The following is a MATLAB problem.
Suppose I define an function f(x,y).
I want to calculate the partial derivative of f with respect to y, evaluated at a specific value of y, e.g., y=6. Finally, I want to integrate this new function (which is only a function of x) over a range of x.
As an example, this is what I have tried
syms x y;
f = #(x, y) x.*y.^2;
Df = subs(diff(f,y),y,2);
Int = integral(Df , 0 , 1),
but I get the following error.
Error using integral (line 82)
First input argument must be a function
handle.
Can anyone help me in writing this code?
To solve the problem, matlabFunction was required. The solution looks like this:
syms x y
f = #(x, y) x.*y.^2;
Df = matlabFunction(subs(diff(f,y),y,2));
Int = integral(Df , 0 , 1);
Keeping it all symbolic, using sym/int:
syms x y;
f = #(x, y) x.*y.^2;
Df = diff(f,y);
s = int(Df,x,0,1)
which returns y. You can substitute 2 in for y here or earlier as you did in your question. Not that this will give you an exact answer in this case with no floating-point error, as opposed to integral which calculated the integral numerically.
When Googling for functions in Matlab, make sure to pay attention what toolbox they are in and what classes (datatypes) they support for their arguments. In some cases there are overloaded versions with the same name, but in others, you may need to look around for a different method (or devise your own).
Related
I am in interested in passing extra arguments to nlinfit function in Matlab
beta = nlinfit(X,Y,modelfun,beta0)
and let the modelfun is
function y = modelfun(beta, c, X)
y = beta(1)*x.^(beta2) + c;
My interest is estimate beta and also to provide c externally. X and Y have their obvious meanings.
Can it be done?
If c is a value generated before you call nlinfit (i.e. its value is fixed while nlinfit is running), then you can use an anonymous function wrapper to pass the extra parameter like so:
beta = nlinfit(X, Y, #(beta, X) modelfun(beta, c, X), beta0);
The equation is 4*x^2-2*y^2==9. Using implicit differentiation, I can find that the second derivative of y with respect to x is -9/y^3, which requires a substitution in the final step.
I am trying to duplicate this answer using Matlab's symbolic toolbox. I did find some support for the first derivative here, and was successful finding the first derivative.
clear all
syms x y f
f=4*x^2-2*y^2-9
sol1=-diff(f,x)/diff(f,y)
But I am unable to continue onward to find the second derivative with the final simplification (replacing 4*x^2-2*y^2 with 9).
Can someone show me how to do this in Matlab?
To my knowledge there is no direct way to obtain an implicit second derivative in Matlab. And working with implicit functions in Matlab can be fairly tricky. As a start, your variable y is implicitly a function of x s you should define it as such:
clear all
syms y(x) % defines both x and y
f = 4*x^2-2*y^2-9
Here y(x) is now what is called an arbitrary or abstract symbolic function, i.e., one with no explicit formula. Then take the derivative of f with respect to x:
s1 = diff(f,x)
This returns a function in terms of the implicit derivative of y(x) with respect to x, diff(y(x), x) (in this case diff(y) is shorthand). You can solve this function for diff(y) algebraically with subs and solve:
syms dydx % arbitrary variable
s2 = subs(s1,diff(y),dydx)
s3 = solve(s2,dydx)
This yields the first implicit derivative. You can then take another derivative of this expression to obtain the second implicit derivative as a function of the first:
s4 = diff(s3,x)
Finally, substitute the expression for the first implicit derivative into this and simplify to obtain the final form:
s5 = simplify(subs(s4,diff(y),s3))
This yields (2*(y(x)^2 - 2*x^2))/y(x)^3. And then you can eliminate x using the original expression for f with further substitution and solving:
syms x2
f2 = subs(f,x^2,x2)
x2 = solve(f2,x2)
s6 = subs(s5,x^2,x2)
Finally, you can turn this back into an explicit algebraic expression with a final substitution, if desired:
s7 = subs(s6,y,'y')
This yields your solution of -9/y^3.
This whole process can be written more concisely (but very unclearly) as:
clear all
syms y(x) dydx x2
f = 4*x^2-2*y^2-9;
s1 = solve(subs(diff(f,x),diff(y),dydx),dydx)
s2 = simplify(subs(subs(subs(diff(s1,x),diff(y),s1),x^2,solve(subs(f,x^2,x2),x2)),y,'y'))
There are many other ways to achieve the same result. See also these two tutorials: [1], [2].
It's been almost four years since I asked this question, and got brilliant help from horchler, but I've discovered another method using the chain rule.
syms x y
f=4*x^2-2*y^2-9
dydx=-diff(f,x)/diff(f,y)
d2ydx2=diff(dydx,x)+diff(dydx,y)*dydx
d2ydx2=simplifyFraction(d2ydx2,'Expand',true)
s1=solve(f,x)
subs(d2ydx2,x,s1(2))
I have an expression with three variables x,y and v. I want to first integrate over v, and so I use int function in MATLAB.
