I have a vector of symbolic expressions like this in matlab:
eqn = k*r*t0*Lp+k*t0*(r*Li+La)+In -Ic;
I'd like to find the expression(k*r).Any idea how can I do that? Thanks!
simple example：
a+b+c+d = 0;
I'd like to find the expression(a+c),then:
a+c = -(b+d);
You want to isolate k*r, so.... do that!
syms k r t0 Lp Li La In Ic
eqn = k*r*t0*Lp+k*t0*(r*Li+La)+In -Ic==0;
isolate(expand(eqn),k*r) % for some reason this case did not work without expand()
ans =
k*r == -(In - Ic + La*k*t0)/(Li*t0 + Lp*t0)
Related
I tried various methods to simplify the solution to the following differential equation, but was unable to fully simplify it to 0.01e^(-0.15t)sin(9.999t+1.556) and expressions with radicals were not properly simplified either. Can someone please explain how the solution can be fully simplified with the number of terms reduced as much as possible?
syms y(t) m k x c
Dy = diff(y,t);
Dy2 = diff(y,t,2);
m = 10; c = 3; k = 1000;
ode = m*Dy2 +c*Dy + k*y == 0;
eqns = [ode]
cond = [y(0) == 0.01,Dy(0) == 0];
ySol(t) = dsolve(eqns,cond)
ySol(t) = simplify(ySol(t),'steps',500)
pretty(ySol(t))
vpa(ySol(t), 5)
simplify(ySol(t))
Pass output of vpa function into simplify function, as follows:
vpa(simplify(vpa(ySol(t), 3)),3)
>> syms x v(x) w(x);
>> eq1 = 2*v + 3*w == 4;
>> eq2 = 5*v + 4*w == 3;
>> sol = solve([eq1,eq2],[v,w])
I tried to implement this code in MATLAB, but error flashes out as "The second argument must be a vector of symbolic variables." I have tried similar things in Python using SymPy, but never such error comes. How to correct out this?
Have a look at the help file for the multivariate case of solve and the example in the help file
openExample('symbolic/SolveMultivariateEquationsAndAssignOutputsToStructureExample')
Applied to your problem
syms v w;
eq1 = [2*v + 3*w == 4;5*v + 4*w == 3];
sol = solve(eq1)
sol.v
sol.w
but if you just want to solve for v w then you could use e.g.
[2 3;5 4]\[4;3]
I am trying to solve an equation for x in Matlab, but keep getting the error:
Empty sym: 0-by-1
The equation has an integral, where x is the upper bound and also part of the integrand 1. The code I use is the following:
a = 0.2; b= 10; c = -10; d = 15; mu = 3; sig = 1;
syms x t
eqn = 0 == a + b*normcdf(x,mu,sig)+c*int( normcdf(d + x - t,mu,sig)*normpdf(t,mu,sig),t,0,x);
A = vpasolve(eqn,x)
Any hints on where I am wrong?
I believe that the symbolic toolbox may not be good enough to solve that integral... Maybe some assume or some other trick can do the job, I personally could not find the way.
However, to test if this is solvable, I tried Wolfram Alpha. It gives a result, that you can use.
eq1=a + b*normcdf(x,mu,sig);
resint=c*(t^3*(d - t + x)*erfc((mu - x)/(sqrt(2)*sig)))/(4*sig*exp((-mu + x)^2/(2*sig^2))*sqrt(2*pi));
A=vpasolve(eq1+subs(resint,t,x)-subs(resint,t,0) ==0)
gives 1.285643225712432599485355373093 in my PC.
