Differentiability of a Series of Functions

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The series of functions f(x) = Ʃ (xn)/n2 converges uniformly on the interval [-1,1], establishing its continuity and integrability. The challenge lies in proving the differentiability of f(x), which requires showing that the series of derivatives converges uniformly. The differentiated series is confirmed to converge on the open interval (-1,1), indicating that f(x) is differentiable there. However, it diverges at x=1 and converges conditionally at x=-1, suggesting that f(x) may not be differentiable at these endpoints. This analysis highlights the complexities of differentiability in the context of series of functions.
luke8ball
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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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luke8ball said:
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.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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