Concrete Mathematics: A Portfolio of Problems

Transcription

1 Texas A&M University - San Antonio Concrete Mathematics Concrete Mathematics: A Portfolio of Problems Author: Sean Zachary Roberson Supervisor: Prof. Donald Myers July, 204

2 Chapter : Recurrent Problems Many problems in mathematics are recurrent problems, meaning that they are defined in terms of themselves. For example, one can find the greatest common divisor of two integers a and b by the following relation: gcd(a, a) = ; gcd(a, b) = gcd(b, a mod b) Each successive step reduces the problem into a simpler case. Another recursive function is the factorial: a 0 = ; a = ; a n = na n Solving these recurrences can be done in many ways. One may be able to deduce a closed form after finding a pattern in the first few terms. Another method is to unfold the recurrence by successively plugging in previous terms until the base case is reached. Other methods include using the characteristic polynomial for linear recurrences, or the use of generating functions for general recurrences. Generating functions will be examined in depth in Chapter 7 (see the section titled Generating Functions). In this section, we examine one well-known recurrent problem, known as the Josephus Problem. According to legend, Flavius Josephus and a group of Jewish rebels were captured by Romans. The rebels formed a circle and killed every third person in the circle until all were dead. The main question of the Josephus problem is as follows: Suppose n people, numbered from to n, stand in a circle. If every other person is killed, which person is the last to survive? To investigate this problem, suppose there are six people in the circle. The first person to die is numbered 2, then next is 4, followed by 6, 3, and. The person numbered 5 is the last one surviving. Let J(n) be the person left standing after all other n persons have killed themselves. We can then say J(6) = 5. Now, let us suppose that we have 2n people in this circle at the start. The first person to die is number 2, then 4, and every even-numbered person until the sword (or whatever object is used to commit suicide with) returns to person number. Now, only odd-numbered persons are left in the circle. This situation is similar to starting with n people, only this time their labels have been multiplied by 2 and had subtracted. For example, person 5 would become person 9. So, a form for the survivor amongst an even-numbered group of people is J(2n) = 2J(n). Using the fact that J(6) = 5, we can deduce that J(2) = 2J(6) = 9. We can then conclude that J(24) = 7, and J(48) = 33.

3 But what if there are an odd number of people from the start, say, 2n +? Execution begins as normal, with persons 2, 4, 6, 8,..., 2n dying in the first trip around the circle, but now the next to die after 2n is person. The people left in the circle are 3, 5, 7,..., 2n +. This is similar to the case with n people, as before, but now numbers are increased by after they are doubled, not decreased by. Again, we have a form for determining the survivor in a group of 2n + people; that is, J(2n + ) = 2J(n) +. For example, J(5) = 2J(7) + = 2(2J(3) + ) + = 5. Combining the previous two general forms with the base case J() =, we have the following recursion: J() = ; J(2n) = 2J(n) ; J(2n + ) = 2J(n) +. For this recursion, it is sufficient to divide the number of people in the circle by 2 and round down to the nearest integer. Repeated application will give the survivor s number. What if we desire a closed-form solution? Such a solution is possible to construct. First, let us generate a short table of values of J(n). n J(n) Notice that the table is blocked in certain areas. This sparks some curious thought. 2

