Consider a function y(x) sampled in an array of values, represented by the arrays x and y. If I have another x value x0, I can evaluate y(x0) using spline
y0 = spline(x,y,x0);
Now, I can also write
pp = spline(x,y);
y0 = ppval(pp,x0);
MY QUESTION: If I already have the coefficient and x matrices, my_coefs (size(my_coefs) = [length(y),4]) and x, how can I create a piecewise polynomial My_pp such that pp.coefs = my_coefs and that y0 = ppval(My_pp,x0)?
OK, There is no "spline object", but rather a piecewise polynomial object. So, if my_coefs was attained by break-points my_x then the code needed is
my_spline = mkpp(my_x,my_coefs);
y0 = ppval(my_spline, x0);
In case that dimensions are dazzeling here, which they are, then
my_coefs is 4*n
my_x is n
y0 is N
x0 is N
Related
This is my code for finding the centered coefficients for lagrange polynomial interpolation:
% INPUT
% f f scalar - valued function
% interval interpolation interval [a, b]
% n interpolation order
%
% OUTPUT
% coeff centered coefficients of Lagrange interpolant
function coeff = lagrangeInterp (f, interval , n)
a = interval(1);
b = interval(2);
x = linspace(a,b,n+1);
y = f(x);
coeff(1,:) = polyfit(x,y,n);
end
Which is called in the following script
%Plot lagrangeInterp and sin(x) together
hold on
x = 0:0.1*pi:2*pi;
for n = 1:1:4
coeff = lagrangeInterp(#(x)sin(x),[0,2*pi],n);
plot(x,polyval(coeff,x,'-'));
end
y = sin(x);
plot(x,y);
legend('1st order','2nd order','3rd order','4th order','sin(x)');
To check for stability I would like to perturb the function (eg g(x) = f(x) + epsilon). How would I go about this?
Well, a little trick for you.
You know randn([m,n]) in matlab generate a m*n random matrix. The point is to generate a random vector, and interp1 to a function of x. Like this:
x = linspace(a,b,n+1); % Your range of input
g = #(ep,xx)f(xx)+interp1(x,ep*randn([length(x),1]),xx);
I have an equation like this:
dy/dx = a(x)*y + b
where a(x) is a non-constant (a=1/x) and b is a vector (10000 rows).
How can I solve this equation?
Let me assume you would like to write a generic numerical solver for dy/dx = a(x)*y + b. Then you can pass the function a(x) as an argument to the right-hand side function of one of the ODE solvers. e.g.
a = #(x) 1/x;
xdomain = [1 10];
b = rand(10000,1);
y0 = ones(10000,1);
[x,y] = ode45(#(x,y,a,b)a(x)*y + b,xdomain,y0,[],a,b);
plot(x,y)
Here, I've specified the domain of x as xdomain, and the value of y at the bottom limit of x as y0.
From my comments, you can solve this without MATLAB. Assuming non-zero x, you can use an integrating factor to get a 10000-by-1 solution y(x)
y_i(x) = b_i*x*ln(x) + c_i*x
with 10000-by-1 vector of constants c, where y_i(x), b_i and c_i are the i-th entries of y(x), b and c respectively. The constant vector c can be determined at some point x0 as
c_i = y_i(x0)/x_0 - b_i*ln(x0)
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:
Given a system of the form y' = A*y(t) with solution y(t) = e^(tA)*y(0), where e^A is the matrix exponential (i.e. sum from n=0 to infinity of A^n/n!), how would I use matlab to compute the solution given the values of matrix A and the initial values for y?
That is, given A = [-2.1, 1.6; -3.1, 2.6], y(0) = [1;2], how would I solve for y(t) = [y1; y2] on t = [0:5] in matlab?
I try to use something like
t = 0:5
[y1; y2] = expm(A.*t).*[1;2]
and I'm finding errors in computing the multiplication due to dimensions not agreeing.
Please note that matrix exponential is defined for square matrices. Your attempt to multiply the attenuation coefs with the time vector doesn't give you what you'd want (which should be a 3D matrix that should be exponentiated slice by slice).
One of the simple ways would be this:
A = [-2.1, 1.6; -3.1, 2.6];
t = 0:5;
n = numel(t); %'number of samples'
y = NaN(2, n);
y(:,1) = [1;2];
for k =2:n
y(:,k) = expm(t(k)*A) * y(:,1);
end;
figure();
plot(t, y(1,:), t, y(2,:));
Please note that in MATLAB array are indexed from 1.
I'm trying to get Matlab to take this as a function of x_1 through x_n and y_1 through y_n, where k_i and r_i are all constants.
So far my idea was to take n from the user and make two 1×n vectors called x and y, and for the x_i just pull out x(i). But I don't know how to make an arbitrary sum in MATLAB.
I also need to get the gradient of this function, which I don't know how to do either. I was thinking maybe I could make a loop and add that to the function each time, but MATLAB doesn't like that.
I don't believe a loop is necessary for this calculation. MATLAB excels at vectorized operations, so would something like this work for you?
l = 10; % how large these vectors are
k = rand(l,1); % random junk values to work with
r = rand(l,1);
x = rand(l,1);
y = rand(l,1);
vals = k(1:end-1) .* (sqrt(diff(x).^2 + diff(y).^2) - r(1:end-1)).^2;
sum(vals)
EDIT: Thanks to #Amro for correcting the formula and simplifying it with diff.
You can solve for the gradient symbolically with:
n = 10;
k = sym('k',[1 n]); % Create n variables k1, k2, ..., kn
x = sym('x',[1 n]); % Create n variables x1, x2, ..., xn
y = sym('y',[1 n]); % Create n variables y1, y2, ..., yn
r = sym('r',[1 n]); % Create n variables r1, r2, ..., rn
% Symbolically sum equation
s = sum((k(1:end-1).*sqrt((x(2:end)-x(1:end-1)).^2+(y(2:end)-y(1:end-1)).^2)-r(1:end-1)).^2)
grad_x = gradient(s,x) % Gradient with respect to x vector
grad_y = gradient(s,y) % Gradient with respect to y vector
The symbolic sum and gradients can be evaluated and converted to floating point with:
% n random data values for k, x, y, and r
K = rand(1,n);
X = rand(1,n);
Y = rand(1,n);
R = rand(1,n);
% Substitute in data for symbolic variables
S = double(subs(s,{[k,x,y,r]},{[K,X,Y,R]}))
GRAD_X = double(subs(grad_x,{[k,x,y,r]},{[K,X,Y,R]}))
GRAD_Y = double(subs(grad_y,{[k,x,y,r]},{[K,X,Y,R]}))
The gradient function is the one overloaded for symbolic variables (type help sym/gradient) or see the more detailed documentation online).
Yes, you could indeed do this with a loop, considering that x, y, k, and r are already defined.
n = length(x);
s = 0;
for j = 2 : n
s = s + k(j-1) * (sqrt((x(j) - x(j-1)).^2 + (y(j) - y(j-1)).^2) - r(j-1)).^2
end
You should derive the gradient analytically and then plug in numbers. It should not be too hard to expand these terms and then find derivatives of the resulting polynomial.
Vectorized solution is something like (I wonder why do you use sqrt().^2):
is = 2:n;
result = sum( k(is - 1) .* abs((x(is) - x(is-1)).^2 + (y(is) - y(is-1)).^2 - r(is-1)));
You can either compute gradient symbolically or rewrite this code as a function and make a standard +-eps calculation. If you need a gradient to run optimization (you code looks like a fitness function) you could use algorithms that calculate them themselves, for example, fminsearch can do this