I have a symbolic function exp(a+b), and would like to factor out A=exp(a) to produce exp(a+b) = A*exp(b), but I cannot figure out how to do this in MATLAB. Below is my attempt:
syms a b A
X = exp(a+b);
Y = subs(X,exp(a),A) % = A*exp(b)
however, Y = exp(a+b). For some reason, MATLAB cannot determine:
exp(a+b) = exp(a) * exp(b) = A*exp(b).
Any help is greatly appreciated.
First, expand the expression so that the exponents are separated then do the substitution. By default, when writing out an expression for the first time (before running it through any functions), MATLAB will try and simplify your expression and so exp(a)*exp(b) can be much better expressed using exp(a+b). This is why your substitution had no effect. However, if you explicitly want to replace a part of the expression that is encompassed by an exponent with a base, expand the function first, then do your substitution:
>> syms a b A;
>> X = exp(a+b);
>> Xexpand = expand(X)
Xexpand =
exp(a)*exp(b)
>> Y = subs(Xexpand, exp(a), A)
Y =
A*exp(b)
Related
I'm having trouble with implementing double integral in Matlab.
Unlike other double integrals, I need the result of the first (inside) integral to be an expression of the second variable, before going through the second (outside) integral, as it must be powered by k.
For example:
In the example above, I need the result of the inside integral to be expressed as 2y, so that I can calculate (2y)^k, before doing the second (outside) integral.
Does anyone know how to do this in Matlab?
I don't like doing things symbolically, because 99.9% of all problems don't have a closed form solution at all. For 99.9% of the problems that do have a closed-form solution, that solution is unwieldy and hardly useful at all. That may be because of my specific discipline, but I'm going to assume that your problem falls in one of those 99.9% sets, so I'll present the most obvious numerical way to do this.
And that is, integrate a function which calls integral itself:
function dbl_int()
f = #(x,y) 2.*x.*y + 1;
k = 1;
x_limits = [0 1];
y_limits = [1 2];
val = integral(#(y) integrand(f, y, k, x_limits), ...
y_limits(1), y_limits(2));
end
function val = integrand(f, y, k, x_limits)
val = zeros(size(y));
for ii = 1:numel(y)
val(ii) = integral(#(x) f(x,y(ii)), ...
x_limits(1), x_limits(2));
end
val = val.^k;
end
I have the following series
I tried this code but it does not print the final result...instead gives a long line of numbers!
syms n
y = symsum(1/sqrt(n),[1,100])
Result:
y =
2^(1/2)/2 + 3^(1/2)/3 + 5^(1/2)/5 + 6^(1/2)/6 + % and so on....
So the question is how to produce a final number as answer?!
Should I go with a script like this instead?
y = 0;
for i = 1:1:100
y = y + (1/sqrt(i));
end
disp(y);
To answer the original question, you can convert the symbolic expression you initially got using double, to convert from a symbolic to a numeric value:
y = double(y)
Or actually:
syms n
y = double(symsum(1/sqrt(n),[1,100]))
and you get 18.5896.
Additionally, you can use eval to evaluate the symbolic expression (thanks Luis Mendo).
Yay!
how about dropping the loop and use this instead:
n=1:100
result = sum(1./sqrt(n))
>> result =
18.5896
I'm not sure if you want to use the symbolic sum of series function in your case since you are only dealing with a simple function.
I have the following series
I tried this code but it does not print the final result...instead gives a long line of numbers!
syms n
y = symsum(1/sqrt(n),[1,100])
Result:
y =
2^(1/2)/2 + 3^(1/2)/3 + 5^(1/2)/5 + 6^(1/2)/6 + % and so on....
So the question is how to produce a final number as answer?!
Should I go with a script like this instead?
y = 0;
for i = 1:1:100
y = y + (1/sqrt(i));
end
disp(y);
To answer the original question, you can convert the symbolic expression you initially got using double, to convert from a symbolic to a numeric value:
y = double(y)
Or actually:
syms n
y = double(symsum(1/sqrt(n),[1,100]))
and you get 18.5896.
Additionally, you can use eval to evaluate the symbolic expression (thanks Luis Mendo).
Yay!
how about dropping the loop and use this instead:
n=1:100
result = sum(1./sqrt(n))
>> result =
18.5896
I'm not sure if you want to use the symbolic sum of series function in your case since you are only dealing with a simple function.
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}