Related
I must write using Laguerre's method a piece of code to find the real and complex roots of poly:
P=X^5-5*X^4-6*X^3+6*X^2-3*X+1
I have little doubt. I did the algorithm in the matlab, but 3 out of 5 roots are the same and I don't think that is correct.
syms X %Declearing x as a variabl
P=X^5-5*X^4-6*X^3+6*X^2-3*X+1; %Equation we interest to solve
n=5; % The eq. order
Pd1 = diff(P,X,1); % first differitial of f
Pd2 = diff(P,X,2); %second differitial of f
err=0.00001; %Answear tollerance
N=100; %Max. # of Iterations
x(1)=1e-3; % Initial Value
for k=1:N
G=double(vpa(subs(Pd1,X,x(k))/subs(P,X,x(k))));
H=G^2 - double(subs(Pd2,X,x(k))) /subs(P,X,x(k));
D1= (G+sqrt((n-1)*(n*H-G^2)));
D2= (G-sqrt((n-1)*(n*H-G^2)));
D = max(D1,D2);
a=n/D;
x(k+1)=x(k)-a
Err(k) = abs(x(k+1)-x(k));
if Err(k) <=err
break
end
end
output (roots of polynomial):
x =
0.0010 + 0.0000i 0.1434 + 0.4661i 0.1474 + 0.4345i 0.1474 + 0.4345i 0.1474 + 0.4345i
What you actually see are all the values x(k) which arose in the loop. The last one, 0.1474 + 0.4345i is the end result of this loop - the approximation of the root which is in your given tolerance threshold. The code
syms X %Declaring x as a variable
P = X^5 - 5 * X^4 - 6 * X^3 + 6 * X^2 - 3 * X + 1; %Polynomial
n=5; %Degree of the polynomial
Pd1 = diff(P,X,1); %First derivative of P
Pd2 = diff(P,X,2); %Second derivative of P
err = 0.00001; %Answer tolerance
N = 100; %Maximal number of iterations
x(1) = 0; %Initial value
for k = 1:N
G = double(vpa(subs(Pd1,X,x(k)) / subs(P,X,x(k))));
H = G^2 - double(subs(Pd2,X,x(k))) / subs(P,X,x(k));
D1 = (G + sqrt((n-1) * (n * H-G^2)));
D2 = (G - sqrt((n-1) * (n * H-G^2)));
D = max(D1,D2);
a = n/D;
x(k+1) = x(k) - a;
Err(k) = abs(x(k+1)-x(k));
if Err(k) <=err
fprintf('Initial value %f, result %f%+fi', x(1), real(x(k)), imag(x(k)))
break
end
end
results in
Initial value -2.000000, result -1.649100+0.000000i
If you want to get other roots, you have to use other initial values. For example one can obtain
Initial value 10.000000, result 5.862900+0.000000i
Initial value -2.000000, result -1.649100+0.000000i
Initial value 3.000000, result 0.491300+0.000000i
Initial value 0.000000, result 0.147400+0.434500i
Initial value 1.000000, result 0.147400-0.434500i
These are all zeros of the polynomial.
A method for calculating the next root when you have found another one would be that you divide through the corresponding linear factor and use your loop for the resulting new polynomial. Note that this is in general not very easy to handle since rounding errors can have a big influence on the result.
Problems with the existing code
You do not implement the Laguerre method properly as a method in complex numbers. The denominator candidates D1,D2 are in general complex numbers, it is inadvisable to use the simple max which only has sensible results for real inputs. The aim is to have a=n/D be the smaller of both variants, so that one has to look for the D in [D1,D2] with the larger absolute value. If there were a conditional assignment as in C, this would look like
D = (abs(D_1)>abs(D2)) ? D1 : D2;
As that does not exist, one has to use commands with a similar result
D = D1; if (abs(D_1)<abs(D2)) D=D2; end
The resulting sequence of approximation points is
x(0) = 0.0010000
x(1) = 0.143349512707684+0.466072958423667i
x(2) = 0.164462212064089+0.461399841949893i
x(3) = 0.164466373475316+0.461405404094130i
There is a point where one can not expect the (residual) polynomial value at the root approximation to substantially decrease. The value close to zero is obtained by adding and subtracting rather large terms in the sum expression of the polynomial. The accuracy lost in these catastrophic cancellation events can not be recovered.
