Advanced Calculus Math 127B, Winter 2005 Solutions: Final. nx2 1 + n 2 x, g n(x) = n2 x

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1 . Define f n, g n : [, ] R by f n (x) = Advanced Calculus Math 27B, Winter 25 Solutions: Final nx2 + n 2 x, g n(x) = n2 x 2 + n 2 x. 2 Show that the sequences (f n ), (g n ) converge pointwise on [, ], and determine their pointwise limits. Determine (with proof) whether or not each sequence converges uniformly on [, ]. As n, we have f n and g n g pointwise, where { g(x) = /x if < x, if x =. Given ɛ >, choose N = /ɛ. Then n > N implies that f n (x) = ( ) nx 2 < ɛ for all x [, ]. n /n + nx 2 n Therefore f n converges uniformly to. The functions g n are continuous, and their pointwise limit g is discontinuous. Since the uniform limit of continuous functions is continuous, (g n ) does not converge uniformly.

2 2. Find all points x R where the following power series converges: n= + n2 n xn. According to the ratio test, the radius of convergence R of the power series a n x n is given by R = lim a n n a n+ (provided that this limit exists). Hence the radius of convergence of the given power series is When x = 2, the series is R = lim n + (n + )2 n+ + n2 n /(n2 n ) + ( + /n)2 = lim n /(n2 n ) + = 2. n= 2 n + n2 = n n + 2. n Since n + 2 n n + this series diverges by comparison with the divergent harmonic series When x = 2, the series is n= n= n= n +. ( ) n 2 n + n2 = ( ) n n n + 2, n 2 n=

3 which converges by the alternating series test, since n + 2 n as n and is decreasing in n. The power series therefore converges for 2 x < 2. 3

4 3. (a) Prove that the following series converge on R to continuous functions: f(x) = n= cos nx n 2, g(x) = n= (b) Prove that g is differentiable on R, and g = f. sin nx n 3. (a) Since for all x R and cos nx n 2 n, 2 n= n 2 <, sin nx n 3 n 3 n= n 3 < the Weierstrass M-test implies that both series converge uniformly on R. Since the terms in the series are continuous, and the uniform limit of continuous functions is continuous, the sums f, g are continuous. (b) Since the uniform convergence of Riemann integrable functions implies convergence of their Riemann integrals, we can integrate the series for f term-by-term over the interval [, x] (or [x, ] if x < ) to obtain x f(t) dt = = n= n= = g(x). x sin nx n 3 cos nt n 2 Since f is continuous, the fundamental theorem of calculus implies that g is differentiable and g = f. dt 4

5 4. Let a >. Give a definition of the following improper Riemann integral as a limit of Riemann integrals: 2 x(log x) a dx. For what values of a does this integral converge? We define 2 b dx = lim x(log x) a b 2 x(log x) dx. a Let b I(b) = 2 x(log x) dx. a Making the substitution u = log x, we get For a, we have I(b) = I(b) = [ u a log b a log 2 ] log b u a du. log 2 = (log b) a (log 2) a, a which diverges as b if a <. If a >, then If a =, then I(b) (log 2) a a as b. I(b) = [log u] log b log 2 = log(log b) log(log 2) as b. The improper integral therefore converges when a >, and then 2 (log 2) a dx =. x(log x) a a 5

6 5. Define f : [, ] R by f(x) = { x if x Q, if x / Q. Is f Riemann integrable on [, ]? Prove your answer. The function f is not Riemann integrable. Suppose that P = {t, t,..., t n } is any partition of [, ] (so t =, t n =, and t k < t k ). Since every interval [t k, t k ] contains irrational numbers, we have m (f, [t k, t k ]) = inf {f(x) : x [t k, t k ]} =. The lower Darboux sum of f is therefore given by L(f, P ) = n m (f, [t k, t k ]) (t k t k ) =, k= and the lower Darboux integral of f is L(f) = sup {L(f, P ) : P is a partition of [, ]} =. Since the rational numbers are dense in any interval, we have M (f, [t k, t k ]) = sup {f(x) : x [t k, t k ]} = t k. Define l : [, ] R by l(x) = x. Then Therefore U(f, P ) = = n M (f, [t k, t k ]) (t k t k ) k= n t k (t k t k ) k= = U(l, P ). U(f) = inf {U(f, P ) : P is a partition of [, ]} = U(l). 6

