IB Mathematics HL Year 2 Unit 11: Completion of Algebra (Core Topic 1)

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1 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.

2 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 ( ) 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.

3 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?

4 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? ) )

5 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 x x T n(x) = det 0 0 x x

6 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: 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

7 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 = 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?

8 Exericse 5. Prove the following: (i) (n ) = n (n =,,...) (ii) n 3 = 4 n (n + ) (n =,,...) (iii) (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., n 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 },

9 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.

10 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 + ) ), 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!

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