# How to calculate non linear function

So I need to calculate `x-sin(x)=y` where y is known. It isn’t possible to solve for `x` so I need to approximate.
My first try is

``````using Optim
f(x) = abs(x-sin(x) - y)
optimize(f, 0, 2π).minimizer
``````

But to do this for a matrix of `y`s I need to

``````using Optim
function cacly(y)
f(x) = abs(x-sin(x) - y)
optimize(f, 0,2π).minimizer
end
x = cacly.(y)
``````

I feel like there has to be a better way. To me, this is extremely ugly.

What you want to do is called “root finding”. Check out this library:

Thank you for the response but, sorry, I don’t immediately see anything here that solves my problem.
Sorry for being unclear, my problem is that I need to recompile the function `x-sin(x)` for every y, which results in 99% compilation time everytime.
What I need is a general method of calculating `x` for a given `y`

Never really used Roots.jl before but this should work for you?

``````julia> using Roots

julia> f(x, p=0.0) = x-sin(x) - p
f (generic function with 2 methods)

julia> Z = ZeroProblem(f, 0.01) # 0.01 is an arbitrary start value
ZeroProblem{typeof(f), Float64}(f, 0.01)

julia> @time solve(Z, p=0.1)
0.614689 seconds (6.49 M allocations: 312.507 MiB, 21.92% gc time, 99.99% compilation time)
0.8537501566408657

julia> @time solve(Z, p=0.2)
0.000012 seconds (1 allocation: 16 bytes)
1.083691880314489

julia> @time solve(Z, p=0.3)
0.000011 seconds (1 allocation: 16 bytes)
1.2485154675427026

julia> @time solve(Z, p=0.4)
0.000011 seconds (1 allocation: 16 bytes)
1.3822841337179512
``````
1 Like

Thank you very much