I have the following anonymous function:
f = #(x)x^2+2*x+1
I'm using this so that I use it in the following way:
f(0) = 1
But what if I want to find the derivative of such a function while still keeping it's anonymous function capability? I've tried doing the following but it doesn't work:
f1 = #(x)diff(f(x))
but this just returns
[]
Any thoughts on how to accomplish this?
Of course I could manually do this in 3 seconds but that's not the point...
If you have symbolic math toolbox, you can use symbolic functions to achieve the desired as follows:
syms x
myFun=x^2+2*x+1;
f=symfun(myFun,x);
f1=symfun(diff(f),x);
%Check the values
f(2)
f1(2)
You should get 9 and 6 as answers.
When you do diff of a vector of n elements it just outputs another vector of n-1 elements with the consecutive differences.. so when you put a 1 element vector you get an empty one.
A way to go would be to decide an epsilon and use the Newton's difference quotient:
epsilon = 1e-10;
f = #(x) x^2+2*x+1;
f1 = #(x) (f(x+epsilon) - f(x)) / epsilon;
or just do the math and write down the formula:
f1 = #(x) 2*x+2;
http://en.wikipedia.org/wiki/Numerical_differentiation
#jollypianoman this works to me. Actually you need to say that the symfun has to be evaluate using eval command, then you get all the features of an anonymous function. the best is to read the example below...
clear
N0=1;N1=5;
N=#(t) N0+N1*sin(t);
syms t
Ndot=symfun(diff(N(t)),t);
Ndot_t=#(t) eval(Ndot);
Ndot_t(0)
ans = 5
Ndot_t(1)
ans = 2.7015
[tstop] = fsolve(Ndot_t,pi/3)
tstop =
1.5708
cheers,
AP
Related
I have a function F which takes as an input a vector a. Both the output of the function and a are vectors of length N, where N is arbitrary. Each component Fn is of the form g(a(n),a(n-k)), where g is the same for each component.
I want to implement this function in matlab using its symbolic functionality and calculate its Jacobian (and then store both the function and its jacobian as a regular .m file using matlabFunction). I know how to do this for a function where each input is a scalar that can be handled manually. But here I want a script that is capable of producing these files for any N. Is there a nice way to do this?
One solution I came up with is to generate an array of strings "a0","a1", ..., "aN" and define each component of the output using eval. But this is messy and I was wondering if there is a better way.
Thank you!
[EDIT]
Here is a minimal working example of my current solution:
function F = F_symbolically(N)
%generate symbols
for n = 1:N
syms(['a',num2str(n)]);
end
%define output
F(1) = a1;
for n = 2:N
F(n) = eval(sprintf('a%i + a%i',n,n-1));
end
Try this:
function F = F_symbolically(N)
a = sym('a',[1 N]);
F = a(1);
for i=2:N
F(i) = a(i) + a(i-1);
end
end
Note the use of sym function (not syms) to create an array of symbolic variables.
let say that from given function f(t), we want to construct new function which is given from existed function by this way
where T is some constant let say T=3; of course k can't be from -infinity to infinity in reality because we can't do infinity summation using computer,so it is first my afford
first let us define our function
function y=f(t);
y=-1/(t^2);
end
and second program
k=-1000:1:999;
F=zeros(1,length(k));
T=3;
for t=1:length(k)
F(t)=sum(f(t+k*T));
end
but when i am running second program ,i am getting
>> program
Error using ^
Inputs must be a scalar and a square matrix.
To compute elementwise POWER, use POWER (.^) instead.
Error in f (line 2)
y=-1/(t^2);
Error in program (line 5)
F(t)=sum(f(t+k*T));
so i have two question related to this program :
1.first what is error why it shows me mistake
how can i do it in excel? can i simplify it somehow? thanks in advance
EDITED :
i have changed my code by this way
k=-1000:1:999;
F=zeros(1,length(k));
T=3;
for t=1:length(k)
result=0;
for l=1:length(k)
result=result+f(t+k(l)*T);
end
F(t)=result;
end
is it ok?
