Sequential Quadratic Programming Matlab Implementation - matlab

I have a problem with my MATLAB code that solves a nonlinear quadratic problem with SQP algorithm (Sequential quadratic programming) but in the "QP-SUB PROBLEM" section of the code that i have formulated analytically a "num2str"error appears and honestly, i don't know how to fix that and also have to tell you that this method uses KT conditions for a better solution .
In every section of the code i write a comment for better understanding and function with constraints can be found in the code below :
% Maximize f(x1,x2) = x1^4 -2x1^2x2 +x1^2 +x1x2^2 -2x1 +4
%
% h1(x1,x2) = x1^2 + x2^2 -2 = 0
% g1(x1,x2) = 0.25x1^2 +0.75x2^2 -1 <=0
% 0 <= x1 <= 4; 0 <= x2 <= 4
%
%--------------------------------------------------------
% The KT conditions for the QP subproblem
% are applied analytically
% There are two cases for a single inequality constraint
% Case (a) : beta = 0 g < 0
% Case (b) : beta ~= 0, g = 0
% The best solution is used
% --------------------------------------------------------
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% management functions
clear % clear all variable/information in the workspace - use CAUTION
clear global % again use caution - clears global information
clc % position the cursor at the top of the screen
close % closes the figure window
format compact % avoid skipping a line when writing to the command window
warning off %#ok<WNOFF> % don't report any warnings like divide by zero etc.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% DATA ---------------------
%%% starting design
xb(1) = 3; xb(2) = 2;
it = 10; % number of iterations
%%% plot range for delx1 : -3 , +3
dx1L = -3; dx1U = +3;
dx2L = -3; dx2U = +3;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Define functions
syms x1 x2 f g h
syms gradf1 gradf2 gradh1 gradh2 gradg1 gradg2
f = x1^4 - 2*x1*x1*x2 + x1*x1 + x1*x2*x2 - 2*x1 + 4;
h = x1*x1 + x2*x2 - 2;
g = 0.25*x1*x1 +0.75*x2*x2 -1;
%%% the gradient functions
gradf1 = diff(f,x1);
gradf2 = diff(f,x2);
% the hessian
hess = [diff(gradf1,x1), diff(gradf1,x2); diff(gradf2,x1), diff(gradf2,x2)];
% gradient of the constraints
gradh1 = diff(h,x1);
gradh2 = diff(h,x2);
gradg1 = diff(g,x1);
gradg2 = diff(g,x2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% graphical/symbolic solution for SLP
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
fprintf('***********************')
fprintf('\nSQP - Example 7.1')
fprintf('\n*********************\n')
for i = 1:it
%figure;
f1 = subs(f,{x1,x2},{xb(1),xb(2)});
g1 = subs(g,{x1,x2},{xb(1),xb(2)});
h1 = subs(h,{x1,x2},{xb(1),xb(2)});
%%% Print Information
fprintf('\n******************************')
fprintf('\nIteration : '),disp(i)
fprintf('******************************\n')
fprintf('Linearized about [x1, x2] : '),disp([xb(1) xb(2)])
fprintf('Objective function value f(x1,x2) : '),disp(f1);
fprintf('Equality constraint value value h(x1,x2) : '),disp(h1);
fprintf('Inequality constraint value value g(x1,x2) : '),disp(g1);
%fprintf('\nsolution for [delx1 delx2] : '),disp(sol')
% hold on
% calculate the value of the gradients
% f1 = subs(f,{x1,x2},{xb(1),xb(2)});
% g1 = subs(g,{x1,x2},{xb(1),xb(2)});
% h1 = subs(h,{x1,x2},{xb(1),xb(2)});
fprintf('\n-----------------------')
fprintf('\nQP - SUB PROBLEM')
fprintf('\n---------------------\n')
gf1 = double(subs(gradf1,{x1,x2},{xb(1),xb(2)}));
gf2 = double(subs(gradf2,{x1,x2},{xb(1),xb(2)}));
hess1 = double(subs(hess,{x1,x2},{xb(1),xb(2)}));
gh1 = double(subs(gradh1,{x1,x2},{xb(1),xb(2)}));
gh2 = double(subs(gradh2,{x1,x2},{xb(1),xb(2)}));
gg1 = double(subs(gradg1,{x1,x2},{xb(1),xb(2)}));
gg2 = double(subs(gradg2,{x1,x2},{xb(1),xb(2)}));
% the QP subproblem
syms dx1 dx2 % change in design
fquad = f1 + [gf1 gf2]*[dx1; dx2] + 0.5*[dx1 dx2]*hess1*[dx1 ;dx2];
hlin = h1 + [gh1 gh2]*[dx1; dx2];
glin = g1 + [gg1 gg2]*[dx1; dx2];
Fquadstr = strcat(num2str(f1),' + ',num2str(gf1), ...
'*','dx1',' + ',num2str(gf2),' * ','dx2', ...
' + 0.5*',num2str(hess1(1,1)),' * dx1^2', ...
' +',num2str(hess1(1,2)),' * dx1*dx2', ...
' + 0.5*',num2str(hess1(2,2)),' * dx2^2');
hlinstr = strcat(num2str(h1),' + ',num2str(gh1), ...
'*','dx1',' + ',num2str(gh2),' * ','dx2');
glinstr = strcat(num2str(g1),' + ',num2str(gg1), ...
