I want to draw a line on a graph to find the intersection point with another line. However, there's no response after I executed the script below. May I know what is the problem and how can I solve it?
x=1:2^20;
y2=2^24;
plot(x,y2);
Thanks!
What you want is to plot a line on 2^24. However, there are too many points for you computer probably, and you run out of memory
I am guessing that you'll need to plot your other inequality as well.
Something like
x=1:100:2^20;
% As Zoran and others suggested, You may not want all the points!
% It is too much memory
y2=2^24*ones(size(x)); % This ones is optional, but its good to know what you are doing (personal opinion)
plot(x,y2);
hold on
y1=(x+1).*log(x);
plot(x,y1);
However, you are still not there!
Another solution, which does not rely on plotting:
>> f = #(x) (x+1)*log(x)-2^24;
>> soln = fzero(f,1e6)
soln = 1.1987e+006
>> f(soln)
ans = 3.7253e-009
So your intersection point is at 1.1987e6.
Apparently,you have too many points for x, 2^20
Have to wait program to calculate, or plot,for example, every 100th point
This solution works for Matlab
x=1:100:2^20;
y2=2^2;
plot(x,y2,'o');
There is one more and maybe a bit smarter way: if you want to solve ((k+1)(ln k)<2^24) as you've commented above, use fsolve function to get just solution of an equation(!). Then use that solution to specify the area of your interest, so you won't have to plot the domain of 2^20.
(All functions are continuous so you don't have to worry about any wild singularities. Just examine the neigborhood of ks for which (k+1)(ln k)-2^24=0.)
Related
I want to plot just the first 50 points of the following function in MATLAB and I'm having difficulty finding out how to do this, I genuinely cannot find the command or much that is similar online. Can anyone help me out with this? Thank you in advance for all the help, anything is appreciated.
>> p=pi;
>> x=[0:0.1:4*p];
>> y = exp(-0.4*x).*sin(x);
>> plot(x,y), plot(x(1:50)) %%%this is one of my attempts
>> title ('MATLAB PRACTICE');
>> xlabel('x-Axis');
>> ylabel('y-Axis');
>> grid on
You are close, but it is instructive to break the problem into steps.
Think of it this way: you want to first select the first 50 values of x and y, and then plot those.
You can do that like so:
xSubset = x(1:50);
ySubset = y(1:50);
plot(xSubset, ySubset);
The syntax x(1:50) can be used to select a portion of x (or any vector).
We can accomplish the same thing without creating the temporary vectors, which may be desirable because it is more concise (and less typing):
plot(x(1:50), y(1:50));
I am writing a report for a class and am having some issues with the lines of an unstable plot going beyond the boundary of the graph and overlapping the title and xlabel. This is despite specifying a ylim from -2 to 2. Is there a good way to solve this issue?
Thanks!
plot(X,u(:,v0),X,u(:,v1),X,u(:,v2),X,u(:,v3),X,u(:,v4))
titlestr = sprintf('Velocity vs. Distance of %s function using %s: C=%g, imax=%g, dx=%gm, dt=%gsec',ICFType,SDType,C,imax,dx,dt);
ttl=title(titlestr);
ylabl=ylabel("u (m/s)");
xlabl=xlabel("x (m)");
ylim([-2 2])
lgnd=legend('t=0','t=1','t=2','t=3','t=4');
ttl.FontSize=18;
ylabl.FontSize=18;
xlabl.FontSize=18;
lgnd.FontSize=18;
EDIT: Minimum reproducible example
mgc=randi([-900*10^10,900*10^10], [1000,2]);
mgc=mgc*1000000;
plot(mgc(:,1),mgc(:,2))
ylim([-1,1])
This is odd. It really looks like a Bug... partly
The reason is probably that the angle of the lines are so narrow that MATLAB runs into rounding errors when calculating the points to draw for your limits very small limits given very large numbers. (You see that you don't run into this problem when you don't scale the matrix mgc.
mgc = randi([-900*10^10,900*10^10], [1000,2]);
plot(mgc(:,1),mgc(:,2))
ylim([-1,1])
but if you scale it further, you run into this problem...
mgc = randi([-900*10^10,900*10^10], [1000,2]);
plot(mgc(:,1)*1e6,mgc(:,2)*1e6)
ylim([-1,1])
While those numbers are nowhere near the maximum number a double can represent (type realmax in the command window to see that this is a number with 308 zeros!); limiting the plot to [-1,1] on one of the axes -- note that you obtain the same phenom on the x-axis -- let MATLAB run into precision problems.