The threshold for polynomial values that are effectively zero can be estimated as the machine constant of the double type times the polynomial value where all coefficients and the evaluation point are replaced by their absolute values. This test serves in the code primarily to avoid divisions by zero or near-zero.
Finding all roots
One approach is to apply the method to a sufficiently large number of initial points along some circle containing all the roots, with some strict rules for early termination at too slow convergence. One would have to make the list of the roots found unique, but keep the multiplicity,...
The other standard method is to apply deflation, that is, divide out the linear factor of the root found. This works well in low degrees.
There is no need for the slower symbolic operations as there are functions that work directly on the coefficient array, such as polyval and polyder. Deflation by division with remainder can be achieved using the deconv function.
For real polynomials, we know that the complex conjugate of a root is also a root. Thus initialize the next iteration with the deflated polynomial with it.
Other points:
There is no point in the double conversions as at no point there is a conversion into the single type.
If you don't do anything with it, it makes no sense to create an array, especially not for Err.
Roots of the example
Implementing all this I get a log of
x(0) = 0.001000000000000+0.000000000000000i, |Pn(x(0))| = 0.99701
x(1) = 0.143349512707684+0.466072958423667i, |dx|= 0.48733
x(2) = 0.164462212064089+0.461399841949893i, |dx|=0.021624
x(3) = 0.164466373475316+0.461405404094130i, |dx|=6.9466e-06
root found x=0.164466373475316+0.461405404094130i with value P0(x)=-2.22045e-16+9.4369e-16i
Deflation
x(0) = 0.164466373475316-0.461405404094130i, |Pn(x(0))| = 2.1211e-15
root found x=0.164466373475316-0.461405404094130i with value P0(x)=-2.22045e-16-9.4369e-16i
Deflation
x(0) = 0.164466373475316+0.461405404094130i, |Pn(x(0))| = 4.7452
x(1) = 0.586360702193454+0.016571894375927i, |dx|= 0.61308
x(2) = 0.562204173408499+0.000003168181059i, |dx|=0.029293
x(3) = 0.562204925474889+0.000000000000000i, |dx|=3.2562e-06
root found x=0.562204925474889+0.000000000000000i with value P0(x)=2.22045e-16-1.33554e-17i
Deflation
x(0) = 0.562204925474889-0.000000000000000i, |Pn(x(0))| = 7.7204
x(1) = 3.332994579372812-0.000000000000000i, |dx|= 2.7708
root found x=3.332994579372812-0.000000000000000i with value P0(x)=6.39488e-14-3.52284e-15i
Deflation
x(0) = 3.332994579372812+0.000000000000000i, |Pn(x(0))| = 5.5571
x(1) = -2.224132251798332+0.000000000000000i, |dx|= 5.5571
root found x=-2.224132251798332+0.000000000000000i with value P0(x)=-3.33067e-14+1.6178e-15i
for the modified code
P = [1, -2, -6, 6, -3, 1];
P0 = P;
deg=length(P)-1; % The eq. degree
err=1e-05; %Answer tolerance
N=10; %Max. # of Iterations
x=1e-3; % Initial Value
for n=deg:-1:1
dP = polyder(P); % first derivative of P
d2P = polyder(dP); %second derivative of P
fprintf("x(0) = %.15f%+.15fi, |Pn(x(0))| = %8.5g\n", real(x),imag(x), abs(polyval(P,x)));
for k=1:N
Px = polyval(P,x);
dPx = polyval(dP,x);
d2Px = polyval(d2P,x);
if abs(Px) < 1e-14*polyval(abs(P),abs(x))
break % if value is zero in relative accuracy
end
G = dPx/Px;
H=G^2 - d2Px / Px;
D1= (G+sqrt((n-1)*(n*H-G^2)));
D2= (G-sqrt((n-1)*(n*H-G^2)));
D = D1;
if abs(D2)>abs(D1) D=D2; end % select the larger denominator
a=n/D;
x=x-a;
fprintf("x(%d) = %.15f%+.15fi, |dx|=%8.5g\n",k,real(x),imag(x), abs(a));
if abs(a) < err*(err+abs(x))
break
end
end
y = polyval(P0,x); % check polynomial value of the original polynomial
fprintf("root found x=%.15f%+.15fi with value P0(x)=%.6g%+.6gi\n", real(x),imag(x),real(y),imag(y));
disp("Deflation");
[ P,R ] = deconv(P,[1,-x]); % division with remainder
x = conj(x); % shortcut for conjugate pairs and clustered roots
end
I'd like to create an anonymous function that does something like this:
n = 5;
x = linspace(-4,4,1000);
f = #(x,a,b,n) a(1)*exp(b(1)^2*x.^2) + a(2)*exp(b(2)^2*x.^2) + ... a(n)*exp(b(n)^2*x.^2);
I can do this as such, without passing explicit parameter n:
f1 = #(x,a,b) a(1)*exp(-b(1)^2*x.^2);
for j = 2:n
f1 = #(x,a,b) f1(x,a,b) + a(j)*exp(b(j)^2*x.^2);
end
but it seems, well, kind of hacky. Does someone have a better solution for this? I'd like to know how someone else would treat this.
