An idea how to solve some of the problems. diverges the same must hold for the original series. T 1 p T 1 p + 1 p 1 = 1. dt = lim

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1 An idea how to solve some of the problems (a) Does not converge: By multiplying across we get Hence 2k 2k 2 /2 k 2k2 k 2 /2 k 2 /2 2k 2k 2 /2 k. As the series diverges the same must hold for the original series. k (b) Converges: We have (k )/(k2 k ) 2 k and the series 2 k converges. (c) Divergent: In this case /(2k ) > /(2k) (multiply in cross) and the series /k diverges. (d) Divergent: Assume first that p > and take f(x) = x p. Then f is monotonically decreasing to zero. Furthermore T f(t)dt t p dt T T p T p + p = p <. The claim follows then from Theorem Let now p =. We have T x dx = log T as T. It follows that /x dx does not exists and hence k does not converge according to Theorem If p then /k p /k and hence k p diverges Suppose x k for all k N, and suppose that lim k k x k = L exists. (a) If L > then x k diverges: Let < r < L. Then there exists N N such that for all n N we have r k x k. Hence x k r k. The claim follows now because k=n rk does not exists. (b) If L < then x k converges: Let L < r <. Then there exists N N such that k x k r for all n N. This implies that x k r k and hence x k = x x N + x x N + k=n x k r k <. k=n

2 Hence the series converges. (c) If L = there is no information: Let x k = for all k. Then k x k = and the series x k diverges. On the other hand, if x k = k 2 then lim k k x k = as we will see in a moment and this time the series x k converges. Let n N and consider the sequence x k = k k n. Taking the log we see that (using L Hospital) Hence lim log x n log k k k k k lim x k = e =. k n k k =. 5.2-: Test for convergence: (a) k!/kk : Convergent because with x k = k!/k k we have x k+ x k = (k + )!kk k!(k + ) k+ = ( + /k) k /e <. (b) k/e k2 : Convergent because (k + )e k2 ke (k+)2 = ( + /k)e 2k as k. (c) k=2 /(log k)k : Convergent. Use the root test (fill in the details) We have 3 k 4 = k 3 k 4 k = 2 3 = If the sequence {c k } is summable then it follows that c k is bounded, i.e., there exists a C > such that c k C for all k (use that lim c k = ). Hence c k x k C x k <

3 for x <. If x = then c k x k = c k is summable by our assumption on c k We do (a) Let ɛ > be given. Let N > 2/ɛ. Then, if n > m N there exists µ (/n, /m) such that ( f(/n) f(/m) = f (µ) n ) m As f (µ) < it follows that f(/n) f(/m) < n m < 2 N < ɛ. It follows that {f(/n)} is a Cauchy sequence and hence exists. lim f(/n) = L n (b) Let now {x k } be an arbitrary sequence x k. Then, by the same argument as above it follows that {f(x k )} is a Cauchy sequence and hence lim f(x k ) = L exists. Define a new sequence y 2k = /k and y 2k+ = x k. Then y k and the above argument show that lim k f(y k ) exists. Add the details to show that this implies that L = L (use subsequence) It was shown that all the limits exists, so we will not do it here (on an exam you would have to do the details). Let v, w V and c R. Then T n (cv + w) = ct n (v) + T n (w) because T n is linear. As all the limits exists we have: which shows that T is linear. T (cv + w) (T n (cv + w)) n (ct n (v) + T n (w)) n = c lim T n (v) + lim T n (w) n n = ct (v) + T (w) First we have to show that is a norm on l. Let x = {x k }, y = {y k } l and c R. Note first that cx k + y k cx k + y k = c x k + y k.

4 Hence cx + y = sup cx k + y k k ( cx k + y k ) sup k c sup k = c x + y x k + sup y k k Furthermore x = if and only if all x k = which happen if and only if x =. Next we have to show that l is complete. Let {x n } be a Cauchy sequence in l. Let ɛ > be given. Then there exists N N such that for all n, m N we have x n x m = sup x n k x m k < ɛ/2. k It follows that the sequence {x n k } n is a Cauchy sequence in R and hence there exists a x k R such that x n k x k. Let x = {x k } we have to show that x n x and that x l. Let N be as above. Then x n k xm k < ɛ/2. Letting m this implies that Thus and x n k x k ɛ/2 < ɛ. ( n N) x n x < ɛ x = x x N + x N x x N + x N < ɛ + x N <. This proves both statements : Recall that the sequence x n l is defined by x n k = (n + )/(n2k ). a) Show that x n l : We have x n = n + n 2 k 2 2 k < because n + n + /n 2.

