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

Your help is very appreciated!

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- April 28th 2008, 07:31 AMhasanbalkanTaylor series
Find the Taylor series of the given function at a = 0

Your help is very appreciated! - April 28th 2008, 07:44 AMhasanbalkan
I got to the point where I found that

But I am not sure how to proceed from here... - April 28th 2008, 07:49 AMJameson
It's been a while since I've done these, but let's give it a try.

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

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,

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...

Can you take it from here? - April 28th 2008, 07:53 AMJameson
- April 28th 2008, 07:54 AMcolby2152
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 - April 28th 2008, 08:06 AMhasanbalkan
Thank you for the quick response! Now I know what I need to do.

- April 28th 2008, 11:49 AMMathstud28
Use the fact that the maclaurin series for

Now notice that

Now that we have shown this relationship between and we can make the logical next step

Since the derivative of is

and we know that

THen it is logical to assume that the power series for is the derivative of

Which is

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 because that is the radisu of convergence of

But now we must check -1 and 1

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

and using the n-th term test we see that

so this new series diverges at x=1

Now we must check when x=-1

Maknig the series

But we dont even have to check the Alternating series test because as we showed

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 and is convergent on

Just for fun and to check

and

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

Mathstud