Thread: Problem 30
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Old July 10th, 2007, 05:12 PM
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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 \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}

Can you prove the following?
Let f(x) = \sum_{n=0}^{\infty} a_nx^n with radius of convergence R>0

Then, f is differenciable on (-R,R) and furthermore f'(x) = \sum_{n=1}^{\infty} na_nx^{n-1}.

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