I need to create a symbolic matrix in MATLAB. It can be done statically as
syms a11 a12 a21 a22;
A = [a11 a12; a21 a22];
or using compact syntax as
A = sym('A%d', [2 2]);
However i don't see how any of these syntax's should allow dynamic initialization. I would like to initialize each matrix element individually using two loops. One shot at a possible syntax (it's NOT valid MATLAB syntax) could be
syms a x1;
P = sym(zeros(n,n));
for l = 1:n
for m = 1:n
kl = l; km = m;
P(l,m)(x1) = a*sin(kl*x1)*sin(km*x1);
end
end
where I used the (invalid) syntax P(l,m)(x1) to specify that each element in P is a function of x1. Hence P(2) would be an (n,n) matrix with elements P(l,m) = a*sin(kl*2)*sin(km*2). I know it's possible to construct the P by building the matrix string (A=[...]) on run time and evaluating using eval. Something like
syms a x1;
command = [];
for i = 1:n
for j = 1:n
kl = i; km = j;
command = [command ' + a*sin(' num2str(kl) '*x1)*sin(' num2str(km) '*x1)'];
if(j < n) command = [command ',']; end
end
if(i < n) command = [command ';']; end
end
eval(['P(x1) = [' command '];'])
However, using eval is bad practice, so I see this solution only as a last resort. Is there any correct way to achieve my goal?
NB: I have written the elements P(l,m) = a*sin(kl*x1)*sin(km*x1) to simplify the code. The actual expression is P(l,m) = sin(kl*x1)*sin(km*x1)*epsilon + kl*km*cos(kl*x1)*cos(km*x1).*b + km*cos(km*x1)*sin(kl*x1)-kl*cos(kl*x1)*sin(km*x1)/(kl^2-km^2).
If you're just trying to avoid the for loops, you can use meshgrid, which hides them from you:
syms a x1
n = 3;
x = meshgrid(1:n)*x1; % multiplying by the symbolic x1 makes x symbolic
P = a*sin(x).*sin(x.');
which returns
P =
[ a*sin(x1)^2, a*sin(2*x1)*sin(x1), a*sin(3*x1)*sin(x1)]
[ a*sin(2*x1)*sin(x1), a*sin(2*x1)^2, a*sin(2*x1)*sin(3*x1)]
[ a*sin(3*x1)*sin(x1), a*sin(2*x1)*sin(3*x1), a*sin(3*x1)^2]
I'm fairly new to MATLAB. Normal matrix multiplication of a M x K matrix by an K x N matrix -- C = A * B -- has c_ij = sum(a_ik * b_kj, k = 1:K). What if I want this to be instead c_ij = sum(op(a_ik, b_kj), k = 1:K) for some simple binary operation op? Is there any nice way to vectorize this in MATLAB (or maybe even a built-in function)?
EDIT: This is currently the best I can do.
% A is M x K, B is K x N
% op is min
C = zeros(M, N);
for i = 1:M:
C(i, :) = sum(bsxfun(#min, A(i, :)', B));
end
Listed in this post is a vectorized approach that persists with bsxfun by using permute to create singleton dimensions as needed by bsxfun to let the singleton-expansion do its work and thus essentially replacing the loop in the original post. Please be reminded that bsxfun is a memory hungry implementation, so expect speedup with it only until it is stretched too far. Here's the final solution code -
op = #min; %// Edit this with your own function/ operation
C = sum(bsxfun(op, permute(A,[1 3 2]),permute(B,[3 2 1])),3)
NB - The above solution was inspired by Removing four nested loops in Matlab.
if the operator can operate element-by-element (like .*):
if(size(A,2)~=size(B,1))
error(blah, blah, blah...);
end
C = zeros(size(A,1),size(B,2));
for i = 1:size(A,1)
for j = 1:size(B,2)
C(i,j) = sum(binaryOp(A(i,:)',B(:,j)));
end
end
You can always write the loops yourself:
A = rand(2,3);
B = rand(3,4);
op = #times; %# use your own function here
C = zeros(size(A,1),size(B,2));
for i=1:size(A,1)
for j=1:size(B,2)
for k=1:size(A,2)
C(i,j) = C(i,j) + op(A(i,k),B(k,j));
end
end
end
isequal(C,A*B)
Depending on your specific needs, you may be able to use bsxfun in 3D to trick the binary operator. See this answer for more infos: https://stackoverflow.com/a/23808285/1121352
Another alternative would be to use cellfun with a custom function:
http://matlabgeeks.com/tips-tutorials/computation-using-cellfun/