The Legendre polynomials are implemented in MATLAB as vectors, where you also get all the associated Legendre polynomials evaluated at a particular point x. Thus, I don't know how I can use these functions inside an integral. My question is:
How can I evaluate the (NUMERICALLY CALCUALTED(!)) integral from -1 to 1 over the n-th Legendre polynomial in Matlab?
EDIT: As I received an answer that is really not what I want: I want to use the implementation of the Legendre polynomials in MATLAB cause other suggestions may be highly unstable.
n=3 % degree of Legendre polynomial
step=0.1 % integration step
trapz(legendre(n,-1:step:1)')*step
this should do what you want
As #thewaywewalk mentionned, you can use trapz to numerically integrate.
Legendre polynomials of degree n are defined as:
Therefore you can define them in Matlab like so:
sym x % You probably have to define x as being symbolic since you integrate as a function of x.
x = -1:0.1:1;
n = 1; Change according to the degree of the polynomial.
F = (x.^2)-1).^n;
Pol_n = (1./((2.^n).*factorial(n))).*diff(F,x,n) % n is the degree of the polynomial
Then using trapz :
Output = trapz(x,Pol_n)
That should get you going.
Related
I have to fit the dots, results of measurements, by an exponential function on Matlab. My profesor asked me to use only
fminsearch
polyval
polyfit
One of them or both. I have to find the parameters a and b (the value) which are fitting it.
There is the lines I wrote :
x=[1:10:70]
y=[0:10:70]
x=[12.5,11.8,10.8,10.9,6.5,6.2,6.1,5.423,4.625]
y=[0,0.61,1.3,1.4,14.9,18.5,20.1,29.7,58.2]
xlabel('Conductivité')
ylabel('Inductance')
The function has the form a*e^(-b*x) +c
Well polyfit and polyval are only usefull for working with polynomials. So you would have to write a minimization problem of the form min(f(x)).
functionToMinimize = #(pars, x, y)(norm(pars(1).*exp(-pars(2).*x) - y));
targetFunctionForFminseardch = #(pars)(functionToMinimize(pars, x, y));
minPars = fminsearch(targetFunctionForFminseardch, [0, 1])
Read up on anonymous functions and the use of vector norms if you have questions how to construct such a minimization problem.
Your code also has some flaws. Why are you defining x and y twice when you only want to use the actual measured data?
For the linear regression, I want to generate the matrix for polynomials of n degree.
if n is 1
X=[x(:), ones(length(x),1)]
if n is 2
X=[x(:).^2 x(:) ones(length(x),1)]
...
if n is 5
X=[x(:).^5 x(:).^4 x(:).^3 x(:).^2 x(:) ones(length(x),1)]
I do not know how to code with matlab if I set n=6 and it will automatically generate the wanted X matrix. Hope you can help me.
This can be easily done with bsxfun:
X = bsxfun(#power, x(:), n:-1:0);
Or, in Matlab versions from R1016b onwards, you can use implicit expansion:
X = x(:).^(n:-1:0);
Check out the polyval function. I believe that will do what you’re looking for.
To get increasing the polynomial to increase in degree, you can increase the length of your p argument using a loop.
If you write edit polyfit you can see how MATLAB have implemented the polyfit command, which is similar to what you are trying to do. In there you will find the code
% Construct the Vandermonde matrix V = [x.^n ... x.^2 x ones(size(x))]
V(:,n+1) = ones(length(x),1,class(x));
for j = n:-1:1
V(:,j) = x.*V(:,j+1);
end
Which constructs the matrix you are interested in. The benefit of this method over the bsxfun is that you only calculate x(:).^n and then saves the intermediary results. Instead of treating all powers as seperate problems, e.g. x(:)^(n-1) as a seperate problem to x(:).^n.
I want to integrate p(x)*f(x) where p(x) is a polynomial and f(x) is a function. I am working in MATLAB.
I have the coefficients of the polynomial in a vector.
p=[2,3,4,5];
funct=#(x) xˆ2;
I know how to integrate the function by itself, as well as how to integrate the polynomial by itself. However, I just can't find any info on how to take the integral of the product.
