I have a use case as follows:
Inside F.m I have a function F that takes as its argument a 2 x 1 matrix x. F needs to matrix multiply the matrix kmat by x. kmat is a variable that is generated by a script.
So, what I did was set kmat to be global in the script:
global kmat;
kmat = rand(2);
In F.m:
function result = F(x)
global kmat;
result = kmat*x;
end
Then finally, in the script I have (x_0 has already been defined as an appropriate 2 x 1 matrix, and tstart and tend are positive integers):
xs = ode45(F, [tstart, tend], x_0);
However, this is causing the error:
Error using F (line 3)
Not enough input arguments.
Error in script (line 12)
xs = ode45(F, [tstart, tend], x_0);
What is going on here, and what can I do to fix it? Alternatively, what is the right way to pass kmat to F?
Firstly, the proper way to handle kmat is to make it an input argument to F.m
function result = F(x,kmat)
result = kmat*x;
end
Secondly, the input function to ode45 must be a function with inputs t and x (possibly vectors, t is the dependent variable and x is the dependent). Since your F function doesn't have t as an input argument, and you have an extra parameter kmat, you have to make a small anonymous function when you call ode45
ode45(#(t,x) F(x,kmat),[tstart tend],x_0)
If your derivative function was function result=derivative(t,x), then you simply do ode45(#derivative,[tstart tend],x_0) as Erik said.
I believe F in ode45(F,...) should be a function handle, i.e. #F. Also, you can have a look at this page of the MATLAB documentation for different methods to pass extra parameters to functions.
Related
I have a question about using the lsqnonlin function.
In my case I have two functions:
f_1=#(t,x)sin(t+x.^2);
f_2=#(t,x)cos(x.^2)+3.*t.^2;
f = {f_1, f_2};
I want to find the values of the arguments t and x which would result in the least square error, defined as: f_1(t,x)^2+f_2(t,x)^2. In other words, argmin for LSE.
My code is as follow with initial guess [1,2]:
lsqnonlin(f,[1,2])
And I'm getting the error:
Error in lsqnonlin (line 196)
initVals.F = feval(funfcn{3},xCurrent,varargin{:});
Caused by:
Failure in initial objective function evaluation. LSQNONLIN cannot continue.
lsqnonlin can be used for vector function and vector input according to the documentation. I wonder how to prepare corresponding codes for it. Could anyone suggest the solution?
You are getting an error because lsqnonlin expects a scalar function handle that maps a vector to a vector, whereas you specify a cell array of function handles. To fix this, instead of a vector of functions outputting one scalar each, you need to rewrite it into a single function that accepts a vector of inputs and also outputs a vector:
f = #(xt)[sin(xt(2)+xt(1).^2), cos(xt(1).^2)+3.*xt(2).^2];
% xt = [x,t]
% f = [f_1(xt), f_2(xt)]
so f and xt are both vectors of 2 elements.
Then, the solver works:
lsqnonlin(f,[1,2])
ans =
1.6144 0.5354
I have a non-linear equation with a constant 'X'.
I want to obtain the solution of the equation for 10 different values of 'X'.
What is the best way of doing it in MATLAB?
Employing fsolve, I thought of doing this using a loop which runs 10 times. But the problem is that it's not possible to send value of 'X' as a parameter to the function which is called by fsolve (as according to its syntax, fsolve can send only the initial guess value) and which contains the non-linear equation.
This is my MATLAB code:
function f = crd(m)
X=0.1; %Paramter for whose different values I want to solve the NLE using a loop
t=(1/(0.8/3600))*log(1/(1-X));
U=350; P=0.1134; T=165; L=21.415; %Constants
a=0.00102*820*2200/(U*P); %Constant
Q=(0.8/3600)*900*exp(-(0.8/3600)*t)*0.9964*(347.3*1000);
Tmi=60; %Constant
b=m*2200;
q=(a/t)+(b/L);
f = ( b - (Q/(T-Tmi)) ) * (b/(L*L*q*q)) - exp(-1/q);
end
Changing the value of the parameter 'X' each time, I can use "fsolve(#crd,10)" from the Command Window multiple times. But I want to do this using a loop.
I want to get solution for X=0.1,0.2,...,0.9,1.0
One possible way to do this is to change your function to accept two parameters: m and X:
function f = crd(m,X)
....
end
Then, when you want to call fsolve, you can create a temporary function that accepts a single parameter m, and pass an actual value for X to it. That is:
for X = [0.1, 0.2, 0.3, 0.4]
f = #(m) crd(m,X); % a function that accepts m as input
sol = fsolve(f, 10);
...
end
So i'm a little confounded by how to structure my problem.
So the assignment states as following:
Type a m-file numerical_derivative.m that performs numerical derivation. Use
it to calculate f'(-3) when f(x) = 3x^2 /(ln(1-x))
In the m-file you to use h = 10^-6 and have the following mainfunction:
function y = numericalderivative (f, x)
% Calculates the numerical value in the case of f in punk x.
% --- Input ---
% f: function handle f(x)
% x: the point where the derivative is calculated
% --- output ---
% y: the numerical derivative of f on the point x
If I want to save it as a file and run the program in matlab, does't it make it redundant to use handles then?
I won't give you the answer to your homework, but perhaps a simpler example would help.
Consider the following problem
Write a function named fdiff which takes the following two arguments:
A function f represented by a function handle which takes one argument,
and a point x which can be assumed to be in the domain of the f.
