I apologize for asking this, I believe this is a simple task, but I don't know how to do it.
Suppose I have a formula y = (exp(-x) + x^2)/sqrt(pi(x) and I want to plot it as y versus x^2.
How does one do this?
Like this:
X = 0:0.1:5; %// Get the x values
x = X.^2; %// Square them
%// Your formula had errors, I fixed them but I could have misinterpreted here, please check
y = (exp(-x) + x.^2)./sqrt(pi*x); %// Calculate y at intervals based on the squared x. This is still y = f(x), I'm just calculating it at the points at which I want to plot it.
plot(x,y) %//Plot against the square X.
At this point this is no different to having just plotted it normally. What you want is to make the tickmarks go up in values of X.^2. This does not change the y-values nor distort the function, it just changes what it looks like visually. Similar to plotting against a log scale:
set(gca, 'XTick', X.^2) %//Set the tickmarks to be squared
The second method gives you a plot like
edit:
Actually I think you were asking for this:
x = 0:0.1:5;
y = x.^2; %// Put your function in here, I'm using a simple quadratic for illustrative purposes.
plot(x.^2,y) %//Plot against the square X. Now your y values a f(x^2) which is wrong, but we'll fix that later
set(gca, 'XTick', (0:0.5:5).^2) %//Set the tickmarks to be a nonlinear intervals
set(gca, 'XTickLabel', 0:0.5:5) %//Cahnge the labels to be the original x values, now accroding to the plot y = f(x) again but has the shape of f(x^2)
So here I'm plotting a simple quadratic, but if I plot it against a squared x it should become linear. However I still want to read off the graph that y=x^2, not y=x, I just want it to look like y=x. So if I read the y value for the x value of 4 on that graph i will get 16 which is still the same correct original y value.
Here's my answer: it is similar to Dan's one, but fundamentally different. You can calculate the values of y as a function of x, but plot them as a function of x^2, which is what the OP was asking, if my understanding is correct:
x = 0:0.1:5; %// Get the x values
x_squared = x.^2; %// Square them
%// Your formula had errors, I fixed them but I could have misinterpreted here, please check
y = (exp(-x) + x.^2)./sqrt(pi*x); %// Calculate y based on x, not the square of x
plot(x_squared,y) %//Plot against the square of x
As Dan mentioned, you can always change the tickmarks:
x_ticks = (0:0.5:5).^2; % coarser vector to avoid excessive number of ticks
set(gca, 'XTick', x_ticks) %//Set the tickmarks to be squared
Related
I need to plot a parabola and ellipse. However the ellipse is giving me trouble. Can anyone help? The equations are: y = -5*x^2 + 2 and (x^2/16) + (y^2/2) = 4
I've tried this code but obviously I feel like like it isn't right.
x = linspace(-5, 5);
y1 = (x.^2/16) + (y.^2/2) - 1;
y2 = -5*x.^2 +2;
figure(1)
plot(x, y1)
hold on
plot(x, y2)
hold off
Firstly, you did not define a range variable x. Secondly, the ellipse won't pass the vertical line test and can't be plotted like a regular function f(x). Thirdly, your equation y1 = (x.^2/16) + (y.^2/2) - 1; is non-sensical because you have y on each side.
You could correct your method by defining a range variable x1 and x2 that each have appropriate ranges for the functions your plotting. What I mean by this is that you probably don't want the same range for each function, because the ellipse is undefined over most of the range that the parabola is defined. To plot the ellipse using f(x) you could observe that there are + and - values that are identical, using this fact you could plot your ellipse by two functions one to represent the top half and one to represent the bottom half, each of these would pass the vertical line test.
OR
You could utilize ezplot and have a nice time with it because it makes your life easier. Here is a solution.
ezplot('x^2/16+y^2/2-4'); axis equal; hold on
ezplot('-5*x^2+2-y')
There are multiple ways to plot an ellipse, e.g. you could also use a parametric representation of the equation.
