I'm using DocPolynom a lot at the moment (see here if unfamiliar: http://www.mathworks.co.uk/help/techdoc/matlab_oop/f3-28024.html)
I have a polynomial f = DocPolynom(v) where v is a vector of coefficients. I really would like to be able to convert f to the polynomial corresponding to f(x-a), where a is a pre-determined constant. Does anyone know if/how I can do this?
Thanks!
Although not a direct answer, since you know the coefficients of the polynomial, you can evaluate the polynomial by polyval at the inputs x-a and using the resultant output you can use polyfit to get the coefficients of the polynomial that passes through your data.
v=[1 2 3];
x=1:3;
a=2;
y=polyval(v,x-a);
polyfit(x,y,2) % 2 here is the order of your polynomial (i.e. length(v)-1)
ans =
1.0000 -2.0000 3.0000
To do this, you need at least N+1 data points, where N is the order of your polynomial.
I'm not sure what this object you are writing is supposed to do, but you might play with my sympoly toolbox, which allows symbolic computation on polynomials. It is on the file exchange.
If all you have are simple polynomials, you can always use conv to compute powers of (x - a), adding them together. Thus, if we have the polynomial
P(x) = 3*x^2 + 2*x + 1
and we wish to form the polynomial Q(x) = P(x-3), it takes only a few operations.
Q = 3*conv([1 -3],[1 -3]) + 2*conv([0 1],[1 -3]) + 1*conv([0 1],[0 1])
Q =
3 -16 22
Related
How can I implement the function lu(A) in MATLAB so that L*U is directly A and I also get the real L matrix?
When I use [L,U] = lu(A), MATLAB doesn't give me the right L matrix. When I use [L,U,P] = lu(A), I need to implement P*A = L*U, but I only want to multiply L*U to receive A.
MATLAB's lu always performs pivoting by default. If you had for example a diagonal coefficient that was equal to 0 when you tried to do the conventional LU decomposition algorithm, it will not work as the diagonal coefficients are required when performing the Gaussian elimination to create the upper triangular matrix U so you would get a divide by zero error. Pivoting is required to ensure that the decomposition is stable.
However, if you can guarantee that the diagonal coefficients of your matrix are non-zero, it is very simple but you will have to write this on your own. All you have to do is perform Gaussian elimination on the matrix and reduce the matrix into reduced echelon form. The result reduced echelon form matrix is U while the coefficients required to remove the lower triangular part of L in Gaussian elimination would be placed in the lower triangular half to make U.
Something like this could work, assuming your matrix is stored in A. Remember that I'm assuming a square matrix here. The implementation of the non-pivoting LU decomposition algorithm is placed in a MATLAB function file called lu_nopivot:
function [L, U] = lu_nopivot(A)
n = size(A, 1); % Obtain number of rows (should equal number of columns)
L = eye(n); % Start L off as identity and populate the lower triangular half slowly
for k = 1 : n
% For each row k, access columns from k+1 to the end and divide by
% the diagonal coefficient at A(k ,k)
L(k + 1 : n, k) = A(k + 1 : n, k) / A(k, k);
% For each row k+1 to the end, perform Gaussian elimination
% In the end, A will contain U
for l = k + 1 : n
A(l, :) = A(l, :) - L(l, k) * A(k, :);
end
end
U = A;
end
As a running example, suppose we have the following 3 x 3 matrix:
>> rng(123)
>> A = randi(10, 3, 3)
A =
7 6 10
3 8 7
3 5 5
Running the algorithm gives us:
>> [L,U] = lu_nopivot(A)
L =
1.0000 0 0
0.4286 1.0000 0
0.4286 0.4474 1.0000
U =
7.0000 6.0000 10.0000
0 5.4286 2.7143
0 0 -0.5000
Multiplying L and U together gives:
>> L*U
ans =
7 6 10
3 8 7
3 5 5
... which is the original matrix A.
You could use this hack (though as already mentioned, you might lose numerical stability):
[L, U] = lu(sparse(A), 0)
You might want to consider doing LDU decomposition instead of unpivoted LU. See, LU without pivoting is numerically unstable - even for matrices that are full rank and invertible. The simple algorithm provided above shows why - there is division by each diagonal element of the matrix involved. Thus, if there is a zero anywhere on the diagonal, decomposition fails, even though the matrix could still be non-singular.