To solve your problem in a vectorized way, you'll have to change the function f such that it can be called with vectors as input. This is, as #patrik suggested, achieved by using the element-wise operators .* ./ .^ (Afaik, no .+ .- exist). Unfortunately the comment of #rayryeng is not entirely correct, which may have lead to confusion. The correct way is to use the element-wise operators for both the division ./ and the square .^:
function y = f(t)
y = -1 ./ (t.^2);
end
Your existing code (first version)
k = -1000:1:999;
F = zeros(1,length(k));
T = 3;
for t=1:length(k)
F(t) = sum(f(t+k*T));
end
then works as expected (and is much faster then the version you posted in the edit).
You can even eliminate the for loop and use arrayfun instead. For simple functions f, you can also use function handles instead of creating a separate file. This gives
f = #(t) -1 ./ (t.^2);
k = -1000:1:999;
t = 1:2000;
T = 3;
F = arrayfun(#(x)sum(f(x+k*T)), t);
and is even faster and a simple one-liner. arrayfun takes any function handle as first input. We create a function handle which takes an argument x and does the sum over all k: #(x) sum(f(x+k*T). The second argument, the vector t, contains all values for which the function handle is evaluated.
As proposed by #Divakar in comments, you can also use the bsxfun function:
f = #(t) -1 ./ (t.^2);
k = -1000:1:999;
t = 1:2000;
T = 3;
F = sum(f(bsxfun(#plus,k*T,t.')),2);
where bsxfun creates a matrix containing all combinations between t and k*T, they are all evaluated using f(...) and last, the sum along the second dimension sums over all k's.
Benchmarking
Lets compare these solutions:
Combination of for loop and sum (original question):
Elapsed time is 0.043969 seconds.
Go through all combinations in 2 for loops (edited question):
Elapsed time is 1.367181 seconds.
Vectorized approach with arrayfun:
Elapsed time is 0.063748 seconds.
Vectorized approach with bsxfun as proposed by #Divakar:
Elapsed time is 0.099399 seconds.
So (sadly) the first solution including a for loop beats both vectorized approaches. For larger k vectors (-10000:1:9999), this behavior can be reproduced. The conclusion seems to be that MATLAB has indeed learned how to optimize for loops.
I have a problem with symbolic functions. I am creating function of my own whose first argument is a string. Then I am converting that string to symbolic function:
f = syms(func)
Lets say my string is sin(x). So now I want to calculate it using subs.
a = subs(f, 1)
The result is sin(1) instead of number.
For 0 it works and calculates correctly. What should I do to get the actual result, not only sin(1) or sin(2), etc.?
You can use also use eval() to evaluate the function that you get by subs() function
f=sin(x);
a=eval(subs(f,1));
disp(a);
a =
0.8415
syms x
f = sin(x) ;
then if you want to assign a value to x , e.g. pi/2 you can do the following:
subs(f,x,pi/2)
ans =
1
You can evaluate functions efficiently by using matlabFunction.
syms s t
x =[ 2 - 5*t - 2*s, 9*s + 12*t - 5, 7*s + 2*t - 1];
x=matlabFunction(x);
then you can type x in the command window and make sure that the following appears:
x
x =
#(s,t)[s.*-2.0-t.*5.0+2.0,s.*9.0+t.*1.2e1-5.0,s.*7.0+t.*2.0-1.0]
you can see that your function is now defined by s and t. You can call this function by writing x(1,2) where s=1 and t=1. It should generate a value for you.
Here are some things to consider: I don't know which is more accurate between this method and subs. The precision of different methods can vary. I don't know which would run faster if you were trying to generate enormous matrices. If you are not doing serious research or coding for speed then these things probably do not matter.
We have an equation similar to the Fredholm integral equation of second kind.
To solve this equation we have been given an iterative solution that is guaranteed to converge for our specific equation. Now our only problem consists in implementing this iterative prodedure in MATLAB.
