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^xx^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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