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.
Related
Firstly, I'm sure a simple answer exists for this, maybe I'm just not wording it right in searching for an answer online.
I'm trying to solve an equation that looks like this:
a*x*cot(a*x) == b
Where a and b are constants. Using
solve(a*x*cot(a*x) == b, x)
I'm getting a result I know is wrong (with the values I'm using for the constants, I'm getting like -227, and it should be something around +160.) I plotted it up in Mathematica as two separate functions, and they do cross each other right around there, but since the cot part is periodic, they do so many times.
I want to constrain Matlab's search for the solution to a specific interval, such as 0 to 200; how do I do that?
I'm pretty new to Matlab (rather more experienced in Mathematica).
You can specify the bounds on x using fzero with only two requirements
The function must be in a "residual" form (i.e., r(x) = 0)
The residual values at the two bounds must have opposite sign (this guarantees that a root exists within the interval for continuous functions).
So we re-write the function in residual form:
r = #(x) a*x*cot(a*x) - b;
define the interval
% These are just random numbers; the actual bounds should come
% from the graph the ensures r has different signs a xL and xR
xL = 150;
xR = 170;
and solve
x = fzero(r,[xL,xR]);
I see you were trying to use the Symbolic Toolbox for a solution, but since the equation is a non-linear combination of a polynomial and a trigonometric function, there is more than likely no closed form solution. So I differed to a non-linear, numeric root-finder.
I tried some values and it seems solve returns a numeric solution. This is the documented behaviour if no analytic solution is found.
In this case, you may directly call the numeric solver with a matching start value
vpasolve(a*x*cot(a*x) == b, x,160)
It's not exactly what you asked for, but using your reading from the plot as a start value should do it.
I have two matrices J1 (sparse) = J2(full).
The dimension of the matrices are ~ 5200x2600
Then when I do:
hlm1 = (J1'*J1 + u*I)\g, I = eye(n);
and
hlm2 = (J2'*J2 + u*I)\g, I = eye(n);
i have after that: norm(hlm1 - hlm12, Inf) is 4.8625e-05 ...
That difference is my problem, is correct the way to use the matrice sparse ?.
Thx.
This is not a complete answer, but I think it could be useful. I can partially reproduce this difference using some random data:
H1=sprand(1000,1000,.4);
g=sprand(1000,1,.5);
x=H1*g;
H2=full(H1);
x2=full(x);
g1=H1\x;
g2=H2\x2;
difference=norm(g1-g2,Inf)
errorSparse=norm(g1-g,Inf)
errorFull=norm(g2-g,Inf)
the norm ends up roughly O(1e-12). I think the difference is due to the method used to solve sparse system of equations. Solving the sparse system will be using sparse function, and solving the full matrix will be using a different set of functions. Naturally these functions will be different, and I think this is probably causing some differences. I can't explain why the errors are that large though.
See the documentation for mldivide which includes some short discussion about sparse matrices as well as some methods use to solve them.
I have a linear system of equations AX = B to solve in MATLAB. What I have known is A is sparse, positive-definite and symmetric. I know the command x = A \ b works yet I am not sure MATLAB takes full advantage of A's good properties so as to maximize the efficiency. Is there any way to specify the algorithm to solve it, for example Conjugate Gradient algorithm in MATLAB?
If your matrix is sparse, you can use all these iterative functions, for example bicg for a biconjugate gradients method.
MATLAB's mldivide operator does indeed take advantage of properties of A. See the documentation for details - expand the "Algorithm" section.
I have a curve IxV. I also have an equation that I want to fit in this IxV curve, so I can adjust its constants. It is given by:
I = I01(exp((V-R*I)/(n1*vth))-1)+I02(exp((V-R*I)/(n2*vth))-1)
vth and R are constants already known, so I only want to achieve I01, I02, n1, n2. The problem is: as you can see, I is dependent on itself. I was trying to use the curve fitting toolbox, but it doesn't seem to work on recursive equations.
Is there a way to make the curve fitting toolbox work on this? And if there isn't, what can I do?
Assuming that I01 and I02 are variables and not functions, then you should set the problem up like this:
a0 = [I01 I02 n1 n2];
MinFun = #(a) abs(a(1)*(exp(V-R*I)/(a(3)*vth))-1) + a(2)*(exp((V-R*I)/a(4)*vth))-1) - I);
aout = fminsearch(a0,MinFun);
By subtracting I and taking the absolute value, the point where both sides are equal will be the point where MinFun is zero (minimized).
No, the CFTB cannot fit such recursively defined functions. And errors in I, since the true value of I is unknown for any point, will create a kind of errors in variables problem. All you have are the "measured" values for I.
The problem of errors in I MAY be serious, since any errors in I, or lack of fit, noise, model problems, etc., will be used in the expression itself. Then you exponentiate these inaccurate values, potentially casing a mess.
You may be able to use an iterative approach. Thus something like
% 0. Initialize I_pred
I_pred = I;
% 1. Estimate the values of your coefficients, for this model:
% (The curve fitting toolbox CAN solve this problem, given I_pred)
I = I01(exp((V-R*I_pred)/(n1*vth))-1)+I02(exp((V-R*I_pred)/(n2*vth))-1)
% 2. Generate new predictions for I_pred
I_pred = I01(exp((V-R*I_pred)/(n1*vth))-1)+I02(exp((V-R*I_pred)/(n2*vth))-1)
% Repeat steps 1 and 2 until the parameters from the CFTB stabilize.
The above pseudo-code will work only if your starting values are good, and there are not large errors/noise in the model/data. Even on a good day, the above approach may not converge well. But I see little hope otherwise.
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)