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March 30, 2015

March 30, 2015

Posted by **Janet** on Sunday, April 17, 2011 at 3:19pm.

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 -
**MathMate**, Sunday, April 17, 2011 at 3:49pmOn 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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