Use Newton's method to find all solutions of $5\cos(x) = x + 1$ correct to six decimal places.

Question

Use Newton's method to find all solutions of the equation correct to six decimal places. (Enter your answers as a comma-separated list.)

$5\cos(x) = x + 1$

I got $1.130597$ as an answer but apparently I am suppose to have 3 different answers. Also how do I find the initial approximation? Thank you

Answer 1

For Newton use e.g. the start values $1$, $-2$ and $-4$ because the start values have to be near by the solution. (It's not difficult to create your list, you can easily create it yourself.)

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Note without Newton, only a recursion:

It's $\cos 2x=2\cos^2 x-1$.

To avoid the cyclic property of cosine, the variable of cosine should be less than $\pi/2$ . For the first solution with start value $1$ you can use $5\cos x -x-1$ but for the second solution with start value $-2$ you should use e.g. $10\cos^2(\frac{x}{2})-x-6$ and for the third solution with the start value $-4$ it's senseful to use e.g. $40\cos^2(\frac{x}{4})-40\cos(\frac{x}{4})-x+4$ .

Answer 2

Let $$ f(x)=5\cos(x)-x-1 $$ Then $f(x)=0$ can be solved using the recursion $$ \begin{align} x_{n+1} &=x_n-\frac{f(x_n)}{f'(x_n)}\\ &=x_n-\frac{5\cos(x_n)-x_n-1}{-5\sin(x_n)-1}\\ &=\frac{5x_n\sin(x_n)+5\cos(x_n)-1}{5\sin(x_n)+1} \end{align} $$ where $x_1$ is near some root of $f$.

Plotting $f(x)$,

we see that there are roots near $-4,-2,+1$. Apply the recursion to $x_1\in\{-4,-2,+1\}$.