Math 61CM - Solutions to homework 6

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1 Math 61CM - Solutions to homework 6 Cédric De Groote November 5 th, 2018 Problem 1: (i) Give an example of a metric space X such that not all Cauchy sequences in X are convergent. (ii) Let X be a metric space, and K a compact subset of X. Let x n be a Cauchy sequence such that each x n K. Show that x n is a convergent sequence. Solution: (i) We can take X to be the open interval (0, 1), with the usual Euclidean distance d E (x, y) = x y. Then, the sequence x n = 1/n is Cauchy (because it converges in R), but is not convergent in X. (ii) Since K is compact, we know that there exists a subsequence x jn of x n that converges in K; call the limit x. We will use the fact that x n is Cauchy to show that, in fact, the full sequence x n converges to x. Let ε > 0. Since x n is Cauchy, we know that there exists N 1 N such that x m x n < ε/2 if m, n N 1. Since x jn converges to x, we know that there exists N 2 N such that x jn x < ε/2 if n N 2. Let N = max(n 1, N 2 ). Let n N. Then, pick k such that k N (which implies j k N, since these are the indices of a subsequence). Then, we have where we x n x x n x jk + x jk x < ε/2 + ε/2 = ε, used the fact that j k N 1 and n N 1, along with the Cauchyness of x n to bound the first term, and used the fact that k N 2, along with the fact that x jn converges to x to bound the second term. This means that x n converges to x. 1

2 Problem 2: Let f n (x) be a sequence of functions on [0, 1]. We say that the sequence f n converges pointwise to a function f if f n (x) f(x) as n + for each x fixed. We say that f n converges uniformly to a function f if sup f n (x) f(x) 0 as n. x [0,1] (i) Show that f n converges to f uniformly if and only if for any ε > 0 there exists N so that f n (x) f(x) < ε for all n N and x [0, 1]. (ii) Give an example of a sequence of functions f n (x) and a function f(x) such that f n converges to f pointwise but not uniformly. (iii) Show that if all f n are continuous functions on [0, 1] and f n converges uniformly to a function f then the function f is also continuous on [0, 1]. (iv) Give an example of a sequence f n of continuous functions on [0, 1] that converges pointwise to a function f that is discontinuous at some point x [0, 1]. Solution: (i) The fact that sup x [0,1] f n (x) f(x) converges to 0 means that for any ε > 0, there exists N N such that sup x [0,1] f n (x) f(x) < ε for all n N (here we use that this supremum is nonnegative, as it s a supremum of a family of non-negative numbers, to get rid of the absolute value). In particular, for any given x, we have f n (x) f(x) sup f n (x) f(x) < ε x [0,1] whenever n N, since the supremum of a set is in particular an upper bound for that set. (ii) Take f n (x) = { 1/(nx) if x 0 0 else and f to be identically 0. Then, given any x 0, the sequence f n (x) = 1/(nx) converges to f(x) = 0, and also f n (0) = 0 converges to f(0) = 0; this shows that f n (x) converges pointwise to the function f. However, taking ε = 1, there can not exist N such that f n (x) f(x) < ε for all n N and x [0, 1]. Indeed, assume that N exists. Then, take x = 1/2N; then we have f N (x) = 2 which is larger than 1; contradiction. Therefore, f n does not converge uniformly to f. (iii) Let ε > 0, and a [0, 1]. Pick N N such that f N (x) f(x) < ε/3 for all x [0, 1]; this exists since we assume that f n converges uniformly to f. Since f N is continuous at a, there exists δ > 0 such that f N (x) f N (a) < ε/3 whenever x a < δ. Therefore, we have f(x) f(a) f(x) f N (x) + f N (x) f N (a) + f N (a) f(x) < ε/3 + ε/3 + ε/3 = ε, whenever x a < δ. This shows that f is continuous at a. Since a was arbitrary, we conclude that f is continuous., 2

