For example , if I want to plot Sin(z) where z is a complex variable , how I will achieve it in either Octave or Maxima?
I don't know about Octave, but here is a message about that, with some code you can try in Maxima: https://www.ma.utexas.edu/pipermail/maxima/2007/006644.html
There may be more specific information for wxMaxima -- you can try their user forum: https://sourceforge.net/p/wxmaxima/discussion/435775/
(referring Octave 4.0.0)
How do you want to try to represent the output of the function? Plotting either the real or imaginary parts of the output can be done fairly simply using a 3-dimensional graph, where the x and y axes are the real and imaginary components of z, and the vertical axis is either the real or imaginary values of sin(z). Producing those are fairly simple in Octave. Here's a link to a script you can save and run to show an example.
Simply change the g = exp(f) line to g = sin(f).
Octave-help mailing list example
Note that the imaginary part plot is commented out. Just switch the # between the different plot commands if you want to see that part.
Now, are you instead looking for options to map the Z plane (z=x+iy) to the W plane (w=u+iv) and represent closed contours mapped by w=sin(z)? in that case you'll need to do parametric plotting as described on this FIT site. There is a link to his Matlab program at the bottom of the explanation that provides one method of using color coding to match z->w plane contour mapping.
Those m-files are written for Matlab, so a few things do not work, but the basic plotting is compatible with Octave 4.0.0. (the top level ss13.m file will fail on calls to flops and imwrite)
But, if you put your desired function in myfun13.m for f, df and d2f, (sin(z), cos(z), -sin(z) respectively), then run cvplot13, you'll get color maps showing the correspondence between z and w planes.
wxMaxima has a plot3d that can do it. Since the expression to plot is in terms of x and y, I plotted the function's magnitude with abs(f(x+%i*y)):
plot3d(abs((x+%i*y-3)*(x+%i*y-5)*(x+%i*y-6)), [x,2,7], [y,-1,1], [grid,100,100], [z,0,5])$
Related
So, I have a vector that corresponds to a given feature (same dimensionality). Is there a package in Julia that would provide a mathematical function that fits these data points, in relation to the original feature? In other words, I have x and y (both vectors) and need to find a decent mapping between the two, even if it's a highly complex one. The output of this process should be a symbolic formula that connects x and y, e.g. (:x)^3 + log(:x) - 4.2454. It's fine if it's just a polynomial approximation.
I imagine this is a walk in the park if you employ Genetic Programming, but I'd rather opt for a simpler (and faster) approach, if it's available. Thanks
Turns out the Polynomials.jl package includes the function polyfit which does Lagrange interpolation. A usage example would go:
using Polynomials # install with Pkg.add("Polynomials")
x = [1,2,3] # demo x
y = [10,12,4] # demo y
polyfit(x,y)
The last line returns:
Poly(-2.0 + 17.0x - 5.0x^2)`
which evaluates to the correct values.
The polyfit function accepts a maximal degree for the output polynomial, but defaults to using the length of the input vectors x and y minus 1. This is the same degree as the polynomial from the Lagrange formula, and since polynomials of such degree agree on the inputs only if they are identical (this is a basic theorem) - it can be certain this is the same Lagrange polynomial and in fact the only one of such a degree to have this property.
Thanks to the developers of Polynomial.jl for leaving me just to google my way to an Answer.
Take a look to MARS regression. Multi adaptive regression splines.
I was asked to create a leg follower robot (I already did it) and in the second part of this assignment I have to develop a Kalman filter in order to improve the following process of the robot. The robot gets from the person the distance where she is to the robot and also the angle (it is a relative angle, because the reference is the robot itself, not absolute x-y coordinates)
About this assignment I have a serious doubt. Everything I have read, every sample I have seen about kalman filter has been in one dimension (a car running distance or a rock falling from a building) and according to the task I would have to apply it in 2 dimensions. Is it possible to apply a kalman filter like this?
If it is possible to calculate kalman filter in 2 dimensions then I would understand that what is asked to do is to follow the legs in a linnearized way, despite a person walks weirdly (with random movements) --> About this I have the doubt of how to establish the function of the state matrix, could anyone please tell me how to do it or to tell me where I can find more information about this?
thanks.
Well you should read up on Kalman Filter. Basically what it does is estimate a state through its mean and variance separately. The state can be whatever you want. You can have local coordinates in your state but also global coordinates.
Note that the latter will certainly result in nonlinear system dynamics, in which case you could use the Extended Kalman Filter, or to be more correct the continuous-discrete Kalman Filter, where you treat the system dynamics in a continuous manner and the measurements in discrete time.
Example with global coordinates:
Assuming you have a small cubic mass which can drive forward with velocity v. You could simply model the dynamics in local coordinates only, where your state s would be s = [v], which is a linear model.
But, you could also incorporate the global coordinates x and y, assuming we are moving on a plane only. Then you would have s = [x, y, phi, v]'. We need phi to keep track of the current orientation since the cube can only move forward in respect to its orientation of course. Let's define phi as the angle between the cube's forward direction and the x-axis. Or in other words: With phi=0 the cube would move along the x-axis, with phi=90° it would move along the y-axis.
The nonlinear system dynamics with global coordinates can then be written as
s_dot = [x_dot, y_dot, phi_dot, v_dot]'
with
x_dot = cos(phi) * v
y_dot = sin(phi) * v
phi_dot = ...
v_dot = ... (Newton's Law)
In EKF (Extended Kalman Filter) Prediction step you would use the (discretized) equations above to predict the mean of the state in the first step of and the linearized (and discretized) equations for prediction of the Variance.
