MA3110S Mathematical Analysis II (Version S) AY 2010/2011 Sem 2 NATIONAL UNIVERSITY OF SINGAPORE MATHEMATICS SOCIETY PAST YEAR PAPER SOLUTIONS solutions prepared by Joseph Andreas and Chang Hai Bin MA3110S Mathematical Analysis II (Version S) AY 2010/2011 Sem 2 Question 1 Note that for any open set I containing a, consider γ : I → R3 satisfying γ(x) = a for any x ∈ I. Then, f (γ(t)) = f (a) = 0 = g(a) = g(γ(t)) as required. Now, since f (γ(t)) = g(γ(t)) = 0, we have D(f ◦ γ)(t) = D(g ◦ γ)(t) = 0 for all t ∈ I. By chain rule, we have   u 0 = D(f ◦ γ)(t0 ) = Df (γ(t0 ))D(γ(t0 )) = Df (a)D(γ(t0 )) = 0 1 −3 v  1 and   u 0 = D(g ◦ γ)(t0 ) = Dg(γ(t0 ))D(γ(t0 )) = Dg(a)D(γ(t0 )) = 1 −2 6 v  1 Solving two equations, we have u = 0 and v = 3. Question 2 We shall use two lemmas : Lemma 1 : T : Rn → Rm is a linear transformation, and h ∈ Rn , then |T h| ≤ |T ||h|. Lemma 2: If a, b ∈ Rn , then ||a| − |b|| ≤ |a − b| Letting X = T x, H = T h, we have : T x · T h |T (x + h)|(x + h) − |T x|x − x − |T x|h |T x| |h| T x · T h |T x + T h|x − |T x|x − x + (|T x + T h| − |T x|) h |T x| = |h| T x · T h |T x + T h|x − |T x|x − x | |T x + T h| − |T x| | · |h| |T x| ≤ + |h| |h| T x · T h |T x + T h| − |T x| − |x| |T x| = |T | + | |T x + T h| − |T x| | |T | |h| T x · T h |T x + T h| − |T x| − |T x| ≤ |T ||x| + | T x + T h − T x | |T h| NUS Math LaTeXify Proj Team Page: 1 of 3 NUS Mathematics Society MA3110S Mathematical Analysis II (Version S) AY 2010/2011 Sem 2 |X + H| − |X| − X · H |X| = |T ||x| + |T h| |H| Note that: |X + H| − |X| − X · H |X| |H| |X + H|2 − |X|2 X · H |X + H| + |X| − |X| = |H| |X + H|2 |X| − |X|3 − (X · H)(|X + H| + |X|) = |H| · |X| · (|X + H| + |X|) |X|2 + |H|2 + 2X · H |X| − |X|3 − (X · H)(|X + H| + |X|) = |H| · |X| · (|X + H| + |X|) |H|2 |X| + (X · H)|X| − (X · H)|X + H| = |H| · |X| · (|X + H| + |X|) ≤ 1 |H|2 |X| | (X · H)|X| − (X · H)|X + H| | + |H| |H| |X| · (|X + H| + |X|) |X · H| · | |X| − |X + H| | 1 = |H||X| + |H| |X| · (|X + H| + |X|) |X · H| · | X − (X + H) | 1 ≤ |H||X| + |H| |X| · (|X + H| + |X|) |X · H| · | − 1||H| 1 = |H||X| + |H| |X| · (|X + H| + |X|) = (|H||X| + |X · H|) 1 |X| · (|X + H| + |X|) So, |T (x + h)|(x + h) − |T x|x − T x · T h x − |T x|h |T x| 0≤ |h| |X + H| − |X| − X · H |X| ≤ |T ||x| + |T h| |H| NUS Math LaTeXify Proj Team Page: 2 of 3 NUS Mathematics Society MA3110S Mathematical Analysis II (Version S) ≤ (|H||X| + |X · H|) AY 2010/2011 Sem 2 1 |T ||x| + |T h| |X| · (|X + H| + |X|) |T ||x| + |T h| − − − (1) |T x| · (|T x + T h| + |T x|) and the whole expression (1) goes to zero as |h| → 0 = (|T h||T x| + |T x · T h|) Question 3 Since f is continuous, so is |f |. By extreme value theorem, there exists α such