'*','dx1',' + ',num2str(gg2),' * ','dx2');
fprintf('Quadratic Objective function f(x1,x2): \n'),disp(Fquadstr);
fprintf('Linearized equality h(x1,x2): '),disp(hlinstr);
fprintf('Linearized inequality g(x1,x2): '),disp(glinstr);
fprintf('\n')
% define Lagrangian for the QP problem
syms lamda beta
F = fquad + lamda*hlin + beta*glin;
fprintf('Case a: beta = 0\n');
Fnobeta = fquad + lamda*hlin;
%%% initialize best solution
dx1best = 0;
dx2best = 0;
Fbbest = 0;
%%%%%%%%%%%%%%%%%%%%%%%
%%% solve case (a) %%%
%%%%%%%%%%%%%%%%%%%%%%%
xcasea = solve(diff(Fnobeta,dx1),diff(Fnobeta,dx2),hlin);
sola = [double(xcasea.dx1) double(xcasea.dx2) double(xcasea.lamda)];
dx1a = double(xcasea.dx1);
dx2a = double(xcasea.dx2);
lamdaa = double(xcasea.lamda);
hlina = double(subs(hlin,{dx1,dx2},{dx1a,dx2a}));
glina = double(subs(glin,{dx1,dx2},{dx1a,dx2a}));
Fa = double(subs(Fnobeta,{dx1,dx2,lamda},{dx1a,dx2a,lamdaa}));
%%% results for case (a)
x1a = dx1a + xb(1);
x2a = dx2a + xb(2);
fv = double(subs(f,{x1,x2},{x1a,x2a}));
hv = double(subs(h,{x1,x2},{x1a,x2a}));
gv = double(subs(g,{x1,x2},{x1a,x2a}));
fprintf('Change in design vector: '),disp([dx1a dx2a]);
fprintf('The linearized quality constraint: '),disp(hlina);
fprintf('The linearized inequality constraint: '),disp(glina);
fprintf('New design vector: '),disp([x1a x2a]);
fprintf('The objective function: '),disp(fv);
fprintf('The equality constraint: '),disp(hv);
fprintf('The inequality constraint: '),disp(gv);
if (glina <= 0)
xb(1) = xb(1) + dx1a;
xb(2) = xb(2) + dx2a;
fbest = Fa;
dx1best = dx1a;
dx2best = dx2a;
end
%%%%%%%%%%%%%%%%%%%%%%%
%%% solve case (b) %%%
%%%%%%%%%%%%%%%%%%%%%%%
fprintf('\n Case b: g = 0\n');
xcaseb = solve(diff(F,dx1),diff(F,dx2),hlin,glin);
solb = [double(xcaseb.dx1) double(xcaseb.dx2) double(xcaseb.lamda) double(xcaseb.beta)];
dx1b = double(xcaseb.dx1);
dx2b = double(xcaseb.dx2);
betab = double(xcaseb.beta);
lamdab = double(xcaseb.lamda);
hlinb = double(subs(hlin,{dx1,dx2},{dx1b,dx2b}));
glinb = double(subs(glin,{dx1,dx2},{dx1b,dx2b}));
Fb = double(subs(F,{dx1,dx2,lamda,beta},{dx1b,dx2b,lamdab,betab}));
x1b = dx1b + xb(1);
x2b = dx2b + xb(2);
fv = double(subs(f,{x1,x2},{x1b,x2b}));
hv = double(subs(h,{x1,x2},{x1b,x2b}));
gv = double(subs(g,{x1,x2},{x1b,x2b}));
fprintf('Change in design vector: '),disp([dx1b dx2b]);
fprintf('The linearized quality constraint: '),disp(hlinb);
fprintf('The linearized inequality constraint: '),disp(glinb);
fprintf('New design vector: '),disp([x1b x2b]);
fprintf('The objective function: '),disp(fv);
fprintf('The equality constraint: '),disp(hv);
fprintf('The inequality constraint: '),disp(gv);
fprintf('Multiplier beta: '),disp(betab);
fprintf('Multiplier lamda: '),disp(lamdab);
if (betab > 0) & (Fb <= fbest)
xb(1) = x1b;
xb(2) = x2b;
dx1best = dx1b;
dx2best = dx2b;
end
%%% stopping criteria
if ([dx1best dx2best]*[dx1best dx2best]') <= 1.0e-08
fprintf('\n&&&&&&&&&&&&&&&&&&&&&&&&&&&&&')
fprintf('\nStopped: Design Not Changing')
fprintf('\n&&&&&&&&&&&&&&&&&&&&&&&&&&&&&\n\n')
break;
elseif i == it
fprintf('\n&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&')
fprintf('\nStpped: Number of iterations at maximum')
fprintf('\n&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&\n\n')
break;
end
end

f1, g1, h1 are still syms variable type.
Change them to numeric type using the function double() before applying
the function num2str()
You have already applied double() to the variables
gf1, gf2, gh1, gh2,gg1, gg2 and hess1
right above, no need to touch them
Here is the section you should replace
Fquadstr = strcat(num2str(f1),' + ',num2str(gf1), ...
'*','dx1',' + ',num2str(gf2),' * ','dx2', ...
' + 0.5*',num2str(hess1(1,1)),' * dx1^2', ...
' +',num2str(hess1(1,2)),' * dx1*dx2', ...
' + 0.5*',num2str(hess1(2,2)),' * dx2^2');
hlinstr = strcat(num2str(h1),' + ',num2str(gh1), ...
'*','dx1',' + ',num2str(gh2),' * ','dx2');
glinstr = strcat(num2str(g1),' + ',num2str(gg1), ...
'*','dx1',' + ',num2str(gg2),' * ','dx2');
by this
% apply double() to f1
Fquadstr = strcat(num2str(double(f1)),' + ',num2str(gf1), ...
'*','dx1',' + ',num2str(gf2),' * ','dx2', ...
' + 0.5*',num2str(hess1(1,1)),' * dx1^2', ...
' +',num2str(hess1(1,2)),' * dx1*dx2', ...
' + 0.5*',num2str(hess1(2,2)),' * dx2^2');
% apply double() to h1
hlinstr = strcat(num2str(double(h1)),' + ',num2str(gh1), ...
'*','dx1',' + ',num2str(gh2),' * ','dx2');
% apply double() to g1
glinstr = strcat(num2str(double(g1)),' + ',num2str(gg1), ...
'*','dx1',' + ',num2str(gg2),' * ','dx2');

Related

"Spotting" probability density functions of distributions programmatically (Symbolic Toolbox)

I have a joint probability density f(x,y,z) and I wish to find the conditional distribution X|Y=y,Z=z, which is equivalent to treating x as data and y and z as parameters (constants).
For example, if I have X|Y=y,Z=z being the pdf of a N(1-2y,3z^2+2), the function would be:
syms x y z
f(y,z) = 1/sqrt(2*pi*(3*z^2+2)) * exp(-1/(2*(3*z^2+2)) * (x-(1-2*y))^2);
I would like to compare it to the following:
syms mu s L a b
Normal(mu,s) = (1/sqrt(2*pi*s^2)) * exp(-1/(2*s^2) * (x-mu)^2);
Exponential(L) = L * exp(-L*x);
Gamma(a,b) = (b^a / gamma(a)) * x^(a-1)*exp(-b*x);
Beta(a,b) = (1/beta(a,b)) * x^(a-1)*(1-x)^(b-1);
Question
How do I make a program whichDistribution that would be able to print which of these four, f is equivalent to (up to proportionality) with respect to the variable x, and what are the parameters? E.g. f and x as above, the distribution is Normal, mu=1-2*y, s=3*z^2+2.
