6.5.E: Problems on Repeated Differentiation and Taylor Expansions
( \newcommand{\kernel}{\mathrm{null}\,}\)
Complete all details in the proof of Theorem 1. What is the motivation for introducing the auxiliary functions ht and gs in this particular way?
Is symbolic "multiplication" in Note 2 always commutative? (See Example (A).) Why was it possible to collect "similar" terms
∂2f∂x∂ydxdy and ∂2f∂y∂xdydx
in Example (B)? Using (5), find the general formula for d3f. Expand it!
Carry out the induction in Theorem 2 and Corollary 2. (Use a suitable notation for subscripts: k1k2… instead of jk….)
Do Example(C) with m=3 (instead of m=2) and with →p=(0,0). Show that Rm→0, i.e., f admits a Taylor series about →p.
Do it in the following two ways.
(i) Use Theorem 2.
(ii) Expand siny as in Problem 6(a) in Chapter 5, §6, and then multiply termwise by x.
Give an estimate for R3.
Use Theorem 2 to expand the following functions in powers of x−3 and y+2 exactly (choosing m so that Rm=0).
(i) f(x,y)=2xy2−3y3+yx2−x3;
(ii) f(x,y)=x4−x3y2+2xy−1;
(iii) f(x,y)=x5y−axy5−x3.
For the functions of Problem 15 in §4, give their Taylor expansions up to R2, with
→p=(1,π4,1)
in case (i) and
→p=(e,π4e)
in (ii). Bound R2.
(Generalized Taylor theorem.) Let →u=→x−→p≠→0 in E′(E′ need not be En or Cn); let I=L[→p,→x]. Prove the following statement:
If f:E′→E and the derived functions Di→uf(i≤m) are relatively continuous on I and have →u-directed derivatives on I−Q (Q countable), then formula (6) and Note 3 hold, with dif(→p;→u) replaced by Di→uf(→p).
[Hint: Proceed as in Theorem 2 without using the chain rule or any partials or components. Instead of (8), prove that h(i)(t)=Di→uf(→p+t→u) on J−Q′,Q′=g−1[Q].]
(i) Modify Problem 7 by setting
→u=→x−→p|→x−→p|.
Thus expand f(→x) in powers of |→x−→p|.
(ii) Deduce Theorem 2 from Problem 7, using Corollary 2.
Given f:E2(C2)→E,f∈CDm on an open set A, and →s∈A, prove that (∀→u∈E2(C2))
dif(→s;→u)=i∑j=0(ij)uj1ui−j2Dk1…kif(→s),1≤i≤m,
where the (ij) are binomial coefficients, and in the jth term,
k1=k2=⋯=kj=2
and
kj+1=⋯=ki=1.
Then restate formula (6) for n=2.
[Hint: Use induction, as in the binomial theorem.]
⇒ Given →p∈E′=En(Cn) and f:E′→E, prove that f∈CD1 at →p iff f is differentiable at →p and
(∀ε>0)(∃δ>0)(∀→x∈G→p(δ))‖d1f(→p;⋅)−d1f(→x;⋅)‖<ε,
with norm ‖ as in Definition 2 in §2. (Does it apply?)
[Hint: If f∈CD1, use Theorem 2 in §3. For the converse, verify that
ε≥|d1f(→p;→t)−d1f(→x;→t)|=|n∑k=1[Dkf(→p)−Dkf(→x)]tk|
if →x∈G→p(δ) and |→t|≤1. Take →t=→ek, to prove continuity of Dkf at →p.]
Prove the following.
(i) If ϕ:En→Em is linear and [ϕ]=(vik), then
‖ϕ‖2≤∑i,k|vik|2.
(ii) If f:En→Em is differentiable at →p, then
‖df(→p;⋅)‖2≤∑i,k|Dkfi(→p)|2.
(iii) Hence find a new converse proof in Problem 10 for f:En→Em.
Consider f:Cn→Cm, too.
[Hints: (i) By the Cauchy-Schwarz inequality, |ϕ(→x)|2≤|→x|2∑i,k|vik|2. (Why?) (ii) Use part (i) and Theorem 4 in §3.]
(i) Find d2u for the functions of Problem 10 in §4, in the "variable" and "mapping" notations.
(ii) Do it also for
u=f(x,y,z)=(x2+y2+z2)−12
and show that D11f+D22f+D33f=0.
(iii) Does the latter hold for u=arctanyx?
Let u=g(x,y),x=rcosθ,y=rsinθ (passage to polars).
Using "variables" and then the "mappings" notation, prove that if g is differentiable, then
(i) ∂u∂r=cosθ∂u∂x+sinθ∂u∂y and
(ii) |∇g(x,y)|2=(∂u∂r)2+(1r∂u∂θ)2.
(iii) Assuming g∈CD2, express ∂2u∂r∂θ,∂2u∂r2, and ∂2u∂θ2 as in (i).
Let f,g:E1→E1 be of class CD2 on E1. Verify (in "variable" notation, too) the following statements.
(i) D11h=a2D22h if a∈E1 (fixed) and
h(x,y)=f(ax+y)+g(y−ax).
(ii) x2D11h(x,y)+2xyD12h(x,y)+y2D22h(x,y)=0 if
h(x,y)=xf(yx)+g(yx).
(iii) D1h⋅D21h=D2h⋅D11h if
h(x,y)=g(f(x)+y)
Find D12h, too.
Assume E′=En(Cn) and E′′=Em(Cm). Let f:E′→E′′ and g:E′′→E be twice differentiable at →p∈E′ and →q=f(→p)∈E′′, respectively, and set h=g∘f.
Show that h is twice differentiable at →p, and
d2h(→p;→t)=d2g(→q;→s)+dg(→q;→v),
where →t∈E′,→s=df(→p;→t), and →v=(v1,…,vm)∈E′′ satisfies
vi=d2fi(→p;→t),i=1,…,m.
Thus the second differential is not invariant in the sense of Note 4 in §4.
[Hint: Show that
Dklh(→p)=m∑j=1m∑i=1Dijg(→q)Dkfi(→p)Dlfj(→p)+m∑i=1Dig(→q)Dklfi(→p).
Proceed.]
Continuing Problem 15, prove the invariant rule:
drh(→p;→t)=drg(→q;→s),
if f is a first-degree polynomial and g is r times differentiable at →q.
[Hint: Here all higher-order partials of f vanish. Use induction.]