# Proof/derivation of the trapezium rule error estimate

• Nov 16th 2012, 07:17 AM
MattBurrows
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!
• Nov 16th 2012, 08:06 AM
emakarov
Re: Proof/derivation of the trapezium rule error estimate
Quote:

Originally Posted by MattBurrows
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).
• Nov 16th 2012, 08:54 AM
MattBurrows
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!!