I'm trying to calculate a double integral but get an error.
My code is:
clear
Gamma = rand([22,1]);
data_i = rand([1,10]);
Y_1 = 1;
fun = #(D_star, Y_0 ) fun1(D_star , Y_1 , Y_0 , Gamma , data_i);
p = integral2 ( fun , 0 , 1 , 0 , 1);
and the function fun1 is defined as
function [p] = fun1(D_star , Y_1 , Y_0 , Gamma , data_i)
gamma=Gamma(1:3);
beta_1=Gamma(4:5);
beta_0=Gamma(6:7);
alpha_D=Gamma(8);
alpha_1=Gamma(9);
alpha_0=Gamma(10);
sigma2_1=Gamma(11);
sigma2_0=Gamma(12);
Lambda=Gamma(13:17);
sigma2_M=Gamma(18:22);
Sigma2_temp = [1 ; sigma2_1 ; sigma2_0 ; sigma2_M];
Alpha = [alpha_D ; alpha_1 ; alpha_0 ; Lambda];
Sigma2 = Alpha * Alpha' + diag(Sigma2_temp);
Z = data_i(3:5);
X = data_i(3:4);
M = data_i(6:10);
Mu = [Z*gamma , X*beta_1 , X*beta_0 , zeros(1,5) ]';
YY = [D_star ; Y_1 ; Y_0 ; M' ];
p = mvnpdf(YY,Mu,Sigma2);
end
I don't think there is a problem in the definition of the functions because I can evaluate them (e.g. fun(1,1) give me an answer). The error message that I get is:
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
Error in fun1 (line 23)
YY = [D_star ; Y_1 ; Y_0 ; M' ];
Error in #(D_star,Y_0)fun1(D_star,Y_1,Y_0,Gamma,data(n,:))
Error in integral2Calc>integral2t/tensor (line 228)
Z = FUN(X,Y); NFE = NFE + 1;
Error in integral2Calc>integral2t (line 55)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 9)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 106)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
I tried also the command -quad2d- instead of -integral2- and got a similar error. Any ideas?
I found the problem. According to the documentation, when using -integral2-, the function inside the integral must get an array of both variables as an input and return an array as an output.
Related
I am trying to minimise the expected value of this function w.r.t the variable Q :
where f(D) is the probability density function of a multivariate normal distribution, c0 and cu parameters.
This is what I have done so far :
syms X1
syms X2
syms X3
c0 = 1 ;
cu = 1 ;
x = [X1; X2; X3] ;
mu = [3; 5; 7 ] ;
sigma = [4,3,6;3,8,5;6,5,10] ;
s = mvnpdf(x,mu,sigma)
syms Q
syms D
f = (Q-D).*s ;
z = c0.*int(f,D,0,Q) ;
anon_z = matlabFunction(z) ;
g = (D-Q).*s ;
t = cu.*int(g,D,Q,inf) ;
anon_t = matlabFunction(t) ;
tot = #(Q) anon_z(Q) + anon_t(Q) ;
[x fx] = fminsearch(tot,0)
It gives me this error :
Error in
symengine>#(Q,X1,X2,X3)Q.^2.*exp(conj(X1).conj(X3).(3.0./2.0)).*exp(conj(X2).conj(X3).(1.0./1.1e1)).exp(conj(X1).-3.0).exp(conj(X2).(3.0./1.1e1)).exp(conj(X3).(2.6e1./1.1e1)).exp(conj(X1).^2.(-5.0./4.0)).exp(conj(X2).^2.(-1.0./1.1e1)).exp(conj(X3).^2.(-2.3e1./4.4e1)).exp(-8.75688228083863).(1.0./2.0)
Error in #(Q)anon_z(Q)+anon_t(Q)
Error in fminsearch (line 189) fv(:,1) = funfcn(x,varargin{:});
Error in multi_3giocatori (line 19) [x fx] = fminsearch(tot,0)
Can you help me fix this?
