MATLAB: Functions for finding values

fzero

how and what MATLAB functions could i use to find/verify the value of 'a', like i have done using the cubic spline method. (i am pretending that i don't know what that value is)

    a=5;
    w=-a:a/1000:a;
    F=2*sin(a*w)./w; 
    %
    % 
    w_int=-5:0.01:5;                              
    F_int=spline(w,F,w_int);
    plot(w_int,F_int,'o');
    plot(w,F, '--', w_int,F_int,'o')
    c = max(F_int); 
    [~, idx] = min(abs(w_int - 0.005));
    d = w_int(idx);
    a1 = (c*d)/(2*sin(d));

Best Answer

Use Curve Fitting Toolbox

a=5;
w=-a:a/1000:a;
F=2*sin(a*w)./w; 
%

w_int=-5:0.01:5;   
w_int(abs(w_int) < eps(1e3)) = [];                          
F_int=spline(w,F,w_int);

solution:

f = fittype('a2*sin(a1*x)./x');
ff = fit(w_int(:),F_int(:),f,'StartPoint',[1 1]);
out = ff.a1

OR use Statistics Toolbox

a=5;
w=-a:a/1000:a;
F=2*sin(a*w)./w; 
%
w_int=-5:0.01:5;                              
F_int=spline(w,F,w_int);
mf = @(b,x)b(2)*sin(b(1)*x)./x;
b_out = nlinfit(w_int,F_int,mf,[1; 1]);
out = b_out(1);

AND variant with Optimization Toolbox

xdata=(-5:0.01:5)';                              
ydata=spline(w,F,xdata);
t = abs(xdata) < eps(1e4);
xdata(t) = [];
ydata(t) = [];
f = @(a,xdata)a(2)*sin(a(1)*xdata)./xdata;
x = lsqcurvefit(f,[1; 1],xdata,ydata);
out = x(1);

Related Solutions

MATLAB: Value search using functions

You cannot do that. fzero() is based upon the assumption that when upper and lower bounds are provided, that they are constant, not values that vary according to the current "x" value. (Your "w" varies from -a to +a rather than being fixed.)

MATLAB: Not enough input arguments

Your main function calls broyden with a seven element vector as its first input and a handle to the F function as its second input.

broyden(x,@F,7,0.001);

The broyden function calls feval to evaluate the F function with one input, the seven element vector. The f on this line does NOT refer to your f function but to the function handle stored in the variable named f.

fr = feval(f, xv);

You may have expected that MATLAB would automatically expand that vector to pass each element into F as a separate input w1 through w7.

function y = F(w1,w2,w3,w4,w5,w6,w7)

It doesn't. In this call w1 is a seven element vector and w2 through w7 do not exist. [You can add a call to the whos function as the first line inside your F function to check this; that whos output will show only w1 in the workspace.] Thus when you try to use w2 on this line inside F:

y1 = f(1/2,w1,w2);

MATLAB realizes you called F with not enough input arguments and so throws an error.

You could modify F so it accepts one input argument and refers to elements of that input argument. The first two lines would then be the following.

function y = F(w)
y1 = f(1/2,w(1),w(2));
end

Actually, I'd recommend preallocating y and filling it in as you go. Then you don't need to assemble the pieces together at the end.

function y = F(w)
y = zeros(7, 1);
y(1) = f(1/2,w(1),w(2));
end

Alternately you could modify your broyden function to pass each element of the vector into F individually. But I'd prefer the former, as among other things it's less typing.

fr = feval(f, xv(1), xv(2), xv(3), xv(4), xv(5), xv(6), xv(7));

Best Answer

Related Solutions

MATLAB: Value search using functions

MATLAB: Not enough input arguments

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