new lag 5_4_6 on min max

scraps Derive Equations (6.4.6) This is amazingly tedious to derive, it took me half of a day to get all the algebra right! Start with ε1 < ε2 < ε3 . and recall u = U/M and ε1 ≤ u ≤ ε3 . (5.4.3) We regard both N1 and N2 as functions of N3 where, according to (5.3.1), N1 = + (ε2- ε1)-1[ ε2M - U + (ε3- ε2)N3] = N1(N3) N2 = - (ε2- ε1)-1[ ε1M - U + (ε3 - ε1)N3] = N2(N3) . (5.3.3) (5.4.4) It seems clear that one must have 0 ≤ N1 ≤ M 0 ≤ N2 ≤ M . (5.4.5) In my original doc, I claim that these 4 inequalities can be summarized in this way N3min ≤ N3 ≤ N3max N3min = max(M, 0) N3max = min( M, M) = M . // since u ≤ ε3 (5.4.6) I want now to verify this claim. Look at each of these four inequalities 0 ≤ N1 : 0 ≤ (ε2- ε1)-1[ ε2M - U + (ε3- ε2)N3] 0 ≤ [ ε2M - U + (ε3- ε2)N3] U - ε2M ≤ (ε3- ε2)N3 N3 ≥ (U-ε2M)/(ε3- ε2) = M (u-ε2)/(ε3- ε2) one form of N3min (1) 0 ≤ N2 : 0 ≤ - (ε2- ε1)-1[ ε1M - U + (ε3 - ε1)N3] 0 ≥ (ε2- ε1)-1[ ε1M - U + (ε3 - ε1)N3] 0 ≥ [ ε1M - U + (ε3 - ε1)N3] (ε3 - ε1)N3 ≤ U - ε1M N3 ≤ (U - ε1M) / (ε3- ε1) N3 ≤ M (u - ε1) / (ε3- ε1) one form of N3max (2) N1 ≤ M : (ε2- ε1)-1[ ε2M - U + (ε3- ε2)N3] ≤ M [ ε2M - U + (ε3- ε2)N3] ≤ (ε2-ε1)M (ε3- ε2)N3 ≤ (ε2-ε1)M + U - ε2M = - ε1M + U = (u-ε1)M N3 ≤ (u-ε1)M/(ε3- ε2) one form for N3 max (3) N2 ≤ M : - (ε2- ε1)-1[ ε1M - U + (ε3 - ε1)N3] ≤ M (ε2- ε1)-1[ ε1M - U + (ε3 - ε1)N3] ≥ -M [ ε1M - U + (ε3 - ε1)N3] ≥ -(ε2- ε1)M (ε3 - ε1)N3 ≥ - (ε2- ε1)M - ε1M + U = (-ε2)M + U = [u -ε2]M N3 ≥ [u - ε2]M/ (ε3 - ε1) one form for N3min (4) So my four inequalities result in : N3 ≥ M (u-ε2)/(ε3- ε2) (1) N3 ≥ M (u-ε2)/(ε3 - ε1) (4) and N3 ≤ M (u-ε1)/(ε3- ε1) (2) N3 ≤ M (u-ε1)/(ε3- ε2) (3) So it would seem that N3min = M max [ (u-ε2)/(ε3- ε2), (u-ε2)/(ε3- ε1) ] N3max = M min [ (u-ε1)/(ε3- ε2), (u-ε1)/(ε3- ε1) ] Now in each line, the second denominator is larger than the first, so the second term is smaller than the first. In the first line we therefore select the first term. In the second line we therefore select the second term term. Therefore N3min = M (u-ε2)/(ε3- ε2) but also need N3min > 0 N3max = M (u-ε1)/(ε3- ε1) but also need N3max < M So final result is N3min = M max[ (u-ε2)/(ε3- ε2), 0 ] N3max = M min[ (u-ε1)/(ε3- ε1), 1 ] . But since u ≤ ε3 the first fraction is always ≤ 1, so it always wins in the min, and then N3min = M max[ (u-ε2)/(ε3- ε2), 0 ] N3max = M (u-ε1)/(ε3- ε1) This replicates my original Lagrange doc results (6.4.6) Note: This result depends on the fact that we assumed ε1 < ε2 < ε3. What happens if you treat N1 or N2 as the independent variable? In the above, I started with N1 = + (ε2- ε1)-1[ ε2M - U + (ε3- ε2)N3] = N1(N3) N2 = - (ε2- ε1)-1[ ε1M - U + (ε3 - ε1)N3] = N2(N3) 0 ≤ N1 ≤ M 0 ≤ N2 ≤ M where N3 is the independent variable. Suppose more generally I want to have Na and Nb be the two dependent variables. I then would start with Na = + (εb- εa)-1[ εbM - U + Σi≠a,b (εi- εb)Ni] = Na(Ni≠a,b) Nb = - (εb- εa)-1[ εaM - U + Σi≠a,b (εi - εa)Ni] = Nb(Ni≠a,b) (5.3.18a) But if there are only 3 items, I can call the last one "c" and these then read Na = + (εb- εa)-1[ εbM - U + (εc- εb)Nc] = Na(Nc) Nb = - (εb- εa)-1[ εaM - U + (εc- εa)Nc] = Nb(Nc) (5.3.18a) 0 ≤ Na ≤ M 0 ≤ Nb ≤ M This then is the system I would have to analyze. In general we don't know the sign of any ε-ε differences, so I think this general case is not each. Let's write the equations for the three cases abc = 123 : N1 = + (ε2- ε1)-1[ ε2M - U + (ε3- ε2)Nc] = N1(N3) N2 = - (ε2- ε1)-1[ ε1M - U + (ε3- ε1)Nc] = N2(N3) (5.3.18a) 0 ≤ N1 ≤ M 0 ≤ N2 ≤ M // this does replicate (5.3.3) abc = 132 : N1 = + (ε3- ε1)-1[ ε3M - U + (ε2- ε3)N2] = N1(N2) N3 = - (ε3- ε1)-1[ ε1M - U + (ε2- ε1)N2] = N3(N2) (5.3.18a) 0 ≤ N1 ≤ M 0 ≤ N3 ≤ M