I have a problem when computing the derivative of a first order function as below:
syms x(t)
xd = diff(x);
y = xd*xd;
how to compute derivative of y by xd;
functionalDerivative(y,xd);
Then, it raises an error as below:
Error using symengine
The variable 'diff(x(t), t)' is invalid.
The result should be:
2*diff(x,t)
I also think about name xd as a symbolic variable, then use diff(y,xd) but this way is not good for some situation. Do we have any method can directly compute the derivative of a differential function?
Please suggest me some solutions.
Thank in advance!
I find it a little odd that the Symbolic Engine can't handle that, lacking a better word I'll borrow one from the TeX world, unexpandable expression. However, you can get around the limitation by subs-ing in a temporary variable, taking the derivative, and subs-ing it back in:
>> syms u(t)
>> dydxd = subs(functionalDerivative(subs(y,xd(t),u(t)),u),u(t),xd(t))
dydxd(t) =
2*diff(x(t), t)
Hopefully this subs approach will work in most cases, but more complex expressions for y may make it not work.
Related
I want to minimize a function like below:
Here, n can be 5,10,50 etc. I want to use Matlab and want to use Gradient Descent and Quasi-Newton Method with BFGS update to solve this problem along with backtracking line search. I am a novice in Matlab. Can anyone help, please? I can find a solution for a similar problem in that link: https://www.mathworks.com/help/optim/ug/unconstrained-nonlinear-optimization-algorithms.html .
But, I really don't know how to create a vector-valued function in Matlab (in my case input x can be an n-dimensional vector).
You will have to make quite a leap to get where you want to be -- may I suggest to go through some basic tutorial first in order to digest basic MATLAB syntax and concepts? Another useful read is the very basic example to unconstrained optimization in the documentation. However, the answer to your question touches only basic syntax, so we can go through it quickly nevertheless.
The absolute minimum to invoke the unconstraint nonlinear optimization algorithms of the Optimization Toolbox is the formulation of an objective function. That function is supposed to return the function value f of your function at any given point x, and in your case it reads
function f = objfun(x)
f = sum(100 * (x(2:end) - x(1:end-1).^2).^2 + (1 - x(1:end-1)).^2);
end
Notice that
we select the indiviual components of the x vector by matrix indexing, and that
the .^ notation effects that the operand is to be squared elementwise.
For simplicity, save this function to a file objfun.m in your current working directory, so that you have it available from the command window.
Now all you have to do is to call the appropriate optimization algorithm, say, the quasi Newton method, from the command window:
n = 10; % Use n variables
options = optimoptions(#fminunc,'Algorithm','quasi-newton'); % Use QM method
x0 = rand(n,1); % Random starting guess
[x,fval,exitflag] = fminunc(#objfun, x0, options); % Solve!
fprintf('Final objval=%.2e, exitflag=%d\n', fval, exitflag);
On my machine I see that the algorithm converges:
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the optimality tolerance.
Final objval=5.57e-11, exitflag=1
Lets say, I have a function 'x' and a function '2sin(x)'
How do I output the intersects, i.e. the roots in MATLAB? I can easily plot the two functions and find them that way but surely there must exist an absolute way of doing this.
If you have two analytical (by which I mean symbolic) functions, you can define their difference and use fzero to find a zero, i.e. the root:
f = #(x) x; %defines a function f(x)
g = #(x) 2*sin(x); %defines a function g(x)
%solve f==g
xroot = fzero(#(x)f(x)-g(x),0.5); %starts search from x==0.5
For tricky functions you might have to set a good starting point, and it will only find one solution even if there are multiple ones.
The constructs seen above #(x) something-with-x are called anonymous functions, and they can be extended to multivariate cases as well, like #(x,y) 3*x.*y+c assuming that c is a variable that has been assigned a value earlier.
When writing the comments, I thought that
syms x; solve(x==2*sin(x))
would return the expected result. At least in Matlab 2013b solve fails to find a analytic solution for this problem, falling back to a numeric solver only returning one solution, 0.
An alternative is
s = feval(symengine,'numeric::solve',2*sin(x)==x,x,'AllRealRoots')
which is taken from this answer to a similar question. Besides using AllRealRoots you could use a numeric solver, manually setting starting points which roughly match the values you have read from the graph. This wa you get precise results:
[fzero(#(x)f(x)-g(x),-2),fzero(#(x)f(x)-g(x),0),fzero(#(x)f(x)-g(x),2)]
For a higher precision you could switch from fzero to vpasolve, but fzero is probably sufficient and faster.
