How do you find the intersection points of two functions?
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My problem tells me to plot and then find and print the points of intersection for x=[2:7]. I already sought help and they explained that I should use 'find' and then the '==' to find where the output match. Sounded simple enough but I'm getting "Empty Matrix: 1-by-0" as the answer. Any one know what I'm doing wrong or a better way?
Also, any idea how I'd then go about using a "for - while" loop to find every intersection? I'm taking it one problem at a time though.
Here's my script so far:
% clear all windows and variables
clc
% Declare an array for the values of x
x=[2:7];
% Input the given functions
f1=@(x) 90.*exp(-x);
f2=@(x) 3*sin(2*pi*x);
% Graph the functions in the same window
fplot(f1,[2,7],'b')
hold on
fplot(f2,[2,7],'r')
grid on
title('Finding Intersections of Functions')
xlabel('Input Values (x)')
ylabel('Ouput Values (f)')
% Find the first intersection
f1a=90.*exp(-x);
f2a=3*sin(2*pi*x);
find(f1a==f2a)
2 个评论
ankit kumar
2018-4-5
is there anyway i can give a txt file including x and y axis value and can find how many times a curve is crossing through a threshold and also to note down the x values.
Elina Nikolopoulou
2021-12-28
How do i keep the Y and the X coordinates of the intersections in separate arrays (one for the X and one for the Y coordinates) ???
回答(3 个)
Andrei Bobrov
2013-11-7
编辑:Andrei Bobrov
2013-11-7
EDIT
f1=@(x) 90.*exp(-x);
f2=@(x) 3*sin(2*pi*x);
f = @(x)f1(x)-f2(x);
xx = 2:.1:7;
t = f(xx) > 0;
i0 = find(diff(t(:))~=0);
i0 = [i0(:)';i0(:)'+1];
n = size(i0,2);
xout = zeros(n,1);
for jj = 1:n
xout(jj) = fzero(f,xx(i0(:,jj)));
end
2 个评论
Mathijs Frenken
2018-11-27
Thank you for your answer. However, you realize you are not the only one that reads your code right? It's actually unreadable.
Alexander Efremov
2013-11-7
编辑:Alexander Efremov
2013-11-7
There is a fancy contribution on "File Exchange" which appeared in the Pick of the week. This should help.
1 个评论
Syed Riza
2016-7-31
编辑:Walter Roberson
2016-7-31
Hello Dan Teep; you can find your answer in this modified code
% Declare an array for the values of x
x=linspace(2,7,1200);
% Input the given functions
f1=@(x) 90.*exp(-x);
f2=@(x) 3*sin(2*pi*x);
% Graph the functions in the same window
fplot(f1,[2,7],'b')
hold on
fplot(f2,[2,7],'r')
grid on
title('Finding Intersections of Functions')
xlabel('Input Values (x)')
ylabel('Ouput Values (f)')
% Find the x-cordinates of intersecting points
f1a=90.*exp(-x);
f2a=3*sin(2*pi*x);
Intersections=find(abs(f1a-f2a)<=(0.05));
X_Values=x(Intersections)
2 个评论
Maya Priveetra
2017-3-9
Hi can I know why you (abs(fla-f2a)<=(0.05))
Like what does it do especially the 0.05
Adeel Yousuf
2019-2-13
He has used a tolerance value of 0.05 as the dataset contains floating point values. We can keep tolerance as small as it seems suitable (like: 1e-4). It's meant to "detect" the closeness of 2 floating point values. Moreover in simple words, we need to make MATLAB determine if 60.2745 is the same as 60.274 or not? So for that we can use threshold/tolerance of 0.001.
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