3 (iv) Take f n (x) = x n, and f(x) = { 0 if x 1 1 if x = 1. Then it is clear that f n converges pointwise fo f. However, the function f is not continuous, therefore by the point (iii) the sequence f n can not converge uniformly, since it consists of continuous functions. Problem 3: (i) Let a n > 0 be a decreasing sequence of positive numbers such that lim n a n = 0. Show that the series ( 1)n+1 a n converges. Hint: use the Cauchy criterion. (ii) We say that a series c n is a rearrangement of a series b n if there exists a one-to-one map J : N N such that for each n we have c J(n) = b n. Let b n be a series that converges but not absolutely. Show that for any s R we can find a rearrangement c n of the series bn such that n c n = s. Hint: start with showing that the series that consists only of the positive b n > 0 diverges, as does the series consisting only of the negative b n. Keep in mind what lim b n is. Solution: (i) (Solution without using the Cauchy criterion) Write s k = k ( 1)n+1 a n the sequence of partial sums. Notice that 2k+2 s 2k+2 s 2k = ( 1) n+1 a n 2k ( 1) n+1 a n = a 2k+2 + a 2k+1 0 2k+3 2k+1 s 2k+3 s 2k+1 = ( 1) n+1 a n ( 1) n+1 a n = a 2k+3 a 2k+2 0 since the sequence a n is decreasing. It follows that the sequence (s 2k ) k of even partial sums is increasing, while the sequence (s 2k+1 ) k of odd partial sums is decreasing. Furthermore, we have s 2k = s 2k+1 a 2k+1 s 2k+1 s 1, where we used a n > 0 and the decreasingness of the sequence of odd partial sums. Similarly, s 2k+1 = s 2k + a 2k+1 s 2k s 2, where we used a n > 0 and the increasingness of the sequence of even partial sums. From this, it follows that the sequence (s 2k ) k is increasing and bounded, and therefore converges to some limit t. Similarly, the sequence (s 2k+1 ) k is decreasing and bounded, and therefore converges to some limit s. Let us show that s = t: s t = lim k s 2k+1 s 2k = lim k a 2k+1 = 0. 3

4 It is now easy to conclude that (s n ) n converges to s = t. Let ε > 0. Since (s 2k ) k converges to s, there exists N 1 N such that s s 2k < ε when k N 1. Since (s 2k+1 ) k converges to s, there exists N 2 N such that s s 2k+1 < ε. when k N 2. It follows that for every n max(2n 1, 2N 2 + 1), we have s n s < ε. (i) (Solution using the Cauchy criterion) Given n > m 1, note that any sums ( 1) m+1 a m + + ( 1) n+1 a n is always in the interval whose bounds are ( 1) m+1 a m and ( 1) m+2 a m+1. This is clear from a picture, and also follows from the same kind of arguments as in the first half of the proof above (but it s annoying to write down). Notice that the length of this interval is a m a m+1. Let ε > 0. Choose N 1 such that a m a m+1 < ε/2 for all m N 1 ; this exists since the sequence a n is Cauchy (as it converges to 0). Choose N 2 such that a m < ε/2 for m N 2 ; this exists since a n converges to 0. Let N = max(n 1, N 2 ). Then, for any m and n larger than N, we have that ( 1) m+1 a m + + ( 1) n+1 a n a m + am ( 1) m+1 a m + + ( 1) n+1 a n a m + a m a m+1 ε/2 + ε/2 = ε. The first inequality is the triangle inequality, and the second one follows from that fact that the distance between a point in an interval and an extremity of the interval is always smaller than the length of the interval. This shows that the Cauchy criterion is satisfied, hence our series converges. (ii) Outline: We first prove that both the series of the positive terms, and the series of the negative terms, must diverge. This is done by using the fact that the series converges but not absolutely, by passing to finite sums. Then, given any s R (that we assume non-negative for simplicity; the other case is similar), add the positive terms until we get something larger than s. Then add the negative terms until we get something smaller than s. Then add the positive terms until we get something larger than s. Keep going this way; it is always possible because both the series of positive terms and of negative terms diverge. Since lim b n = 0, at some point we only add b n very small, so we will stay in a very small interval around s, proving that the series obtained this way converges to s. Let b + n = max(b n, 0) and b n = min(b n, 0). Notice that b n = b + n + b n and that b n = b + n b n. Also, b n = b + n if it is positive, and b n = b n if it is negative. Lemma 0.1 Both n b+ n and n b n diverge. Proof. Suppose that they both converge, say to b + and b. Then M b n = M M b + n b n b + b would also converge, which contradicts the fact that n b n does not converge absolutely. If n b+ n converges but not n b n, then M b n = M M b + n + b n, 4