Here is what I tried:
p2=poly2sym(p)
integral(funct*p2,-1,1)
but p2 is not a function handle.
Help is appreciated!
Yes, p2 is not a function handle – it's a symbolic expression, but integral performs numerical integration and requires a function handle that returns floating point values. Even if p2 was a function handle, multiply function handles (e.g., funct*p2) is not valid. Additionally, funct needs to be vectorized.
Instead of poly2sym, you can evaluate your integral numerically with polyval like this:
p = [2,3,4,5];
funct = #(x)x.^2; % use element-wise power to vectorize
p2 = #(x)polyval(p,x);
integral(#(x)funct(x).*p2(x),-1,1) % evaluate two handles into one
which returns 4.533333333333333. Or you could compute this particular integral symbolically using int:
p = [2,3,4,5];
syms x;
funct = x^2;
p2 = poly2sym(p,x);
int(funct*p2,x,-1,1)
which returns the exact rational value of 68/15 (use vpa or double to convert to decimal or floating point, respectively).
I have a matrix z(x,y)
This is an NxN abitary pdf constructed from a unique Kernel density estimation (i.e. not a usual pdf and it doesn't have a function). It is multivariate and can't be separated and is discrete data.
I wan't to construct a NxN matrix (F(x,y)) that is the cumulative distribution function in 2 dimensions of this pdf so that I can then randomly sample the F(x,y) = P(x < X ,y < Y);
Analytically I think the CDF of a multivariate function is the surface integral of the pdf.
What I have tried is using the cumsum function in order to calculate the surface integral and tested this with a multivariate normal against the analytical solution and there seems to be some discrepancy between the two:
% multivariate parameters
delta = 100;
mu = [1 1];
Sigma = [0.25 .3; .3 1];
x1 = linspace(-2,4,delta); x2 = linspace(-2,4,delta);
[X1,X2] = meshgrid(x1,x2);
% Calculate Normal multivariate pdf
F = mvnpdf([X1(:) X2(:)],mu,Sigma);
F = reshape(F,length(x2),length(x1));
% My attempt at a numerical surface integral
FN = cumsum(cumsum(F,1),2);
% Normalise the CDF
FN = FN./max(max(FN));
X = [X1(:) X2(:)];
% Analytic solution to a multivariate normal pdf
p = mvncdf(X,mu,Sigma);
p = reshape(p,delta,delta);
% Highlight the difference
dif = p - FN;
error = max(max(sqrt(dif.^2)));
% %% Plot
figure(1)
surf(x1,x2,F);
caxis([min(F(:))-.5*range(F(:)),max(F(:))]);
xlabel('x1'); ylabel('x2'); zlabel('Probability Density');
figure(2)
surf(X1,X2,FN);
xlabel('x1'); ylabel('x2');
figure(3);
surf(X1,X2,p);
xlabel('x1'); ylabel('x2');
figure(5)
surf(X1,X2,dif)
xlabel('x1'); ylabel('x2');
Particularly the error seems to be in the transition region which is the most important.
Does anyone have any better solution to this problem or see what I'm doing wrong??
Any help would be much appreciated!
EDIT: This is the desired outcome of the cumulative integration, The reason this function is of value to me is that when you randomly generate samples from this function on the closed interval [0,1] the higher weighted (i.e. the more likely) values appear more often in this way the samples converge on the expected value(s) (in the case of multiple peaks) this is desired outcome for algorithms such as particle filters, neural networks etc.
Think of the 1-dimensional case first. You have a function represented by a vector F and want to numerically integrate. cumsum(F) will do that, but it uses a poor form of numerical integration. Namely, it treats F as a step function. You could instead do a more accurate numerical integration using the Trapezoidal rule or Simpson's rule.
The 2-dimensional case is no different. Your use of cumsum(cumsum(F,1),2) is again treating F as a step function, and the numerical errors resulting from that assumption only get worse as the number of dimensions of integration increases. There exist 2-dimensional analogues of the Trapezoidal rule and Simpson's rule. Since there's a bit too much math to repeat here, take a look here:
http://onestopgate.com/gate-study-material/mathematics/numerical-analysis/numerical-integration/2d-trapezoidal.asp.
