MATH NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS TOMASZ PRZEBINDA. Final project, due 0:00 am, /0/208 via e-mail.. State the Fundamental Theorem of Algebra. Recall that a subset K R n is called compact if and only if every finite covering of K contains a finite subcovering. 2. Prove that a closed subset of a compact set is compact. 3. For j =, 2, 3,..., let I j = [a j, b j ] [c j, d j ] be rectangles in R 2 such that Prove that I I 2 I 3.... I j. () j= 4. Prove that any rectangle [a, b ] [a 2, b 2 ] R 2 is compact. 5. Show that a disc of radius > R > 0 in the complex plane C is compact. D R = {z C z R} 6. Let f : R n R m be a continuous function and let K R n be a compact set. Show that the image f(k) R m is compact. 7. Let f : C R be a continuous function and let D R = {z C z R} be a closed bounded disc in C. Show that there is z 0 D R such that f(z 0 ) = inf{f(z); z D R }. 8. Let a 0, a,..., a n C and let P (z) = z n + a n z n +... + a z + a 0 (z C)

2 TOMASZ PRZEBINDA be a polynomial of degree n. Show that P (z) z n ( a n z... a z n a 0 z n ) (z C). 9. With the notation of Problem 8 show that lim P (z) =. z 0. With the notation of Problem 8, suppose inf{ P (z) ; z C} > 0. (2) Show that there is z 0 C such that inf{ P (z) ; z C} = P (z 0 ).. With the notation and assumptions of Problem 0 let Check that Q(0) = and that Q(z) = P (z + z 0) P (z 0 ) (z C). Q(z) (z C). 2. With the notation of Problem 0, check that there is 0 < k n and complex numbers b k 0, b k+,..., b n such that Q(z) = + b k z k +... + b n z n. 3. Check that for a nonzero complex number b there is t R such that e it b = b. 4. With the notation of Problem show that there is t R such that Q(e it z) = b k z k + b k+ (e it z) k+ +... + b n (e it z) n. 5. With the the notation of Problem 3 show that for 0 < r, Q(re it ) r k ( b k r b k+... r n k b n ).

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 3 6. With the the notation of Problem 5 show that there is ɛ > 0 such that b k r b k+... r n k b n > 0 for 0 < r < ɛ. 7. Show that the result of Problem 6 contradicts the statement of Problem. 8. What is the reason for the contradiction in Problem 7? 9. Prove the Fundamental Theorem of Algebra (stated in Problem ).

4 TOMASZ PRZEBINDA. Compute the integral with the accuracy 0.0000. 2. Homework 0 e x2 dx, 2. Compute the integral π with the accuracy 0.00. 0 sin(x) x dx,

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 5. Let Show that 3. Solutions to homework 0 f(x) = { e /x if x > 0, 0 if x 0. f (n) (0) = 0 (n = 0,, 2,...). For a polynomial P (x) consider { P (/x)e f P (x) = 0 if x 0. if x > 0, By l Hospital Rule, this function is continuous. The derivative for x 0 is of the same form with P (/x) replaced by Q(/x) = (P (/x) P (/x)/x 2. Therefore f (0) exists and is equal zero. We apply the same argument to f Q and deduce that the second derivative exists and is equal to zero if x = 0. We continue... till we get tired of it. 2. Show that the function defined in the previous problem is not equal to its Taylor series centered at x = 0. The equality f(x) = n=0 f (n) (0) x n n! is impossible in any open interval containing 0, because the right hand side is zero wile the left hand side isn t. 3. Solve for x + x 2! + x2 4! + x3 6! + x4 xn +... + +... = 0. 8! (2n)! Clearly there are no solution x 0. Assume x < 0. Then the above series is equal to x n (2n)! = ( x ) n ( ) n ( x ) 2n = = cos( x ). (2n)! (2n)! n=0 n=0 Thus the equation to be solved is The slotions are given by n=0 cos( x ) = 0. π x = + kπ (k Z). 2

6 TOMASZ PRZEBINDA Hence the solutions of the original equation are x = ( π 2 + kπ)2 (k Z). 4. Let Compute the derivative f(x) = (n 2 + n + )x n ( x < ). n=0 f (5) (0). Since f is given in terms of a convergent power series, In particular f (n) (0) n! = (n 2 + n + ). f (5) (0) = 5!(5 2 + 5 + ).

