Could you please help me with the following question:
I want to solve a second order equation with two unknowns and use the results to plot an ellipse.
Here is my function:
fun = #(x) [x(1) x(2)]*V*[x(1) x(2)]'-c
V is 2x2 symmetric matrix, c is a positive constant and there are two unknowns, x1 and x2.
If I solve the equation using fsolve, I notice that the solution is very sensitive to the initial values
fsolve(fun, [1 1])
Is it possible to get the solution to this equation without providing an exact starting value, but rather a range? For example, I would like to see the possible combinations for x1, x2 \in (-4,4)
Using ezplot I obtain the desired graphical output, but not the solution of the equation.
fh= #(x1,x2) [x1 x2]*V*[x1 x2]'-c;
ezplot(fh)
axis equal
Is there a way to have both?
Thanks a lot!
you can take the XData and YData from ezplot:
c = rand;
V = rand(2);
V = V + V';
fh= #(x1,x2) [x1 x2]*V*[x1 x2]'-c;
h = ezplot(fh,[-4,4,-4,4]); % plot in range
axis equal
fun = #(x) [x(1) x(2)]*V*[x(1) x(2)]'-c;
X = fsolve(fun, [1 1]); % specific solution
hold on;
plot(x(1),x(2),'or');
% possible solutions in range
x1 = h.XData;
x2 = h.YData;
or you can use vector input to fsolve:
c = rand;
V = rand(2);
V = V + V';
x1 = linspace(-4,4,100)';
fun2 = #(x2) sum(([x1 x2]*V).*[x1 x2],2)-c;
x2 = fsolve(fun2, ones(size(x1)));
% remove invalid values
tol = 1e-2;
x2(abs(fun2(x2)) > tol) = nan;
plot(x1,x2,'.b')
However, the easiest and most straight forward approach is to rearrange the ellipse matrix form in a quadratic equation form:
k = rand;
V = rand(2);
V = V + V';
a = V(1,1);
b = V(1,2);
c = V(2,2);
% rearange terms in the form of quadratic equation:
% a*x1^2 + (2*b*x2)*x1 + (c*x2^2) = k;
% a*x1^2 + (2*b*x2)*x1 + (c*x2^2 - k) = 0;
x2 = linspace(-4,4,1000);
A = a;
B = (2*b*x2);
C = (c*x2.^2 - k);
% solve regular quadratic equation
dicriminant = B.^2 - 4*A.*C;
x1_1 = (-B - sqrt(dicriminant))./(2*A);
x1_2 = (-B + sqrt(dicriminant))./(2*A);
x1_1(dicriminant < 0) = nan;
x1_2(dicriminant < 0) = nan;
% plot
plot(x1_1,x2,'.b')
hold on
plot(x1_2,x2,'.g')
hold off
Related
In MATLAB, I am trying to write a program that will take 3 coordinates on a graph, (x,y), use those values to solve a system of equations that will find the coefficients of a polynomial equation, y = ax^2 + bx + c, which I can then use to plot a parabola.
To test my code, I figured I could start with a polynomial, graph it, find the minimum location of the parabola, use its immediate neighbors for my other 2 locations, then run those 3 locations through my code which should spit out the coefficients of my original polynomial. But for some reason, my resulting parabola is right shifted and my values for b and c are incorrect.
Does anyone see where my issue is? I am out of ideas
clear all; close all;
x = -10:10;
%Original Polynomial
y = 6.*x.^2 + 11.*x -35;
% Find 3 Locations
[max_n, max_i] = min(y)
max_il = max_i - 1 % left neighbor of max_ni
max_nl = y(max_il) % value at max_il
max_ir = max_i + 1 % left neighbor of max_ni
max_nr = y(max_ir) % value at max_ir
% Solve for coefficients
syms a b c
equ = (a)*(max_i)^2 + (b)*(max_i) + (c) == (max_n);
equ_l = (a)*(max_il)^2 + (b)*(max_il) + (c) == (max_nl);
equ_r = (a)*(max_ir)^2 + (b)*(max_ir) + (c) == (max_nr);
sol = solve([equ, equ_l, equ_r],[a, b, c]);
Sola = sol.a
Solb = sol.b
Solc = sol.c
% New Polynomial
p = (sol.a).*(x).^2 + (sol.b).*(x) +(sol.c);
%Plot
plot(x,y); grid on; hold on;
plot(x, p);
axis([-10 10 -41 40])
[max_np, max_ip] = min(p)
legend('OG', 'New')
You are confusing the index into your array y, and the corresponding x coordinate.
