Is showing that a series of functions is differentiable as simple as it appears?

  • Thread starter jdinatale
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  • #1
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Here is the problem and my want. I think I might be overlooking something because it seems rather simple...

prime.png


primeprime.png
 

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  • #2
mathwonk
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"Hence"? how do you know you can differentiate this infinite sum of differentiable functions? what rule tells you that? do you know the weierstrass "M test"?
 
  • #3
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"Hence"? how do you know you can differentiate this infinite sum of differentiable functions? what rule tells you that? do you know the weierstrass "M test"?

Yes, I have access to the Weierstrass M-test. But that seems to deal with uniform convergence, and it doesn't quite seem to deal with differentiability or continuity. Unless I'm misunderstanding it.
 
  • #4
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Straight from a book:

"If, on differentiating a convergent infinite series [itex] \sum_{v = 0}^{\infty} G_v(x) = F(x) [/itex] term by term, we obtain a uniformly convergent series of continuous terms [itex] \sum_{v = 0}^{\infty} g_v(x) = f(x) [/itex], then the sum of this last series is equal to the derivative of the sum of the first series"
 

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