Posted by Michael on Thursday, November 15, 2012 at 9:44pm.
Use Newton's method to find a value of x that satisfies: e^x- (x^2/2) = 0
Use the starting value x0=0. Display answer in 4 significant figures
- Math - Reiny, Thursday, November 15, 2012 at 10:07pm
e^x - x^2/2=0
2e^x - x^2 = 0
let y = 2e^x - x^2
y' = 2e^x - 2x
newx = x - y/y'
= x -(2e^x-x^2)/(2e^x - 2x)
which simplified to
(2xe^x - x^2 - 2e^x)/(2e^x - 2x)
make a chart with
x ---- newx , then replacing x with newx in the next line
x = -.901201 correct to 4 decimals after 3 iterations, not bad
LS = e^x - x^2/2
= .406081662 - .406081621
- Math - Michael, Friday, November 16, 2012 at 1:29am
Could u please show how u got -.9034 too? Thanks!!!
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