x = -10:10;
y = 6.*x.^2 + 11.*x -35;
[max_n, max_i] = min(y)
Here. max_i is the index into the y array, the corresponding x coordinate would be x(max_i).
I suggest you find three data points to fit your curve to as follows:
[~, max_i] = min(y);
pts_x = x(max_i + (-1:1));
pts_y = y(max_i + (-1:1));
then use pts_x(i) and pts_y(i) as your x and y values:
syms a b c
equ = a * pts_x.^2 + b * pts_x + c == pts_y;
sol = solve(equ, [a, b, c]);
I updated the question to clarify it more. Here is a graph:
For the curve in the attached photo, I hope to draw the curve. I have its equation and it is after simplification will be like this one
% Eq-2
(b*Y* cos(v) + c - k*X*sin(v))^2 + ...
sqrt(k*X*(cos(v) + 1.0) + b*Y*sin(v))^2) - d = 0.0
Where:
v = atan((2.0*Y)/X) + c
and b, c, d and k are constants.
from the attached graph,
The curve is identified in two points:
p1 # (x=0)
p2 # (y=0)
I a new on coding so accept my apologize if my question is not clear.
Thanks
So, after your edit, it is a bit more clear what you want.
I insist that your equation needs work -- the original equation (before your edit) simplified to what I have below. The curve for that looks like your plot, except the X and Y intercepts are at different locations, and funky stuff happens near X = 0 because you have numerical problems with the tangent (you might want to reformulate the problem).
But, after checking your equation, the following code should be helpful:
function solve_for_F()
% graininess of alpha
N = 100;
% Find solutions for all alphae
X = zeros(1,N);
options = optimset('Display', 'off');
alpha = linspace(0, pi/2, N);
x0 = linspace(6, 0, N);
for ii = 1:numel(alpha)
X(ii) = fzero(#(x)F(x, alpha(ii)), x0(ii), options);
end
% Convert and make an X-Y plot
Y = X .* tan(alpha);
plot(X, Y,...
'linewidth', 2,...
'color', [1 0.65 0]);
end
function fval = F(X, alpha)
Y = X*tan(alpha);
% Please, SIMPLIFY in the future
A = 1247745517111813/562949953421312;
B = 4243112111277797/4503599627370496;
V = atan2(2*Y,X) + A;
eq2 = sqrt( (5/33*( Y*sin(V) + X/2*(cos(V) + 1) ))^2 + ...
(5/33*( Y*cos(V) - X/2* sin(V) ))^2 ) - B;
fval = eq2;
end
Results:
So, I was having fun with this (thanks for that)!
Different question, different answer.
The solution below first searches for the constants causing the X and Y intercepts you were looking for (p1 and p2). For those constants that best fit the problem, it makes a plot, taking into account numerical issues.
In fact, you don't need eq. 1, because that's true always for any curve -- it's just there to confuse you, and problematic to use.
So, here it is:
function C = solve_for_F()
% Points of interest
px = 6;
py = 4.2;
% Wrapper function; search for those constants
% causing the correct X,Y intercepts (at px, py)
G = #(C) abs(F( 0, px, C)) + ... % X intercept at px
abs(F(py, 0, C)); % Y intercept at py
% Initial estimate, based on your original equation
C0 = [5/33
1247745517111813/562949953421312
4243112111277797/4503599627370496
5/66];
% Minimize the error in G by optimizing those constants
C = fminsearch(G, C0);
% Plot the solutions
plot_XY(px, py, C);
end
function plot_XY(xmax,ymax, C)
% graininess of X
N = 100;
% Find solutions for all alphae
Y = zeros(1,N);
X = linspace(0, xmax, N);
y0 = linspace(ymax, 0, N);
options = optimset('Display', 'off',...,...