4 First, let s see what happens to powers of two. Suppose k is an integer that is at least zero. Then what is J(2 k )? For k = 0, we have J() =, and for k =, we have J(2) = (from the table). Now, assume that, for all integers k up to j, that J(2 j ) =. Now, for k = j +, we have: J(2 j+ ) = 2J(2 j ) = 2() = as needed. Hence J(2 k ) =, where k is an integer greater than or equal to zero. What s next to show? Let s try to see what happens between powers of two, say, from 8 to 6. We see that J(8) =, J(9) = 3, J(0) = 5, and J() = 7. What s happening? The value of the function increases by 2 as n increases by. But, we see that J(5) = 5, and J(6) =. So, something happens between n = 2 j and n = 2 j. Observe that 5 = , and 6 = 2 4. From the table, this is a transition between new blocks. But what happens at this transition? We see that 8 is the largest power of two that does not exceed 5. While 8 is not larger than 6, it is not the biggest power of two that is less than 6 (here, the desired number is 6). We almost see a pattern. Let l = n 2 j, where 2 j = max{2 p 2 p < n}. We conjecture that J(n) = 2l +. For n =, that is, = , we have J() =, as needed. The case n = 2 is trivial, by the previous derivation, so we move to n = 3. Here, we write 3 = 2 + and so J(3) = 2() + = 3. Now assume this holds for every n = 2 j, inducting on j. For the case j = q +, we must first consider even l. So, suppose l is even. Then J(2 q+ + l) = 2(J(2 q + l/2)) = 2(2(l/2) + ) = 2l + as needed. Otherwise, when l is odd (say, l = 2p + ), the induction step is as follows: J(2 q+ + 2p + ) = 2J(2 q + p) + = 2(2p + ) + = 2(l ) + 3 = 2l + From the first to the second line, we assumed that p was even. However, it can be shown that any even number can be divided by 2 a finite number of times to produce an odd number. So, for odd l, J(2 q + l) = 2l +. 3

5 We combine this with the previous result to give a suitable closed form solution for the Josephus problem: where n = 2 q + l, and l < 2 q. 8. Solve the recurrence J(n) = 2l + Q 0 = α; Q = β; Q n = + Q n Q n 2. Solution. The recurrence doesn t seem to have any recognizable pattern at first, so the best way to proceed is to start finding terms using the initial conditions. Q 2 = + β α + + β Q 3 = α β = + α + β αβ + + α + β αβ Q 4 = + β α = + α β + + α β Q 5 = + α + β αβ = α But, note that Q 5 = Q 0. This suggests that for every n, Q n = Q n 5. This recurrence is linear, and it can be solved. To solve this recurrence, let r n = Q n. Then we have r n = r n 5, or, equivalently, r 5 =. Solving this equation in the complex numbers gives the solution r = cos ( ) ( 2π 5 + i sin 2π ) 5. This value of r can actually be written as exp ( ) 2πi 5. To account for the initial condition, it suffices to solve ( ) 2πi 0 α = c exp, 5 whose solution gives c = α. Hence, the solution to the recurrence is ( ) 2nπi Q n = α exp. 5 4

6 2 Chapter 2: Sums In mathematics, sums appear in almost every branch. For example, they appear in linear algebra when computing the (i, j) entry of a matrix product: n c ij = a ik b kj k= where the sum is across k. Also, they appear in calculus and analysis when approximating a function with a polynomial: T (x) = j=0 f (j) (a) (x a) j j! But how do sums really work? How can one manipulate a sum and get a closed form? There are many ways to view a sum. First, one can view a sum as a recurrence. Suppose, for the sake of example, that S n = n j=0 a j, where the terms a j come from a sequence. Observe that the sum can also be written as the following recurrence: S 0 = a 0 ; S n = S n + a n. This recurrence says that each successive value of the sum depends on the previous terms. There are many well-known sums to keep in mind. Suppose one wishes to sum the first n integers in succession; that is, what is n? There are many ways to solve this problem. Perhaps one of the most well-known ways to find this sum is by adding a copy of the sum to itself. This technique is attributed to Gauss, and is told through a story. Following Gauss method, we see that, after first calling our sum S, 2S = n n = (n + ) + (n + ) (n + ) }{{} n terms = n(n + ) And from this, we deduce that S = n(n + ) 2. These are called the triangular numbers. Using the notation above, we can also write the sum as S = n n(n + ) j=0 j =. 2 Working with sums may require trickery sometimes. For example, it may be advantageous to change the index of summation to force a form already known. In the sum 7 j can make the substitution j = n + 2 so that the sum transforms to j=3 5 n + 3 n= 5