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 7 4. Solutions to Exam 2.. Construct a bounded set E R which has exactly three limit points. E = {0} { ; n =, 2, 3,...} {} {+ ; n =, 2, 3,...} {2} {2+ ; n =, 2, 3,...}. n n n 2. Give an example of (a) a compact subset of R that is not connected, (b) a connected subset of R that is not compact, (c) a subset of R that is both compact and connected. (a) [0, ] [2, 3], (b) (0, ), (c) [0, ]. 3. Let f, g : R R be two continuous functions. Show that the set of points where they agree, {x R : f(x) = g(x)}, is closed. Let E = {x R : f(x) = g(x)} and let h(x) = f(x) g(x). Then E = h ({0}) is the preimage of zero via the function h. The complement of E in R is equal to the the preimage of the complement of {0} in R: R \ E = h (R \ {0}). Since h is continuous and since R \ {0} is open, this set is open. Hence E = h ({0}) is closed. Another argument looks as follows. We need to show that E contains all its limit points. Let x n E and let x = lim n x n be a limit point of E. Then Hence x E. f(x) = lim n f(x n ) = lim n g(x n ) = g(x). 4. Let f : R R be a continuous odd function (f( x) = f(x)). Show that there is x 0 R such that f(x 0 ) = 0. If f(x) = 0 for all x R, then there is nothing to prove. If not, then there is x R such that f(x ) > 0. Then f( x ) < 0. From the Intermediate Value Theorem we know that f takes all the values between f(x ) and f( x ). In particular there is x 0 R such that f(x 0 ) = 0. There is also an argument that does not assume the continuity of f. Since we see that f(0) = 0. f(0) = f( 0) = f(0) = f(0)

8 TOMASZ PRZEBINDA 5. Give an example of a Riemann integrable function f : [, ] R. The constant function f(x) = 0, x [, ] is Riemann integrable. 6. Give an example of a function f : [, ] R which is not Riemann integrable. As explained in class is not Riemann integrable. f(x) = { if x is rational, 0 if x is irrational 7. A set S of real valued functions defined on the interval [ π, π] is called orthonormal if for any f, g S, π { if f = g, f(x)g(x) = 0 if f g. Check that the functions 2π, form an orthonormal set. Recall that Hence, Therefore Similarly, Obviously π π Since cosine times sin is odd π π π cos(nx), π sin(nx) (n =, 2, 3,...) (3) cos 2 (t) = (cos(2t) + ). 2 cos 2 (nx) dx = 2 π π π π π ( ) π n sin(nx) + = π π ( π cos(nx)) 2 dx =. ( π sin(nx)) 2 dx =. π π ( 2π ) 2 dx =. π cos(nx) π sin(mx) dx = 0.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 9 Furthermore, for n m, Hence Similarly 2 π π cos(nx) cos(mx) dx = π π π π π π (cos((n m)x) + cos((n + m)x) dx = 0. π cos(nx) π cos(mx) dx = 0. π sin(nx) π sin(mx) dx = 0. 8. Find a continuous function f such that f(x) = x 0 f(t) dt +. Since f is continuous, the right hand side is differentiable. and f (x) = f(x). Hence f(x) = e x will do. Hence f is differentiable 9. Find a continuous function f : R (0, ) such that for x > 0, f(x) t dt = x The left hand side is equal to ln(f(x)). Hence ln(f(x)) = x Thus f is the inverse of the logarithm. Hence f(x) = e x.

0 TOMASZ PRZEBINDA 5. Solutions to homework 9.. Let { if x = for some m N f(x) = m 0 otherwise. Show that f is Riemann integrable on [0, ] and compute the integral 0 f(x) dx. Let P be an arbitrary partition of the interval [0, ]. Since every interval [x i, x i ] contains points outside the sequence (m N) m where the function is zero, we see that L(P, f) = 0. Fix ɛ > 0. Let N > 0 be an integer such that N < ɛ 2 and let 0 < δ < N(N + ). Suppose the partition P has diameter (maximum of x i x i ) smaller than δ. Then each m, m =, 2, 3,..., N is in a single interval [x i, x i ], because m m + = m(m + ) N(N + ) > δ. Let A denote the set of all such i. Then the cardinality of A is at most N. Hence sup{f(x); x [x i, x i ]}(x i x i ) (x i x i ) < δ Nδ < N +. i A i A i A On the other hand sup{f(x); x [x i, x i ]}(x i x i ) i/ A i/ A Hence N (x i x i ) i U(P, f) N + + N < ɛ. N (x i x i ) = N. 2. Decide whether each statement is true or false, providing a short justification for

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS each conclusion. (a) If g = h for some h on [a, b], then g is continuous on [a, b]. (b) If g is continuous on [a, b], then g = h for some h on [a, b]. (c) If H(x) = x h(t) dt is differentiable at c (a, b), then h is continuous at c. a (a) Yes, we verified it in class. (b) Yes, this is the FTC: h(x) = x g(t) dt. a (c) No, take h to be zero everywhere except one point. Then H(x) = for all x, hence it is differentiable, but h is not continuous. 3. Show that if f : [a, b] R is continuous and x f(t) dt = 0 for all x [a, b], then a f(x) = 0 everywhere on [a, b]. By the FTC for all x. f(x) = d dx x a f(t) dt = d dx 0 = 0 4. Let γ n = + 2 +... + n ln(n). Prove that γ n converges. (The constant γ = lim n γ n is called Eulers constant.) We compute Thus γ n n γ n = + 2 +... + n n = n n + k= = n n + k= k+ k ( k ) t k(k + ) n n + t dt = n n + n k dt n n + k= k= k= k+ k k < 2 n + k= k= k+ k ( k k + k < 2 2 k= is a bounded increasing sequence. Hence the limit exists. t dt ) dt k 2 <. 5. Verify the Cauchy - Schwartz inequality for Riemann integrable functions: b b b f(x)g(x) dx f(x) 2 dx g(x) 2 dx. a a a Set (f, g) = b a f(x)g(x) dx.