Your hacky solution is definitely not the best, as recursive function calls in MATLAB are not very efficient, and you can quickly run into the maximum recursion depth (500 by default).
You can introduce a new dimension along which you can sum up your arrays a and b. Assuming that x, a and b are row vectors:
f = #(x,a,b,n) a(1:n)*exp((b(1:n).^2).'*x.^2)
This will use the first dimension as summing dimension: (b(1:n).^2).' is a column vector, which produces a matrix when multiplied by x (this is a dyadic product, to be precise). The resulting n * length(x) matrix can be multiplied by a(1:n), since the latter is a matrix of size [1,n]. This vector-matrix product will also perform the summation for us.
Mini-proof:
n = 5;
x = linspace(-4,4,1000);
a = rand(1,10);
b = rand(1,10);
y = 0;
for k=1:n
y = y + a(k)*exp(b(k)^2*x.^2);
end
y2 = a(1:n)*exp((b(1:n).^2).'*x.^2); %'
all(abs(y-y2))<1e-10
The last command returns 1, so the two are essentially identical.
I want to solve:
I use following MATLAB code, but it does not work.
Can someone please guide me?
function f=objfun
f=-f;
function [c1,c2,c3]=constraint(x)
a1=1.1; a2=1.1; a3=1.1;
c1=f-log(a1)-log(x(1)/(x(1)+1));
c2=f-log(a2)-log(x(2)/(x(2)+1))-log(1-x(1));
c3=f-log(a3)-log(1-x(1))-log(1-x(2));
x0=[0.01;0.01];
[x,fval]=fmincon('objfun',x0,[],[],[],[],[0;0],[1;1],'constraint')
You need to flip the problem around a bit. You are trying to find the point x (which is (l_1,l_2)) that makes the minimum of the 3 LHS functions the largest. So, you can rewrite your problem as, in pseudocode,
maximise, by varying x in [0,1] X [0,1]
min([log(a1)+log(x(1)/(x(1)+1)) ...
log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
log(a3)+log(1-x(1))+log(1-x(2))])
Since Matlab has fmincon, rewrite this as a minimisation problem,
minimise, by varying x in [0,1] X [0,1]
max(-[log(a1)+log(x(1)/(x(1)+1)) ...
log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
log(a3)+log(1-x(1))+log(1-x(2))])
So the actual code is
F=#(x) max(-[log(a1)+log(x(1)/(x(1)+1)) ...
