The well-known Fibonacci sequence is defined by a recurrence. Specifically,letF i denote the ith number. Then, we have:

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1 Week 4 Recurrences (I) 4.1 The Fibonacci Sequence The well-known Fibonacci sequence is defined by a recurrence. Specifically,letF i denote the ith number. Then, we have: F 0 =0 (4.1) F 1 =1 (4.2) F n = F n 1 + F n 2,foralln 2. (4.3) Notice that the definition has 2 parts: the first 2 equations specify initial conditions while the last equation shows us how to generate the next term of the sequence, given the terms we have so far. This is quite similar to the Principle of Mathematical Induction, in which we show something is true explicitly for a base case, thenuseinductiontoshowthatitholdsinthenextcase,giventhatitholds for the cases we have considered so far. Such recurrences can be useful for a variety of counting problems. Here is one surprising example from Sanskrit poetry. Example 4.1. Suppose that each line of a poem is made up of some fixed number of beats. There are 2 kinds of syllables, heavy and light. Heavy syllables take up 2 beats, while light syllables take up 1 beat. How many di erent patterns of syllables are there for a line with n beats? Solution. There is one line with no beats (the empty line) and one line with a single beat (it has 1 light syllable). Think now about an arbitrary line with n 2 beats. How many ways can we make such a line? Well, we can end with either a heavy or light syllable (but obviously not both). If we end with a heavy syllable, 36

2 4.1. The Fibonacci Sequence MTH6109 Combinatorics (Fall 2017) the rest of the line must be made up of n 2beats. Thenumberoflinesthat end with a heavy syllable is then just the number of ways to make up an arbitrary line of n 2beats. Ifweendwithalightsyllable,therestofthelinemustbe made up of n 1beats. Thus,thenumberoflinesthatendwithalightsyllable is the number of ways to make up an arbitrary line with n 1beats.Bytherule of sum, we get that the number of lines with n beats is the number of lines with n 1 beats plus the number of lines with n 2 beats. By now, we recognise that this is exactly the Fibonacci sequence. 1 The recurrence gives us a way to compute F n iteratively starting from F 0.But what if we wanted to know F 10000? For this, it would be nice to have a direct formula for F n as a function of n alone. Over the next few lectures, we will see a general recipe for solving recurrences and turning recurrences into functions of n. Fornow,let sproceedaccordingtointuition Solving the Fibonacci Recurrence If we look at the first several terms of the Fibonacci sequence, we observe that they seem to be growing roughly exponentially. Based on this, we adopt the hypothesis that F n = x n,forsomeappropriatelychosenconstantx. Buthowdowedetermine x? Well,let slookattherecurrence.iff n = x n,then(4.3)becomes x n = x n 1 + x n 2. Dividing by x n 2 we get an order 2 polynomial: x 2 = x +1. Then, we only need to solve for x. Ignoringthetrivialsolutionx =0,wegetthe following pair of solutions (via the quadratic formula): For convenience, define: x = 1 ± p 5 2. = 1+p 5 2 = 1 p 5 2 Then, we must have that both F n = n and F n = n satisfy (4.3). We now need to make sure that the initial conditions (4.1) and (4.2) are valid. Here things get 1 There is admittedly some disagreement over whether we start the sequence with 0, 1, 1 or just with 1, 1, but you can verify the sequence of remaining terms is the same 37

3 4.2. Solving Linear Recurrences MTH6109 Combinatorics (Fall 2017) a little tricky. Notice that if F n = n and F n = n are both solutions to (4.3) then so F n = A n + B n for any constants A and B. Indeed, in that case we would have: F n = A n + B n = A( n 1 + n 2 )+B( n 1 + n 2 ) (Assume n and n satisfy (4.3)) =(A n 1 + B n 1 )+(A n 2 + B n 2 ) (Rearrange terms) = F n 1 + F n 2. So, let s choose A and B so that our initial conditions hold. If F n = A n + B n then equations (4.1) and (4.2) become: A 0 + B 0 =0 A 1 + B 1 =1 A + B =0 A + B =1. After a bit of algebraic wrangling, we find that A = p and B = p5. solution is then: Our F n = 1 p 5 n 1 p5 n = 1 p 5 1+ p 5 2! n 1 p5 1 2 p! n 5.. We can now observe another interesting property of the Fibonacci numbers. The number above is also known as the Golden Ratio, which occurs throughout nature and art. Notice that > 1while < 1. So, lim n n!1 =0,andthefirsttermisagoodapproximationforlargen. Intuitively, this means that for large n, the ratio between two consecutive Fibonacci numbers F is approximately the Golden Ratio. Indeed, one can show that lim n n!1 F n+1 =. 4.2 Solving Linear Recurrences We now generalise our previous discussion into a general recipe for solving recurrences of the following particular form. cients). We call a re- Definition 4.2 (Linear Recurrences with Constant Coe currence of the form: a n = b 1 a n 1 + b 2 a n b d a n d + f(n), where b 1,b 2,...,b d are constants and f : N! R, alinear recurrence of order d with constant coe cients. Iff(n) =0foralln, wesaythattherecurrence is homogeneous. Otherwise,wesaythatitisnon-homogeneous. 38