I get an error when taking the gradient of a symbolic function. Can anyone tell me why I get this error?
syms x1 x2 R
L0=0; %[m]
k1=8; %[N/m]
k2=4; %[N/m]
F1=5; %[N]
F2=10; %[N]
F = 0.5*k1*(sqrt(x1^2 + (L0-x2)^2) - L0)^2 + 0.5*k2*(sqrt(x1^2 + (L0+x2)^2)
- L0)^2 - F1*x1 - F2*x2;
f = matlabFunction(F);
R = 0.1;
GI = x1^2 + x2^2 - R^2;
gi = matlabFunction(GI);
epsilon=0.001;
xo=[0,0]';
k = 0;
r = (sqrt(5)-1) / 2;
rpe=0.01;
Merit_pe = #(x1,x2) f(x1,x2) + rpe*(max(0,gi(x1,x2)))^2;
g = gradient(Merit_pe, [x1 x2]);
Error:
Error using sym/max (line 97)
Input arguments must be convertible to
floating-point numbers.
Error in
Ex>#(x1,x2)f(x1,x2)+rpe*(max(0,gi(x1,x2)))^2
Error in sym>funchandle2ref (line 1249)
S = x(S{:});
Error in sym>tomupad (line 1154)
x = funchandle2ref(x);
Error in sym (line 163)
S.s = tomupad(x);
Error in sym/gradient (line 17)
args = privResolveArgs(sym(f));
Error in Ex (line 31)
g = gradient(Merit_pe, [x1 x2])
I think the max part is causing me trouble, but I still need to determine the gradient of this function. Any advise? (I could do it by hand I guess, but I'd rather not if I don't have to)
Directly taking the gradient of max(0,gi(x1,x2)) is not possible in matlab. Instead the function should be defined according to the following definition.
The function Merit_pe can then be defined as follows:
if gi(xo(1,1),xo(2,1)) > 0
Merit_pe = #(x1,x2) f(x1,x2) + rpe*(gi(x1,x2))^2;
else
Merit_pe = #(x1,x2) f(x1,x2);
end
The gradient can then be determined using:
g = gradient(Merit_pe, [x1 x2]);
I got 2 questions here.
1. If I have y = a*x^2 + 5. What function can make it into y = a.*x.^2 +5. As you seen, dot was inserted.
It's easy, but kinda diffcult to describe, but please have patience with me. Thank you so much.
First, let me make a very simple example of my problem.
If I want to calucate Y = F(x=1)+ 2.^2, and I know F(x=1) = a+b,(a and b are syms). This means Y = a + b + 4. Problem is here,
Matlab give me error if I write down as.
F = function( .... ); <==== output of function is F(X=1), and = F(x=1) = a+b
Y = integral2( F + 2.^2, .. , .. ,..)
However, if I just copy the output of F as
Y = integral2( a+b + 2.^2, .. , .. ,..)
Now it works!!!
Ok. Please allow me to talk about my code here. I am trying to find a double interation by using integral2. One part of my equcation(which is Y) is from another int output(which is F). Matlab will give ERROR for code below:
clear all;
a=4;
la1=1/(pi*500^2); la2= la1*5;
p1=25; p2=p1/25;
sgma2=10^(-11);
index=1;
g=2./a;
syms r u1 u2
powe= -2 ;
seta= 10^powe;
xNor = ( (u2./u1).^(a./2) + 1 ).^(2./a);
x = (xNor).^(0.5) * seta^(-1/a);
fun1 = r./(1+ r.^a );
out1 = int(fun1, x, Inf) ; %== This is my F in my example
q=pi.*(la1.*p1.^(2./a)+la2.*p2.^(2./a));
yi = #(u2,u1) exp(-u2.*(1+2.*...
( out1 )./... %=== out1 is the problem here.
( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...
exp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...
((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );
maxF =#(u2) u2;
out2 = integral2(yi,0,Inf,0 ,maxF) % == this is Y in my previous example.
