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VBBV on 6 Dec 2022
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Commented: Jan on 10 Dec 2022
Accepted Answer: Star Strider
I tried the following code to use intersect function available in Matlab f=@(x)1+cos(x); g=@(x)sin(x); x=-8:.01:8; A = intersect(f(x),g(x)) % this returns empty plot(x,f(x),x,g(x)); but the intersect function returns empty, even though it appears that there are common values for both functions defined over same x-interval. Can anyone clarify this ?
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Stephen23 on 6 Dec 2022
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Edited: Stephen23 on 6 Dec 2022
"Can anyone clarify this ?"
The two arrays do not share any data points.
"How can i use intersect function correctly ?"
To do what exactly?:
- return common data points from two sets of data (just as the function is documented)
- find intersections of two functions (perhaps what you really want to achieve)
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Accepted Answer
Star Strider on 6 Dec 2022
Use calls to interp1 in a loop to get each intersection —
f=@(x)1+cos(x);
g=@(x)sin(x);
fg = @(x) f(x)-g(x);
x=-8:.01:8;
A = intersect(f(x),g(x)) % this returns empty
A = 1×0 empty double row vector
ixx = find(diff(sign(fg(x)))) % Approximate Indices Of Intersection
ixx = 1×5
329 486 958 1115 1586
for k = 1:numel(ixx)
idxrng = max(1,ixx(k)-1) : min (numel(x),ixx(k)+1); % Index Range
xv(k) = interp1(fg(x(idxrng)),x(idxrng),0); % X-Coordinate
yv(k) = f(xv(k)); % Y-Coordinate
end
Intersections = [xv; yv]
Intersections = 2×5
-4.7124 -3.1416 1.5708 3.1416 7.8540 1.0000 0.0000 1.0000 0.0000 1.0000
figure
plot(x,f(x),x,g(x), 'DisplayName','Function');
hold on
plot(xv, yv, 'sr', 'DisplayName','Intersections')
hold off
legend('Location','best')
Using the loop and the ‘idxrng’ vector avoids problems with non-monotonic regions of the functions.
.
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VBBV on 6 Dec 2022
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Thank you for the Nice explanation. are there any builtin function/s available in Matlab which can detect the region of crossovers for curves that overlap?
Star Strider on 6 Dec 2022
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As always, my pleasure!
Not that I’m aware of.
The best it is possible to do would be to put something like this into its own function. This version works with vectors, so the vectors would have to be passed to it.
A different version could also work with funcitons, however that would have to be a separate function.
It might be possible to use one function for both vectors and functions, however it would require internal logic to evaluate them and use them correctly.
Using this version as a function —
f=@(x)1+cos(x);
g=@(x)sin(x);
x=-8:.01:8;
fx = f(x);
gx = g(x);
[xv,yv] = functionIntersections(fx,gx,x)
xv = 1×5
-4.7124 -3.1416 1.5708 3.1416 7.8540
yv = 1×5
1.0000 0.0000 1.0000 0.0000 1.0000
function [xv,yv] = functionIntersections(f,g,x)
ixx = find(diff(sign(f-g)));
if isempty(ixx)
error('No intersections exist between these two functions.')
return
end
for k = 1:numel(ixx)
idxrng = max(1,ixx(k)-1) : min (numel(x),ixx(k)+1); % Index Range
xv(k) = interp1(f(idxrng)-g(idxrng),x(idxrng),0); % X-Coordinate
yv(k) = interp1(x(idxrng),f(idxrng),xv(k)); % Y-Coordinate % Y-Coordinate
end
end
There are File Exchange contributions that can handle much more extensive sets of line intersections. This one works for relatively straightforward problems of two functions. I have incorporated logic in it for failed intersections (the two function arguments having no intersections).
.
Jan on 10 Dec 2022
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Open in MATLAB Online
@Star Strider: The accuracy of the interpolation method is limited:
f = @(x)1+cos(x);
g = @(x)sin(x);
x = -8:.01:8;
[xv,yv] = functionIntersections(f(x), g(x), x)
xv = 1×5
-4.7124 -3.1416 1.5708 3.1416 7.8540
yv = 1×5
1.0000 0.0000 1.0000 0.0000 1.0000
format long g
xv(3) - pi/2
ans =
3.69037267589079e-06
This is rough, but maybe sufficient for a graphical display. But if the points of the polygon is all you have and the functions f() and g() are not available, your linear interpolation method is trustworthy. Including further assumptions like cubic or spline interpolations can increase and decrease the accuracy, so this is a fragile option.
If f() and g() are available, this interpolation method ist a good method to find start points for e.g. fzero().
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More Answers (1)
Jan on 6 Dec 2022
Edited: Jan on 6 Dec 2022
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There are no overlapping points:
f = @(x) 1 + cos(x);
g = @(x) sin(x);
x = 1.54:.01:1.6;
plot(x, f(x), 'r.', x, g(x), 'b.');
The same matters the other points, where the graphs intersect, but not at any of the fixed raster points x=-8:.01:8. A real intersection:
format long g
xi = fzero(@(x) f(x) - g(x), [1.54, 1.6])
xi =
1.5707963267949
A symbolic solution of the intersections:
syms x
eq = 1 + cos(x) - sin(x) == 0
eq=
solve(eq)
ans=
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VBBV on 6 Dec 2022
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Thank you for the Nice explanation. are there any builtin function/s available in Matlab which can detect the region of crossovers for curves that overlap?
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