By Vinogradov V.
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Extra resources for A cookbook of mathematics
Recipe 13 – What are the Necessary Conditions for the Solution of (4)? Equate all partials of L with respect to x1 , . . , xn , λ1 , . . , λm to zero: m ∂L(x1 , . . , xn , λ1 , . . , λm ) ∂f (x1 , . . , xn ) ∂g j (x1 , . . , xn ) = − λj = 0, ∂xi ∂xi ∂xi j=1 ∂L(x1 , . . , xn , λ1 , . . , λm ) = bj − g j (x1 , . . , xn ) = 0, ∂λj i = 1, 2, . . , n, j = 1, 2, . . , m. Solve these equations for x1 , . . , xn and λ1 , . . , λm . In the end we will get a set of stationary points of the Lagrangian.
Xn ) = lim = lim ∂xi ∆xi →0 ∆xi ∆xi →0 ∆xi for all i = 1, 2, . . , n. If these limits exist, our function is differentiable with respect to all ∂f arguments. g. ∂x = fxi (= fxi ). i 32 Example 54 If y = 4x31 + x1 x2 + ln x2 then ∂y ∂x1 = 12x21 + x2 , ∂y ∂x2 = x1 + 1/x2 . Definition 38 The total differential of a function f is defined as dy = ∂f ∂f ∂f dx1 + dx2 + . . + dxn . ∂x1 ∂x2 ∂xn The total differential can be also used as an approximation device. For instance, if y = f (x1 , x2 ) then by definition y − y0 = ∆f (x01 , x02 ) ≈ df (x01 , x02 ).
Xn , y1 , . . , ym ) = 0 2 f (x1 , . . , xn , y1 , . . , ym ) = 0 ... ... m f (x1 , . . , xn , y1 , . . , ym ) = 0 (3) ∂yj , i = 1, . . , n, j = 1, . . , m. ∂xi If we take the total differentials of all f j , we get Our objective is to find ∂f 1 ∂f 1 dy1 + . . + dym ∂y1 ∂ym ∂f m ∂f m dy1 + . . + dym ∂y1 ∂ym ∂f 1 ∂f 1 = − dx1 + . . + dxn , ∂x1 ∂xn ... ∂f m ∂f m = − dx1 + . . + dxn . e. dxi = 0, dxl = 0, l = 1, 2, . . , i−1, i+1, . . , n) and divide each remaining term in the system above by dxi .
A cookbook of mathematics by Vinogradov V.