4 4.3. The Homogeneous Case MTH6109 Combinatorics (Fall 2017) 4.3 The Homogeneous Case First, we consider the easy case, in which f(n) =0. recurrence of the form: Then,weinfacthavea a n = b 1 a n 1 + b 2 a n b d a n d (4.4) where b 1,b 2,...,b d are constants. Suppose that we are also given d initial values a 0, a 1,..., a d.weproceedusingthefollowingrecipe. Step 1: Substitute x i for each term a i in the recurrence. Simplify to obtain adegreed polynomial in x. This is called the characteristic polynomial of the recurrence. We then solve the characteristic polynomial to obtain a set of roots for x. Suppose that we get k apple d di erent roots r 1,r 2,...r k with multiplicities m 1,m 2,...,m k.inotherwords,thecharacteristicpolynomialis equivalent to: (x r 1 ) m 1 (x r 2 ) m2 (x r k ) m k. Step 2: The general solution of the recurrence relation will be of the form: a n = r n 1 (A 1 + A 2 n + A 3 n A m1 n m 1 1 ) + r n 2 (B 1 + B 2 n + B 3 n B m2 n m 2 1 ) (4.5). + r n k (K 1 + K 2 n + K 3 n K mk n m k 1 ), where each capital letter represents an unknown coe cient. Notice that we will have exactly d unknown coe cients altogether. We also have d initial values a n for 0 apple n<d.pluggingthesedvalues into (4.5) together with the corresponding value of n gives a system of d linear equations in the unknown coe cients. We can therefore solve this system to find the correct coe cients. This gives the solution to the recurrence under the given initial conditions. This is all a bit abstract, so let s look at a concrete example. Example 4.3. Solve a n =3a n 1 4a n 3 with initial values a 0 =1,a 1 =2, a 2 =3. Solution. Let s go through our recipe 39

5 4.4. The Non-Homogeneous Case MTH6109 Combinatorics (Fall 2017) Step 1: We construct x n = 3x n 1 4x n 3. We simplify this to obtain the polynomial: x 3 3x 2 +4=0. We can factor this as So, we have k =2roots: (x +1)(x 2) 2 =0. r 1 = 1 m 1 =1 r 2 =2 m 2 =2. Step 2: Using the above roots, we form the general solution: a n =( 1) n A 1 +2 n (B 1 + B 2 n). (4.6) We combine this with our initial values to get the system: 1=a 0 =( 1) 0 A (B 1 +0 B 2 )=A 1 + B 1 2=a 1 =( 1) 1 A (B 1 +1 B 2 )= A 1 +2B 1 +2B 2 3=a 2 =( 1) 2 A (B 1 +2 B 2 )=A 1 +4B 1 +8B 2 Eliminating A 1 and simplifying, we get: 3=3B 1 +2B 2 5=6B 1 +10B 2 Finally, eliminating B 1 gives us B 2 = 1/6, and so B 1 =10/9, and A 1 = 1/9. Now, we substitute the values of these 3 coe cients into the general solution (4.6) to get our answer: a n = ( 1)n +2 n 9 n The Non-Homogeneous Case Now, let s turn to the more complicated case in which f(n) isnotidentically0. In this case, we need to guess an additional solution of a particular form in order to deal with f(n). Luckily, if f(n) issomepolynomialinn, thenthereisanapproach that is guaranteed to always work: guess a general polynomial in n of the same degree as f(n), then substitute into the recurrence to determine the coe cients.. 40

