MATLAB strange error Gamma function numerical integration - matlab

i try to run the following in order to integrate numerically:
nu = 8;
psi=-0.2;
lambda = 1;
git = #(u) tpdf((0 - lambda * skewtdis_inverse(u, nu, psi)), nu);
g(t,i) = integral(git,1e-10,1-1e-10,'AbsTol',1e-16);
where tpdf is a matlab function and skewtdis:inverse looks like this:
function inv = skewtdis_inverse(u, nu, lambda)
% PURPOSE: returns the inverse cdf at u of Hansen's (1994) 'skewed t' distribution
c = gamma((nu+1)/2)/(sqrt(pi*(nu-2))*gamma(nu/2));
a = 4*lambda*c*((nu-2)/(nu-1));
b = sqrt(1 + 3*lambda^2 - a^2);
if (u<(1-lambda)/2);
inv = (1-lambda)/b*sqrt((nu-2)./nu)*tinv(u/(1-lambda),nu)-a/b;
elseif (u>=(1-lambda)/2);
inv = (1+lambda)/b*sqrt((nu-2)./nu).*tinv(0.5+1/(1+lambda)*(u-(1-lambda)/2),nu)-a/b;
end
What i get out is:
Error in skewtdis_inverse (line 6)
c = gamma((nu+1)/2)/(sqrt(pi*(nu-2))*gamma(nu/2));
Output argument "inv" (and maybe others) not assigned during call to "F:\Xyz\skewtdis_inverse.m>skewtdis_inverse".
Error in #(u)tpdf((0-lambda*skewtdis_inverse(u,nu,psi)),nu)
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 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 89)
Q = integralCalc(fun,a,b,opstruct);
If i , however call the function in thr handle directly there are no Problems:
tpdf((0 - lambda * skewtdis_inverse(1e-10, nu, psi)), nu)
ans =
1.4092e-11
tpdf((0 - lambda * skewtdis_inverse(1-1e-10, nu, psi)), nu)
ans =
7.0108e-10
Your effort is highly appreciated!

By default, integral expects the function handle to take a vector input.
In your code, the if-statement creates a complication since the condition will evaluate to true only if all elements of u satisfy it.
So, if u is a vector that has elements both greater than and less than (1-lambda)/2, inv will never be assigned.
There are two options:
Put the if-statement in a for-loop and iterate over all of the elements of u.
Use logical indexes for the assignment.
The second option is faster for large element count and, in my opinion, cleaner:
inv = u; % Allocation
IsBelow = u < (1-lambda)/2; % Below the threshold
IsAbove = ~IsBelow ; % Above the threshold
inv(IsBelow) = (1-lambda)/b*sqrt((nu-2)./nu)*tinv(u(IsBelow)/(1-lambda),nu)-a/b;
inv(IsAbove) = (1+lambda)/b*sqrt((nu-2)./nu)*tinv(0.5+1/(1+lambda)*(u(IsAbove)-(1-lambda)/2),nu)-a/b;