7 Since l is Riemann integrable, U(l) = x dx = 2. So U(f) = /2. Thus U(f) > L(f), and f is not Riemann integrable. 7

8 6. Suppose that F (x) = Evaluate the Riemann-Stieltjes integral { x 2 for x <, x for x. Briefly justify your computations. We write F = F + F 2, where e x2 df (x). F (x) = F 2 (x) = { for x <, 2 for x, { x 2 for x <, x 2 for x. Using standard properties of the Riemann-Stieltjes integral, and its expression for jump and continuously differentiable integrators, we get e x2 df (x) = = = e 2 + = 2 e x2 df (x) + e x2 df (x) + = 2 [e x2] e x2 d ( x 2) + 2xe x2 dx + + [e x2] = 2 ( e) + (e ) = 2e. e x2 df 2 (x) e x2 df 2 (x) + 2xe x2 dx e x2 d ( x 2) e x2 df 2 (x) 8

9 7. (a) Find the Taylor series of e x (at x = ). (b) Give an expression for the remainder R n (x) between e x and its Taylor polynomial of degree n involving an intermediate point y between and x. (c) Prove from your expression in (b) that the Taylor series for e x converges to e x for every x R. (Don t use general theorems.) (a) Let f(x) = e x. Then The kth Taylor coefficient of f is f (k) (x) = ( ) k e x. a k = f (k) () k! = ( )k. k! The Taylor series of e x is therefore ( ) k x k = x + k! 2! x2 3! x k= (b) By the Taylor remainder theorem, e x = n k= where R n (x) = ( )n e y n! for some y between and x. (c) If x >, then < y < x and e y <. Hence R n (x) < xn n! ( ) k x k + R n (x), () k! x n as n. (Note that if c n = x n /n! then c n+ /c n = x/(n + ) < /2 for n > 2x, so c n as n for every x >.) Taking the limit as n in (), we obtain that e x ( ) k = x k. k! k= 9

10 If x <, then e y < e x, and the Taylor series also converges, since x x n R n (x) < e n! as n.

11 8. Define f : R R by f(x) = { x 2 [sin(/x) 2] for x, for x =. (a) Prove that f(x) has a strict maximum at x = (i.e. f() > f(x) for all x ). (b) Prove that f is differentiable on R. (c) Prove that f is not increasing on the interval ( ɛ, ) and f is not decreasing on the interval (, ɛ) for any ɛ >. (a) We have f() =. If x, then since sin(/x) f(x) x 2 [ 2] x 2 <. (b) The function f is differentiable at any nonzero x since it is a product and composition of differentiable functions. At x = the function is differentiable, with f () =, since { } f(x) f() lim x x = lim x { [ x sin ( ) ]} 2 =. x (c) For x, we compute using the chain and product rules that ( ) [ ( ) ] f (x) = cos + 2x sin 2. x x If x /2 then so cos ( ) ] [sin 2x 2 6 x < x 2, ( ) ( ) x 2 < f (x) < cos + x 2. It follows that f < (hence f is strictly decreasing) in any interval where cos(/x) > /2, and f > (hence f is strictly increasing) in any interval where cos(/x) < /2. Since there exist such intervals arbitrarily close to, the function f is not increasing throughout any interval ( ɛ, ), nor is it decreasing throughout any interval (, ɛ).

12 This example shows that a differentiable function may attain a maximum at a point even though it s not increasing on any interval to the left of the point or decreasing on any interval to the right. 2