6 4.4. The Non-Homogeneous Case MTH6109 Combinatorics (Fall 2017) Let s formulate a general recipe for solving non-homogeneous linear recurrences with constant coe cients when f(n) is polynomial. In this case, we are looking at recurrences like: a n = b 1 a n 1 + b 2 a n b d a n d + f(n), (4.7) and assuming that f(n) isapolynomialofsomedegree`. As usual, to uniquely determine the solution, we must also be given d initial values of the form a 0,a 1,...a d. We now have a 3 step recipe: Step 1: Ignore the term f(n) intherecurrence(4.7)toobtainahomogeneous recurrence of the form: a n = b 1 a n 1 + b 2 a n b d a n d. Use Step 1 from Section 4.3 to get a general solution to this recurrence. That is, find the characteristic polynomial, determine its roots, and then construct a solution with d unknown coe cients, as in (4.5). Call this expression the general (homogeneous) solution a (h). Step 2: We assume that f(n)isapolynomialofdegreeh. Considerageneral degree ` polynomial p(n) with` + 1 unknown coe cients P i,0appleiapple`: p(n) =P`n` + P` 1 n` P 1 n + P 0. Substitute a n = p(n) intotheoriginalnon-homogeneousrecurrence. This gives a single equation, relating 2 degree ` polynomials. Two polynomials are equal for all n if and only if each term n i has the same coe cient for 0 apple i apple `. Thus, we now get ` +1 equations: one for each 0apple i apple ` that involves all the terms like Xn i.solvingthesewecanfindthevaluesofthe`+1 coe cients P i for p(n). If we set the coe cients to these values, the resulting polynomial will be a solution of our recurrence (4.7). This polynomial is a particular solution, which we call a (p) n. Step 3: We now find the remaining coe cients from Step 1, using our initial values. We take the solution: a n = a (h) n + a (p) n. We substitute each initial value into this equation (together with the corresponding value of n). We get d equations over d unknown coe cients as before. We then solve to find the solution. 41

7 4.4. The Non-Homogeneous Case MTH6109 Combinatorics (Fall 2017) Notice that we proceed essentially as in the homogeneous case, ignoring f(n), except that we carry out an intermediate step. Specifically, in Step 2 we find a particular solution corresponding to the term f(n). We then add this solution onto the homogeneous solution before finding our coe cients using the given initial values in the last step. As before, let s see an example. Example 4.4. Solve the recurrence a n =3a n 1 4a n 3 +6n 9withinitial values a 0 = 1, a 1 =5,anda 2 =18. Solution. Here we have f(n) =6n apply the recipe: 9, which is a polynomial of degree 1. Let s Step 1: Ignoring the term 6n 9correspondingtof(n) wegetahomogeneous recurrence: a n =3a n 1 4a n 3. We form the characteristic polynomial, find its roots, and construct a general solution. This is done exactly as in the previous example. We find: a (h) n =( 1) n A 1 +2 n (B 1 + B 2 n) (4.8) Step 2: Here f(n) =6n 9isapolynomialofdegree` =1. Wethusconstruct ageneraldegree1polynomial: p(n) =P 1 n + P 0. Substituting p(n) infora n in our original (non-homogeneous) recurrence for this problem we get: P 1 n + P 0 =3P 1 (n 1) + 3P 0 4P 1 (n 3) 4P 0 +6n 9. Now, equating the coe cients for each power of n above gives us 2 equations: one for terms involving n and one for constant terms: P 1 n =3P 1 n 4P 1 n +6n P 0 = 3P 1 +3P 0 +12P 1 4P 0 9. Solving the first we get P 1 = 3, and so we must then have P 0 = 9. So, our particular solution is: a (p) n =3n +9. (4.9) 42

8 4.4. The Non-Homogeneous Case MTH6109 Combinatorics (Fall 2017) Step 3: solution: We add the general solution (4.8) and particular solution (4.9) to get a a n = a (h) n + a (p) n =( 1) n A 1 +2 n (B 1 + B 2 n)+3n +9. This gives us an equation corresponding to each of the initial values a 0, a 1, a 2. Specifically, we have: 1=A 1 + B 1 +9 (n =0) 5= A 1 +2B 1 +2B (n =1) 18 = A 1 +4B 1 +8B (n =2) Just as before, we have a linear system of 3 equations in 3 unknowns. Solving this system gives A 1 = 1, B 1 = 9andB 2 =5. Thus,oursolutionis: a n =( 1) n ( 1) + 2 n ( 9+5n)+3n +9 =( 1) n+1 +2 n (5n 9) + 3n +9. Finally, let s see an example of using recurrences to solve real problems. Example 4.5. You take out a 1000 loan, with an interest rate of 10%, compounded monthly. How long will it take you to pay it o, if you make a monthly payment of 150? Proof. We can model this as a recurrence. Let a n be the balance you owe n months after you take the loan. Then, this is equal to the previous balance plus 10% interest, minus the payment you make at the end of the month. That is: a n = a n a n = 1.1 a n This is a degree 1, non-homogeneous linear recurrence with constant coe cients. Also, the term f(n) isjustadegree0polynomialf(n) =100,sowecanusethe machinery we ve seen to solve it. Step 1: We form the homogeneous recurrence: a n =1.1 a n 1. This has characteristic polynomial x 1.1 =0 which has 1 root r 1 =1.1 of multiplicity 1. So our general solution is: a n =(1.1) n A 0. (4.10) 43

