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Magazine Chapter 8: Problem 34
Use generating functions to solve the recurrence relation \(a_{k}=7 a_{k-1}\) with the initial condition \(a_{0}=5\)
Short Answer
Expert verified
The solution to the recurrence relation is \(a_k = 5 \times 7^k\).
Step by step solution
01
Define the Generating Function
Let the generating function be defined as \[ A(x) = \sum_{k=0}^{\infty} a_k x^k \]. \Using the given recurrence relation \[a_k = 7 a_{k-1} \] with initial condition \[a_0 = 5 \].
02
Express Terms Using Generating Function
We can express the generating function as: \[ A(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... \]. Using the recurrence relation: \[a_k = 7 a_{k-1}\], we get \[a_1 = 7a_0, a_2 = 7a_1 = 7^2 a_0, a_3 = 7a_2 = 7^3 a_0, ...\].
03
Substitute Recurrence in the Generating Function
Substitute the values of \[a_k \]: \[ A(x) = a_0 + 7a_0 x + 7^2 a_0 x^2 + 7^3 a_0 x^3 + ...\]. This simplifies to \[A(x) = a_0 (1 + 7x + (7x)^2 + (7x)^3 + ... )\]
04
Recognize the Geometric Series
Recognize that the series in the parentheses is a geometric series with the first term 1 (\[a_0 = 5 \]) and the common ratio \[7x \]. The sum of an infinite geometric series \[1 + r + r^2 + r^3 + ... \] is given by \[ \frac{1}{1-r} \] where |r| < 1.
05
Apply the Sum Formula
Since the common ratio \[r = 7x \], we get the sum: \[ \sum_{k=0}^{\infty} (7x)^k = \frac{1}{1-7x} \].
06
Combine Results
Combine this with the factor of \[a_0\] from Step 3: \[ A(x) = 5 \frac{1}{1-7x} \].
07
Extract the Coefficients
Extract the coefficients of \[A(x)\] to find \[a_k \]. This is done using the expansion \[\frac{1}{1-7x} = \sum_{k=0}^{\infty} (7x)^k = \sum_{k=0}^{\infty} 7^k x^k \]. Multiply by 5 to get: \[A(x) = \sum_{k=0}^{\infty} 5 \times 7^k x^k \].
08
Write the Final Solution
Thus, \[a_k \] can be written as: \[ a_k = 5 \times 7^k \].
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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 based on previous terms. In this exercise, we use the relation \(a_k = 7 a_{k-1}\) to generate subsequent terms from a given initial term \(a_0 = 5\). This type of relation is common in discrete mathematics and helps in finding closed-form expressions. Breaking down the steps:
- We start from \(a_0\) and use the recurrence relation to find \(a_1 = 7a_0\), \(a_2 = 7a_1 = 7^2 a_0\), and so on.
- The relation adjusts according to the pattern of multiplication by 7 in this case, yielding a geometric pattern in the sequence.
geometric series
A geometric series is a series with a constant ratio between successive terms. This exercise uses the geometric series formula in the generating function. The general form of the infinite geometric series is: \[1 + r + r^2 + r^3 + \ldots = \frac{1}{1 - r}\]. Here, \ r \ is the common ratio.
- In our problem, the common ratio is \ 7x \.
- By recognizing the series inside the generating function as geometric, we convert it into the fraction \frac{1}{1-7x}\.
discrete mathematics
Discrete mathematics involves study of mathematical structures that are countable or otherwise distinct and separable. Generating functions and recurrence relations are essential tools in this field. They help solve problems involving sequences and series and provide a bridge to continuous mathematics.
- Using generating functions in this example simplifies finding the sequence terms.
- We handle discrete sets of numbers and their relations, as seen with \(a_k = 7 a_{k-1}\).
coefficients extraction
Extracting coefficients from generating functions is a key step in determining the sequence terms. The goal is to express the function in a form where coefficients are explicitly clear. In this exercise:
- We start with the generating function \ A(x) = \frac{5}{1-7x} \.
- We use the expansion \[ \frac{1}{1-7x} = \sum_{k=0}^{\infty} (7x)^k \] to identify the pattern.
- The coefficients of x^k in this expansion are terms of the sequence: \(a_k = 5 \times 7^k\).
