clc
clear all
x= -2:0.1:4;
y= (3*x.^3)-(26*x)+10;
y1= diff(y)./diff(x);
y2= diff(y1)./diff(x);
plot(x,y,'R')
hold on
plot(x,y1,'G')
hold on
plot(x,y2,'B')
Is this right way to find second derivative?
It's not finding the second derivative and giving an error that matrix dimensions must agree in line 6.
You have values y that correspond to the values x. While calculating the approximation to the first derivatives y1 of y, you use finite differences. These kind of finite differences approximate the derivative best at the points inbetween x. This means your calculation of the approximation of the second derivative should involve (x(1:end-1)+x(2:end))/2:
x1 = (x(1:end-1) + x(2:end))/2;
y2 = diff(y1) / diff(x1);
Note that these approximations y2 best approximate the second derivative at the positions x2 = (x1(1:end-1) + x1(2:end))/2.
All in all your code might look like this:
%// Clear command window, delete all variables, close all figures
clc;
clear all;
close all;
%// Define function
f = #(x) (3*x.^3)-(26*x)+10;
%// Define positions
x = -2:0.1:4;
%// Calculate f(x)
y= f(x);
%// Calculate approximation to first derivative and their positions
x1 = (x(1:end-1) + x(2:end))/2;
y1 = diff(y) ./ diff(x);
%// Calculate approximation to second derivative and their positions
x2 = (x1(1:end-1) + x1(2:end))/2;
y2 = diff(y1) ./ diff(x1);
%// Plot
plot(x, y, 'R',...
x1,y1,'G',...
x2,y2,'B');
legend('f','f''', 'f''''');
Related
Interpolate the Runge function of Example 10.6 at Chebyshev points for n from 10 to 170
in increments of 10. Calculate the maximum interpolation error on the uniform evaluation
mesh x = -1:.001:1 and plot the error vs. polynomial degree as in Figure 10.8 using
semilogy. Observe spectral accuracy.
The runge function is given by: f(x) = 1 / (1 + 25x^2)
My code so far:
x = -1:0.001:1;
n = 170;
i = 10:10:170;
cx = cos(((2*i + 1)/(2*(n+1)))*pi); %chebyshev pts
y = 1 ./ (1 + 25*x.^2); %true fct
%chebyshev polynomial, don't know how to construct using matlab
yc = polyval(c, x); %graph of approx polynomial fct
plot(x, yc);
mErr = (1 / ((2.^n).*(n+1)!))*%n+1 derivative of f evaluated at max x in [-1,1], not sure how to do this
%plotting stuff
I know very little matlab, so I am struggling on creating the interpolating polynomial. I did some google work, but I was confused with the current functions as I didn't find one that just simply took in points and the polynomial to be interpolated. I am also a bit confused in this case of whether I should be doing i = 0:1:n and n=10:10:170 or if n is fixed here. Any help is appreciated, thank you
Since you know very little about MATLAB, I will try explain everything step by step:
First, to visualize the Runge function, you can type:
f = #(x) 1./(1+25*x.^2); % Runge function
% plot Runge function over [-1,1];
x = -1:1e-3:1;
y = f(x);
figure;
plot(x,y); title('Runge function)'); xlabel('x');ylabel('y');
The #(x) part of the code is a function handle, a very useful feature of MATLAB. Notice the function is properly vecotrized, so it can receive as an argument a variable or an array. The plot function is straightforward.
To understand the Runge phenomenon, consider a linearly spaced vector of [-1,1] of 10 elements and use these points to obtain the interpolating (Lagrange) polynomial. You get the following:
% 10 linearly spaced points
xc = linspace(-1,1,10);
yc = f(xc);
p = polyfit(xc,yc,9); % gives the coefficients of the polynomial of degree 10
hold on; plot(xc,yc,'o',x,polyval(p,x));
The polyfit function does a polynomial curve fitting - it obtains the coefficients of the interpolating polynomial, given the poins x,y and the degree of the polynomial n. You can easily evaluate the polynomial at other points with the polyval function.
