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The curvature calculation PhL 10.14.16 I have this all cleaned up and installed into new Lagrange. The result is the same as my original result. Keeping this only for the record, no need to read what is below. Here is how I start off: _________________________________________________________ 5.3 More details of the solution Start with f and its partial derivatives, f = lnΩ = Σi [(1+lngi)Ni - NilnNi ] = Σi Ni [1 + ln(gi/Ni)] (5.2.3) fi = ln(gi/Ni) // fi ≡ ∂f/∂Ni (5.2.6) The constraints Σi=1mNi= M and Σi=1Niεi= U can be regarded as two equation in the two unknowns N1 and N2 which are easily solved to obtain, N1 = + (ε2- ε1)-1[ ε2M - U + Σi=3m (εi- ε2)Ni] = N1(N3,...Nm) N2 = - (ε2- ε1)-1[ ε1M - U + Σi=3m (εi - ε1)Ni] = N2(N3,...Nm) . (5.3.1) Notice that, = + (ε2- ε1)-1 [ (εi- ε2) ] = i = 3,4..m = - (ε2- ε1)-1 [ (εi- ε1) ] = - . i = 3,4..m (5.3.2) Now, using (5.3.1), think of f as function F of m-2 independent variables of N3,,,,Nm as follows, F(N3,N4.....Nm) ≡ f(N1(N3,...Nm), N2(N3,...Nm), N3, .....Nm) . (5.3.3) F has the same value as f but has a different functional form. ________________________________________________________________________________ What are the implications of this last definition? How do you rewrite earlier equations in terms of F? F(N3,N4.....Nm) = N1(N3,...Nm) [ 1 + ln(g1) - lnN1(N3,...Nm)] + N2(N3,...Nm) [ 1 + ln(g2) - lnN1(N3,...Nm)] + Σj=3m Nj [1 + ln(gj) - lnNj] where now I expose functional dependency. Now derivatives are very complicated. Letting i = 3,4.... get = [ 1 + ln(g1) - lnN1(N3,...Nm)] - N1(N3,...Nm) [ 1/N1(N3,...Nm)] + [ 1 + ln(g2) - lnN2(N3,...Nm)] - N2(N3,...Nm) [ 1/N2(N3,...Nm)] + Σj=3m { δi,j [ 1 + ln(gj) - lnNj] - Nj [ 1/Nj] δi,j } I can compact this down a bit = [ 1 + ln(g1) - lnN1(N3,...Nm)] - + [ 1 + ln(g2) - lnN2(N3,...Nm)] - + { [ 1 + ln(gi) - lnNi] - 1 } and then still more = [ 0 + ln(g1) - lnN1(N3,...Nm)] + [ 0 + ln(g2) - lnN2(N3,...Nm)] + [ 0 + ln(gi) - lnNi] i = 3,4.... Now what happens if I insert the two derivatives shown = [ 0 + ln(g1) - lnN1(N3,...Nm)] - [ 0 + ln(g2) - lnN2(N3,...Nm)] + [ 0 + ln(gi) - lnNi] i = 3,4.... which I rewrite as = [ 0 + ln(g1/N1(N3,...Nm))] - [ 0 + ln(g2/N2(N3,...Nm))] + [ 0 + ln(gi/Ni)] Now recall this earlier result ln(Ni/gi) = λ1+ λ2εi i = 1,2....m (5.2.9) This is valid only at the solution Ni point, and it is valid for all values of i. So insert into the above = [ 0 - λ1- λ2ε1] - [ 0 - λ1- λ2ε2] +[ 0 - λ1- λ2εi] = λ1 [ - + - 1 ] + λ2 [ - ε1 + ε2 - εi ] + [ 0 - 0 + 0 ] Now add a factor 1 = into the last term in each square bracket, = λ1 [ - + - ] + λ2 [ - ε1 + ε2 - εi ] + [ 0 - 0 + 0 ] = [ - (εi-ε2) + (εi-ε1) - (ε2-ε1) ] + [ -(εi-ε2)ε1 + (εi-ε1)ε2 - (ε2-ε1)εi ] + [ (εi-ε2) - (εi-ε1) + (ε2-ε1)] 0 = [ - εi+ε2 + εi- ε1 - ε2+ε1) ] + [ -εiε1 +ε2ε1 + εiε2-ε1ε2 - ε2εi+ε1εi ] + [ εi-ε2 - εi+ε1 + ε2-ε1] 0 = [ 0 ] + [ ] + [ 0] 0 = 0 Wow. This finally confirms what I want which is that the partial vanishes at the stationary point solution N. Now we can go on to attempt the second derivative starting with this result valid for i = 3,4.... = [ 2 + ln(g1) - lnN1(N3,...Nm)] - [ 2 + ln(g2) - lnN2(N3,...Nm)] + [ 2 + ln(gi) - lnNi] i = 3,4.... Then : = [ -1/N1 * ] - [ -1/N2 * ] + [ -1/Ni ] Now insert (5.3.2) to get = [ -1/N1 * ] - [ + 1/N2 * ] + [ -1/Ni ] = – ( )2 /N1 – ( )2 /N2 - 1/Ni - [ ( )2 /N1 + ( )2 /N2 + 1/Ni ] and this is exactly the result I got in the original Lagrange doc!!! But this time I have shown an actual derivation, not just some words. _______________________________________________________ Try to compact down the algebra shown abovwe F(N3,N4.....Nm) = N1[ 1 + ln(g1) - lnN1] + N2 [ 1 + ln(g2) - lnN2] + Σj=3m Nj [1 + ln(gj) - lnNj] where now I expose functional dependency. Now derivatives are very complicated. Letting i = 3,4.... get = [ 1 + ln(g1) - lnN1] + N1 [ 1/N1] + [ 1 + ln(g2) - lnN2] + N2 [ 1/N2] + Σj=3m { δi,j [ 1 + ln(gj) - lnNj] + Nj [ 1/Nj] δi,j } = [ 2 + ln(g1) - lnN1] + [ 2 + ln(g2) - lnN2] + [ 2 + ln(gi) - lnNi] = [ 2 - ln(N1/g1)] - [ 2 - ln(N2/g)] + [ 2 - ln(Ni/gi) ] // (5.3.2) As a check on the algebra, we evaluate the above expression at the stationary point N where we know that ln(Ni/gi) = λ1+ λ2εi (6.2.9) Then at the stationary point we find (adding 1 = to the third term) = [ 2 - λ1- λ2ε1] - [ 2 - λ1- λ2ε2] + [ 2 - λ1- λ2εi] = { (εi- ε2) [ 2 - λ1- λ2ε1] - (εi- ε1) [ 2 - λ1- λ2ε2] + (ε2- ε1)[ 2 - λ1- λ2εi] } Inside the Curly Bracket all terms cancel, as Maple verifies with the final result = 0 i = 3,4...m Now compact down the second derivative algebra = [ 0 + ln(g1) - lnN1(N3,...Nm)] - [ 0 + ln(g2) - lnN2(N3,...Nm)] + [ 0 + ln(gi) - lnNi] i = 3,4.... Then : = [ -1/N1 * ] - [ -1/N2 * ] + [ -1/Ni ] Now insert (5.3.2) to get = [ -1/N1 * ] - [ + 1/N2 * ] + [ -1/Ni ] = – ( )2 /N1 – ( )2 /N2 - 1/Ni - [ ( )2 /N1 + ( )2 /N2 + 1/Ni ] I can compact this down a bit = [ 1 + ln(g1) - lnN1(N3,...Nm)] + + [ 1 + ln(g2) - lnN2(N3,...Nm)] +