Related

Matlab Pso Algorithm

I'm trying to solve a symbolic optimization problem using PSO optimizer in MATLAB. The variables r x a c n theta z are symbolic and CD is calculated by integrating r.
The CD is the objective function with free variables a,n, theta and lb and ub are bounds. Full code is as follows:
syms r x a c n theta z
assume(n,'positive');
D=0.24;
L=2;
f=L/D;
b=.8;
a0=0.02;
db=0.05;
V=1;
Re=(V*(D/2))/0.000001;
Cf=(0.075/(((log10(Re))-2)^2))+0.00025;
% Define r(x)
c=L-a-b-a0;
r1=0.5*D*(2*x/a)^(1/n);
I1=simplify(int(2*pi*r1,x,a0,a));
r2=D/2;
I2=simplify(int(2*pi*r2,x,a,a+b));
r3=(0.5*D)-((((3*D)/(2*(c)^2))-(tan(theta)/c))*(x-a-b)^2)+(((D/c^3 ...
(tand(theta)/c^2))*(x-a-b)^3);
I3=simplify(int(2*pi*r3,x,a+b,L));
A=simplify(I1+I2+I3);
Sn=pi*(D^2/4);
Cdstar=Cf*(1+(60*f^-3 )+(0.0025*f))*(A/(L^2));
Cdb=0.029*((db/D)^3)*(Cdstar^-0.5)*(Sn/(L^2));
CD=simplify(Cdstar+Cdb);
%optimization problem
objective=matlabFunction(CD,'Vars',[a,n,theta])
nVar=3;
lb = [deg2rad(5),0.25,a0];
ub = [deg2rad(60),5,L/2];
options =
optimoptions('particleswarm','SwarmSize',100,'HybridFcn',#fmincon);
[z,fval,exitflag,output] = particleswarm(objective,nVar,lb,ub,options)
And this is the error I get:
#(a,n,theta)pi.*4.404634153141517e-4+pi.*1.0./sqrt(pi.4.404634153141517e-4-pi.(a.*5.0-6.0).*1.0./(a.5.0e+1-5.9e+1).^3.(a.*2.32335e+6+tan((theta.*pi)./1.8e+2).*4.779e+6-tan(theta).*6.2658e+6-a.^2.*tan(theta).*2.655125e+7+a.^3.*tan(theta).*1.4875e+7-a.^4.*tan(theta).*3.125e+6-a.*tan((theta.*pi)./1.8e+2).*1.59975e+7+a.*tan(theta).*2.1063e+7+a.^2.*tan((theta.*pi)./1.8e+2).*2.008125e+7-a.^3.*tan((theta.*pi)./1.8e+2).*1.1203125e+7+a.^4.*tan((theta.*pi)./1.8e+2).*2.34375e+6-a.^2.*1.98e+6+a.^3.*5.625e+5-9.08811e+5).*1.223509486983755e-5-(n.pi.((a.*2.5e+1).^(-1.0./n)-2.0.^(1.0./n+1.0).*a.*2.5e+1).*1.101158538285379e-5)./(n+1.0)).9.440104166666668e-7-pi.(a.*5.0-6.0).*1.0./(a.5.0e+1-5.9e+1).^3.(a.*2.32335e+6+tan((theta.*pi)./1.8e+2).*4.779e+6-tan(theta).*6.2658e+6-a.^2.*tan(theta).*2.655125e+7+a.^3.*tan(theta).*1.4875e+7-a.^4.*tan(theta).*3.125e+6-a.*tan((theta.*pi)./1.8e+2).*1.59975e+7+a.*tan(theta).*2.1063e+7+a.^2.*tan((theta.*pi)./1.8e+2).*2.008125e+7-a.^3.*tan((theta.*pi)./1.8e+2).*1.1203125e+7+a.^4.*tan((theta.*pi)./1.8e+2).*2.34375e+6-a.^2.*1.98e+6+a.^3.*5.625e+5-9.08811e+5).*1.223509486983755e-5-(n.pi.((a.*2.5e+1).^(-1.0./n)-2.0.^(1.0./n+1.0).*a.*2.5e+1).*1.101158538285379e-5)./(n+1.0)
Not enough input arguments.
Error in
symengine>#(a,n,theta)pi.*4.404634153141517e-4+pi.*1.0./sqrt(pi.4.404634153141517e-4-pi.(a.*5.0-6.0).*1.0./(a.5.0e+1-5.9e+1).^3.(a.*2.32335e+6+tan((theta.*pi)./1.8e+2).*4.779e+6-tan(theta).*6.2658e+6-a.^2.*tan(theta).*2.655125e+7+a.^3.*tan(theta).*1.4875e+7-a.^4.*tan(theta).*3.125e+6-a.*tan((theta.*pi)./1.8e+2).*1.59975e+7+a.*tan(theta).*2.1063e+7+a.^2.*tan((theta.*pi)./1.8e+2).*2.008125e+7-a.^3.*tan((theta.*pi)./1.8e+2).*1.1203125e+7+a.^4.*tan((theta.*pi)./1.8e+2).*2.34375e+6-a.^2.*1.98e+6+a.^3.*5.625e+5-9.08811e+5).*1.223509486983755e-5-(n.pi.((a.*2.5e+1).^(-1.0./n)-2.0.^(1.0./n+1.0).*a.*2.5e+1).*1.101158538285379e-5)./(n+1.0)).9.440104166666668e-7-pi.(a.*5.0-6.0).*1.0./(a.5.0e+1-5.9e+1).^3.(a.*2.32335e+6+tan((theta.*pi)./1.8e+2).*4.779e+6-tan(theta).*6.2658e+6-a.^2.*tan(theta).*2.655125e+7+a.^3.*tan(theta).*1.4875e+7-a.^4.*tan(theta).*3.125e+6-a.*tan((theta.*pi)./1.8e+2).*1.59975e+7+a.*tan(theta).*2.1063e+7+a.^2.*tan((theta.*pi)./1.8e+2).*2.008125e+7-a.^3.*tan((theta.*pi)./1.8e+2).*1.1203125e+7+a.^4.*tan((theta.*pi)./1.8e+2).*2.34375e+6-a.^2.*1.98e+6+a.^3.*5.625e+5-9.08811e+5).*1.223509486983755e-5-(n.pi.((a.*2.5e+1).^(-1.0./n)-2.0.^(1.0./n+1.0).*a.*2.5e+1).*1.101158538285379e-5)./(n+1.0)
Error in particleswarm>makeState (line 694)
firstFval = objFcn(state.Positions(1,:));
Error in particleswarm>pswcore (line 169) state =
makeState(nvars,lbMatrix,ubMatrix,objFcn,options);
Error in particleswarm (line 151) [x,fval,exitFlag,output] =
pswcore(objFcn,nvars,lbRow,ubRow,output,options);
Error in MYRING_SYMS_optimisation_K (line 56) [z,fval,exitflag,output]
= particleswarm(objective,nVar,lb,ub,options)
Caused by:
Failure in initial objective function evaluation. PARTICLESWARM cannot continue.
The fun takes only one argument, which is a vector with nvars elements. From particleswarm doc:
x = particleswarm(fun,nvars) attempts to find a vector x that achieves a local minimum of fun. nvars is the dimension (number of design variables) of fun.
So you need to declare a new objective function that only takes 1 argument:
[z,fval,exitflag,output] = particleswarm( ...
#(x) objective(x(1), x(2), x(3)), ...
nVar,lb,ub,options)

