I have two 5th order polynomial fits, one of which I have taken the derivative of. I want to divide to the two polynomials and plot the results. From what I can the only way to do this is to use the "deconv" function, but how do I plot the results? Or is there an easier way to do this?
cdpdz=coeffvalues(fitresult{2});
%where pn are the coefficients of the first derivative of the fitted polynomial
pp=[p1,p2,p3,p4,p5,p6];
[divpp.r]=deconv(pp/cdpdz);
Since you want to plot the results, I am assuming that you actually want to calculate the values of the polynomials at many points, and then divide those values. That is different than the polynomial division that is performed by deconv.
I would suggest that you must make a range of values x over which to evaluate the polynomials. Then use polyval to get the values over that range, and divide the two arrays (being careful of zeros in the denominator!)
ppval = polyval(pp,x);
cdpdzval = polyval(cdpdz,x);
plot(x,ppval./cdpdzval);
Related
I've got an arbitrary probability density function discretized as a matrix in Matlab, that means that for every pair x,y the probability is stored in the matrix:
A(x,y) = probability
This is a 100x100 matrix, and I would like to be able to generate random samples of two dimensions (x,y) out of this matrix and also, if possible, to be able to calculate the mean and other moments of the PDF. I want to do this because after resampling, I want to fit the samples to an approximated Gaussian Mixture Model.
I've been looking everywhere but I haven't found anything as specific as this. I hope you may be able to help me.
Thank you.
If you really have a discrete probably density function defined by A (as opposed to a continuous probability density function that is merely described by A), you can "cheat" by turning your 2D problem into a 1D problem.
%define the possible values for the (x,y) pair
row_vals = [1:size(A,1)]'*ones(1,size(A,2)); %all x values
col_vals = ones(size(A,1),1)*[1:size(A,2)]; %all y values
%convert your 2D problem into a 1D problem
A = A(:);
row_vals = row_vals(:);
col_vals = col_vals(:);
%calculate your fake 1D CDF, assumes sum(A(:))==1
CDF = cumsum(A); %remember, first term out of of cumsum is not zero
%because of the operation we're doing below (interp1 followed by ceil)
%we need the CDF to start at zero
CDF = [0; CDF(:)];
%generate random values
N_vals = 1000; %give me 1000 values
rand_vals = rand(N_vals,1); %spans zero to one
%look into CDF to see which index the rand val corresponds to
out_val = interp1(CDF,[0:1/(length(CDF)-1):1],rand_vals); %spans zero to one
ind = ceil(out_val*length(A));
%using the inds, you can lookup each pair of values
xy_values = [row_vals(ind) col_vals(ind)];
I hope that this helps!
Chip
I don't believe matlab has built-in functionality for generating multivariate random variables with arbitrary distribution. As a matter of fact, the same is true for univariate random numbers. But while the latter can be easily generated based on the cumulative distribution function, the CDF does not exist for multivariate distributions, so generating such numbers is much more messy (the main problem is the fact that 2 or more variables have correlation). So this part of your question is far beyond the scope of this site.
Since half an answer is better than no answer, here's how you can compute the mean and higher moments numerically using matlab:
%generate some dummy input
xv=linspace(-50,50,101);
yv=linspace(-30,30,100);
[x y]=meshgrid(xv,yv);
%define a discretized two-hump Gaussian distribution
A=floor(15*exp(-((x-10).^2+y.^2)/100)+15*exp(-((x+25).^2+y.^2)/100));
A=A/sum(A(:)); %normalized to sum to 1
%plot it if you like
%figure;
%surf(x,y,A)
%actual half-answer starts here
%get normalized pdf
weight=trapz(xv,trapz(yv,A));
A=A/weight; %A normalized to 1 according to trapz^2
%mean
mean_x=trapz(xv,trapz(yv,A.*x));
mean_y=trapz(xv,trapz(yv,A.*y));
So, the point is that you can perform a double integral on a rectangular mesh using two consecutive calls to trapz. This allows you to compute the integral of any quantity that has the same shape as your mesh, but a drawback is that vector components have to be computed independently. If you only wish to compute things which can be parametrized with x and y (which are naturally the same size as you mesh), then you can get along without having to do any additional thinking.
You could also define a function for the integration:
function res=trapz2(xv,yv,A,arg)
if ~isscalar(arg) && any(size(arg)~=size(A))
error('Size of A and var must be the same!')
end
res=trapz(xv,trapz(yv,A.*arg));
end
This way you can compute stuff like
weight=trapz2(xv,yv,A,1);
mean_x=trapz2(xv,yv,A,x);
NOTE: the reason I used a 101x100 mesh in the example is that the double call to trapz should be performed in the proper order. If you interchange xv and yv in the calls, you get the wrong answer due to inconsistency with the definition of A, but this will not be evident if A is square. I suggest avoiding symmetric quantities during the development stage.
I am trying to calculate an integral using spline interpolation with matlab (version R2014a on windows 8).
I have the 3 values of the function (for x=0,0.5,1).
so I have 2 vectors - x and y that contain the values of the function, and I'm executing
cube_spline = spline(x,y);
coefficients = qube_spline.coefs
And I'm expecting to get 2 polynomials, each of degree 3, i.e I'm expecting coefficients to be a matrix of size 2*4, but somewhy I'm getting a matrix that is 1*4, which means only 1 polynomial for 2 panels.
On the other hand, if for example I'm using 4 dots, (i.e 3 panels) then I'm getting that coefficients's size is 3*4 as expected, which means 3 polynomials for 3 panels.
