1 A first example: detailed calculations

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1 MA4 Handout # Linear recurrence relations Robert Harron Wednesday, Nov 8, 009 This handout covers the material on solving linear recurrence relations It starts off with an example that goes through detailed calculations It then explains how to extract a basic method from this example Subsequent examples are solved by applying a simple set of steps The first part deals with recurrence relations that give rise to matrices that have distinct real eigenvalues The case of distinct complex eigenvalues is dealt with in section 3 in some sense it is the same method, but some simplifications can be used We don t have the machinery required to derive the method in the case of eigenvalues with multiplicity greater than Nonetheless, I have included a section that explains how to solve such recurrence relations section 4 you are not required to know the material of that section though A first example: detailed calculations Let and for n a, a 5 a n+ 6a n+ 8a n Find a formula for a n You could think of this as saying: solve equation for a n, given that a and a 5 In case you re interested, the beginning of this sequence is, 5,, 9, 376, 50, 6, The first idea behind solving this problem is to write the recurrence relation a n+ 6a n+ 8a n in terms of matrix multiplication: find α β A γ δ such that a n+ A a n+ a n+ a n The right-hand side is A a n+ αa n+ + βa n a n γa n+ + δa n so A 6 8 0

2 Then, we can write a n+ a n a n+ a n a n a n n a a so if we can find a formula for A n, then we will have a formula for a n+ This is where the second idea comes in: the best way to take a power of a matrix is to diagonalize it If we have A P DP with D λ 0, 0 λ then A n P DP P DP P DP P D n P and D n λ n 0 0 λ n The characteristic polynomial of A is ch A x det 6 x 8 x x 6x whose roots are the eigenvalues λ 4 and λ You can find that v 4 is a eigenvector corresponding to λ and v is an eigenvector corresponding to λ So P 4 and P detp 4

3 Therefore, a n+ a n+ A n a a P D n P 4 a a 4 n 0 0 n 4 n+ n+ 5 4 n n 4 n+ n+ 3 4 n n 3 4n+ n n n 9 Taking the first component, say, of this equality gives a n+ 3 4n+ n+ or, equivalently, a n 3 4n n 0 From the example to the method By studying the above example, we can remove a lot of the work out of the method of solution In this section, I ll explain how we can, firstly, avoid having to find the eigenvectors and P Then, I ll describe how we can even avoid writing down the matrix A Looking at equation 6, we can see that only the entries of D n depend on n So doing this multiplication, whatever it is, will give something that looks like a n+ a n+ a 4 n+ + b n+ c 4 n + d n for some constants a, b, c, d Taking the first component of this equality gives a n a 4 n + b n We know this without having to know what P is How do we figure out what a and b are without computing P? Simply plug n and n into equation and use : n : a a b 0 a + b n : 5 a a 4 + b 3

4 This gives us two linear equations for the two unknowns a and b We can solve this using an augmented matrix of course, this is a rather simple system, so you could do this without matrices: Thus, a 3 and b so a n 3 4n n which agrees with the answer obtained above in 0 So, we didn t need to find v, v, P, P, we just needed to know about the 4 n and the n What are 4 and? They were the entries of D, ie the eigenvalues of A Now, how do we avoid having to find A? We just saw that all we need from A are its eigenvalues, so we just need the characteristic polynomial of A Looking at equation 3 you may or may not notice that it s similar to the recurrence relation we started with This is no coincidence The recurrence relation has the form a n+ pa n+ + qa n The matrix A for this recurrence relation is A p q 0 whose characteristic polynomial is ch A x det p x q x x px q Thus, the characteristic equation is x px q 0 This is equation if you replace a n+ by x, a n+ by x, and a n by So, we don t need to find A or compute its characteristic polynomial because we can read off its characteristic polynomial from the definition of the recurrence relation The method Let s introduce a bit of terminology The recurrence relation we used as an example in section is referred to as a linear recurrence relation of order with initial conditions a and a 5 or a second order linear recurrence relation with initial conditions Second order refers to the fact that a n+ is defined in relation to the two previous values a n+ and a n Since every value is defined in relation to the previous two, you need to start off by specifying two consecutive values eg a, a which are then called initial conditions You could give different initial conditions for the same recurrence relation and thus 4

