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Material Type: Assignment; Class: ADVANCED CALCULUS; Subject: Mathematics; University: University of Pennsylvania; Term: Spring 2007;
Typology: Assignments
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Due: Friday, April 13, 2007
Math 360 - Advanced Calculus / Problem Set 10 (two pages)
b) If f, g are differentiable at x 0 , then so is f g and f /g, provided g(x) 6 = 0 on D in the last case.
differentiable on (a, b), and f ′(x) 6 = 0 for all x ∈ (a, b). Then one has: a) f ′^ is either strictly positive, or strictly negative on I. b) f is either strictly increasing, or strictly decreasing on I.
d) f −^1 : J → I is continuous. e) f −^1 is differentiable and (f −^1 )′(y) = 1/f ′(x), provided y = f (x).
[Hint: To d): Let [c, d] ⊂ I be a closed interval. Then [c, d] is compact, hence f : [c, d] → f ([c, d]) is a continuous bijection of compact topological spaces, hence a homeomorphism (WHY?). Deduce from this that f is a homeomorphism, hence f −^1 is continuous. To e): Take yn → y 0 in J, and xn → x 0 the preimages in I. Then yn = f (xn), y 0 = f (x 0 ), and proceed as in the proof of the “chain rule”.]
a) (^) xlim→∞(1 + (^1) x )x^ = lim x→ 0 (1 + x) 1 x^ = lim x→ 1 x x−^11. b) log′ a(x 0 ) := (^) xlim→x 0 loga(x x)−−logx 0 a(x^0 ) = (^) xlim→x 0 x^ log 0 (x/xa(x/x 0 −^0 1)) = (^) x^10 ulim→ 1 u−^11 · loga(u) = (^) x^10 loga(e) = (^) ln(^1 a) x^10 (WHY?). c) Deduce from this exp′ a using the Inverse Function Theorem.
[Hint: xα^ = exp
log(xα)
= exp
α log(x)
, and use the chain rule, etc.]
ulim→ 0 sinu^ u= 1 (WHY?), and then show: a) sin′(x 0 ) := (^) xlim→x 0 sin(x x)−−sin(x 0 x^0 )= lim u→ 0 sin(x^0 +u u)− sin(x^0 )= cos(x 0 ) (WHY?). b) Deduce from this that cos′(x) = − sin(x) using cos(x) = sin( π 2 − x) and the chain rule. c) Deduce that tan and cot are differentiable using the definitions and a), b), above.
which is at least (n + 1)-times differentiable on I. Let a < b in I be fixed. The polynomial
Pf,n(X) := f (a) + f^
′(a) 1! (X^ −^ a) +^...^ +^
f (n)(a) n! (a)(X^ −^ a)
n (^) = ∑^ n k=
f (k)(a) k! (a)(X^ −^ a)
k
is called the Taylor polynomial of degree n attached to f about x = a. If f is k-times differentiable for all k, and a = 0, then the (formal) power series
Σf (X) := f (0) + f^
f (n)(0) n! (a)X
n (^) +... =
k=
f (k)(0) k! (0)X
k
is called the Maclaurin series attached to f (about x = 0).
a) For every x ∈ [a, b] there exist ξ ∈ [a, x] such that f (x) = Pf,n(x) + f^ ((nn+1)+1)!(ξ )(x − a)n+1. b) In particular, if f (n+1)^ is bounded, say by M > 0 on [a, b], then ‖f − Pf,n‖sup ≤ M^ (b−a) n+ (n+1)!. c) In particular, if f is k-times differentiable for all k, and f (k)^ is bounded, say by M > 0 on [a, b], for all k, then the Maclaurin series Σf (x) is absolutely convergent for x ∈ [a, b], and f (x) = Σf (x) for x ∈ [a, b].
c) cos(x) = 1 − x 2!^2 + x 4!^4 −... =
k=0(−1)k x
2 k
all k, and compute its Maclaurin series. What do you conclude?