9 4.4. The Non-Homogeneous Case MTH6109 Combinatorics (Fall 2017) Step 2: Since f(n) isadegree0polynomial,weconsideranarbitrarydegree0 polynomial p(n) =P 0. We substitute p(n) =P 0 into our recurrence for a n. We get: P 0 =1.1 P Then, we must have 0.1 P 0 = 150, and so P 0 =1500. Thisgivesusaparticular solution of: a (p) n =1500. (4.11) Step 3: Now, we combine (4.10) and (4.11) to get: a n = a (h) n + a (p) n =(1.1) n A Substituting in our initial value a 0 =1000forn =0weget: 1000 = A and so A 0 = 500. Our final equation is then: a n = (1.1) n. (4.12) Let s sanity check our work before we go further. We can do this using mathematical induction. We want to show that (4.12) is a solution for all n. Thebasecasecorresponds to our initial value a 0 =1000. Indeed,wehavea 0 = (1.1 0 )= For the induction step, let s assume that a n = (1.1) n and consider a n+1.fromtherecurrence,weget: a n+1 =(1.1)a n 150 =(1.1) ( (1.1) n ) 150 = (1.1) n = (1.1) n+1 as required. Note that we used the induction hypothesis in the second line. To complete the problem we need only figure out how big n has to be before a n apple 0. This is precisely when (1.1) n apple 0 So it takes 12 months to repay the loan. n 500 (1.1) n apple 1500 (1.1) n 3 n ln 1.1 ln 3 ln 3 ln

10 4.5. A Recurrence for Derangements MTH6109 Combinatorics (Fall 2017) 4.5 A Recurrence for Derangements In order to get a feel for how to model problems by recurrences, let us reconsider derangements, as first discussed in Section 3.3. Remember that a derangement on n elements is just a permutation f :[n]! [n]withthepropertythatf(i) 6= i for all i 2 [n]. In Theorem 3.14 we derived a formula for the number of derangements of n elements by using Inclusion-Exclusion. However, we can also solve the problem with recurrences A Second-Order Recurrence Let D n be the number of derangements on [n]. In order to apply recurrences to solve a di cult counting problem, we usually proceed similarly to Example 4.1. First, we work out a few small terms by brute force. By thinking carefully, we find that D 0 =1,andD 1 =0. Thereisonefunctiononnoelements: itisthe empty function f : ;!;. Clearly there is then no element i such that f(i) =i, because the domain of f is empty. Thus D 0 =1. Forasingleelement,theonly function f : {1}!{1} is the identity function f(1) = 1 and clearly this is not a derangement, so D 1 =0. For{1, 2} we have 2 permutations, but only one of them is a derangement: the one that sets f(1) = 2 and f(2) = 1. Now, let s see if we can do the di cult part. The trick here is to think about an object of size n, andthenimagineallofthedi erentwaysthatwecouldmakesome final choice when constructing this object. For each of these final choices, we try to show that the number of objects made with this choice is somehow related to the total number of objects of size less than n. In Example 4.1 we thought about the last syllable chosen for the line. It could have 1 or 2 beats. We then saw that in the case the number of possible ways to choose the syllables in each case was the number of ways to fill the first n 1and n 2beats,respectively.Ingeneral,weworkbystartingwithanobjectofsizen, then think backwards. Weimagineeachlaststep,thenfigureouthowmanyways we could have gotten to that step. Here, we imagine a derangement on [n] andthinkaboutchoosingthemapping f(n). Since we cannot have f(n) =n, therearen 1forthiselement,ingeneral. For each possible choice f(n) =i, howmanywayscouldwehaveconstructedsuch aderangement? Unfortunately, things are now not so obvious, but here is an idea: f maps n to i; wheredoesitmapi to? We consider 2 cases. First, suppose that f(i) = n. Then, f swaps i and n. We have n 2otherelementsin[n] \{i, n} and they must map to elements of [n] \{i, n}. So, in this case each derangement must take each element in j 2 [n] \{i, n} to some other element of f(j) 2 [n] \{i, n}, with f(j) 6= j. Butthesemappingscorrespondexactlytoderangementsonn 2 45

