I want to take a polynomial p as input from user in matlab for a given degree (specified by the user each time) such that the polynomial is input one element at a time into a matrix at each index from 1 to n. where n is the polynomial degree. was tryin to do something like this but m stuck
for M = 1:n
p[n] = input('polynomial')
p
end
How should I input a polynomial coefficient at each index of matrix i.e. how to reach each index position?
Instead of using a loop, you can take a polynomial as input using the following method:
p = input('Enter a polynomial in [] brackets');
Now the user should enter the polynomial like this:
[2, 4, 3, 8];
Then you can calculate its degree using the length command:
n = length(p);
Related
Assume we want to determine the coefficients of a polynomial equation that is approximating the tangent function between 0 to 1, as follow:
-A is m×n vandermonde matrix. The entries are populated using m value between 0 to 11(given as input).
-The corresponding vector b is calculated using tangent function.
-x is calculated by typing x= A\b in MATLAB.
Now, using MATLAB, the computed x are subsittued in Ax. The result is plotted and it is pretty close to tangent function. But if I use polyval function of n−1 degree (in MATLAB) to calculate b, the resulting plot is significantly different from the original b. I cannot understand the reason for such a significant difference between the results of these two methods.
Here is the code:
clear all;
format long;
m = 60;
n = 11;
t = linspace(0,1,m);
A= fliplr(vander(t));
A=A(:,1:n);
b=tan(t');
x= A\b;
y=polyval(x, t);
plot(t,y,'r')
y2= A*x
hold on;
plot(t,y2,'g.');
hold on;
plot(t,tan(t),'--b');
Any insight would be appreciated. Thank you.
After A= fliplr(vander(t)) the A matrix is equal to
1 t(1) t(1)^2 ...
1 t(2) t(2)^2 ...
...
1 t(m) t(m)^2 ...
It is not correct because polyval accepts the coefficients in descending powers. You don't need to flip the columns of A:
A= vander(t);
A= A(:,end-n+1:end);
So my question refers to the regress() function in matlab. Click here for the Matlab documentation
If I want to run multiple regressions using this function and output both the coefficients and the confidence intervals, what's the best way to do this in a For loop?
Matlab's own syntax for this is [b,bint] = regress(y,X). But when I try to implement this in a for loop it tells me that the dimension mismatch. My code is the following:
for i=1:6
[a, b]=regress(Dataset(:,i),capm_factors);
capm_coefs(i,:)=a;
capm_ci(i,:)=b;
end
Please help, thanks!
regress outputs a column vector of coefficients that minimize the least squared error between your input data (capm_factors) and your predicted values (Dataset(:,i)). However, in your for loop, you are assuming that a and b are row vectors.
Also, the first output of regress is the solution to your system, but the second output contains a matrix of confidence values where the first column denotes the lower end of the confidence interval for each variable and the second column denotes the upper end of the confidence interval.
Specifically, your input capm_factors should be a M x N matrix where M is the total number of input samples and N is the total number of features. In your code, a would thus give you a N x 1 vector and b would give you a N x 2 matrix.
If you'd like use a loop, make sure capm_coefs is a N x l matrix where l is the total number of times you want to loop and capm_ci should either be a N x 2 x l 3D matrix or perhaps a l element cell array. Either way is acceptable.... but I'll show you how to do both.
Something like this comes to mind:
Confidences as a 3D matrix
l = 6; %// Define # of trials
[M,N] = size(capm_factors); %// Get dimensions of data
capm_coefs = zeros(N, l);
capm_ci = zeros(N, 2, l);
for ii = 1 : l
[a,b] = regress(Dataset(:,i), capm_factors);
capm_coefs(:,ii) = a;
capm_ci(:,:,ii) = b;
end
You'd then access the coefficients for a trial via capm_coefs(:,ii) where ii is the iteration you want. Similarly, the confidence matrix can be accessed via capm_ci(:,:,ii)
Confidences as a cell array
l = 6; %// Define # of trials
[M,N] = size(capm_factors); %// Get dimensions of data
capm_coefs = zeros(N, l);
capm_ci = cell(l); %// Cell array declaration
for ii = 1 : l
[a,b] = regress(Dataset(:,i), capm_factors);
capm_coefs(:,ii) = a;
capm_ci{ii} = b; %// Assign confidences to cell array
end
Like above, you'd access the coefficients for a trial via capm_coefs(:,ii) where ii is the iteration you want. However, the confidence matrix can be accessed via capm_ci{ii} as we are now dealing with cell arrays.