First of all, you see that it plots much less lines than before (in my case)... although, I just said to zoom on the y-axis. The thing is, that MATLAB does not recalculate the lines for the section but it really zooms into it (I guess that this may cause resolution errors with regard to pixels?)
Well, lets have a look at the data (Pro-tip, you can get the data of a line from a MATLAB figure by calling this snippet
datObj = findobj(gcf,'-property','YData','-property','XData');
X = datObj.XData;
Y = datObj.YData;
xlm = get(gca,'XLim'); % get the current x-limits
) We see that it represents the original data set, which is not surprising as you can also zoom out again.
Note that his only occurs if you have such a chaotic, jagged line. If you sort it, it does not happen.
quick fix:
Now, what happens, if we calculate the exact points for this section?
m = diff(Y)./diff(X); % slope
n = Y(1:end-1)-m.*X(1:end-1); % offset
x = [(-1-n); (1-n)]./m;
y = ones(size(x))./[-1 1].';
% plot
plot([xMinus1;xPlus1],(ones(length(xMinus1),2).*[-1 1]).')
xlim(xlm); % limit to exact same scale as before
The different colors indicate that they are now individual lines and not a single wild chaos;)
It seems Max pretty much hit the nail on the head as it pertains to the reason for this error is occurring. Per Enrico's advice I went ahead and submitted a bug report. MathWorks responded saying they were unsure it was "unexpected behavior" and would look into it more shortly. They also did suggest a temporary workaround (which, in my case, may be permanent).
This workaround is to put
set(gca,'ClippingStyle','rectangle');
directly after the plotting line.
Below is a modified version of the minimum reproducible example with this modification.
mgc=randi([-900*10^10,900*10^10], [1000,2]);
mgc=mgc*1000000;
plot(mgc(:,1),mgc(:,2))
set(gca,'ClippingStyle','rectangle');
ylim([-1,1])
I have a Matlab project in which I need to make a GUI that receives two mathematical functions from the user. I then need to find their intersection point, and then plot the two functions.
So, I have several questions:
Do you know of any algorithm I can use to find the intersection point? Of course I prefer one to which I can already find a Matlab code for in the internet. Also, I prefer it wouldn't be the Newton-Raphson method.
I should point out I'm not allowed to use built in Matlab functions.
I'm having trouble plotting the functions. What I basically did is this:
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
cla % To clear axes when plotting new functions
ezplot(fun_f);
hold on
ezplot(fun_g);
axis ([-20 20 -10 10]);
The problem is that sometimes, the axes limits do not allow me to see the other function. This will happen, if, for example, I will have one function as log10(x) and the other as y=1, the y=1 will not be shown.
I have already tried using all the axis commands but to no avail. If I set the limits myself, the functions only exist in certain limits. I have no idea why.
3 . How do I display numbers in a static text? Or better yet, string with numbers?
I want to display something like x0 = [root1]; x1 = [root2]. The only solution I found was turning the roots I found into strings but I prefer not to.
As for the equation solver, this is the code I have so far. I know it is very amateurish but it seemed like the most "intuitive" way. Also keep in mind it is very very not finished (for example, it will show me only two solutions, I'm not so sure how to display multiple roots in one static text as they are strings, hence question #3).
function [Sol] = SolveEquation(handles)
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
f = inline(fun_f);
g = inline(fun_g);
i = 1;
Sol = 0;
for x = -10:0.1:10;
if (g(x) - f(x)) >= 0 && (g(x) - f(x)) < 0.01
Sol(i) = x;
i = i + 1;
end
end
solution1 = num2str(Sol(1));
solution2 = num2str(Sol(2));
set(handles.roots1,'string',solution1);
set(handles.roots2,'string',solution2);
The if condition is because the subtraction will never give me an absolute zero, and this seems to somewhat solve it, though it's really not perfect, sometimes it will give me more than two very similar solutions (e.g 1.9 and 2).