In your approach though, when plotting functions using plot(x,y) command, you need to express your dependent variable (y) through independent variable (x). You defined the range for x, which is what you substitute into your equations in order to find y's. While for the parabola, the dependency of y from x is obvious, you forgot to derive such a relationship for the ellipse. In this case it will be +-sqrt((1 - x^2/16)*2). So in your approach, you'll have to take into account both negative and positive y's for the same value of x. Also there's discrepancy in your written equation for the ellipse (=4) and the one in Matlab code (=1).
x = linspace(-5, 5);
y1 = sqrt((1 - x.^2/16)*2);
y2 = -5*x.^2 +2;
figure(1)
plot(x, real(y1), 'r', x, -real(y1), 'r')
hold on
plot(x, y2)
hold off
Since the ellipse has real y's not on the whole x domain, if you want to plot only real parts, specify real(y1) or abs(y1) (even though Matlab does it for you, too). You can also dismiss complex numbers for certain x when computing y1, but you'll need a for-loop for that.
In order to make things simpler, you can check the function fimplicit, ezplot is not recommended according to Matlab's documentation. Or if you want to plot the ellipse in a parametric way, fplot will work, too.
Another (more classic) approach for parametric plotting is given here already, then you don't need any other functions than what you already use. I think it is the simplest and most elegant way to plot an ellipse.
You will not be able to generate points for the ellipse using a function f(x) from a Cartesian linspace range. Instead, you can still use linspace but for the angle in a polar notation, from 0 to 2*pi. You should also be able to easily adjust radius and offset on both axis on the cos and sin expressions.
x = linspace(-5, 5);
y2 = -5*x.^2 +2;
figure(1)
clf;
plot(x, y2)
hold on
a = linspace(0,2*pi);
x2 = 4*cos(a);
y2 = sqrt(2)*sin(a);
plot(x2, y2)
xlim([-5,5]);
ylim([-5,5]);
hold off
MATLAB's surf command allows you to pass it optional X and Y data that specify non-cartesian x-y components. (they essentially change the basis vectors). I desire to pass similar arguments to a function that will draw a line.
How do I plot a line using a non-cartesian coordinate system?
My apologies if my terminology is a little off. This still might technically be a cartesian space but it wouldn't be square in the sense that one unit in the x-direction is orthogonal to one unit in the y-direction. If you can correct my terminology, I would really appreciate it!
EDIT:
Below better demonstrates what I mean:
The commands:
datA=1:10;
datB=1:10;
X=cosd(8*datA)'*datB;
Y=datA'*log10(datB*3);
Z=ones(size(datA'))*cosd(datB);
XX=X./(1+Z);
YY=Y./(1+Z);
surf(XX,YY,eye(10)); view([0 0 1])
produces the following graph:
Here, the X and Y dimensions are not orthogonal nor equi-spaced. One unit in x could correspond to 5 cm in the x direction but the next one unit in x could correspond to 2 cm in the x direction + 1 cm in the y direction. I desire to replicate this functionality but drawing a line instead of a surf For instance, I'm looking for a function where:
straightLine=[(1:10)' (1:10)'];
my_line(XX,YY,straightLine(:,1),straightLine(:,2))
would produce a line that traced the red squares on the surf graph.
I'm still not certain of what your input data are about, and what you want to plot. However, from how you want to plot it, I can help.
When you call
surf(XX,YY,eye(10)); view([0 0 1]);
and want to get only the "red parts", i.e. the maxima of the function, you are essentially selecting a subset of the XX, YY matrices using the diagonal matrix as indicator. So you could select those points manually, and use plot to plot them as a line:
Xplot = diag(XX);
Yplot = diag(YY);
plot(Xplot,Yplot,'r.-');
The call to diag(XX) will take the diagonal elements of the matrix XX, which is exactly where you'll get the red patches when you use surf with the z data according to eye().
Result:
Also, if you're just trying to do what your example states, then there's no need to use matrices just to take out the diagonal eventually. Here's the same result, using elementwise operations on your input vectors:
datA = 1:10;
datB = 1:10;
X2 = cosd(8*datA).*datB;
Y2 = datA.*log10(datB*3);
Z2 = cosd(datB);
XX2 = X2./(1+Z2);
YY2 = Y2./(1+Z2);
plot(Xplot,Yplot,'rs-',XX2,YY2,'bo--','linewidth',2,'markersize',10);
legend('original','vector')
Result:
Matlab has many built-in function to assist you.
