Consider the equation $\cos x = c$, where $c$ is constant. Find the value of $c$ so that the approximate solution by Newton-Raphson method at the $k$th iteration is $x_k = (-1)^k$, where $k = 0, 1, 2, \cdots$.
Can anyone explain this problem? I can not understand what it mean or how to solve it.
On
Newton-Rhapson method is a numerical method that approximates the solutions to an equation $f(x)=0$ with a sequence $x_n$ defined recursively as $x_{n+1}=G_f(x_n)$, where $G_f$ is a function that can be written in terms of $f$ and $f'$ and $x_0$ is given. Its convergence theorem guarantees that, if $x_0$ is chosen inside some interval $[a,b]$ such that $f$ satisifies a few properties in it, then the sequence $x_n$ will converge to the only zero of $f$ in $[a,b]$.
Now, in your case, $x_0=1$ is given implicitly. Moreover, $f$ depends on the parameter $c$, and so do $G_{f_c}$ and the recursivey defined sequence $x^c_n$. The problem asks you to find $c$ such that $x^c_n=(-1)^n$.
The Newton-Raphson method solves $\cos x -c=0$ by iterating $x\mapsto x+\frac{\cos x -c}{\sin x}$. We require this to map $1$ to $-1$, so $-1=1+\frac{\cos 1 -c}{\sin 1}$. Rearranging gives $c=\cos 1+2\sin 1$.