There are two things to keep in mind when you decide what your state vector s should look like:
You might be tempted to use my linear example s = [v] and then integrate the velocity outside of the Kalman Filter in order to obtain the global coordinate estimates. This would work, but you would lose the awesomeness of the Kalman Filter since you would only integrate the mean of the state, not its variance. In other words, you would have no idea what the current uncertainties for your global coordinates are.
The second step of the Kalman Filter, the measurement or correction update, requires that you can describe your sensor output as a function of your states. So you may have to add states to your representation just so that you can express your measurements correctly as z[k] = h(s[k], w[k]) where z are measurements and w is a noise vector with Gaussian distribution.
What it the best way to plot a vertical line using Octave?
So, I have two methods for this. One, I found, and the other I made up.
Method 1: From here.
%% Set x value where verticle line should intersect the x-axis.
x = 0;
%% plot a line between two points using plot([x1,x2],[y1,y2])
plot([x,x],[-10,10]);
Method 2: A slightly different approach, exact same result
%% Setup a vector of x values
x = linspace(0,0,100);
%% Setup a vector of y values
y = linspace(0,10,100);
%% Plot the paired points in a line
plot(x,y);
I think Method 2 may write more information to memory before the plot process and it's a line longer, so in my eyes, Method 1 should be the better option. If you prefer Method 2, make sure your x and y vectors are the same dimension or you'll end up with a bunch of dots where you're line should be.
Unfortunately the Octave documentation for doing obvious things can be ridiculously lousy with no working examples. Drawing a simple line on top of a plot is one.
As already mentioned, it's is very silly to plot straight lines in octave. It's a waste of memory and processing. Instead use the line() function, to draw on top of your plot.
The line() function require 2 non-standard x-values and y-values vectors, instead of the standard point-slope arguments for point A and point B, normally represented by (x1,y1) and (x2,y2). Instead, you need to write this as: X=(x1,x2) and Y=(y1,y2). Thus confusing every living soul!
Here is an example of the correct way to do this in Octave language:
pkg load statistics % Need to load the statistics package
x = randn (1,1000); % Normal Distribution of random numbers
clf; histfit(x) % Make a histogram plot of x and fit the data
ylim ([-20,100]) % Change the plot limits (to lift graph up)
% Draw the (vertical) line between (0,-10) and (0,90)
line ("xdata",[0,0], "ydata",[-10,90], "linewidth", 3)
With the result:
notation (x1,x2),(y1,y2) is really confusing and against textbooks.
Anyway, this is my way:
figure;
hold on;
% vertical line x=0
plot([0,0],[0,10]);
%horizontal line y=0
plot([0,10],[0,0]);
% vertical line x=2
plot([2,2],[0,10]);
hold off;
I would like to make a prediction by using Least Squares Support Vector Machine for Regression, which is proposed by Suykens et al. I am using LS-SVMlab, which you can find the MATLAB toolbox here. Let's consider I have an independent variable X and a dependent variable Y, that both are simulated. I am following the instructions in the tutorial.
>>X = linspace(-1,1,50)’;
>>Y = (15*(X.^2-1).^2.*X.^4).*exp(-X)+normrnd(0,0.1,length(X),1);
>>type = ’function estimation’;
>>[gam,sig2] = tunelssvm({X,Y,type,[], [],’RBF_kernel’},’simplex’,...’leaveoneoutlssvm’,’mse’});
>>[alpha,b] = trainlssvm({X,Y,type,gam,sig2,’RBF_kernel’});
>>plotlssvm({X,Y,type,gam,sig2,’RBF_kernel’},{alpha,b});
The code above finds the best parameters using simplex method and leave-one-out cross validation and trains the model and give me alphas (support vector values for all the data points in the training set) and b coefficients. However, it does not give me the predictions of the variable Y. It only draws the plot. In some articles, I saw plots like the one below,
As I said before, the LS-SVM toolbox does not give me the predicted values of Y, it only draws the plot but no values in the workspace. How can I get these values and draw a graph of predicted values together with actual values?
There is one solution that I think of. By using X values in the training set, I re-run the model and get the prediction of values Y by using simlssvm command but it does not seem reasonable to me. Any solution that you can offer? Thanks in advance.
I am afraid you have answered your own question. The only way to obtain the prediction for the training points in LS-SVMLab is by simulating the training points after training your model.
[yp,alpha,b,gam,sig2,model] = lssvm(x,y,'f')
when u use this function yp is the predicted value
Started learning octave recently. How do I generate a matrix from another matrix by applying a function to each element?
eg:
Apply 2x+1 or 2x/(x^2+1) or 1/x+3 to a 3x5 matrix A.
The result should be a 3x5 matrix with the values now 2x+1
if A(1,1)=1 then after the operation with output matrix B then
B(1,1) = 2.1+1 = 3
My main concern is a function that uses the value of x like that of finding the inverse or something as indicated above.
regards.
You can try
B = A.*2 + 1
The operator . means application of the following operation * to each element of the matrix.
You will find a lot of documentation for Octave in the distribution package and on the Web. Even better, you can usually also use the extensive documentation on Matlab.
ADDED. For more complex operations you can use arrayfun(), e.g.
B = arrayfun(#(x) 2*x/(x^2+1), A)