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Math Help - |r(x)| < Ce^-x

  1. #1
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    |r(x)| < Ce^-x

    Given: f(x) = \int_x^{x+1} \sin e^t dt. Show: e^x f(x) = \cos (e^x) - e^{-1} \cos (e^{x+1}) + r(x), where |r(x)| < Ce^{-x}, C \in \mathbb{R}.


    I was fairly easily able to prove that $e^x f(x) = \cos (e^x) - e^{-1} \cos (e^{x+1}) - e^x \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} du,$ so I won't go through all those steps.

    Where I'm having trouble is the last part, namely, showing that | $e^x \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} du| < Ce^{-x}$. How do I do this? I tried | $\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} du| < |\int_{e^x}^{e^{x+1}} \frac{du}{u^2} |$, but that did not produce the desired result.
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  2. #2
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    Re: |r(x)| < Ce^-x

    Hey phys251.

    There is a norm result that states that |<A,B>| <= |A|*|B| and its called the Cauchy-Schwarz inequality. If you can show that the integral is a valid inner product and that the integral in the LHS term is bounded by a constant, then you have proved the result.
    Thanks from phys251 and mash
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    Re: |r(x)| < Ce^-x

    Quote Originally Posted by chiro View Post
    Hey phys251.

    There is a norm result that states that |<A,B>| <= |A|*|B| and its called the Cauchy-Schwarz inequality. If you can show that the integral is a valid inner product and that the integral in the LHS term is bounded by a constant, then you have proved the result.
    I don't get it. Can you walk me through this procedure?
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    Re: |r(x)| < Ce^-x

    Basically if the integral follows an inner product then you have the Cauchy-Schwartz Inequality.

    Cauchy?Schwarz inequality - Wikipedia, the free encyclopedia

    Now if an integral has that over a space of functions then <f,g> = Integral [Some Domain] f(x)g(x)dx.

    In your example, I suggested to find a suitable f(x) and g(x) for <f,g> |where Integral f(x)dx| <= |C| and |Integral g(x)(dx)| <= |D| for some constants/variables C,D.

    Once you can write the LHS in terms <f,g> where <f,g> = Integral [Fixed Domain] f(x)g(x) dx (or relate the integral to the bound of this integral) then you are done.

    This idea of norms and inner products used on functions, matrices, and many other abstract objects (not just your normal vectors) is a common thing in advanced mathematics.
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    Re: |r(x)| < Ce^-x

    Hi,
    Presumably, you used integration by parts to arrive at your first formula. (BTW, the limits on your integral should be ln(x) and ln(x+1).) For the inequality, just use integration by parts again. If need be, you can look at the attached detailed solution.

    |r(x)| &lt; Ce^-x-mhfcalc42.png

    Chiro, as far as I can, see your suggested use of the Cauchy Schwartz inequality for this problem leads nowhere. I'd be quite interested in your detailed solution using said inequality.
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  6. #6
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    Re: |r(x)| < Ce^-x

    The variable for integration is u with respect to du. This means we need a function g(u) such that <f,g> = Integral (e^(x),e^(x+1)) [cos(u)/u^2]du * e^(x) = Integral (e^(x),e^(x+1)) [cos(u)/u^2] * g(u) du.

    For our g(u) we will have to evaluate both integrals and solve for g(u). Once we do this we evaluate Integral [e^(x),e^(x+1)] |cos(u)/u^2)|^2 du and Integral [e^x,e^(x+1)] [g(u)^2]du and take the square root of the multiple of these two integrals or compare the product with the inner product all squared.

    At the moment I'm a little busy (and I'm sorry for not clarifying this further), but I would get g(u) as a function by isolating it (probably have to integrate by parts and get an expression in terms of the integral of cos(u)/u^2 du much in the same way you did) and once that is done, use the Cauchy-Schwartz inequality for g(u).

    Again I'm sorry I haven't been able to answer things more specifically, but unfortunately my time is not as free as it once was.
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