Wikipedia talks a little about LDU decomposition here:
https://en.wikipedia.org/wiki/LU_decomposition#LDU_decomposition
without citing an algorithm. It cites the following textbook for proof of existence:
Horn, Roger A.; Johnson, Charles R. (1985), Matrix Analysis, Cambridge University Press, ISBN 978-0-521-38632-6. See Section 3.5.
LDU is guaranteed to exist (at least for an invertible matrix), it is numerically stable, and it is also unique (provided that both L and U are constrained to have unit elements on the diagonal).
Then, if for any reason "D" gets in your way, you can absorb the diagonal matrix D into either L (L:=LD) or U (U:=DU), or split it symmetrically between L and U (such as L:=L*sqrt(D) and U:=sqrt(D)*U), or however you want to do it. There is an infinite number of ways to split LDU into LU, and this is why LU decomposition is not unique.
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);
I have an equation like this
y = a*x+b;
I have sets of y and x
y = [1 2 3 4 5]
x = [6 7 8 9 10]
I want to find a and b, but not one solution; all solutions. I guess, I have to use polyfit, but I don't know how to do this and I don't understand why I have to use polyfit? Can you explain this to me?
From the polyfit documentation:
p = polyfit(x,y,n) finds the coefficients of a polynomial p(x) of
degree n that fits the data, p(x(i)) to y(i), in a least squares
sense. The result p is a row vector of length n+1 containing the
polynomial coefficients in descending powers:
So, you have data y at x-coordinates x, and you want to fit a first-degree polynomial to it. So use
p=polyfit(x,y,1);
and then p(1)=a and p(2)=b, or y=p(1)*x+p(2).
There are other ways to do this, but polyfit is very simple.
I need help fitting data in a least square sense to a nonlinear function. Given data, how do I proceed when I have following equation?
f(x) = 20 + ax + b*e^(c*2x)
So I want to find a,b and c. If it was products, I would linearize the function by takin the natural logaritm all over the function but I can not seem to do that in this case.
Thanks
You can use the nlinfit tool, which doesn't require the Curve Fitting Toolbox (I don't think...)
Something like
f = #(b,x)(20 + b(1)*x + b(2)*exp(b(3)*2*x));
beta0 = [1, 1, 1];
beta = nlinfit(x, Y, f, beta0);
When MATLAB solves this least-squares problem, it passes the coefficients into the anonymous function f in the vector b. nlinfit returns the final values of these coefficients in the beta vector. beta0 is an initial guess of the values of b(1), b(2), and b(3). x and Y are the vectors with the data that you want to fit.
Alternatively, you can define the function in its own file, if it is a little more complicated. For this case, you would have something like (in the file my_function.m)
function y = my_function(b,x)
y = 20 + b(1)*x + b(2)*exp(b(3)*2*x);
end
and the rest of the code would look like
beta0 = [1, 1, 1];
beta = nlinfit(x, Y, #my_function, beta0);
See also: Using nlinfit in Matlab?
You can try the cftool which is an interactive tool for fitting data. The second part I don't quite understand. It may help if you describe it in more detail.
First off, I didn't know what to put as title, since the question is not easy to formulate shortly.
I need to convolve a matrix-valued function (k) with a vector-valued function (X), each of which is defined on R^3. I need to to this in MATLAB, so naturally I will do the discretized version. I plan on representing k and X by 5- and 4-dimensional arrays, respectively. This seems a bit heavy though. Do you know if there are any better ways to do it?
Instead of doing the convolution directly, I will go to Fourier space by fft'ing both k and X, pad with zeros, multiply them and then use ifft. That should produce the same result and run much, much faster.
My question here is whether there is any way to multiply these arrays/matrices easily? I.e. is there any way to do k(i,j,k,:,:)*X(i,j,k,:) for all i,j,k, without using three nested loops?
Do you need to discretize? Matlab is perfectly capable of taking functions as input and output. For examples, you could define a convolution function:
>> convolve = #(fm,fv) #(x) fm(x) * fv(x); %fm matrix valued, fv vector valued
and define some matrix-valued and vector-valued functions (this assumes the input is a column vector)
>> f = #(x) [x x x];
>> g = #(x) cos(x);
now their convolution:
>> h = convolve(f,g);
and try applying it to a vector:
>> h([1;2;3])
ans =
-0.8658
-1.7317
-2.5975
You get the same answer as if you did the operation manually:
>> f([1;2;3]) * g([1;2;3])
ans =
-0.8658
-1.7317
-2.5975
You perform element-by-element operation by using . together with the operator of choice. For example:
Element-by-element multiplication: .*
Element-by-element division: ./
and so on... is that what you mean?