CODEWITHSUNDEEP
matlabintersection
scadda
scadda

Reputation: 31

How to find the intersections of two functions in MATLAB?

Lets say, I have a function 'x' and a function '2sin(x)'

How do I output the intersects, i.e. the roots in MATLAB? I can easily plot the two functions and find them that way but surely there must exist an absolute way of doing this.

Upvotes: 3

Views: 12514

Answers (2)

Daniel
Daniel

Reputation: 36710

When writing the comments, I thought that 

syms x; solve(x==2*sin(x))

would return the expected result. At least in Matlab 2013b solve fails to find a analytic solution for this problem, falling back to a numeric solver only returning one solution, 0.

An alternative is

s = feval(symengine,'numeric::solve',2*sin(x)==x,x,'AllRealRoots')

which is taken from this answer to a similar question. Besides using AllRealRoots you could use a numeric solver, manually setting starting points which roughly match the values you have read from the graph. This wa you get precise results:

[fzero(@(x)f(x)-g(x),-2),fzero(@(x)f(x)-g(x),0),fzero(@(x)f(x)-g(x),2)]

For a higher precision you could switch from fzero to vpasolve, but fzero is probably sufficient and faster.

Upvotes: 3

Andras Deak -- Слава Україні
Andras Deak -- Слава Україні

Reputation: 35080

If you have two analytical (by which I mean symbolic) functions, you can define their difference and use fzero to find a zero, i.e. the root:

f = @(x) x;        %defines a function f(x)
g = @(x) 2*sin(x); %defines a function g(x)

%solve f==g
xroot = fzero(@(x)f(x)-g(x),0.5); %starts search from x==0.5

For tricky functions you might have to set a good starting point, and it will only find one solution even if there are multiple ones.

The constructs seen above @(x) something-with-x are called anonymous functions, and they can be extended to multivariate cases as well, like @(x,y) 3*x.*y+c assuming that c is a variable that has been assigned a value earlier.

Upvotes: 3

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