I would like to plot the roots of the cubic equation x^{3}+Ax^{2}+1=0 in matlab. I know that there are 3 real roots for A<-1.88 and 1 if A>-1.88. I would like to plot the 3 real roots as a function of A and when it switches to 1 real root and 2 complex to plot the real root and the real part of the complex conjugate solutions all in the same plot (perhaps as 2-3 graphs).
I am a matlab beginner though. I tried
syms x A
r = solve(x^3 + A*x^2+1 == 0, x);
ezplot(vpa(r(1)),[-10,10])
ezplot(vpa(r(2)),[-10,10])
ezplot(vpa(r(3)),[-10,10])
but vpa doesnt know how to numerically evaluate r.
There's no need to do symbolic math for this,
A = (-3:0.01:0)'; % create a vector of values for A
r = arrayfun(#(A)real(roots([1 A 0 1])),A,'uni',false); % calculate the polynomial roots for all values of A
r = [r{:}]; % convert result to a numeric array
plot(A,r'); % plot the result
grid on;
title('Real parts of Polynomial');
xlabel('A');
I have an nx2 matrix r in Matlab reporting n draws from a bivariate normal distribution
n=1000;
m1=0.3;
m2=-m1;
v1=0.2;
n=10000;
v2=2;
rho=0.5;
mu = [m1, m2];
sigma = [v1,rho*sqrt(v1)*sqrt(v2);rho*sqrt(v1)*sqrt(v2),v2];
r = mvnrnd(mu,sigma,n);
I want to normalise these draws to the unit square [0,1]^2
First option
rmax1=max(r(:,1));
rmin1=min(r(:,1));
rmax2=max(r(:,2));
rmin2=min(r(:,2));
rnew=zeros(n,2);
for i=1:n
rnew(i,1)=(r(i,1)-rmin1)/(rmax1-rmin1);
rnew(i,2)=(r(i,2)-rmin2)/(rmax2-rmin2);
end
Second option
rmin1, rmax1, rmin2, rmax2 may be quite variable due to the sampling process. An alternative is applying the 68–95–99.7 rule (here) and I am asking for some help on how to generalise it to a bivariate normal (in particular Step 1 below). Here's my idea
%Step 1: transform the draws in r into draws from a bivariate normal
%with variance-covariance matrix equal to the 2x2 identity matrix
%and mean equal to mu
%How?
%Let t be the transformed vector
%Step 2: apply the 68–95–99.7 rule to each column of t
tmax1=mu(1)+3*1;
tmin1=mu(1)-3*1;
tmax2=mu(2)+3*1;
tmin2=mu(2)-3*1;
tnew=zeros(n,2);
for i=1:n
tnew(i,1)=(t(i,1)-tmin1)/(tmax1-tmin1);
tnew(i,2)=(t(i,1)-tmin2)/(tmax2-tmin2);
end
%Step 3: discard potential values (very few) outside [0,1]
In your case the x and y coordinates of the random vector are correlated, so it's not just a transformation in x and in y independently. You first need to rotate your samples so that x and y become uncorrelated (then the covariance matrix will be diagonal. You don't need it to be the identity, since anywya you normalize later). Then you can apply the transformation you call "2nd option" to the new x and y independently. Shortly, you need to diagonalize the covariance matrix.
As a side note, your code adds/subtracts 3 times 1, instead of 3 times the standard deviation. Also, you can avoid the for loop, using (e.g) Matlab's bsxfun which applies an operation between matrix and vector:
t = bsxfun(#minus,r,mean(r,1)); % center the data
[v, d] = eig(sigma); % find the directions for projection
t = t * v; % the projected data is uncorrelated
sigma_new = sqrt(diag(d)); % that's the std in the new coordinates
% now transform each coordinate independently
tmax1 = 3*sigma_new(1);
tmin1 = -3*sigma_new(1);
tmax2 = 3*sigma_new(2);
tmin2 = -3*sigma_new(2);
tnew = bsxfun(#minus, t, [tmin1, tmin2]);
tnew = bsxfun(#rdivide, tnew, [tmax1-tmin1, tmax2-tmin2]);
You still need to discard the few samples which are out of [0,1], as you wrote.
I'm trying to create a power law (x3) contour plot in the code.
Please look at the following question and the accepted answer: (logarithmic vector :temp)
How to create nonlinear spaced vector in Matlab?
I want a power law vector for temp instead of logarithmic.