7 and this sum is easier to evaluate. Because sums arise so often, particular sums have been given names, like the triangular numbers. One other special sum gives the square pyramidal numbers, which come from the sum of the first n integer squares: P n = n j 2 = j= n(n + )(2n + ) 6 There is also a name for the sum of the first n unit fractions, the harmonic numbers. H n = n j= So, sums do appear often in mathematics, but sometimes more tools are needed, other than expanding the sum, adding it to itself, and re-indexing. What else can be used to evaluate a sum? Fortunately, there is a powerful tool available, involving a discrete analogue to calculus. In calculus, one can differentiate functions by the following definition: j f (x) = lim h 0 f(x + h) f(x) h The finite calculus has a similar operator, the difference operator. Just like the derivative of infinite calculus, the difference operator shows change. It is defined as: f(x) = f(x + ) f(x) The difference operator can be thought of the differential operator in which h =. So, how can we use the difference operator? It works just like differentiation. For example, what is (x 2 )? By the definition, it is (x + ) 2 x 2 = 2x +. In infinite calculus, polynomials differentiate well by the power rule. Finite calculus doesn t do the same sort of thing with ordinary polynomials, as seen above. A new type of power can be defined that transforms well under the difference operator. This new power is a falling power, defined as x n = n factors {}}{ x(x )(x 2)... (x m + ) It is left to the reader to verify that (x n ) = nx n. Just as the differential operator D in infinite calculus has an inverse operator, the difference operator has an inverse, the sum (or anti-difference) operator. So, just as we say g(x) = D(f(x)) g(x) dx = f(x) + C we can say g(x) = f(x) g(x) δx = f(x) + C where g(x) δx describes all functions whose difference is g(x). Also, the C that comes as a result of indefinite summation need not be a constant; it can be a function φ(x + ) = φ(x). 6

8 Definite summation can be defined like definite integration. We have b g(x) δx = f(x) a b a = f(b) f(a). Many of the usual differentiation rules carry over to the finite calculus. The difference operator is linear, so it preserves addition and scalar multiplication. That is, A product rule also exists: (αf + βg) = α (f) + β (g). (u(x)v(x)) = u(x + )v(x + ) u(x)v(x) = u(x + )v(x + ) u(x)v(x + ) + u(x)v(x + ) u(x)v(x) = v(x + )(u(x + ) u(x)) + u(x)(v(x + ) v(x)) = v(x + ) u(x) + u(x) v(x) = Ev(x) u(x) + u(x) v(x) where Ef(x) = f(x + ). Now, just as integration by parts can be derived by the product rule of infinite calculus, a discrete analogue of this method exists in the finite calculus. The following steps are true: (u(x)v(x)) = Ev(x) u(x) + u(x) v(x) (u(x)v(x)) δx = Ev(x) u(x) + u(x) v(x) δx u(x)v(x) = u(x) v(x) + Ev(x) u(x) u(x) v(x) = u(x)v(x) Ev(x) u(x) The last line gives the summation by parts formula: u(x) v(x) = u(x)v(x) Ev(x) u(x) A proof of this formula using rules of summation is given at the end of this section. What follows are example sums that can be evaluated with summation by parts. First, examine the sum Hx δx. This is an indefinite sum of harmonic numbers. The harmonic numbers are the discrete analogue of natural logarithms. To see this, consider the integral ln x dx. 7