The command that I use is the following:
g =int((1-fxyz)*pv, v, y,+inf)%
PS I haven't given you what the function fxyv is but it is very complicated and so int is taking so long and I am afraid after waiting it might not solve it.
I know one option for me is to integrate numerically using for example integrate, however I want to note that the second part of this problem requires me to integrate exp[g(x,y)] over x and y from 0 to infinity and from x to infinity respectively. So I can't take numerical values of x and y when I want to integrate over v I think or maybe not ?
Thanks
Since the question does not contain sufficient detail to attempt analytic integration, this answer focuses on numeric integration.
It is possible to solve these equations numerically. However, because of complex dependencies between the three integrals, it is not possible to simply use integral3. Instead, one has to define functions that compute parts of the expressions using a simple integral, and are themselves fed into other calls of integral. Whether this approach leads to useful results in terms of computation time and precision cannot be answered generally, but depends on the concrete choice of the functions f and p. Fiddling around with precision parameters to the different calls of integral may be necessary.
I assume that the functions f(x, y, v) and p(v) are defined in the form of Matlab functions:
function val = f(x, y, v)
val = ...
end
function val = p(v)
val = ...
end
Because of the way they are used later, they have to accept multiple values for v in parallel (as an array) and return as many function values (again as an array, of the same size). x and y can be assumed to always be scalars. A simple example implementation would be val = ones(size(v)) in both cases.
First, let's define a Matlab function g that implements the first equation:
function val = g(x, y)
val = integral(#gIntegrand, y, inf);
function val = gIntegrand(v)
% output must be of the same dimensions as parameter v
val = (1 - f(x, y, v)) .* p(v);
end
end
The nested function gIntegrand defines the object of integration, the outer performs the numeric integration that gives the value of g(x, y). Integration is over v, parameters x and y are shared between the outer and the nested function. gIntegrand is written in such a way that it deals with multiple values of v in the form of arrays, provided f and p do so already.
Next, we define the integrand of the outer integral in the second equation. To do so, we need to compute the inner integral, and therefore also have a function for the integrand of the inner integral:
function val = TIntegrandOuter(x)
val = nan(size(x));
for i = 1 : numel(x)
val(i) = integral(#TIntegrandInner, x(i), inf);
end
function val = TIntegrandInner(y)
val = nan(size(y));
for j = 1 : numel(y)
val(j) = exp(g(x(i), y(j)));
end
end
end
Because both function are meant to be fed as an argument into integral, they need to be able to deal with multiple values. In this case, this is implemented via an explicit for loop. TIntegrandInner computes exp(g(x, y)) for multiple values of y, but the fixed value of x that is current in the loop in TIntegrandOuter. This value x(i) play both the role of a parameter into g(x, y) and of an integration limit. Variables x and i are shared between the outer and the nested function.
Almost there! We have the integrand, only the outermost integration needs to be performed:
T = integral(#TIntegrandOuter, 0, inf);
This is a very convoluted implementation, which is not very elegant, and probably not very efficient. Again, whether results of this approach prove to be useful needs to be tested in practice. However, I don't see any other way to implement these numeric integrations in Matlab in a better way in general. For specific choices of f(x, y, v) and p(v), there might be possible improvements.
The problem in the link:
can be integrated analytically and the answer is 4, however I'm interested in integrating it numerically using Matlab, because it's similar in form to a problem that I can't integrate analytically. The difficulty in the numerical integration arises because the function in the two inner integrals is a function of x,y and z and z can't be factored out.
by no means, this is elegant. hope someone can make better use of matlab functions than me. i have tried the brute force way just to practice numerical integration. i have tried to avoid the pole in the inner integral at z=0 by exploiting the fact that it is also being multiplied by z. i get 3.9993. someone must get better solution by using something better than trapezoidal rule
function []=sofn
clear all
global x y z xx yy zz dx dy
dx=0.05;
x=0:dx:1;
dy=0.002;
dz=0.002;
y=0:dy:1;
z=0:dz:2;
xx=length(x);
yy=length(y);
zz=length(z);
s1=0;
for i=1:zz-1
s1=s1+0.5*dz*(z(i+1)*exp(inte1(z(i+1)))+z(i)*exp(inte1(z(i))));
end
s1
end
function s2=inte1(localz)
global y yy dy
if localz==0
s2=0;
else
s2=0;
for j=1:yy-1
s2=s2+0.5*dy*(inte2(y(j),localz)+inte2(y(j+1),localz));
end
end
end
function s3=inte2(localy,localz)
global x xx dx
s3=0;
for k=1:xx-1
s3=s3+0.5*dx*(2/(localy+localz));
end
end
Well, this is strange, because on the poster's similar previous question I claimed this can't be done, and now after having looked at Guddu's answer I realize its not that complicated. What I wrote before, that a numerical integration results in a number but not a function, is true – but beside the point: One can just define a function that evaluates the integral for every given parameter, and this way effectively one does have a function as a result of a numerical integration.
