I have N functions in MATLAB and I can define them using strcat, num2str and eval in a for loop. So without defining by hand I am able to define N functions. Let N=4 and let them be given as follows:
f1=#(x) a1*x+1;
f2=#(x) a2*x+1;
f3=#(x) a3*x+1;
f4=#(x) a4*x+1;
Now I add these four functions and I can do this by hand as follows:
f=#(x)(f1(x)+f2(x)+f3(x)+f4(x));
Here I can do it by hand because I know that N=4. However, in general I never know how many functions I will have. For all cases I cannot write a new function.
Is there any way to do this automatically? I mean if I give N=6 I am expecting to see MATLAB giving me this:
f=#(x)(f1(x)+f2(x)+f3(x)+f4(x)+f5(x)+f6(x));
Whenever I give N=2 then I must have the function f, defined as follows:
f=#(x)(f1(x)+f2(x));
How can we do this?
First of all, you should read this answer that gives a series of reasons to avoid the use of eval. There are very few occasions where eval is necessary, in all other cases it just complicates things. In this case, you use to dynamically generate variable names, which is considered a very bad practice. As detailed in the linked answer and in further writings linked in that answer, dynamic variable names make the code harder to read, harder to maintain, and slower to execute in MATLAB.
So, instead of defining functions f1, f2, f3, ... fN, what you do is define functions f{1}, f{2}, f{3}, ... f{N}. That is, f is a cell array where each element is an anonymous function (or any other function handle).
For example, instead of
f1=#(x) a1*x+1;
f2=#(x) a2*x+1;
f3=#(x) a3*x+1;
f4=#(x) a4*x+1;
you do
N = 4;
a = [4.5, 3.4, 7.1, 2.1];
f = cell(N,1);
for ii=1:N
f{ii} = #(x) a(ii) * x + 1;
end
With these changes, we can easily answer the question. We can now write a function that outputs the sum of the functions in f:
function y = sum_of_functions(f,x)
y = 0;
for ii=1:numel(f)
y = y + f{ii}(x);
end
end
You can put this in a file called sum_of_functions.m, or you can put it at the end of your function file or script file, it doesn't matter. Now, in your code, when you want to evaluate y = f1(x) + f2(x) + f3(x)..., what you write is y = sum_of_functions(f,x).
Related
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.
In the latest MATLAB release, I am using the feature of local functions in the script to simplify and shorten the code.
Right now I have a simple script that does mathematical calculations, up to millions of iterations in a for loop. From the loop, I extracted the mathematical part as it is used in multiple independent loops. However, I have the problem with speed and constant variables that need to be in the function. If I write the code without the local function, it is fast, but in the case of using the function, it is 10x slower.
I think it is due to the repetitive read-in of variables in the function. If the read-in happens from a file, it takes forever. The best I have achieved is by using evalin function to call in variables from the base workspace. It works, but as I mentioned, it is 10x slower than without a separate local function.
How to overcome this?
The structure looks like this:
load('Data.mat'); r=1; %Reads variables L,n
for Lx = L1:Ld:L2
for x = x1:d:x2
Yx(r) = myF(x); r=r+1
end
end
%A few more different loops...
function y = myF(x)
L = evalin('base','L');
n = evalin('base','n');
...
a = (x-L)^2; b = sin(a)-n/2; ... y=b*a/L;
end
You will want to explicitly pass the variables as additional input arguments to myF so that you don't have to load them every time. This way you don't have to worry about calling evalin or loading them from the file every iteration through the loop.
load('Data.mat'); r=1; %Reads variables L,n
for Lx = L1:Ld:L2
for x = x1:d:x2
Yx(r) = myF(x, L, n); r=r+1
end
end
function y = myF(x, L, n)
a = (x-L)^2; b = sin(a)-n/2;
y=b*a/L;
end
If for some reason you can't modify myF, just turn the entire thing into a function and make myF a subfunction and it will therefore automatically have access to any variable within the parent function. In this case you'll also want to assign the output of load to a variable to prevent strange behavior and not have issues with adding a variable to a static workspace
function myfunction()
data = load('Data.mat');
L = data.L;
n = data.n;
for Lx = L1:Ld:L2
for x = x1:d:x2
Yx(r) = myF(x);
r=r+1
end
end
function y = myF(x)
a = (x-L)^2; b = sin(a)-n/2;
y=b*a/L;
end
end
A third alternative is to take a hard look at myF and consider whether you can vectorize the contents so that you don't actually have to continuously call it millions of times.
You could vectorize it by doing something like
[Lx, x] = ndgrid(L1:Ld:L2, x1:d:x2);
a = (x - L).^2;
b = sin(a) - (n / 2);
Yx = (b .* a) ./ L;
Is it possible in Matlab to specify at function declaration that this specific function is to be applied to indiviual arguments (as opposed to vectors)? An if so, what about anonymous functions?
What I mean is, instead of writing something like
f = #(x, y) x.^2+2*x.*y.^2-x./(y.^3)...
One might be able to write something like
f = .#(x, y) x^2+2*x*y^2-x/y^3
This would help a lot light-headed people like me who tend to forget dots.
Sorry if it's a duplicate, I searched but as you can see I have trouble putting it in clear words so I cannot find an answer.
not sure if that does the trick for you, but you can but an arrayfun inside the anonymous function like:
f = #(x,y) x.^2+2*x.*y.^2-x./(y.^3);
g = #(x,y) arrayfun(#(A,B) A^2+2*A*B^2-A/B^3 ,x,y) %for matrix output
h = #(x,y) arrayfun(#(A,B) A^2+2*A*B^2-A/B^3 ,x,y,'UniformOutput',false); % for cell output
with A being an element of x and B being an element of y
but i want to make a program in which i can generate a function of multiple variables that depend on the number of rows of a matrix.
for k = 1:sizel;
f(k)=(alpha(k,1)+(beta(k,1)*p(k))+(gamma(k,1)*p(k)^2));
end
cost=(sum(f))
this is for the purpose of optimization so i need that at the end the variables are declares as p(1),p(2),p(3)... these will be the input for my function.