log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
log(a3)+log(1-x(1))+log(1-x(2))])
[L,fval]=fmincon(F,[0.5 0.5])
which returns
L =
0.3383 0.6180
fval =
1.2800
Can also solve this in the convex optimization package CVX with the following MATLAB code:
cvx_begin
variables T(1);
variables x1(1);
variables x2(1);
maximize(T)
subject to:
log(a1) + x1 - log_sum_exp([0, x1]) >= T;
log(a2) + x2 - log_sum_exp([0, x2]) + log(1 - exp(x1)) >= T;
log(a3) + log(1 - exp(x1)) + log(1 - exp(x2)) >= T;
x1 <= 0;
x2 <= 0;
cvx_end
l1 = exp(x1); l2 = exp(x2);
To use CVX, each constraint and the objective function has to be written in a way that is proveably convex using CVX's ruleset. Making the substitution x1 = log(l1) and x2 = log(l2) allows one to do that. Note that: log_sum_exp([0,x1]) = log(exp(0) + exp(x1)) = log(1 + l1)
This also returns the answers: l1 = .3383, l2 = .6180, T = -1.2800
I'm trying to create a matrix such that if I define a random number between 0 and 1 and a random location in the matrix, I want all the values around that to "diffuse" out. Here's sort of an example:
0.214 0.432 0.531 0.631 0.593 0.642
0.389 0.467 0.587 0.723 0.654 0.689
0.421 0.523 0.743 0.812 0.765 0.754
0.543 0.612 0.732 0.843 0.889 0.743
0.322 0.543 0.661 0.732 0.643 0.694
0.221 0.321 0.492 0.643 0.521 0.598
if you notice, there's a peak at (4,5) = 0.889 and all the other numbers decrease as they move away from that peak.
I can't figure out a nice way to generate a code that does this. Any thoughts? I need to be able to generate this type of matrix with random peaks and a random rate of decrease...
Without knowing what other constraints you want to implement:
Come up with a function z = f(x,y) whose peak value is at (x0,y0) == (0,0) and whose values range between [0,1]. As an example, the PDF for the Normal distribution with mu = 0 and sigma = 1/sqrt(2*pi) has a peak at x == 0 of 1.0, and whose lower bound is zero. Similarly, a bivariate normal PDF with mu = {0,0} and determinate(sigma) == [1/(2*pi)]^2 will have similar characteristics.
Any mathematical function may have its domain shifted: f(x-x0, y-y0)
Your code will look something like this:
someFunction = #(x,y) theFunctionYouPicked(x,y);
[x0,y0,peak] = %{ you supply these values %};
myFunction = #(x,y) peak * someFunction(x - x0, y - y0);
[dimX,dimY] = %{ you supply these values %};
mymatrix = bsxfun( myFunction, 0:dimX, (0:dimY)' );
You can read more about bsxfun here; however, here's an example of how it works:
bsxfun( blah, [a b c], [d e f]' )
That should give the following matrix (or its transpose ... I don't have matlab in front of me):
[blah(a,d) blah(a,e) blah(a,f);
blah(b,d) blah(b,e) blah(b,f);
blah(c,d) blah(c,e) blah(c,f)]
Get a toy example working, then you can tinker with it to be more flexible. If the function dictating how it decreases is random (with the constraint that points closer to (x0,y0) are larger than more distant points), it won't be an issue to make a procedural function instead of using strictly mathematical ones.
In response to your answer:
Your equation could be thought of as a model for gravity where an object instantaneously induces a force on another mass, then stops exerting force. Following that logic, it could be modified to a naive vector formulation like this:
% v1 & v2 are vectors that point from the two peak points to the point [ii,jj]
theMatrix(ii,jj) = norm( (r1 / norm( v1 )) * v1 / norm( v1 ) ...
+ (r2 / norm( v2 )) * v2 / norm( v2 ) ...
);
The most extreme type of corner case you'll run into is one where v1 & v2 point in the same direction as in the following row:
[ . . A X1 X2 . . ]
... where you want a value for A w/respect to X1 & X2. Using the above expression it'll boil down to A = X1 / norm(v1) + X2 / norm(v2), which will definitely exceed the peak value at X1 because norm(v1) == 1. You could certainly do some dirty stuff to Band-Aid it, but personally I'd start looking for a different function.
Along those lines, if you used Newton's Law of Universal Gravitation with a few modifications:
You wouldn't need an analogue for G, so you could just assume G == 1
Treat each of the points in the matrix as having mass m2 == 1, so the equation reduces to: F_12 == -1 * (m1 / r^2) * RHAT_12
Sum the "force" vectors and calculate the norm to get each value
... you'll still run into the same problem. The corner case I laid out above would boil down to A = X1/norm(v1)^2 + X2/norm(v2)^2 == X1 + X2/4. Since it's inversely proportional to the square of the distances, it'd be easier to Band-Aid than the linear one, but I wouldn't recommend it.
