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
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]);
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
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
Im new to matlab and I want to plot a ContourPlot together with two inequalities but I dont know how. The function that I want to plot its ContourPlot is something like this:
Z = (X-2).^2 + (Y-1).^2;
and here are the two inequalities:
ineq1 = X.^2 - Y <= 2;
ineq2 = X + Y <= 2;
this is what I have dodne so far:
[X,Y] = meshgrid(-4:.2:4,-4:.2:4);
Z = (X-2).^2 + (Y-1).^2;
[C,h] = contour(X,Y,Z);
clabel(C,h)
ineq1 = X.^2 - Y <= 2;
ineq2 = X + Y <= 2;
range = (-4:.2:4);
hold on
plot(range,ineq1, range, ineq2)
hold off
but this does not feel right.
What I want to do is to visualize an optimization problem. First I want to plot the ContourPlot of the function and then the possible areas in that same plot, It would be great if I could show the intersection areas too.
If you want to draw the boundaries of the inequalities onto the contour plot, you can do this with line.
[X,Y] = meshgrid(-4:.2:4,-4:.2:4);
Z = (X-2).^2 + (Y-1).^2;
[C,h] = contour(X,Y,Z);
clabel(C,h)
x_vect = (-4:.05:4);
y_ineq1 = x_vect.^2 - 2;
y_ineq2 = 2 - x_vect;
line(x_vect, y_ineq1);
line(x_vect, y_ineq2);
Coloring the intersection area is a bit more tricky, and I'm not sure if it's exactly what you want to do, but you could look into using patch, something like
% Indexes of x_vect, y_ineq1 and y_ineq2 for region where both satisfied:
region_indexes = find(y_ineq1<y_ineq2);
region_indexes_rev = region_indexes(end:-1:1);
% To plot this area need to enclose it
% (forward along y_ineq1 then back along y_ineq2)
patch([x_vect(region_indexes),x_vect(region_indexes_rev)],...
[y_ineq1(region_indexes),y_ineq2(region_indexes_rev)],...
[1 0.8 1]) % <- Color as [r g b]
% May or may not need following line depending on MATLAB version to set the yaxes
ylim([-4 4])
% Get the original plot over the top
hold on
[C,h] = contour(X,Y,Z);
clabel(C,h)
hold off
If I have two plots defined by two different equations:
x = 0:0.01:30;
y1 = x .^2 + 2;
y2 = x .^3 ;
and I plot them as
plot(x, y1, x, y2);
How do I get a small ring around the point of intersection programatically (as in the following plot)?
You'll have to find the point of intersection (px, py) manually:
idx = find(y1 - y2 < eps, 1); %// Index of coordinate in array
px = x(idx);
py = y1(idx);
Remember that we're comparing two numbers in floating point representation, so instead of y1 == y2 we must set a tolerance. I've chosen it as eps, but it's up to you to decide.
To draw a circle around this point, you can compute its points and then plot them, but a better approach would be to plot one point with a blown-up circle marker (credit to Jonas for this suggestion):
plot(px, py, 'ro', 'MarkerSize', 18)
This way the dimensions of the circle are not affected by the axes and the aspect ratio of the plot.
Example
x = 0:0.01:30;
y1 = x .^ 2 + 2;
y2 = x .^ 3;
%// Find point of intersection
idx = find(y1 - y2 < eps, 1);
px = x(idx);
py = y1(idx);
figure
plot(x, y1, x, y2, px, py, 'ro', 'MarkerSize', 18)
axis([0 10 0 10])
This should produce the following plot:
In your example, when you have x, y1 and y2
What you can do is
idx = find(abs(y1 - y2) == min(abs(y1 - y2)));
xInter = x(idx)
yInter = y1(idx) % or y2(idx)
If you have x1, y1 and x2, y2, where x1 ~= x2
you could first do 1D interpolation using
yy2 = interp1(x2, y2, x1);
then apply
idx = find(abs(y1 - yy2) == min(abs(y1 - yy2)));
xInter = x1(idx)
yInter = y1(idx) % or yy2(idx)
Excellent post by #EitanT, however I would like to complement this with a different (automated) way to find the intersection (Assuming there is one and the graphs behave nicely).
Here is our starting point:
x = 0:0.01:30;
y1 = x .^2 + 2;
y2 = x .^3 ;
First of all we check whether these values are exactly equal, for non-floating point non-discrete situations this should be sufficient:
idx = find(y1==y2)
If they are never recorded to be exactly equal, an intersection occurs if one surpasses the other, hence we look at the difference:
if isempty(idx)
d = y1-y2;
% At the moment of crossing, the sign will change:
s = diff(sign(d));
% Now just find the point where it changes
f = find(s,1);
end
To summarize this in compact form without additional variables, I would recommend using:
idx = find(y1==y2)
if isempty(idx)
idx = find(diff(sign(y1-y2)),1)
end
Especially when knowing the functions, the symbolic math toolbox can be used.
y1 = x .^2 + 2;
y2 = x .^3 ;
syms x real
intersection=simplify(solve(y1==y2))
Use vpa(intersection) to convert it to a number or double(intersection) to convert it to a floating point value.
Last but not least, perhaps the cleanest way to do this is the command polyxpoly:
[xi,yi] = polyxpoly(x,y1,x,y2)
xi = 1.69560153754948
yi = 4.87508921229275
Good luck!