2 TOMASZ PRZEBINDA We need to check that (f, g) 2 (f, f)(g, g). We may assume that (g, g) 0. Notice that 0 (f tg, f tg) = (g, g)t 2 2(f, g)t + (f, f) (t R). From the theory of quadratic functions we know that this happens if and only if This is the inequality we wanted to verify. (2(f, g)) 2 4(g, g)(f, f) 0. 6. With the notation of previous problem, set b f = a f(x) 2 dx. Show verify the triangle inequality: f + g f + g. It ll suffice to check that ( f + g ) 2 ( f + g ) 2. But this amounts to checking the Cauchy-Schwartz inequality, which we just did.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 3 6. Solutions to homework 8.. Give an example of a continuous function which has continuous derivative F : R R F : R R but the second derivative F (x) does not exist for any x R. Let f : R R be the Weierstrass function which is continuous but has no derivative anywhere. Define F (x) = x 0 f(t) dt (x R). Then the Fundamental Theorem of Calculus implies that F is differentiable and F (x) = f(x). Hence F is continuous, but the second derivative F does not exist. 2. Suppouse f : R R is a function such that Show that f is a constant function. f(x) f(y) (x y) 2 ( < x < y < ). Since f(x) f(y) x y x y, we see that the derivative f (y) exists and is equal to zero for any y. Hence f is constant. 3. Exactly one of the following requests is impossible. Decide which it is, and provide examples for the other three. In each case, let s assume the functions are defined on all of R. (a) Functions f and g not differentiable at zero but where fg is differentiable at zero. (b) A function f not differentiable at zero and a function g differentiable at zero where fg is differentiable at zero. (c) A function f not differentiable at zero and a function g differentiable at zero where f + g is differentiable at zero. (d) A function f differentiable at zero but not differentiable at any other point. (a) This is possible. Let f(x) = x for x 0 and f(x) = 0 for x 0. Then f has no derivative at zero. Similarly, if g(x) = 0 for x 0 and g(x) = x for x 0, then g has no derivative at zero. However f(x)g(x) = 0 for all x and therefore is differentiable everywhere. (b) This is possible. Let f(x) = x and g(x) = 0. Then f does not have the derivative at zero, but the product fg is the constant zero function, so it does have the derivative at every point.

4 TOMASZ PRZEBINDA (c) This is impossible because f(x) = (f(x) + g(x)) g(x) is the difference of two differentiable functions, hence differentiable. (d) This is possible. Let g(x) be the Weierstrass function, which is continuous and has no derivative anywhere. Define f(x) = g(x)x. Then f(x) f(0) = g(x), x 0 which has limit g(0) if x goes to zero. Thus the derivative at zero exists f (0) = g(0). On the other hand g(x) = f(x)x for x 0. If f (x) existed, then the derivative of the product would exist too, but it doesn t.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 5 7. Solutions to homework 7.. Consider the open interval (0, ). For each point x (0, ), let O x be the open interval ( x 2, ). Taken together, the infinite collection {O x; x (0, )} forms an open cover for the open interval (0, ). Show that this collection does not contain a finite subcover. See Example 3.3.7 in the text. 2. Let K and L be nonempty compact sets. Define d = inf{ x y : x K and y L}. If K and L are disjoint, show d > 0 and that d = x 0 y 0 for some x 0 K and y 0 K. Since the set { x y : x K and y L} is bounded, the infimum exists and there are sequences x n K and y n L such that lim x n y n = d. n Since K is bounded, there is a subsequence x nk K which has a limit x 0 K. Also, lim x n k y nk = d. k Since L is compact, there is a subsequence y nkl lim x n kl y nkl = d. l But lim x n kl y nkl = x 0 y 0, l so x 0 y 0 = d. In particular d = 0 if and only if x 0 = y 0, i.e. K L. K which has a limit y 0 L. Also, 3. Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample. (a) The arbitrary intersection of compact sets is compact. (b) The arbitrary union of compact sets is compact. (c) Let A be arbitrary, and let K be compact. Then, the intersection A K is compact. (a) Yes, because it is closed and bounded. (b) No, R itself is a union of single points and each of them is a compact. (c) No, for example Q [, ] is not closed, hence not compact. 4. Show that the limit lim sin(x ) x 0