Obseve that, close to the end domains, you get an oscilatting polynomial and the interpolation is not a good approximation of the function. As a matter of fact, you can plot the absolute error, comparing the value of the function f(x) and the interpolating polynomial p(x):
plot(x,abs(y-polyval(p,x))); xlabel('x');ylabel('|f(x)-p(x)|');title('Error');
This error can be reduced if, instead of using a linearly space vector, you use other points to do the interpolation. A good choice is to use the Chebyshev nodes, which should reduce the error. As a matter of fact, notice that:
% find 10 Chebyshev nodes and mark them on the plot
n = 10;
k = 1:10; % iterator
xc = cos((2*k-1)/2/n*pi); % Chebyshev nodes
yc = f(xc); % function evaluated at Chebyshev nodes
hold on;
plot(xc,yc,'o')
% find polynomial to interpolate data using the Chebyshev nodes
p = polyfit(xc,yc,n-1); % gives the coefficients of the polynomial of degree 10
plot(x,polyval(p,x),'--'); % plot polynomial
legend('Runge function','Chebyshev nodes','interpolating polynomial','location','best')
Notice how the error is reduced close to the end domains. You don't get now that high oscillatory behaviour of the interpolating polynomial. If you plot the error, you will observe:
plot(x,abs(y-polyval(p,x))); xlabel('x');ylabel('|f(x)-p(x)|');title('Error');
If, now, you change the number of Chebyshev nodes, you will get an even better approximation. A little modification on the code lets you run it again for different numbers of nodes. You can store the maximum error and plot it as a function of the number of nodes:
n=1:20; % number of nodes
% pre-allocation for speed
e_ln = zeros(1,length(n)); % error for the linearly spaced interpolation
e_cn = zeros(1,length(n)); % error for the chebyshev nodes interpolation
for ii=1:length(n)
% linearly spaced vector
x_ln = linspace(-1,1,n(ii)); y_ln = f(x_ln);
p_ln = polyfit(x_ln,y_ln,n(ii)-1);
e_ln(ii) = max( abs( y-polyval(p_ln,x) ) );
% Chebyshev nodes
k = 1:n(ii); x_cn = cos((2*k-1)/2/n(ii)*pi); y_cn = f(x_cn);
p_cn = polyfit(x_cn,y_cn,n(ii)-1);
e_cn(ii) = max( abs( y-polyval(p_cn,x) ) );
end
figure
plot(n,e_ln,n,e_cn);
xlabel('no of points'); ylabel('maximum absolute error');
legend('linearly space','chebyshev nodes','location','best')
I am trying to fit a 3D surface polynomial of n-degrees to some data points in 3D space. My system requires the surface to be monotonically increasing in the area of interest, that is the partial derivatives must be non-negative. This can be achieved using Matlab's built in lsqlin function.
So here's what I've done to try and achieve this:
I have a function that takes in four parameters;
x1 and x2 are my explanatory variables and y is my dependent variable. Finally, I can specify order of polynomial fit. First I build the design matrix A using data from x1 and x2 and the degree of fit I want. Next I build the matrix D that is my container for the partial derivatives of my datapoints. NOTE: the matrix D is double the length of matrix A since all datapoints must be differentiated with respect to both x1 and x2. I specify that Dx >= 0 by setting b to be zeroes.
Finally, I call lsqlin. I use "-D" since Matlab defines the function as Dx <= b.
function w_mono = monotone_surface_fit(x1, x2, y, order_fit)
% Initialize design matrix
A = zeros(length(x1), 2*order_fit+2);
% Adjusting for bias term
A(:,1) = ones(length(x1),1);
% Building design matrix
for i = 2:order_fit+1
A(:,(i-1)*2:(i-1)*2+1) = [x1.^(i-1), x2.^(i-1)];
end
% Initialize matrix containing derivative constraint.
% NOTE: Partial derivatives must be non-negative
D = zeros(2*length(y), 2*order_fit+1);
% Filling matrix that holds constraints for partial derivatives
% NOTE: Matrix D will be double length of A since all data points will have a partial derivative constraint in both x1 and x2 directions.
for i = 2:order_fit+1
D(:,(i-1)*2:(i-1)*2+1) = [(i-1)*x1.^(i-2), zeros(length(x2),1); ...
zeros(length(x1),1), (i-1)*x2.^(i-2)];
end
% Limit of derivatives
b = zeros(2*length(y), 1);
% Constrained LSQ fit
options = optimoptions('lsqlin','Algorithm','interior-point');
% Final weights of polynomial
w_mono = lsqlin(A,y,-D,b,[],[],[],[],[], options);
end
So now I get some weights out, but unfortunately they do not at all capture the structure of the data. I've attached an image so you can just how bad it looks. .
I'll give you my plotting script with some dummy data, so you can try it.
