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March 30, 2017

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You are using Newton's method to solve

e^x - 2. If you first guess is x1=1, what value will you calculate for the next approximation x2?

...I don't know what Newton's method is, so I looked it up...and I'm still confused. Help plz?

Thx

  • Calculus - ,

    first of all, to "solve" we need an equation.
    You have the expression e^x - 2
    Are you solving e^x - 2 = 0 ?

    I will assume you are.

    let f(x) = e^x - 2
    f ' (x) = e^x

    Newton's method says ...

    xn+1 = xn - f(xn)/f ' (xn)

    so x2 = 1 - (e^1 - 2)/e^1
    = .73576

    x3 = .73576 - (e^.73576 - 2)/e^.73576 = .69404

    x4 = .69404 - (e^.69404 - 2)/e^.69404 = .69314758

    notice we are rapidly approaching some kind of fixed value.
    If the value of x of our input returns as output, we have the solution.
    A most remarkable procedure.

    Using logs I get x = .69314718 , so our method was accurate to 6 digits after only 3 iterations.

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