Find the Taylor series of the given function at a = 0
Your help is very appreciated!
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?
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