I'm new to programming in Matlab. I'm trying to figure out how to calculate the following function:
I know my code is off, I just wanted to start with some form of the function. I have attempted to write out the sum of the function in the program below.
function [g] = square_wave(n)
g = symsum(((sin((2k-1)*t))/(2k-1)), 1,n);
end
Any help would be much appreciated.
Update:
My code as of now:
function [yout] = square_wave(n)
syms n;
f = n^4;
df = diff(f);
syms t k;
f = 1; %//Define frequency here
funcSum = (sin(2*pi*(2*k - 1)*f*t) / (2*k - 1));
funcOut = symsum(func, v, start, finish);
xsquare = (4/pi) * symsum(funcSum, k, 1, Inf);
tVector = 0 : 0.01 : 4*pi; %// Choose a step size of 0.01
yout = subs(xsquare, t, tVector);
end
Note: This answer was partly inspired by a previous post I wrote here: How to have square wave in Matlab symbolic equation - However, it isn't quite the same, which is why I'm providing an answer here.
Alright, so it looks like you got the first bit of the question right. However, when you're multiplying things together, you need to use the * operator... and so 2k - 1 should be 2*k - 1. Ignoring this, you are symsuming correctly given that square wave equation. The input into this function is only one parameter only - n. What you see in the above equation is a Fourier Series representation of a square wave. A bastardized version of this theory is that you can represent a periodic function as an infinite summation of sinusoidal functions with each function weighted by a certain amount. What you see in the equation is in fact the Fourier Series of a square wave.
n controls the total number of sinusoids to add into the equation. The more sinusoids you have, the more the function is going to look like a square wave. In the question, they want you to play around with the value of n. If n becomes very large, it should start approaching what looks like to be a square wave.
The symsum will represent this Fourier Series as a function with respect to t. What you need to do now is you need to substitute values of t into this expression to get the output amplitude for each value t. They define that for you already where it's a vector from 0 to 4*pi with 1001 points in between.
Define this vector, then you'll need to use subs to substitute the time values into the symsum expression and when you're done, cast them back to double so that you actually get a numeric vector.
As such, your function should simply be this:
function [g] = square_wave(n)
syms t k; %// Define t and k
f = sin((2*k-1)*t)/(2*k-1); %// Define function
F = symsum(f, k, 1, n); %// Define Fourier Series
tVector = linspace(0, 4*pi, 1001); %// Define time points
g = double(subs(F, t, tVector)); %// Get numeric output
end
The first line defines t and k to be symbolic because t and k are symbolic in the expression. Next, I'll define f to be the term inside the summation with respect to t and k. The line after that defines the actual sum itself. We use f and sum with respect to k as that is what the summation calls for and we sum from 1 up to n. Last but not least, we define a time vector from 0 to 4*pi with 1001 points in between and we use subs to substitute the value of t in the Fourier Series with all values in this vector. The result should be a 1001 vector which I then cast to double to get a numerical result and we get your desired output.
To show you that this works, we can try this with n = 20. Do this in the command prompt now:
>> g = square_wave(20);
>> t = linspace(0, 4*pi, 1001);
>> plot(t, g);
We get:
Therefore, if you make n go higher... so 200 as they suggest, you'll see that the wave will eventually look like what you expect from a square wave.
If you don't have the Symbolic Math Toolbox, which symsum, syms and subs relies on, we can do it completely numerically. What you'll have to do is define a meshgrid of points for pairs of t and n, substitute each pair into the sequence equation for the Fourier Series and sum up all of the results.
As such, you'd do something like this:
function [g] = square_wave(n)
tVector = linspace(0, 4*pi, 1001); %// Define time points
[t,k] = meshgrid(tVector, 1:n); %// Define meshgrid
f = sin((2*k-1).*t)./(2*k-1); %// Define Fourier Series
g = sum(f, 1); %// Sum up for each time point
end
The first line of code defines our time points from 0 to 4*pi. The next line of code defines a meshgrid of points. How this works is that for t, each column defines a unique time point, so the first column is 200 zeroes, up to the last column which is a column of 200 4*pi values. Similarly for k, each row denotes a unique n value so the first row is 1001 1s, followed by 1001 2s, up to 1001 1s. The implications with this is now each column of t and k denotes the right (t,n) pairs to compute the output of the Fourier series for each time that is unique to that column.