9 With the choices u = ln x, dv = dx, integration by parts gives x ln x x + C. The same reasoning is used for this sum: let u = H x, and v =. We then have: Hx δx = xh x δx = xh x x + C Another example involves exponential functions. What is xc x δx, where c is an integer other than? Again, consider this the discrete analogue of the integral xe x dx, whose result is xe x e x + C. Again, proceed by summation by parts: let u = x and v = c x. Then the sum is now: xc x δx = xcx c (c + ) x δx c = xcx (c + )x c c 2 + C What follows is an alternative proof of summation by parts.. The general rule (2.56) for summation by parts is equivalent to (a k+ a k ) b k = a n b n a 0 b 0 a k+ (b k+ b k ), 0 k<n 0 k<n for n 0. Prove this formula directly by using the distributive, associative, and commutatitve laws. Solution. Before proceeding, note that the procedure is similar to finding an expression for the derivative of a product. That is, a cross-term (in this case, a k+ b k+ ) must be added and subtracted. Now, the summand can be writen as: (a k+ a k ) b k = a k+ b k a k b k = a k+ b k a k b k + a k+ b k+ a k+ b k+ = a k+ (b k b k+ ) a k+ + a k+ b k+ a k b k = a k+ (b k+ b k ) Summing from 0 to n, observe that the last two terms form a telescoping sum, so what remains is a n b n a 0 b 0. Combining this with the new sum formed gives the result. 8

10 3 Chapter 3: Integer Functions There are two functions that are used in computer science for estimates. These functions, known as the floor and ceiling, allow a person to force any real number to an integer. Both these functions are related by inequalities; this relationship will be explained below. The floor function maps a real number x to the largest integer that does not exceed x. In short, it rounds the number down. The floor of x is denoted x. For example, 3.5 = 3 and = 2. The ceiling function behaves similarly. It maps a real number x to the smallest integer that exceeds x. The ceiling of x is denoted by x. In short, it rounds a number up. For example, 3.4 = 3 and = 3. The floor and ceiling functions are related by a series of inequalities. The first should be easy to see: x x x This inequality can actually be extended in both directions by adding x on the lower end and x + on the upper end: x x x x < x + Observe that there is strict inequality in the ends, while equality can occur in the middle terms. This equality is met when x is an integer. There are a series of equivalences that involve floors and ceilings; these can be used to simplify expressions in computations: x = m m x < m + x = n n < x n x = m x < m x x = n x n < x + Manipulating sums with floors and ceilings takes extra work. For example, consider the sum m k k=0 If a table of values of k were to be created, one would find that there are repeated occurrences of k in each interval [n 2, (n + ) 2 ]. In order to find the desired sum, we must first count the number of times k appears in each interval. Each interval, except possibly the last, sums to k(2k + ). The last interval may not contain this number, so it is enough to call the sum in the last interval m ( m m 2 + ). The remaining sum is as follows: m k=0 k(2k + ) = n(n + )(2n + ) 3 + n(n + ) 2 9

11 where n = m. The final sum is then m k = k=0 n(n + )(2n + ) 3 4 Chapter 4: Number Theory + n(n + ) 2 + (n + )(m (n + ) 2 + ). The study of number theory is centered around the integers. There are many functions whose domain is the set of integers. In number theory, some important functions include the Euler phi function φ(n), and the Möbius mu function µ(n). This section will focus on these two functions and their relationship to one another. The Euler phi function counts the number of positive integers less than n that are relatively prime to n. So, if S = {m N gcd(m, n) = and m < n}, then φ(n) = S. For small values of n, one can find φ(n) by counting the integers that satisfy the condition. For example, φ(6) = 2 and φ() = 0. For larger numbers, it may be more difficult to count the integers relatively prime to n and less than n (unless a computer is programmed to do so). Before attempting to find a way to compute the number of integers less than and relatively prime to n, it would be wise to first observe what happens when n is prime, or a prime power. So, suppose p is a prime. The integers less than p are {, 2, 3,..., p }. But which of these are relatively prime to p? Recall that a prime has only two divisors, and itself (these are called the trivial divisors). Since p is prime, none of the integers from to p divide evenly into p. So, φ(p) = p. What about φ(p 2 )? A similar argument can be used. Construct the set S = {, 2,..., p, p,..., 2p,..., p(p ),..., p 2 } Note that p divides p 2, so any multiple of p not exceeding p 2 will divide p 2. In S, there are p multiples of p that divide p 2. The phi function counts the integers relatively prime to p 2, so we must have φ(p 2 ) = p 2 p. In general, we have, for prime p, φ(p n ) = p n p n. There are two ways to use the phi function without directly counting the integers that meet the requirement. One way is to use a product definition: φ(n) = p n ( ) p p where the product is taken over all primes p that divide n. A derivation of this formula is omitted. A second way is to use the fact that the phi function is multiplicative. relatively prime, then φ(mn) = φ(m)φ(n). That is, when m and n are 0

