IB Mathematics HL Year Unit : Completion of Algebra (Core Topic ) Homewor for Unit Ex C:, 3, 4, 7; Ex D: 5, 8, 4; Ex E.: 4, 5, 9, 0, Ex E.3: (a), (b), 3, 7. Now consider these: Lesson 73 Sequences and series (A) Define the function f(x) = n=0 x n n. (i) What is the domain of the function f? ( ) ( ) (ii) Compute f, f 3 (B) Consider the infinite series (i) Is this arithmetic, geometric? n= n(n + ). (ii) Compute the above series. (Note that =.) n(n+) n n+ 00 (C) Compute log a n + n= (D) Compute 4n. n= (E) Compute the following infinite continued fraction + + + +... (Hint: Let x be this quantity and solve for x.) (F) Prove that if x <, then ( )x = = x ( x) 3. (Hint: Doesn t this loo almost lie the second derivative of a geometric series?) (G) MPOS (Algebra), #, #5, #7, #9, #, #4, #7, #8, #. Unless specified otherwise, the homewor problems will come from the text, Mathematics for the international student, Mathematics HL (Core), by Paul Urban et al.
Homewor for Unit, continued Lesson 74 The Binomial Theorem; integral exponents; Ex 9G:, 3, 4, 6, 7. (A) Compute 0!! +!0! +! 9! + 3! 8! + (B) Show that for any positive integer n, = n. =0 (C) Show that for any positive integer n, ( ) = 0. =0 (D) Show that for any positive integer n, ( ) n =. =0 (E) Compute (Hint: expand (x+) n, tae the derivative, and set x ( =.) ) ( ) ( ) n n n =0 (F) Show that + = (Note: this is exactly what maes Pascal s triangle wor!) (G) MPOS (Algebra), #4, #6, #6. + 4!7! 5! 6! (A) Determine the first four terms of the series expansion of + x. For what values of x is this expansion valid? Lesson 74.5 The Binomial Theorem; fractional and negative exponents; (B) Compute n=0 nx n explicitly. For what values of x is yourexpansion valid? (C) Using the binomial expansion for ( + x) /, compute to 7 decimal places. 0.998 ( ) x (D) Compute in increasing powers of x, through + x the x 4 term. (E) Find the values of the constants a and b for which the expansions, in ascending powers of x, of the two expressions ( + x) / and + ax + bx up to and including x term, are the same.
Lesson 75 Mathematical induction Homewor from Unit, continued Ex 0B: (all parts), (a) (compare with (B) of Lesson ), 5 (all parts), 7 (a), (b). 3 (A) Recall the definition of the binomial coefficients = n!!(n )!. (i) Using Exercise (E) of Lesson, together with induction show that the number of subsets of size in a set of size n is exactly ( n ). (ii) Do you see how (i) maes the Binomial Theorem (a + b) n = a b n absolutely trivial? =0 (B) Try these: (i) Show that (ii) Show that (iii) Show that = 5 j= j= j(j + ) = n(n + )(n + ) 3 j(j + )(j + ) = n(n + )(n + )(n + 3) 4 j(j + )(j + )(j + 3) j= n(n + )(n + )(n + 3)(n + 4) (iv) Do you see any patterns? (C) Prove that for all n that n + n + + n + + + n >. (E) See appended discussion on sequences and finite differences, as well as for extra practice problems. I personally probably wouldn t use induction to obtain these results. For example, for part (a), assume that n = 3 + r; thus, r is the remainder when we divide 3 into n. Then (thin about it!) r is also the remainder when we divide 3 into n 3 ; that is, n 3 = 3 + r, for some integer (which doesn t really matter to us). Finally, n 3 + n = 3 + r + (3 + r) = 3 something, so we re done! 3 Suppose that you relax the hypothesis of the problem. What would the correcponding result be?
