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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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