Re: Empirically solving complex problems

There's an easier way. Integrals are additive. The integral of the sum is equal to the sum of the integrals.

Abstraction is not useless.

After Compline,
Zaxo

Comment on Re: Empirically solving complex problems

Replies are listed 'Best First'.
Re^2: Empirically solving complex problems by spurperl (Priest) on Mar 05, 2005 at 07:50 UTC
Hmm... If he wants to find the area of a union between two graphs, what you say is not entirely applicable. The area under the union is NOT the areas of each of the graphs summed, because there are overlapping portions that will be added twice. What you said is true, if g is one function and f is another, then I(f+g) = I(f) + I(g) but he needs a union - not a sum. And for linear functions, F(union of A and B) = F(A) + F(B) - F(intersection of A and B)	[reply]

Replies are listed 'Best First'.

Re^2: Empirically solving complex problems
by spurperl (Priest) on Mar 05, 2005 at 07:50 UTC

What you said is true, if g is one function and f is another, then I(f+g) = I(f) + I(g) but he needs a union - not a sum. And for linear functions, F(union of A and B) = F(A) + F(B) - F(intersection of A and B)

[reply]