This is Taylor series of function

My question is:

What is form of ?

Thank you very much!

Results 1 to 15 of 15

- June 25th 2010, 02:54 PM #1

- June 25th 2010, 05:37 PM #2

- June 25th 2010, 05:49 PM #3

- June 25th 2010, 06:11 PM #4
Aha. So it is a nice little factorial in the denominator. Well, the recurrence relation idea sounds nice, but I'm not sure how it would work in principle. Why don't you show all the details of the computations for those first few terms? Maybe something'll pop out.

- June 25th 2010, 06:15 PM #5

You ask for computations ?

Here it is: Function calculator

Moreover... look here...

1 3 11 50 - OEIS Search Results

- June 25th 2010, 06:25 PM #6

- June 25th 2010, 06:39 PM #7

- June 25th 2010, 08:32 PM #8

- June 25th 2010, 08:41 PM #9

- June 25th 2010, 09:05 PM #10

- June 25th 2010, 09:39 PM #11

- Joined
- Jan 2009
- Posts
- 715

- June 26th 2010, 01:11 AM #12

- June 26th 2010, 09:56 AM #13

- June 26th 2010, 10:01 AM #14

- June 26th 2010, 10:42 AM #15

- Joined
- Oct 2009
- Posts
- 340

If exists, then does as well, and they are equal. Since the first expression exists, you can use it to find the radius of convergence, which will work out to be 1.

If you think to calculate from the terms you listed out then it becomes pretty easy to figure out the form of the , though this isn't exactly a rigorous solution.