Many thanks
In general, you can choose your spacing based on any function. For instance, here y is a function of x. But x is not linearly spaced. Instead, it is a function of another variable, t, which is linearly spaced. x(t) defines the spacing:
t = 0:.02:2;
x = #(u) u.^3; % or any other function
y = #(u) sin(u); % or any other function
figure; plot(x(t), y(x(t)), '*b')
With this output:
I am trying to simulate a network of mobile robots that uses artificial potential fields for movement planning to a shared destination xd. This is done by generating a series of m-files (one for each robot) from a symbolic expression, as this seems to be the best way in terms of computational time and accuracy. However, I can't figure out what is going wrong with my gradient computation: the analytical gradient that is being computed seems to be faulty, while the numerical gradient is calculated correctly (see the image posted below). I have written a MWE listed below, which also exhibits this problem. I have checked the file generating part of the code, and it does return a correct function file with a correct gradient. But I can't figure out why the analytic and numerical gradient are so different.
(A larger version of the image below can be found here)
% create symbolic variables
xd = sym('xd',[1 2]);
x = sym('x',[2 2]);
% create a potential function and a gradient function for both (x,y) pairs
% in x
for i=1:size(x,1)
phi = norm(x(i,:)-xd)/norm(x(1,:)-x(2,:)); % potential field function
xvector = reshape(x.',1,size(x,1)*size(x,2)); % reshape x to allow for gradient computation
grad = gradient(phi,xvector(2*i-1:2*i)); % compute the gradient
gradx = grad(1);grady=grad(2); % split the gradient in two components
% create function file names
gradfun = strcat('GradTester',int2str(i),'.m');
phifun = strcat('PotTester',int2str(i),'.m');
% generate two output files
matlabFunction(gradx, grady,'file',gradfun,'outputs',{'gradx','grady'},'vars',{xvector, xd});
matlabFunction(phi,'file',phifun,'vars',{xvector, xd});
end
clear all % make sure the workspace is empty: the functions are in the files
pause(0.1) % ensure the function file has been generated before it is called
% these are later overwritten by a specific case, but they can be used for
% debugging
x = 0.5*rand(2);
xd = 0.5*rand(1,2);
% values for the Stackoverflow case
x = [0.0533 0.0023;
0.4809 0.3875];
xd = [0.4087 0.4343];
xp = x; % dummy variable to keep x intact
% compute potential field and gradient for both (x,y) pairs
for i=1:size(x,1)
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
% preallocate the gradient and potential matrices
gradx = zeros(length(xGrid),length(yGrid));
grady = zeros(length(xGrid),length(yGrid));
phi = zeros(length(xGrid),length(yGrid));
% generate appropriate function handles
fun = str2func(strcat('GradTester',int2str(i)));
fun2 = str2func(strcat('PotTester',int2str(i)));
% compute analytic gradient and potential for each position in the xGrid and
% yGrid vectors
for ii = 1:length(yGrid)
for jj = 1:length(xGrid)
xp(i,:) = [xGrid(ii) yGrid(jj)]; % select the position
Xvec = reshape(xp.',1,size(x,1)*size(x,2)); % turn the input into a vector
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd); % compute gradients
phi(jj,ii) = fun2(Xvec,xd); % compute potential value
end
end
[FX,FY] = gradient(phi); % compute the NUMERICAL gradient for comparison
%scale the numerical gradient
FX = FX/0.005;
FY = FY/0.005;
% plot analytic result
subplot(2,2,2*i-1)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-gradx,-grady)
contour(xGrid,yGrid,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-FX,-FY)
contour(xGrid,yGrid,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
end
The potential field I am trying to generate is defined by an (x,y) position, in my code called xd. x is the position matrix of dimension N x 2, where the first column represents x1, x2, and so on, and the second column represents y1, y2, and so on. Xvec is simply a reshaping of this vector to x1,y1,x2,y2,x3,y3 and so on, as the matlabfunction I am generating only accepts vector inputs.
The gradient for robot i is being calculated by taking the derivative w.r.t. x_i and y_i, these two components together yield a single derivative 'vector' shown in the quiver plots. The derivative should look like this, and I checked that the symbolic expression for [gradx,grady] indeed looks like that before an m-file is generated.