5 obtain a different sequence in example below, we use the same recurrence relation as in the example in section, but different initial conditions A general k th order linear recurrence relation has the form a n+k p a n+k + p a n+k + + p k a n, where p i R For such a recurrence relation, an initial condition is specified by k consecutive values eg a, a,, a k I ll begin by outlining the method for solving second order linear recurrence relations restricting to the case where the corresponding eigenvalues are real and distinct, then I ll move on to higher order ones Order Let a n+ p a n+ + p a n 3 be a second order linear recurrence relation Step : Find the fundamental solutions The fundamental solutions are the 4 n and n in the example we did above The fundamental solutions are obtained by finding the solutions of the quadratic equation coming from 3, ie in 3, replace a n+ by x, a n+ by x, and a n by to obtain x p x p 0 and solve Call the roots you get λ and λ in this section, we are assuming these are distinct and real Then the fundamental solutions are λ n and λ n Step : The general solution of 3 is a n aλ n + bλ n 4 with a, b R Now suppose we have initial conditions a r, a s then the solution to the recurrence relation 3 with initial conditions a r and a s is given by: Step 3: Plug in n and n into equation 4 and solve for a and b: a r a + b a s aλ + bλ 5

6 which can be solved by row reduction of r λ λ s Remark Note that the system of equations giving a and b has a unique solution since det since we are assuming λ and λ are distinct λ λ 0 λ λ Higher order Let a n+k p a n+k + p a n+k + + p k a n 5 be a k th order linear recurrence relation Step : Find the fundamental solutions The fundamental solutions are obtained by finding the solutions of the equation coming from 5, ie in 5, replace a n+i by x i to obtain x k p x k p x k p k 0 and solve Call the roots you get λ, λ,, λ k in this section, we are assuming these are distinct and real Then the fundamental solutions are λ n, λ n,, λ n k Step : The general solution of 5 is with b i R a n b λ n + b λ n + + b k λ n k 6 Now suppose we have initial conditions a r, a r,, a k r k then the solution to the recurrence relation 5 with initial conditions a i r i is given by: It s not too hard to prove that the characteristic polynomial of the matrix A of interest is ch A x k x k p x k p x k p k This is a standard fact in the theory of the rational canonical form 6

7 Step 3: Plug in n, n,, n k into equation 6 and solve for the b i : which can be solved by row reduction of a r b + b + + b k a r b λ + b λ + + b k λ k a 3 r 3 b λ + b λ + + b k λ k a k r k b λ k + b λ k + + b k λ k k r λ λ λ k r λ λ λ k r 3 λ k λ k λ k k r k Remark Note that the system of equations giving the b i has a unique solution since λ λ λ k λ λ λ k λ k λ k λ k k since we are assuming the λ i are distinct λ λ λ 3 λ λ k λ λ 3 λ λ 4 λ λ k λ k 0 Examples of applying the method Example Consider the recurrence relation with initial conditions a n+ 6a n+ 8a n 7 a, a 4 8 Find a formula for a n Note that this is the same recurrence relation as the example in section, but with different initial conditions Solution: Step : Find the fundamental solutions This determinant is called the Vandermonde determinant 7

8 Replacing a n+i with x i in equation 7, we get x 6x The solutions of this equation are given by the quadratic formula We get λ 4 and λ The fundamental solutions of 7 are thus 4 n and n Step : The general solution is a4 n + b n where a, b R Step 3: Plug in n and n a a + b a 4 4a + b so a, b 0 The answer is a n 4 n + 0 n 4 n Remark 3 Because b ends up equaling zero here, the formula for a n is particularly simple Note how changing the initial condition can really change the formula for a n Question: So, I could give you the above recurrence relation with initial conditions and ask: What is the 00th term? Now, you can answer 4 99 Example Fibonnaci sequence Consider the recurrence relation with initial conditions a n+ a n+ + a n 9 a, a 0 Find a formula for a n Solution: 8

9 Step : Find the fundamental solutions Replacing a n+i with x i in equation 9, we get x x 0 The solutions of this equation are given by the quadratic formula We get λ + 5 and λ 5 Let us introduce the common notation for these particular numbers The fundamental solutions of 9 are thus ϕ + 5, so that ϕ 5 ϕ n and ϕ n Step : The general solution is aϕ n + b ϕ n where a, b R Step 3: Plug in n and n a a + b a aϕ + b ϕ ϕ ϕ 0 ϕ ϕ 0 ϕ ϕ 0 ϕ 0 ϕ ϕ ϕ Note that so ϕ a ϕ ϕ ϕ ϕ ϕ ϕ ϕ 5 5 b ϕ ϕ ϕ 5 The answer is a n ϕ ϕ n + ϕ ϕ n ϕn ϕ n Remark 4 The sequence a n,,, 3, 5, 8, 3,, in this example is called the Fibonacci sequence, named for Leonardo da Pisa aka Fibonacci, a mathematician of the middle ages who introduced the arabic numeral system we use today into western 9

