x=[1;2;3]
x =
1
2
3
y=[4;5;6]
y =
4
5
6
x\y
ans =
2.2857
How did Matlab find that result ? (I searched many forums but I did not understand what they told.I would like to know the algorithm which gave this result.)
From MATLAB documentation of \:
If A is an M-by-N matrix with M < or > N and B is a column vector with M components, or a matrix with several such columns, then X = A\B is the solution in the least squares sense to the under- or overdetermined system of equations A*X = B.
Here your system is not under/over-determined. Since both have 3 rows. So you can visualize your equation as:
xM=y
M=inv(x)*y
Now, since your matrix is not square, it will calculate the pseudo-inverse using SVD. Therefore,
M=pinv(x)*y;
You will get value of M as 2.2857.
Another explanation can be: It will give you the solution of xM=y in the sense of least squares. You can verify this as follows:
M=lsqr(x,y)
This will give you the value of M = 2.2857.
You can always do help \ in MATLAB command window to get more information.
You are encouraged to check more details about the least squares and pseudo-inverse.
This documentation should explain it
http://www.mathworks.com/help/matlab/ref/mrdivide.html
Here is a link to the algorithm
http://www.maths.lth.se/na/courses/NUM115/NUM115-11/backslash.html
You can see the source inside matlab much more easily though. (I don't have it locally so I can't check but the source of a lot of matlab functions is available inside matlab)
Related
I want to fit a curve to my data points (x;y) that will have a formula as such:
1/y = (x^-1)*a + b
At first I wanna do this using Octave but later I have to code this into microcontroller using c.
A quick search on google and matlab documentation don't give an anwesr I can't find a function that do polyfit with elements with negative order.
Is there a special set of function for such operation or do I have to somehow transfer my formula to fit into standard math problem ?
Your unknowns are aand b which are both linear in your problem. So you can use the 1st order polynomial fitting. It is already in the form of a standard math problem. To see just rename
Y = a*X + b
with the known data vectors (or points)
Y = 1/y
X = 1/x
Thats all.
I have a matrix that is a function of some parameter A=A(x). I would like to find the points x where this matrix becomes singular. Example (I have a large matrix though):
syms x
A=[x sin(x); cos(x^2) 2.5];
So far I have been symbolically computing the determinant of the matrix and then used fzero or newtzero to find the roots of that characteristic equation. I.e.
detA = det(A);
fzero(matlabFunction(detA),startingGuess)
Then I found this: How to find out if a matrix is singular?, where it is advocated to not use the determinant under any circumstances.
Indeed the symbolic determinant calculation is terribly slow. However I tried to use rank(A) instead as suggested in the link and it does not seem to work for symbolic matrices.
Is there any way to implement the suggestions in the link for finding the roots of a characteristic equation of a matrix that is given symbolically?
A possible approach would be the following: a square matrix A is singular if and only if the homogeneous linear (with respect to the vector y) system A*y = 0 has nontrivial solutions y <> 0 (which is equivalent to det(A) = 0 and rank(A) = 0 among others. So a more or less standard, as I recall from the past, technique to compute such points x is to solve the nonlinear system
A(x)*y = 0 (1)
||y|| = 1 (2)
This way you can compute a point x* and a vector y* such that A(x*) is singular and y* is an eigenvector corresponding to the zero eigenvalue of A(x*).
If I remember correctly, you can also solve the somewhat easier system
A(x)*y = 0 (1)
<y,c> = 1 (2a)
where c is "almost" any nonzero random vector (normalize it to 1 to avoid numerical problems).
As a matter of fact there is an enormous bibliography on the subject - you can look for saddle-node bifurcation computations (in case A(x) is the Jacobian of a vector field), or for "distance to instability".
From a discussion with Ander Biguri it seems that the determinant is actually a perfectly fine method of approaching this problem. The problem seems to be to solve the final equation in a stable manner, which would be a different question.
I am trying to solve some equations on Matlab using Binary Integer Programming.
I have 3 sets of equations:
Ma.X=1
Mp.X<=1
Mr.X<=m*
Where, Ma is a known matrix with size 5*12
X is unknown set with size 12*1
Also Mp is known matrix with size 5*12
and Mr is a known matrix with size 4*12.
1 in the equations is an unity matrix with size 5*1 in both sets(1&2)
m* is a given known matrix with size 4*1
I'm trying to use command bintprog but how to put 1 equality and 2 inequalities
to get values of X. Also I don't have Function f to insert, I just have set of equations. Given that X unknown values with values 1 or 0.
I tried this command bintprog([],Ma,One51,Mp,One51)
but it gives me The problem is infeasible. with zeros answer matrix.
Please help me to solve this on Matlab
The correct syntax for bintprog is X = bintprog(f,A,b,Aeq,beq).
If you don't have f (which means you just want any feasible point), you can set it to []. However, for the others, your syntax is slightly wrong.
I am assuming that the + in your constraints is actually a * since otherwise, the matrix algebra doesn't quite make sense.
Try this:
X = bintprog([],[Mp;Mr],[ones(5,1);mstar],Ma,ones(5,1))
If even then it tells you that the problem is infeasible; it might as well be true that there are no Xs that can satisfy all your constraints.
I have two matrices D and Y.
I want to find the matrix G according to this:
G*D = Y
Note that all of these matrices are not square matrices.
According to Matlab's documentation, if you want to solve an equation of the form
xA = b
you can solve it by doing
x = b/A
Note that your system is underdetermined, and you cannot simply find a single solution without additional constraints. An example:
A=[1;2;3];
b=[14;32];
x=b/A;
x*A==b % check if solution is correct
[1,2,3;4,5,6]*A==b % another, equally correct solution
It "works", but without restating the problem you're not going to get at anything better.
Note this is quite extensively explained in the same documentation.
Suppose I have a weight matrix W nxm where m is the number of variables and the n is the number of instances. Also I have data matrix X of the same size. I try to find the closest weight vector to each instance in X. However both matrices are so dimensional therefore plain methods are not sufficient enough. I have tried some GPU trick at MATLAB but it does not work well since it was sequential approach that was calculating the closest weight for each instance sequentially. I am now looking for efficient one shot code. That takes all the W and X and find the winner with some MATLAB tricks with possibly some GPU addition. Is there any one that can suggest any code snippet in the MATLAB?
This is the thing that I wrote for sequential
x_in_d = gpuArray(x_in); % take input instance to device
W_d = gpuArray(W); % take weight matrix to device
Dx = W_d - x_in_d(ones(size(W_d,1),1),logical(ones(1,length(x_in_d))));
[d_min,winner] = min(sum((Dx.^2)'));
d_min = gather(d_min); %gather results
winner = gather(winner);
What do you mean by so dimensional? It's just an m x n matrix right?
It would be really helpful if you could provide some sample data, based off your description (which isn't the clearest), here is what I think your data looks like.
weights=
[1 4 2
5 3 1]
data=
[2 5 1
1 2 2]
And you want to figure out which row of weights is closest to the row of data? Which in this case would be the first row of weights for both rows of data.
Please edit your question to clarify what your asking for and consider using some examples.
EDIT:
I like Rody's Dup. Comment, if I am correct, check out: Link Here