power = 2;
sigma=0.1;
a = 1 /(sigma*sqrt(2*pi));
c= (sigma^2)*2;
syms x y
f = exp(-(x.^power)./c);
dfdx = diff(f,x);
c1 = diff(dfdx,x);
f = exp(-(y.^power)./c);
dfdy = diff(f,y);
c2 = diff(dfdy,y);
meancurvature = (c1 + c2)./ 2;
gaussiancuravture =(c1 .* c2);
mean_curv = integral2(meancurvature, -Inf,+Inf, -Inf, +Inf)
gauss_curv = integral2(gaussiancurvature, -Inf,+Inf, -Inf, +Inf)
I've tried everything in my limited matlab knowledge and google searching to find an answer to the error that comes up:
Error using integral2
First input argument must be a function handle.
You need pass in anonymous functions for the first argument. Create them using the # sign. For more details, check out these links
http://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html
Using integral2 in Matlab with vectors
Related
I've noticed some weird facts about integral2. These are probably due to my limitations in understanding how it works. I have some difficulties in integrating out variables when I have particular functions. For instance, look at the following Code:
function Output = prova(p,Y)
x = p(1);
y = p(2);
w = p(3);
z = p(4);
F1 = #(Data,eta_1,eta_2,x,y,w,z) F2(eta_1,eta_2,Data) .* normpdf(eta_1,x,y) .* normpdf(eta_2,w,z);
Output = integral2(#(eta_1,eta_2)F1(Y,eta_1,eta_2,0,1,10,2),-5,5,-5,5);
end
function O = F2(pp1,pp2,D)
O = pp1 + pp2 + sum(D);
end
In this case the are no problems in evaluating the integral. But if I change the code in this way I obtain some errors, although the output of F2 is exactly the same:
function Output = prova(p,Y)
x = p(1);
y = p(2);
w = p(3);
z = p(4);
F1 = #(Data,eta_1,eta_2,x,y,w,z) F2(eta_1,eta_2,Data) .* normpdf(eta_1,x,y) .* normpdf(eta_2,w,z);
Output = integral2(#(eta_1,eta_2)F1(Y,eta_1,eta_2,0,1,10,2),-5,5,-5,5);
end
function O = F2(pp1,pp2,D)
o = sum([pp1 pp2]);
O = o + sum(D);
end
The problems increase if F2 for example have some matrix multiplication in which "eta_1" and "eta_2", which I want to integrate out, are involved. This problems makes practically impossible to solve computations in which, for instance, we have to integrate out a variable X which is inside a Likelihood Function (whose calculation could require some internal Loop, or Sum, or Prod involving our variable X). What is the solution?
I am trying to solve this optimization problem with fmincon function in MATLAB:
Where all H are complex matrices, g and Pdes are complex column vectors, D0 and E0 are numbers.
I expect to get complex column vector of g (and in general its should be complex), So I divided problem in two parts: real and imag, but it does not work, MATLAB returning me a message:
Not enough input arguments.
Error in temp>nonlincon (line 17)
c(1) = norm ( H_d*([g(1)+1i*g(4); g(2)+1i*g(5); g(3)+1i*g(6)]) )^2 - D_0;
Error in temp (line 12)
= fmincon(objective,x0,[],[],[],[],[],[],nonlincon);
Where I am wrong?
And in general am I right in writing given problem in the following way:?
% For example:
D_0 = 2*10^(-5)*10^(60/10);
E_0 = 50;
H_b = rand(15,3) + 1i*rand(15,3);
P_des = rand(15,1) + 1i*rand(15,1);
H_d = rand(10,3) + 1i*rand(10,3);
objective = #(g) (norm ( H_b*([g(1)+1i*g(4); g(2)+1i*g(5); g(3)+1i*g(6)]) - P_des ))^2;
x0 = ones(1,3*2)';
options = optimoptions('fmincon','MaxFunctionEvaluations',10e3);
[X,FVAL,EXITFLAG,OUTPUT,LAMBDA,GRAD,HESSIAN]...
= fmincon(objective,x0,[],[],[],[],[],[],nonlincon);
% So I expect to get a column vector g: 1st 3 elements - Real part, next 3 - Imag
function [c,ceq] = nonlincon(g, H_d, E_0, D_0)
c(1) = (norm ( H_d*([g(1)+1i*g(4); g(2)+1i*g(5); g(3)+1i*g(6)]) ))^2 - D_0;
c(2) = (norm ([g(1)+1i*g(4); g(2)+1i*g(5); g(3)+1i*g(6)]))^2 - E_0;
ceq = [];
end
You just need to specify that nonlincon is a function of variable g only when calling fmincon
The code is as follow
[X,FVAL,EXITFLAG,OUTPUT,LAMBDA,GRAD,HESSIAN]...
= fmincon(objective,x0,[],[],[],[],[],[],#(g)nonlincon(g, H_d, E_0, D_0));
Solution X
X = [0.2982; 0.1427; 0.3597; -0.0729; -0.1187; 0.2090]
I am trying to feed a function handle into the function I created below. I'm not exactly sure how to do this.