%% Plot different order polynomials to data with constraints
x1 = [-5;12;4;9;18;-1;-8;13;0;7;-5;-8;-6;14;-1;1;9;14;12;1;-5;9;-10;-2;9;7;-1;19;-7;12;-6;3;14;0;-8;6;-2;-7;10;4;-5;-7;-4;-6;-1;18;5;-3;3;10];
x2 = [81.25;61;73;61.75;54.5;72.25;80;56.75;78;64.25;85.25;86;80.5;61.5;79.25;76.75;60.75;54.5;62;75.75;80.25;67.75;86.5;81.5;62.75;66.25;78.25;49.25;82.75;56;84.5;71.25;58.5;77;82;70.5;81.5;80.75;64.5;68;78.25;79.75;81;82.5;79.25;49.5;64.75;77.75;70.25;64.5];
y = [-6.52857142857143;-1.04736842105263;-5.18750000000000;-3.33157894736842;-0.117894736842105;-3.58571428571429;-5.61428571428572;0;-4.47142857142857;-1.75438596491228;-7.30555555555556;-8.82222222222222;-5.50000000000000;-2.95438596491228;-5.78571428571429;-5.15714285714286;-1.22631578947368;-0.340350877192983;-0.142105263157895;-2.98571428571429;-4.35714285714286;-0.963157894736842;-9.06666666666667;-4.27142857142857;-3.43684210526316;-3.97894736842105;-6.61428571428572;0;-4.98571428571429;-0.573684210526316;-8.22500000000000;-3.01428571428571;-0.691228070175439;-6.30000000000000;-6.95714285714286;-2.57232142857143;-5.27142857142857;-7.64285714285714;-2.54035087719298;-3.45438596491228;-5.01428571428571;-7.47142857142857;-5.38571428571429;-4.84285714285714;-6.78571428571429;0;-0.973684210526316;-4.72857142857143;-2.84285714285714;-2.54035087719298];
% Used to plot the surface in all points in the grid
X1 = meshgrid(-10:1:20);
X2 = flipud(meshgrid(30:2:90).');
figure;
for i = 1:4
w_mono = monotone_surface_fit(x1, x2, y, i);
y_nr = w_mono(1)*ones(size(X1)) + w_mono(2)*ones(size(X2));
for j = 1:i
y_nr = w_mono(j*2)*X1.^j + w_mono(j*2+1)*X2.^j;
end
subplot(2,2,i);
scatter3(x1, x2, y); hold on;
axis tight;
mesh(X1, X2, y_nr);
set(gca, 'ZDir','reverse');
xlabel('x1'); ylabel('x2');
zlabel('y');
% zlim([-10 0])
end
I think it may have something to do with the fact that I haven't specified anything about the region of interest, but really I don't know. Thanks in advance for any help.
Alright I figured it out.
The main problem was simply an error in the plotting script. The value of y_nr should be updated and not overwritten in the loop.
Also I figured out that the second derivative should be monotonically decreasing. Here's the updated code if anybody is interested.
%% Plot different order polynomials to data with constraints
x1 = [-5;12;4;9;18;-1;-8;13;0;7;-5;-8;-6;14;-1;1;9;14;12;1;-5;9;-10;-2;9;7;-1;19;-7;12;-6;3;14;0;-8;6;-2;-7;10;4;-5;-7;-4;-6;-1;18;5;-3;3;10];
x2 = [81.25;61;73;61.75;54.5;72.25;80;56.75;78;64.25;85.25;86;80.5;61.5;79.25;76.75;60.75;54.5;62;75.75;80.25;67.75;86.5;81.5;62.75;66.25;78.25;49.25;82.75;56;84.5;71.25;58.5;77;82;70.5;81.5;80.75;64.5;68;78.25;79.75;81;82.5;79.25;49.5;64.75;77.75;70.25;64.5];
y = [-6.52857142857143;-1.04736842105263;-5.18750000000000;-3.33157894736842;-0.117894736842105;-3.58571428571429;-5.61428571428572;0;-4.47142857142857;-1.75438596491228;-7.30555555555556;-8.82222222222222;-5.50000000000000;-2.95438596491228;-5.78571428571429;-5.15714285714286;-1.22631578947368;-0.340350877192983;-0.142105263157895;-2.98571428571429;-4.35714285714286;-0.963157894736842;-9.06666666666667;-4.27142857142857;-3.43684210526316;-3.97894736842105;-6.61428571428572;0;-4.98571428571429;-0.573684210526316;-8.22500000000000;-3.01428571428571;-0.691228070175439;-6.30000000000000;-6.95714285714286;-2.57232142857143;-5.27142857142857;-7.64285714285714;-2.54035087719298;-3.45438596491228;-5.01428571428571;-7.47142857142857;-5.38571428571429;-4.84285714285714;-6.78571428571429;0;-0.973684210526316;-4.72857142857143;-2.84285714285714;-2.54035087719298];
