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 February 1st, 2010, 01:19 PM #1 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Term by term differentiation Suppose $f_{n}(x) \geq 0$ and is nondecreasing $\forall x \in [a,b]$ n = 1,2,3,... and that if $S_{N}(x)= \sum^{N}_{n=1} f_{n}(x)$ then $S_{N}(x)$ converges to a finite limit $S(x)= \sum^{\infty}_{n=1} f_{n}(x)$. Prove that S'(x) exists a.e. and that term-by-term differentiation is valid a.e. in [a,b]. The a.e. existence of S'(x) follows from the monotonicity of S(x), but I don't know how to show term by term differentiation holds a.e. I know that uniform convergence would allow this, and I know term-by-term integration holds a.e. but I don't know how to show that I can exchange the limiting operation of differentiation with the limit of the sum. Any hints/suggestions would be greatly appreciated!
February 1st, 2010, 03:53 PM   #2
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Re: Term by term differentiation

Quote:
 Originally Posted by forcesofodin but I don't know how to show that I can exchange the limiting operation of differentiation with the limit of the sum. Any hints/suggestions would be greatly appreciated!
I don't completely understand what you're trying to say here... Could you write it more formally?

 February 2nd, 2010, 02:23 AM #3 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Term by term differentiation You're asking how to prove that the derivative of a finite sum is the sum of the derivative of the terms? Start with two terms. You may or may not need to prove that the (f + g)'(x) = f'(x) + g'(x), depending on what level your course is. Then proceed by induction.
February 2nd, 2010, 11:57 AM   #4
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Re: Term by term differentiation

Quote:
 Originally Posted by cmusick You're asking how to prove that the derivative of a finite sum is the sum of the derivative of the terms?
S(x) is the limit of the partial sums. I.e., it is an infinite series (of functions) he's trying to differentiate. He has some special conditions on the terms of the series. (which may be necessary?)

I really should know how to do this...

 August 14th, 2010, 06:50 PM #5 Newbie   Joined: Aug 2010 Posts: 2 Thanks: 0 Re: Term by term differentiation For the term by term differentiation, you need to show that the series $\sum_{n=1}^\infty{f_n^#39;}$ converges uniformly (on $[a,b]-D$, where $D=\bigcup_{n=1}^\infty{D_n}$ and $D_n$ is the set of dicontinuities of $f_n$). Try to use the wierstrass M test.

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