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Chapter 10: Problem 1

Solve the following recurrence relations by the method of generating functions. a) \(a_{n+1}-a_{n}=3^{n}, n \geq 0, a_{0}=1\) b) \(a_{n+1}-a_{n}=n^{2}, n \geq 0, a_{0}=1\) c) \(a_{n+2}-3 a_{n+1}+2 a_{n}=0, n \geq 0, a_{0}=1, a_{1}=6\) d) \(a_{n+2}-2 a_{n+1}+a_{n}=2^{n}, n \geq 0, a_{0}=1, a_{1}=2\)

Short Answer

Expert verified
a) \(A(x)= \frac{1}{1-3x}\), b) \(A(x)= \frac{2x^{2}+x^{2}(x^{2}-3x+3)}{(1-x)^{4}}\),c) \(A(x) = \frac{1+x}{1-x+x^{2}}\) ,d) \(A(x)=\frac{1-x}{(1-x)^{2}(1-2x)}\).

Step by step solution

01

Establish the Generating Series

Start by expressing the sequence \(a_n\) as a generating function \(A(x)=\sum_{n=0}^{\infty} a_n x^{n} \). Using the given initial conditions and by shifting the index of summation, you express the various terms in the recurrence relation as functions of \(x\).
02

Solve the Recurrence Relation for a), b), c) and d)

a) Given that \(a_{n+1}-a_{n}=3^{n}\), the generating functions would be: \[ A(x) - \frac{1}{x}\sum_{n=0}^{\infty} a_{n}x^{n+1}= \sum_{n=0}^{\infty} 3^{n}x^{n}\]Solve that equation for A(x) and you'll get the generating function for the sequence a).b) For \(a_{n+1}-a_{n}=n^{2}\), the generating function would be \[A(x)- \frac{1}{x}\sum_{n=0}^{\infty} a_{n}x^{n+1}=\sum_{n=0}^{\infty}n^{2}x^{n}\]Again, solve this equation for A(x) to obtain the generating function for the sequence a).c) Given that \(a_{n+2}-3 a_{n+1}+2 a_{n}=0\), the generating function becomes \[A(x) - \frac{1}{x}\sum_{n=1}^{\infty} a_{n}x^{n+1} +\frac{2}{x}\sum_{n=0}^{\infty} a_{n}x^{n+1}\]Solve this for A(x) to obtain the generating function for the sequence a).d) Finally, for \(a_{n+2}-2 a_{n+1}+a_{n}=2^{n}\), the equation becomes:\[A(x) - \frac{1}{x}\sum_{n=0}^{\infty} a_{n}x^{n+2} - \frac{2}{x}\sum_{n=0}^{\infty} a_{n}x^{n+1}=\sum_{n=0}^{\infty}2^{n}x^{n}\]. Solve that equation for A(x) and you'll get the generating function for the sequence a).
03

Convert the Generating Function to a Closed-Form Solution

Finally, calculating the coefficients of x in the generating function will provide the explicit solution for the recurrence relation.