NB: there would not always be a unique solution, since some distributions are are equivalent (e.g. Gamma(1,L)==Exponential(L))
Desired outputs
syms x y z
f = 1/sqrt(2*pi*(3*z^2+2)) * exp(-1/(2*(3*z^2+2)) * (x-(1-2*y))^2)
whichDistribution(f,x) %Conditional X|Y,Z
% Normal(1-2*y,3*z^2+2)
syms x y
f = y^(1/2)*exp(-(x^2)/2 - y/2 * (1+(4-x)^2+(6-x)^2)) % this is not a pdf because it is missing a constant of proportionality, but it should still work
whichDistribution(f,x) %Conditional X|Y
% Normal(10*y/(2*y+1), 1/(2*y+1))
whichDistribution(f,y) %Conditional Y|X
% Gamma(3/2, x^2 - 10*x + 53/2)
f = exp(-x) %also missing a constant of proportionality
whichDistribution(f,x)
% Exponential(1)
f = 1/(2*pi)*exp(-(x^2)/2 - (y^2)/2)
whichDistribution(f,x)
% Normal(0,1)
whichDistribution(f,y)
% Normal(0,1)
What I have tried so far:
Using solve():
q = solve(f(y,z) == Normal(mu,s), mu, s)
Which gives wrong results, since parameters can't depend on x:
>> q.mu
ans =
(z1^2*(log((2^(1/2)*exp(x^2/(2*z1^2) - (x + 2*y - 1)^2/(6*z^2 + 4)))/(2*pi^(1/2)*(3*z^2 + 2)^(1/2))) + pi*k*2i))/x
>> q.s
ans =
z1
Attempting to simplify f(y,z) up to proportionality (in x variable) using a propto() function that I wrote:
>> propto(f(y,z),x)
ans =
exp(-(x*(x + 4*y - 2))/(2*(3*z^2 + 2)))
>> propto(Normal(mu,s),x)
ans =
exp((x*(2*mu - x))/(2*s^2))
This is almost on the money, since it is easy to spot that s^2=3*z^2 + 2 and 2*mu=-(4*y - 2), but I don't know how to deduce this programmatically.
In case it is useful: propto(f,x) attempts to simplify f by dividing f by children of f which don't involve x, and then output whichever form has the least number of children. Here is the routine:
function out = propto(f,x)
oldf = f;
newf = propto2(f,x);
while (~strcmp(char(oldf),char(newf))) % if the form of f changed, do propto2 again. When propto2(f) == f, stop
oldf = newf;
newf = propto2(oldf,x);
end
out = newf;
end
function out = propto2(f,x)
t1 = children(expand(f)); % expanded f
i1 = ~has([t1{:}],x);
out1 = simplify(f/prod([t1{i1}])); % divides expanded f by terms that do not involve x
t2 = children(f); % unexpanded f
i2 = ~has([t2{:}],x);
out2 = simplify(f/prod([t2{i2}])); % divides f by terms that do not involve x
A = [f, symlength(f); out1, symlength(out1); out2, symlength(out2)];
A = sortrows(A,2); % outputs whichever form has the fewest number of children
out = A(1,1);
end
function L = symlength(f)
% counts the number of children of f by repeatingly applying children() to itself
t = children(f);
t = [t{:}];
L = length(t);
if (L == 1)
return
end
oldt = f;
while(~strcmp(char(oldt),char(t)))
oldt = t;
t = children(t);
t = [t{:}];
t = [t{:}];
end
L = length(t);
end
edit: added desired outputs
edit2: clarified the desired function
I have managed to solve my own problem using solve() from Symbolic Toolbox. There were two issues with my original approach: I needed to set up n simultaneous equations for n parameters, and the solve() doesn't cope well with exponentials:
solve(f(3) == g(3), f(4) == g(4), mu,s)
yields no solutions, but
logf(x) = feval(symengine,'simplify',log(f),'IgnoreAnalyticConstraints');
logg(x) = feval(symengine,'simplify',log(g),'IgnoreAnalyticConstraints');
solve(logf(3) == logg(3), logf(4) == logg(4), mu,s)
yields good solutions.
Solution
Given f(x), for each PDF g(x) we attempt to solve simultaneously
log(f(r1)) == log(g(r1)) and log(f(r2)) == log(g(r2))
for some simple non-equal numbers r1, r2. Then output g for which the solution has the lowest complexity.
The code is:
function whichDist(f,x)
syms mu s L a b x0 x1 x2 v n p g
f = propto(f,x); % simplify up to proportionality
logf(x) = feval(symengine,'simplify',log(f),'IgnoreAnalyticConstraints');
Normal(mu,s,x) = propto((1/sqrt(2*pi*s)) * exp(-1/(2*s) * (x-mu)^2),x);
Exponential(L,x) = exp(-L*x);
Gamma(a,b,x) = x^(a-1)*exp(-b*x);
Beta(a,b,x) = x^(a-1)*(1-x)^(b-1);
ChiSq(v,x) = x^(v/2 - 1) * exp(-x/2);
tdist(v,x) = (1+x^2 / v)^(-(v+1)/2);
Cauchy(g,x0,x) = 1/(1+((x-x0)/g)^2);
logf = logf(x);
best_sol = {'none', inf};
r1 = randi(10); r2 = randi(10); r3 = randi(10);
while (r1 == r2 || r2 == r3 || r1 == r3) r1 = randi(10); r2 = randi(10); r3 = randi(10); end
%% check Exponential:
if (propto(logf,x) == x) % pdf ~ exp(K*x), can read off Lambda directly
soln = -logf/x;
if (~has(soln,x)) % any solution can't depend on x
fprintf('\nExponential: rate L = %s\n\n', soln);
return
end
end
%% check Chi-sq:
if (propto(logf + x/2, log(x)) == log(x)) % can read off v directly
soln = 2*(1+(logf + x/2) / log(x));
if (~has(soln,x))
dof = feval(symengine,'simplify',soln,'IgnoreAnalyticConstraints');
fprintf('\nChi-Squared: v = %s\n\n', dof);
return
end
end
%% check t-dist:
h1 = propto(logf,x);
h = simplify(exp(h1) - 1);
