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Re: Empirically solving complex problems

by Zaxo (Archbishop)
on Mar 05, 2005 at 06:53 UTC ( [id://436865]=note: print w/replies, xml ) Need Help??


in reply to 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

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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)

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