10 civilization In his 0 book Liber Abaci, he used the Fibonacci sequence as a model for the population growth of rabbits a n is the number of pairs of rabbits after n months According to wikipedia, the Fibonacci sequence was well-known in ancient India as early as 00 BC, and was used to count patterns in a form of Sanskrit poetry The number ϕ is called the Golden ratio or Golden mean, etc It has been studied since ancient Greece and is considered to be an aesthetically pleasing ratio used in designing buildings, drawing spirals, etc Certain people might say that the fact that there s a relationship between the Fibonacci numbers and the Golden ratio is cool Others might not 3 Example 3 Third order example Consider the third order recurrence relation a n+3 a n+ + 4a n+ + 4a n with initial conditions a, a 6, a 3 8 Find a formula for a n Solution: Step : Find the fundamental solutions Replacing a n+i with x i in equation, we get x 3 + x 4x 4 0 The solutions of this equation can be obtained by the rational roots test We get λ, λ, λ 3 The fundamental solutions of are thus n, n, and n Step : The general solution is b n + b n + b 3 n where b, b, b 3 R 0

11 Step 3: Plug in n, n, and n 3 a b + b + b 3 a 6 b b + b 3 a 3 8 4b + b + 4b so b, b 4, and b 3 The answer is a n n + 4 n + n n + 4 n + n Question: So, I could give you the above recurrence relation with initial conditions and ask: What is the 00th term? Now, you can answer Complex eigenvalues The solution in the case of distinct complex eigenvalues can be carried out the same as with real eigenvalues, but a shortcut can be taken I ll illustrate this with an example 3 Example 4 Consider the recurrence relation with initial conditions a n+ a n+ 5a n 3 a 4, a 8 4 Find a formula for a n Solution:

12 Step : Find the fundamental solutions Replacing a n+i with x i in equation 3, we get x x The solutions of this equation are given by the quadratic formula We get λ + i and λ i The fundamental solutions of 3 are thus + i n and i n Step : The general solution is a + i n + b i n where, now, a, b C Trick: If a x + iy, then b x iy this is because a + i n + b i n must be a real number, so a and b must be complex conjugates Step 3: Plug in n and n a 4 a + b x + iy + x iy x so x a 8 a + i + b i + iy + i + iy i + 4i + iy y + 4i iy y 4 4y so y The answer is a n i + i n + + i i n 4 Repeated eigenvalues To derive the method for solving linear recurrence relations that have eigenvalues with multiplicity >, we would need the theory of the Jordan canonical form This is a generalization of the theory of diagonalization that allows you to find a similar almost diagonal matrix specifically a Jordan block even for non-diagonalizable matrices this requires finding generalized eigenvectors Since we don t have this knowledge, I figured I d just tell you how to solve the recurrence relations in case you re interested

13 Let s focus on the second order case Suppose you have a second order linear recurrence relation, say a n+ 4a n+ 4a n whose eigenvalues are the roots of ie x 4x + 4 x λ with multiplicity The problem you run into is that you need two fundamental solutions, but if your two eigenvalues λ λ are the same, then λ n λ n so you can only obtain one fundamental solution of that form What you get out of the theory of the Jordan canonical form is that the two fundamental solutions are n and n n, so the general solution is a n + bn n with a, b R For example, suppose our initial conditions are a and a 0 Plugging in n and n gives a a + b a 0 a + 4b This can be solved by row reducing So, the general term of the sequence is given by a n 3 n + 4n n n 4n 3 For a k th order linear recurrence relation all of whose eigenvalues are the same λ with multiplicity k, you still need k fundamental solutions These are λ n, nλ n, n λ n,, n k λ n When there are several eigenvalues λ,, λ q with respective multiplicities m,, m q, the k fundamental 3

14 solutions are λ n, nλ n,, n m λ n, λ n, nλ n,, n m λ n, λ n q,, n mq λ n Remark 5 If you think this looks similar to how you solve linear differential equations, you re not wrong The theory behind solving linear differential equations is exactly the same q 4