In 2D the easiest way to do this is polar that allows you to make a graph using theta and rho vectors:
theta = linspace(0,2*pi,100);
r = sin(2*theta);
figure(1)
polar(theta, r), grid on
So, you would get this.
There also is pol2cart function that would convert your data into x and y format:
[x,y] = pol2cart(theta,r);
figure(2)
plot(x, y), grid on
This would look slightly different
Then, if we extend this to 3D, you are only left with plot3. So, If you have data like:
theta = linspace(0,10*pi,500);
r = ones(size(theta));
z = linspace(-10,10,500);
you need to use pol2cart with 3 arguments to produce this:
[x,y,z] = pol2cart(theta,r,z);
figure(3)
plot3(x,y,z),grid on
Finally, if you have spherical data, you have sph2cart:
theta = linspace(0,2*pi,100);
phi = linspace(-pi/2,pi/2,100);
rho = sin(2*theta - phi);
[x,y,z] = sph2cart(theta, phi, rho);
figure(4)
plot3(x,y,z),grid on
view([-150 70])
That would look this way
Can someone share a technique using MATLAB to plot the surface f(x,y)=(21/4)x^2y over the region x^2 <= y <= 1?
Also, if anyone is aware of some tutorials or links that would help with this type of problem, could you please share them?
Thanks.
Here is another approach:
%%
close all
x=linspace(-1,1,40);
g1=x.^2;
g2=ones(1,40);
y=[];
n=20;
for k=0:n
y=[y;g1+(g2-g1)*k/n];
end
x=x(ones(1,n+1),:);
z=21/4*x.^2.*y;
meshz(x,y,z)
axis tight
xlabel('x-axis')
ylabel('y-axis')
view(136,42)
And the result:
And finally, you can map the region (-1,1)x(0,1) in the uv-plane into the region bounded by $y=x^2 and y=1 in the xy-plane with the parametrization:
f(u,v) = (u\sqrt{v},v)
Capture from: https://math.stackexchange.com/questions/823168/transform-rectangular-region-to-region-bounded-by-y-1-and-y-x2
This code produces the same image shown above:
close all
[u,v]=meshgrid(linspace(-1,1,40),linspace(0,1,20));
x=u.*sqrt(v);
y=v;
z=21/4*x.^2.*y;
meshz(x,y,z)
axis tight
xlabel('x-axis')
ylabel('y-axis')
view(136,42)
First off, let's look at your valid region of values. This is telling us that y >= x^2 and also y <= 1. This means that your y values need to be on the positive plane bounded by the parabola x^2 and they also must be less than or equal to 1. In other words, your y values must be bound within the area dictated from y = x^2 to y = 1. Pictorially, your y values are bounded within this shape:
As such, your x values must also be bound between -1 and 1. Therefore, your actual boundaries are: -1 <= x <= 1 and 0 <= y <= 1. However, this only locates our boundaries for x and y but it doesn't handle where the plot has valid values. We'll tackle that later.
Now that we have that established, you can use ezsurf to plot surface plots in MATLAB that are dictated by a 2D equation.
You call ezsurf like so:
ezsurf(FUN, [XMIN,XMAX,YMIN,YMAX]);
FUN is a function or a string that contains the equation you want, and XMIN,XMAX,YMIN,YMAX contain the lowest and highest x and y values you want to plot. Plotting without these values assumes a span from -2*pi to 2*pi in both dimensions. As such, let's create a new function that will handle when we have valid values, and when we don't. Use this code, and save it to a new file called myfun.m. Make sure you save this to your current Working Directory.
function z = myfun(x,y)
z = (21/4)*x.^2.*y;
z(~(x.^2 <= y & y <= 1)) = nan;
end
This will allow you to take a series of x and y values and output values that are dictated by the 2D equation that you have given us. Any values that don't satisfy the condition of x^2 <= y <= 1, you set them to NaN. ezsurf will not plot NaN values.
