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CTK Exchange
japam
Member since Feb-15-04
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Feb-25-06, 07:01 PM (EST) |
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"rational function maximus"
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suppose you have two functions A(X)concave, and B(X)convexe,both interesect in just 2 points and you want to find the maxim of B(X)/A(X) (Sorry i dont know how to draw a diagram here, but hope is clear)Now suppose that the line that joins the 2 intersection points is CX D Probe or disprobe: I say that you can find the maximus calculating (MAX (CX +D)/A(X))*(MAX B(X)/(CX + D)) I made some tests with 2n degree equations and it works, but i dont know how to proof general ( if true) Regards JP |
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sfwc
Member since Jun-19-03
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Feb-26-06, 09:09 AM (EST) |
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1. "RE: rational function maximus"
In response to message #0
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>suppose you have two functions A(X)concave, and >B(X)convexe,both interesect in just 2 points and you want to >find the maxim of B(X)/A(X) >(Sorry i dont know how to draw a diagram here, but hope is >clear) > >Now suppose that the line that joins the 2 intersection >points is CX >D > >Probe or disprobe: I say that you can find the maximus >calculating (MAX (CX +D)/A(X))*(MAX B(X)/(CX + D)) I shall disprove the result, by giving a counterexample. Let A(X) be given by X for X >= 1 and by 3-2X for X < 1. Let B(X) be given by X + 3 for X < 2 and by 9 - 2X for X >= 2. The points of intersection are at (0, 3) and (3, 3). So C = 0, D = 3. Then Max((CX + D)/A(X)) = 3/Min(A(X)) = 3/1 = 3. Also, Max(B(X)/(CX+D)) is Max(B(X))/3 = 5/3. But Max((B(X)/A(X)) is achieved at X = 1, where it is 4 != 5 = 3 * 5/3.Thankyou sfwc <>< |
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