# Math

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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 -

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 -

Could u please show how u got -.9034 too? Thanks!!!

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