6 TOMASZ PRZEBINDA does not exist. See Example 4.2.6 in the text. 5. (Extreme Value Theorem). If f : K R is continuous on a compact set K R, then f attains a maximum and minimum value. In other words, there exist x 0, x K such that f(x 0 ) f(x) f(x ) for all x K. See Theorem 4.4.2 in the text. 6. Let f, g : R R be continuous functions. Show that the composition f g is continuous. Let U R be an open set. Then f (U) is open because f is continuous. Since g is continuous, g (f (U)) is open. Thus is open. (f g) (U) = g (f (U)) 7. Show that if f : R R is continuous function and E R a connected set, then f(e) is connected. Suppose f(e) is not connected. Then there are open sets U, V R such that U f(e), V f(e), f(e) (U V ) = and f(e) U V. Then E f (U) f (V ), f (U) E, f (U) E and E (f (U) f (V )) =. Thus E is not connected, a contraction. 8. (Intermediate Value Theorem). Let f : [a, b] R be continuous. If L is a real number satisfyingf(a) < L < f(b) or f(a) > L > f(b), then there exists a point c (a, b) where f(c) = L. We know from Problem 7 that f([a, b]) is connected. Hence it is an interval, thus the claim follows. 9. Provide an example of each of the following or explain why the request is impossible. (a) A continuous function defined on an open interval with range equal to a closed interval. (b) A continuous function defined on a closed interval with range equal to an open interval. (c) A continuous function defined on an open interval with range equal to an unbounded closed set different from R. (d) A continuous function defined on all of R with range equal to Q.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 7 (a) Let f(x) =, for < x <. Then the range of f is the single point {} = [, ]. (b) This is impossible because the image by a continuous function of a compact set is compact, in particular it is closed. (c) Let f(x) =, for < x <. Then the range of f is equal to [, ). x 2 (d) This is impossible because the range must be connected, i.e. an interval, and Q is not.

8 TOMASZ PRZEBINDA 8. Solutions to homework 6.. Prove that the only sets that are both open and closed in R are R and the empty set. Suppose B R is a non-empty set, which is both open and closed. Then B = R \ A is open and closed. Also R = A B and A B = set. As we checked in class, this is impossible. 2. Only one of the following three descriptions can be realized. Provide an example that illustrates the viable description, and explain why the other two cannot exist. (i) A countable set contained in [0, ] with no limit points. (ii) A countable set contained in [0, ] with no isolated points. (iii) A set with an uncountable number of isolated points. The statement (i) is impossible because any bounded sequence has a limit point. For (ii) the subset of the rational numbers in the interval is countable and has no isolated points. Suppose S R and A S in uncountable and consists of the isolated points of S. Thus for each a A, there is ɛ a > 0 such that V 2ɛa (a) S =. In particular V ɛa (a) b A V ɛb (b) =. Thus we have an uncountable number of mutually disjoint non-empty open intervals Each set V ɛa (a) R (a A). V ɛa Q (a A) is countable, but there are uncountably many of them. Since the union of them is contained in Q V ɛa Q Q, a A which is countable, this is a contradiction. 3. Decide whether the following statements are true or false. Provide counterexamples for those that are false, and supply proofs for those that are true. (a) An open set that contains every rational number must necessarily be all of R. (b) The Nested Interval Property remains true if the term closed interval is replaced by closed set. (c) Every nonempty open set contains a rational number. (d) Every bounded infinite closed set contains a rational number. (e) The Cantor set is closed.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 9 (a) This false. The set ( 2) ( 2, ) is open and contains Q. (b) This is false: [n, ) =. n= (c) This is true. We checked it in class: between any two real numbers there is a rational number. (d) This is false. The set 2 { n + 2; n =, 2, 3,...} { 2} has the required properties but contains no rational numbers. (e) This is true. As an intersection of closed sets, Cantor set is closed. 4. Decide whether the following sets are open, closed, or neither. If a set is not open, find a point in the set for which there is no ɛ-neighborhood contained in the set. If a set is not closed, find a limit point that is not contained in the set. (a) Q. (b) N. (c) {x R : x 0}. (d) { + + + + : n N}. 4 9 n 2 (e) { + + + + : n N}. 2 3 n (a) This set is neither open nor closed. For every point in x Q there is no ɛ-neighborhood contained in the Q. Every irrational number is a limit point of rational numbers. These are the limit points of Q that are not in Q. (b) This set is closed, but not open. For every x N there is an ɛ-neighborhood not contained in the N. (c) This set is open but not closed. Here 0 is the limit point which is not in this set. (d) This set is neither open nor closed. The point is in the set but every ɛ neighborhood of contains points from outside of the set. Hence this set is not open. Also, the number n= is a limit point which does not belong to the set. Thus this set is not closed. n 2 (e) This set is not open, but it is closed. The point is in the set but every ɛ neighborhood of contains points from outside of the set. Hence this set is not open. Since the series diverges, this set has no limit points. Hence it is closed. n= n 5. Show that if K is compact and nonempty, then supk and infk both exist and are elements of K. Since K is bounded, both supk and infk exist. They are both limits of elements of K. Since K is closed, they belong to K.

