しおり's question at Yahoo Answers regarding Newton's method

[MarkFL]

Here is the question:

Use Newton's Method to find all roots of the equation sinx= x^2-3x+1 correct to six decimal places.?


I think first thing that I need to do is
let x^2-3x+1-sinx=0
then f(x)=x^2-3x+1-sinx

But I am stuck here...
I don't know how to get the interval and how to get x1

Please please explain how to solve this question for me.

Thank you!

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Hello しおり,

The first thing I would do is plot:

\(\displaystyle \sin(x)=x^2-3x+1\)

in order to get an idea where the roots are:

View attachment 1670

So, we see the smaller root is about $0.25$ and the larger root is about $2.75$.

Now, as you did, I would define:

\(\displaystyle f(x)=x^2-3x+1-\sin(x)\)

Hence:

\(\displaystyle f'(x)=2x-3-\cos(x)\)

Newton's method gives us the recursive algorithm:

\(\displaystyle x_{n+1}=x_n-\frac{f\left(x_n \right)}{f'\left(x_n \right)}\)

Using our function $f(x)$, we have:

\(\displaystyle x_{n+1}=x_n-\frac{x_n^2-3x_n+1-\sin\left(x_n \right)}{2x_n-3-\cos\left(x_n \right)}=\frac{2x_n^2-3x_n-x_n\cos\left(x_n \right)-\left(x_n^2-3x_n+1-\sin\left(x_n \right) \right)}{2x_n-3-\cos\left(x_n \right)}\)

\(\displaystyle x_{n+1}=\frac{x_n^2-x_n\cos\left(x_n \right)+\sin\left(x_n \right)-1}{2x_n-3-\cos\left(x_n \right)}\)

i) The smaller root: \(\displaystyle x_0=0.25\)

\(\displaystyle x_1\approx0.253025027391\)

\(\displaystyle x_2\approx0.268799448273\)

\(\displaystyle x_3\approx0.268881342724\)

\(\displaystyle x_4\approx0.268881344942\)

\(\displaystyle x_5\approx0.268881344942\)

ii) The larger root: \(\displaystyle x_0=2.75\)

\(\displaystyle x_1\approx2.77019710423\)

\(\displaystyle x_2\approx2.77005756932\)

\(\displaystyle x_3\approx2.77005756269\)

\(\displaystyle x_4\approx2.77005756269\)

Hence, the two roots of the given equation, rounded to 6 decimal places, are:

\(\displaystyle x\approx 0.268881,\,2.770058\)