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Chapter 4: Problem 215

Find the generating function for the solutions to the recurrence $$ a_{i}=5 a_{i-1}-6 a_{i-2}+2^{i} $$ The recurrence relations we have seen in this section are called second order because they specify \(a_{i}\) in terms of \(a_{i-1}\) and \(a_{i-2},\) they are called linear because \(a_{i-1}\) and \(a_{i-2}\) each appear to the first power, and they are called constant coefficient recurrences because the coefficients in front of \(a_{i-1}\) and \(a_{i-2}\) are constants.

Short Answer

Expert verified
The generating function is \[ A(x) = \frac{a_0(1-5x) + a_1 x + \frac{1}{1-2x}}{1 - 5x + 6x^2} \]

Step by step solution

01

Write the recurrence relation

The recurrence relation given is: \[ a_i = 5a_{i-1} - 6a_{i-2} + 2^i \]
02

Define the generating function

Let the generating function be defined as \[ A(x) = \sum_{i=0}^{\infty} a_i x^i \]
03

Find expressions for shifted generating functions

Multiplying the recurrence relation by \(x^i\) and summing for each term: \[ \sum_{i=0}^{\infty} a_i x^i = 5 \sum_{i=0}^{\infty} a_{i-1} x^i - 6 \sum_{i=0}^{\infty} a_{i-2} x^i + \sum_{i=0}^{\infty} 2^i x^i \] Recognize that the sums involving \(a_{i-1}\) and \(a_{i-2}\) can be shifted to involve \(a_i\): \[ A(x) - a_0 - a_1 x = 5(xA(x) - a_0) - 6(x^2 A(x)) + \sum_{i=0}^{\infty} (2x)^i \]
04

Simplify the expression

Combine the series and solve for \(A(x)\): \[ A(x) - a_0 - a_1 x = 5xA(x) - 5a_0 - 6x^2 A(x) + \sum_{i=0}^{\infty} (2x)^i \] The sum term is a geometric series that sums to \(\frac{1}{1-2x}\): \[ A(x) - a_0 - a_1 x = 5xA(x) - 5a_0 - 6x^2A(x) + \frac{1}{1-2x} \] Rearrange to isolate \(A(x)\): \[ A(x)(1 - 5x + 6x^2) = a_0(1-5x) + a_1 x + \frac{1}{1-2x} \] Thus, \[ A(x) = \frac{a_0(1-5x) + a_1 x + \frac{1}{1-2x}}{1 - 5x + 6x^2} \]
05

Present the generating function

The final form of the generating function is: \[ A(x) = \frac{a_0(1-5x) + a_1 x + \frac{1}{1-2x}}{1 - 5x + 6x^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recurrence Relations
A recurrence relation is a way of defining sequences where future terms are derived from past terms. These relations help in understanding how a sequence progresses. For example, in our exercise, the given recurrence relation is \[ a_i = 5a_{i-1} - 6a_{i-2} + 2^i \]. This means each term \(a_i\) is expressed using the previous two terms \(a_{i-1}\) and \(a_{i-2}\), along with an additional term \(2^i\). This is a very powerful method in mathematics and computer science to solve sequences in an efficient manner.
  • First, we identify how each term depends on the others.
  • Then, we use algebraic manipulations or generating functions to simplify and solve these relations.
Understanding recurrence relations allows us to predict subsequent elements without directly computing all previous terms.
Second-Order Recurrence
Second-order recurrence relations involve sequences where each term is defined using the previous two terms. In our case, the relation is \[ a_i = 5a_{i-1} - 6a_{i-2} + 2^i \]. This type of relation requires the knowledge of two initial conditions, \(a_0\) and \(a_1\), to specify the sequence completely. These types of recurrences can be solved using characteristic equations, but they can also be approached using generating functions for more complex cases.
Some key points about second-order recurrences:
  • They provide a broader perspective on the sequence by involving two prior terms.
  • They often appear in various mathematical problems like dynamic programming.
  • They can typically be reduced to simpler linear problems or explored using functions for more insight.
These recurrences add a richer structure compared to first-order relations, making them both challenging and interesting to work with.
Linear Recurrence
Linear recurrence relations are recurrences where each term is a linear combination of previous terms. In our example, \[ a_i = 5a_{i-1} - 6a_{i-2} + 2^i \], each term \(a_i\) is linearly dependent on \(a_{i-1}\) and \(a_{i-2}\). The term linear implies that the coefficients and terms are simply added or subtracted, rather than being multiplied together or raised to powers.
Important features include:
  • Simplicity: The terms are combined in a straightforward manner.
  • Predictability: Given the initial conditions, future terms can be determined precisely.
  • Well-suited for generating function analysis, allowing for elegant solutions.
Linear recurrences like this one provide a clear and manageable way to analyze sequences.
Constant Coefficient Recurrences
In constant coefficient recurrences, the coefficients multiplying the previous terms are constants. Our example is \[ a_i = 5a_{i-1} - 6a_{i-2} + 2^i \], and here, the coefficients (5 and -6) are constant. This stability in coefficients provides a structure that can be exploited to find solutions more easily.
Why constant coefficients matter:
  • Simplification: The consistent coefficients make it easier to apply generating functions or other algebraic methods.
  • Stability: These sequences have predictable patterns due to the unchanging nature of the coefficients.
  • Analytic solutions: This consistency often allows for closed-form solutions, making it easier to handle large sequences.
In essence, constant coefficient recurrences offer a sturdy foundation to build on when solving and analyzing sequences.