'TolX' , 1e-10);
% Solve the nonlinear equation for each X
for ii = 1:numel(X)
% Wrapper function for fzero()
fcn1 = #(y)F(y, X(ii), C);
% fzero() is probably the fastest and most intuitive
% solver for this problem
[Y(ii),~,flag] = fzero(fcn1, y0(ii), options);
% However, it uses an algorithm that easily diverges
% when the function slope is large. For those cases,
% solve with fminsearch()
if flag ~= 1
% In this case, the minimum of the absolute value
% is searched for (which should be zero)
fcn2 = #(y) abs(fcn1(y));
Y(ii) = fminsearch(fcn2, y0(ii), options);
end
end
% Now plot the X,Y solutions
plot(X, Y,...
'linewidth', 2,...
'color', [1 0.65 0]);
xlabel('X'), ylabel('Y')
axis([0 xmax+.1 0 ymax+.1])
end
function fval = F(Y, X, C)
% Unpack constants
b = C(1); d = C(3);
c = C(2); k = C(4);
% pre-work
V = atan2(2*Y, X) + c;
% Eq. 2
fval = sqrt( (b*Y*sin(V) + k*X*(cos(V) + 1))^2 + ...
(b*Y*cos(V) - k*X* sin(V) )^2 ) - d;
end
I am trying to implement the finite difference method in matlab. I did some calculations and I got that y(i) is a function of y(i-1) and y(i+1), when I know y(1) and y(n+1). However, I don't know how I can implement this so the values of y are updated the right way. I tried using 2 fors, but it's not going to work that way.
EDIT
This is the script and the result isn't right
n = 10;
m = n+1;
h = 1/m;
x = 0:h:1;
y = zeros(m+1,1);
y(1) = 4;
y(m+1) = 6;
s = y;
for i=2:m
y(i) = y(i-1)*(-1+(-2)*h)+h*h*x(i)*exp(2*x(i));
end
for i=m:-1:2
y(i) = (y(i) + (y(i+1)*(2*h-1)))/(3*h*h-2);
end
The equation is:
y''(x) - 4y'(x) + 3y(x) = x * e ^ (2x),
y(0) = 4,
y(1) = 6
Thanks.
Consider the following code. The central differential quotient is discretized.
% Second order diff. equ.
% y'' - 4*y' + 3*y = x*exp(2*x)
% (y(i+1)-2*y(i)+y(i-1))/h^2-4*(y(i+1)-y(i-1))/(2*h) + 3*y(i) = x(i)*exp(2*x(i));
The solution region is specified.
x = (0:0.01:1)'; % Solution region
h = min(diff(x)); % distance
As said in my comment, using this method, all points have to be solved simultaneously. Therefore, above numerical approximation of the equation is transformed in a linear system of euqations.
% System of equations
% Matrix of coefficients
A = zeros(length(x));
A(1,1) = 1; % known solu for first point
A(end,end) = 1; % known solu for last point
% y(i) y'' y
A(2:end-1,2:end-1) = A(2:end-1,2:end-1)+diag(repmat(-2/h^2+3,[length(x)-2 1]));
% y(i-1) y'' -4*y'
A(1:end-1,1:end-1) = A(1:end-1,1:end-1)+diag(repmat(1/h^2+4/(2*h),[length(x)-2 1]),-1);
% y(i+1) y'' -4*y'
A(2:end,2:end) = A(2:end,2:end)+diag(repmat(1/h^2-4/(2*h),[length(x)-2 1]),+1);
With the rhs of the differential equation. Note that the known values are calculated by 1 in the matrix and the actual value in the solution vector.