My question is Why does matlab return only 1 polynomial for 2 panels spline, and 3 polynomial for 3 panels spline (or any number that is greater then 2)?
There are multiple possible boundary conditions for splines, e.g.:
second derivatives equals zero on the boundary
given first derivatives on the boundary
periodic conditions, i.e. same first and second derivatives on the boundary
not-a-knot: take the outermost three points to specify the boundary conditions.
It seems spline is using the not-a-knot condition by default. So for three points only a single cubic polynomial is necessary to interpolate your data (a quadratic one would be enough too if it weren't for the not-a-knot condition), so there's no reason for spline to return one spline for each of the two intervals. This is however not a bad thing.
By the way: If all you want is to interpolate the values and don't need the polynomial coefficients, you could go with interp1 instead. You can specify in a simpler way which kind of discontinuities you want. You have the options to go with:
'pchip': C^1 continuity.
Shape-preserving piecewise cubic interpolation. The interpolated value at a query point is based on a shape-preserving piecewise cubic interpolation of the values at neighboring grid points.
integral(#(xs) interp1(x, y, xs, 'pchip'), xmin, xmax)
'spline': C^2 continuity. (Seems to be using the same not-a-knot end conditions as spline.)
Spline interpolation using not-a-knot end conditions. The interpolated value at a query point is based on a cubic interpolation of the values at neighboring grid points in each respective dimension.
integral(#(xs) interp1(x, y, xs, 'spline'), xmin, xmax)
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).
In MATLAB I need to generate a second derivative of a gaussian window to apply to a vector representing the height of a curve. I need the second derivative in order to determine the locations of the inflection points and maxima along the curve. The vector representing the curve may be quite noise hence the use of the gaussian window.
What is the best way to generate this window?
Is it best to use the gausswin function to generate the gaussian window then take the second derivative of that?
Or to generate the window manually using the equation for the second derivative of the gaussian?
Or even is it best to apply the gaussian window to the data, then take the second derivative of it all? (I know these last two are mathematically the same, however with the discrete data points I do not know which will be more accurate)
The maximum length of the height vector is going to be around 100-200 elements.
Thanks
Chris
I would create a linear filter composed of the weights generated by the second derivative of a Gaussian function and convolve this with your vector.
The weights of a second derivative of a Gaussian are given by:
Where:
Tau is the time shift for the filter. If you are generating weights for a discrete filter of length T with an odd number of samples, set tau to zero and allow t to vary from [-T/2,T/2]
sigma - varies the scale of your operator. Set sigma to a value somewhere between T/6. If you are concerned about long filter length then this can be reduced to T/4
C is the normalising factor. This can be derived algebraically but in practice I always do this numerically after calculating the filter weights. For unity gain when smoothing periodic signals, I will set C = 1 / sum(G'').
In terms of your comment on the equivalence of smoothing first and taking a derivative later, I would say it is more involved than that. As which derivative operator would you use in the second step? A simple central difference would not yield the same results.
You can get an equivalent (but approximate) response to a second derivative of a Gaussian by filtering the data with two Gaussians of different scales and then taking the point-wise differences between the two resulting vectors. See Difference of Gaussians for that approach.
Starting from the plot of one curve, it is possible to obtain the parametric equation of that curve?
In particular, say x={1 2 3 4 5 6....} the x axis, and y = {a b c d e f....} the corresponding y axis. I have the plot(x,y).
Now, how i can obtain the equation that describe the plotted curve? it is possible to display the parametric equation starting from the spline interpolation?
Thank you
If you want to display a polynomial fit function alongside your graph, the following example should help:
x=-3:.1:3;
y=4*x.^3-5*x.^2-7.*x+2+10*rand(1,61);
p=polyfit(x,y,3); %# third order polynomial fit, p=[a,b,c,d] of ax^3+bx^2+cx+d
yfit=polyval(p,x); %# evaluate the curve fit over x
plot(x,y,'.')
hold on
plot(x,yfit,'-g')
equation=sprintf('y=%2.2gx^3+%2.2gx^2+%2.2gx+%2.2g',p); %# format string for equation
equation=strrep(equation,'+-','-'); %# replace any redundant signs
text(-1,-80,equation) %# place equation string on graph
legend('Data','Fit','Location','northwest')
Last year, I wrote up a set of three blogs for Loren, on the topic of modeling/interpolationg a curve. They may cover some of your questions, although I never did find the time to add another 3 blogs to finish the topic to my satisfaction. Perhaps one day I will get that done.
The problem is to recognize there are infinitely many curves that will interpolate a set of data points. A spline is a nice choice, because it can be made well behaved. However, that spline has no simple "equation" to write down. Instead, it has many polynomial segments, pieced together to be well behaved.
You're asking for the function/mapping between two data sets. Knowing the physics involved, the function can be derived by modeling the system. Write down the differential equations and solve it.
Left alone with just two data series, an input and an output with a 'black box' in between you may approximate the series with an arbitrary function. You may start with a polynomial function
y = a*x^2 + b*x + c
Given your input vector x and your output vector y, parameters a,b,c must be determined applying a fitness function.
There is an example of Polynomial Curve Fitting in the MathWorks documentation.
Curve Fitting Tool provides a flexible graphical user interfacewhere you can interactively fit curves and surfaces to data and viewplots. You can:
Create, plot, and compare multiple fits
Use linear or nonlinear regression, interpolation,local smoothing regression, or custom equations
View goodness-of-fit statistics, display confidenceintervals and residuals, remove outliers and assess fits with validationdata
Automatically generate code for fitting and plottingsurfaces, or export fits to workspace for further analysis