% Used to plot the surface in all points in the grid
X1 = meshgrid(-10:1:20);
X2 = flipud(meshgrid(30:2:90).');
figure;
for i = 1:4
w_mono = monotone_surface_fit(x1, x2, y, i);
% NOTE: Should only have 1 bias term
y_nr = w_mono(1)*ones(size(X1));
for j = 1:i
y_nr = y_nr + w_mono(j*2)*X1.^j + w_mono(j*2+1)*X2.^j;
end
subplot(2,2,i);
scatter3(x1, x2, y); hold on;
axis tight;
mesh(X1, X2, y_nr);
set(gca, 'ZDir','reverse');
xlabel('x1'); ylabel('x2');
zlabel('y');
% zlim([-10 0])
end
And here's the updated function
function [w_mono, w] = monotone_surface_fit(x1, x2, y, order_fit)
% Initialize design matrix
A = zeros(length(x1), 2*order_fit+1);
% Adjusting for bias term
A(:,1) = ones(length(x1),1);
% Building design matrix
for i = 2:order_fit+1
A(:,(i-1)*2:(i-1)*2+1) = [x1.^(i-1), x2.^(i-1)];
end
% Initialize matrix containing derivative constraint.
% NOTE: Partial derivatives must be non-negative
D = zeros(2*length(y), 2*order_fit+1);
for i = 2:order_fit+1
D(:,(i-1)*2:(i-1)*2+1) = [(i-1)*x1.^(i-2), zeros(length(x2),1); ...
zeros(length(x1),1), -(i-1)*x2.^(i-2)];
end
% Limit of derivatives
b = zeros(2*length(y), 1);
% Constrained LSQ fit
options = optimoptions('lsqlin','Algorithm','active-set');
w_mono = lsqlin(A,y,-D,b,[],[],[],[],[], options);
w = lsqlin(A,y);
end
Finally a plot of the fitting (Have used a new simulation, but fit also works on given dummy data).
I have 2 vectors and a scalar:
grid which is (N x 1);
Value which is (N x 1);
sval which is (1,1);
If I want to interpolate sval on grid I know that I can simply do:
intervalue = interp1(grid, Value, sval, 'PCHIP');
What if now I want the derivatives, i.e. the slope of the function Value at that particular point sval?
As mentioned in the comments, you can approximate the derivative by forward finite difference approximation:
slope = diff(Value) ./ diff(grid);
Alternatively:
slope = gradient(Value(1:end-1),grid);
This is a simple method of numerical differentiation. For a detailed guide on numerical differentiation in MATALB, see this answer.
Here is an example of the finite difference method with the desired interpolation:
% Define function y = x^3
grid = 1:100;
Value = grid .^ 3;
% Approximate derivative via the local slope
slope = diff(Value) ./ diff(grid);
% Or: slope = gradient(Value(1:end-1),grid);
slope_grid = grid(1:end-1);
% Interpolate derivative
sval = 33.5;
sval_slope = interp1(slope_grid, slope, sval, 'PCHIP');
We can visualize the result:
figure;
plot(grid, 3*grid.^2)
hold on
plot(slope_grid, slope)
legend('Reference', 'Approximation', 'Location', 'NorthWest')
I want to test the Akaike criterion (it is a criterion that gives where do you get a significant change in a time series), but to do that I need to generate data that for example follow a sinusoidal trend, a linear trend with positive or negative slope, a constant trend, etc. So far I have done this but with random numbers, this is:
%Implementation of the Akaike method for Earth sciences.
N=100;
data=zeros(N,1);
for i=1:N
data(i,1)=unifrnd(1,N);
end
%AIC=zeros(N-1,1);
data=rand(1,N);
for k=1:N
%y=datasample(data,k);
AIC(k,1)=k*log(var(data(1:k),1))+(N-k-1)*log(var(data(k+1:N),1));
end
AIC(1)=NaN;
%AIC(N-1)=[];AIC(N)=[];
%disp(AIC)
%plot(AIC)
subplot(2,1,1)
plot(data,'Marker','.')
subplot(2,1,2)
plot(AIC,'Marker','.')
