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Math Help - Proof/derivation of the trapezium rule error estimate

  1. #1
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    Proof/derivation of the trapezium rule error estimate

    Hi, I'm trying to derive the error estimate in the trapezium rule but can't find any derivations on the internet and don't understand my lecturers notes after one point.


    Using one strip for the trapezium rule, between a and (a+h). So the approximate integral is 0.5h[f(a) + f(a+h)]


    so the error on this approximation is:


    E(h) = (between a and a+h)int(f(x))dx - 0.5h[f(a) + f(a+h)]


    Differentiating both sides twice wrt h gives:


    E''(h) = -0.5h*f''(a+h) (this is in my notes and I understand it up to here, but it's the next part I don't understand)


    let


    M = max[abs(f''(x))] between a and (a+h)


    Then


    -0.5hM < E''(h) < 0.5hM (note I use < for less than or equal to, as I couldn't find the proper symbol)


    Where have the last two steps come from? After these two steps we integrate twice to get


    abs(E) < (1/12)M*h^3 , but I understand that part. It's just I don't get why -f''(a+h) < max[abs(f''(x))]


    Thanks!
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  2. #2
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    Re: Proof/derivation of the trapezium rule error estimate

    Quote Originally Posted by MattBurrows View Post
    It's just I don't get why -f''(a+h) < max[abs(f''(x))]
    For any x we have -|x| <= x <= |x|. So, -f''(a+h) <= |-f''(a+h)| = |f''(a+h)|. Further, |f''(a+h)| <= max[abs(f''(x))] between a and (a+h) because the left-hand side is just one element of the set over which the maximum is taken (this set is {|f''(x)| : a <= x <= a + h}). Therefore, -f''(a+h) <= max[abs(f''(x))]. Similarly, -|f''(a+h)| = -|-f''(a+h)| <= -f''(a+h) and -max[abs(f''(x))] <= -|f''(a+h)|, so -max[abs(f''(x))] <= -f''(a+h).
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  3. #3
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    Re: Proof/derivation of the trapezium rule error estimate

    Thank you so much! Took me a while to decipher (inequalities confuse me), but I fully understand it now Thanks!!
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