Consider fixed point iterations $x_{n+1} = 2 \sin x_n+ \tan x_n − 3x_n $, where $g(x) = 2 \sin x + \tan x − 3x$. Find its order of convergence at fixed point $\alpha = 0$. (Hint: Using Tayor’s series might be easier.)
What have I tried?
I tried the idea of taking the derivative of g(x).
Because this is an fixed point iteration, $g(\alpha)$ will affect the convergence of the iteration. If $g(\alpha) < 1$, the iteration will converge with linear order. If $g(\alpha) = 1$, we have no clue whether it converges or not, and if it converges, it will converge very slow. if $g(\alpha) = 0$, it will converge with higher order.
I realize that $\alpha=0$ is the fixed point. So I tried to calculate $g', g'', g''', g^{(4)}$.
I came up with $g'(\alpha) = 0, g''(\alpha) = 0, g'''(\alpha)=g^{(4)}(\alpha) = 0$, and $g^{(5)}(\alpha) \neq 0$. So I guess the order of convergence is 5. But it take a lot of steps to evaluate until the fifth derivatives of $g(x)$. According to the hint, there might be a better and faster solution by using the Taylor' series.
$\sin x = x-\frac {x^3}{3!}+\frac {x^5}{5!} -\frac{x^7}{7!}+...$
$\tan x = x+\frac {x^3}{3}+\frac {2x^5}{15}+\frac {17x^7}{315}+...$
Therefore,
$g(x) = 2(x-\frac {x^3}{3!}+\frac {x^5}{5!}+-\frac{x^7}7+...) + (x+\frac {x^3}{3}+\frac {2x^5}{15}+\frac {17x^7}{315}+...)-3x $
$g(x) = x^5(\frac {1}{5!}+\frac{2}{15})+x^7(\frac{-1}{7!}+\frac{17}{315})+...$
The leading term of the Taylor's expansion of $g(x)$ is $x^5$ which means $g'(\alpha)=g''(\alpha)=g'''(\alpha)=g^{(4)}(\alpha)=0$, $g^{(5)}(\alpha) \neq 0$,where $\alpha$ is the fixed point of $x=g(x)$
By applying Taylor's theorem to $g(x_n)$, we have:
$g(x_n)=g(\alpha) +\frac {g^{(5)}(c)}{5!}(x_n-\alpha)^5 $, where $c\in[\alpha,x_n]$
$g(x_n)\approx g(\alpha) +\frac {g^{(5)}(\alpha)}{5!}(x_n-\alpha)^5 $
$g(x_n)-g(\alpha)\approx \frac {g^{(5)}(\alpha)}{5!}(x_n-\alpha)^5 $
$|\alpha - x_{n+1}|\approx\frac {g^{(5)}(\alpha)}{5!}|\alpha-x_n|^5$
Therefore, the convergence order is 5