I am trying to minimize with respect to a variable "y" a function that contains a parameter which must be calculated as a solution of an equality that contains "y" as well (say, y=-3; in my complete problem it is an equation with no analytic closed form solution, so I really need fzero).
Because of this, I include the fzero function in the argument of fminsearch:
fminsearch( #(y) 10*fzero(#(y) y+3, 0)) ;
I get the error:
Error using fminsearch (line 85)
The input to FMINSEARCH should be either a structure with
valid fields or consist of at least two arguments.
I obviously get the same error with:
f = fzero(#(y) y+3, 0);
fminsearch(#(y) 10*f);
Apparently the problem is that I cannot "nest" a fzero inside fminsearch.
Any idea about how to turn around this problem?
If you read the error message you got and look at the documentation of fminsearch you'll see that you need to call it with two input arguments. You call it with only one.
fminsearch( #(y) 10*fzero(#(x) x+3, 0), 0 )
Related
I have to plot the solution for masses between 4 and 8kg, which I wrote using the code below:
weight=linspace(4,8,100)
Then I use a for loop to find the solutions for the weights within this vector, while also declaring the variable a, and a guess of the time required:
for z=1:length(weight)
a=(weight./(1360*pi)).^(1/3)
tguess=14400
timeinseconds=fsolve(#(t)cookingtimes(t,a),tguess)
timeinhours=timeinseconds/(60*60)
end
The function that I mentioned previously is below:
function F=cookingtimes(t,a)
k=1:22;
alpha=0.7*10^(-7);
F(1)=(2./(pi*0.464))*sum(((-1).^(k-1))./k).*sin(k*pi*0.464).*exp(((-(k.^2)).*(pi.^2).*((alpha.*t))./(a.^2)))-57/80
end
Where F(1) represents the following equation 1, which will be equal to 57/80. Rather than adding until infinity, I chose the value of 22 as the equation converges to this value. r/a is equal to 0.464, which is why this is present within my function.
When running this code, I am given errors that I cant manage to solve.
Arrays have incompatible sizes for this operation.
Error in Homework1>cookingtimes (line 58)
F(1)=(2./(pi*0.464))*sum(((-1).^(k-1))./k).*sin(k*pi*0.464).*exp(((-(k.^2)).*(pi.^2).*.((alpha.*t))./(a.^2)))-57/80
Error in Homework1 (line 28)
timeinseconds=fsolve(#(t)cookingtimes(t,a),tguess)
Error in fsolve (line 264)
fuser = feval(funfcn{3},x,varargin{:});
Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue.
The arrays that I have produced are all of the same length, which is why I am very stuck on this issue.
You are trying to perform an operation on two arrays that have different sizes. In your case, the issue may be caused by the fact that you are defining the variable a as a vector inside the for loop, but using it in the function cookingtimes as a scalar. To fix this, you can modify your function to accept a vector of a values and return a corresponding vector of function values.
I modified the cookingtimes function to accept a vector of a values, and return a corresponding vector of function values.
function F=cookingtimes(t,a)
k=1:22;
alpha=0.7*10^(-7);
F=(2./(pi*0.464))*sum(((-1).^(k-1))./k).*sin(k*pi*0.464).*exp(((-(k.^2)).* (pi.^2).*((alpha.*t))./(a.^2)))-57/80;
end
In the for loop, You need to compute a for each weight value, and use a vector of a values in the call to fsolve. Also store the solutions in a vector timeinseconds instead of a scalar, as there are multiple solutions to be found.
for z=1:length(weight)
a=(weight(z)./(1360*pi)).^(1/3); % compute a for each weight value
tguess=14400;
timeinseconds(z)=fsolve(#(t)cookingtimes(t,a),tguess);
timeinhours=timeinseconds/(60*60);
end
I am trying to solve the following function with fminunc:
Σ((x(i)-1).^2) for i = 1 to 20
My code is below:
fun4 = #(x) sum((x(i)-1).^2, i, 1, 20)
[x,fval,exitflag,output]= fminunc(fun4,[1,1])
And, it gives the following error:
??? Error using ==> sym.sym>notimplemented at 2514
Function 'subsindex' is not implemented for MuPAD symbolic objects.
Could you help me, please?
You should check out the sum documentation for details about how to use it to sum over a vector.
In your case, if you have a vector x of at least 20 elements, you can get the subvector of "x(i) for i in the range 1 to 20" by x(1:20).
You can subtract scalars from a vector just by using the standard subtract sign, but to then square all of the elements individually you should use an "element-wise" operator (having a dot before the operator, like .^). As it happens, you were already doing this anyway.