The range of x is arbitrary, chosen by me.
I know this is a long question, so I really appreciate your patience.
Thank you very much in advance!
Question 1
I think this is a more robust method for finding the roots given data at discrete points. Looking for when the difference between the functions changes sign, which corresponds to them crossing over.
S=sign(g(x)-f(x));
h=find(diff(S)~=0)
Sol=x(h);
If you can evaluate the function wherever you want there are more methods you can use, but it depends on the size of the domain and the accuracy you want as to what is best. For example, if you don't need a great deal of accurac, your f and g functions are simple to calculate, and you can't or don't want to use derivatives, you can get a more accurate root using the same idea as the first code snippet, but do it iteratively:
G=inline('sin(x)');
F=inline('1');
g=vectorize(G);
f=vectorize(F);
tol=1e-9;
tic()
x = -2*pi:.001:pi;
S=sign(g(x)-f(x));
h=find(diff(S)~=0); % Find where two lines cross over
Sol=zeros(size(h));
Err=zeros(size(h));
if ~isempty(h) % There are some cross-over points
for i=1:length(h) % For each point, improve the approximation
xN=x(h(i):h(i)+1);
err=1;
while(abs(err)>tol) % Iteratively improve aproximation
S=sign(g(xN)-f(xN));
hF=find(diff(S)~=0);
xN=xN(hF:hF+1);
[~,I]=min(abs(f(xN)-g(xN)));
xG=xN(I);
err=f(xG)-g(xG);
xN=linspace(xN(1),xN(2),15);
end
Sol(i)=xG;
Err(i)=f(xG)-g(xG);
end
else % No crossover points - lines could meet at tangents
[h,I]=findpeaks(-abs(g(x)-f(x)));
Sol=x(I(abs(f(x(I))-g(x(I)))<1e-5));
Err=f(Sol)-g(Sol)
end
% We also have to check each endpoint
if abs(f(x(end))-g(x(end)))<tol && abs(Sol(end)-x(end))>1e-12
Sol=[Sol x(end)];
Err=[Err g(x(end))-f(x(end))];
end
if abs(f(x(1))-g(x(1)))<tol && abs(Sol(1)-x(1))>1e-12
Sol=[x(1) Sol];
Err=[g(x(1))-f(x(1)) Err];
end
toc()
Sol
Err
This will "zoom" in to the region around each suspected root, and iteratively improve the accuracy. You can tweak the parameters to see whether they give better behaviour (the tolerance tol, the 15, number of new points to generate, could be higher probably).
Question 2
You would probably be best off avoiding ezplot, and using plot, which gives you greater control. You can vectorise inline functions so that you can evaluate them like anonymous functions, as I did in the previous code snippet, using
f=inline('x^2')
F=vectorize(f)
F(1:5)
and this should make plotting much easier:
plot(x,f(x),'b',Sol,f(Sol),'ro',x,g(x),'k',Sol,G(Sol),'ro')
Question 3
I'm not sure why you don't want to display your roots as strings, what's wrong with this:
text(xPos,yPos,['x0=' num2str(Sol(1))]);
In Matlab, I've got several points in the 3D space. Those points represent a rope and I would like to draw a line linking all those points. Here is my problem: How the oraganization of those points should be done to have a "simple" and "more or less straigth line". In other words I would like to draw a line linking all the points from the first to the last one but not "going back". Maybe with a simple image i can explain better my problem:
This is what the code should do:
This is what the code shouldn't do:
Some idea of how can I achieve the intended result? How can I organize the points? I'm working with Matlab but if you know any paper where I can read how to do this it will be fine. Thank you.
If you just don't want to go back in the upper direction, the solution that #Dan suggested should be fine. Sort them in that direction before connecting.