Y = x.*exp(2*x);
Y(1) = 4; % known solu for first point
Y(end) = 6; % known solu for last point
y = A\Y;
Having an equation to approximate the first order derivative (see above) you can verify the solution. (note, ddx2 is an own function)
f1 = ddx2(x,y); % first derivative (own function)
f2 = ddx2(x,f1); % second derivative (own function)
figure;
plot(x,y);
saveas(gcf,'solu1','png');
figure;
plot(x,f2-4*f1+3*y,x,x.*exp(2*x),'ko');
ylim([0 10]);
legend('lhs','rhs','Location','nw');
saveas(gcf,'solu2','png');
I hope the solution shown below is correct.
I try to use contour to plot this function
3y + y^3 - x^3 = 5
I try contour(3*y+y^3-x^3-5) but it doesn't work.
How can I use contour to plot this function?
Are x and y properly defined as 2x2 matrices? If so then the "power" operator needs to be done on a component-wise basis (.^3 instead of ^3).
This works:
[x,y] = meshgrid(-2:.2:2,-2:.2:2);
contour(3*y+y.^3-x.^3-5)
Maybe you can try fcontour, which plots the contour lines of the function z = f(x,y) for constant levels of z over the default interval [-5 5] for x and y.
f = #(x,y) 3*y + y.^3 - x.^3 - 5;
fcontour(f)
Output:
I'm not convinced this addresses all parts of your question but it's a start. If you absolutely want contour to call a function, you can adjust my example to contour(X,Y,fh(X,Y)).
Better Approach
fh=#(x,y) 3*y + y.^3 - x.^3 -5; % <--- This is your function
x = (-4:.25:4)';
y = (-2:.25:2)';
[X,Y] = meshgrid(x,y);
Z = fh(X,Y);
contour(X,Y,fh(X,Y))
The Direct Approach (not preferred but works)
Notice the Z is transposed to make this work.
fh=#(x,y) 3*y + y.^3 - x.^3 -5; % <----this is your function
X = (-4:.25:4)';
Y = (-2:.25:2)';
Z = zeros(length(X),length(Y));
for i = 1:length(X)
for j = 1:length(Y)
xi = X(i);
yj = Y(j);
Z(i,j) = fh(xi,yj);
end
end
contour(X,Y,Z','LevelList',-60:10:60,'ShowText','on','LineWidth',1.4) % Fancied it up a bit
I would like to fit a curve on the form y=a+b*sin(2*pi*x)+c*cos(2*pi*x) to some data points in Matlab. I've been trying to use 'fit' but then I only get this message 'if isa( fittypeobj, 'fittype' )'
This is my code:
L = load('file.mat');
x = filedata(:,1);
ft = fittype('a+b*sin(2*pi*x)+c*cos(2*pi*x)');
fit(x, filedata(:,3), ft)
Can somebody please tell me what I'm doing wrong?
Here is how to do the fit 'by hand' in a least-squares way:
x = x(:); %make sure we have column vectors
y = y(:);
f0 = 1;
M = [ones(size(x)), sin(2*pi*f0*x), cos(2*pi*f0*x)];
%design matrix, columns are base vectors
% least square approximation of x = k(1)*M(:,1) + k(2)*M(:,2) + k(3)*M(:,3);
% see help mldivide
k = M \ y;
a = k(1);
b = k(2);
c = k(3);
Quick test to see if it works:
>> x = linspace(0,10,1000)'; % note transpose to make column
>> y = 3 + 1.5 * sin(2*pi*x) + 8 * cos(2*pi*x) + randn(size(x)); % add some noise
>> f0 = 1;
>> M = [ones(size(x)), sin(2*pi*f0*x), cos(2*pi*f0*x)];
>> k = M \ y
k =
3.0383
1.5264
7.9385
>> plot(x, y, x, M*k, 'r'); legend('measurement', 'fit')