5 contradicting the convergence of n b n. Similarly, if n b n converges but not n b+ n, then M b n = contradicting the convergence of n b n. M M b + n + b n, In particular, the series b + n and b n both have infinitely many non-zero terms. Denote x n the n th term in the sequence b + n that is non-negative, and y n the absolute value of the n th term in the sequence b n that is negative. Notice that this defines a bijection between {b n n N} and {x n n N} {y n n N}, and also b + n = x n and b n = y n. n n n Therefore, n x n and n y n also both diverge. n Now, let µ 1 be the smallest element of N such that µ 1 x n > s; this exists since n x n diverges (if s 0, then set µ 1 = 0). Let ν 1 be the smallest element of N such that µ 1 x n ν 1 y n < s; again, it exists because n y n diverges. Let µ 2 be the smallest element of N such that µ 2 x n ν 1 y n > s; this exists since n x n diverges. Let ν 2 be the smallest element of N such that µ 2 x n ν 2 y n < s; again, it exists because n y n diverges. Here is the general step. Suppose that µ 1,, µ k, ν 1,, ν k have been chosen such that µ k is the least element of N such that µ k x n ν k 1 y n > s and ν k is the least element of N such that µk x n ν k y n < s, and such that µ k > µ k 1 and ν k > ν k 1. Let µ k+1 be the least element of N such that µ k+1 x n ν k y n > s and let ν k be the least element of N such that µ k+1 x n ν k+1 y n < s. It is straightworward that µ k+1 > µ k and ν k+1 > ν k. Now, define c n as the n th term of the sequence x 1,, x µ1, y 1,, y ν1, x µ1 +1,, x µ2, y ν1 +1,, y ν2, x µ2 +1, (again, if s < 0 this sequence starts with y 1 ). By construction, since the sequence (µ k ) k and (ν k ) k are increasing and since there is a bijection between {b n n N} and {x n n N} {y n n N}, this sequence c n is a rearrangement of the sequence b n. Notice in particular that this rearrangement keeps the relative order of the non-negative terms of (b n ) n, as well as the relative order of the negative terms of (b n ) n. It remains to show that n c n = s; this is where we use the fact that lim n b n = 0, which follows from the convergence of n b n. Let ε > 0. Since the sequence c n must alternate between terms from the x n s and terms from the y n s, and because lim n b n = 0, there exists N N such that c n < ε for n N. Notice that K c n differs from s by at most the last x n we added if this sum is larger than s, and by at most the last y n we added if this sum is smaller than s. Therefore, s r c n < ε as soon as r is larger than the first µ k or ν k that is larger than N. 5

6 Problem 4: A ball that falls from height h bounces back up to height qh, with some fixed 0 < q < 1. Find the time until it comes to rest, and the total distance it travels through the air. Hint: yes, you should use equations of motion for a ball falling through vacuum. Solution: First, we compute the total distance it travels. From the time it is dropped to the first time it touches the ground, it travels distance h. From the first time it touches the ground to the second time, it travels from height 0 to height qh, back to height 0, therefore travelling 2qh. Generalizing, from the k th time it touches the ground to the (k + 1) st time, it will travel from height 0 to height q k h, and back to 0, therefore travelling a distance of 2q k h. Therefore, the total distance travelled is h + 2q k h = 2h q k h = k=1 k=0 2h 1 q h, using the fact that if q < 1, we have k=0 qk = 1/(1 q). Now, let us compute the total time it takes. Recall that the height at time t of a ball that is dropped from height h 0 at time t = 0, without initial velocity, on earth, is h 0 t2 g 2. So, the time t for which this is equal to 0 (i.e. the time it touches the ground for the first time) is 2h0 g. If it starts at height qk h, then this time would be equal to 2q k h g. Similarly, since the laws of mechanics are symmetric under reversal of time, this is also the time it takes from the time the ball rebounds to the time it reaches height q k h with 0 velocity. So, the total time this process takes is 2h g + 2q 2 k h 2h = 2 ( q) k 2h g g g = 2 2h 2h g(1 q) g. k=1 k=0 Problem 5: (i) If S is the 2-dimensional subspace of R 4 spanned by the vectors (1, 1, 0, 0), (0, 0, 1, 1), find an orthonormal basis for S and find the matrix of the orthogonal projection of R 4 onto S. (ii) If S is the subspace of R 4 spanned by the vectors (1, 0, 0, 1), (1, 1, 0, 0), (0, 0, 1, 1), find an orthonormal basis for S, and find the matrix of the orthogonal projection of R 4 onto S. Solution: We recall the Gram-Schmidt process, which is explained in the Lemmas 5.2 and 5.3 of Simon s book (and also here: Given a set of vectors v 1,, v n in an inner product space V, we always can create an orthonormal set of vector u 1,, u n whose span is the same as the span of v 1,, v n. Notice in particular that if v 1,, v n form a basis of V, then the u 1,, u n must generate V (since they have the same span), and by dimension they must also form a basis. Here is the algorithm: i=1 ũ 1 = v 1, k 1 ũ i, v k ũ k = v k ũ i, ũ i ũi, k 2. 6