5 b) By the above we have that lim n infty xn k = 2 k = x k exists and the sequence x = {x k } is in l because 2 k <. c) We have x n k x k =. Furthermore n2 k 2 k =. Hence x n x = x n k x k = n. Let ɛ >. Let N N be such that N > /ɛ. Then, if n N we have x n x = n N < ɛ In this problem we define x n by x n k = if k n and xn k = k 2 if k > n. (a) We have Hence x n l. x n = x n k = n + k=n+ k 2 <. () (b) Let k N, then for all n k we have x n k =. Hence x k n x n k = for all k. In particular x = {x k } l. (c) The sequence {x n } can not be a Cauchy sequence because otherwise lim x n = x l would exists If α < show that xk conveges uniformly on [, α]. Solution: We have M k = sup x [,α] x k = α k and hence the series M k converges. The claim follows by the Weierstrass M-test : If f k converges uniformly on D, prove that f n as n. Is the converse true? Solution: As f k converges uniformly it follows that the sequence of partial sums s n = n f k is a Cauchy sequence in the supremum norm. Let ɛ >. Then there exists a N N such that n, m N s n s m < ɛ. In particular for n > N: f n = s n s n < ɛ.

6 The converse is not true. For that let f k (x) = k on [, ]. Then f k = /k, but f k(x) does not even converge at x = : (a) The sequence e kx converges uniformly on [, ). For that note that on this interval we have e kx e k = (/e) k and the series (/e)k converges. The claim follows then by the Weierstrass M-test. (b) sin(kx) converges uniformly on R because k 3 sin(kx) k 3 k 3 and the series /k3 converges. The claim follows then by the Weierstrass M-test. (c) The series sink (x) converges uniformly on [, π/4] because on this interval sin k (x) (/ 2) k and the series (/ 2) k converges. (d) No, the series tank x does not even converge at x = π/ : (a) We have to show that ( ) k 2k + x2k+ = x ( ) k 2k + x2k converges uniformly on [, ]. We note that for all x [, ] the series ( ) k 2k+ x2k is alternating and x k = x2k monotonically. Hence 2k+ k x k exists and by Theorem 5..2 we we have with s n (x) = n Hence which proves the claim. ( ) k 2k+ x2k and s(x) = ( ) k 2k+ x2k : xs n (x) xs(x) = x s n (x) s(x) x n+ = x n ( ) k 2k + x2k+ s(x) x 2n 2n + 2 2(n + ). 2(n + ) (b) Define g(x) = ( ) k 2k+ x2k then it follows by (a) and Theorem 5.5., part a, it follows that g(x) is continuous on [, ]. As g(x) = tan (x) for x (, ) and tan x is continuous, it follows that g(±) = tan (±).

7 (c) We know that tan () = π 4. Hence π = 4 tan () = 4 ( ) k 2k : We have tk = t Taking t = /2 we get if t <. Hence, by Theorem 5.6. and Theorem 5.5.: t k+ k + = t k k = t du = log( u). u = log(/2) = log 2. k2k 5.6-5: Find the interval of convergence of the series c k x k. We use the ratio test: In case lim c k+ k c k = L exists, then R = L. (In case L = this reads R = and L = reads R =.) (a) c k = /(k!). Then c k+ = c k k +. Hence the power series converges for all x R. (b) a = and c k = ( ) k+ /(k + ). Then lim c k+ k c k = and hence R =. If x =, then we have a alternating series so the power series converges at x =. If x = 2 then we are looking at the series ( ) 2k+ k +

