Intersections between two discretised functions

Hi Everybody, I have the following problem:

I have two vectors:

• v1 = list of values representing a discretised non monotonic function
• v2 = list of values representing a constant and horizontal line

v1 has no analytical model. It's just a vector containing the values a function assumes in a given interval.

To make everything clearer, let's consider the following case:

• v1 = list of 100 values representing a parabola: y= x^2, with x=linspace(-5, +5, 100)
• v2 = ones(1,100)

f = @(x) x.^2;
values = linspace(-5, 5, 100);
v1 = f(values);
v2 = ones(1,100);

Clearly, the intersection points are two, for x1=-1 and x2=+1.

Now let's forget about the fact that we had the analytical expression for v1 and let's just consider the vector itself.

How can I implement this on Matalb for a generical vector v1 which is not monotonic and is supposed to have two intersection points with v2? I tried doing something with interp1 but couldn't figure it out.

Thanks a lot to whoever will help me!

[Walter Roberson]

If you look in the File Exchange you will find at least 4 contributions for finding curve intersections.

Answer 1

[Dyuman Joshi]

You can use FEX submissions for this, I have run these two submissions below - InterX, intersections as example -

You can also find many answers on this forum, which utilize interp1 to get intersection points of two curves.

x = linspace(-5, 5);
v1 = x.^2;
v2 = ones(size(x));

P1 = InterX([x;v1],[x;v2])

P1 = 2×2

-0.9988 0.9988 1.0000 1.0000

[x0, y0] = intersections(x, v1, x, v2)

x0 = 2×1

-0.9988 0.9988

y0 = 2×1

1 1

[EDOARDO GIOVANNI]

I didn't know of the existance of these submissions, thank you Dyuman.

[Dyuman Joshi]

You're Welcome!

Answer 2

[Matt J]

f = @(x) x.^2;
x = linspace(-5, 5, 100);
v1 = f(x);
v2 = ones(size(v1));

X0=x(diff(sign(v1-v2))~=0);
X=nan(size(X0));

for i=1:numel(X)
    
    X(i)=fzero(@(t)interp1(x,v1-v2,t,'cubic'),  X0(i));
    
end

X %solutions

X = 1×2

-1.0000 1.0000