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Recursive function, diff under the integral sign

  1. May 9, 2010 #1
    Hey

    i stumbled over a problem which i can't solve:

    f is a continous function on [a,b] and F_n is recursive given through:

    [tex]F_1(x)=\int_a^x f(t) \ dt[/tex]
    and
    [tex]F_{n+1}=\int_a^x F_n(t) \ dt[/tex]

    I have to proof that F_n looks like this:

    [tex]F_n(x)=\frac{1}{(n-1)!}\int_a^x (x-t)^{n-1} f(t) \ dt[/tex]

    For n=1 it's no big deal but then i have no clue. I think it must have something to do with differentiation under the integral sign because thats where i found it.

    Hope anybody can help me! thx
     
  2. jcsd
  3. May 9, 2010 #2

    D H

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    Hint: Use induction. You have the base case since "For n=1 it's no big deal."
     
  4. May 9, 2010 #3
    ;)

    actually i used induction but for n+1 im stuck:

    n=1:
    [tex]\frac{1}{0!} \int_a^x (x-t)^{1-1} f(t) \ dt = \int_a^x f(t) \ dt[/tex]

    n=n+1
    [tex]\frac{1}{(n+1-1)!} \int_a^x (x-t)^{n+1-1} f(t) \ dt = \frac{1}{n!}\int_a^x (x-t)^{n} f(t) \ dt[/tex]
     
  5. May 9, 2010 #4

    D H

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    Since you found this under differentiation under the sign, don't you think you should be doing that?
     
  6. May 9, 2010 #5
    sure which i did:

    [tex]\frac{dF_{n+1}}{dx}=\cdots =\frac{1}{(n-1)!}\int_a^x (x-t)^(n-1) f(t) \ dt[/tex]

    but if i look at how F_n+1 is defined i should get:

    [tex]F_n+1=\int_a^x F_n(t) \ dt= \int_a^x \frac{1}{(n-1)!}*\int_a^t (t-t)^{n-1} f(t) \ dt = \int_a^x \frac{1}{(n-1)!}[/tex]
     
  7. May 9, 2010 #6

    D H

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    First, what is [itex]F_{n+1}(x)[/itex] given

    [tex]F_n(x)=\frac{1}{(n-1)!}\int_a^x (x-t)^{n-1} f(t) \ dt[/tex]

    (Just rewrite with [itex]n\to n+1[/itex]).

    Now try both differentiations again. In particular, what is [tex]\frac{dF_{n+1}(x)}{dx}[/tex] given
    1. The formulation of [itex]F_{n+1}(x)[/itex] from the above, and

    2. [tex]F_{n+1}=\int_a^x F_n(t)\, dt[/tex]
     
  8. May 9, 2010 #7
    i'm sorry i have absolutly no idea, :-(

    i thought I did the n->n+1 and the differentiation correct.
     
  9. May 9, 2010 #8

    D H

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    Try doing what I asked you to do in post #6, and be explicit in your steps. You skipped over a lot of steps, so it is hard to say where you went wrong.
     
  10. May 9, 2010 #9
    So heres what i did:

    [tex]F_n(x)=\\frac{1}{(n-1)!}\\int_a^x (x-t)^{n-1} f(t) \\ dt [/tex]

    so n+1:
    [tex]\frac{1}{(n+1-1)!} \\int_a^x (x-t)^{n+1-1} f(t) \\ dt = \\frac{1}{n!}\\int_a^x (x-t)^{n} f(t) \\ dt [/tex]

    then [tex]\frac{dF_{n+1}(x)}{dx}[/tex] :

    [tex]\frac{1}{n!} \int_a^x \frac{d}{dx} f(t) (x-t)^n dt - f(a)(x-a)^n \frac{d}{dx}(a)+0^n f(x) \frac{d}{dx}(x) = \frac{1}{n!} \int_a^x n*(x-t)^{n-1} f(t) dt = \frac{1}{(n-1)!} \int_a^x (x-t)^(n-1)f(t) dt[/tex]
     
  11. May 9, 2010 #10

    D H

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    That needs some help.

    Ahh! You are using a lot of double backslashes.

    What are you trying to do here, and how does the second line follow from the first?

    Please try following my suggestions in post #6.
     
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