So, How can I generate different data with different trend in MATLAB?
Thanks a lot in advance.
What you can do is first start off with a known curve, then add some noise or random values so that the signal does follow a trend but it is noisy. Given a set of independent values, use these to generate values for your sinusoidal curves, a line with a positive or negative slope and a constant value.
Something like this comes to mind:
X = 1 : N; % N is defined in your code
Y1 = sin(X) + rand(1, N); % Sinusoidal
slope1 = 2; intercept = 3;
Y2 = slope1*X + intercept + rand(1, N); % Line with a positive slope
slope2 = -1; intercept2 = 0.5;
Y3 = slope2*X + intercept2 + rand(1, N); % Line with a negative slope
B = 2;
Y4 = B*ones(1, N) + rand(1, N); % Constant line
rand is a function in MATLAB that uniformly generates floating-point values between [0,1]. Y1, Y2, Y3 and Y4 are the trends you desire where they follow the curve defined but they add a bit of random values so that you don't completely get the trend you want and the noise is designed to decrease how similarity those curves are to the curve you defined. Increase the magnitude of the random values to decrease the similarity.
I need some help with finding solution to Cauchy problem in Matlab.
The problem:
y''+10xy = 0, y(0) = 7, y '(0) = 3
Also I need to plot the graph.
I wrote some code but, I'm not sure whether it's correct or not. Particularly in function section.
Can somebody check it? If it's not correct, where I made a mistake?
Here is separate function in other .m file:
function dydx = funpr12(x,y)
dydx = y(2)+10*x*y
end
Main:
%% Cauchy problem
clear all, clc
xint = [0,5]; % interval
y0 = [7;3]; % initial conditions
% numerical solution using ode45
sol = ode45(#funpr12,xint,y0);
xx = [0:0.01:5]; % vector of x values
y = deval(sol,xx); % vector of y values
plot(xx,y(1,:),'r', 'LineWidth',3)
legend('y1(x)')
xlabel('x')
ylabel('y(x)')
I get this graph:
ode45 and its related ilk are only designed to solve first-order differential equations which are of the form y' = .... You need to do a bit of work if you want to solve second-order differential questions.
Specifically, you'll have to represent your problem as a system of first-order differential equations. You currently have the following ODE:
y'' + 10xy = 0, y(0) = 7, y'(0) = 3
If we rearrange this to solve for y'', we get:
y'' = -10xy, y(0) = 7, y'(0) = 3
Next, you'll want to use two variables... call it y1 and y2, such that:
y1 = y
y2 = y'
The way you have built your code for ode45, the initial conditions that you specified are exactly this - the guess using y and its first-order guess y'.
Taking the derivative of each side gives:
y1' = y'
y2' = y''
Now, doing some final substitutions we get this final system of first-order differential equations:
y1' = y2
y2' = -10*x*y1
If you're having trouble seeing this, simply remember that y1 = y, y2 = y' and finally y2' = y'' = -10*x*y = -10*x*y1. Therefore, you now need to build your function so that it looks like this:
function dydx = funpr12(x,y)
y1 = y(2);
y2 = -10*x*y(1);
dydx = [y1 y2];
end
Remember that the vector y is a two element vector which represents the value of y and the value of y' respectively at each time point specified at x. I would also argue that making this an anonymous function is cleaner. It requires less code:
funpr12 = #(x,y) [y(2); -10*x*y(1)];
Now go ahead and solve it (using your code):
%%// Cauchy problem
clear all, clc
funpr12 = #(x,y) [y(2); -10*x*y(1)]; %// Change
xint = [0,5]; % interval
y0 = [7;3]; % initial conditions
% numerical solution using ode45
sol = ode45(funpr12,xint,y0); %// Change - already a handle
xx = [0:0.01:5]; % vector of x values
y = deval(sol,xx); % vector of y values
plot(xx,y(1,:),'r', 'LineWidth',3)
legend('y1(x)')
xlabel('x')
ylabel('y(x)')
Take note that the output when simulating the solution to the differential equation by deval will be a two column matrix. The first column is the solution to the system while the second column is the derivative of the solution. As such, you'll want to plot the first column, which is what the plot syntax is doing.
I get this plot now: