Tuesday

September 30, 2014

September 30, 2014

Posted by **kwack** on Tuesday, September 7, 2010 at 3:45am.

Let x1=1 be the initial approximation.

The second approximation x2 is

and the third approximation x3 is

- calculus -
**drwls**, Tuesday, September 7, 2010 at 6:09amf(x) = 3 sinx - x

f'(x) = 3 cosx -1

if xo = 1 is the first approximation, the second approximation is

x1 = xo - f(1)/f'(1)

= 1 - 1.524/0.621 = -1.454

x2 = x1 - f(x1)/f'(x1)

= -1.454 -1.525/1.804 = -2.299

x3 = x2 - f(x2)/f'(x2)

= -2.299 -0.0599/(-1.997)

= -2.269

- calculus -
**drwls**, Tuesday, September 7, 2010 at 6:32amI tried to follow the format at

http://en.wikipedia.org/wiki/Newton's_method

My xo may be your x1, etc.

- calculus -
**kwack**, Tuesday, September 7, 2010 at 5:45pmI put -1.454 for x2 but when I put -2.299 for x3 I got it wrong

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