As such, you'd simply use the sequence equation and do element-wise multiplication and division, then sum along each individual column to finally get the square wave output. With the above code, you will get the same result as above, and it'll be much faster than symsum because we're doing it numerically now and not doing it symbolically which has a lot more computational overhead.
Here's what we get when n = 200:
This code with n=200 ran in milliseconds whereas the symsum equivalent took almost 2 minutes on my machine - Mac OS X 10.10.3 Yosemite, 16 GB RAM, Intel Core i7 2.3 GHz.
Hi everybody I have this problem:
I have Dataset of n vectors each has D dimensions.
I also have a covariance matrix of size D*D, Let It be C.
I perform the following action:
I choose K vectors from the dataset, and also choose E dimensions randomly. Let M be the sample covariance of the selected data on the selected dimensions.so M is a E*E matrix.
let P be the partial covariance matrix corresponding to the dimensions E of C, ie. C(E,E) in matlab
is the following matrix positive semi definite?:
X = (1-a)P + aM
where a is a constant like 0.2.
I sometimes get the following error when using mvnrnd(mean,X) :
SIGMA must be a symmetric positive semi-definite matrix
My code is:
%%%Dims are randomly choosen dimensions
%%%Inds are randomly choosen Indexes form {1, 2, ...,n}
%%% PP are n D dimensional vectors, composing my data set PP is n*D
%%% Sigmaa is a D*D covariance matrix
co = cov(PP(Inds,Dims));
me = mean(PP(Inds,Dims));
Bettaa = 0.2;
sigmaaDims = sigmaa(Dims,Dims);
sigmaaDims = (1-Bettaa)*sigmaaDims + (co)*Bettaa;
Tem = mvnrnd(me,sigmaaDims);
Simply looking at the matrix dimensions It is not possible to tell if a matrix is positive semi-definite.
To find out if a given matrix is positive semi-definite, you must check if It's eigenvalues are non-negative and it's symmetry:
symmetry = issymmetric(X);
[~,D]=eig(X);
eigenvalues = diag(D);
if all(eigenvalues>0) & symmetry
disp('Positive semi-definite matrix.')
else
disp('Non positive semi-definite matrix.')
end
Where X is the matrix you are interested in.
Note that if you use the weaker definition of a positive definite matrix (see Extention for non symmetric matrices section), X does not need to be symmetric and you would end up with:
[~,D]=eig(X);
eigenvalues = diag(D);
if all(eigenvalues>=0)
disp('Positive semi-definite matrix.')
else
disp('Non positive semi-definite matrix.')
end
I am trying to solve a second order differential using ODE45 in Matlab with matrix as inputs. I am struck with couple of errors that includes :
"In an assignment A(I) = B, the number of elements in B and
I must be the same."
Double order differential equations given below:
dy(1)= diag(ones(1,100) - 0.5*y(2))*Co;
dy(2)= -1 * Laplacian(y(1)) * y(2);
Main function call is:
[T,Y] = ode45(#rigid,[0.000 100.000],[Co Xo]);
Here, Co is Matrix of size 100x100 and Xo is a column matrix of size 100x1. Laplacian is a pre-defined function to compute matrix laplacian.
I will appreciate any help in this. Should I reshape input matrices and vectors to fall in same dimensions or something?
Your guess is correct. The MATLAB ode suite can solve only vector valued ode, i.e. an ode of the form y'=f(t,y). In your case you should convert y, and dy, back and forth between a matrix and an array by using reshape.
To be more precise, the initial condition will be transformed into the array
y0 = reshape([Co Xo], 100*101, 1);
while y will be obtained with
y_matrix = reshape(y, 100, 101);
y1 = y_matrix(:,1:100);
y2 = y_matrix(:,101);
After having computed the matrices dy1 and dy2 you will have to covert them in an array with
dy = reshape([dy1 dy2], 100*101, 1);
Aside from the limitations of ode45 your code gives that error because, in MATLAB, matrices are not indexed in that way. In fact, if you define A = magic(5), A(11) gives the eleventh element of A i.e. 1.