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Chapter 8: Problem 36

Use generating functions to solve the recurrence relation \(a_{k}=3 a_{k-1}+4^{k-1}\) with the initial condition \(a_{0}=1\)

Short Answer

Expert verified
The solution to the recurrence relation is a combination of terms in the power series related to the coefficients of \(3^k\) and \((3-4)^k\).

Step by step solution

01

Identify the components of the recurrence relation

The recurrence relation given is: \[a_{k} = 3a_{k-1} + 4^{k-1}\] with the initial condition \(a_0 = 1\).
02

Form the generating function

Let the generating function be \(A(x) = \sum_{k=0}^{\infty} a_k x^k\). We will use this to find a closed form for the sequence \(\{a_k\}\).
03

Express the recurrence relation using generating functions

Multiply both sides of the recurrence relation by \(x^k\) and sum from \(k=1\) to infinity: \[\sum_{k=1}^{\infty} a_k x^k = \sum_{k=1}^{\infty} 3a_{k-1} x^k + \sum_{k=1}^{\infty} 4^{k-1} x^k\]
04

Simplify the series sums

Shift the indices for the sums: \[\sum_{k=1}^{\infty} a_k x^k = x\sum_{k=0}^{\infty} a_k x^{k} - a_0 x = 3x\sum_{k=0}^{\infty} a_k x^{k} + \sum_{k=0}^{\infty} 4^k x^{k+1}\] We recognize these as generating functions.
05

Apply initial condition and find generating function

Substitute \(a_0 = 1\), and simplify: \[A(x) = 1 + x (3A(x) + \sum_{k=0}^{\infty} 4^k x^k)\] Use the sum of geometric series for \(\sum_{k=0}^{\infty} 4^k x^k\), which is \(\frac{1}{1-4x}\).
06

Solve for the generating function

Express \(A(x)\) as a closed form: \[A(x) = 1 + 3x A(x) + x \frac{1}{1-4x}\] Solve for \(A(x)\): \[A(x) - 3x A(x) = 1 + x \frac{1}{1-4x}\] \[A(x)(1 - 3x) = 1 + \frac{x}{1-4x}\] Combine terms and solve: \[A(x) = \frac{1}{1-3x} + \frac{x}{(1-3x)(1-4x)}\]
07

Find the coefficient of the power series

Expand each term to find \(a_k\): \[A(x) = \sum_{k=0}^{\infty} 3^k x^k + x \sum_{k=0}^{\infty} \sum_{m=0}^k 3^{k-m} 4^m x^k\] Combine the sums to identify the pattern in \(a_k\), leading to the solution.

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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 an equation that defines a sequence where each term is based on the previous ones. In the given problem, we have the recurrence relation \[a_{k} = 3a_{k-1} + 4^{k-1}\].
  • The term \(a_k\) represents the current term in the sequence.
  • The term \(a_{k-1}\) is the preceding term, which influences the current term.
  • \(4^{k-1}\) is an additional component that depends on the index and adds extra terms to the sequence.
Recurrence relations are useful in defining sequences with specific patterns. By using them, we can create complex sequences from simple initial values, and they often appear in algorithm analysis, combinatorics, and computer science.
initial conditions
Initial conditions are the starting points of a sequence defined by a recurrence relation. For the relation given, the initial condition is \(a_0 = 1\).
  • Initial conditions set the foundation for the sequence.
  • With \(a_0 = 1\), we know the first term of the sequence, allowing us to compute subsequent values using the recurrence relation.
They are crucial because without them, the sequence cannot be uniquely determined. It's like knowing the initial position and velocity of an object to predict its future motion in physics.
geometric series
A geometric series is a sum of terms where each term is a constant multiple of the previous one. In the context of generating functions, it helps simplify complex series.
For example, the sum \[\sum_{k=0}^{\text{鈭瀩} 4^k x^k = \frac{1}{1-4x}\]
  • This formula applies when the absolute value of the common ratio is less than 1.
  • We used this to transform the series sum in the generating function solution.
By recognizing parts of generating functions as geometric series, we can solve them more efficiently. This technique transforms infinite sums into manageable expressions.

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Most popular questions from this chapter

Find a closed form for the exponential generating function for the sequence \(\left\\{a_{n}\right\\},\) where \(\begin{array}{ll}{\text { a) } a_{n}=(-2)^{n} .} & {\text { b) } a_{n}=-1} \\\ {\text { c) } a_{n}=n .} & {\text { d) } a_{n}=n(n-1)} \\ {\text { e) } a_{n}=1 /((n+1)(n+2))}\end{array}\)

a) Find a recurrence relation for the number of ternary strings of length \(n\) that contain two consecutive symbols that are the same. b) What are the initial conditions? c) How many ternary strings of length six contain consecutive symbols that are the same?

(Calculus required) Let \(\left\\{C_{n}\right\\}\) be the sequence of Catalan numbers, that is, the solution to the recurrence relation \(C_{n}=\sum_{k=0}^{n-1} C_{k} C_{n-k-1}\) with \(C_{0}=C_{1}=1\) (see Example 5 in Section 8.1\()\) a) Show that if \(G(x)\) is the generating function for the sequence of Catalan numbers, then \(x G(x)^{2}-G(x)+\) \(1=0 .\) Conclude (using the initial conditions) that \(G(x)=(1-\sqrt{1-4 x}) /(2 x)\) b) Use Exercise 42 to conclude that $$ G(x)=\sum_{n=0}^{\infty} \frac{1}{n+1}\left(\begin{array}{c}{2 n} \\\ {n}\end{array}\right) x^{n} $$ so that $$ C_{n}=\frac{1}{n+1}\left(\begin{array}{c}{2 n} \\ {n}\end{array}\right) $$ c) Show that \(C_{n} \geq 2^{n-1}\) for all positive integers \(n\)

Use generating functions to solve the recurrence rela- tion \(a_{k}=2 a_{k-1}+3 a_{k-2}+4^{k}+6\) with initial conditions \(a_{0}=20, a_{1}=60\)

Show that if \(X\) and \(Y\) are independent random variables on a sample space \(S\) such that \(X(s)\) and \(Y(s)\) are nonneg- ative integers for all \(s \in S,\) then \(G_{X+Y}(x)=G_{X}(x) G_{Y}(x)\)
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