I am testing MATLAB capabilities in solving equations for a project that I intend to do, so I gave it a test run with something simple, but the results that it gives me are incorrect. I tried to solve two non-linear equations with two unknowns, one of the solutions is correct the other is not.
syms theta d x y
eq1 = d * cos(theta) == x;
eq2 = d * sin(theta) == y;
sol = solve(eq1, eq2, theta, d)
sol.theta
sol.d
The solutions for d are correct, but for theta I get:
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*atan((x + (x^2 + y^2)^(1/2))/y)
And the correct answer for theta is simply atan(y/x)
Then when I evaluate these solutions with x = 1, y = 0, I get:
eval(sol.d)
eval(sol.theta)
d = 1, -1
theta = NaN, -3.1416
Solutions for d are correct, but theta in that scenario should be 0.
What am I doing wrong?
EDIT: solving it by hand it looks like this: Divide the y equation by the x equation
y/x = (d * sin(theta)) / (d * cos(theta))
y/x = sin(theta)/cos(theta)
y/x = tan(theta)
theta = atan(y/x)
Even if matlab solves it in some other way and gets a different expression, it should still yield the same final result when I use numbers and it PARTIALLY does.
For x = 1 and y = 0, theta should be 0, => this doesnt work, it gives NaN (explanation bellow)
for x = 1 and y = 1, theta should be 45 degrees => this works
for x = 0 and y = 1 theta should be 90 degrees => this works
And I just checked it again with the 45 and 90 degree values for x and y and it works, but for x = 1 and y = 0 it still gives NaN as one of the answers and that is because it gets a 0/0 from the way it is expressing it
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*(1 - (1^2 + 0^2))^(1/2)/0
-2*(1 - 1)^(1/2)/0
0/0
but if its in the form of atan(y/x) the result is
theta = atan(0/1)
theta = atan(0)
theta = 0
Did you mean to solve this:
syms a b theta d real
eq1 = a==d * cos(theta) ;
eq2 = b==d * sin(theta) ;
[sol] = solve([eq1 eq2],[d theta] ,'IgnoreAnalyticConstraints', true,'Real',true,'ReturnConditions',true);
When solving the equations with symbolic x and y, the solver will find a solution with a certain condition, which can be obtained using the argument 'ReturnCondition':
syms x y theta d real
eq1 = d*cos(theta) == x;
eq2 = d*sin(theta) == y;
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
This gives the following result for sol
>> sol.d
(x^2 + y^2)^(1/2)
-(x^2 + y^2)^(1/2)
>> sol.theta
2*pi*k - 2*atan((x - (x^2 + y^2)^(1/2))/y)
2*pi*k - 2*atan((x + (x^2 + y^2)^(1/2))/y)
>> sol.parameters
k
>> sol.conditions
y ~= 0 & in(k, 'integer')
y ~= 0 & in(k, 'integer')
As you can see, y = 0 does not fulfill this general solution given by the solver, resulting in your problem for y = 0. You can find solutions for y = 0 by either making y numeric instead of symbolic, or by adding an assumption:
syms x y theta d real
assume(y==0)
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
I guess its easier to just set y=0 numeric, for this one condition, since there are already 4 possible solutions and conditions for the three lines above.
Related
I am trying to use the fixed point iteration method with initial approximation x(1)=0 to obtain an approximation to the root of the equation f(x)=3x+sin(x)e^x=0.
The stopping criterion is
|x(k+1)-x(k)|<0.0001
x(1) = 0;
n = 100;
for k = 1:n
f(k) = 3*x(k) +sin(x(k))-exp(x(k));
if (abs(f(k))<0.0001)
break;
end
syms x
diff(f(k));
x(k+1) = x(1)- (f(k))/(diff(f(k)));
end
[x' f']
This is the error I am getting: Error using / Matrix dimensions must
agree. Error in prac2Q2 (line 15)
x(k+1) = x(1)- (f(k))/(diff(f(k)));
I would suggest to calculate the derivative by hand and use that term as denominator or to save the derivative in another variable and use this as the denominator.
Derivative as Variable
f(k) = ...;
df(k) = diff(f(k));
x(k+1) = x(k) - f(k) / df(k);
PS: I cannot test this, because I do not have access to the Symbolic Toolbox right now.
