Erro in calling function using horner()

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Lucas
Lucas 2022-11-28
编辑： Torsten 2022-11-28
I wrote the following code to fild the roots of a given function. In program, the example function ig given by f(x) = x^3-6x^3+11x-6. When I compile the code the following messages appears:
Incorrect number or types of inputs or outputs for function 'horner'.
Error in test>newtonhorner (line 34)
[pz,b] = horner(a,x);
Error in test (line 4)
a = [1 -6 11 -6]; [x,niter]=newtonhorner(a,0,1.e-15,100);
How can I solve this? Thanks if anyone can help!
My code:
a = [1 -6 11 -6]; [x,niter]=newtonhorner(a,0,1.e-15,100);
function [roots , iter]= newtonhorner(a,x0,tol,nmax)
%NEWTONHORNER Newton - Horner method
% [roots , ITER]= NEWTONHORNER(A,X0) computes the roots of
% polynomial
% P(X) = A(1)*X^N + A(2)*X^(N-1) + ... + A(N)*X +
% A(N+1)
% using the Newton - Horner method starting from the
% initial datum X0. The method stops for each root
% after 100 iterations or after the absolute value of
% the difference between two consecutive iterates is
% smaller than 1.e-04.
% [roots , ITER]= NEWTONHORNER(A,X0 ,TOL , NMAX) allows to
% define the tolerance on the stopping criterion and
% the maximal number of iterations.
if nargin == 2
tol = 1.e-04; nmax = 100;
elseif nargin == 3
nmax = 100;
end
n=length(a)-1;
roots = zeros (n,1);
iter = zeros(n,1);
for k = 1:n
% Newton iterations
niter = 0; x = x0; diff = tol + 1;
while niter <= nmax & diff >= tol
[pz,b] = horner(a,x);
[dpz,b]= horner(b,x);
xnew = x - pz/dpz;
diff = abs(xnew -x);
niter = niter + 1;
x = xnew;
end
if niter >= nmax
fprintf('Fails to converge within maximum ',...
'number of iterations\n ');
end
% Deflation
[pz ,a] = horner(a,x); roots(k) = x; iter(k) = niter;
end
return
end

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采纳的回答

Torsten
Torsten 2022-11-28
编辑：Torsten 2022-11-28
I don't understand what you intend by using the lines
[pz,b] = horner(a,x);
[dpz,b]= horner(b,x);
[pz ,a] = horner(a,x)
They will error because of two reasons:
a and b must be symbolic polynomials.
"horner" has only one output, namely the input polynomial in horner form.

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