Oh dear me, I would have to get my numerical analysis book out of the attic for this error analysis.
I guess I would suggest trying it with a few values of n and seeing how quickly it converges, unless you have a text with the error bound rules handy.
how big must n be so that the absolute value of the error in using the trapezoid rule is less than .00001?
for the integral (square root of (1+x^4))
where a=0 and b=1
3 answers
OK, I found it online
|E| /= (1/[12 n^2])(b-a)^3 df/dx^3
use the maximum value of the second derivative of the function on the interval a to b, here 0 to 1
|E| /= (1/[12 n^2])(b-a)^3 df/dx^3
use the maximum value of the second derivative of the function on the interval a to b, here 0 to 1
typo repair
|E| /= (1/[12 n^2])(b-a)^3 d^2f/dx^2
f = (1+x^4)^.5
df/dx = .5 (1+x^4)^-.5 (4 x^3)
= 2 x^3(1+x^4)^-.5
so
d^2/dx^2 = 2 x^3 [ -.5)(4x^3)(1+x^4)^-1.5] + (1+x^4)^-.5 [ 6 x^2]
find the maximum of that between x = 0 and x = 1
and then your |
n^2 = (1/|12 E|)1^3 times that maximum
|E| /= (1/[12 n^2])(b-a)^3 d^2f/dx^2
f = (1+x^4)^.5
df/dx = .5 (1+x^4)^-.5 (4 x^3)
= 2 x^3(1+x^4)^-.5
so
d^2/dx^2 = 2 x^3 [ -.5)(4x^3)(1+x^4)^-1.5] + (1+x^4)^-.5 [ 6 x^2]
find the maximum of that between x = 0 and x = 1
and then your |
n^2 = (1/|12 E|)1^3 times that maximum
0 answers