12 This helps in breaking the argument into smaller factors for which the values of the function are easily known. For example: φ(78) = φ(2)φ(39) = φ(2)φ(3)φ(3) = 2 2 = 24. The Möbius function is a function that depends on the prime factorization of an integer n. It is is defined as follows:, n = ; µ(x) = ( ) r, if n is a product of r distinct primes; 0, if n is not squarefree. For example, µ(6) = and µ(36) = 0. It is important to note that if one sums values of µ(n) over the factors of n, the sum will be zero except when n =. That is, For example, µ(d) = [n = ]. d n µ(d) = µ() + µ(2) + µ(4) + µ(8) + µ(6) d 6 = = 0. Just like the phi function, the mu function is multiplicative. Again, when m and n are relatively prime, µ(mn) = µ(m)µ(n). Both the phi and mu functions are examples of arithmetic functions. These functions preserve either addition or multiplication, much like a homomorphism. A special product can be defined with arithmetic functions, using a sum. This product is called the Dirichlet convolution, written as (f g)(n), where f and g are arithmetic functions. The convolution is defined as follows: (f g)(n) = d n ( n ) f(d)g d Two equivalent convolutions are of great importance. The Möbius inversion formula, due to Richard Dedekind and Joseph Liouville, states that g(n) = d n f(d) f(n) = d n ( n ) µ(d)g. d

13 In the notation of the Dirichlet convolution, the inversion formula can be written as g(n) = (f )(n) f(n) = (µ g)(n), where, as a function, is the constant function in which every integer is mapped to the number. Now, the phi function has a special property such that when a sum is taken over all factors d of the argument n, one finds that the value of the sum is n. That is, φ(d) = n d n Now, suppose that f(n) = n, the identity map. Then one could apply the Möbius inversion formula (or, equivalently, the Dirichlet convolution) to see that φ(n) = d n µ(d) n d. This gives a relationship between both functions under the convolution. 5 Chapter 5: Binomial Coefficients The binomial coefficients come from combinatorics, a branch of mathematics involving counting. They arise in the problem of taking a certain number of objects from a larger set. We define them as follows: ( ) n n! = k k!(n k)! where n! is the factorial of n. The left side is read n choose k. By convention, this is defined where n k, even if k is allowed to be negative. Otherwise, if k exceeds n, then the value of the coefficient is zero. There are many identities involving the binomial coefficients. What if k were replaced with n k? Then we have: ( ) n = n k so the binomial coefficients admit symmetry. n! (n k)!(n (n k))! = n! (n k)!k! = ( ) n k There is also an identity involving the sum of two binomial coefficients, but requires a bit of thought: ( ) n + k ( ) n = k ( ) n + This is Pascal s identity. The identity says the number of ways to choose k objects from n plus the number of ways to choose k from those same n is equal to the number of ways to select k objects from n +. This identity can be expressed in fewer words by referring to Pascal s triangle, a visual representation of the binomial coefficients. When written out, Pascal s identity says that an entry is equal to the sum of the two entries above it. There are many more identities involving binomial coefficients; too many to be explained in this paper. They will be referenced in problems, and it is left to the reader to prove the truth of these identities. k 2