However, since I know the out1 = pi/4 - atan(10*(u2^2/u1^2 + 1)^(1/2))/2 (no dot,1/2, not 1./2). Instead of writing down out1, I will just type the equaction and add dot in the
yi = #(u2,u1) exp(-u2.*(1+2.*...
( pi./4 - atan(10.*(u2.^2./u1.^2 + 1).^(1./2))./2 )./... %=== not "out1"
( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...
exp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...
((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );
Now the code is working!!!! The final output is = 0.9957.
Dear friends, I already spend a long time on this, but I still can not find out the problem. Could you please take a deeper look for me. Please copy the code to you matlab and test. Thank you so much.
Below is the error given by matlab, if I just use "out1" in yi = #(u2,u1) ......
Error using integralCalc/finalInputChecks (line 511)
Input function must return 'double' or 'single' values. Found 'sym'.
Error in integralCalc/iterateScalarValued (line 315)
finalInputChecks(x,fx);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in
integral2Calc>#(xi,y1i,y2i)integralCalc(#(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions)
(line 18)
innerintegral = #(x)arrayfun(#(xi,y1i,y2i)integralCalc( ...
Error in
integral2Calc>#(x)arrayfun(#(xi,y1i,y2i)integralCalc(#(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions),x,ymin(x),ymax(x))
(line 18)
innerintegral = #(x)arrayfun(#(xi,y1i,y2i)integralCalc( ...
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 84)
[q,errbnd] = vadapt(#AToInfInvTransform,interval);
Error in integral2Calc>integral2i (line 21)
[q,errbnd] = integralCalc(innerintegral,xmin,xmax,opstruct.integralOptions);
Error in integral2Calc (line 8)
[q,errbnd] = integral2i(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 107)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Error in ref7_equ11n2 (line 129)
out2 = integral2(yi,0,Inf,0 ,maxF)
The documentation of the function integral2 states:
integral2 - Numerically evaluate double integral
You won't be able to use integral2 with symbolic expressions, unless you convert them to function handles using matlabfunction.
I am not so much experiences with Matlab. I just need it for the sake of solving some lengthy non-linear equations. Instead of using fzero, I wanna use Newton-Raphson's to solve the equation.
newton.m file contains the following code.
function [ x, ex ] = newton( f, df, x0, tol, nmax )
if nargin == 3
tol = 1e-4;
nmax = 1e1;
elseif nargin == 4
nmax = 1e1;
elseif nargin ~= 5
error('newton: invalid input parameters');
end
x(1) = x0 - (f(x0)/df(x0));
ex(1) = abs(x(1)-x0);
k = 2;
while (ex(k-1) >= tol) && (k <= nmax)
x(k) = x(k-1) - (f(x(k-1))/df(x(k-1)));
ex(k) = abs(x(k)-x(k-1));
k = k+1;
end
end
And in the main file, I have called this function as follows:
ext_H = newton( exp(x) + x^3, diff(exp(x) + x^3), 9, 0.5*10^-5, 10);
When I run this function, it gives me the following error.
Error using sym/subsref (line 9)
Error using maplemex
Error, (in MTM:-subsref) Array index out of range
Error in newton (line 37)
x(1) = x0 - (f(x0)/df(x0));
Error in main (line 104)
ext_H = newton( exp(x) + x^3, diff(exp(x) + x^3), 9, 0.5*10^-5, 10);
Could anyone please help me to get through this?
You can probably use Matlab's fsolve (http://uk.mathworks.com/help/optim/ug/fsolve.html) with a couple of options
x0 = 9;
options = optimoptions('fsolve','Algorithm','levenberg-marquardt','TolFun',5*10^-6,'MaxIter',100);
x = fsolve(#(x)(exp(x) + x^3), x0,options);
or just fzero (http://uk.mathworks.com/help/optim/ug/fzero.html) which implements BD algorithm
x = fzero(#(x)(exp(x) + x^3), x0);
I want to solve equations in matlab, eg.