7 Then, we simply do u k = ũk ũ k, k 1, unless ũ k is 0, in which case we define u k = 0. What the first part with the ũ i does, is the following. It keeps the first vector v 1. Then, it projects v 2 onto the orthogonal complement of the span of v 1, making sure that ũ 1 and ũ 2 are orthogonal. Then, it keeps doing so: it projects the k th vector onto the orthogonal complement of the k 1 first vectors, ensuring that everything is orthogonal. Then, the second part simply normalizes each vector, so that they have length 1. Remark 0.2 A few remarks: 1. You can change the order of the vectors, if you notice it will make things simpler (this is what I ll do in (ii) below). 2. If you notice that the vectors are already orthogonal to each other, then you can simply normalize them, and it will give an orthonormal set (this is what I ll do in (i) below). 3. If we throw away all the u k s which are 0, we always get a set of vectors which are linearly independent. Now, we explain the solution to the problem. (i) Notice that these two vectors are already orthogonal, so all we need to do is normalize them. Since they both have norm 2, an orthonormal basis of S is given by ( ) ( ) u 1 = 2,, 0, 0, u 2 = 0, 0,, To find the matrix of the orthogonal projection, we use the Lemma 5.2 in Simon s book, which says that the orthogonal projection of a vector x onto S is P S (x) = x, u 1 u 1 + x, u 2 u 2. Applying it to the basis vectors e 1, e 2, e 3, e 4 of R 4, we get that the matrix of the orthogonal projection onto S is 1/2 1/ /2 1/ /2 1/ /2 1/2 (ii) Here, since the last two vectors are already orthogonal, we will change their order to (1, 1, 0, 0), (0, 0, 1, 1), (1, 0, 0, 1). Of course it does not change anything, except that the first step of the Gram-Schmidt process leaves the first two vectors unchanged now, instead of messing everything around. For the third one: (1, 1, 0, 0) (1, 0, 0, 1) 0, 1, 1) (1, 0, 0, 1) ũ 3 = (1, 1, 0, 0) (1, 1, 0, 0) (0, (0, 0, 1, 1) = (1/2, 1/2, 1/2, 1/2). (1, 1, 0, 0) (1, 1, 0, 0) (0, 0, 1, 1) (0, 0, 1, 1) 7

8 Finally, we just need to normalize them. For the first two it s as before. For the third one, it already has norm 1, so it remains unchanged. Therefore, we get the following orthonormal basis for S: ( ) ( ) ( u 1 = 2,, 0, 0, u 2 = 0, 0,,, u 3 = , 1 2, 1 ) 2, 1. 2 To find the matrix of the projection onto S, one can use the same formula as in the previous part of the problem. But, out of laziness, I ll just use part (ii) of Problem 1 of homework 5. Since S is 3-dimensional, it suffices to find a vector orthogonal to S with norm 1, and then we can use the result in the previous homework. Notice that all three vectors u 1, u 2, u 3 satisfy x 1 x 2 +x 3 x 4 = 0, so they all lie in the hyperplan defined by this equation. A vector orthogonal to that plane is given by the coefficients of the variables in that equation, so the vector (1, 1, 1, 1) will do it. We divide it by its norm (which is 2) to make it have length 1: we obtain v := (1/2, 1/2, 1/2, 1/2). The the projection of R 4 onto S is the projection of R 4 onto span (v), which is the following by problem 1.(ii) of homework 5: 3/4 1/4 1/4 1/4 1/4 3/4 1/4 1/4 1/4 1/4 3/4 1/4 1/4 1/4 1/4 3/4 Problem 6: Showing all row operations, calculate the determinant of Solution: We start by substracting the first column from the second, third, and fourth ones: det Then we substract the second row from the fifth one: det Now, we can develop with respect to the fifth row. The only element will be the coming coming 8

9 from the 3 in row-column (5, 1). The associated sign is ( 1) 1+5 = +1. So, this determinant is equal to det Now, substract the second row from the first one: det We develop with respect to the first row. Again, the only non-zero term is the one corresponding to the 5 in row-column (1, 4), coming with sign ( 1) 1+4 = 1. So, this determinant is equal to det Now, add the second colum to the third one: det Develop with respect to the third row; the only nonzero term is the one corresponding to the 1 in row-column (1, 3), coming with sign ( 1) 3+1 = +1: ( ) det. 1 2 Now, add twice the second row to the first one: ( ) det. 1 2 Develop with respect to the first row: the only nonzero term is the one corresponding to the 1 in row-column (1, 2), coming with sign ( 1) 1+2 = 1: 15 det( 1). By linearity of the determinant, det( 1) = ( 1) det(1), and det(1) = 1 by the normalization property of the determinant. Therefore, this determinant is equal to 15. 9