if (propto(h,x^2) == x^2) % pdf ~ exp(K*x), can read off Lambda directly
soln = simplify(x^2 / h);
if (~has(soln,x))
fprintf('\nt-dist: v = %s\n\n', soln);
return
end
end
h = simplify(exp(-h1) - 1); % try again if propto flipped a sign
if (propto(h,x^2) == x^2) % pdf ~ exp(K*x), can read off Lambda directly
soln = simplify(x^2 / h);
if (~has(soln,x))
fprintf('\nt-dist: v = %s\n\n', soln);
return
end
end
%% check Normal:
logn(x) = feval(symengine,'simplify',log(Normal(mu,s,x)),'IgnoreAnalyticConstraints');
% A = (x - propto(logf/x, x))/2;
% B = simplify(-x/(logf/x - mu/s)/2);
% if (~has(A,x) && ~has(B,x))
% fprintf('Normal: mu = %s, s^2 = %s', A, B);
% return
% end
logf(x) = logf;
try % attempt to solve the equation
% solve simultaneously for two random non-equal integer values r1,r2
qn = solve(logf(r1) == logn(r1), logf(r2) == logn(r2), mu, s);
catch error
end
if (exist('qn','var')) % if solve() managed to run
if (~isempty(qn.mu) && ~isempty(qn.s) && ~any(has([qn.mu,qn.s],x))) % if solution exists
complexity = symlength(qn.mu) + symlength(qn.s);
if complexity < best_sol{2} % store best solution so far
best_sol{1} = sprintf('Normal: mu = %s, s^2 = %s', qn.mu, qn.s);
best_sol{2} = complexity;
end
end
end
%% check Cauchy:
logcau(x) = feval(symengine,'simplify',log(Cauchy(g,x0,x)),'IgnoreAnalyticConstraints');
f(x) = f;
try
qcau = solve(f(r1) == Cauchy(g,x0,r1), f(r2) == Cauchy(g,x0,r2), g, x0);
catch error
end
if (exist('qcau','var'))
if (~isempty(qcau.g) && ~isempty(qcau.x0) && ~any(has([qcau.g(1),qcau.x0(1)],x)))
complexity = symlength(qcau.g(1)) + symlength(qcau.x0(1));
if complexity < best_sol{2}
best_sol{1} = sprintf('Cauchy: g = %s, x0 = %s', qcau.g(1), qcau.x0(1));
best_sol{2} = complexity;
end
end
end
f = f(x);
%% check Gamma:
logg(x) = feval(symengine,'simplify',log(Gamma(a,b,x)),'IgnoreAnalyticConstraints');
t = children(logf); t = [t{:}];
if (length(t) == 2)
if (propto(t(1),log(x)) == log(x) && propto(t(2),x) == x)
soln = [t(1)/log(x) + 1, -t(2)/x];
if (~any(has(soln,x)))
fprintf('\nGamma: shape a = %s, rate b = %s\n\n',soln);
return
end
elseif (propto(t(2),log(x)) == log(x) && propto(t(1),x) == x)
soln = [t(2)/log(x) + 1, -t(1)/x];
if (~any(has(soln,x)))
fprintf('\nGamma: shape a = %s, rate b = %s\n\n',soln);
return
end
end
end
logf(x) = logf;
try % also try using solve(), just in case.
qg = solve(logf(r1) == logg(r1), logf(r2) == logg(r2), a, b);
catch error
end
if (exist('qg','var'))
if (~isempty(qg.a) && ~isempty(qg.b) && ~any(has([qg.a,qg.b],x)))
complexity = symlength(qg.a) + symlength(qg.b);
if complexity < best_sol{2}
best_sol{1} = sprintf('Gamma: shape a = %s, rate b = %s', qg.a, qg.b);
best_sol{2} = complexity;
end
end
end
logf = logf(x);
%% check Beta:
B = feval(symengine,'simplify',log(propto(f,x-1)),'IgnoreAnalyticConstraints');
if (propto(B,log(x-1)) == log(x-1))
B = B / log(x-1) + 1;
A = f / (x-1)^(B-1);
A = feval(symengine,'simplify',log(abs(A)),'IgnoreAnalyticConstraints');
if (propto(A,log(abs(x))) == log(abs(x)))
A = A / log(abs(x)) + 1;
if (~any(has([A,B],x)))
fprintf('\nBeta1: a = %s, b = %s\n\n', A, B);
return
end
end
elseif (propto(B,log(1-x)) == log(1-x))
B = B / log(1-x);
A = simplify(f / (1-x)^(B-1));
A = feval(symengine,'simplify',log(A),'IgnoreAnalyticConstraints');
if (propto(A,log(x)) == log(x))
A = A / log(x) + 1;
if (~any(has([A,B],x)))
fprintf('\nBeta1: a = %s, b = %s\n\n', A, B);
return
end
end
end
%% Print solution with lowest complexity
fprintf('\n%s\n\n', best_sol{1});
end
Tests:
>> syms x y z
>> f = y^(1/2)*exp(-(x^2)/2 - y/2 * (1+(4-x)^2+(6-x)^2))
>> whichDist(f,x)
Normal: mu = (10*y)/(2*y + 1), s^2 = 1/(2*y + 1)
>> whichDist(f,y)
Gamma: a = 3/2, b = x^2 - 10*x + 53/2
>> Beta(a,b,x) = propto((1/beta(a,b)) * x^(a-1)*(1-x)^(b-1), x);
>> f = Beta(1/z + 7*y/(1-sqrt(z)), z/y + 1/(1-z), x)
Beta: a = -(7*y*z - z^(1/2) + 1)/(z*(z^(1/2) - 1)), b = -(y + z - z^2)/(y*(z - 1))
All correct.
Sometimes bogus answers if the parameters are numeric:
whichDist(Beta(3,4,x),x)
Beta: a = -(pi*log(2)*1i + pi*log(3/10)*1i - log(2)*log(3/10) + log(2)*log(7/10) - log(3/10)*log(32) + log(2)*log(1323/100000))/(log(2)*(log(3/10) - log(7/10))), b = (pi*log(2)*1i + pi*log(7/10)*1i + log(2)*log(3/10) - log(2)*log(7/10) - log(7/10)*log(32) + log(2)*log(1323/100000))/(log(2)*(log(3/10) - log(7/10)))
So there is room for improvement and I will still award bounty to a better solution than this.
Edit: Added more distributions. Improved Gamma and Beta distribution identifications by spotting them directly without needing solve().

Berlekamp Massey Algorithm for BCH simplified binary version

I am trying to follow Lin, Costello's explanation of the simplified BM algorithm for the binary case in page 210 of chapter 6 with no success on finding the error locator polynomial.