20 TOMASZ PRZEBINDA 6. Decide which of the following sets are compact and explain why. (a) N. (b) Q [0, ]. (c) The Cantor set. (d) { + + + + : n N}. 4 9 n 2 (e) {,, 2 n,...,,...}. 2 3 n (a) This set is not compact because it is not bounded. (b) This set is not compact because it is not closed. (c) This set is compact because it is bounded and closed. (d) This set is not compact because it is not closed. (e) This set is not compact because it is not closed.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 2 9. Solutions to homework 5.. Find all complex numbers z such that z 3 = 8. Let z = re iθ. Then our equation looks as follows r 3 e 3iθ = 8. Hence, r 3 = 8. Thus r = 2. Therefore the equation reduces to However where k is any integer. Thus From this we see that Thus the distinct values of e iθ are e 3iθ =. = e i(π+2kπ), e 3iθ = e i(π+2kπ). θ = π 3 + 2kπ 3. e i π 3, e i( π 3 + 2π 3 ), e i( π 3 + 4π 3 ). Hence, the solutions of the original equation are z = 2e i π 3, z2 = 2e i( π 3 + 2π 3 ), z 3 = 2e i( π 3 + 4π 3 ). 2. Prove the identity + z + z 2 +... + z n = zn+ z (z C; z, n = 0,, 2,..). It suffices to multiply both sides by z: ( z)( + z + z 2 +... + z n ) = + z + z 2 +... + z n (z + z 2 +... + z n + z n+ ) = z n+. 3. Check that under the correspondence of complex numbers and matrices ( ) x y x + iy y x

22 TOMASZ PRZEBINDA the complex conjugation coincides with the transposition of matrices. This is obvious, because x + iy = x iy and ( ) t ( x y x y = y x y x ). 4. Check that under the correspondence of complex numbers and matrices ( ) x y x + iy y x the radius squared r 2 coincides with the determinant of matrix. Again, this is straightforward: and ( x y det y x r 2 = x 2 + y 2 ) = x 2 + y 2. 5. Deduce from the formula of Problem 2 that z n = (z C, z < ). z n=0 The point is that, if z <, then lim n z n+ z = z 6. Derive the trigonometric formulas cos(a + b) = cos(a) cos(b) sin(a) sin(b) (4) and sin(a + b) = cos(a) sin(b) + sin(a) cos(b) (5) from the identity e α+β = e α e β (α, β C). (6) We compute e ia+ib = cos(a + b) + i sin(a + b)

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 23 and e ia e ib = (cos(a) + i sin(a))(cos(b) + i sin(b)) = (cos(a) cos(b) sin(a) sin(b)) + i(cos(a) sin(b) + sin(a) cos(b)) and then equate the real and the imaginary parts. 7. Which of the following sets are open in R, A = {x R; x < 5} B = {x R; x 5} C = {x R; 5 < x 5},. Only A is open 8. Give an example of an infinite close subset F R which has no limit points. The set of all the integers has the property.

24 TOMASZ PRZEBINDA 0. Solutions to exam I.. Fix a non-zero number a. For n = 0,, 2,... define I n = ( x 2 ) n cos(ax) dx. Use integration by parts to show that I n = 2n(2n )a 2 I n 4n(n )a 2 I n 2 (n = 2, 3, 4,...). We compute ( x 2 ) n cos(ax) dx = ( x 2 ) n a sin(ax) = = n( x 2 ) n ( 2x)a sin(ax) dx n( x 2 ) n 2xa sin(ax) dx = n( x 2 ) n 2xa 2 ( cos(ax)) = = = = = = = d dx n( x 2 ) n ( 2x)a sin(ax) dx d ( n( x 2 ) n 2x ) a 2 ( cos(ax)) dx dx ( n( x 2 ) n 2x ) a 2 cos(ax) dx ( 2n((2n )x 2 )( x 2 ) n 2 )a 2 cos(ax) dx ( 2n((2n )x 2 )( x 2 ) n 2 )a 2 cos(ax) dx d ( n( x 2 ) n 2x ) a 2 ( cos(ax)) dx dx ( 2n((2n )((x 2 ) + ) )( x 2 ) n 2 )a 2 cos(ax) dx (2n(2n )( x 2 ) n 4n(n )( x 2 ) n 2 )a 2 cos(ax) dx 2n(2n )( x 2 ) n a 2 cos(ax) dx 4n(n ) = 2n(2n )a 2 I n 4n(n )a 2 I n 2. ( x 2 ) n 2 a 2 cos(ax) dx

