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Induction and integral

  1. Apr 23, 2006 #1
    Hi, I'm having trouble with the following question.

    Q. Prove that

    [tex]
    \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{n - 1} } {f\left( {x_n } \right)} } } dx_n ...dx_2 dx_1 = \frac{1}{{\left( {n - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt
    [/tex]

    The hint says to use induction. So I tried showing that it's true for n = 1. For that particular case the statement is a single integral with terminals 0 (lower) and x (upper) I think. But that doesn't matter too much for my purposes. After assuming that the result is true for some integer k, I'm having trouble verifying the result for n = k + 1. I wrote down

    [tex]
    I = \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)dx_{k + 1} dx_k } } } } ...dx_2 dx_1
    [/tex]...(1)

    Since the result that I've assumed has f(x_k) as the integrand rather than
    f(x_(k+1)), I tried to use parts to get rid of the f(x_(k+1)). I did this by differenting f(x_(k+1)) and integrating "1". But doing this leads to integrals which cancel, leading me back to equation (1) above. Can someone help me out? Thanks.
     
  2. jcsd
  3. Apr 23, 2006 #2

    HallsofIvy

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    Yes, in case n= 1, that becomes
    [tex]\int_0^x f(x)dx= \int_0^x f(t)dt[/itex]
    which is true since it is the same integral with different "dummy" variables.
    So your k+1 integral is just the k integral with
    [tex]fx_(x_k)= \int_{o}^{x_k}f(x_{k+1})dx_{k+1}[/tex].
    How about integration by parts with
    [tex]u=f_(x_k)= \int_{o}^{x_k}f(x_{k+1})dx_{k+1}[/tex].
    and dv= dxk?
     
    Last edited: Apr 23, 2006
  4. Apr 23, 2006 #3
    I'm not sure what you mean by

    [tex]
    fx\left( {x_k } \right) = \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1}
    [/tex]

    Is f applied to x_k or x(x_k)?

    Your suggestion looks like what I tried before.

    [tex]
    \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} = \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } - \int\limits_0^{x_k } {x_{k + 1} f'\left( {x_{k + 1} } \right)dx_{k + 1} }
    [/tex]

    [tex]
    = \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } - \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } + \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1}
    [/tex]

    Which is the "n = k+1" integral I started with. Unless I missed something.
     
  5. Apr 23, 2006 #4

    HallsofIvy

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    Sorry, I had an extra "x" in that formula!

    No, my suggestion is not exactly what you did before. You tryed to integrate
    [tex]\int_0^{x_k}f(x_{k+1})dx_{k+1}[/tex]
    by parts. My suggestion was to integrate
    [tex]\int_0^{x_{k-1}}\int_0^{x_k}f(x_{k+1})dx_{k+1}dx_k[/tex]
    taking
    [tex]u= \int_0^{x_k}f(x_{k+1})[/tex]
    and du= dxk+1.

    (If you use integration by parts twice swapping what you use for u and dv in the middle, of course you just get back to what you started with- I'm saying use integration by parts once to get rid of that innermost integral.)

    Look at a simple example:
    [tex]\int_0^x\int_0^y f(t)dt dy[/tex]
    Let [itex]u= \int_0^y f(t)dt[/itex] and dv= dy. Then du= f(y) and v= y. We have
    [tex]uv\|_0^x- \int_0^xvdu= y\int_0^xf(t)dt- \int_0^x yf(y)dy[/tex]
    If we change the dummy variable y in the second integral to t, we have
    [tex]y\int_0^xf(t)dt- \int_0^x tf(t)dt= \int_0^x(y-t)f(t)dt[/tex]
    Do you see how that fits into your induction?
     
    Last edited: Apr 23, 2006
  6. Apr 23, 2006 #5
    Oh ok, I can see that what I did was different to what you suggested. I misread it before.

    I'm not aware of any other requirements. The question doesn't refer to other conditions which are required.

    I'll see what I can come up with, thanks.
     
  7. Apr 24, 2006 #6
    I've thought about this and I still can't get it out.

    Firstly, if [itex]u = \int\limits_0^y {f\left( t \right)dt} ,dv = dy[/itex] then [itex]du = f\left( y \right)dy,v = y[/itex] by FTC. Ok I get that part. But what about uv evaluated at 0 and x?

    In the example you gave, the outer most integral was done wrty so I would've thought that would mean in evaluating uv at 0 and x, where the variable 'y' appear in uv, it would be replaced by x. That is,

    [tex]
    \left[ {uv} \right]_0^x = \left[ {y\int\limits_0^y {f\left( t \right)dt} } \right]_0^x = x\int\limits_0^x {f\left( t \right)dt}
    [/tex]

    In any case, considering the following integral "I" and integrating by parts as you suggested, with

    [tex]u = \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} ,dv = dx_k [/tex] I obtain:

    [tex]
    I = \int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} } dx_{k + 1} dx_k
    [/tex]

    [tex]
    = \left[ {x_k \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} } \right]_0^{x_{k - 1} } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)} dx_k
    [/tex]

    [tex]
    = x_{k - 1} \int\limits_0^{x_{k - 1} } {f\left( {x_{k + 1} } \right)dx_{k + 1} } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)dx_k }
    [/tex]

    [tex]
    = x_{k - 1} \int\limits_0^{x_{k - 1} } {f\left( {x_k } \right)dx_k } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)dx_k }
    [/tex]

    [tex]
    = \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - x_k } \right)f\left( {x_k } \right)dx_k }
    [/tex]

    I'm not sure how I can use the above and the assumption (that the statement is true for n = k) to show that the n = k + 1 case is true.
     
    Last edited: Apr 24, 2006
  8. Apr 24, 2006 #7

    HallsofIvy

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    Remember that you can change "dummy" variables at will. You need to change that xk+1 to "t".
     
  9. Apr 25, 2006 #8
    I can rewrite the last line as:

    [tex]
    \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - x_k } \right)} f\left( {x_k } \right)dx_k = \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - t} \right)} f\left( t \right)dt
    [/tex]

    But I don't see how I can use the the assumption to prove the result from the above.
     
  10. Apr 25, 2006 #9

    HallsofIvy

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    All those xk are confusing the devil out of me! That's why earlier I gave an example with x and y. I showed earlier that
    [tex]\int_0^x\int_0^y f(t)dt dy= \int_0^x(y-t)f(t)dt[/tex]
    That can be put in terms of xk and xk+1 by simply make x= xk-1 and y= xk:
    [tex]\int_0^{x_{k-1}}\int_0^{x_k}f(t)dt dy= \int_0^{x_{k-1}(x_k-t)f(t)dt[/tex]

    Now apply your induction hypothesis, that
    [tex]\int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {f\left( {x_n } \right)} } } dx_k ...dx_2 dx_1 = \frac{1}{{\left( {k - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt[/tex]
    with f(x) replaced by (xk-x)f(x).
     
  11. Apr 25, 2006 #10
    Ok I'll try that thanks.
     
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