Please give full detailed math proof. Thank you. Suppose that f and g are r^th-o
ID: 3109604 • Letter: P
Question
Please give full detailed math proof. Thank you.
Suppose that f and g are r^th-order differentiable and that the composite h = g compositefunction f makes sense. A partition divides a set into nonempty disjoint subsets. Prove the Higher Order Chain Rule, (D^r h)_p = sigma^r_k = 1 sigma_mu elementof P(k, r) (D^k g)_q compositefunction (D^mu f)_p where mu partitions {1,. .., r} into k subsets, and q = f(p). In terms of r-linear transformations, this notation means (D^r h)p(v_1,..., v_r) = sigma^r_k = 1 sigma_mu (D^k)_q((D^|mu_1| f)_p (v_mu_1),. .., (D^|mu_k| f)_p(v_mu_k)) where |mu_i| = #mu_i and v_mu_i is the |mu_i|-tuple of vectors v_j with j elementof mu_i. (Symmetry implies that the order of the vectors v_j in the |mu_i|-tuple v_mu i and the order in which the partition blocks mu_1,..., mu_k occur are irrelevant.)Explanation / Answer
Suppose f:RnRm is differentiable at a, and g:RmRp is differentiable at f(a).
This means that
(1) limh0|f(a+h)f(a)Df(a)(h)| / |h|=0,
and that
(2) limh0|g(f(a)+h)g(f(a))[Dg(f(a))](h)| / |h|=0.
Note then that
|(gf)(a+h)(gf)(a)[Dg(f(a))Df(a)](h)| /|h|=
|g(f(a+h))g(f(a))Dg(f(a))[Df(a)(h)]| /|h|=
|g(f(a+h))g(f(a))Dg(f(a))[f(a+h)f(a)f(a+h)+f(a)+Df(a)(h)]| /|h|
|g(f(a+h))g(f(a))Dg(f(a))[f(a+h)f(a)]|/|h|+|Dg(f(a))[f(a+h)f(a)Df(a)(h)]|/|h|.
But by (2), for every >0
, we can find >0 such that 0<|k|< implies that |g(f(a)+k)g(f(a))[Dg(f(a))](k)|<|k|; in particular, if 0<|f(a+h)f(a)|< for any h satisfying 0<|h|< for a suitable >0, then
|g(f(a+h))g(f(a))Dg(f(a))[f(a+h)f(a)]| /|h|<
|f(a+h)f(a)||h|= |f(a+h)f(a)Df(a)(h)+Df(a)(h)| / |h|
|f(a+h)f(a)Df(a)(h)|/|h|+|Df(a)(h)| /|h|.Now, since Df(a)
and Dg(f(a)) are linear transformations, we also have that for certain M,NR,
|Df(a)(h)|<M|h|
and
|Dg(f(a))[f(a+h)f(a)Df(a)(h)]|<N|f(a+h)f(a)Df(a)(h)|.
This in turn gives first that
|Df(a)(h)| /|h|<M,
and also, by using (1), that for every >0 we can find >0 such that |h|< implies that |f(a+h)f(a)Df(a)(h)|<|h|, from which we also obtain
|f(a+h)f(a)Df(a)(h)| /|h|< and
|Dg(f(a))[f(a+h)f(a)Df(a)(h)]|/|h|< N|f(a+h)f(a)Df(a)(h) /||h|< N|h|/ |h|=N.
Finally, piecing all of the above together (from the last three inequalities we can conclude that both terms on the right hand side of the original inequality are arbitrarily small) gives the conclusion:
limh0|(gf)(a+h)(gf)(a)[Dg(f(a))Df(a)](h)| /|h|=0.