I have an integral that looks like that:
I want to the values of ai that minimize H.
I am trying to replicate the example from a book (example 8.1), where the authors say they used GA/fminsearch in MATLAB.
I am not very familiar with MATLAB, but I tried (with no success) to solve it there. I am not sure if I should (or if I can) pass a symbolic equation into fminsearch.
Solution found!
int_term = #(a) integral2(#(u,v) exp(-a(1)*u - a(2)*u.^2 - a(3)*v ...
- a(4)*v.^2 - a(5)*u.*v),0,1,0,1);
a0 = [1,1,1,1,1];
obj_fun = #(a) int_term2(a);
a_sol = fminsearch(obj_fun2, a0)
Related
I'm trying to calculate the phase2 angle/value in the y2 equation of a signal given at a specific frequency if I know the other values. Is this possible? Example below: along with picture example:
y1=A1*cos*(2*pi*f1*t+phase1) we know A1,f1,t=1,phase1
y1=0.00720858*cos*(2*pi*6+6.33)
y2=A2*cos*(2*pi*f2*t+phase2) we know A2,f2,t=1, trying to find **phase2**
y2=.4*cos*(2*pi*6.4951+phase2)
y3=A3*cos*(2*pi*f3*t+phase3) we know A3,f3,t=1,phase3
y3=0.0135274*cos(2*pi*7+.786473)
I'm using maxima 13.04.2, octave 3.8.1.
I tried to solve the y2 equation for phase2 in maxima but it got rid of the cos function
kill(all);
A:A; phase:phase; solve(A*cos*(2*pi*t+phase)=0,phase);
the answer came back as phase=-2pi*t
Is this possible? or should I go about this another way?
Thanks
The weird result might stem from the fact that you multiply the cos function with what is supposed to be its argument (by the way, this is mathematically unsound). What you might want is to apply the cos function to the argument. To illustrate what I mean, compare:
A*cos*(2*pi*t+phase)
with:
A*cos(2*pi*t+phase)
On another hand, why not solve the equation pen-on-paper style?
y2 = A2×cos(2πf2t + φ2) ⇒
y2/A2 = cos(2πf2t + φ2) ⇒
arccos(y2/A2) = 2πf2t + φ2 ⇒
arccos(y2/A2) - 2πf2t = φ2
With the values that you provided:
A2 = 0.4, f2 = 6.4951, t = 1.
you can calculate the phase φ2 as function of your level y2 (left as exercise to you).
I'm working on Matlab Code for Lagrange Interpolation.
My problem is working with equations.
For solving Lagrange you need to find the Li(x)'s ( L0(x) to Ln(x) ) first:
e.g. for L0(x) we've got:
L0(x) = (x-x1) (x-x2) (x-x3) / (x0-x1) (x0-x2) (x0-x3)
which for an example it would be:
l0(x) = ( x^3 - ( 2 * x^2 ) + 2x ) / 6
which is an equation.
I write the code but it won't accept the variable z (which i used instead of X ) and it tells:
"Undefined function or variable 'z'."
Which is certainly correct cause it's a variable and not a data but how could i use or write this?
MATLAB by default does not like undefined variables (as do most programming languages!)
You have two ways of solving your problem: numerically solving Lagrange "by hand" -- i.e. with functions like ode45, or trying to do it symbolically. This means that you either have to explicitly define x as a symbolic variable -- with syms x and use the symbolic maths toolbox, or, alternatively, use a numerical scheme and something like ode45.
If you want to define a function in MATLAB, you need to plonk a definition into a separate file and save it.
For example:
function out = ellZero( x, a )
x0 = a(1); x1 = a(2); x2 = a(3); x3 = a(4); % SET Constants
out = (x-x1).*(x-x2).*(x-x3)./((x0-x1).*(x0-x2).*(x0-x3)); % RET Expression
end
If you want to learn more about how MATLAB handles functions and 'sub-functions', have a look at the documentation. You might also be interested in the Partial Differential Equation Toolbox.
Hope that helps!
I am trying to solve equations with this code:
a = [-0.0008333 -0.025 -0.6667 -20];
length_OnePart = 7.3248;
xi = -6.4446;
yi = -16.5187;
syms x y
[sol_x,sol_y] = solve(y == poly2sym(a), ((x-xi)^2+(y-yi)^2) == length_OnePart^2,x,y,'Real',true);
sol_x = sym2poly(sol_x);
sol_y = sym2poly(sol_y);
The sets of solution it is giving are (-23.9067,-8.7301) and (11.0333,-24.2209), which are not even satisfying the equation of circle. How can I rectify this problem?
If you're trying to solve for the intersection of the cubic and the circle, i.e., where y==poly2sym(a) equals (x-xi)^2+(y-yi)^2==length_OnePart^2 it looks like solve may be confused about something when the circle is represented parametrically rather than as single valued functions. It might also have to do with the fact that x and y are not independent solutions, but rather that the latter depends on the former. It also could depend on the use of a numeric solver in this case. solve seems to work fine with similar inputs to yours, so you might report this behavior to the MathWorks to see what they think.
In any case, here is a better, more efficient way to to tackle this as a root-solving problem (as opposed to simultaneous equations):
a = [-0.0008333 -0.025 -0.6667 -20];
length_OnePart = 7.3248;
xi = -6.4446;
yi = -16.5187;
syms x real
f(x) = poly2sym(a);
sol_x = solve((x-xi)^2+(f(x)-yi)^2==length_OnePart^2,x)
sol_y = f(sol_x)
which returns:
sol_x =
0.00002145831413371390464567553686047
-13.182825373861454619370838716408
sol_y =
-20.000014306269544436430325843024
-13.646590348358951818881695033728
Note that you might get slightly more accurate results (one solution is clearly at 0,-20) if you represent your coefficients and parameters more precisely then just four decimal places, e.g., a = [-1/1200 -0.025 -2/3 -20]. In fact, solve might be able to find one or more solutions exactly, if you provide exact representations.
Also, in your code, the calls to sym2poly are doing nothing other than converting back to floating-point (double can be used for this) as the inputs are not in the form of symbolic polynomial equations.
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).
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}