The 2nd, 3rd and 4th arguments you were passing to sum are causing an error, because sum doesn't expect the arguments you are passing to it. For further details on the inputs it does expect see the linked docs.
All of this means your function can be simplified:
fun4 = #(x) sum((x(1:20)-1).^2);
I am trying to numerically solve an equation using fzero in Matlab. It is part of a bigger exercise. I haven't posted much here so not sure how much background information you need about this exercise so will try to keep it short.
This is my code:
fun = #(a)log(a/xBar) + (1/n) * log(dataProd) + diff(gamma(a))/gamma(a);
x0 = 0.8014;
x = fzero(fun,x0)
These are the values:
n = 209
xBar is 0.6078
dataProd = 3.1554e-77
I get the following error message in Matlab:
Operands to the || and && operators must be convertible to logical
scalar values.
Error in fzero (line 306)
elseif ~isfinite(fx) || ~isreal(fx)
Any idea why I get this error message?
I would guess that, during the search for a solution, fzero tries to evaluate your function at a=0, leading to an infinity. To check if this is the case, either look and see if you can set the optimization parameter 'display' to 'iter', or something similar in your version of Matlab. Otherwise, you can simply move your function to a separate m-file and call disp(a) on the first line. That way you will be able to see what fzero is doing and which value of a is causing the problem.
As part of a group project we have a system of 2 non linear differential equations and we have to draw the S=S(t) , I=I(t) graphic using the midpoint method.
And I'm getting the following error when trying to insert the matrix with the corresponding differential equations:
"Error in inline expression ==> matrix([[-(IS)/1000], [(IS)/1000 - (3*I)/10]])
Undefined function 'matrix' for input arguments of type 'double'.
Error in inline/subsref (line 23)
INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr);"
The code I have done is the following:
syms I S
u=[S;I];
F=[-0.001*S*I;0.001*S*I-0.3*I];
F1=inline(char(F),'I','S');
h=100; %Valores aleatórios
T=100000;
ni=(T/h);
u0=[799;1];
f=zeros(1,2);
k=zeros(1,2);
i=1;
while i<=ni
f(1)=F1(u0(1));
f(2)=F1(u0(2));
dx=h*f;
k(1)=F1((u0(1)+h*(1/2)),(u0(2)+h*(1/2)));
k(2)=F1((u0(1)+h*(1/2)),(u0(2)+h*(1/2)));
u1=u0+h*k;
disp('i:'),disp(i)
disp('u= '),disp(u1)
u0=u1;
i=i+1;
end
I'm new to this so the algorithm it's very likely to be wrong but if someone could help me with that error I'd apreciate it. Thank you!
The problem that specifically creates the error is that you are putting two symbolic functions into a matrix and then calling char (which outputs matrix([[-(IS)/1000], [(IS)/1000 - (3*I)/10]]) rather than converting nicely to string).
The secondary problem is that you are trying to pass two functions simultaneously to inline. inline creates a single function from a string (and using anonymous functions instead of inline is preferred anyway). You cannot put multiple functions in it.
You don't need sym here. In fact, avoid it (more trouble than it's worth) if you don't need to manipulate the equations at all. A common method is to create a cell array:
F{1} = #(I,S) -0.001*S*I;
F{2} = #(I,S) 0.001*S*I-0.3*I;
You can then pass in I and S as so:
F{1}(500,500)
Note that both your functions include both I and S, so they are always necessary. Reconsider what you were expecting when passing only one variable like this: f(1)=F1(u0(1));, because that will also give an error.
I am trying to make my own function in matlab to solve for a system of two nonlinear equations, while using a nested function to share some some parameters, here is a sample code:
function y=solve(a,x0)
a;
y=fsolve(nle,x0); % this is line 3
function f=nle(x)
f(1)=x(1)-a*x(1)^2-x(1)*x(2); % this is line 6
f(2)=2*x(2)-x(2)+3*x(1)*x(2);
end
end
Here a is the parameter I want to pass from command line to the function, and x0 is the start point for the fsolve.
However, when I call the function in malab after specifying a=4 and x0=[1 1]', it gave me the following error:
Error using solve/nle (line 6)
Not enough input arguments.
Error in solve (line 3)
y=fsolve(nle,x0);
I'm quite a newbie for matlab, can anybody tell me where I am doing wrong?
Thanks in advance.
EDIT:
I tried substituting the nle with a function handle #nle, but seems something else went wrong:
Undefined function 'fsolve' for input arguments of type 'function_handle'.
Error in solve (line 3)
y=fsolve(#nle,x0);
Doesn't seem to make sense since I checked the documentation for fsolve, and it says it should indeed use a function handle there...
You miss the '#' in front of nle, i.e.
y = fsolve(#nle,x0);
should work.