Now, call ezsurf like so:
ezsurf(#myfun, [-1,1,0,1]);
You thus get:
This will spawn a new figure for you, and there are some tools at the top that will allow you interact with your 3D plot. For instance, you can use the rotation tool that's at the top bar beside the hand to rotate your figure around and see what this looks like. Click on this tool, then left click your mouse and hold the left mouse button anywhere within the surface plot. You can drag around, changing the azimuth and the latitude to get the perspective that you want.
Edit: June 4th, 2014
Noting your comments, we can decrease the jagged edges by increasing the number of points in the plot. As such, you can append a final parameter to ezsurf which is N, the number of points to add in each dimension. Increasing the number of points will decrease the width in between each point and so the plot will look smoother. The default value of N is 60 in both dimensions. Let's try increasing the amount of points in each dimension to 100.
ezsurf(#myfun, [-1,1,0,1], 100);
Your plot will look like:
Hope this helps!
Try the following to make the required function, compute the values, and plot only the region that is desired:
% Make the function. You could put this in a file by itself, if you wanted.
f = #(x,y) (21/4)*x.^2.*y;
[X Y] = meshgrid(linspace(0,1));
Z = f(X,Y);
% compute the values we want to plot:
valsToPlot = (X.^2 <= Y) & (Y <= 1);
% remove the values that we don't want to plot:
X(~valsToPlot) = nan;
Y(~valsToPlot) = nan;
Z(~valsToPlot) = nan;
% And... plot.
figure(59382);
clf;
surf(X,Y,Z);
For this given example:
a=rand(100,1)
ecdf(a)
The obtained plot will have on x proportion of bin values, and on y the proportion of points. I wanted to add a line y=0.5 (50% percent of values) and when cross the line in the plot gather and shows the predicted x value.
It is possible but it suppresses my knowledge.
Thanks in advance
Not sure I interpret your question correctly - but could it be as simple as
a=rand(100,1)
ecdf(a)
hold on
plot([0 1],[1 1]*0.5);
This adds a line at a height of 0.5 from 0 to 1 (which I believe are the limits of the plot that ecdf produced for you).
If you want to find the point where these two lines intersect, you need to obtain the points in the plot using a different form of the ecdf function:
a = rand(100,1)
[f x] = ecdf(a);
figure
plot(x, f); % now you have to make the plot yourself...
hold all
plot(x, 0.5 * ones(size(x))); % add the line at y=0.5
title 'cumulative probability of rand()'
xest = interp1(f, x, 0.5); % interpolate - find x where f would be 0.5
fprintf(1, 'The intercept is at x=%.2f\n", xest);
I've got 3 vectors
x =[10^(-10) 10^(-9) 10^(-8) 10^(-7) 10^(-6) 10^(-5) 10^(-4) 10^(-3) 10^(-2)]
y =[10^(-10) 10^(-9) 10^(-8) 10^(-7) 10^(-6) 10^(-5) 10^(-4) 10^(-3) 10^(-2)]
and z which is the error that takes values from 5 to 0.01.
I need to create a contour plot. However, I cannot use meshgrid, as the step on my x and y axis should be very small and Matlab crashes when I'm trying that.
Any ideas about resolving it?
You can use meshgrid to produce the required mesh and replace the labels later on. The properties you are looking for are called XTick and XTickLabel. See for example: http://www.mathworks.com/matlabcentral/answers/92565
You need to use a logarithmic scale to see the results in any sensible way. Practically, this means you should probably do
x = logspace(-10, -2, 9);
y = logspace(-10, -2, 9);
logX = log(x);
logY = log(y);
[lxx lyy]=meshgrid(logX, logY);
and then plot contours as usual with
contour(lxx, lyy, reshape(z, [9 9]));
Finally, you would want to annotate your axes to show you are using a log scale - or you can turn off the tick labels and replace them with your own values.
Alternatively, I wonder if you could use the grid you have
[xx yy] = meshgrid(x, y);
and after plotting the contour, doing
set(gca, 'xscale', 'log');
set(gca, 'yscale', 'log');
I can't test this right now, so this may be bad advice. Let me know how it works for you.