14 Why are these numbers called binomial coefficients? They arise in the binomial theorem, the expansion of binomials to integer powers. The binomial theorem is as follows: For example, (x + y) n = n j= ( ) n x j y n j j (x + y) 2 = x 2 + 2xy + y 2 (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3 (x + y) 4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 The coefficients on each term represent a binomial coefficient, or, from a counting perspective, the number of ways to arrange x and y, where the exponent on each determines how many of each variable there are. Here s one example of a sum involving binomial coefficients. Example. Find the sum: Solution. ( n m ) ( k n k) k=0 The summand is a quotient of binomial coefficients. To simplify the summand, we use the trinomial revision identity: ( )( ) n m m k = ( )( ) n n k k m k This almost matches the form of the summand. Dividing both sides of the identity by ( n k)( n m) gives the desired form. So, the sum can now be written as n k=0 ( n k m k ) ( n m). In fact, the denominator can be taken outside the summand, since it does not depend on the counter k. The indices of summation can be changed to one simple condition, k 0, since when k exceeds m, the binomial coefficient will be zero. We now evaluate the simpler sum k 0 ( ) n k. m k This summand still needs to be simplified. The next step is to replace k with m k, and the sum is now m k 0 ( ) n (m k) m (m k) or, equivalently, ( ) n m + k. k k m 3

15 The sum is almost evaluated fully. The last trick uses the parallel summation rule: ( ) ( ) r + k r + n + = k n k n So, letting r = n m, we obtain and now we must find ( n+ m ) ( n m). ( ) ( ) n m + k n + = k m k m ( n+ m ) ( n m) = (n + )! m!(n m)! m!(n m + )! n! = n + n m + so the desired value for the sum is n + n m +.. What is 4? Why is this number easy to compute, for a person who knows binomial coefficients? Solution. Observe that = 0+, so we may expand using the binomial theorem. We see that (0+) 4 expands as follows: (0 + ) 4 = ( ) 4 (0 3 ) + ( ) 4 (0 2 ) + 2 ( ) 4 (0) + = This is easy for anybody that knows binomial coefficients since one could expand by the binomial theorem instead of multiplying by itself 4 times. Also, observe that the result is the fifth row of Pascal s triangle. One may conjecture that this pattern continues for the sixth row, but it fails, since 5 = Evaluate ( ) k by negating (actually un-negating) its upper index. Solution. By the upper negation identity, ( ) k ( ) k ( ) = ( ) k k ( ) k = ( ) k k () (2) = ( ) k. (3) (4) 4

16 But, this is actually incomplete. Note that the binomial coefficients are undefined when the lower index is negative. To complete the simplification, we must add the Iversonian bracket for the test condition k 0. So, ( ) k = ( ) k [k 0]. 30. What hypergeometric series F satisfies Solution. zf (z) + F (z) = z? We solve this differential equation and translate the solution as a hypergeometric series. Observe that the left side of this equation is already in a suitable form to write it as the derivative of the product of two functions. Of course, one may divide both sides of the equation by z and observe that the integrating factor is the function µ(z) = z. Upon multiplication by z, one is returned to the original equation. This equation can then be written as Integrating gives (zf (z)) = z. zf (z) = ln ( z) from which we see that the solution to the differential equation (ignoring constants of integration) is F (z) = ln ( z). z We must now express this solution as a hypergeometric series. Note that So, after replacing z with z, we have ln ( + z) = z k 0 ( z) k (k + ). ln ( z) = z k z (k + ) k 0 which, more compactly, is the hypergeometric function F (, ; 2; z). 5