8 which does not converge. So the power series converges on ( 2, ]. (c) c k = k!/k k so ( ) k (k + )!kk c k+ /c k = k!(k + ) = /e. k+ + /k What about the endpoint? (d) c k = /k k. Then c k+ /c k = k k (k + ) = ( ) k k k+ k + (k + ) k +. Hence the power series converges for all x R, i.e, R = : The function e x is analytic at and so is tan (x). It follows by Theorem that e x tan x is analytic at. There are two ways to find the coefficient of x 4. First, just differentiate the function four times and use that c k = f (k) ()/k!. The other way is to use that if f(x) = a kx k and g(x) = b kx k for x R, then ( k ) fg(x) = a k b j x j+k = a j b k j x k. Hence the coefficient of x k is j= k a j b k j. It follows then from formula (5.2) p. 4 that the coefficient of x 4 is j= j= j= 4 ( ) 4 j j! 2(4 j) + = = simplify : We have tan (x) = ( ) k 2k + x2k+ = f (k) () k! Note, that the coefficients for the even powers of x are all zero. Hence f (even) () =. In particular f () () =. We have = 2 5 +, so k = 5, and hence f () () =! =!. x k.

9 5.7-4: (a) The function f(x) = x can not be analytic at zero, because it is not differentiable at zero (recall: analytic functions are smooth!). (b) The function can not be analytic at zero because we have { k!x, x > f (k ) (x) =, x and this function is not differentiable at zero : (a) True, the function is given by f(x) = x 4 on the interval (, ). (b) True, we have f(x) = on the interval (, ). (c) No (see problem with k = Let f(x) = { e /x 2, x, x =. Note that f is -times differentiable at all points x as that holds for the exponential function and the function x /x 2. (a) To see if f () we need to see if the limit f(h) f() lim h h e /h2 h h exists. Note that this limit is of the form so we can use L Hospital. We set u = /h and consider the limit u : e /h2 lim h h u = u e u2 2ue u2 u Hence, f () exists and is equal to zero, f () =. Before we do the next parts let us note the following: Let k N, then e /h2 u k lim h h k u e u2

10 ku k u 2ue u2 u k(k )u k 2 2e u2 + 4u 2 e u2 = q(u)e u2 u k! where q(u) = 2 k u k +... is a polynomial of degree k. (b) We have { f 2e /x 2 /x (x) = 3, x, x =. Hence, by the above argument f (h) f () h = 2e /h2 h 4 h. Hence the derivative at zero exists and f () =. (c) Use induction to show that there exists an n N and constants c j, j =,..., n such that f (k) (x) = { n j= c j e /x 2, x x j, x = Hence, the above argument shows, that f (k+) (x) exists for all x R and f (k+) () = : We have Hence sin(x) x is analytic and (Fill in the details.) sin(x) = sin(x) = ( ) k (2k + )! x2k+. ( ) k (2k + )! x2k. 5.8-: The function f(x) = /x is unbounded around, whereas every polynomial is bounded. Hence, assume that p(x) is a polynomial. Then f(x) p(x) =. x (,)

11 5.8-2: The function f(x) = e x is unbounded on R. Even more holds. Let p(x) = n j= a jx j be a polynomial with a n. Then for x big, we have e x p(x) = x n e x x a n n a n /x... a /x n as x : (a) Assume that f C([, ]) and that f(x)xk dx = for all k =,,.... Assume that f, Then Let p(x) be a polynomial. Then and f(x) 2 dx = A >. f(x)p(x) dx = f p 2 (f(x) p(x)) 2 dx = f(x) 2 dx 2 f(x)p(x) dx + p(x) 2 dx A >. Let < ɛ < A. Then, by Weierstrass Approximation Theorem, there exists a polynomial p such that f p < ɛ < A a contradiction. (b) Define T k (f) = f(x)xk dx, k =,,.... Then T k (af + g) = at k (f) + T k (g) because the Riemann integral is linear. Furthermore T k (f) = f(x)x k dx f(x) x k dx f x k dx = f k +.

12 Hence T k is bounded. (c) Assume that f, g C([, ]) and that T k (g) = T k (f). Then T k (g f) = for all k and hence by (a) g f = or g = f : Let k(x) = 2 χ [,](x) where χ [,] denotes the indicator function of the interval [, ]. Then k n (x) = nk(nx) = n 2 χ [ /n,/n] (fill in the detail) and k n (x) dx = n 2 /n /n dx =. If δ >, then there exists an n N such that n < δ and hence k n(x) = for δ x.