I have an integral expression which I defined on Matlab using
x = 0:1/1000:1;
g = #(x) (exp(-1./x.^2).*heaviside(x)).*(exp(-1./(1-x).^2).*heaviside(1-x));
t = 0:1/1000:1;
f = zeros(size(t));
for i = 1:length(t)
f(i) = integral(g,0,t(i));
end
I can plot it, for example, using plot(t,f), but for other purposes I would like to attach a function handle to f, i.e. something like f = #(t) zeros(size(t)). I have not been able to figure it out thus far. f = #(t) integral(#(x)g(x),0,t) is also not sufficient.
Sorry, I can't comment yet. But does this work?
funcHand= #(t) integral(g,0,t);
You don't have to define x in your code above, since the input to integral is a function handle.
Then to check it's the same:
f2 = zeros(size(t));
for i = 1:length(t)
f2(i) = funcHand(t(i));
end
Whoops, the other answer said all the above (just replaced the for loop with arrayfun. I didn't see it while writing the answer.
Edit
If you want to build-in the for loop, try:
funcHand= #(t) arrayfun(#(u) integral(g, 0, u),t);
And test:
plot(funcHand(t))
Try
f = #(u) integral(g, 0, u)
The additional level of indirection in g seems superfluous. Note that I have called the input u. Keep in mind that f will not accept vectors as its inputs. So doing something like f(t) in your current workspace will not create the same array as your for loop is doing. You will have to iterate through the array. The convenience function arrayfun will do this for you:
o = arrayfun(f, t)
It is roughly equivalent to the loop you have now:
o = zeros(size(t));
for i = 1:length(o)
o(i) = f(t(i));
end
arrayfun can actually be incorporated into your function handle to allow it to process vector arguments:
h = #(t) arrayfun(f, t)
To avoid the proliferation of unnecessary function handles, you can do
f = #(t) arrayfun(#(u) integral(g, 0, u), t)
Related
I have a pulse shape that is a function of t, call it h(t), that is in the form of:
h = #(t) function of t
I want to create a pulse train that is consisted of N pulses h(t).
I did this by:
for n=0:N-1
comb = comb + h(t - n);
end
However, once I do that I can't change t in the train.
How can I create this train so that it will also be a function of t?
Thanks.
You just need to make comb an anonymous function too. You can initialise it to some trivial function (i.e. it always outputs 0), and then repeatedly modify it. Since variables declared before an anonymous function declaration, including anonymous functions, are kind of "frozen" to the definition at that point this will work.
h = #(t) _______ % some function of 't'
comb = #(t) 0;
for n = 0:N-1
comb = #(t) comb(t) + h(t - n);
end
We can test this with h = #(t) sin(t) and N=3:
>> comb(pi/2)
ans = 1.1242
>> h(pi/2) + h(pi/2-1) + h(pi/2-2)
ans = 1.1242
Note that just displaying comb can be a bit misleading, since information about the recursive definition is somewhat lost
disp(comb)
#(t)comb(t)+h(t-n)
I've noticed some weird facts about integral2. These are probably due to my limitations in understanding how it works. I have some difficulties in integrating out variables when I have particular functions. For instance, look at the following Code:
function Output = prova(p,Y)
x = p(1);
y = p(2);
w = p(3);
z = p(4);
F1 = #(Data,eta_1,eta_2,x,y,w,z) F2(eta_1,eta_2,Data) .* normpdf(eta_1,x,y) .* normpdf(eta_2,w,z);
Output = integral2(#(eta_1,eta_2)F1(Y,eta_1,eta_2,0,1,10,2),-5,5,-5,5);
end
function O = F2(pp1,pp2,D)
O = pp1 + pp2 + sum(D);
end
In this case the are no problems in evaluating the integral. But if I change the code in this way I obtain some errors, although the output of F2 is exactly the same:
function Output = prova(p,Y)
x = p(1);
y = p(2);
w = p(3);
z = p(4);
F1 = #(Data,eta_1,eta_2,x,y,w,z) F2(eta_1,eta_2,Data) .* normpdf(eta_1,x,y) .* normpdf(eta_2,w,z);
Output = integral2(#(eta_1,eta_2)F1(Y,eta_1,eta_2,0,1,10,2),-5,5,-5,5);
end
function O = F2(pp1,pp2,D)
o = sum([pp1 pp2]);
O = o + sum(D);
end
The problems increase if F2 for example have some matrix multiplication in which "eta_1" and "eta_2", which I want to integrate out, are involved. This problems makes practically impossible to solve computations in which, for instance, we have to integrate out a variable X which is inside a Likelihood Function (whose calculation could require some internal Loop, or Sum, or Prod involving our variable X). What is the solution?
I have a vectorized function which is an objective function for an optimizer (genetic algorithm).