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 25 2. Check that I 0 = 2a sin(a) and I = 4a 2 cos(a) + 4a 3 sin(a). We compute and I = = 2 I 0 = a sin(ax) = a (sin(a) sin( a)) = 2a sin(a) ( x 2 ) cos(ax) dx = ( x 2 )a sin(ax) ( 2x)a sin(ax) dx xa sin(ax) dx = 2xa 2 cos(ax) + 2a 2 cos(ax) dx = 4a 2 cos(a) + 2a 3 sin(ax) = 4a 2 cos(a) + 4a 3 sin(a) 3. Using the results of Problems and 2 compute I 2 and I 3. We compute I 2 = 2a 2 I 8a 2 I 0 = 2a 2 ( 4a 2 cos(a) + 4a 3 sin(a)) 8a 2 2a sin(a) = 48a 4 cos(a) + (48a 5 6a 3 ) sin(a) and I 3 = 30a 2 I 2 24a 2 I = 30a 2 ( 48a 4 cos(a) 2a 3 sin(a)) 24a 2 ( 4a 2 cos(a) + 4a 3 sin(a)) = ( 440a 6 + 96a 4 ) cos(a) + (440a 7 576a 5 ) sin(a). 4. Show that there are polynomials P n (a) and Q n (a) of degree at most 2n such that a 2n+ I n = P n (a) cos(a) + Q n (a) sin(a). For n = 0 ai 0 = 2 sin(a), so P 0 (a) = 0 and Q 0 (a) = 2. For n = a 3 I = a 3 (4a 2 cos(a) + 4a 3 sin(a)) = 4a cos(a) + 4 sin(a), so P (a) = 4a and Q (a) = 4.

26 TOMASZ PRZEBINDA Suppose the claim holds for n. We ll show that it holds for n. By assumption a 2n+ I n = a 2n+ (2n(2n )a 2 I n 4n(n )a 2 I n 2 ) = 2n(2n )a 2n I n 4n(n )a 2n I n 2 = 2n(2n )a 2n (P n (a) cos(a) + Q n (a) sin(a)) 4n(n )a 2n (P n 2 (a) cos(a) + Q n 2 (a) sin(a)) = (2n(2n )a 2n P n (a) 4n(n )a 2n P n 2 (a)) cos(a) + (2n(2n )a 2n Q n (a) 4n(n )a 2n Q n 2 (a)) sin(a) is of degree at most 2(n ). Hence a 2(n )+ P n (a) a 2n P n (a) is of degree at most 2(n ) which is smaller than 2n. Similarly, a 2n P n 2 (a), a 2n Q n (a) and a 2n Q n 2 (a) are of degree at most 2n. 5. With the notation of Problem 4 let p n (a) = P n(a) n!, p n (a) = Q n(a) n! Show that the polynomials p n and q n have integer coefficients. From the proof of the claim of Problem 4 we see that this is true if n = 0 or n =. Also p n (a) = n! (2n(2n )a2n P n (a) 4n(n )a 2n P n 2 (a)) = n! (2n(2n )a2n (n )!p n (a) 4n(n )a 2n (n 2)!p n 2 (a)) 2n(2n )(n )! = p n (a)a 2n 4n(n )(n 2)! p n 2 (a)a 2n n! n! = 2(2n )p n (a)a 2n 4p n 2 (a)a 2n. Since 2(2n ) and 4 are integers the claim follows.. 6. Assume that for some integers a and b, π = b a. Take a = π 2 in the formula proven in Problems 4 nd 5 and show that a 2n+ I n = n!p n (a) cos(a) + n!q n (a) sin(a) (7) b 2n+ I n n!

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 27 is an integer. By substituting a = π 2 in the formula (0) we see that ( π ) 2n+ In = n!q n ( π 2 2 ). Since we assume that π = b, this translates to a ( ) 2n+ b I n = n!q n ( b a a ). Thus b 2n+ I n = q n ( b n! a )a2n+. Since the polynomial q n is of degree at most 2n, the right hand side is an integer. 7. Show that b 2n+ I n lim n n! = 0. Notice that I n cos(ax) dx 2. Also, (b 2 ) n lim = 0 n n! because (b 2 ) n n=0 = e b2 converges. Hence the claim follows from the law of two man in n! black. 8. Deduce from Problems 6 and 7 that, under the assumption that π is a rational number, there is N such that for n > N I n = 0. Since the integers b2n+ I n n! have limit 0, they must be zero beginnig at some N. 9. Show that I n > 0 for all n. The function under the integral defining I n is non-negative and positive on some open interval. Hence the area under it is positive. 0. Deduce from Problems 8 and 9 that π is not a rational number.

