Multivariable Advanced Calculus

6.11. TAYLOR’S FORMULA 143where (x, y) → G 1 (x, y) is C k−1 because it is assumed f is C k and one derivative hasbeen taken to write the above. If k ≥ 2, then another Gateaux derivative can be taken.D wj w kx i (y) ≡G 1 (x (y + tw k ) , y + tw k ) − G 1 (x (y) , y)limt→0 t= D 1 G 1 (x (y) , y) Dx (y) w k + D 2 G 1 (x (y) , y)≡G 2 (x (y) , y,Dx (y))Since a similar result holds for all i and any choice of w j , w k , this shows x is at leastC 2 . If k ≥ 3, then another Gateaux derivative can be taken because then (x, y, z) →G 2 (x, y, z) is C 1 and it has been established Dx is C 1 . Continuing this way showsD wj1 w j2 ···w jrx i (y) exists and is continuous for r ≤ k. This proves the following corollaryto the implicit and inverse function theorems.Corollary 6.10.5 In the implicit and inverse function theorems, you can replaceC 1 with C k in the statements of the theorems for any k ∈ N.6.10.2 The Case Of R nIn many applications of the implicit function theorem,f : U ⊆ R n × R m → R nand f (x 0 , y 0 ) = 0 while f is C 1 . How can you recognize the condition of the implicitfunction theorem which says D 1 f (x 0 , y 0 ) −1 exists? This is really not hard. You recallthe matrix of the transformation D 1 f (x 0 , y 0 ) with respect to the usual basis vectors is⎛⎜⎝f 1,x1 (x 0 , y 0 ) · · · f 1,xn (x 0 , y 0 )..f n,x1 (x 0 , y 0 ) · · · f n,xn (x 0 , y 0 )and so D 1 f (x 0 , y 0 ) −1 exists exactly when the determinant of the above matrix isnonzero. This is the condition to check. In the general case, you just need to verifyD 1 f (x 0 , y 0 ) is one to one and this can also be accomplished by looking at the matrixof the transformation with respect to some bases on X and Z.⎞⎟⎠6.11 Taylor’s FormulaFirst recall the Taylor formula with the Lagrange form of the remainder. It will onlybe needed on [0, 1] so that is what I will show.Theorem 6.11.1 Let h : [0, 1] → R have m + 1 derivatives. Then there existst ∈ (0, 1) such thath (1) = h (0) +m∑k=1h (k) (0)k!+ h(m+1) (t)(m + 1)! .Proof:Let K be a number chosen such that()m∑ h (k) (0)h (1) − h (0) ++ K = 0k!k=1