Let

We have (1).

Multiplying (1) with we have

(2)

Substracting (2) from (1) yields

Then

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- Jul 9th 2007, 12:34 PM #1

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- Jul 9th 2007, 12:55 PM #2

- Jul 10th 2007, 03:12 PM #3

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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.

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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.

- Jul 11th 2007, 09:36 AM #4

- Jul 11th 2007, 09:46 AM #5

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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.