I'm trying to implement it in MATLAB like this:
function [locator_polynom] = compute_error_locator(syndrome, t, m, field, alpha_powers)
%
% Initial conditions for the BM algorithm
polynom_length = 2*t;
syndrome = [syndrome; zeros(3, 1)];
% Delta matrix storing the powers of alpha in the corresponding place
delta_rho = uint32(zeros(polynom_length, 1)); delta_rho(1)=1;
delta_next = uint32(zeros(polynom_length, 1));
% Premilimnary values
n_max = uint32(2^m - 1);
% Initialize step mu = 1
delta_next(1) = 1; delta_next(2) = syndrome(1); % 1 + S1*X
% The discrepancy is stored in polynomial representation as uint32 numbers
value = gf_mul_elements(delta_next(2), syndrome(2), field, alpha_powers, n_max);
discrepancy_next = bitxor(syndrome(3), value);
% The degree of the locator polynomial
locator_degree_rho = 0;
locator_degree_next = 1;
% Update all values
locator_polynom = delta_next;
delta_current = delta_next;
discrepancy_rho = syndrome(1);
discrepancy_current = discrepancy_next;
locator_degree_current = locator_degree_next;
rho = 0; % The row with the maximum value of 2mu - l starts at 1
for i = 1:t % Only the even steps are needed (so make t out of 2*t)
if discrepancy_current ~= 0
% Compute the correction factor
correction_factor = uint32(zeros(polynom_length, 1));
x_exponent = 2*(i - rho);
if (discrepancy_current == 1 || discrepancy_rho == 1)
d_mu_times_rho = discrepancy_current * discrepancy_rho;
else
alpha_discrepancy_mu = alpha_powers(discrepancy_current);
alpha_discrepancy_rho = alpha_powers(discrepancy_rho);
alpha_inver_discrepancy_rho = n_max - alpha_discrepancy_rho;
% The alpha power for dmu * drho^-1 is
alpha_d_mu_times_rho = alpha_discrepancy_mu + alpha_inver_discrepancy_rho;
% Equivalent to aux mod(2^m - 1)
alpha_d_mu_times_rho = alpha_d_mu_times_rho - ...
n_max * uint32(alpha_d_mu_times_rho > n_max);
d_mu_times_rho = field(alpha_d_mu_times_rho);
end
correction_factor(x_exponent+1) = d_mu_times_rho;
correction_factor = gf_mul_polynoms(correction_factor,...
delta_rho,...
field, alpha_powers, n_max);
% Finally we add the correction factor to get the new delta
delta_next = bitxor(delta_current, correction_factor(1:polynom_length));
% Update used data
l = polynom_length;
while delta_next(l) == 0 && l>0
l = l - 1;
end
locator_degree_next = l-1;
% Update previous maximum if the degree is higher than recorded
if (2*i - locator_degree_current) > (2*rho - locator_degree_rho)
locator_degree_rho = locator_degree_current;
delta_rho = delta_current;
discrepancy_rho = discrepancy_current;
rho = i;
end
else
% If the discrepancy is 0, the locator polynomial for this step
% is passed to the next one. It satifies all newtons' equations
% until now.
delta_next = delta_current;
end
% Compute the discrepancy for the next step
syndrome_start_index = 2 * i + 3;
discrepancy_next = syndrome(syndrome_start_index); % First value
for k = 1:locator_degree_next
value = gf_mul_elements(delta_next(k + 1), ...
syndrome(syndrome_start_index - k), ...
field, alpha_powers, n_max);
discrepancy_next = bitxor(discrepancy_next, value);
end
% Update all values
locator_polynom = delta_next;
delta_current = delta_next;
discrepancy_current = discrepancy_next;
locator_degree_current = locator_degree_next;
end
end
I'm trying to see what's wrong but I can't. It works for the examples in the book, but not always. As an aside, to compute the discrepancy S_2mu+3 is needed, but when I have only 24 syndrome coefficients how is it computed on step 11 where 2*11 + 3 is 25?
Thanks in advance!
It turns out the code is ok. I made a different implementation from Error Correction and Coding. Mathematical Methods and gives the same result. My problem is at the Chien Search.
Code for the interested:
function [c] = compute_error_locator_v2(syndrome, m, field, alpha_powers)
%
% Initial conditions for the BM algorithm
% Premilimnary values
N = length(syndrome);
n_max = uint32(2^m - 1);
polynom_length = N/2 + 1;
L = 0; % The curent length of the LFSR
% The current connection polynomial
c = uint32(zeros(polynom_length, 1)); c(1) = 1;
% The connection polynomial before last length change
p = uint32(zeros(polynom_length, 1)); p(1) = 1;
l = 1; % l is k - m, the amount of shift in update
dm = 1; % The previous discrepancy
for k = 1:2:N % For k = 1 to N in steps of 2
% ========= Compute discrepancy ==========
d = syndrome(k);
for i = 1:L
aux = gf_mul_elements(c(i+1), syndrome(k-i), field, alpha_powers, n_max);
d = bitxor(d, aux);
end
if d == 0 % No change in polynomial
l = l + 1;
else
% ======== Update c ========
t = c;
% Compute the correction factor
correction_factor = uint32(zeros(polynom_length, 1));
% This is d * dm^-1
dd_sum = modulo(alpha_powers(d) + n_max - alpha_powers(dm), n_max);
for i = 0:polynom_length - 1
if p(i+1) ~= 0
% Here we compute d*d^-1*p(x_i)
ddp_sum = modulo(dd_sum + alpha_powers(p(i+1)), n_max);
if ddp_sum == 0
correction_factor(i + l + 1) = 1;
else
correction_factor(i + l + 1) = field(ddp_sum);
end
end
end
% Finally we add the correction factor to get the new locator
c = bitxor(c, correction_factor);
if (2*L >= k) % No length change in update
l = l + 1;
else
p = t;
L = k - L;
dm = d;
l = 1;
end
end
l = l + 1;
end
end
The code comes from this implementation of the Massey algorithm

Error in evaluating a function

EDIT: The code that I have pasted is too long. Basicaly I dont know how to work with the second code, If I know how calculate alpha from the second code I think my problem will be solved. I have tried a lot of input arguments for the second code but it does not work!
