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.
Related
I have this code:
function main()
a = 1.0e+04 * [0.005055052938847,0.010897917816899,0.022355965424711,0.043081981439108,0.077074049394439,0.127074049394439,0.193081981439108,0.272355965424711,0.360897917816899,0.455055052938847,0.552256864601221,0.650978664311931,0.750415022011379,0.850172973479352,0.950071110045004,1.050028912038499,1.150011648830086]
B = 1.0e+04 * [1.215101736363023,0.697166188613023,0.400000000000000,0.229500515964941,0.131676217070435,0.075549399394941,0.043346565354951,0.024870147785673,0.014269279372341,0.008187017446000,0.004697311820178,0.002695088715947,0.001546310627203,0.000887197716963,0.000509030834515,0.000292057097908,0.000167568136653
m = timeloop(a,B);
end
function m = timeloop(a,B)
st = zeros(49,4);
t = 0:0.001:0.05;
for i = (1:49)
st(i+1,1:4) = next_state(st(i,1:4),a,B,1e-4);
end
m = mean(prod(state,2))
end
function next_state = next_state(state,alpha,beta,dt)
nch = size(state,2);
p01 = rand(1,nch);
alphadt = repmat(alpha,1,nch)*dt;
betadt = repmat(beta,1,nch)*dt;
next_state1 = (p01<alphadt) .* (state==0);
next_state0 = (p01<betadt) .* (state==1);
next_state = state + next_state1 - next_state0;
end
but it gives me the following error:
Matrix dimensions must agree.
Error in q3>next_state (line 59)
next_state1 = (p01<alphadt) .* (state==0);
Error in q3>timeloop (line 49)
st(i+1,1:4) = next_state(st(i,1:4),a,B,1e-4);
Error in q3 (line 4)
m = timeloop(a,B);
Error in run (line 86)
evalin('caller', [script ';']);
I've tried changing the multiplication to just * with no improvement. What I've looked up online hasn't seemed to be able to help me. I don't understand MATLAB very well so you will have to be very specific in your explanation.
If you debugged your code you would quickly realise the issue.
alpha in next_state is the same as a in main, meaning it is of size 1x3.
You then create alphadt using repmat(alpha,1,4), so it is of size 1x12.
Then you try and do the following
(p01<alphadt) .* (state==0)
% 1x12 .* 1x3
As the error clearly states, your matrix dimensions don't agree.
Because you haven't given any context for what you're trying to achieve, it's not obvious what the solution should be. Perhaps you want to also use repmat on state, or loop over single values from a.
I am new to Matlab. I am trying to solve a non-linear equation using this inbuilt Matlab function called fzero() but it's not giving me the results.
The main file goes like
A = 5;
B = 6;
C = 10;
eq = equation (A, B, C);
fzero(#(x)eq);
The other function file is:
function eq = equation (A, B, C)
syms x;
eq = A*x.^2 + B*x + C*(asinh(x)) ;
When I run this code, I get the following error:
Error using fzero (line 118)
The input should be either a structure with valid fields or at least two arguments to
FZERO.
Error in main (line 7)
fzero(#(x)eq);
Could someone help me with this?
EDIT:
WHen I specify the check point as 0, it returns me the following error.
Undefined function 'isfinite' for input arguments of type 'sym'.
Error in fzero (line 308)
elseif ~isfinite(fx) || ~isreal(fx)
Error in main (line 7)
fzero(#(x)eq, 0);
There are several mistakes in your code. For a start, fzero is for finding numerical roots of a non-linear equation, it is not for symbolic computations (check the documentation), so get rid of syms x. The correct way to call fzero in your case is a as follows:
A = 5;
B = 6;
C = 10;
eq = #(x) A*x^2 + B*x + C*(asinh(x));
x0 = 0; % or whatever starting point you want to specify
x = fzero(eq,x0)
You need to specify a guess, x0 point
fun = #sin; % function
x0 = 3; % initial point
x = fzero(fun,x0)
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);
I'm trying to run a global optimization in Matlab using the global optimization toolbox. I am using the fmincon local optimizer as an input. I am able to successfully run the fmincon local optimizer, as well as the MultiStart global optimizer with this problem. However, when I try to set it up as a GlobalSearch, I get an error message.
Below is the code I am trying to use for the GlobalSearch:
%the functions is f(w) rather than f(x)
%constraints:
A = [1,0,0; 0,1,0; 0,0,1];
b = [.80, .80, .80];
Aeq = [1 1 1];
beq = 1;
lb = .10 * [1 1 1];
ub = .8 * [1 1 1];
w = [weight1, weight2, weight3];
wstart = randn(3,1);
options = optimset('Algorithm','interior-point');
% function handle for the objective function (note that variables
% aa through hh are additional parameters that the solver does not modify):
h = #(w)Difference_in_Returns(w(1),w(2),w(3), aa, bb, cc, dd, ee, ff, gg, hh);
% problem structure:
problem = createOptimProblem('fmincon','x0',wstart,'objective',h,'Aineq',A,...'
'bineq',b,'Aeq',Aeq,'beq',beq,'options',options,'lb',lb,'ub',ub);
gs = GlobalSearch;
run(gs,problem)
When I try to run this, Matlab bugs out and prints:
Error using -
Matrix dimensions must agree.
Error in C:\Program Files\MATLAB\R2012a\toolbox\globaloptim\globaloptim\private\
globalsearchnlp.p>i_calcConstrViolation (line 593)
Error in C:\Program Files\MATLAB\R2012a\toolbox\globaloptim\globaloptim\private\
globalsearchnlp.p>i_calcPenalty (line 627)
Error in C:\Program Files\MATLAB\R2012a\toolbox\globaloptim\globaloptim\private\
globalsearchnlp.p>globalsearchnlp (line 343)
Error in GlobalSearch/run (line 330)
[x,fval,exitflag,output] = ...
Error in Optimization_Setup (line 62)
run(gs,problem)
I believe the problem is outlined by the lines:
Error using -
Matrix dimensions must agree.
Error in GlobalSearch/run (line 330)
[x,fval,exitflag,output] = ...
Any tips are highly appreciated. Please let me know if you have any questions.