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March 30, 2017

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Use Newton's Method to approximate the root of the function f(x)=e^x-5x when the intial point chosene is x1=1.5. Do two iterations.

This is what I did, but not working.
f'(x)=e^x-5
x2=1.5 - (e^1.5-(5(1.5))/e^1.5 -(5))
=1.5-(4.4817-7.5/4.4817-5)=1.5-5.8234= -4.3234

x3=-4.3234-(e^-4.3234-(5(-4.3234))/ e^-4.3234-(5)) = -4.3234-(21.630/-4.987)= -4.3234-(-4.337)=.0136

I did another one for x4, but the number was no where near either of these two. Aren't they suppose to be somewhat close? Did I make a dumb algebra mistake or am I not doing this right at all?

  • Calculus - ,

    On the contrary, your equations and calculations are good.

    For X1=1.5
    I get
    x2=-4.32
    x3=0.01415
    x4=0.250868
    Your x4 is quite close to the exact answer of 0.25917110181907.
    In fact, by x7, you will get the answer accurate to 14 decimal places.

    This exercise is probably to illustrate that the Newton's method converges "no matter what", at least in this case where the function has the same concavity within the range.

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