For example, how do I get:
conjugate_gradient(#(y) ABC(y), column_vector, initial_guess)
to not error?
If I use matlab's pcg function in the same way it will work:
pcg(#(y) ABC(y),b,tol).
I tried reading the pcg function, and they do take about this in the function description, however I'm still super inexperienced with MATLAB and had shall we say some difficulty understanding what they did.Thank You!
function [x] = conjugate_gradient(matrix, column_vector, initial_guess)
y = [];
col_size = length(column_vector);
temp = size(matrix);
mat_row_len = temp(2);
% algorithm:
r_cur = column_vector - matrix * initial_guess;
p = r_cur;
k = 0;
x_approx = initial_guess;
for i=1:mat_row_len
alpha = ( r_cur.' * r_cur ) / (p.' *(matrix* p));
x_approx = x_approx + alpha * p;
r_next = r_cur - alpha*(matrix * p);
fprintf(num2str(r_next'*r_next), num2str(i))
y = [y; i, r_next'*r_next];
%exit condition
if sqrt(r_next'*r_next) < 1e-2
y
break;
end
beta = (r_next.'* r_next )/(r_cur.' * (r_cur) );
p = r_next + beta * p;
k = k+1;
r_cur = r_next;
end
y
[x] = x_approx;
end
When you do
f = #(y) ABC(y)
You create a function handle. (Note that in this case it's the same as f=#ABC). This handle is a variable, and this can be passed to a function, but is otherwise the same as the function. Thus:
f(1)
is the same as calling
ABC(1)
You pass this handle to a function as the first argument, which you have called matrix. This seems misleading, as the variable matrix will now be a function handle, not a matrix. Inside your function you can do matrix(y) and evaluate the function for y.
However, reading your function, it seems that you treat the matrix input as an actual matrix. This is why you get errors. You cannot multiply it by a vector and expect a result!
I am trying to compute the value of this integral using Matlab
Here the other parameters have been defined or computed in the earlier part of the program as follows
N = 2;
sigma = [0.01 0.1];
l = [15];
meu = 4*pi*10^(-7);
f = logspace ( 1, 6, 500);
w=2*pi.*f;
for j = 1 : length(f)
q2(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(2));
q1(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(1));
C2(j)= 1/(q2(j));
C1(j)= (q1(j)*C2(j) + tanh(q1(j)*l))/(q1(j)*(1+q1(j)*C2(j)*tanh(q1(j)*l)));
Z(j) = sqrt(-1)*2*pi*f(j)*C1(j);
Apprho(j) = meu*(1/(2*pi*f(j))*(abs(Z(j))^2));
Phi(j) = atan(imag(Z(j))/real(Z(j)));
end
%integration part
c1=w./(2*pi);
rho0=1;
fun = #(x) log(Apprho(x)/rho0)/(x.^2-w^2);
c2= integral(fun,0,Inf);
phin=pi/4-c1.*c2;
I am getting an error like this
could anyone help and tell me where i am going wrong.thanks in advance
Define Apprho in a separate *.m function file, instead of storing it in an array:
function [ result ] = Apprho(x)
%
% Calculate f and Z based on input argument x
%
% ...
%
meu = 4*pi*10^(-7);
result = meu*(1/(2*pi*f)*(abs(Z)^2));
end
How you calculate f and Z is up to you.
MATLAB's integral works by calling the function (in this case, Apprho) repeatedly at many different x values. The x values called by integral don't necessarily correspond to the 1: length(f) values used in your original code, which is why you received errors.
Following code throws out an error.
syms z positive;
syms n;
syms m;
N = 10;
Ms = 10;
Es = 1;
pd = 0.9;
pd_dash = 1-pd;
pf = 0.1;
pf_dash = 1-pf;
pr = 0.1;
qr = 1-pr;
p = 0.005
pi = pf_dash*p;
pb = pd_dash*p;
qi = 1-pi;
qb = 1-pb;
sm = symsum( z^((n+1)*Es), n, 0, N-1 );
temp_sum = symsum(z^((n+m+1)*Es)*qr^(n+m)*pr, m, 0, N-1);
z=1; %assume a value of z
x = eval(sm); %works fine
y = eval(temp_sum);
% Error:The expression to the left of the equals sign is not a valid target for an assignment.
Please suggest a way to resolve this.
The problem that I suspect is: the temp_sum comes out to be in piecewise(...) which eval is not capable of evaluating.
What you actually did:
Create a symbolic expression
Create a variable z which is unused
Call a undocumented function sym/eval
I assume you wanted to:
Create a symbolic expression
Substitute z with 1: temp_sum=subs(temp_sum,z,1)
get the result. Here I don't know what you really have because I don't know which variables are symbolic unknowns and which constants. Try simplify(temp_sum). If you substituted all unknowns it should return a number.