However, if that is not a requirement but you are actually looking for a solution without ugly crossings that is as short as possible, you may want to look into solutions for the travelling salesman problem.
If you define the distance between 1 and 9 to be zero, the solution you find to the travelling salesman problem should give you a nice and short rope.
If you want to go for the TSP approach but get a little lost, just try a 'furthest insertion' start position or '2 opt' improvements as that is likely going to give a big improvement over random path selection already with minimal effort.
OK. I've found the right answer (by Gunther Struyf): Sort Coordinates Points in Matlab
In order to visualize the final result just add to Gunther's code the following lines:
row=data(:,1);
col=data(:,2);
figure
scatter(row,col,'r');
hold on
line(row(result),col(result));
The following code shows the same algorithm but for a 3D implementation:
close all;
data = [2,2, 2; 2,3,2 ; 1,2,3 ; 1,3,3 ; 2,1,3 ; 1,1,3 ; 3,2,4 ; 3,3,4 ; 3,1,5; 5,4,6; 7,3,8; 8,9,7; 9,4,7; 6,2,5; 5,8,6; 9,3,8;6,9,2];
row=data(:,1);
col=data(:,2);
frame=data(:,3);
dist = pdist2(data,data);
N = size(data,1);
result = NaN(1,N);
result(1) = 1; % first point is first row in data matrix
for ii=2:N
dist(:,result(ii-1)) = Inf;
[~, closest_idx] = min(dist(result(ii-1),:));
result(ii) = closest_idx;
end
figure
%Ploting the points directly (without sorting)
subplot(2,1,1);
scatter3(row,col,frame,'r');
hold on;
line(row,col,frame);
%Ploting after sorting
subplot(2,1,2);
scatter3(row,col,frame,'r');
hold on
line(row(result),col(result),frame(result));
Hey, I've got a problem plotting a function in Matlab.
I first run this:
format long
f = inline('-x.^2');
for i = 0:10
[I(i+1) h(i+1) tid(i+1)] = trapets(f,0,1,2^i);
end
trunk = I - log(2);
hold on
grid on
plot(log(h),log(trunk),'r+')
t = -7:0;
c = polyfit(log(h),log(trunk),1);
yy = polyval(c,t);
plot(t,yy)
grid off
hold off
koefficienter = real(c)
and after that I run this file:
hold on
plot(h,trunk,'r+:','linewidth',2)
axis([0 0.6 0 0.0014])
Thing is, I don't get any errors, and the plot windows pops up with axes and all, but there is no graph to be found. It's just an empty window with two axes.
Anyone got any ideas?
Edit:
Okay, so I'm new to this site and couldn't find the reply button, so I add a reply here instead.
#woodchips :
I just realized that I hadn't given you all the information for this problem.. Sorry about that, anyhow I would really appreciate it if someone had the time to help me with this, it would seriously make my week.
This is the part I accidentally left out:
function [ I,h,tid ] = trapets(
f,a,b,n )
h=(b-a)/n;
tic; I=(f(a)+f(b));
for k=2:2:n-2
I = I+2*f(a+k*h);
end
for k = 1:2:n-1
I = I + 4*f(a+k*h);
end
I = I * h/3;
tid = toc;
end
Edit 2: So, I think that the graph I'm seeking is actually getting plotted in the first code that I wrote, the problem is that the variabe 'I' is not changing, which I expect it to do, although the variabels 'n' and 'h' do change. If 'I' was working correctly, I would probably get the right graph (hopefully). Any ideas, anyone?
Unfortunately the home computer I had with Matlab on it died the other day so I can't test anything. First thing I can think of if to simply run step by step through the code and see if the results of the math are what you are expecting. For instance Matlab was primarily made and runs as a matrix calculator if I recall correctly. As such most of the simple math doesn't function as it would punching it in a calculator. An example would be that 2^i needs to be 2.^i to operate correctly in some cases. Same with .* and ./ to use the singular scalar verses the matrices math.
The best way to find out what is going wrong is to iterate through the math a few times to ensure that it is being performed as expected. Once that is verified then you can move on to looking at plotting formatting.