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Math Help - Give the quadratic approx. to sqrt(25 + h)...

  1. #1
    Senior Member sfspitfire23's Avatar
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    Give the quadratic approx. to sqrt(25 + h)...

    Hi guys, am wondering if my reasoning is correct -

    Give the quadratic approx. to sqrt(25 + h) and discuss how this approx compares to the first order approximation-

    I start the problem by Taylor expanding f(x) = sqrt(x) around the initial starting point x_0 = 25. I then get

    First order: f(x_0 + (x - x_0)) = f(x_0) + f ' (x_0) * (x - x_0)

    This is the same as f(x_0 + h) = f(x_0) + f ' (x_0) * h

    Second order: f(x_0 + h) = f(x_0) + f ' (x_0) * h + (f '' (x_0) / 2 ) * h^2

    The second order approximation is much closer to the actual value x as the error is o(h^2) which implies that the error goes to 0 faster than o(h) as h goes to 0




    Thanks
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  2. #2
    Super Member
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    Re: Give the quadratic approx. to sqrt(25 + h)...

    I think if you look at Taylor's Theorem, you'll find that the error in the linear approximation is O(h^2) and the error in the quadratic approximation is O(h^3). Otherwise it sounds like you have the right idea.

    You'll have to give the actual Taylor expansion (5 + (number)h + (number)h^2) - that's what "Give the quadratic approx." means. You probably know that, but it wasn't clear from your post.

    - Hollywood
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