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

Newton said

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

0 -1

-1 -.9034

-.9034 -.9012

-.9012 -.901201

x = -.901201 correct to 4 decimals after 3 iterations, not bad

check:

LS = e^x - x^2/2

= .406081662 - .406081621

= .000000041

fantastic result!!!

- Math -
**Michael**, Friday, November 16, 2012 at 1:29am
Could u please show how u got -.9034 too? Thanks!!!

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