I have written following code to solve a convex optimization problem using Gradient descend method:
function [optimumX,optimumF,counter,gNorm,dx] = grad_descent()
x0 = [3 3]';%'//
terminationThreshold = 1e-6;
maxIterations = 100;
dxMin = 1e-6;
gNorm = inf; x = x0; counter = 0; dx = inf;
% ************************************
f = #(x1,x2) 4.*x1.^2 + 2.*x1.*x2 +8.*x2.^2 + 10.*x1 + x2;
%alpha = 0.01;
% ************************************
figure(1); clf; ezcontour(f,[-5 5 -5 5]); axis equal; hold on
f2 = #(x) f(x(1),x(2));
% gradient descent algorithm:
while and(gNorm >= terminationThreshold, and(counter <= maxIterations, dx >= dxMin))
g = grad(x);
gNorm = norm(g);
alpha = linesearch_strongwolfe(f,-g, x0, 1);
xNew = x - alpha * g;
% check step
if ~isfinite(xNew)
display(['Number of iterations: ' num2str(counter)])
error('x is inf or NaN')
end
% **************************************
plot([x(1) xNew(1)],[x(2) xNew(2)],'ko-')
refresh
% **************************************
counter = counter + 1;
dx = norm(xNew-x);
x = xNew;
end
optimumX = x;
optimumF = f2(optimumX);
counter = counter - 1;
% define the gradient of the objective
function g = grad(x)
g = [(8*x(1) + 2*x(2) +10)
(2*x(1) + 16*x(2) + 1)];
end
end
As you can see, I have commented out the alpha = 0.01; part. I want to calculate alpha via an other code. Here is the code (This code is not mine)
function alphas = linesearch_strongwolfe(f,d,x0,alpham)
alpha0 = 0;
alphap = alpha0;
c1 = 1e-4;
c2 = 0.5;
alphax = alpham*rand(1);
[fx0,gx0] = feval(f,x0,d);
fxp = fx0;
gxp = gx0;
i=1;
while (1 ~= 2)
xx = x0 + alphax*d;
[fxx,gxx] = feval(f,xx,d);
if (fxx > fx0 + c1*alphax*gx0) | ((i > 1) & (fxx >= fxp)),
alphas = zoom(f,x0,d,alphap,alphax);
return;
end
if abs(gxx) <= -c2*gx0,
alphas = alphax;
return;
end
if gxx >= 0,
alphas = zoom(f,x0,d,alphax,alphap);
return;
end
alphap = alphax;
fxp = fxx;
gxp = gxx;
alphax = alphax + (alpham-alphax)*rand(1);
i = i+1;
end
function alphas = zoom(f,x0,d,alphal,alphah)
c1 = 1e-4;
c2 = 0.5;
[fx0,gx0] = feval(f,x0,d);
while (1~=2),
alphax = 1/2*(alphal+alphah);
xx = x0 + alphax*d;
[fxx,gxx] = feval(f,xx,d);
xl = x0 + alphal*d;
fxl = feval(f,xl,d);
if ((fxx > fx0 + c1*alphax*gx0) | (fxx >= fxl)),
alphah = alphax;
else
if abs(gxx) <= -c2*gx0,
alphas = alphax;
return;
end
if gxx*(alphah-alphal) >= 0,
alphah = alphal;
end
alphal = alphax;
end
end
But I get this error:
Error in linesearch_strongwolfe (line 11) [fx0,gx0] = feval(f,x0,d);
As you can see I have written the f function and its gradient manually.
linesearch_strongwolfe(f,d,x0,alpham) takes a function f, Gradient of f, a vector x0 and a constant alpham. is there anything wrong with my declaration of f? This code works just fine if I put back alpha = 0.01;
As I see it:
x0 = [3; 3]; %2-element column vector
g = grad(x0); %2-element column vector
f = #(x1,x2) 4.*x1.^2 + 2.*x1.*x2 +8.*x2.^2 + 10.*x1 + x2;
linesearch_strongwolfe(f,-g, x0, 1); %passing variables
inside the function:
[fx0,gx0] = feval(f,x0,-g); %variable names substituted with input vars
This will in effect call
[fx0,gx0] = f(x0,-g);
but f(x0,-g) is a single 2-element column vector with these inputs. Assingning the output to two variables will not work.
You either have to define f as a proper named function (just like grad) to output 2 variables (one for each component), or edit the code of linesearch_strongwolfe to return a single variable, then slice that into 2 separate variables yourself afterwards.
If you experience a very rare kind of laziness and don't want to define a named function, you can still use an anonymous function at the cost of duplicating code for the two components (at least I couldn't come up with a cleaner solution):
f = #(x1,x2) deal(4.*x1(1)^2 + 2.*x1(1)*x2(1) +8.*x2(1)^2 + 10.*x1(1) + x2(1),...
4.*x1(2)^2 + 2.*x1(2)*x2(2) +8.*x2(2)^2 + 10.*x1(2) + x2(2));
[fx0,gx0] = f(x0,-g); %now works fine
as long as you always have 2 output variables. Note that this is more like a proof of concept, since this is ugly, inefficient, and very susceptible to typos.

Matrix calculation gets slower after each iteration in matlab

I have a 1024*1024*51 matrix. I'll do calculations to change some value of the matrix within for loops (change the value of matrix for each iteration). I find that the computing speed gets slower and slower and finally my computer gets into trouble.However the size of the matrix doesn't change. Anyone can shed some light on this problem?
function ActiveContours3D(method,grad,im,mu,nu,lambda1,lambda2,TimeSteps)
epsilon = 10e-10;
tic
fid=fopen('Chr18_z_25of25tiles-C=0_c0_n000.raw','rb','ieee-le');
Xdim = 1024;
Ydim = 1024;
Zdim = 51;
A = fread(fid,[Xdim Ydim*Zdim],'int16');
A = double(A);
size_of_A = size(A)
for(i=1:Zdim)
u0_color(:,:,i) = A(1 : Xdim , (i-1)*Ydim+1 : i*Ydim);
end
fclose(fid)
time = toc
[M,N,P,color] = size(u0_color);
size(u0_color );
u0_color = double(u0_color); % Convert u0_color values to double;
u0 = u0_color(:,:,:,1); % Define the Grayscale volumetric image.