To fix the particular problem given in the question, you were actually calculating phi in such a way that meant you doing gradient(phi) was not giving the correct results compared to the symbolic gradient. I'll try and explain. Here is how you created xGrid and yGrid:
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
But then in the for loop, ii and jj were used like phi(jj,ii) or gradx(ii,jj), but corresponding to the same physical position. This is why your results were different. Another problem you had was you used gradient incorrectly. Matlab assumes that [FX,FY]=gradient(phi) means that phi is calculated from phi=f(x,y) where x and y are matrices created using meshgrid. You effectively had the elements of phi arranged differently to that, an so gradient(phi) gave the wrong answer. Between reversing the jj and ii, and the incorrect gradient, the errors cancelled out (I suspect you tried doing phi(jj,ii) after trying phi(ii,jj) first and finding it didn't work).
Anyway, to sort it all out, on the line after you create xGrid and yGrid, put this in:
[X,Y]=meshgrid(xGrid,yGrid);
Then change the code after you load fun and fun2 to:
for ii = 1:length(xGrid) %// x loop
for jj = 1:length(yGrid) %// y loop
xp(i,:) = [X(ii,jj);Y(ii,jj)]; %// using X and Y not xGrid and yGrid
Xvec = reshape(xp.',1,size(x,1)*size(x,2));
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd);
phi(ii,jj) = fun2(Xvec,xd);
end
end
[FX,FY] = gradient(phi,0.005); %// use the second argument of gradient to set spacing
subplot(2,2,2*i-1)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))]) %// use axis rather than xlim/ylim
quiver(X,Y,gradx,grady)
contour(X,Y,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))])
quiver(X,Y,FX,FY)
contour(X,Y,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
I have some other comments about your code. I think your potential function is ill-defined, which is causing all sorts of problems. You say in the question that x is an Nx2 matrix, but you potential function is defined as
norm(x(i,:)-xd)/norm(x(1,:)-x(2,:));
which means if N was three, you'd have the following three potentials:
norm(x(1,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(2,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(3,:)-xd)/norm(x(1,:)-x(2,:));
and I don't think the third one makes sense. I think this could be causing some confusion with the gradients.
Also, I'm not sure if there is a reason to create the .m file functions in your real code, but they are not necessary for the code you posted.
I have a problem with numerical derivative of a vector that is x: Nx1 with respect to another vector t (time) that is the same size of x.
I do the following (x is chosen to be sine function as an example):
t=t0:ts:tf;
x=sin(t);
xd=diff(x)/ts;
but the answer xd is (N-1)x1 and I figured out that it does not compute derivative corresponding to the first element of x.
is there any other way to compute this derivative?
You are looking for the numerical gradient I assume.
t0 = 0;
ts = pi/10;
tf = 2*pi;
t = t0:ts:tf;
x = sin(t);
dx = gradient(x)/ts
The purpose of this function is a different one (vector fields), but it offers what diff doesn't: input and output vector of equal length.
gradient calculates the central difference between data points. For an
array, matrix, or vector with N values in each row, the ith value is
defined by
The gradient at the end points, where i=1 and i=N, is calculated with
a single-sided difference between the endpoint value and the next
adjacent value within the row. If two or more outputs are specified,
gradient also calculates central differences along other dimensions.
Unlike the diff function, gradient returns an array with the same
number of elements as the input.
I know I'm a little late to the game here, but you can also get an approximation of the numerical derivative by taking the derivatives of the polynomial (cubic) splines that runs through your data:
function dy = splineDerivative(x,y)
% the spline has continuous first and second derivatives
pp = spline(x,y); % could also use pp = pchip(x,y);
[breaks,coefs,K,r,d] = unmkpp(pp);
% pre-allocate the coefficient vector
dCoeff = zeroes(K,r-1);
% Columns are ordered from highest to lowest power. Both spline and pchip
% return 4xn matrices, ordered from 3rd to zeroth power. (Thanks to the
% anonymous person who suggested this edit).
dCoeff(:, 1) = 3 * coefs(:, 1); % d(ax^3)/dx = 3ax^2;
dCoeff(:, 2) = 2 * coefs(:, 2); % d(ax^2)/dx = 2ax;
dCoeff(:, 3) = 1 * coefs(:, 3); % d(ax^1)/dx = a;
dpp = mkpp(breaks,dCoeff,d);
dy = ppval(dpp,x);
The spline polynomial is always guaranteed to have continuous first and second derivatives at each point. I haven not tested and compared this against using pchip instead of spline, but that might be another option as it too has continuous first derivatives (but not second derivatives) at every point.
The advantage of this is that there is no requirement that the step size be even.
There are some options to work-around your issue.
First: you can make your domain larger. Instead of N, use N+1 gridpoints.
Second: depending on the end-point of interest, you can use
Forward difference: F(x + dx) - F(x)
Backward difference: F(x) - F(x - dx)