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Key Concepts

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

Recurrence Relations
Recurrence relations are equations that define sequences recursively using earlier terms. It's a way to express each term of a sequence as a function of its predecessors. This method comes in handy when trying to predict the behavior of sequences without explicitly enumerating each term.
For example, if you have a recurrence like \( a_{n+1} - a_n = 3^n \), it means that each term is built upon the previous one with an additional component determined by the expression \( 3^n \). Recurrence relations can represent both simple arithmetic sequences and more complex structures.
Understanding the initial conditions is crucial, as they serve as the anchor or starting point from which the entire sequence unravels. In such problems, initial values like \( a_0 = 1 \) or \( a_1 = 6 \) allow us to apply the recursive rule, establishing a baseline for generating further terms.
Generating Series
Generating series, or generating functions, are a significant mathematical tool that transforms sequences into series. They convert a sequence \( a_n \) into a function \( A(x) = \sum_{n=0}^{\infty} a_n x^n \).
This approach provides a bridge between recursion (a step-by-step method) and functions. It's like having a magical toolkit that helps analyze sequences using algebraic manipulations. By treating series equations algebraically, one can solve many extensive problems with less effort.
Take, for instance, a sequence derived from the recurrence relation \( a_{n+1} - a_n = n^2 \). This recurrence challenges us to think creatively. By expressing the sequence as a generating series, the original problem transforms into an algebraic problem where you solve for \( A(x) \).
*Note on Practice:*
  • Formulate the generating series based on the sequence.
  • Perform algebra to find \( A(x) \).
  • Once solved, converting back helps derive explicit terms.
Closed-Form Solution
Once you have a generating series like \( A(x) \), find a closed-form solution, which is a single expression. This form provides a clean, easy-to-understand formula for the terms of the sequence.
Closed-form solutions eliminate the need for recursive calculations. They are handy since they give the term \( a_n \) directly without needing all previous terms in the sequence. Often, they appear as a neat formula, such as \( a_n = 2^n \) or \( a_n = 3^n + n^2 \). Finding such solutions means deriving a formula from the generating function that explicitly writes \( a_n \) in terms of \( n \) only.
While arriving at this from generating series involves some mathematical manipulations, such a form allows for direct and simple calculations of sequence terms. This is particularly useful for computer algorithms, where speed and efficiency are vital.

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Most popular questions from this chapter

In each of the following, \(f: \mathbf{Z}^{+} \rightarrow \mathbf{R}\). Solve for \(f(n)\) relative to the given set \(S\), and determine the appropriate "big- Oh" form for \(f\) on \(S\). a) \(f(1)=0\) \(f(n)=2 f(n / 5)+3, \quad n=5,25,125, \ldots\) \(S=\left\\{5^{\prime} \mid i \in \mathbf{N}\right\\}\) b) \(f(1)=1\) $$ \begin{aligned} &f(n)=f(n / 2)+2, \quad n=2,4,8, \ldots \\ &S=\left\\{2^{\prime} \mid i \in \mathbf{N}\right\\} \end{aligned} $$

Consider ternary strings - that is, strings where \(0,1,2\) are the only symbols used. For \(n \geq 1\), let \(a_{n}\) count the number of ternary strings of length \(n\) where there are no consecutive 1 's and no consecutive \(2^{\prime} s\). Find and solve a recurrence relation for \(a_{n}\) -

Find the general solution for the recurrence relation \(a_{n+3}-3 a_{n+2}+3 a_{n+1}-a_{n}=3+5 n, n \geq 0\)

For \(n>1\), a permutation \(p_{1}, p_{2}, p_{3}, \ldots, p_{n}\) of the integers \(1,2,3, \ldots, n\) is called orderly if, for each \(i=1,2,3, \ldots\) \(n-1\), there exists a \(j>i\) such that \(\left|p_{J}-p_{l}\right|=1\). [If \(n=2\), the permutations 1,2 and 2,1 are both orderly. When \(n=3\) we find that \(3,1,2\) is an orderly permutation, while \(2,3,1\) is not. (Why not?)] (a) List all the orderly permutations for \(1,2,3\). (b) List all the orderly permutations for \(1,2,3,4\). (c) If \(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}\) is an orderly permutation of \(1,2,3,4,5\), what value(s) can \(p_{1}\) be? (d) For \(n>1\), let \(a_{n}\) count the number of orderly permutations for \(1,2,3, \ldots, n\). Find and solve a recurrence relation for \(a_{n}\).

Show that for all \(n \geq 2\) $$ \left(\begin{array}{c} 2 n-1 \\ n \end{array}\right)-\left(\begin{array}{c} 2 n-1 \\ n-2 \end{array}\right)=\frac{1}{(n+1)}\left(\begin{array}{c} 2 n \\ n \end{array}\right) $$
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