11 4.5. A Recurrence for Derangements MTH6109 Combinatorics (Fall 2017) elements. Summarising, we have the following: Claim 4.6. For any i 6= n, thenumberofpossiblederangementson[n] withf(n) = i and f(i) =n is D n 2. Now, suppose that f(i) 6= n. Then, there must be some other element j 6= i such that f(j) =n. We construct a derangement g :[n 1]! [n 1] from f as follows. For all x 2 [n] \{n, j}, setg(x) =f(x). Then, since f was a derangement we have g(x) 6= x for all of such x. Moreover,g(x) 6= n for all x 6= j, sotherange of g is contained in [n 1] as required. So far so good, but our problem is that g does not yet map the element j to anything and (since f(n) =i) g does not yet map anything to element i. Wefixthisbysimplysettingg(j) = i. Now, the range and domain of g are [n 1] as required. Moreover, since j 6= i by assumption, g is now a derangement. We now claim that this operation produces a one-to-one correspondence between the derangements f on [n] withf(n) =i and f(i) 6= n and the derangements g on [n 1]. First we show that it is surjective. Note that every derangement g on [n 1] must have g(j) =i for some j 6= i. Wecouldindeedobtainsuchag from the derangement f on [n] thatagreeswithg on all x 62 {n, j} but has f(j) =n, and f(n) =i. Notice that, as required, we also have f(i) 6= n since f(j) =n and j 6= i. Next, we show that our operation is injective. Suppose that we have two derangements f 1 6= f 2 on [n] withf 1 (n) =i, f 1 (i) 6= n and f 2 (n) =i, f 2 (i) 6= n. Consider the derangements g 1 and g 2 on [n 1] produced by applying our operation to f 1 and f 2,respectively. Wewillshowthatg 1 6= g 2. Let j 1 = f1 1 (n) and j 2 = f2 1 (n). Then, our procedure sets g 1 (j 1 )=f 1 (n) =i and g 2 (j 2 )=f 2 (n) =i. If j 1 6= j 2,wegetthatg 1 6= g 2 (they map di erent elements to i). Suppose that j 1 = j 2.Theremustbesomexsuch that f 1 (x) 6= f 2 (x), since f 1 6= f 2.Wecannot have x = n, sincef 1 (n) =i and f 2 (n) =i. Also, we cannot have x = j 1 (= j 2 ) since f 1 (j 1 )=n and f 2 (j 2 )=n. But, this means that we set g 1 (x) =f 1 (x) and g 2 (x) =f 2 (x) soalsog 1 (x) 6= g 2 (x). Again g 1 and g 2 are di erent as required. By the Correspondence Principle we then have: Claim 4.7. For any i 6= n, thenumberofpossiblederangementson[n] withf(n) = i and f(i) 6= n is D n 1. Putting everything together we get our main recurrence Theorem 4.8. For any i 0, let D i be the number of derangements on [i]. Then, D 0 =1,D 1 =0,andforalln 2: D n =(n 1)(D n 1 + D n 2 ) Proof. Consider a derangement f on [n]. As we discussed there are n 1di erent 46

12 4.5. A Recurrence for Derangements MTH6109 Combinatorics (Fall 2017) choices for f(n). For each such choice of i, thereared n 1 derangements with f(n) =i and f(i) =n and D n 2 derangements with f(n) =i and f(i) 6= n. Bythe Rule of Sum there are then altogether D n 1 + D n 2 derangements with f(n) =i. Since there are n 1choicesfori, eachleadingtoauniquesetofd n 1 + D n 2 derangements, the Rule of Product gives that D n =(n 1)(D n 1 + D n 2 ). Notice that this recurrence ended up looking very similar to the Fibonacci recurrence. It is a linear recurrence, but unfortunately it has non-constant coe - cients, so our recipes cannot be applied. 47

CDM. Recurrences and Fibonacci. 20-fibonacci 2017/12/15 23:16. Terminology 4. Recurrence Equations 3. Solution and Asymptotics 6.

CDM. Recurrences and Fibonacci. 20-fibonacci 2017/12/15 23:16. Terminology 4. Recurrence Equations 3. Solution and Asymptotics 6. CDM Recurrences and Fibonacci 1 Recurrence Equations Klaus Sutner Carnegie Mellon University Second Order 20-fibonacci 2017/12/15 23:16 The Fibonacci Monoid Recurrence Equations 3 Terminology 4 We can