17 6 Chapter 6: Special Numbers There are particular sequences of numbers that appear frequently in mathematics. For example, the Fibonacci numbers {,, 2, 3, 5, 8, 3,...} and the harmonic numbers show up in different problems. This section will focus on one particular number sequence, the Bernoulli numbers. These numbers will be used in the Euler- Maclaurin summation formula, given in Chapter 9. The Bernoulli numbers were discovered by Jacob Bernoulli (known in other sources as James or Jacques). These numbers occur in formulas used to sum powers of integers. The first five of these sums are shown below. n j = 2 n2 2 n j= n j 2 = 3 n3 2 n2 + 6 n j= n j 3 = 4 n4 2 n3 + 4 n2 j= n j 4 = 5 n5 2 n4 + 3 n3 30 n j= n j= j 5 = 6 n6 2 n n4 2 n2 A pattern emerges in the coefficients in these sums. First, the leading coefficient is a unit fraction whose denominator is one more than the leading power of n. Second, the coefficient of n k, where k is the degree of the polynomial representing the sum, is always 2. A series of patterns continues for successive terms. In general, when summing mth powers, the form of the sum is: S m (n) = m ( ) m + B k n m+ k, m + k k=0 where B k represents the Bernoulli numbers. The sums show that the Bernoulli numbers are multiplied by a binomial coefficient, so they are hidden in a simplified form of the sum. The Bernoulli numbers can be generated by using their exponential generating function (see Chapter 7): z e z = k 0 where the coefficients are the Bernoulli numbers. Any specific Bernoulli number can be found by taking derivatives of the generating function with respect to z and taking the limit as z tends to 0. In short, d n ( ) z B n = lim z 0 dt n e z. B k z k An implicit recurrence also defines the Bernoulli numbers: m ( ) m + B j = [m = 0], j j=0 k! 6

18 where the recurrence is defined for all integers m greater than or equal to zero. The first few Bernoulli numbers are listed below. n B n In Chapter 9, the Bernoulli numbers will be used in the Euler-Maclaurin summation formula to give estimates for definite summations. What follows are problems involving other special numbers. 4. Express in terms of harmonic numbers n + Solution. This sum of unit fractions includes only those with an odd-numbered denominator. We can easily get the desired sum by looking at the sum of the first 2n + unit fractions. This sum is the harmonic number H 2n+. If we subtract the terms with even-numbered denominators - that is, subtract the sum n, then we are done. Note, however, that this string of terms simplifies to 2 H n. So, the desired sum is H 2n+ 2 H n. 0. What is the continued fraction representation of φ? Solution. Note that + φ = φ. To see this, clear the fractions and solve for φ, ignoring the negative solution. Now, to get the continued fraction expansion, replace φ in the denominator by + φ, giving φ implicitly. This process continues to infinity, so the desired expansion is 43. Prove that the infinite sum φ = converges to a rational number. 7

19 Solution. Observe that each term contains a Fibonacci number, However, each term is multiplied by a new power of ten. We conclude that each term is of the form F n 0 n, where F n is the nth Fibonacci number. There is a generating function for the Fibonacci numbers, namely the function G(x) = Now, let x = 0. Then we must compute the sum x x x 2. n 0 F n 0 n which, by the above generating function, sums to /0 89/00 = Chapter 7: Generating Functions Generating functions allow us to view a sequence in a different way. The idea of a generating function stems from needing to view a sequence s terms and possibly giving an interpretation to them. To first understand a generating function, one must understand a power series. A power series is a polynomial (possibly of infinite degree) whose coefficients are created by some sort of rule. In general, a power series looks like where a j is one term from a sequence, say, A. a j x j j 0 Perhaps the most basic power series is the geometric series, j 0 which converges to x if x <. All the coefficients on each xj are, so we can say that the sequence generated by the function is {,,,...}. x Of course, there are many more series than this. One can create series for any differentiable function (possibly of class C ) by taking derivatives and dividing by a necessary factorial. This creates a Taylor series, whose form is as follows: j 0 x j f (j) (a) (x a) j j! where f is differentiable and a is in the domain of f. All the generating functions in the remainder of this paper will be centered at a = 0; these are Maclaurin series. The following table lists some functions, their power series representations, and the sequence their coefficients generate. 8

