Let

We have (1).

Multiplying (1) with we have

(2)

Substracting (2) from (1) yields

Then

Results 1 to 5 of 5

- July 9th 2007, 01:34 PM #1

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- July 9th 2007, 01:55 PM #2

- July 10th 2007, 04:12 PM #3

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Since you answered so quickly, here is a second challenge. I am not sure if is a fair question to ask but I post it anyway.

---

Having shown that

Can you prove the following?

Let with radius of convergence

Then, is differenciable on and furthermore .

Note: This works even if is a complex number, but the proofs are completely anagolous.

- July 11th 2007, 10:36 AM #4

- July 11th 2007, 10:46 AM #5

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

I am not sure what you mean. In my Analysis class we proved differenciation term-by-term by working backwards with the Fundamental Theorem of Calculus after proving integration term-by-term. And that was just Real fucntions.

By my Complex Analysis book had a really nice approach to this proof. Without using more advandeced techiqnues. It was able to prove term-by-term differenciation by using the series posted above. It was a long but straightforward derivation. That is what I am asking to show.