I have to find the root of such equation with accuracy $0.2$ by fixed-point iteration method: $$ x-\cot{x} = 0 $$ So I have: $$ f(x) = x-\cot{x} $$ $$ g(x) = \cot{x} $$ For $x_0$ I choose $-0.9$ and then I have $x_1=-0,794$ and that's okay, but if I will continue the values will rising rapidly without convergation. What am I doing wrong?
2026-04-07 02:07:15.1775527635
How to find the root of $x-\cot{x} = 0$ by fixed-point iteration?
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By $x\tan x=1$ one can see that at $x=1$ the left side is above $1$, and at $x=\frac\pi4$ it is below, giving a bracketing interval for the solution.
Set $x=\frac\pi4+y$ with $y$ small. Then $$ \frac\pi4+y=\frac{\cos y -\sin y}{\cos y+\sin y}\approx\frac{1-y}{1+y}\approx 1-2y. $$ So taking the last variant gives $3y\approx1-\frac\pi4$ so $x\approx \frac{\pi}{6}+\frac13$. This gives numerical values of $x=0.857$ and $\cot x=0.866$, thus within the desired accuracy.
If $x$ is close but below the target value, then $\cot x$ will be above and vice versa. So the weighted midpoint formula cancelling the linear terms at about $x=\frac\pi4$, $$ x_{+1}=g(x)=\frac{2x+\cot x}3, $$ has also good chances to work well. This indeed converges linearly with factor about $0.087$.