Lesson 76 The geometry of complex numbers Homewor from Unit, continued Ex 6A.: 4; Ex 6A.: 4; Ex 6B.: 4, 7, 9, 0,. Ex 6B.3:, 3, 4, 5 (thin about this result geometrically!) Ex 6B.4: 6; Ex 6B.5:,, 4; Ex 6C:,, 7, 9, 0,, 3. (A) Define the Chebyshev polynomial (of the first ind), T n (x), to be the polynomial with real coefficients satisfying T n (cos θ) = cos nθ. In other words, this polynomial is determined by woring out the multiple-angle formula for cos nθ as a polynomial in cos θ. Clearly, we have T 0 (x) =, since cos 0 θ = ; T (x) = x, since cos θ = cos θ; T (x) = x, since cos θ = cos θ. (i) Compute T 3 (x), and (ii) T 4 (x). (iii) Using induction, together with DeMoivre s Theorem, show that T n (x) is a polynomial of degree n. 4, 5 (B) Consider the complex number ζ = cos π π + i sin (i) Show that ζ = (so ζ is a root of unity.) (ii) Show that ζ 6 = (so ζ is a zero of the polynomial x 6 +). (iii) Find a polynomial with integer coefficients of lesser degree (than 6) which has ζ as a zero. (iv) Factor x completely (over the integers). (C) Let n 3 be an integer, and set α n = cis( π ) + n cis( π ) Show that α n =, α = +,..., α n = + + + (n iterations). To do this, consider showing that α n = + α n, n =, 3,.... (D) MPOS (Algebra), #, #3, #8, #0, #, #5, #9, #. 4 I ll give you a few hints on this one. First of all, note that if z = cos θ + i sin θ, then by DeMoivre s theorem z n + z n = cos nθ. ( n ( n ( n Next, note that n cos n θ = (z + z ) n = z ) n = (z n + z n ) + (z ) n + z n ) + (z ) n 4 + z 4 n ) + = ( =0 n ( n cos nθ + cos(n )θ + cos(n 4)θ +. Now what? ) )
Additional problems from Sadler and Thorning: Lesson 73: (Page ) 3, 4, 9,, 6; (page 6), 3, 4, 8, 4, 6,, 4, 5. Lesson 74: (Page 6) (a), (b), 6 (under what circumstances would you thin that the higher powers of x might be safely neglected?) 8, 9 (a), (b),, 3, 5, 8. Lesson 74.5: (Page 3) (a) (d), (all parts), 3, 4, 7 (a), (b) (Hint: ( x) 3 = x 3 ( x) 3), 9, 0. Lesson 76: Using mathematical induction, show that (i) ( + ) = 3 = (ii) ( + ) = n (Do you really need induction for this one? Now compute n + = ( + ).) = (iii)! = (n + )! = d n (iv) dx (f(x)g(x)) = n (F) of Lesson 74.) Also, (page 3), 3. =0 f () (x)g (n (x). (You may need the result of exercise Lesson 76: (Page 463),, 4 (do a few of these), 8, ; (page 467),, 3, 6, 7, 8, 9,, 3; (page 47),, 6. 5 One can show that T n(x) can actually be evaluated through the determinant of a matrix: x 0 0 0 0 x 0... 0 0 0 x... 0 0 T n(x) = det 0 0 x... 0 0. 0 0 0... 0................ 0 0 0 0 x
Discussion: Induction, Sequences and Finite Differences. We have already encountered arithmetic sequences; these have the form {a n } n=, where a n = an + d, where a and d are fixed real numbers and where n =,,.... As a result of this, we see immediately that for all n > that a n a n = d, a real constant. Furthermore, the converse of this is true: if {a n } n= is a sequence of real numbers whose successive differences a n a n is constant, then {a n } n= is an arithmetic sequence. Another way to describe arithmetic sequences is simply by noting that the terms a n of an arithmetic sequence are linear functions of n; indeed, such terms are of the form a n = an + d, for real constants a and d. Suppose, instead that we had the following non-arithmetic sequence:, 4, 7,, 6,.... What is the general recipe, i.e., can we find a formula for a n as a function of the index n? Well, what we notice that that the first-order differences are 4 =, 7 4 = 3, 7 = 4, 6 = 5 which means that the second-order differences are now constant: 3 =, 4 3 =, 5 4 =, and so on. You have no doubt encountered the differences summarized as follows: 4 7 6 3 4 5 Just as constant first-order differences imply a linear expression for the quantities a n, constant second-order differences imply a quadratic expression for the quantities a n. To see this, assume that we do have a quadratic relation of the form, a n = an + bn + c, where a, b, and c are fixed real constants, and where n =,,.... We wish to show that the second order differences (a n a n ) (a n a n ), n = 3, 4,..., are constant. Since the linear portions of a n will have zero second-order differences (since the firstorder differences are constant), we only need to worry about the quadratic parts, i.e., we only need to show that (an a(n ) ) (a(n ) a(n ) ) is constant. One easily checs that the above quantity reduces to a, which is constant, and so we re done. Conversely, assume that we have a sequence {a n } n= such that the second-order differences are constant; call this constant a. We shall show that there exist real numbers b, and c such that for each n =,, 3,..., we have a n = an + bn + c. We start by letting α, b, and
c be the unique real numbers satisfying αn + bn + c = a n for n =,, 3. We now that this is possible since there is a unique parabola passing through the points (, a ), (, a ), and (3, a 3 ). Next, note that a = (a 3 a ) (a a ) = 3 α + 3b + c ( α + b + c) + (α + b + c) = α which implies already that we must have α = a. We shall now use mathematical induction to prove that for all n, we have a n = an + bn + c. Assume that for for every < n, we have a = a + b + c. We shall show that it is also true that a n = an + bn + c. We have that and so a = (a n a n ) (a n a n ) a n = a + a n a n = a + a(n ) + b(n ) + c (a(n ) + b(n ) + c) (by induction) = an + bn + c; by mathematical induction, we conclude that for all n, a n = an + bn + c. Exercise. Return to the sequence given above, viz.,, 4, 7,, 6, and obtain a general quadratic recipe for the n-th term of this sequence. Exercise. Suppose that the third-order differences of a sequence are constant. What would you expect to happen? Can you prove this? Exercise 3. Consider the sequence P, P, P 3,..., where P n = + + + n. Compute the first-, second-, and third-order differences for P (),..., P (6). What do you observe? Can you use this to conjecture a general formula for P n? Can you prove this formula? Exercise 4. As in Exercise 3 above, consider the sequence P, P, P 3,..., where this time P n = 3 + 3 + + n 3. Compute the first-, second-, third-, and fourth-order differences for P (),..., P (6). What do you observe? Can you use this to conjecture a general formula for P n? Can you prove this formula?
