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March 4, 2015

March 4, 2015

Posted by **HELLLLLPPPPPPP** on Wednesday, September 8, 2010 at 1:47am.

Let x1=1 be the initial approximation.

The second approximation x2 is

and the third approximation x3 is

I got x2=-1.454

but can't get x3 :(

- calculus -
**MathMate**, Wednesday, September 8, 2010 at 8:53amYou are on the right track!

x1=1,

x2=-1.455,

x3=-3.787,

x4=-2.140,

...

eventually it will settle on x=-2.28.

Remember that in the case of multiple roots, the one obtained by Newton's method is very dependent on the initial approximation. I assumed x1=1 is in radians.

- calculus -
**Reiny**, Wednesday, September 8, 2010 at 8:57amI agree with drwls's setup in

Newton's method

starting with x1=1 I also got

x2= -41.454 but then I got the next values as

x3 = -3.787

x4 = -2.1404

x5 = -2.2878

x6 = -2.1813

x7 = -2.283

etc

appears to converge to around x = -2.28

check:

3sin(-2.28) = -2.2766

- calculus -
**Reiny**, Wednesday, September 8, 2010 at 9:01amthere was supposed to be a reference to

drwls setup in

http://www.jiskha.com/display.cgi?id=1283845535

As you can see in MathMate's reply both of our results coincide.

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