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Most popular questions from this chapter

Suppose we deposit \(\$ 5000\) in a savings certificate that pays ten percent interest and also participate in a program to add \(\$ 1000\) to the certificate at the end of each year (from the end of the first year on) that follows (also subject to interest.) Assuming we make the \(\$ 5000\) deposit at the end of year 0 , and letting \(a_{i}\) be the amount of money in the account at the end of year \(i,\) write a recurrence for the amount of money the certificate is worth at the end of year \(n\). Solve this recurrence. How much money do we have in the account (after our year-end deposit) at the end of ten years? At the end of 20 years?

Suppose you are going to choose a snack of between zero and three apples, between zero and three pears, and between zero and three bananas. Write down a polynomial in one variable \(x\) such that the coefficient of \(x^{n}\) is the number of ways to choose a snack with \(n\) pieces of fruit. (w)

(Used in Chapter 6.) Notice that when we used \(A^{2}\) to stand for taking two apples, and \(P^{3}\) to stand for taking three pears, then we used the product \(A^{2} P^{3}\) to stand for taking two apples and three pears. Thus we have chosen the picture of the ordered pair ( 2 apples, 3 pears) to be the product of the pictures of a multiset of two apples and a multiset of three pears. Show that if \(S_{1}\) and \(S_{2}\) are sets with picture functions \(P_{1}\) and \(P_{2}\) defined on them, and if we define the picture of an ordered pair \(\left(x_{1}, x_{2}\right) \in S_{1} \times S_{2}\) to be \(P\left(\left(x_{1}, x_{2}\right)\right)=P_{1}\left(x_{1}\right) P_{2}\left(x_{2}\right),\) then the picture enumerator of \(P\) on the set \(S_{1} \times S_{2}\) is \(E_{P_{1}}\left(S_{1}\right) E_{P_{2}}\left(S_{2}\right) .\) We call this the product principle for picture enumerators.

In Problem 216 you see that when we added numerical multiples of the reciprocals of first degree polynomials we got a fraction in which the denominator is a quadratic polynomial. This will always happen unless the two denominators are multiples of each other, because their least common multiple will simply be their product, a quadratic polynomial. This leads us to ask whether a fraction whose denominator is a quadratic polynomial can always be expressed as a sum of fractions whose denominators are first degree polynomials. Find numbers \(c\) and \(d\) so that $$ \frac{5 x+1}{(x-3)(x+5)}=\frac{c}{x-3}+\frac{d}{x+5} $$

Suppose that we have two sets \(S_{1}\) and \(S_{2}\). Let \(v_{1}\) (v stands for value) be a function from \(S_{1}\) to the nonnegative integers and let \(v_{2}\) be a function from \(S_{2}\) to the nonnegative integers. Define a new function \(v\) on the set \(S_{1} \times S_{2}\) by \(v\left(x_{1}, x_{2}\right)=v_{1}\left(x_{1}\right)+v_{2}\left(x_{2}\right) .\) Suppose further that \(\sum_{i=0}^{\infty} a_{i} x^{i}\) is the generating function for the number of elements \(x_{1}\) of \(S_{1}\) of value \(i,\) that is with \(v_{1}\left(x_{1}\right)=i\). Suppose also that \(\sum_{j=0}^{\infty} b_{j} x^{j}\) is the generating function for the number of elements \(x_{2}\) of \(S_{2}\) of value \(j,\) that is with \(v_{2}\left(x_{2}\right)=j .\) Provethat the coefficient of \(x^{k}\) in $$ \left(\sum_{i=0}^{\infty} a_{i} x^{i}\right)\left(\sum_{j=0}^{\infty} b_{j} x^{j}\right) $$ is the number of ordered pairs \(\left(x_{1}, x_{2}\right)\) in \(S_{1} \times S_{2}\) with total value \(k\), that is with \(v_{1}\left(x_{1}\right)+v_{2}\left(x_{2}\right)=k .\) This is called the product principle for generating functions. (h) Problem 191 may be extended by mathematical induction to prove our next theorem.Theorem 4.1.1. If \(S_{1}, S_{2}, \ldots, S_{n}\) are sets with a value function \(v_{i}\) from \(S_{i}\) to the nonnegative integers for each \(i\) and \(f_{i}(x)\) is the generating function for the number of elements of \(S_{i}\) of each possible value, then the generating function for the number of \(n\) -tuples of each possible value is \(\prod_{i=1}^{n} f_{i}(x)\)
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