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-- Oct 18 In-Class Exercise

X1=4.4

X2= 4.4 - (11.2896/25.536) = 3.957

X3=3.63

X4=3.4

X1=4.4 X2= 4.4 - (11.2896/25.536) = 3.957 X3=3.63 X4=3.4

-- Oct 18 In-Class Exercise

derivative of f, fp(x) = 2(x-3)(x-2)(2x-5)

x0 = 5 f(x0) = 36 and fp(x0) = 60

x1 = x0 - f(x0)/fp(x0) = 4.4

f(x1) = 11.289 and fp(x1) = 25.536

x2 = x1 - f(x1)/fp(x1) = 3.957

f(x2) = 3.507 and fp(x2) = 10.914

x3 = x2 - f(x2)/fp(x2) = 3.636

x4 = 3.407

- Yeshwanth Bashyam

(Edited: 2017-10-18)

derivative of f, fp(x) = 2(x-3)(x-2)(2x-5) x0 = 5 f(x0) = 36 and fp(x0) = 60 x1 = x0 - f(x0)/fp(x0) = '''4.4''' f(x1) = 11.289 and fp(x1) = 25.536 x2 = x1 - f(x1)/fp(x1) = '''3.957''' f(x2) = 3.507 and fp(x2) = 10.914 x3 = x2 - f(x2)/fp(x2) = '''3.636''' '''x4 = 3.407''' - Yeshwanth Bashyam

-- Oct 18 In-Class Exercise

`f(x) = (x-2)^2(x-3)^2`

`f'(x) = 4x^3-30x^2+74x-60`

`x_0 = 5`

`x_1 = 4.4`

`x_2 = 3.9`

`x_3 = 3.6`

`x_4 = 3.4`

(Edited: 2017-10-18)

@BT@f(x) = (x-2)^2(x-3)^2@BT@ @BT@f'(x) = 4x^3-30x^2+74x-60@BT@ @BT@x_0 = 5@BT@ @BT@x_1 = 4.4@BT@ @BT@x_2 = 3.9@BT@ @BT@x_3 = 3.6@BT@ @BT@x_4 = 3.4@BT@

-- Oct 18 In-Class Exercise

x0 = 5

f(x) = ((x-3)**2 ) * ((x-2)**2)

f'(x) = 2*(x-2) * (x-3) * (2x-5)

x1 = 4.4

x2 = 3.958

x3 = 3.636

x4 = 3.407

(Edited: 2017-10-18)

x0 = 5 f(x) = ((x-3)**2 ) * ((x-2)**2) f'(x) = 2*(x-2) * (x-3) * (2x-5) x1 = 4.4 x2 = 3.958 x3 = 3.636 x4 = 3.407

-- Oct 18 In-Class Exercise

If we take x0 = 5, we can get the next point x1 like this:

x1 = x0 - f(x)/f'(x)

so we get: x1 = 4.4 x2 = 3.95 x3 = 3.63 x4 = 3.40

If we take x0 = 5, we can get the next point x1 like this: x1 = x0 - f(x)/f'(x) so we get: x1 = 4.4 x2 = 3.95 x3 = 3.63 x4 = 3.40

-- Oct 18 In-Class Exercise

`x_0 = 5`

`f(x) = (x-2)^2(x-3)^2`

`f'(x) = 4x^3-30x^2+74x-60`

`x_1` = 4.4

`x_2` = 3.958

`x_3` = 3.636

`x_4` = 3.407

(Edited: 2017-10-18)

@BT@x_0 = 5@BT@ @BT@f(x) = (x-2)^2(x-3)^2@BT@ @BT@f'(x) = 4x^3-30x^2+74x-60@BT@ @BT@x_1@BT@ = 4.4 @BT@x_2@BT@ = 3.958 @BT@x_3@BT@ = 3.636 @BT@x_4@BT@ = 3.407

-- Oct 18 In-Class Exercise

(Edited: 2017-10-19)

((resource:IMG_20171020_245411185_HDR.jpg|Resource Description for IMG_20171020_245411185_HDR.jpg))

-- Oct 18 In-Class Exercise

Given x0 = 5 & f(x) = ((x-2)^2 )*((x-3)^2).

First derivative of f(x), f_der(x): f'(x) = 2*(x-2)*(x-3)*(2x-5)

Newton-Raphson: x_n+1 = x_n - f(x)/f_der(x)

Computing...

x0 = 5.0 x1 = 4.4 x2 = 4.0 x3 = 3.6 x4 = 3.4

(Edited: 2017-10-22)

Given x0 = 5 & f(x) = ((x-2)^2 )*((x-3)^2). First derivative of f(x), f_der(x): f'(x) = 2*(x-2)*(x-3)*(2x-5) Newton-Raphson: x_n+1 = x_n - f(x)/f_der(x) Computing... x0 = 5.0 x1 = 4.4 x2 = 4.0 x3 = 3.6 x4 = 3.4

-- Oct 18 In-Class Exercise

X1=4.4 X2= 4.4 - (11.2/25.5) = 3.96 X3=3.63 X4=3.4

(Edited with corrections)

(Edited: 2017-10-30)

X1=4.4 X2= 4.4 - (11.2/25.5) = 3.96 X3=3.63 X4=3.4 (Edited with corrections)

-- Oct 18 In-Class Exercise

f(x) = (x-2)**2 * (x - 3)**2

f'(x) = 2*((x-3)**2)*(x-2)+2*(x-3)*((x-2)**2) using xn+1 = xn + f(x) / f'(x)

x1 = 4.4

x2 = 3.95

x3 = 3.63

x4 = 3.40

(Edited: 2017-10-30)

f(x) = (x-2)**2 * (x - 3)**2 f'(x) = 2*((x-3)**2)*(x-2)+2*(x-3)*((x-2)**2) using xn+1 = xn + f(x) / f'(x) x1 = 4.4 x2 = 3.95 x3 = 3.63 x4 = 3.40