Note: i dont want to assign values to the variables because this will be done be the optimization algorithm in the optimization toolbox.
here is the complete code
function cost = cost(p) ;
clc
clear
costfunctionconstantsmatrix;
sizel=size(CostFormulaconstants);
alpha=CostFormulaconstants(:,1);
beta=CostFormulaconstants(:,2);
gamma=CostFormulaconstants(:,3);
for k = 1:sizel;
f(k)=(alpha(k,1)+(beta(k,1)*p(k))+(gamma(k,1)*p(k)^2));
end
cost=(sum(f))
end
i used the symbolic approach and i got the correct answer for the cost indeed, i got something like this: (53*p(1))/10 + (11*p(2))/2 + (29*p(3))/5 + p(1)^2/250 + (3*p(2)^2)/500 + (9*p(3)^2)/1000 + 1100. But when i try to specify my function to be optimized in the optimization toolbox it tells me that the variables p are sym and cannot be converted to double. the trouble is how to convert this expression to double so that the optimization algorithm can input values for the variable p(1), p(2) and p(3)
Can you pass the matrix as an argument to the function?
function cost = fcn(my_mat)
[m,n] = size(my_mat);
f = zeros(m,1);
for k = 1:m % for example
f(k)=(alpha(k,1)+(beta(k,1)*p(k))+(gamma(k,1)*p(k)^2));
end
cost = sum(f);
end
Your problem is not entirely clear to me but I believe you wish to generate a series of functions in which the variables alpha, beta, gamma are constants with different values for each function, and the vector p is an argument.
What confuses me in your question is that you use the index k for both the constants and the arguments, which I think is not what you intended to write. Assuming I understand your goal, a solution may make use of function handles.
The notation f(k) = #(p) p(1)+p(2) for example, generates a function that adds p(1) and p(2). Abbreviating CostFormulaconstants to cf, the following would generate a series of functions, one for each row in cf.
for k = 1 : size(cf, 1)
f{k} = #(p) cf(k,1) + cf(k,2)*p(1) + cf(k,3)*p(2)^2;
end
You can supply individual function handles to callers from the optimization toolbox simply with f{3} for the third function, for example. A call to f{3} would look like
a = f{3}([3,4]);
If your functions are indeed all polynomials, polyval may be worth a look as well.
EDIT: After clarification, the problem seems a bit simpler, no need for function handles. Why not simply
function c = cost(p)
c = 0;
cf = [...]; % your coefficients here.
for k = 1 : size(cf, 1)
c = c + cf(k,1) + cf(k,2)*p(k) + cf(k,3)*p(k)^2;
end
I have one file with the following code:
function fx=ff(x)
fx=x;
I have another file with the following code:
function g = LaplaceTransform(s,N)
g = ff(x)*exp(-s*x);
a=0;
b=1;
If=0;
h=(b-a)/N;
If=If+g(a)*h/2+g(b)*h/2;
for i=1:(N-1)
If=If+g(a+h*i)*h;
end;
If
Whenever I run the second file, I get the following error:
Undefined function or variable 'x'.
What I am trying to do is integrate the function g between 0 and 1 using trapezoidal approximations. However, I am unsure how to deal with x and that is clearly causing problems as can be seen with the error.
Any help would be great. Thanks.
Looks like what you're trying to do is create a function in the variable g. That is, you want the first line to mean,
"Let g(x) be a function that is calculated like this: ff(x)*exp(-s*x)",
rather than
"calculate the value of ff(x)*exp(-s*x) and put the result in g".
Solution
You can create a subfunction for this
function result = g(x)
result = ff(x) * exp(-s * x);
end
Or you can create an anonymous function
g = #(x) ff(x) * exp(-s * x);
Then you can use g(a), g(b), etc to calculate what you want.
You can also use the TRAPZ function to perform trapezoidal numerical integration. Here is an example:
%# parameters
a = 0; b = 1;
N = 100; s = 1;
f = #(x) x;
%# integration
X = linspace(a,b,N);
Y = f(X).*exp(-s*X);
If = trapz(X,Y) %# value returned: 0.26423
%# plot
area(X,Y, 'FaceColor',[.5 .8 .9], 'EdgeColor','b', 'LineWidth',2)
grid on, set(gca, 'Layer','top', 'XLim',[a-0.5 b+0.5])
title('$\int_0^1 f(x) e^{-sx} \,dx$', 'Interpreter','latex', 'FontSize',14)
The error message here is about as self-explanatory as it gets. You aren't defining a variable called x, so when you reference it on the first line of your function, MATLAB doesn't know what to use. You need to either define it in the function before referencing it, pass it into the function, or define it somewhere further up the stack so that it will be accessible when you call LaplaceTransform.
Since you're trying to numerically integrate with respect to x, I'm guessing you want x to take on values evenly spaced on your domain [0,1]. You could accomplish this using e.g.
x = linspace(a,b,N);
EDIT: There are a couple of other problems here: first, when you define g, you need to use .* instead of * to multiply the elements in the arrays (by default MATLAB interprets multiplication as matrix multiplication). Second, your calls g(a) and g(b) are treating g as a function instead of as an array of function values. This is something that takes some getting used to in MATLAB; instead of g(a), you really want the first element of the vector g, which is given by g(1). Similarly, instead of g(b), you want the last element of g, which is given by g(length(g)) or g(end). If this doesn't make sense, I'd suggest looking at a basic MATLAB tutorial to get a handle on how vectors and functions are used.