Exericse 5. Prove the following: (i) + 3 + 5 + + (n ) = n (n =,,...) (ii) 3 + 3 + 3 3 + + n 3 = 4 n (n + ) (n =,,...) (iii) 3 + 3 5 + (n )(n + ) = n (n =,,...). (Do you really need n + mathematical induction? Try partial fractions!) ( ) ( ) ( ) (iv) + + + + < (n =, 3,...) 3 n n Exercise 6. (a) Prove ( ) that ( for any ) positive ( ) integer n and any integer with 0 n, n n n = + ( ) ( ) l n + (b) Show that for integers n, with 0 n that =. + (c) Loo again at Exercise (B), Lesson 57 (Unit 9). As you can see, the general problem boils down to counting the number of sequences (a, a,..., a ), where a, a,..., a are integers satisfying a a a n. Using induction, ( together ) with (b) above, show that the number of such sequences is equal n + to. (This immediately implies that the probability ased for in the above-mentioned problem (B) is simply ( 9 4) /6 4.) Exercise 7. Recall that the Fibonacci sequence is given by,,, 3, 5,.... In other words, if a n is the n-th element of this sequence, then we have the recurrence relation a n+ = a n+ + a n, n =,,.... Prove that for all n =,,..., for n =,,.... ( a n = 5 + ) n ( 5 5 l= ) n 5, Exercise 8. Prove that for all n., 3 + 3 + n 3 = ( + + 3 + + n). Exercise 9. Prove that for all n, and for all x 0, that ( + x) n > + nx. (Is induction really needed?) Exercise 0. Prove the classical inequality x + x + + x n n It s actually possible to arrive at this count directly by noticing that every nondecreasing sequence a a a determines uniquely the subset where b = a, b = a +, b 3 = a 3 +,..., b = a +. {b, b,..., b },
whenever x, x,... x n > 0 and x + x + x n =. (Hint: this is not really an induction problem! Indeed, show that if x, x,... x n > 0, then ( + + + ) (x + x + x n ) n. x x x n This result relies on the easily proved fact that x i x j + x j x i.) Exercise. Prove that for all integers n, sin x cos j x = sin nx. j= Exercise. Prove that for all integers n 0, sin x n j=0 cos j x = sin (n+ x) n+. Exercise 3. Prove that for all integers n 0, that sin(j )x = j= cos nx. sin x Exercise 4. (This is a bit harder.) Prove the partial fraction decomposition x(x + )(x + ) (x + n) = n! where n is a non-negative integer. ( ) x +, Exercise 5. We shall use mathematical induction to prove that all positive integers are equal. Let P (n) be the proposition =0 P (n) : If the maximum of two positive integers is n then the integers are equal. Clearly P () is true. Assuming that P (n) is true, assume that u and v are positive integers such that the maximum of u and v is n +. Then the maximum of u and v is n, forcing u = v by the validity of P (n). Therefore, u = v. What s wrong with this argument? Exercise 6. If A is a finite subset of real numbers, let π(a) be the product of the elements of A. If A =, set π(a) =. Let S n = {,, 3,..., n}, n and show that (a) A S n π(a) = n +, and that (b) A S n ( ) A π(a) = 0 Due to T.I. Ramsamujh, THE MATHEMATICAL GAZETTE, Vol. 7, No. 460 (Jun., 988), p. 3.
Exercise 7. If A is a finite subset of real numbers, let σ(a) be the sum of the elements of A. Let n, and set S n = {,, 3,..., n}, as above. Show that (a) ( 3 σ(a) π(a) = (n + n) (n + ) + + 3 + + ), and that n A S n (b) A S n ( ) A σ(a) π(a) = n 3 This is Problem # on the 0th USA Mathematical Olympiad, April 3, 99. It s really not that hard!