If you're looking for the root of 3*x +sin(x)-exp(x) you want to resolve this equation:
3*x + sin(x) - exp(x) = 0
The easiest way will be to isolate x in one side of the equation:
x = (exp(x) - sin(x))/3 % now iterate until x = (exp(x) - sin(x))/3
Now I would recommand to use an easier fixed point method: x(k+1) = (x(k)+f(x(k)))/2
x = 1 % x0
while 1
y = (exp(x)-sin(x))/3; % we are looking for the root not for a fixed point !!! y = f(x)
x = (x+y)/2 % after a few iterations x == y, so x = (x+y)/2 or x = 2x/2
if abs(x-y) < 1e-10
break
end
end
And you obtain the correct result:
x = 0.36042
No need of symbolic math.
How explicitly does Matlab solve the rightmost equation in c1, specifically ((x-1)\y)?
I am well aware what happens when you use a matrix, e.g. A\b (where A is a matrix and b is a column vector), but what happens when you use backslash on two vectors with equal rows?
The problem:
x = (1:3)';
y = ones(3,1);
c1 = ((x-1)\y) % why does this =0.6?
You're doing [0;1;2]\[1;1;1]. Essentially x=0.6 is the least-squares solution to
[0;1;2]*x=[1;1;1]
The case you have from the documentation is the following:
If A is a rectangular m-by-n matrix with m ~= n, and B is a matrix with m rows, then A\B returns a least-squares solution to the system of equations A*x= B.
(Specifically, you have m=3, n=1).
A = (1:3).' - 1; % = [0;1;2]
B = ones(3,1); % = [1;1;1]
x = A\B; % = 0.6
Algebraically, it's easy to see this is the solution to the least-squares minimisation
% Calculate least squares
leastSquares = sum( ((A*x) - B).^2 )
= sum( ([0;1;2]*x - [1;1;1]).^2 )
= sum( [-1; x-1; 2x-1].^2 )
= 1 + (x-1)^2 + (2x-1)^2
= 1 + x^2 - 2*x + 1 + 4*x^2 - 4*x + 1
= 5*x^2 - 6*x + 3
% Minimum least squares -> derivative = 0
d(leastSquares)/dx = 10*x - 6 = 0
10*x = 6
x = 0.6
I have no doubt that MATLAB uses a more sophisticated algorithm to come to the same conclusion, but this lays out the mathematics in a fairly plain way.
You can see experimentally that there is no better solution by testing the following for various values of x... 0.6 gives the smallest error:
sum( ([0;1;2]*x - [1;1;1]).^2 )
Assume we have three equations:
eq1 = x1 + (x1 - x2) * t - X == 0;
eq2 = z1 + (z1 - z2) * t - Z == 0;
eq3 = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1 == 0;
while six of known variables are:
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
So we are looking for three unknown variables that are:
X , Z and t
I wrote two method to solve it. But, since I need to run these code for 5.7 million data, it become really slow.
Method one (using "solve"):
tic
S = solve( eq1 , eq2 , eq3 , X , Z , t ,...
'ReturnConditions', true, 'Real', true);
toc
X = double(S.X(1))
Z = double(S.Z(1))
t = double(S.t(1))
results of method one:
X = 316190;
Z = 234060;
t = -2.9280;
Elapsed time is 0.770429 seconds.
Method two (using "fsolve"):
coeffs = [a,b,x1,x2,z1,z2]; % Known parameters
x0 = [ x2 ; z2 ; 1 ].'; % Initial values for iterations
f_d = #(x0) myfunc(x0,coeffs); % f_d considers x0 as variables
options = optimoptions('fsolve','Display','none');
tic
M = fsolve(f_d,x0,options);
toc
results of method two:
X = 316190; % X = M(1)
Z = 234060; % Z = M(2)
t = -2.9280; % t = M(3)
Elapsed time is 0.014 seconds.
Although, the second method is faster, but it still needs to be improved. Please let me know if you have a better solution for that. Thanks
* extra information:
if you are interested to know what those 3 equations are, the first two are equations of a line in 2D and the third equation is an ellipse equation. I need to find the intersection of the line with the ellipse. Obviously, we have two points as result. But, let's forget about the second answer for simplicity.
