Posted by **Blair** on Saturday, July 19, 2008 at 5:45pm.

Use Newton's method to approximate a root of the equation 5sin(x)=x as follows.

Let x1=2 be the initial approximation.

The second approximation x2 is:

and the third approximation x3 is:

- Math -
**Damon**, Saturday, July 19, 2008 at 6:11pm
5 sin x = x

let y = x - 5 sin x, search for y = 0

dy/dx = y' = 1 - 5 cos x

Xn+1 = Xn + y(Xn)/y'at Xn

X1 = 2

y = 2 - 5 sin 2 = 2 - 4.54 = -2.54

y'=1 - 5 cos 2 = 3.08

X2 = 2 -2.54/3.08 = 1.17

y = 1.17 - 5 sin 1.17 = -3.43

y' = 1 - 5 cos 1.17 = -.951

X3 = 1.17 -3.43/-.951 = 4.77

This is unlikely to work the way you want because you are jumping from cycle to cycle of the original sine wave

- Math -
**Blair**, Saturday, July 19, 2008 at 6:52pm
The answers are wrong for this one.

- Math - sign wrong -
**Damon**, Saturday, July 19, 2008 at 7:47pm
sorry, sign wrong. I drew my picture wrong

5 sin x = x

let y = x - 5 sin x, search for y = 0

dy/dx = y' = 1 - 5 cos x

Xn+1 = Xn - y(Xn)/y'at Xn

X1 = 2

y = 2 - 5 sin 2 = 2 - 4.54 = -2.54

y'=1 - 5 cos 2 = 3.08

X2 = 2 + 2.54/3.08 = 2.82

y = 2.82 - 5 sin 2.82 = 1.24

y' = 1 - 5 cos 2.82 = 5.74

X3 = 2.82 -1.24/5.74 = 2.60

- Math -
**Damon**, Sunday, July 20, 2008 at 1:40pm
check my arithmetic carefully

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