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.
Related
I am trying to solve the equation below for array d. I have used the snippet below:
channel_size = 9e-3;
d = [11e-3, 12e-3];
sigma = 0.49;
ee = 727/9806.65;
alpha = d-channel_size;
sym('p',[1 2])
for i = 1:2
eqn = alpha == (4.3^(1/7))*(p^(3/11))*(((1-(sigma^2))/ee)^(7/5))/(d.^(1/6))
S = solve(eqn, p)*0.015;
vpa(S/13e-12)
end
In fact, I should get two number corresponding to d(1) and d(2), but it does not work and this error appears:
Error using mupadmex
Error in MuPAD command: Operands are invalid. [linalg::matlinsolve]
Error in sym/privBinaryOp (line 1693)
Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});
Error in sym/mrdivide (line 232)
X = privBinaryOp(A, B, 'symobj::mrdivide');
Declaring (d.^(1/6)) is wrong and instead (d(i)^(1/6)) should be used. In addition, as alpha = d-channel_size, in the equation, I should also declare alpha(i) instead of simple alpha.
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 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;
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);