100+a/2=173*cos(b)
sqrt(3)*a/2=173*sin(b）
and the code would be:
[a,b]=solve('100+a/2=173*cos(b)','sqrt(3)*a/2=173*sin(b)','a','b')
However, if I want to take 100 as a variable, like
for k=1:100
[a,b]=solve('k+a/2=173*cos(b)','sqrt(3)*a/2=173*sin(b)','a','b')
end
There would be an error, how to make it?
degree=140/1000000;
p=42164000;
a=6378136.5;
b=6356751.8;
x_1=0;
y_1=p;
z_1=0;
for i=451:550
for j=451:550
alpha=(1145-i)*degree;
beta=(1145-j)*degree;
x_2=p/cos(alpha)*tan(beta);
y_2=0;
z_2=p*tan(alpha);
syms x y z x_1 x_2 y_1 y_2 z_1 z_2 a b
eq = [(x-x_1)*(y2-y_1)-(x_2-x_1)*(y-y_1),(x-x_1)*(z_2-z_1)-(x_2-x_1)*(z-z_1), b^2*(x^2+y^2)+a^2*(y^2)-a^2*b^2 ];
sol = solve(eq(1),x,eq(2),y, eq(3),z);
sol.x
sol.y
sol.z
end
end
I got the expression value, how do I get the numeric value of x,y,z?
[['x(1)=';'x(2)='],num2str(double(sol.x))]
not work ,shows
??? Error using ==> mupadmex
Error in MuPAD command: DOUBLE cannot convert the input expression into a double array.
If the input expression contains a symbolic variable, use the VPA function instead.
Error in ==> sym.sym>sym.double at 927
Xstr = mupadmex('mllib::double', S.s, 0);
Error in ==> f2 at 38
[['x(1)=';'x(2)='],num2str(double(sol.x))]
If you have access to the Symbolic Toolkit then you do the following:
syms a b k
eq = [k+a/2-173*cos(b), sqrt(3)*a/2-173*sin(b)];
sol = solve(eq(1),a,eq(2),b);
sol.a = simplify(sol.a);
sol.b = simplify(sol.b);
% There are two solutions for 'a' and 'b'
% check residuals for example k=20
subs(subs(eq,{a,b},{sol.a(1),sol.b(1)}),k,20)
% ans = 0.2e-13
subs(subs(eq,{a,b},{sol.a(2),sol.b(2)}),k,20)
% ans = 0.2e-13
Edit 1
Based on new code by OP the matlab script to solve this is:
clear all
clc
syms alpha beta
degree=140/1000000;
p=42164000;
a=6378136.5;
b=6356751.8;
x_1=0;
y_1=p;
z_1=0;
x_2 = p/cos(alpha)*tan(beta);
y_2 = 0;
z_2 = p*tan(alpha);
syms x y z
eq = [(x-x_1)*(y_2-y_1)-(x_2-x_1)*(y-y_1);...
(x-x_1)*(z_2-z_1)-(x_2-x_1)*(z-z_1); ...
b^2*(x^2+y^2)+a^2*(y^2)-a^2*b^2 ];
sol = solve(eq(1),x,eq(2),y,eq(3),z);
sol.x = simplify(sol.x);
sol.y = simplify(sol.y);
sol.z = simplify(sol.z);
pt_1 = [sol.x(1);sol.y(1);sol.z(1)] % First Solution Point
pt_2 = [sol.x(2);sol.y(2);sol.z(2)] % Second Solution Point
x = zeros(100,100);
y = zeros(100,100);
z = zeros(100,100);
for i=451:550
disp(['i=',num2str(i)])
for j=451:550
res = double(subs(pt_1,{alpha,beta},{(1145-i)*degree,(1145-j)*degree}));
x(i-450, j-450) = res(1);
y(i-450, j-450) = res(2);
z(i-450, j-450) = res(3);
end
end
disp('x=');
disp(x);
disp('y=');
disp(x);
disp('z=');
disp(x);
I would try
for i=1:100
k=num2str(i)
[a,b]=solve('100+a/2=173*cos(b)','sqrt(3)*a/2=173*sin(b)','a','b')
end
and then solve the equation