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Math Help - Integrals problem

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Integrals problem

    Let f be a continuous function on open interval I.

    Let h>0 and a<b so that: (a-h,b+h) c I.

    Now we choose d: 0 < d< h

    and now for every x in(a,b) we define the following function:

    G(x) = 1/2d * {int(-d --> d) f(x+t)dt}

    Prove:
    1. G is differentiable functions whose derivative is continuous.
    2. lim(d-->0+) G(x) = f(x)





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  2. #2
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let f be a continuous function on open interval I.

    Let h>0 and a<b so that: (a-h,b+h) c I.

    Now we choose d: 0 < d< h

    and now for every x in(a,b) we define the following function:

    G(x) = 1/2d * {int(-d --> d) f(x+t)dt}

    Prove:
    1. G is differentiable functions whose derivative is continuous.
    2. lim(d-->0+) G(x) = f(x)
    Notice that \frac1\delta\int_d^{d+\delta}\kern-12pt f(x+t)\,dt -f(x+d) = \frac1\delta\int_d^{d+\delta}\kern-12pt \bigl(f(x+t)-f(x+d)\bigr)\,dt \to0 as \delta\searrow0. This is because of the continuity of f, which means that we can take |f(x+t)-f(x+d)|<\varepsilon for all t in the interval [d,d+\delta]. So \int_d^{d+\delta}\kern-12pt \bigl|f(x+t)-f(x+d)\bigr|\,dt < \delta\varepsilon.

    For 1, \frac{G(x+\delta)-G(x)}\delta = \frac1{2d\delta}\biggl(\int_d^{d+\delta}\kern-12pt f(x+t)\,dt - \int_{-d-\delta}^{-d}\kern-12pt f(x+t)\,dt\biggr) \to \frac1{2d}\bigl(f(x+d)-f(x-d)\bigr) as \delta\searrow0.
    For 2,  G(x) - f(x) = \frac1{2d}\int_{-d}^d\!\!\!\bigl(f(x+t)-f(x)\bigr)\,dt, which you can estimate in the same way.
    Last edited by Opalg; April 30th 2010 at 03:04 AM. Reason: corrected several errors
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