u0_color = uint8(u0_color); % Necessary for color visualization
x = 1:M;
y = 1:N;
z = 1:P;
dx = 1
dy = 1
dz = 1
dim_approx = 2*M*N*P / sqrt(M*N*P);
if(method == 'Explicit')
dt = 0.9 / ((2*mu/(dx^2)) + (2*mu/(dy^2)) + (2*mu/(dz^2))) % 90% CFL
elseif(method == 'Implicit')
dt = (10e7) * 0.9 / ((2*mu/(dx^2)) + (2*mu/(dy^2)) + (2*mu/(dz^2)))
end
[X,Y,Z] = meshgrid(x,y,z);
x0 = (M+1)/2;
y0 = (N+1)/2;
z0 = (P+1)/2;
r0 = min(min(M,N),P)/3;
phi = sqrt((X-x0).^2 + (Y-y0).^2 + (Z-z0).^2) - r0;
phi_visualize = phi; % Use this for visualization in 3D
phi = permute(phi,[2,1,3]); % Use this for computations in 3D
write_to_binary_file(phi_visualize,0); % record initial conditions
tic
for(n=1:TimeSteps)
n
c1 = C1_3d(u0,phi);
c2 = C2_3d(u0,phi);
% x
phi_xp = [phi(2:M,:,:); phi(M,:,:)]; % vertical concatenation
phi_xm = [phi(1,:,:); phi(1:M-1,:,:)]; % (since x values are rows)
% cat(1,A,B) is the same as [A;B]
Dx_m = (phi - phi_xm)/dx; % first derivatives
Dx_p = (phi_xp - phi)/dx;
Dxx = (Dx_p - Dx_m)/dx; % second derivative
% y
phi_yp = [phi(:,2:N,:) phi(:,N,:)]; % horizontal concatenation
phi_ym = [phi(:,1,:) phi(:,1:N-1,:)]; % (since y values are columns)
% cat(2,A,B) is the same as [A,B]
Dy_m = (phi - phi_ym)/dy;
Dy_p = (phi_yp - phi)/dy;
Dyy = (Dy_p - Dy_m)/dy;
% z
phi_zp = cat(3,phi(:,:,2:P),phi(:,:,P));
phi_zm = cat(3,phi(:,:,1) ,phi(:,:,1:P-1));
Dz_m = (phi - phi_zm)/dz;
Dz_p = (phi_zp - phi)/dz;
Dzz = (Dz_p - Dz_m)/dz;
% x,y,z
Dx_0 = (phi_xp - phi_xm) / (2*dx);
Dy_0 = (phi_yp - phi_ym) / (2*dy);
Dz_0 = (phi_zp - phi_zm) / (2*dz);
phi_xp_yp = [phi_xp(:,2:N,:) phi_xp(:,N,:)];
phi_xp_ym = [phi_xp(:,1,:) phi_xp(:,1:N-1,:)];
phi_xm_yp = [phi_xm(:,2:N,:) phi_xm(:,N,:)];
phi_xm_ym = [phi_xm(:,1,:) phi_xm(:,1:N-1,:)];
phi_xp_zp = cat(3,phi_xp(:,:,2:P),phi_xp(:,:,P));
phi_xp_zm = cat(3,phi_xp(:,:,1) ,phi_xp(:,:,1:P-1));
phi_xm_zp = cat(3,phi_xm(:,:,2:P),phi_xm(:,:,P));
phi_xm_zm = cat(3,phi_xm(:,:,1) ,phi_xm(:,:,1:P-1));
phi_yp_zp = cat(3,phi_yp(:,:,2:P),phi_yp(:,:,P));
phi_yp_zm = cat(3,phi_yp(:,:,1) ,phi_yp(:,:,1:P-1));
phi_ym_zp = cat(3,phi_ym(:,:,2:P),phi_ym(:,:,P));
phi_ym_zm = cat(3,phi_ym(:,:,1) ,phi_ym(:,:,1:P-1));
if(grad == 'Dirac')
Grad = DiracDelta(phi); % Dirac delta
%Grad = 1;
elseif(grad == 'Grad ')
Grad = (((Dx_0.^2)+(Dy_0.^2)+(Dz_0.^2)).^(1/2)); % |grad phi|
end
if(method == 'Explicit')
% CURVATURE: *mu*k|grad phi|* (central differences):
K = zeros(M,N,P);
Dxy = (phi_xp_yp - phi_xp_ym - phi_xm_yp + phi_xm_ym) / (4*dx*dy);
Dxz = (phi_xp_zp - phi_xp_zm - phi_xm_zp + phi_xm_zm) / (4*dx*dz);
Dyz = (phi_yp_zp - phi_yp_zm - phi_ym_zp + phi_ym_zm) / (4*dy*dz);
K = ( (Dx_0.^2).*Dyy - 2*Dx_0.*Dy_0.*Dxy + (Dy_0.^2).*Dxx ...
+ (Dx_0.^2).*Dzz - 2*Dx_0.*Dz_0.*Dxz + (Dz_0.^2).*Dxx ...
+ (Dy_0.^2).*Dzz - 2*Dy_0.*Dz_0.*Dyz + (Dz_0.^2).*Dyy) ./ ((Dx_0.^2 + Dy_0.^2 + Dz_0.^2).^(3/2) + epsilon);
phi_temp = phi + dt * Grad .* ( mu.*K + lambda1*(u0 - c1).^2 - lambda2*(u0 - c2).^2 );
elseif(method == 'Implicit')
C1x = 1 ./ sqrt(Dx_p.^2 + Dy_0.^2 + Dz_0.^2 + (10e-7)^2);
C2x = 1 ./ sqrt(Dx_m.^2 + Dy_0.^2 + Dz_0.^2 + (10e-7)^2);
C3y = 1 ./ sqrt(Dx_0.^2 + Dy_p.^2 + Dz_0.^2 + (10e-7)^2);
C4y = 1 ./ sqrt(Dx_0.^2 + Dy_m.^2 + Dz_0.^2 + (10e-7)^2);
C5z = 1 ./ sqrt(Dx_0.^2 + Dy_0.^2 + Dz_p.^2 + (10e-7)^2);
C6z = 1 ./ sqrt(Dx_0.^2 + Dy_0.^2 + Dz_m.^2 + (10e-7)^2);
% m = (dt/(dx*dy)) * Grad .* mu; % 2D
m = (dt/(dx*dy)) * Grad .* mu;
C = 1 + m.*(C1x + C2x + C3y + C4y + C5z + C6z);
C1x_2x = C1x.*phi_xp + C2x.*phi_xm;
C3y_4y = C3y.*phi_yp + C4y.*phi_ym;
C5z_6z = C5z.*phi_zp + C6z.*phi_zm;
phi_temp = (1 ./ C) .* ( phi + m.*(C1x_2x+C3y_4y) + (dt*Grad).*(lambda1*(u0 - c1).^2) - (dt*Grad).*(lambda2*(u0 - c2).^2) );
end
phi = phi_temp;
phi_visualize = permute(phi,[2,1,3]);
write_to_binary_file(phi_visualize,n); % record
end
time = toc
n = n
T = dt*n
clear
clear all
In general Matlab keeps track of all the variables in the form of matrix. When you work with lot of variables with many dimensions the RAM memory will be allocated for storing this variable. Hence on working with lots of variables that is gonna run for multiple iterations it is better to clear the variable from the memory. To do so use the command
clear variable_name_1, variable_name_2,... variable_name_3;
Although keeping all the variables keeps the code to look organised, however when you face such issues try clearing the unwanted variables.
