Question

How do you solve `cot(x) = x` ?

Answer 1

There is no algebraic closed form solution, but you can use Newton's method to find numeric solutions.

Explanation:

Let `f(x) = cot(x)-x`

Then `f'(x) = -csc^2(x) - 1`

Newton's method starts with an initial approximation `a_0`, then iterates using the formula:

`a_(i+1) = a_i - f(a_i)/(f'(a_i))`

So in our case:

`a_(i+1) = a_i - (cot(a_i)-a_i)/(-csc^2(a_i)-1)=a_i + (cot(a_i)-a_i)/(csc^2(a_i)+1)`

Looking at the graphs of `y = cot(x)` and `y = x` we see the following:

graph{(cot(x)-y)(x-y)=0 [-20, 20, -10, 10]}

We see that there is one solution for each interval `(0, pi/2)`, `(pi, pi+pi/2)`, `(2pi, 2pi+pi/2)`, etc. and for each interval `(-pi/2, 0)`, `(-pi-pi/2, -pi)`, `(-2pi-pi/2, -2pi)`, etc. In addition note that the further a solution is from the origin, the closer it is to the asymptote of `cot(x)`.

The only algebraic relationship between these solutions is that they occur in `+-` pairs.

In order to converge to the desired solution, it seems that for larger values of `x` you need to choose a starting value only slightly larger than `n pi` (e.g. `2pi+0.01`), otherwise it will tend to converge to the smallest solution (`~~ +-0.860`)

With suitable initial approximations it only takes a few iterations to find approximate solutions:

`x ~~ +-0.860333589019380`

`x ~~ +-3.425618459481728`

`x ~~ +-6.437298179171947`

`x ~~ +-9.529334405361963`