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March 6, 2015

March 6, 2015

Posted by **sarah** on Sunday, February 3, 2008 at 7:05pm.

for the integral (square root of (1+x^4))

where a=0 and b=1

- calc -
**Damon**, Sunday, February 3, 2008 at 9:28pmOh 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.

- calc -
**Damon**, Sunday, February 3, 2008 at 9:43pmOK, 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

- calc -
**Damon**, Sunday, February 3, 2008 at 9:52pmtypo 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

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