Similarly, if you use polynomials it won't scale well; you can design one that won't ever exceed your chosen peaks, but there wouldn't be a lower bound.
You could use the logistic function to help with this:
1 / (1 + E^(-c*x))
Here's an example of using the logistic function on a degree 4 polynomial with peaks at points 2 & 4; you'll note I gave the polynomial a scaling factor to pull the polynomial down to relatively small values so calculated values aren't so close together.
I ended up creating a code that wraps the way I want based on a dimension, which I provide. Here's the code:
dims = 100;
A = zeros(dims);
b = floor(1+dims*rand(1));
c = floor(1+dims*rand(1));
d = rand(1);
x1 = c;
y1 = b;
A(x1,y1) = d;
for i = 1:dims
for j = i
k = 1-j;
while k <= j
if x1-j>0 && y1+k>0 && y1+k <= dims
if A(x1-j,y1+k) == 0
A(x1-j,y1+k) = eqn(d,x1-j,y1+k,x1,y1);
end
end
k = k+1;
end
end
for k = i
j = 1-k;
while j<=k
if x1+j>0 && y1+k>0 && y1+k <= dims && x1+j <= dims
if A(x1+j,y1+k)==0
A(x1+j, y1+k) = eqn(d,x1+j,y1+k,x1,y1);
end
end
j = j+1;
end
end
for j = i
k = 1-j;
while k<=j
if x1+j>0 && y1-k>0 && x1+j <= dims && y1-k<= dims
if A(x1+j,y1-k) == 0
A(x1+j,y1-k) = eqn(d,x1+j,y1-k,x1,y1);
end
end
k=k+1;
end
end
for k = i
j = 1-k;
while j<=k
if x1-j>0 && y1-k>0 && x1-j <= dims && y1-k<= dims
if A(x1-j,y1-k)==0
A(x1-j,y1-k) = eqn(d,x1-j,y1-k,x1,y1);
end
end
j = j+1;
end
end
end
colormap('hot');
imagesc(A);
colorbar;
If you notice, the code calls a function (I called it eqn), which provided the information for how to changes the values in each cell. The function that I settled on is d/distance (distance being computed using the standard distance formula).
It seems to work pretty well. I'm now just trying to develop a good way to have multiple peaks in the same square without one peak completely overwriting the other.
Hello I am relatively new to MATLAB and have received and assignment in which we could use any programming language. I would like to continue MATLAB and have decided to use it for this assignment. The questions has to do with the following formula:
x(t) = A[1+a1*E(t)]*sin{w[1+a2*E(t)]*t+y}(+/-)a3*E(t)
The first question we have is to develop an appropriate discretization of x(t) with a time step h. I think i understand how to do this using step but because there is a +/- in the end I am running into errors. Here is what I have (I have simplified the equation by assigning arbitrary values to each variable):
A = 1;
E = 1;
a1 = 1;
a2 = 2;
a3 = 3;
w = 1;
y = 0;
% ts = .1;
% t = 0:ts:10;
t = 1:1:10;
x1(t) = A*(1+a1*E)*sin(w*(1+a2*E)*t+y);
x2(t) = a3*E;
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
plot(y)
The problem is I keep getting the following error because of the +/-:
In an assignment A(I) = B, the number of elements in B and I must be the same.
Error in Try1 (line 21)
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
Any help?? Thanks!
You can remove the (t) from the left-hand side of all three assignments.
y = [x1+x2, x1-x2]
MATLAB knows what to do with vectors and matrices.
Or, if you want to write it out the long way, tell MATLAB there will be two columns:
y(t, 1:2) = [x1(t)'+x2(t)', x1(t)'-x2(t)']
or two rows:
y(1:2, t) = [x1(t)+x2(t); x1(t)-x2(t)]
But this won't work when you have fractional values of t. The value in parentheses is required to be the index, not a dependent variable. If you want the whole vector, just leave it out.