Inside this function, there is a fast optimization which is part of the computation of this function as follow:
function error = ObjectiveFunction(a, b, c)
x = a.*b;
y = c.*b;
z = patternsearch(#fun, [x, y]);
error = x+y.*z;
end
solution = ga(#ObjectiveFunction, 'vectorized', true);
ObjectiveFunction accepts vector of solution. This makes ga works faster. However, since inside ObjectiveFunction there is patternsearch process, this vectorization will be useless because patternsearch (as an optimizer) does not work in vectorized manner.
So I had to edit my function to:
function error = ObjectiveFunction(a, b, c)
x = a.*b;
y = c.*b;
for i = 1:size(x,1)
z(i) = patternsearch(#fun, [x(i), y(i)]);
end
error = x+y.*z;
end
Is there anyway to replace the loop with a vectorized call to patternsearch?
Please consider using arrayfun as follow
function error = ObjectiveFunction(a, b, c)
x = a.*b;
y = c.*b;
z = arrayfun(#(x1,y1) patternsearch(#fun, [x1, y1]),x,y);
error = x+y.*z;
end
I hope this may help
This is a part of my code.
clear all;
clc;
p = 50;
t = [-6 : 0.01 : 6];
f = inline('(t+2).*sin(t)', 't')
v = inline('3*f(p*t+2)','t','f','p')
plot(t,f(t));
v(t,f,p);
figure;
plot(t,v(t,f,p));
Here I have two questions.
Why I have to pass p into the function v even though p is a constant which has already declared ?
How I can get an expression for v completely in terms of t as 3*[(50*t+2)*sin(50*t+2)] or in its simplified form ?
Update
This is an update for the second question
Let
f(x) = 1 + x - x^2
g(x) = sin(x)
If I give f(g(x)), I wanna get the output in words, like this
f(g(x)) = (cos(X))^2 + sin(x)
not in numerical value. Is there any function capable to do that?
1) Why do I have to pass p to v even though p is a constant which has already been declared?
Well, a MATLAB's inline function object has an eval wrapper, so the only variables in its scope are those which were automatically captured from the expression or explicitly specified.
In other words, if you want v to recognize p, you have no other option but declaring it when creating the inline object and passing it to v explicitly. The same goes for f as well!
2) How I can get an expression for v completely in terms of t as 3*[(50*t+2)*sin(50*t+2)] or in its simplified form?
Use anonymous functions, like Shai suggested. They are more powerful, more elegant and much faster. For instance:
v = #(t)(3*(50*t+2)*sin(50*t+2))
Note that if you use a name, which is already in use by a variable, as an argument, the anonymous function will treat it as an argument first. It does see other variables in the scope, so doing something like g = #(x)(x + p) is also possible.
EDIT #1:
Here's another example, this time a function of a function:
x = 1:5;
f = #(x)(x .^ 3); %// Here x is a local variable, not as defined above
g = #(x)(x + 2); %// Here x is also a local variable
result = f(g(x));
or alternatively define yet another function that implements that:
h = #(x)f(g(x)); %// Same result as h = #(x)((x + 2) .^ 3)
result = h(x);
The output should be the same.
EDIT #2:
If you want to make an anonymous function out of the expression string, concatenate the '#(x)' (or the correct anonymous header, as you see fit) to the beginning and apply eval, for example:
expr = '(x + 2) .^ 3';
f = eval(['#(x)', expr]) %// Same result as f = #(x)((x + 2) .^ 3)
Note that you can also do char(f) to convert it back into a string, but you'll have to manually get rid of the '#(...)' part.
EDIT #3:
If you're looking for a different solution, you can explore the Symbolic Toolbox. For example, try:
syms x
f(x) = x + 2
g(x) = x ^ 3
or can also use sym, like so:
f(x) = sym('x + 2');
g(x) = sym('x ^ 3');
Use subs to substitute values and evaluate the symbolic expression.
How about using anonymous functions:
p = 50;
t = -6:0.01:6;
f = #(x) (x+2).*sin(x);
v = #(x) 3*f(p*x+2);
figure;
subplot(1,2,1); plot( t, f(t) ); title('f(t)');
subplot(1,2,2); plot( t, v(t) ); title('v(t)');
Is this what you wanted?
Adding a constant into an inline can be done during its definition.
Instead of
p = 50;
v = inline('3*f(p*t+2)','t','f','p')
You can write
p = 50;
v = inline( sprintf('3*f(%f*t+2)', p), 't','f')
I want to create a function handle to the function:
f = #(x) (x-1)*(x-2)*...*(x-50);
How can I do this in MATLAB without typing all 50 terms?
Here is a vectorized solution:
y = prod((x-[1:50]))
Or if you want an anonymous function:
f = #(x) ( prod((x-[1:50])) )
By the way, it might not be faster than #Chris solution (which is good, and I upvoted it), because of Matlab JIT-Accelerator.
You could wrap it in a function. For example,
function y = myfunc(x, n)
y = 1.;
for i = 1:n
y = y*(x-i);
end
end
The function you defined is basically the product of a sequence, which are trivially written as for loops.
In your case you want to compute this result for 50 terms, so you could just use y = myfunc(x, 50) or, if you want this to be a function handle, you could define
f = #(x) myfunc(x, 50);