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?
Related
Basically I would like to use the fsolve command in order to find the roots of an equation.
I think I should create a function handle that evaluates this equation in the form "right hand side - left hand side =0", but I've been struggling to make this work. Does anyone know how to do this?
The equation itself is 1/sqrt(f) = -1.74log((1.254/((1.27310^8)sqrt(f)))+((110^-3)/3.708)). So I would like to find the point of intersection of the left and right side by solving for 1/sqrt(f)+(1.74log((1.254/((1.27310^8)sqrt(f)))+((110^-3)/3.708))) = 0 using fsolve.
Thanks a lot!
The code so far (not working at all)
f = #(x) friction(x,rho,mu,e,D,Q, tol, maxIter) ;
xguess = [0, 1];
sol = fsolve(x, xguess ) ;
function y = friction(x,rho,mu,e,D,Q, tol, maxIter)
D = 0.1;
L = 100
rho = 1000;
mu = 0.001;
e = 0.0001;
Q = 0.01;
U = (4*Q)/(pi*D^2);
Re = (rho*U*D)/mu ;
y = (1/sqrt(x))-(-1.74*log((1.254/(Re*sqrt(x)))+((e/D)/3.708)))
end
Error message:
Error using lsqfcnchk (line 80)
FUN must be a function, a valid character vector expression, or an inline function object.
Error in fsolve (line 238)
funfcn = lsqfcnchk(FUN,'fsolve',length(varargin),funValCheck,gradflag);
Error in Untitled (line 6)
sol = fsolve(x, xguess ) ;
opt = optimset('Display', 'Iter');
sol = fsolve(#(x) friction(x), 1, opt);
function y = friction(x)
D = 0.1;
L = 100; % note -- unused
rho = 1000;
mu = 0.001;
e = 0.0001;
Q = 0.01;
U = (4*Q)/(pi*D^2);
Re = (rho*U*D)/mu ;
y = (1/sqrt(x))-(-1.74*log((1.254/(Re*sqrt(x)))+((e/D)/3.708)));
end
sol = 0.0054
the first argument of fsolve should be the function not variable. Replace:
sol = fsolve(x, xguess );
with
sol = fsolve(f, xguess );
And define Rho, mu, e etc before you define f (not inside the friction function).
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.
i am trying to solve a system of nonlinear differential equation using ODE45 MATLAB , i did that many times successfully , but this time i get the following error and i really don't know what is wrong i am confused compeletly. here are the codes.
%% this is the error:
Subscript indices must either be real positive integers or logicals.
Error in non_L_ss (line 6)
(-Fk*(ds0+x(3)-x(1))+Fk*ds0-Fc(x(4)-x(2)))/ms +Fa/ms ] ;
Error in odearguments (line 87)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode45 (line 115)
odearguments(FcnHandlesUsed,solver_name,ode, tspan,y0,optio varargin);
Error in solve (line 50)
[t X]=ode45(#non_L_ss,t_span,IC);
%% the equations be defined in a function:
function dX=non_L_ss(t,x)
global Fk Fc kt Fa q ds0 ms mu
dX=[ x(2);
(Fk*(ds0+x(3)-x(1))-Fk*ds0+Fc*(x(4)-x(2))-kt*x(1))/mu-Fa/m-kt*q/mu ;
x(4);
(-Fk*(ds0+x(3)-x(1))+Fk*ds0-Fc(x(4)-x(2)))/ms +Fa/ms ] ;
end
%% and here the function be called to solve by ODE45:
clear
clc
global Fk Fc kt Fa q ds0 ms mu qdot v2
mu = 100 ;
ms = 1242 ;
k1s = 80000 ;
k2s = 32000 ;
kt = 405000 ;
c1s = 4000 ;
c2s = 1600 ;
v = 20 ;
Gq = 256e-6 ;
ds0 = 0.1538 ;
a = 1 ;
b = 0.001 ;
n0 = 0.1 ;
f0 = 0.011*v ;
w = 0.5 ;
Fa = 2000 ;
q = 0.05 ;
xs = 0.1 ;
xu = 0.1 ;
dxs = 0.1 ;
dxu = 0.2 ;
Fk = k1s+k2s*(ds0+xs-xu).^2 ;
if dxs >= dxu
Fc = c1s ;
elseif dxs < dxu
Fc = c2s ;
end
t_span=[0 1];
IC=[2 3 2 2];
[t X]=ode45(#non_L_ss,t_span,IC);
Subscript indices must either be real positive integers or logicals.
Error in non_L_ss (line 6)
(-Fk*(ds0+x(3)-x(1))+Fk*ds0-Fc(x(4)-x(2)))/ms +Fa/ms ] ;
means that you use something as an indexed array with a non-integer index. Which probably means that the object is no array at all. I'd propose to replace
Fc(x(4)-x(2))
by
Fc*(x(4)-x(2))
to try to solve this issue.
**I want to plot an array vector having symbolic variables in it. But I am getting an error which is given below.
??? Error using ==> plot
Conversion to double from sym is not possible.
Error in ==> dsolvenew at 36
plot(t2,qn2) **
The code is given below.
syms A Epsilon0 k1 k2 e s d R s f T c0 v0 v
An = 0.5*10^(-9);
Epsilon0n = 8.85*10^-12;
k1n = 10;
k2n = 10;
en = 1.6*10^-19;
sn = 1.8*10^(-9);%m gap size%
dn = 3*sn; % total distance
Rn = 1;
c0n=(An*Epsilon0n*k1n)/dn;
t1=0:0.1:5;
fn=1000;
v0n=7.5;
Tn=5/f;
vn=v0n*sin(2*3.14*fn*t1);
plot(t1,vn);
inits = 'q(0)=(20*10^(-20)),Q(0)=0';
[q,Q] = dsolve('Dq=((v/R)-(1/R)*(((d*q)+(s*Q))/(c0*d)))','DQ= (((q+Q)/(A*Epsilon0*k1))*s)',inits);
qn1=vpa(subs(q,{A,Epsilon0,k1,k2,e,s,d,R,c0}, {An,Epsilon0n,k1n,k2n,en,sn,dn,Rn,c0n}),3);
i=1;
while i<length(t1);
qn2(i)=subs(qn1, v, vn(i));
i=i+1;
end
t2=t1(1:50);
plot(t2,qn2)
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