fsolve issue with initial condition for ODE

I am trying to solve the following ODE:
function [eta, sol] = compressible_similarity_wo
global Gamm Ma Pr omega;
Gamm = 1.4;
Ma = 2;
Pr = 0.7;
omega=0.76;
global eta_max_ode;
eta_max_ode = 20;
opt = optimset('Display','off','TolFun',1E-20);
F = fsolve(#(F) eval_boundary(F),[0,0,0.4,1,0],opt);
[eta_ode, fg_ode] = solve_ode(F);
sol = [fg_ode];
eta = eta_ode;
end
function [eta_ode, fg_ode] = solve_ode(F)
global eta_max_ode
options = odeset('RelTol',1e-9,'AbsTol',1e-9);
[eta_ode, fg_ode] = ode45(#BLFunc,[0,eta_max_ode],F,options);
end
function [g] = eval_boundary(F)
% Get the solution to the ODE with inital condition F
[eta_ode, fg_ode] = solve_ode(F);
% Get the function values (for BCs) at the starting/end points
f_start = fg_ode(1,1); %f(0) = 0
df_start = fg_ode(1,2); %f'(0) = 0
df_end = fg_ode(end,2); %f'(inf) - 1 = 0
t_end = fg_ode(end,4); %T(inf) - 1 = 0
dt_start = fg_ode(1,5); %T'(0) = 0
% Evaluate the boundary function
g = [f_start
df_start
df_end - 1
t_end - 1
dt_start];
end
function [df] = BLFunc(f)
global Gamm Ma Pr omega;
df = zeros(5,1);
df(1) = f(2);
df(2) = f(3);
df(3) = -f(1)*f(3)/(f(4)^(omega-1))-(omega-1)*f(3)/f(4);
df(4) = f(5);
df(5) = -Pr*f(1)*f(5)/(f(4)^(omega-1)) - Pr*(Gamm - 1.0)*Ma*Ma*f(3)*f(3) - (omega-1)*f(5)/f(4);
end
but fsolve returns the following problem
Error using BLFunc
Too many input arguments.
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode45 (line 115)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in solve_ode (line 5)
[eta_ode, fg_ode] = ode45(#BLFunc,[0,eta_max_ode],F,options);
Error in eval_boundary (line 3)
[eta_ode, fg_ode] = solve_ode(F);
Error in compressible_similarity_wo>#(F)eval_boundary(F) (line 15)
F = fsolve(#(F) eval_boundary(F),[0,0,0.4,1,0],opt);
Error in fsolve (line 230)
fuser = feval(funfcn{3},x,varargin{:});
Error in compressible_similarity_wo (line 15)
F = fsolve(#(F) eval_boundary(F),[0,0,0.4,1,0],opt);
Error in launch (line 3)
[eta, sol] = compressible_similarity_wo;
Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue.
Do you have an idea of what's going on?
I'll cite you the friendly manual page
The function dydt = odefun(t,y), for a scalar t and a column vector y, must return a column vector dydt of data type single or double that corresponds to f(t,y). odefun must accept both input arguments, t and y, even if one of the arguments is not used in the function.
That is, you simply need to change to
function [df] = BLFunc(t,f)
to get a result (no guarantee that it is THE result).
Try to replace BLFunc signature to
function [df] = BLFunc(t, f)
You need to provide odefun to ode45, which takes 2 arguments, as stated in documentation:
The function dydt = odefun(t,y), for a scalar t and a column vector y, must return a column vector dydt of data type single or double that corresponds to f(t,y). odefun must accept both input arguments, t and y, even if one of the arguments is not used in the function.

