008, 07:54 AM
colby2152
Quote:

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

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

Your help is very appreciated!

Do you how to find a Taylor series?

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

http://upload.wikimedia.org/math/6/c...8f7fec25ea.png
• Apr 28th 2008, 08:06 AM
hasanbalkan
Thank you for the quick response! Now I know what I need to do.
• Apr 28th 2008, 11:49 AM
Mathstud28
Quote:

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

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

Your help is very appreciated!

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