Anyways, here it goes:
function q = outer
f = #(z) (z .* exp(inner(z)));
q = quad(f, eps, 2);
end
function qs = inner(zs)
% compute \int_0^1 1 / (y + z) dy for given z
qs = nan(size(zs));
for i = 1 : numel(zs)
z = zs(i);
f = #(y) (1 ./ (y + z));
qs(i) = quad(f, 0 , 1);
end
end
I applied the simplification suggested by myself in a comment, eliminating x. The function inner calculates the value of the inner integral over y as a function of z. Then the function outer computes the outer integral over z. I avoid the pole at z = 0 by letting the integration run from eps instead of 0. The result is
4.00000013663955
inner has to be implemented using a for loop because a function given to quad needs to be able to return its value simultaneously for several argument values.
How can I make a function from a symbolic expression? For example, I have the following:
syms beta
n1,n2,m,aa= Constants
u = sqrt(n2-beta^2);
w = sqrt(beta^2-n1);
a = tan(u)/w+tanh(w)/u;
b = tanh(u)/w;
f = (a+b)*cos(aa*u+m*pi)+a-b*sin(aa*u+m*pi); %# The main expression
If I want to use f in a special program to find its zeroes, how can I convert f to a function? Or, what should I do to find the zeroes of f and such nested expressions?
You have a couple of options...
Option #1: Automatically generate a function
If you have version 4.9 (R2007b+) or later of the Symbolic Toolbox you can convert a symbolic expression to an anonymous function or a function M-file using the matlabFunction function. An example from the documentation:
>> syms x y
>> r = sqrt(x^2 + y^2);
>> ht = matlabFunction(sin(r)/r)
ht =
#(x,y)sin(sqrt(x.^2+y.^2)).*1./sqrt(x.^2+y.^2)
Option #2: Generate a function by hand
Since you've already written a set of symbolic equations, you can simply cut and paste part of that code into a function. Here's what your above example would look like:
function output = f(beta,n1,n2,m,aa)
u = sqrt(n2-beta.^2);
w = sqrt(beta.^2-n1);
a = tan(u)./w+tanh(w)./u;
b = tanh(u)./w;
output = (a+b).*cos(aa.*u+m.*pi)+(a-b).*sin(aa.*u+m.*pi);
end
When calling this function f you have to input the values of beta and the 4 constants and it will return the result of evaluating your main expression.
NOTE: Since you also mentioned wanting to find zeroes of f, you could try using the SOLVE function on your symbolic equation:
zeroValues = solve(f,'beta');
Someone has tagged this question with Matlab so I'll assume that you are concerned with solving the equation with Matlab. If you have a copy of the Matlab Symbolic toolbox you should be able to solve it directly as a previous respondent has suggested.
If not, then I suggest you write a Matlab m-file to evaluate your function f(). The pseudo-code you're already written will translate almost directly into lines of Matlab. As I read it your function f() is a function only of the variable beta since you indicate that n1,n2,m and a are all constants. I suggest that you plot the values of f(beta) for a range of values. The graph will indicate where the 0s of the function are and you can easily code up a bisection or similar algorithm to give you their values to your desired degree of accuracy.
If you broad intention is to have numeric values of certain symbolic expressions you have, for example, you have a larger program that generates symbolic expressions and you want to use these expression for numeric purposes, you can simply evaluate them using 'eval'. If their parameters have numeric values in the workspace, just use eval on your expression. For example,
syms beta
%n1,n2,m,aa= Constants
% values to exemplify
n1 = 1; n2 = 3; m = 1; aa = 5;
u = sqrt(n2-beta^2);
w = sqrt(beta^2-n1);
a = tan(u)/w+tanh(w)/u;
b = tanh(u)/w;
f = (a+b)*cos(aa*u+m*pi)+a-b*sin(aa*u+m*pi); %# The main expression
If beta has a value
beta = 1.5;
eval(beta)
This will calculate the value of f for a particular beta. Using it as a function. This solution will suit you in the scenario of using automatically generated symbolic expressions and will be interesting for fast testing with them. If you are writing a program to find zeros, it will be enough using eval(f) when you have to evaluate the function. When using a Matlab function to find zeros using anonymous function will be better, but you can also wrap the eval(f) inside a m-file.
If you're interested with just the answer for this specific equation, Try Wolfram Alpha, which will give you answers like:
alt text http://www4c.wolframalpha.com/Calculate/MSP/MSP642199013hbefb463a9000051gi6f4heeebfa7f?MSPStoreType=image/gif&s=15
If you want to solve this type of equation programatically, you probably need to use some software packages for symbolic algebra, like SymPy for python.
quoting the official documentation:
>>> from sympy import I, solve
>>> from sympy.abc import x, y
Solve a polynomial equation:
>>> solve(x**4-1, x)
[1, -1, -I, I]
Solve a linear system:
>>> solve((x+5*y-2, -3*x+6*y-15), x, y)
{x: -3, y: 1}