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March 1, 2015

March 1, 2015

Posted by **Adam** on Saturday, May 29, 2010 at 8:36pm.

Let x1=–1 be the initial approximation.

The second approximation x2 is _____?

and the third approximation x3 is _____?

- CALC -
**Damon**, Saturday, May 29, 2010 at 9:10pmy = x^3 + x + 3

Xn+1 = Xn - Yn/y'

where here

y' = 3Xn^2 +1

X1 = -1

Y1 = -1-1+3 = 1

y'= 3+1 = 4

X2 = X1 - Y1/(3X1^2+1)

X2 = -1 - 1/4 = -1.25

now do X3

- CALC -
**Adam**, Sunday, May 30, 2010 at 1:27amim not very clear on the steps of newton's method because my professor didnt explain it very well. Do you just take the second derivative of y and then plug in -1.25 to find x3. I dont quite understand the concept...thanks for the help in advance.

- CALC -
**gurbz**, Thursday, June 3, 2010 at 5:41amYou're right Adam, just plug -1.25 to find X3.

So...you know...

y=x^3+x+3

y'=3x^2+1

Xn=-1

The equation you use is:

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

So for:

X2=(-1)- ( ((-1)^3+(-1)+3)/ (3(-1)^2+1))

X2= (-1)- (1/4)

X2= -1.25

Then...

X3=(-1.25)- ( ((-1.25)^3+(-1.25)+3)/ (3(-1.25)^2+1))

X3=(-1.25)-(-1/28)

X3=-1.214285714

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