Trying to compute the following expression,
quad(#(n)quad(#(m)unifpdf(m-n,0,1),-10,10),-10,10)
But I get a message saying:
??? Error using ==> minus
Matrix dimensions must agree.
Error in ==> #(m)unifpdf(m-n,0,1)
Can you please let me know how to fix this?
The problem here is the following requirement:
The function y = fun(x) should accept a vector argument x and return a vector result y, the integrand evaluated at each element of x.
unipdf seems to satisfy this requirement, but quad itself is not.
To fix this problem you need to write a wrapper function that accepts a vector argument x, evaluates inner integral using quad function, and return a vector of results:
function [r] = Test()
r = quad(#(n)InnerIntegral(n),-10,10);
end
function [y] = InnerIntegral(n)
y = zeros(size(n));
for i = 1 : length(n)
y(i) = quad(#(m)unifpdf(m - n(i), 0, 1), -10, 10);
end;
end
Related
I want to evaluate the double integral of my objective function (named myfunction, see below) using the build-in-function integral2.
I have tried to run this script;
f = #(r,theta) myfunction(r,theta);
integral2(f,0,2,0,2*pi);
where myfunction is the following function:
function fkt=myfunction(r,theta)
x=r.*cos(theta);
y=r.*sin(theta);
los(:,1)=x;
los(:,2)=y;
norm = (sum( sqrt( los(:,1).^2 + los(:,2).^2)));
fkt=norm*r;
end
I am making the integral in polar coordinates, that why fkt=norm*r.
Matlab gives me the following error message:
>> untitled2
Subscripted assignment dimension mismatch.
Error in myfunction (line 8)
los(:,1)=x;
I can't figure out, what the problem is.
There are two things that can be improved:
los is undefined
los(:,1) is a column while x is a row, so the assignment has to fail.
You can correct this by defining los and change your assignment. For instance:
los = NaN(numel(r), 2);
los(:,1) = x';
los(:,2) = y';
But, why do you need the variable los? Just remove it and the error will be gone:
function fkt=myfunction(r,theta)
x=r.*cos(theta);
y=r.*sin(theta);
norm = (sum( sqrt(x.^2 + y.^2)));
fkt=norm*r;
end
Best,
x is a matrix and you try to assign it to a column vector.
I so far have the following
y = log(x);
% Ask user for input values for h and M
% M denotes the number of steps of the algorithm.
h = input('Input value h: ');
M = input('Input value M: ');
%Initialize an MxM matrix
D = zeros(M);
phi = (1/(2*h)) * (y(x+h) - y(x-h));
print(phi);
I obtain the error
Error using symengine (line 58) Index exceeds matrix dimensions.
Error in sym/subsref (line 696)
B = mupadmex('symobj::subsref',A.s,inds{:});
Error in RE (line 12) phi = (1/(2*h)) * (y(x+h) - y(x-h));
First, I believe I should be getting an error message about x not being defined. Second, I have no idea what the matrix dimension error is about. Third, and most importantly, how can I declare the function phi so that it becomes what I wrote?
First, I believe I should be getting an error message about x not being defined.
I'm guessing that x is defined, or you would get that error upon the line defining phi. To check whether x is defined, type "who" or "whos".
Second, I have no idea what the matrix dimension error is about.
This is most likely because y is a scalar, x + h is equal to some nonzero integer that is not 1, and you're trying to access y(x + h). For your own edification try setting y equal to a scalar (e.g. y = 5;) and seeing what errors are produced by indexing it in various legitimate and non-legitimate ways (e.g. y(1), y(0), y(3), y(-1), y(1.5)).
Third, and most importantly, how can I declare the function phi so that it becomes what I wrote?
Based on the context it looks like you want y to be defined as a function of x instead of a scalar. In other words:
y = #(x)log(x);
phi = (1/(2*h)) * (y(x+h) - y(x-h));
The code runs without error when you change the definitions to the above.
