Newton's Method

[Silvanoshei]

Let us dig right in, here's my work with us asking to start with x=0.8:

\(\displaystyle f(x)=x^{2}-sinx\)
\(\displaystyle f'(x)=2x-cosx\)
\(\displaystyle x_{n}-\frac{f(x_{n})}{f'(x_{n})}\)
\(\displaystyle x_{n}-\frac{x^{2}_{n}-sinx_{n}}{2x_{n}-cosx_{n}}\)

\(\displaystyle Start=0.8\)
\(\displaystyle 1st=-.243226\)
\(\displaystyle 2nd=-.200571\)
\(\displaystyle 3rd=-.169361\)

So, plugging 0.8=x... i get -.243226, where my prof got 1=0.885638. What do you guys get with \(\displaystyle 0.8-\frac{(0.8^{2}-sin0.8)}{(2(.08)-cos0.8)}\)?

 

[DrPhil]

Let us dig right in, here's my work with us asking to start with x=0.8:

\(\displaystyle f(x)=x^{2}-sinx\)
\(\displaystyle f'(x)=2x-cosx\)
\(\displaystyle x_{n}-\frac{f(x_{n})}{f'(x_{n})}\)
\(\displaystyle x_{n}-\frac{x^{2}_{n}-sinx_{n}}{2x_{n}-cosx_{n}}\)

\(\displaystyle Start=0.8\)
\(\displaystyle 1st=-.243226\)
\(\displaystyle 2nd=-.200571\)
\(\displaystyle 3rd=-.169361\)

So, plugging 0.8=x... i get -.243226, where my prof got 1=0.885638. What do you guys get with \(\displaystyle 0.8-\frac{(0.8-sin0.8)}{(2(.08)-cos0.8)}\)?

You probably did square x in the numerator - that is,
f(x) = x^2 - sin(x)
And you probably did use 0.8 instead of 0.08 in the denominator.

BUT - did you set your calculator for radians?

 

[Silvanoshei]

Yup! Sorry, that was a typo. It was entered correctly into my calculator. Dunno why my answer is different. It should be close to 0.8, not way off into -.24 land.

 

[Silvanoshei]

Omg.... I forgot to turn my calc to radians lol... thanks. The answer is now correct on my calculator.