Rootfinding (the newton method, the bisection method)?

Question

Use the bisection method with a hand calculator or PC to find the indicated roots. Use an error tolerance, of, ε=0.0001

a) x^3 - x^2 - x - 1 = 0

b) x = e^(-x)

10 pts to best, please explain all answers

Answer 1

First pick a number for x where you would like to start, then go to the curve, find both the point and the slope. If you have the point and slope, you can find the equation of the tangent line. Find the zero (x-int) of this tangent line. Now take that zero and repeat the process.. find the point and slope on the curve, find tangent line, find the zero.. repeat. Each zero will get closer and closer to the real zero!

I will do (a) you can do (b)

f(x) = x^3 - x^2 - x - 1 so we must figure f ' (x) = 3x^2 - 2x -1

First pick your start...I like zero

x1 = 0

point(0,1) slope is -1

tangent line is y-1 = -1(x-0) ==> y = -x + 1

Find that zero, which is x=1 (this will be my new x to repeat

x2 = 1

point (1,-2) and slope is zero!!!! OOPS this is the one fault of Newton.. try a different number to start!!!!! OOPS!

x1 = 2

point(2,1) slope is f'(2)=7

y-1 = 7 (x-2) which is y = 7x - 13

the zero is 13/7

x2= 13/7 = 1.85714  Pt (1.85714 , .09913) and slope is 5.63265

y - .09913 = 5.63265(x-1.85714)

the zero is 1.874739

x3 = 1.83954 Pt( 1.83954 , .00139) and m = 5.4726

y - ..00139 = 5.4726(x - 1.83954)

the zero is 1.839286

x4 = 1.839286 Point (1.839286 , 0.00000409)  which is within our error tolerance.

Final answer using Newton's Method is x = 1.839286

=)

which is extremely close to the real zero . . .1.8392868...

Good luck with (b)!

I got some time:

I hope you get it!

(b) set equal to zero  e^-x - x = 0

f(x) = e^-x - x and so then f ' (x) = - e^-x - 1

It does not matter where you start, the further away, the faster you converge....

I like to start with x=1

x1 = 1

f(1) = -.6321 and f'(1) = -1.36788

y + .63212 = -1.36788 ( x - 1)

the zero is x = .53788

x2 = .53788

f(.53788) = .0460995 and f ' = -1.58398

y - .0460995 =  -1.58398 (x - .53788)

the zero is x = .566987

x3= .566987

f(.566987) = .0002512 and f' = -1.56723

y - .0002512 = -1.56723( x - .566987)

the zero is .567143286

f(.567143286) = .0000000069305 very very close!

The REAL ZERO is 0.56714329.... close! =)

GOOD LUCK! =)