Write fdiff so that it returns the value f(x) - f(x-1)
Solution (would be in the file named fdiff.m)
function result = fdiff(f, x)
result = f(x) - f(x-1);
end
Example Use Cases
>> my_function1 = #(x) 3*x^2 /(log(1-x));
>> fdiff(my_function1, -3)
ans =
-10.3477
>> my_function2 = #(x) x^2;
>> fdiff(my_function2, 5)
ans =
9
What you've created with fdiff is a function which takes another function as an input. As you can see it doesn't just work for 3*x^2 /(log(1-x)) but any function you want to define.
The purpose of your assignment is to create something very similar, except instead of computing f(x) - f(x-1), you are asked write a function which approximates f'(x). Your use-case will be nearly identical to the first example except instead of fdiff your function will be named numericalderivative.
Note
In case it's not clear, the second example defines the my_function2 as x^2. The value returned by fdiff(my_function2, 5) is therefore 5^2 - 4^2 = 9.
When you make this as a function file and run this in MATLAB without any input arguments i.e., 'f' and 'x', it will give you the error: 'not enough input arguments'. In order to run the file you have to type something like numericalderivative (3x^2 /(ln(1-x)), 5), which gives the value of the numerical derivative at x = 5.
Functions and, in MATLAB function files are a simple implementation of the DRY programming method. You're being asked to create a function that takes a handle and an x file, then return the derivative of that function handle and that x value. The point of the function file is to be able to re-use your function with either multiple function handles or multiple x values. This is useful as it simply involves passing a function handle and a numeric value to a function.
In your case your script file or command window code would look something like:
func = #(x) (3*x^2)/log(1-x);
x = -3;
num_deriv = numericalderivative(func,x);
You should write the code to make the function numericalderivative work.
I am trying to implement the following code in Matlab:
v0=eps;
t0=0; tf=10;
[t,v]=ode45(#const_pow, [t0 tf], v0);
And the const_pow function is:
function f=const_pow(t,v)
f=(P/v-F0-c*v^2)/m;
end
While I am running the program, the following error shows up:
Error using ==> odearguments at 103
CONST_POW returns a vector of length 0, but the length of initial
conditions vector is 1. The vector returned by CONST_POW and the
initial conditions vector must have the same number of elements.
Any suggestions on where the error might be?
From your comments, you're defining some variables globally. In order to access them inside const_pow, you need to tell that function to use the global variables, e.g.:
function f=const_pow(t,v)
global P F0 c m
f=(P/v-F0-c*v^2)/m;
end
Without doing this, the function will not touch the global variables (this is to stop you accidentally changing their values). For example:
function f=const_pow(t,v)
global P F0 c
m = 23.7; % this does not change the global value of m
f=(P/v-F0-c*v^2)/m;
end
An alternative to passing a number of variables without using globals is to define a structure with your required data:
data.P = % initial values
data.F0 =
data.c =
data.m =
This also gives you a little more flexibility if there is more than one set of options. Or, you could pass the variables in directly. To do this with ode45, you would need to parameterize the function using an anonymous function:
function f=const_pow(t,v,P,F0,c,m)
% and so on
end
With P,F0,c,m pre-defined:
odefun = #(t,v) const_pow(t,v,P,F0,c,m); % anonymous function
Then:
[t,v]=ode45(odefun, [t0 tf], v0);
I want to minimize this function:
function [GCV2]=GCV(y,x,k)
[n, p]=size(x);
A=(x'*x+k*eye(p));
A=A\x';
A=x*A;
I_mat=eye(n);
num2=(I_mat-A);
num2=num2*y;
num2=norm(num2);
num2=num2^2;
num2=num2/n;
%(norm((I_mat-A)*y)^2)/n;
den2=(I_mat-A);
den2=trace(den2);
den2=den2/n;
den2=den2^2;
GCV2=num2/den2;
end
The x and y values are 13-by-4) and 13-by-1 arrays, respectively, and these values are already defined in the Matlab workspace. I want to optimize on the k value so that the function value GCV is minimized.
The parameter being optimized as well as the output are scalar so fminsearch should be appropriate.
But I can't get it to run?
I've tried several methods, the latest being:
k_min = fminsearch(#GCV,(x;y;0));
??? k_min = fminsearch(#GCV,(x;y;0));
|
Error: Unbalanced or unexpected parenthesis or bracket.
What am I doing wrong?
Looks like you're learning about anonymous functions. fminsearch minimizes a single variable (which may be a vector). Your objective function must therefore have only one input. You have a function, GCV, that takes three inputs. Two are static and are defined in the workspace outside of the minimization, while k is the one to be minimized. To create a function with one input from GCV, you can use any anonymous function, taking care to specify which variables are parameters:
x = ...
y = ...
k0 = 0;
k_min = fminsearch(#(k)GCV(y,x,k),k0);
The second input to fminsearch is the starting parameter (i.e. k0), so specify a starting value of k. Then you can define an anonymous helper function and optimize on that:
>> % define x,y
>> GCVk = #(k) GCV(y,x,k);
>> k0 = 0;
>> k_min = fminsearch(GCVk,k0)
There is probably another way to do this, but this is one of the listed ways of passing additional parameters for the optimizer.
And since there are no bonus points for being first, how about humor. Let's have an example:
>> x=1; y=1;
>> GCVk = #(k) x+y+k; k0=0;
>> k_min = fminsearch(GCVk,k0)
Exiting: Maximum number of function evaluations has been exceeded
- increase MaxFunEvals option.
Current function value: -316912650057057490000000000.000000
k_min =
-3.1691e+26
Found it - the lowest number (minus 2) in the world! Profit?