How to take a function_handle's coefficients

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Sam
Sam 2017-11-15
评论： Jan 2017-11-15
I don't think it's too complex but I can't see a solution anywhere.
I have a function_handle f =@(y) (y^3 -5) which clearly has coefficients [1 0 0 -5] but I do not know what I need to write to be able to extract it from function_handle.
i.e. q = [(coefficients extracted)]
I want to be able to take this because then i can work out the roots of the polynomial straight from roots(q) where q is the row vector of coefficients of the function handle.
Is anyone able to help, I would really appreciate it.
  1 个评论
Jan
Jan 2017-11-15
This might not be trivial. What do you expect for:
f = @(y) sin(y)
? Is the function guaranteed to be a polynomial? Then why not defining it as [1,0,0,-5] directly?

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采纳的回答

M
M 2017-11-15
syms y
q=sym2poly(f(y))
q =
1 0 0 -5

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