28 TOMASZ PRZEBINDA Clearly the result of 9 contradicts the claim of 8. rational can t be true. Thus the assumption that π was. Write a joyful message to your grandparents expressing your pleasure of understanding why π is not a rational number.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 29. Compute the limit. Solutions to homework 4. lim n n +. Since < n + we have n + < Therefore Since < n + n + < n +. n + < n +. lim ( n n + ) =, we see from the law of two policemen that lim n n + =. 2. Compute the limit lim n ( n 2 + n n). n2 + n n = ( n 2 + n n)(( n 2 + n + n) n2 + n + n = n2 + n n 2 n2 + n + n = n n2 + n + n = = + + + = 2. n 3. Let x = 2 and for n x n+ = 2 + x n. (8) Show that the sequence x n has a limit and that this limit is an algebraic number. Compute that number using wolframalpha or other computer program.

30 TOMASZ PRZEBINDA Notice that this sequence is increasing. Indeed 2 < 2 + 2 2 + 2 < 2 + 2 + 2 2 + 2 + 2 < 2 + 2 +... Inductively, if x n < x n, then and therefore xn < x n 2 + x n < 2 + x n, 2 + 2 which means that x n < x n+. Also, x n < 2 for all n. This is true for n =. Suppose it is true for n. Then x n = 2 + x n < 2 + 2 < 2 + 2 = 2. Thus x n isd an increasing and bounded sequence. Therefore it has a limit, call it x. Going to the limit with both sides of the equation (8) we see that x = 2 + x. Hence Equivalently (x 2 2) 2 x = 0. x 4 4x 2 x + 4 = 0. Thus x is an algebraic number. In order to find the solution (which has to be real and between and 2, type solve for x, x 3 + x 2 3 x 4 = 0 in wolframalpha.com. The answer is complicated. 4. Let y = 2 and for n define Show that this sequence converges and that the limit equals 2. y n+ = 2y n. (9) As in the previous problem, but easier, we check that the sequence is increasing and bounded by 2. Hence the limit exists, call it y. Taking the limit of both sides of the equation (9) we see that y = 2y.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 3 Hence y = 2. 5. Give an example of each of the following, or argue that such a request is impossible. (a) A Cauchy sequence that is not monotone. (b) A Cauchy sequence with an unbounded subsequence. (c) A divergent monotone sequence with a Cauchy subsequence. (d) An unbounded sequence containing a subsequence that is Cauchy. (a) The sequence ( )n converges to 0. Therefore it is a Cauchy sequence. It is obviously not monotone, because it oscillates. n (b) This is impossible, because a Cauchy converges. Hence it is bounded. Therefore every subsequence is bounded. (c) This is impossible. The Cauchy subsequence has a limit. Hence this subsequence is bounded. Because of the monotonicity of the entire sequence, the entire sequence is bounded. hence it is convergent too. (d) x 2n = 2n, x 2n+ = 2n + (n =, 2, 3,...). 6. Does the series converge? Justify your answer. Yes, because and the geometric series converges. 7. Does the series converge? Justify your answer. No, because n= 2 n + n 2 n + n < 2 n n= n= 2 n ( ) n n + 2n n + lim n 2n = 2 0.

32 TOMASZ PRZEBINDA 8. Is there a sequence x n satisfying 0 x n n, where n= ( )n x n diverges? No, because this is an alternating sequence with the nth term going to zero.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 33 2. Solutions to homework 3.. Write a letter to your grandmother or grandfather explaining what are the real numbers what are their basic properties. Please don t use any mathematical symbols. Take a look at Rudin s book. 2. Divide x 3 2x + by x to obtain quotient q(x) and remainder r(x): x 3 2x + = (x )q(x) + r(x). x 3 2x + = (x )(x 2 + x ) +. 3. Write down an equation with rational coefficients a n x n +... + a x + a 0 = 0 one of whose solutions is x = 5 2. In other words, 5 2 is an algebraic number. x 4 4x 2 + 39 = 0. 4. Let Q( 2) denote the subset of R consisting of all the numbers of the form a + b 2 (A, B Q). Show that if x and y are in Q( 2) then so are x + y and xy. Let x = a + b 2 and y = c + d 2, with a, b, c, d Q. Then x + y = (a + c) + (b + d) 2 and xy = (ac + 2bd) + (ad + bc) 2. Since the numbers a + c, b + d, ac + 2bd and ad + bc are rational the claim follows. 5. Let Q( 2) denote the subset of R consisting of all the numbers of the form a + b 2 (A, B Q). Show that if x is in Q( 2) and x 0, then is in Q( 2). x Let x = a + b 2