that |f (x)| ≥ |f (α)| for all x ∈ R. Since |f (x)| > 0 for all x, it follows that |f (α)| > 0. Setting M = |f (α)|, we have f (x) ≥ M for all x ∈ Rn . Now, consider any > 0. Since (fk ) converges uniformly to f , then there exists N1 , N2 ∈ N such M 2 n n that |fk (x) − f (x)| < M 2 for all k ≥ N1 , x ∈ R , and |fk (x) − f (x)| < 2 for all k ≥ N2 , x ∈ R . M M Now, if k ≥ N1 , we have |f (x) − fk (x)| < 2 , thus leaving us with |fk (x)| − |f (x)| > − 2 . Thus, |fk (x)| > M 2 . 1 k (x) Finally, consider any k ≥ max(N1 , N2 ) we have | fk1(x) − f (x) | = | ff(x)−f |= k (x)f (x) |f (x)−fk (x)| |fk (x)||f (x)| < M2 2 M2 2 = Question 4 Rb Since the set of discontinuity of cos(nx) is a f -null set it follows that a cos nxdf . By integration Rb Rb Rb by parts, we have a f d(cos nx) and hence, a f sin(nx)dx exists. We have, -n a f sin(nx)dx = Rb Rb if and only if a cos nxdf is bounded (that happens by integration a f d cos(nx) which is bounded Rb Rb by parts). But, f (a) − f (b) ≤ a cos(nx)df ≤ f (b) − f (a) since −1 ≤ cos x ≤ 1. Thus, a cos(nx)df Rb is bounded. Hence, n a f sin(nx)dx is bounded. The result follows. QED Question 5 By Taylor’s theorem, we have f (x) = j−1 (k) X f (0) k! k=0 xk + f (j) (a) j x j! for some a ∈ [0, x] which leads to f (x) − Pj (x) 1 xj− 2 = f (j) (a)xj j! − f (j) (0)xj j! 1 xj− 2 and hence, we have f (x) − Pj (x) 1 xj− 2 = √ x f (j) (a) − f (j) (0) j! ! = Z x 1 1 √ xf (j) (a) − xf (j) (0) = √ f j (a) − f j (0)dt j! x j! x 0 Rx Next, clearly 0 f (j) (t) − f (j) (0)dt exists, and hence, Z Z x Z x −1 x (j) 1 1 lim √ |f (t)−f (j) (0)|dt ≤ lim √ f (j) (t)−f (j) (0)dt ≤ lim √ |f (j) (t)−f (j) (0)|dt x→0+ x→0+ x→0+ x 0 x 0 x 0 NUS Math LaTeXify Proj Team Page: 3 of 3 NUS Mathematics Society MA3110S Mathematical Analysis II (Version S) Thus, 1 lim √ x→0+ x x Z f (j) (t) − f (j) (0)dt = 0 0 which implies lim x→0+ x Z 1 √ j! x f (j) (t) − f (j) (0)dt = 0 x Z (1) 0 Now, since for any function g and interval [p, q], 1 lim √ x→0+ j! x AY 2010/2011 Sem 2 Rq gdt = g(c)(q − p) for some c ∈ [p, q], we have p 1 √ (f (j) (a)x − f (j) (b)x) x→0+ j! x f (j) (a) − f (j) (t)dt = lim 0 for some b ∈ [0, x]. Thus, 1 lim √ x→0+ j! x Z x 1√ x(f (j) (a) − f (j) (b)) x→0+ j! f (j) (a) − f (j) (t)dt = lim 0 As f (j) is a continuous function and since a, b both in interval [0, x], we have f (j) (b) − f (j) (a) goes √ to 0 as x → 0+. Since x also goes to 0 when x goes to 0, we have Z x 1 f (j) (a) − f (j) (t)dt = 0 (2) lim √ x→0+ j! x 0 Adding (1) and (2), we get 1 lim √ x→0+ j! x x Z f (j) (a) − f (j) (0)dt = 0 0 Thus, we get lim x→0+ f (x) − Pj (x) 1 xj− 2 =0 as desired. NUS Math LaTeXify Proj Team Page: 4 of 3 NUS Mathematics Society