Differentiability of a Series of Functions


by luke8ball
Tags: differentiability, functions, series
luke8ball
luke8ball is offline
#1
Nov18-12, 11:51 AM
P: 22
I'm working on a problem where I need to show that the series of functions, f(x) = Ʃ (xn)/n2, where n≥1, converges to some f(x), and that f(x) is continuous, differentiable, and integrable on [-1,1].

I know how to show that f(x) is continuous, since each fn(x) is continuous, and I fn(x) converges uniformly. Because each fn(x) is also integrable, I can also show f(x) is integrable.

The trouble I'm having is proving that f(x) is differentiable. I need to show that the series of derivatives converges uniformly. However, I don't think I can use the Weierstrass M-Test in this scenario. Any ideas?
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Erland
Erland is offline
#2
Nov20-12, 07:56 AM
P: 302
Quote Quote by luke8ball View Post
I'm working on a problem where I need to show that the series of functions, f(x) = Ʃ (xn)/n2, where n≥1, converges to some f(x), and that f(x) is continuous, differentiable, and integrable on [-1,1].

I know how to show that f(x) is continuous, since each fn(x) is continuous, and I fn(x) converges uniformly. Because each fn(x) is also integrable, I can also show f(x) is integrable.

The trouble I'm having is proving that f(x) is differentiable. I need to show that the series of derivatives converges uniformly. However, I don't think I can use the Weierstrass M-Test in this scenario. Any ideas?
The differentiated series converges on the open interval (-1,1). Hence the original series is differentiable there and the derivative is the differentiated series. The differentated series diverges at 1 and converges conditionally at -1. Although I cannot prove it, this makes me suspect that the original series is not differentiable at 1 and -1.


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