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Infinite series by integration by parts

  1. Aug 4, 2009 #1

    disregardthat

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    Hi, I wonder if this hypothesis is true:

    Let [tex]f_n[/tex] be an arbitrarily chosen n'th anti-derivative of the function [tex]f_0[/tex]. Similarly, let [tex]g_n[/tex] be the n'th derivative of the function [tex]g_0[/tex].

    Now, [tex]\int^b_a f_0 g_0 \rm{d}x=[f_1g_0]^b_a-\int^b_a f_1g_1 \rm{d}x=[f_1g_0-f_2g_1+...]^b_a+(-1)^n \int^b_a f_{n+1}g_n \rm{d}x[/tex].

    hypothesis:

    If [tex]\lim_{n \to \infty} f_{n+1}g_n =0[/tex] for all continous intervals of and never diverges. Then

    [tex]\int^b_a f_0 g_0 \rm{d}x = [\sum^{\infty}_{n=0} (-1)^n f_{n+1}g_n]^b_a[/tex]

    This seems intuitively correct, but I wonder how to prove it.
     
    Last edited: Aug 4, 2009
  2. jcsd
  3. Aug 4, 2009 #2
    I'm thinking Mean Value Theorem. The integral equals [tex]\epsilon(b-a)(f_{n+1}g_n)|_{x=c}[/tex] for some c in the interval [a, b] and [tex]0 \leq \epsilon \leq 1[/tex] which approaches zero.
     
  4. Aug 4, 2009 #3

    disregardthat

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    this was double post, sorry
     
    Last edited: Aug 4, 2009
  5. Aug 4, 2009 #4

    disregardthat

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    Excellent, thank you.

    We have by induction that

    [tex]\int^b_a f_0g_0 \rm{d}x = [\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(-1)^n\int^b_af_{n+1}g_n \rm{d}x=[\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(b-a)(f_{n+1} \circ g_n)(t)[/tex]

    For some [tex]t \in [a,b][/tex], and any non-negative integer n.

    However, [tex]\lim_{n \to \infty} (f_{n+1} \circ g_{n} )(t)=0[/tex] is given, so

    [tex]\int^b_a f_0g_0 \rm{d}x = \lim_{n \to \infty} \int^b_a f_0g_0 \rm{d}x= \lim_{n \to \infty} [\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(-1)^n\int^b_af_{n+1}g_n \rm{d}x=\lim_{n \to \infty}[\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(b-a)(f_{n+1} \circ g_n)(t)=[\sum^{\infty}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a[/tex]

    But this is not so obvious if [tex]\lim_{n \to \infty} (f_{n+1} \circ g_{n} )(t)[/tex] not always equal 0 i.e. is finite for discrete values of x, or if either of the limits are infinite. Can someone help me there?

    Perhaps if the limits are infinite, we can let n tend towards infinity at a rate which make [tex]f_{n+1}g_n[/tex] dominate a limit, say b i.e. so [tex]\lim_{b,n \to \infty} b \cdot f_{n+1}g_n = 0[/tex] Can we choose it to be like that?
     
    Last edited: Aug 4, 2009
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