Matlab- 1D Numerical integration with function stored in multiple variables

as you can see in the last few lines below, I'm trying to store my function on multiple variables because it gets rather ugly. However, doing this yields the error shown below.
A fix for this would be to manually substitute k and kp, but this is precisely what I'm trying to avoid. Any help would be appreciated. thanks!
clc
clear
hbar = 1.055e-34;
mo = 9.1095e-31;
q = 1.602e-19;
kb=1.38065e-23;
T=298;
Ef = -0.1*q; % -100meV in units Joules
V0 = 1*q;
L = 1e-9;
k = #(E) (2*mo*E/hbar.^2)^.5;
kp = #(E) (2*mo*(V0-E)/hbar.^2)^.5;
fun = #(E) (exp(-2*kp*L) .* ((16*k.^2 .* kp.^2) ./ ((k.^2 + kp.^2).^2))) .* exp(-E./(kb*T));
Q = integral(fun,0,inf);
Error below
Undefined operator '*' for input arguments of type 'function_handle'.
Error in #(E)(exp(-2*kp*L).*((16*k.^2.*kp.^2)./((k.^2+kp.^2).^2))).*exp(-E./(kb*T))
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 83)
[q,errbnd] = vadapt(#AToInfInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in PS3_2 (line 17)
Q = integral(fun,0,inf);
Use this here
k = #(E) (2.*mo.*E./hbar.^2).^.5;
kp = #(E) (2.*mo.*(V0-E)./hbar.^2).^.5;
fun = #(E) (exp(-2*kp(E)*L).*((16*k(E).^2.*kp(E).^2)./((k(E).^2+kp(E).^2).^2))).*exp(-E./(kb*T));
Q = integral(fun,0,inf);
I think you need to pass the argument E, then are you sure kb*T is correct? Maybe kp(E)*T? Then, you forgot the . is the sqrt for k and kp, or if it isn't a sqrt then the dot is on the wrong side of the power symbol.

Using MATLAB integral with anonymous functions

I'm trying to use the MATLAB integral() function to integrate along a single parameter of a piecewise linear function that represents a time-varying value.
I would like to define an anonymous function based on the original function:
t1 = 1;
t2 = 2;
t3 = 4;
t4 = 5;
a0 = 1;
a1 = 2;
f = #(x) accel_profile(t1,t2,t3,t4,a0,a1,x);
and here is the accel_profile.m:
function value = accel_profile(t1,t2,t3,t4,a0,a1, t)
if t <= t1
value = a0;
return
elseif (t <= t2)
value = ((t-t1)/(t2-t1)) * (a1-a0) + a0;
return
elseif (t <= t3)
value = a1;
return
elseif (t <= t4)
value = ((t-t3)/(t4-t3)) * (a0-a1) + a1;
return
else
value = a0;
return
end
The problem is that when I exercise the following script:
t_list = 0:0.1:6;
q = zeros(1,length(t_list))
for i = 1:length(t_list)
q(i) = integral(f,0,t_list(i));
end
plot(t_list, q)
I get the following stack trace:
Error using integralCalc/finalInputChecks (line 515)
Output of the function must be the same size as the input. If FUN is an array-valued integrand, set
the 'ArrayValued' option to true.
Error in integralCalc/iterateScalarValued (line 315)
finalInputChecks(x,fx);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 75)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
515 error(message('MATLAB:integral:FxNotSameSizeAsX'));
I'm running MATLAB 2015b on Windows 7.
The problem is integral uses vectorized arguments for your function, but your function does not support it.
The relevant part from the documentation:
For scalar-valued problems, the function y = fun(x) must accept a vector argument, x, and return a vector result, y. This generally means that fun must use array operators instead of matrix operators. For example, use .* (times) rather than * (mtimes). If you set the 'ArrayValued' option to true, then fun must accept a scalar and return an array of fixed size.
Reference
This means integral will call your function f with arguments like f([1,2,3]) and expects a list with [f(1),f(2),f(3)]
The general techniques to vectorize a piecewise defined function are explained in this question

