CODEWITHSUNDEEP
matlab
Lola Martinez
Lola Martinez

Reputation: 31

What does "too many input arguments" mean?

I'm trying to evaluate a function in matlab using quadtx but my code is giving me an error of "too many input arguments" and I am not sure what is wrong? Am I putting a wrong value in for varargin or is it something else?

I enter this in the MATLAB command prompt:

>f = @(t)(1./sin(sqrt(abs(t))));
>quadtx(f, -1, 2, 1.e-6, 3)

but receive the following:

Error using @(t)(1./sin(sqrt(abs(t))))
Too many input arguments.

Error in quadtx (line 29)
fa = feval(F,a,varargin{:});

Here is the MATLAB quadtx code:

function [Q,fcount] = quadtx(F,a,b,tol,varargin)
%QUADTX  Evaluate definite integral numerically.
%   Q = QUADTX(F,A,B) approximates the integral of F(x) from A to B
%   to within a tolerance of 1.e-6.  F is a string defining a function
%   of a single variable, an inline function, a function handle, or a
%   symbolic expression involving a single variable.
%
%   Q = QUADTX(F,A,B,tol) uses the given tolerance instead of 1.e-6.
%
%   Arguments beyond the first four, Q = QUADTX(F,a,b,tol,p1,p2,...),
%   are passed on to the integrand, F(x,p1,p2,..).
%
%   [Q,fcount] = QUADTX(F,...) also counts the number of evaluations
%   of F(x).
%
%   See also QUAD, QUADL, DBLQUAD, QUADGUI.
% Make F callable by feval.
if ischar(F) & exist(F)~=2
   F = inline(F);
elseif isa(F,'sym')
   F = inline(char(F));
end 
% Default tolerance
if nargin < 4 | isempty(tol)
   tol = 1.e-6;
end
% Initialization
c = (a + b)/2;
fa = feval(F,a,varargin{:});
fc = feval(F,c,varargin{:});
fb = feval(F,b,varargin{:});
% Recursive call 
[Q,k] = quadtxstep(F, a, b, tol, fa, fc, fb, varargin{:});
fcount = k + 3;
% ---------------------------------------------------------
function [Q,fcount] = quadtxstep(F,a,b,tol,fa,fc,fb,varargin)
% Recursive subfunction used by quadtx.
h = b - a; 
c = (a + b)/2;
fd = feval(F,(a+c)/2,varargin{:});
fe = feval(F,(c+b)/2,varargin{:});
Q1 = h/6 * (fa + 4*fc + fb);
Q2 = h/12 * (fa + 4*fd + 2*fc + 4*fe + fb);
if abs(Q2 - Q1) <= tol
   Q  = Q2 + (Q2 - Q1)/15;
   fcount = 2;
else
   [Qa,ka] = quadtxstep(F, a, c, tol, fa, fd, fc, varargin{:});
   [Qb,kb] = quadtxstep(F, c, b, tol, fc, fe, fb, varargin{:});
   Q  = Qa + Qb;
   fcount = ka + kb + 2;
end

Upvotes: 0

Views: 1725

Answers (1)

Adam
Adam

Reputation: 2777

f = @(t)(1./sin(sqrt(abs(t)))) has only one input variable t. Remove the fourth argument in quadtx(f, -1, 2, 1.e-6, 3) ---> quadtx(f, -1, 2, 1.e-6)

Meaning of varargin

Trigonometric functions are usually defined with additional parameter n, namely their periodicity. For example fn(x) = sin(2*n*pi*x). To evaluate this function you need to first fix n, let's say n = 3, then evaluate f3(x).

To sum up

To use the fourth varargin in quadtx, update f, for instance

f = @(x, n) = (1./sin(n*sqrt(abs(t)))) --> quadtx(f, -1, 2, 1.e-6, 3)

As reference you may check Bessel Function.

Upvotes: 1

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