I have been working on solving some equation in a more complicated context. However, I want to illustrate my question through the following simple example.
Consider the following two functions:
function y=f1(x)
y=1-x;
end
function y=f2(x)
if x<0
y=0;
else
y=x;
end
end
I want to solve the following equation: f1(x)=f2(x). The code I used is:
syms x;
x=solve(f1(x)-f2(x));
And I got the following error:
??? Error using ==> sym.sym>notimplemented at 2621
Function 'lt' is not implemented for MuPAD symbolic objects.
Error in ==> sym.sym>sym.lt at 812
notimplemented('lt');
Error in ==> f2 at 3
if x<0
I know the error is because x is a symbolic variable and therefore I could not compare x with 0 in the piecewise function f2(x).
Is there a way to fix this and solve the equation?
First, make sure symbolic math is even the appropriate solution method for your problem. In many cases it isn't. Look at fzero and fsolve amongst many others. A symbolic method is only needed if, for example, you want a formula or if you need to ensure precision.
In such an old version of Matlab, you may want to break up your piecewise function into separate continuous functions and solve them separately:
syms x;
s1 = solve(1-x^2,x) % For x >= 0
s2 = solve(1-x,x) % For x < 0
Then you can either manually examine or numerically compare the outputs to determine if any or all of the solutions are valid for the chosen regime – something like this:
s = [s1(double(s1) >= 0);s2(double(s2) < 0)]
You can also take advantage of the heaviside function, which is available in much older versions.
syms x;
f1 = 1-x;
f2 = x*heaviside(x);
s = solve(f1-f2,x)
Yes, the Heaviside function is 0.5 at zero – this gives it the appropriate mathematical properties. You can shift it to compare values other than zero. This is a standard technique.
In Matlab R2012a+, you can take advantage of assumptions in addition to the normal relational operators. To add to #AlexB's comment, you should convert the output of any logical comparison to symbolic before using isAlways:
isAlways(sym(x<0))
In your case, x is obviously not "always" on one side or the other of zero, but you may still find this useful in other cases.
If you want to get deep into Matlab's symbolic math, you can create piecewise functions using MuPAD, which are accessible from Matlab – e.g., see my example here.
Using the code,
syms x(t)
y=x^2
diff(y,t)
diff(y,x)
I get the following error:
2*D(x)(t)*x(t)
Error using sym/diff (line 26)
All arguments, except for the first one, must not be symbolic functions.
Is there a way to tackle this? Thanks for your time.
I dont know much about the Symbolic Math Toolbox, but taking a derivative wrt to a function does not seem to be supported (at least in a direct fashion) for diff.
You can substitute a variable, compute a derivative, and substitute the function back. Like so:
syms z
subs(diff(subs(y,x,z),z),z,x)
ans(t) = 2*x(t)
I have a code that needs to evaluate the arc length equation below:
syms x
a = 10; b = 10; c = 10; d = 10;
fun = 4*a*x^3+3*b*x^2+2*c*x+d
int((1+(fun)^2)^.5)
but all that returns is below:
ans = int(((40*x^3 + 30*x^2 + 20*x + 10)^2 + 1)^(1/2), x)
Why wont matlab evaluate this integral? I added a line under to check if it would evaulate int(x) and it returned the desired result.
Problems involving square roots of functions may be tricky to intgrate. I am not sure whether the integral exists or not, but it if you look up the integral of a second order polynomial you will see that this one is already quite a mess. What you would have, would you expand the function inside the square root, would be a ninth order polynomial. If this integral actually would exist it may be too complex to calculate.
Anyway, if you think about it, would anyone really become any wiser by finding the analytical solution of this? If that is not the case a numerical solution should be sufficient.
EDIT
As thewaywewalk said in the comment, a general rule to calculate these kinds of integrals would be valuable, but to know the primitive function to the particular integral would probably be overkill (if a solution could be found).
Instead define the function as an anonymous function
fun = #(x) sqrt((4*a*x.^3+3*b*x.^2+2*c*x+d).^2+1);
and use integral to evaluate the function between some range, eg
integral(fun,0,100);
for evaluating the function in the closed interval [0,100].