You DO NOT need to compute the 2-dimensional integral of the probability density function in order to sample from the distribution. If you are computing the 2-d integral, you are going about the problem incorrectly.
Here are two ways to approach the sampling problem.
(1) You write that you have a kernel density estimate. A kernel density estimate is a special case of a mixture density. Any mixture density can be sampled by first selecting one kernel (perhaps differently or equally weighted, same procedure applies), and then sampling from that kernel. (That applies in any number of dimensions.) Typically the kernels are some relatively simple distribution such as a Gaussian distribution so that it is easy to sample from it.
(2) Any joint density P(X, Y) is equal to P(X | Y) P(Y) (and equivalently P(Y | X) P(X)). Therefore you can sample from P(Y) (or P(X)) and then from P(X | Y). In order to sample from P(X | Y), you will need to integrate P(X, Y) along a line Y = y (where y is the sampled value of Y), but (this is crucial) you only need to integrate along that line; you don't need to integrate over all values of X and Y.
If you tell us more about your problem, I can help with the details.
I am doing a project where i find an approximation of the Sine function, using the Least Squares method. Also i can use 12 values of my own choice.Since i couldn't figure out how to solve it i thought of using Taylor's series for Sine and then solving it as a polynomial of order 5. Here is my code :
%% Find the sine of the 12 known values
x=[0,pi/8,pi/4,7*pi/2,3*pi/4,pi,4*pi/11,3*pi/2,2*pi,5*pi/4,3*pi/8,12*pi/20];
y=zeros(12,1);
for i=1:12
y=sin(x);
end
n=12;
j=5;
%% Find the sums to populate the matrix A and matrix B
s1=sum(x);s2=sum(x.^2);
s3=sum(x.^3);s4=sum(x.^4);
s5=sum(x.^5);s6=sum(x.^6);
s7=sum(x.^7);s8=sum(x.^8);
s9=sum(x.^9);s10=sum(x.^10);
sy=sum(y);
sxy=sum(x.*y);
sxy2=sum( (x.^2).*y);
sxy3=sum( (x.^3).*y);
sxy4=sum( (x.^4).*y);
sxy5=sum( (x.^5).*y);
A=[n,s1,s2,s3,s4,s5;s1,s2,s3,s4,s5,s6;s2,s3,s4,s5,s6,s7;
s3,s4,s5,s6,s7,s8;s4,s5,s6,s7,s8,s9;s5,s6,s7,s8,s9,s10];
B=[sy;sxy;sxy2;sxy3;sxy4;sxy5];
Then at matlab i get this result
>> a=A^-1*B
a =
-0.0248
1.2203
-0.2351
-0.1408
0.0364
-0.0021
However when i try to replace the values of a in the taylor series and solve f.e t=pi/2 i get wrong results
>> t=pi/2;
fun=t-t^3*a(4)+a(6)*t^5
fun =
2.0967
I am doing something wrong when i replace the values of a matrix in the Taylor series or is my initial thought flawed ?
Note: i can't use any built-in function
If you need a least-squares approximation, simply decide on a fixed interval that you want to approximate on and generate some x abscissae on that interval (possibly equally spaced abscissae using linspace - or non-uniformly spaced as you have in your example). Then evaluate your sine function at each point such that you have
y = sin(x)
Then simply use the polyfit function (documented here) to obtain least squares parameters
b = polyfit(x,y,n)
where n is the degree of the polynomial you want to approximate. You can then use polyval (documented here) to obtain the values of your approximation at other values of x.
EDIT: As you can't use polyfit you can generate the Vandermonde matrix for the least-squares approximation directly (the below assumes x is a row vector).
A = ones(length(x),1);
x = x';
for i=1:n
A = [A x.^i];
end
then simply obtain the least squares parameters using
b = A\y;
You can clearly optimise the clumsy Vandermonde generation loop above I have just written to illustrate the concept. For better numerical stability you would also be better to use a nice orthogonal polynomial system like Chebyshev polynomials of the first kind. If you are not even allowed to use the matrix divide \ function then you will need to code up your own implementation of a QR factorisation and solve the system that way (or some other numerically stable method).