Use Newton's method to solve cos(3x)-2x+1=0. Show the values of each iteraion until you have reached the "best" solution your calculator can display....? ( Show your work) ???

If anyone knows how to do this please help me out :(

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- Dec 18th 2008, 06:45 AMEmmeyh15@hotmail.comNewton's Method
Use Newton's method to solve cos(3x)-2x+1=0. Show the values of each iteraion until you have reached the "best" solution your calculator can display....? ( Show your work) ???

If anyone knows how to do this please help me out :( - Dec 18th 2008, 07:17 AMgalactus
Newton's method is easy once you know how to apply it.

We have $\displaystyle cos(3x)-2x+1=0$

Find its derivative:

$\displaystyle -3sin(3x)-2$

Now, apply the formula. Make an initial guess of, say, 0.5.

$\displaystyle x_{n+1}=x_{1}-\frac{f(x)}{f'(x)}$

$\displaystyle x_{n+1}=.5-\frac{cos(3(.5))-2(.5)+1}{-3sin(3(.5))-2}=.51417......$

Do it again, using the last value:

$\displaystyle .51417-\frac{cos(3(.51417))-2(.51417)+1}{-3sin(3(.51417))-2}=.51417$

It converges rather quickly if we choose a good initial guess.

Keep going if you wish, but that is pretty good.

See now what is going on?. They probably want you to keep going until your calculator goes as far as it can. With a guess of .5 that will be pretty fast. So continue. - Dec 18th 2008, 07:25 AMgalactus
BTW, Emmey, I would suggest not using your email address as your user name. Think up something clever. Email addresses as user names are not a good idea. Okey-doke.