One other error you will run into: the print command is not what you're looking for - this prints a figure to a file. You're probably looking for:
disp(phi);
I'm trying to implement this integral representation of Bessel function of the first kind of order n.
here is what I tried:
t = -pi:0.1:pi;
n = 1;
x = 0:5:20;
A(t) = exp(sqrt(-1)*(n*t-x*sin(t)));
B(t) = integral(A(t),-pi,pi);
plot(A(t),x)
the plot i'm trying to get is as shown in the wikipedia page.
it said:
Error using * Inner matrix dimensions must agree.
Error in besselfn (line 8) A(t) = exp(sqrt(-1)*(n*t-x*sin(t)));
so i tried putting x-5;
and the output was:
Subscript indices must either be real positive integers or logicals.
Error in besselfn (line 8) A(t) = exp(sqrt(-1)*(n*t-x*sin(t)));
How to get this correct? what am I missing?
To present an anonymous function in MATLAB you can use (NOT A(t)=...)
A = #(t) exp(sqrt(-1)*(n*t-x.*sin(t)));
with element-by-element operations (here I used .*).
Additional comments:
You can use 1i instead of sqrt(-1).
B(t) cannot be the function of the t argument, because t is the internal variable for integration.
There are two independent variables in plot(A(t),x). Thus you can display plot just if t and x have the same size. May be you meant something like this plot(x,A(x)) to display the function A(x) or plot(A(x),x) to display the inverse function of A(x).
Finally you code can be like this:
n = 1;
x = 0:.1:20;
A = #(x,t) exp(sqrt(-1)*(n*t-x.*sin(t)));
B = #(x) integral(#(t) A(x,t),-pi,pi);
for n_x=1:length(x)
B_x(n_x) = B(x(n_x));
end
plot(x,real(B_x))
need to find a set of optimal parameters P of the system y = P(1)*exp(-P(2)*x) - P(3)*x where x and y are experimental values. I defined my function
f = #(P) P(1)*exp(-P(2)*x) - P(3)*x
and
guess = [1, 1, 1]
and tried
P = fminsearch(f,guess)
according to Help. I get an error
Subscripted assignment dimension mismatch.
Error in fminsearch (line 191)
fv(:,1) = funfcn(x,varargin{:});
I don't quite understand where my y values would fall in, as well as where the function takes P from. I unfortunately have no access to nlinfit or optimization toolboxes.
You should try the matlab function lsqnonlin(#testfun,[1;1;1])
But first make a function and save in an m-file that includes all the data points, lets say your y is A and x is x like here below:
function F = testfun(P)
A = [1;2;3;7;30;100];
x = [1;2;3;4;5;6];
F = A-P(1)*exp(-P(2)*x) - P(3)*x;
This minimized the 2-norm end gives you the best parameters.
Im new to MATLAB. Want to use integral2 as follows
function num = numer(x)
fun=#(p,w) prod((p+1-p).*(1-w).*exp(w.*x.*x/2))
num= integral2(fun ,0,1,0,1)
end
I get several errors starting with
Error using .*
Matrix dimensions must agree.
Error in numer>#(p,w)prod(p+(1-w).*exp(w.*x.*x/2)) (line 5)
fun=#(p,w) prod(p+(1-w).*exp(w.*x.*x/2))
Can you please tell me what I do wrong.
Thanks
From the help for integral2:
All input functions must accept arrays as input and operate
elementwise. The function Z = FUN(X,Y) must accept arrays X and Y of
the same size and return an array of corresponding values.
When x was non-scalar, your function fun did not do this. By wrapping everything in prod, the function always returned a scalar. Assuming that your prod is in the right place to begin with and taking advantage of the properties of the exponential, I believe this version will do what you need for vector x:
x = [0 1];
lx = length(x);
fun = #(p,w)(p+1-p).^lx.*(1-w).^lx.*exp(w).^sum(x.*x/2);
num = integral2(fun,0,1,0,1)
Alternatively, fun = #(p,w)(p+1-p).^lx.*(1-w).^lx.*exp(sum(x.*x/2)).^w; could be used.