20 Sequence Generating Function Closed Form {, k, k 2,...} (kz) n kz n 0 { (, c ) (, c ) ( 2, c } ( ) c 3),... z n ( + z) n n n 0 {,, 2, 6, 24 },... z n e z n! n 0 {0,, 2, 3 }, 4,... ( ) n+ z n ln ( + z) n n Other generating functions can be found by variable replacements and elementary operations. For example, the series for ln ( z) can be created by replacing z with z, and the geometric series (the first entry in the table) is related to the logarithm by termwise differentiation and integration. These basic generating functions, along with basic operations, can be used to manipulate a difficult function and express it in terms of simpler functions. We will solve a recurrence using generating functions. Consider the Fibonacci recurrence: F 0 = 0; F = ; F n = F n + F n 2 The first step is to write the recurrence as one single equation. Iversonian brackets are needed here. So, the recurrence can be written as F n = F n + F n 2 + [n = ] Next, the equation must be multiplied by x n and summed across all n. Letting F n x n = G(x), we have G(x) = F n x n + F n 2 x n + x n 0 n 0 where n 0[n = ]x n = x. Each of the sums almost looks like G(x), but with appropriate manipulations, they can be transformed to G(x). The first term must be multiplied and divided by x, and the second must be multiplied and divided by x 2. Doing so gives the equation G(x) = xg(x) + x 2 G(x) + x x Rearranging and solving for G(x) gives the solution G(x) =, so we conclude that the generating x x2 x function for the Fibonacci numbers is x x 2. Many different recurrences can be solved using the same methods. Sometimes the equation to be solved may be a differential equation, while others may be an algebraic equation in G(x). 3. What is n 0 H n/0 n? 9

21 Solution. ln ( z) This sum matches the form of the generating function, with m = 0 (see equation 7.43). z ( ) 0 0 So, with z = /0, we see the desired sum is 9 ln. 9 8 Chapter 9: Asymptotics Sometimes, in mathematics, an estimate is good enough for an answer. Estimates are used in the approximation of integrals, derivatives, and sums. There is one formula that is used to approximate the value of a sum, since it may be difficult to give an exact answer. The formula of interest is the Euler-Maclaurin Formula, stated as follows: ( ) b m B k f(k) = f(x) dx + k! f (k ) (x) b + R m a a x<b a where R m is a remainder term, and B k are Bernoulli numbers. For purposes of this paper, the remainder will not be explained in depth. The formula requires derivatives, integrals, and Bernoulli numbers. An example sum will be computed. Example. Estimate the value of Solution. k<5 k 6. Take f(x) to be x 6. The integral is easy to compute: 5 x 6 dx = 57 7 k= What is left to compute is the sum with Bernoulli numbers. For that sum, take m = 5. Then the sum to evaluate is: ) ( x x5 6 2! 20x ! which, with the aid of a computer, comes to approximately Hence, the total estimation comes to , while the actual sum is The Euler-Maclaurin formula has its power to approximate known sums. Here, that true value and the estimated value were very close together. The next sum is another well-known series. Example. Estimate the value of Solution. k<0 k. 20

22 We begin as before. Of course, the integral is still easy to compute: 0 x dx = ln 0 while the sum still requires a bit of work. Again, with the aid of a computer, and m = 5, the Bernoulli sum comes to: 5 k= d k dx k ( x ) Bk k! while the total is ln The true value for this sum (the ninth harmonic number) is approximately 2.828, an error of As these two examples show, the Euler-Maclaurin formula is a powerful tool to approximate sums to a certain accuracy. Of course, the sums were estimate with error of about O(x 5 ), but the results still agreed to a tolerable number of decimal places. References [] Pete L. Clark. Arithmetical Functions II: Convolution & Inversion. Internet. [2] Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. Concrete Mathematics. Addison-Wesley Profession, 2nd edition, 994. [3] Victor H. Moll. The Bernoulli Numbers. Internet. Contains generating function for Bernoulli numbers, including a way to get each term by differentiation. 2