34 TOMASZ PRZEBINDA with a, b Q. Notice that (a + b 2)(a b 2) = a 2 2b 2. (0) This number is not equal to 0, because a b 2 0, because 2 is not rational. Therefore we may divide by a 2 2b 2 and define Notice that by (0), y = a b 2 a 2 2b 2. Also, Thus y =. Since the numbers x a xy = (a + b 2)y = (a + b 2) a b 2 a 2 2b 2 =. y = a a 2 2b + b 2. 2 a 2 2b 2 are rational we are done. a 2 2b 2 and b a 2 2b 2 6. Let Q( 2) denote the subset of R consisting of all the numbers of the form a + b 2 (A, B Q). and let A R denote the field of the algebraic numbers. Show that A Q( 2). Given x = a + b 2 Q( 2), we need to find an equation with rational coefficients satisfied by x. Here it is x 4 2(a 2 + 2b 2 )x 2 + (a 2 + 2b 2 ) 2 8a 2 b 2 = 0. 7. Conclude that Q( 2) is a field containing Q and contained in A, Q Q( 2) A. The first inclusion is obvious and the second one was verified in Problem 6. The fact that Q( 2) is a field was proven in Problems 4 and 5. 8. Compute the limit lim n (2n) n.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 35 lim n (2n3 ) n = lim n 2 n lim n (n 3 ) n ( ) 3 = lim 2 n lim n n =. n n

36 TOMASZ PRZEBINDA 3. Solutions to homework 2.. Read chapter of Abbott s book. 2. Let A = {, 2, 3, 4, 5} and let B = {4, 5, 6, 7}. Write down the set A B and A B. A B = {, 2, 3, 4, 5, 6, 7}, A B = {4, 5}. 3. Give an example of a function f whose domain is equal to the set A = {, 2, 3, 4, 5} and the range to the set B = {4, 5, 6, 7}. Can f be one to one? Can it be onto? The f could be onto f() = 4, f(2) = 5, f(3) = 6, f(4) = 7, f(5) = 6. Since there are more elements in A than in B there is no one to one function with the domain A and values in B. 4. List all functions f : {, 2, 3} {, 2}. f () =, f (2) =, f (3) = f 2 () = 2, f 2 (2) = 2, f 2 (3) = 2 f 3 () = 2, f 3 (2) =, f 3 (3) = f 4 () =, f 4 (2) = 2, f 4 (3) = f 5 () =, f 5 (2) =, f 5 (3) = 2 f 6 () = 2, f 6 (2) = 2, f 6 (3) = f 7 () = 2, f 7 (2) =, f 7 (3) = 2 f 8 () =, f 8 (2) = 2, f 8 (3) = 2 5. How many functions f : {, 2, 3,..., m} {, 2, 3,..., n} are there? The answer is n m. (In previous example 2 3 = 8.) You can prove it via induction on m as follows.

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 37 If m =, then we are talking about functions f : {} {, 2, 3,..., n}. We can list them all f () = f 2 () = 2... f n () = n and see that there are n of them. Since n = n we are done with the initial step of the inductive proof. Now suppose there are n m functions f : {, 2, 3,..., m} {, 2, 3,..., n}. We need to show that there are n m+ functions F : {, 2, 3,..., m, m + } {, 2, 3,..., n}. Given any f : {, 2, 3,..., m} {, 2, 3,..., n} and any k n there is a unique function F : {, 2, 3,..., m, m + } {, 2, 3,..., n} such that Thus there are n extensions of F (j) = f(j) if j {, 2, 3,..., m} and F (m + ) = k. f : {, 2, 3,..., m} {, 2, 3,..., n} to F : {, 2, 3,..., m, m + } {, 2, 3,..., n}. Since the number of all the f is n m the number of all is n m n = n m+. 6. compute F : {, 2, 3,..., m, m + } {, 2, 3,..., n} sup{, 2, 3, 4, 5}, inf{, 2, 3, 4, 5}, sup (, 2), inf (, 2), sup [, 2), inf [, 2). sup{, 2, 3, 4, 5} = 5, inf{, 2, 3, 4, 5} =, sup (, 2) = 2, inf (, 2) =, sup [, 2) = 2, inf [, 2) =. 7. Show that for any subsets A, B R bounded from above, sup(a + B) = sup(a) + sup(b). See Exercise.3.6 in Abbott s book. 8. Show that R has the same cardinality as the open interval (, ).

38 TOMASZ PRZEBINDA The function R x x + x 2 (, ) is one to one and onto. The inverse is given by (, ) y y R. y 2

MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS 39 4. Solutions to homework.. Decompose 208 as a product of primes. You are welcome to goole it, but try to do it with bare hands first. 208 = 2 009. 2. How many prime numbers are there between 2 and 409? You are welcome to goole it. There are 80 of them. See https://mathcs.clarku.edu/ djoyce/numbers/primes.html. 3. Let p be a prime. Prove that there is no rational number r such that r 2 = p. Follow the proof for p = 2, done in class, word by word. 4. Read the section of Rudin s book, posted on my home page, which explains a construction of real numbers. Try to understand it. This might take a few hours.