Suppose f and g are function_handles, try to define f+g,f-g,f*g and f/g.
e.g.
if f = @(x)x and g = @(x)x+1
then
f+g = @(x)2*x+1
f-g = @(x)-1
f*g = @(x)x*(x+1)
f/g = @(x)x/(x+1)
brilliant solution! I used 'switch...case...' structure to judge 'type'.
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Chebyshev polynomials of the 2nd Kind
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Solution 696726
hardly read it.Never thought a function_handle can be created like this.