The exact question given is as follows:
Use Newton’s method to find the intersection points of the curve y =
cos(x) and the circle of unit radius centred at (x, y) = (0.4, 1.7) correct
to 3 decimal places.
I'm able to actually use Newton's Method, but the part I'm having trouble with is getting the initial equation involving $x_n$.
So clearly we have $y=cos(x)$ and the equation for the circle is $(x-0.4)^2+(y-1.7)^2=1$.
Expanding and rearranging the circle equation results in $$y^2-3.4y=4.05-x^2+0.8x$$ and this is where I'm stuck. I need both equations to be equal to each other in order to use it to find the derivative and thus set up the equation for Newton's Method.
One possible solution I thought of was to use $y^2+3.4y = cos^2(x)+3.4cos(x)$, which gives me:
$$F(x)=cos^2(x)+3.4cos(x)-4.05+x^2-0.8x$$
But then $F'(x)$ ends up being quite ugly to use, and usually the questions are designed to be more user friendly, which gives me doubts as to whether this is the right approach or not.
More specifically, I am looking for help in finding the correct $F(x)$ so that I can find it's derivative and then solve the question I have been given.
Best Answer
You are looking for the intersection of the circle $$\left(x-\frac{2}{5}\right)^2+\left(y-\frac{17}{10}\right)^2-1=0\tag 1$$ and the function $y=\cos(x)$.
In $(1)$, replace $y$ by $\cos(x)$, expand and simplify using double angle formula. This will give $$F(x)=x^2-\frac{4 }{5}x-\frac{17 }{5}\cos (x)+\frac{1}{2} \cos (2 x)+\frac{51}{20}\tag 2$$ $$F'(x)=2 x+\frac{17 }{5}\sin (x)-\sin (2 x)-\frac{4}{5}\tag 3$$ I suppose that you can manage this quite easily.
Moreover, using Taylor expansion around $x=0$, you have $$F(x)=-\frac{7}{20}-\frac{4 }{5}x+\frac{17 }{10}x^2+O\left(x^3\right)$$ which allows to generate quite good estimates of the solutions solving the quadratic equation.
With that, Newton will work like a charm, I bet.