For now, the problematic part of our code looks like this:
function delta = delta(x,a,P,H,E,c,c0,w)
delt = #(x)delta_a(x,a,P,H,E,c0,w);
for i=1:500
delt = #(x)delt(x) - 1/E.*integral(#(xi)((c(1)-c(2)*delt(xi))*ms(xi,x,a,P,H,w)),0,a-0.001);
end
delta=delt;
end
delta_a is a function of x, and represent the initial value of the iteration. ms is a function of x and xi.
As you might see we want delt to depend on both x (before the integral) and xi (inside of the integral) in the iteration. Unfortunately this way of writing the code (with the function handle) does not give us a numerical value, as we wish. We can't either write delt as two different functions, one of x and one of xi, since xi is not defined (until integral defines it). So, how can we make sure that delt depends on xi inside of the integral, and still get a numerical value out of the iteration?
Do any of you have any suggestions to how we might solve this?
Using numerical integration
Explanation of the input parameters: x is a vector of numerical values, all the rest are constants. A problem with my code is that the input parameter x is not being used (I guess this means that x is being treated as a symbol).
It looks like you can do a nesting of anonymous functions in MATLAB:
f =
#(x)2*x
>> ff = #(x) f(f(x))
ff =
#(x)f(f(x))
>> ff(2)
ans =
8
>> f = ff;
>> f(2)
ans =
8
Also it is possible to rebind the pointers to the functions.
Thus, you can set up your iteration like
delta_old = #(x) delta_a(x)
for i=1:500
delta_new = #(x) delta_old(x) - integral(#(xi),delta_old(xi))
delta_old = delta_new
end
plus the inclusion of your parameters...
You may want to consider to solve a discretized version of your problem.
Let K be the matrix which discretizes your Fredholm kernel k(t,s), e.g.
K(i,j) = int_a^b K(x_i, s) l_j(s) ds
where l_j(s) is, for instance, the j-th lagrange interpolant associated to the interpolation nodes (x_i) = x_1,x_2,...,x_n.
Then, solving your Picard iterations is as simple as doing
phi_n+1 = f + K*phi_n
i.e.
for i = 1:N
phi = f + K*phi
end
where phi_n and f are the nodal values of phi and f on the (x_i).
I want to solve two nonlinear equations in MATLAB so i did the following:
part of my script
c=[A\u;A\v];
% parts of code are omitted.
x0=[1;1];
sol= fsolve(#myfunc,x0);
the myfunc function is as follows
function F = myfunc(x)
F=[ x(1)*c(1)+x(2)*c(2)+x(1)*x(2)*c(3)+c(4)-ii;
x(1)*c(5)+x(2)*c(6)+x(1)*x(2)*c(7)+c(8)-jj];
end
i have two unknowns x(1) and x(2)
my question is How to pass a values(c,ii,jj) to myfunc in every time i call it?
or how to overcome this error Undefined function or method 'c' for input arguments of type 'double'.
thanks
Edit: The previous answer was bogus and not contributing at all. Hence has been deleted. Here is the right way.
In your main code create a vector of the coefficients c,ii,jj and a dummy function handle f_d
coeffs = [c,ii,jj];
f_d = #(x0) myfunc(x0,coeffs); % f_d considers x0 as variables
sol = fsolve(f_d,x0);
Make your function myfunc capable of taking in 2 variables, x0 and coeffs
function F = myfunc(x, coeffs)
c = coeffs(1:end-2);
ii = coeffs(end-1);
jj = coeffs(end);
F(1) = x(1)*c(1)+x(2)*c(2)+x(1)*x(2)*c(3)+c(4)-ii;
F(2) = x(1)*c(5)+x(2)*c(6)+x(1)*x(2)*c(7)+c(8)-jj;
I think that should solve for x0(1) and x0(2).
Edit: Thank you Eitan_T. Changes have been made above.
There is an alternative option, that I prefer, if a function handle is not what you are looking for.
Say I have this function:
function y = i_have_a_root( x, a )
y = a*x^2;
end
You can pass in your initial guess for x and a value for a by just calling fsolve like so:
a = 5;
x0 = 0;
root = fsolve('i_have_a_root',x0,[],a);
Note: The [] is reserved for fsolve options, which you probably want to use. See the second call to fsolve in the documentation here for information on the options argument.