10 Problem 7: This problem asks you to prove the factor theorem. (i) Suppose f(x) = a d x d + a d 1 x d a 1 x + a 0 is a degree d polynomial (so a d 0) with coefficients in a field F. Prove that the number of zeros (elements x F with f(x) = 0) is at most d. (ii) Conversely, if E F with E d, then there is a degree d polynomial that vanishes on E, i.e., f(x) = 0 for all x E. Solution: We start by proving a classical lemma: Lemma 0.3 If f(a) = 0 for some a F, then (x a) divides f(x). Proof. Let g(x) = f(x+a). Then g(0) = f(a) = 0, hence g does not have independent term. So, g(x) = b d x d + +b 1 x for some b 1,, b d F. Since f(x) = g(x a), we have f(x) = b d (x a) d + +b 1 (x a), which is clearly divisble by (x a). (i) Suppose that f(x) has k distinct zeroes, which we call x 1,, x k. Then, by the Lemma above, the polynomial (x x 1 )(x x 2 ) (x x k ) must divide f(x). So, f(x) must have degree at least k. Therefore, if f has degree d, we must have d k. (ii) Write E = {e 1,, e k } with k d. Consider f(x) = (x e 1 )(x e 2 ) (x e k ); clearly it vanishes on E. In case k < d, it does not quite have degree d, so let s artificially modify its degree by taking for example f(x) = (x e 1 ) d k+1 (x e 2 ) (x e k ). Problem 8: Suppose f(x) = a d x d + a d 1 x d a 1 x + a 0 is a degree d polynomial with coefficients in R. Prove that f(x) has a nonzero multiple in which all the exponents are prime numbers. For instance, such a multiple of f(x) = x 2 x + 5 is x 5 + 4x 3 + 5x 2 = (x 3 + x 2 )(x 2 x + 5). You may assume that that there are infinitely many prime numbers. Solution: Pick d + 1 different prime numbers p 1,, p d+1. Using the long division for polynomials, write x p i = q i (x) f(x) + r i (x), (1) for polynomials q i (x) and r i (x) such that the degree of r i (x) is smaller or equal to d 1. Then the collection {r 1,, r d+1 } is a collection of d+1 polynomials in the space of polynomials of degree smaller or equal to d 1, which has dimension d. It follows that {r 1,, r d+1 } must be linearly dependent, so there exists c 1,, c d+1 R such that c 1 r 1 (x) + + c d+1 r d+1 (x) = 0. Then, multiplying the Equation (1) by c i and summing over i, we get d+1 c i x p i = i=1 ( d+1 ) d+1 c i q i (x) f(x) + c i r i (x), i=1 which shows that d+1 i=1 c ix p i is a multiple of f(x). 10 i=1 } {{ } =0

11 Problem 9: Let σ be the permutation which sends the ordered 5-tuple (12345) to (54321). (a) Prove that if we write σ as the product of adjacent transpositions, then the minimum possible number of such adjacent transpositions is 10. (b) Prove that if we write σ as the product of transpositions, then the minimum possible number of such transpositions is 2. Solution: (a) We can certainly do it in 10 adjacent transpositions: we bring 5 in the first spot using 4 adjacent transpositions, then 4 (which is now in the fifth spot) on the second spot using 3 adjacent transpositions, then 3 (which is now in the fifth spot) on the third spot using 2 adjacent transpositions, then 2 (which is now in the fifth spot) on the fourth spot using 1 adjacent transpositions. So we used = 10 adjacent transpositions. Now, let s show that we can not do it in less than 10 adjacent transpositions. The idea comes from the first part of the proof of the Lemma 1.5, page 63 of Simon s book (in particular, the sixth line before the end): an adjacent transposition can only change the number of inversions by 1 (this is defined on page 62 of Simon s book). It is proven carefully in the book, but the idea is that if we switch two adjacent elements in a transpositions, only the relative order of these two will be affected, so the number of transpositions can only increase or decrease by 1. Since (54321) has 10 inversions and (12345) has 0 inversions, we need at least ten adjacent transpositions to go from one to the other. (b) A transposition of a set of n elements has n 2 fixed points. Since σ has only 1 fixed point, it can not be written as a single transposition. But we can write it as τ 1,5 τ 2,4, proving that we can write it as the product of 2 transpositions, which is furthermore the minimal number. 11