Query status of the solution when using Roots

[optimist]

Thank you! This solves my problem.

@j_verzani

I’m not sure what functions will not satisfy Schroder, but will satisfy Halley

Schroder’s method fails for root_func([10,10] but Halley’s method would work using the code in the original post, though I am not sure if this a good example

@optimist

function root_func(c;guess=0.01,max_evals=100)
    t = collect(0:length(c)-1)
    f(x) = sum(@views c[i]/(1+x)^t[i] for i in 1:length(c))
    #find derivatives of f(x)
    df(x) = sum(@views -t[i]*c[i]/(1+x)^(t[i]+1) for i in 1:length(c))
    ddf(x) = sum(@views (t[i]^2+t[i])*c[i]/(1+x)^(t[i]+2) for i in 1:length(c))
    return find_zero((f,df,ddf),guess,Roots.Schroder();maxevals=max_evals)
end