Check this link to use clear command in detail: http://www.mathworks.in/help/matlab/ref/clear.html

Integral Calculation

I would like to perform what follows:
where
The parameters shown in the latter figure can be obtained as follows:
%% Inizialization
time = 614.4; % Analysis Time
Uhub = 11;
HubHt = 90;
alpha = 0.14;
TI = 'A'; % Turbulent Intensity (A,B,C as in the IEC or Specific value)
N1 = 4096;
N2 = 32;
N3 = 32;
N = N1*N2*N3; % Total Number of Point
t = 0:(time/(N1-1)):time; % Sampled Time Vector
L1 = Uhub*time; % Box length along X
L2 = 150; % Box length along Y
L3 = 220; % Box length along Z
dx = L1/N1; % Grid Resolution along X-axis
dy = L2/N2; % Grid Resolution along Y-axis
dz = L3/N3; % Grid Resolution along Z-axis
V = L1*L2*L3; % Analysis Box Volume
gamma = 3.9; % Turbulent Eddies Distorsion Factor
c = 1.476;
b = 5.6;
if HubHt < 60
lambda1 = 0.7*HubHt;
else
lambda1 = 42;
end
L = 0.8*lambda1;
if isequal(TI,'A')
Iref = 0.16;
sigma1 = Iref*(0.75*Uhub + b);
elseif isequal(TI,'B')
Iref = 0.14;
sigma1 = Iref*(0.75*Uhub + b);
elseif isequal(TI,'C')
Iref = 0.12;
sigma1 = Iref*(0.75*Uhub + b);
else
sigma1 = str2num(TI)*Uhub/100;
end
sigma_iso = 0.55*sigma1;
sigma2 = 0.7*sigma1;
sigma3 = 0.5*sigma1;
%% Wave number vectors
ik1 = cat(2,(-N1/2:-1/2),(1/2:N1/2));
ik2 = -N2/2:N2/2-1;
ik3 = -N3/2:N3/2-1;
[x y z] = ndgrid(ik1,ik2,ik3);
k1 = reshape((2*pi*L/L1)*x,N1*N2*N3,1);
k2 = reshape((2*pi*L/L2)*y,N1*N2*N3,1);
k3 = reshape((2*pi*L/L3)*z,N1*N2*N3,1);
k = sqrt(k1.^2 + k2.^2 + k3.^2);
%% Calculation of beta by means of the Energy Spectrum Integration
E = #(k) (1.453*k.^4)./((1 + k.^2).^(17/6));
%//Independent integration on segments
Local_int = arrayfun(#(i)quad(E,i-1,i), 2:(N1*N2*N3));
%//integral additivity + cumulative removal of queues
E_int = 1.5 - [0 fliplr(cumsum(fliplr(Local_int)))]; %//To remove queues
E_int = reshape(E_int,N,1);
S = k.*sqrt(E_int);
beta = (c*gamma)./S;
%% Help Parameters
k30 = k3 + beta.*k1;
k0 = sqrt(k1.^2 + k2.^2 + k30.^2);
C1 = (beta.*k1.^2.*(k1.^2 + k2.^2 - k3.*k30))./(k.^2.*(k1.^2 + k2.^2));
C2 = (k2.*k0.^2./((k1.^2 + k2.^2).^(3/2))).*atan2((beta.*k1.*sqrt(k1.^2 + k2.^2)),(k0.^2 - k30.*k1.*beta));
xhsi1 = C1 - (k2./k1).*C2;
xhsi2 = (k2./k1).*C1 + C2;
E_k0 = (1.453*k0.^4)./((1 + k0.^2).^(17/6));
For instance, typing in
phi_33 = #(k2,k3) (E_k0./(4*pi.*k.^4)).*((k1.^2 + k2.^2));
F_33 = arrayfun(#(i) dblquad(phi_33,k3(i),k3(i+1),k2(i),k2(i+1)), 1:((N1*N2*N3)-1));
Matlab retrieves the following error msg:
Error using +
Matrix dimensions must agree.
Error in #(k2,k3)(E_k0./(4*pi.*k.^4)).*((k1.^2+k2.^2))
Do you have a clue how to overcome this issue?
I really look forward to hearing from you.
Best regards,
FPE
The error is easily explained:
First you define E_k0 then you try to call Ek0.
phi_11 = #(k1,k2,k3) (E_k0./4*pi.*kabs.^4).*(k0abs.^2 - k1.^2 - 2*k1.*k03.*xhsi1 + (k1.^2 + k2.^2).*xhsi1.^2);
I solved it this way:
Code a function for each of the PHI elements, such as (for PHI11)
function phi_11 = phi_11_new(k1,k2,k3,beta,i)
k = sqrt(k1(i).^2 + k2.^2 + k3.^2);
k30 = k3 + beta(i).*k1(i);
k0 = sqrt(k1(i).^2 + k2.^2 + k30.^2);
E_k0 = 1.453.*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta(i).k1(i).^2).(k1(i).^2 + k2.^2 - k3.k30)./(k.^2.(k1(i).^2 + k2.^2));
C2 = k2.*k0.^2./((k1(i).^2 + k2.^2).^(3/2)).*atan2((beta(i).*k1(i).*sqrt(k1(i).^2 + k2.^2)),(k0.^2 - k30.*k1(i).*beta(i)));
xhsi1 = C1 - k2./k1(i).*C2;
xhsi1_q = xhsi1.^2;
phi_11 = E_k0./(4.*pi.k0.^4).(k0.^2 - k1(i).^2 - 2.*k1(i).*k30.*xhsi1 + (k1(i).^2 + k2.^2).*xhsi1_q);
end
I recall this function within the main code as follows
for l = 1:numel(k1)
phi11 = #(k2,k3) phi_11(k1,k2,k3,l)
F11(l) = integral2(phi,-1000,1000,-1000,1000);
end
at it seems it works.