My suggestion it's to use the second approce,which it's the recommended by matlab for nonlinear equation system.
Declare a M-function
function Y=mysistem(X)
%X(1) = X
%X(2) = t
%X(3) = Z
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
Y(1,1) = x1 + (x1 - x2) * X(2) - X(1);
Y(2,1) = z1 + (z1 - z2) * X(2) - X(3);
Y(3,1) = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1;
end
Then for solving use
x0 = [ x2 , z2 , 1 ];
M = fsolve(#mysistem,x0,options);
If you may want to reduce the default precision by changing StepTolerance (default 1e-6).
Also for more increare you may want to use the jacobian matrix for greater efficencies.
For more reference take a look in official documentation:
fsolve Nonlinear Equations with Analytic Jacobian
Basically giving the solver the Jacobian matrix of the system(and special options) you can increase method efficency.
I'm using octave's ezmesh to plot a linear regression defined as follows:
f = #(x,y) 1 * theta(1) + x * theta(2) + y * theta(3) + x * y * theta(4)
For some fixed vector theta:
octave:275> theta
theta =
9.4350e+00
1.7410e-04
3.3702e-02
1.6498e-07
I'm using a domain of [0 120000 0 1400], and can evaluate:
octave:276> f(0, 0)
ans = 9.4350
octave:277> f(120000, 1400)
ans = 105.23
However, if I run:
octave:278> ezmesh(f, [0 120000 0 1400])
The resulting mesh has a z value of around 570 for (0, 0) and just under 640 for (120000, 1400). I'm baffled. What could be causing this?
EDIT: Even if I simplify f to the following, similar behavior occurs:
octave:308> f = #(x, y) (x * y)
Why is ezmesh not handling multiplication as expected (by me), so that the function evaluates as I expect, and the values change when the function is used inside of ezmesh?
ezmesh invokes the function handle on a matrix of values (to benefit from vectorization performance). Use .* for multiplication.
I have to solve a system of ordinary differential equations of the form:
dx/ds = 1/x * [y* (g + s/y) - a*x*f(x^2,y^2)]
dy/ds = 1/x * [-y * (b + y) * f()] - y/s - c
where x, and y are the variables I need to find out, and s is the independent variable; the rest are constants. I've tried to solve this with ode45 with no success so far:
y = ode45(#yprime, s, [1 1]);
function dyds = yprime(s,y)
global g a v0 d
dyds_1 = 1./y(1) .*(y(2) .* (g + s ./ y(2)) - a .* y(1) .* sqrt(y(1).^2 + (v0 + y(2)).^2));
dyds_2 = - (y(2) .* (v0 + y(2)) .* sqrt(y(1).^2 + (v0 + y(2)).^2))./y(1) - y(2)./s - d;
dyds = [dyds_1; dyds_2];
return
where #yprime has the system of equations. I get the following error message:
YPRIME returns a vector of length 0, but the length of initial
conditions vector is 2. The vector returned by YPRIME and the initial
conditions vector must have the same number of elements.
Any ideas?
thanks
Certainly, you should have a look at your function yprime. Using some simple model that shares the number of differential state variables with your problem, have a look at this example.
function dyds = yprime(s, y)
dyds = zeros(2, 1);
dyds(1) = y(1) + y(2);
dyds(2) = 0.5 * y(1);
end
yprime must return a column vector that holds the values of the two right hand sides. The input argument s can be ignored because your model is time-independent. The example you show is somewhat difficult in that it is not of the form dy/dt = f(t, y). You will have to rearrange your equations as a first step. It will help to rename x into y(1) and y into y(2).
Also, are you sure that your global g a v0 d are not empty? If any one of those variables remains uninitialized, you will be multiplying state variables with an empty matrix, eventually resulting in an empty vector dyds being returned. This can be tested with
assert(~isempty(v0), 'v0 not initialized');
in yprime, or you could employ a debugging breakpoint.
the syntax for ODE solvers is [s y]=ode45(#yprime, [1 10], [2 2])
and you dont need to do elementwise operation in your case i.e. instead of .* just use *