Create flexible function handle

I am using numerical integration in MATLAB, with one varibale to integrate over but the function also contains a variable number of terms depending on the dimension of my data. Right now this looks like the following for the 2-dimensional case:
for t = 1:T
fxt = #(u) exp(-0.5*(x(t,1)-theta*norminv(u,0,1)).^2) .* ...
exp(-0.5*(x(t,2) -theta*norminv(u,0,1)).^2);
f(t) = integral(fxt,1e-4,1-1e-4,'AbsTol',1e-3);
end
I would like to have this function flexible in the sense that there could be any number of data points in, each in the following term:
exp(-0.5*(x(t,i) -theta*norminv(u,0,1)).^2);
I hope this is understandable.
If x and u have a valid dimension match (vector-vector or array-scalar) for the subtraction, you can put the whole matrix x into the handle and pass it to the integral function using the name-parameter pair ('ArrayValued',true):
fxt = #(u) exp(-0.5*(x - theta*norminv(u,0,1)).^2) .* ...
exp(-0.5*(x - theta*norminv(u,0,1)).^2);
f = integral(fxt,1e-4,1-1e-4,'AbsTol',1e-3,'ArrayValued',true);
[Documentation]
You may need a loop if integral ever passes a vector u into the handle.
But in looking at how the integral function is written, the integration nodes are entered as scalars for array-valued functions, so the loop shouldn't be necessary unless some weird dimension-mismatch error is thrown.
Array-Valued Output
In response to the comments below, you could try this function handle:
fx = #(u,t,k) prod(exp(-0.5*(x(t,1:k)-theta*norminv(u,0,1)).^2),2);
Then your current loop would look like
fx = #(u,t,k) prod(exp(-0.5*(x(t,1:k)-theta*norminv(u,0,1)).^2),2);
k = 2;
for t = 1:T
f(t) = integral(#(u)fx(u,t,k),1e-4,1-1e-4,'AbsTol',1e-3,'ArrayValued',true);
end
The ArrayValued flag is needed since x and u will have a dimension mismatch.
In this form, another loop would be needed to sweep through the k indexes.
However, we can improve this function by skipping the loop altogether since each iterate of the loop is independent by using the ArrayValued mode:
fx = #(u,k) prod(exp(-0.5*(x(:,1:k)-theta*norminv(u,0,1)).^2),2);
k = 2;
f = integral(#(u)fx(u,k),1e-4,1-1e-4,'AbsTol',1e-3,'ArrayValued',true);
Vector-Valued Output
If ArrayValued is not desired, which may be the case if the integration requires a lot of subdivisions and a vector-valued u is preferable, you can also try a recursive version of the handle using cell arrays:
% x has size [T,K]
fx = cell(K,1);
fx{1} = #(u,t) exp(-0.5*(x(t,1) - theta*norminv(u,0,1)).^2);
for k = 2:K
fx{k} = #(u,t) fx{k-1}(u,t).*exp(-0.5*(x(t,k) - theta*norminv(u,0,1)).^2);
end
f(T) = 0;
k = 2;
for t = 1:T
f(t) = integral(#(u)fx{k}(u,t),1e-4,1-1e-4,'AbsTol',1e-3);
end
ThanksTroy but now I run into the follwing:
x = [0.3,0.8;1.5,-0.7];
T = size(x,1);
k = size(x,2);
theta= 1;
fx = #(u,t,k) prod(exp(-0.5*(x(t,1:k) - theta*norminv(u,0,1))^2));
for t = 1,T
f(t) = integral(#(u)fx(u,t,k),1e-4,1-1e-4,'AbsTol',1e-3);
end
Error using -
Matrix dimensions must agree.
Error in #(u,t,k)prod(exp(-0.5*(x(t,1:k)-theta*norminv(u,0,1))^2))
Error in #(u)fx(u,t,k)
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 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 89)
Q = integralCalc(fun,a,b,opstruct);