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Magazine Chapter 8: Problem 6
Find a closed form for the generating function for the sequence \(\left\\{a_{n}\right\\},\) where a) \(a_{n}=-1\) for all \(n=0,1,2, \ldots\) b) \(a_{n}=2^{n}\) for \(n=1,2,3,4, \ldots\) and \(a_{0}=0\) c) \(a_{n}=n-1\) for \(n=0,1,2, \ldots\) d) \(a_{n}=1 /(n+1) !\) for \(n=0,1,2, \ldots\)
Short Answer
Expert verified
a) \( \frac{-1}{1-x} \) b) \( \frac{2x}{1-2x} \) c) \( \frac{2x-1}{(1-x)^2} \) d) \( \frac{e^x - 1}{x} \)
Step by step solution
01
Understanding Part (a)
Given the sequence \{a_{n}\}, where \(a_{n} = -1\) for all \(n = 0, 1, 2, \ldots\). The generating function for a sequence \(\{a_{n}\}\) is given by \(G(a_{n}; x) = \sum_{n=0}^\infty a_n x^n\).
02
Generating Function for Part (a)
Substitute \(a_{n} = -1\): \[G(a_{n}; x) = \sum_{n=0}^\infty (-1) x^n\] This is a geometric series with first term \(-1\) and common ratio \(x\): \[G(a_{n}; x) = \frac{-1}{1-x}\] provided \(|x| < 1\).
03
Understanding Part (b)
Given the sequence \{a_{n}\}, where \(a_{n} = 2^n\) for \(n = 1, 2, 3, 4, \ldots\) and \(a_{0} = 0\). The generating function for this sequence is: \[G(a_{n}; x) = \sum_{n=0}^\infty a_n x^n\].
04
Generating Function for Part (b)
Substitute \(a_{0} = 0\) and \(a_{n} = 2^n\) for \(n \geq 1\): \[G(a_{n}; x) = 0 + \sum_{n=1}^\infty 2^n x^n = \sum_{n=1}^\infty (2x)^n\]This is a geometric series starting from \(n=1\) with first term \((2x)\) and common ratio \(2x\): \[G(a_{n}; x) = \frac{2x}{1-2x}\] provided \(|2x| < 1\).
05
Understanding Part (c)
Given the sequence \{a_{n}\}, where \(a_{n} = n - 1\) for \(n = 0, 1, 2, \ldots\). The generating function for this sequence is: \[G(a_{n}; x) = \sum_{n=0}^\infty (n-1) x^n\].
06
Generating Function for Part (c)
Divide the sum into two parts: \[G(a_{n}; x) = \sum_{n=0}^\infty n x^n - \sum_{n=0}^\infty x^n\] The first sum, using the series differentiation property, is \[ x \sum_{n=0}^\infty n x^{n-1} = x \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{x}{(1-x)^2}\] The second sum is a simple geometric series: \[ \sum_{n=0}^\infty x^n = \frac{1}{1-x}\] Putting these together: \[G(a_{n}; x) = \frac{x}{(1-x)^2} - \frac{1}{1-x}\].
07
Simplifying Part (c)
Combine the two fractions into a common denominator: \[G(a_{n}; x) = \frac{x - (1-x)}{(1-x)^2} = \frac{2x-1}{(1-x)^2}\].
08
Understanding Part (d)
Given the sequence \{a_{n}\}, where \(a_{n} = \frac{1}{(n+1)!}\) for \(n = 0, 1, 2, \ldots\). The generating function for this sequence is: \[G(a_{n}; x) = \sum_{n=0}^\infty \frac{x^n}{(n+1)!}\].
09
Generating Function for Part (d)
Consider the function \(e^x\) and its expansion: \[e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\] The given generating function can be derived by integrating \(e^x\): \[\int_0^x e^t \ dt = e^x - 1\] Therefore, \[G(a_{n}; x) = \sum_{n=0}^\infty \frac{x^n}{(n+1)!} = \frac{e^x - 1}{x}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
geometric series
A geometric series is a series of terms where each term is a fixed multiple of the previous term. This fixed multiple is called the common ratio. For instance, in the series \(\frac{1}{1-x}\), the first term is 1, and the common ratio is x. This means each term is x times the previous one. If you start adding the terms of this series from 0 to infinity, it sums to \(\frac{1}{1-x}\) provided \(|x| < 1\). Geometric series are very useful in deriving generating functions, as seen in the provided solutions. For example, \(\frac{2x}{1-2x}\) is a geometric series with first term 2x and common ratio 2x, making it valid under the condition that \(|2x| < 1\).
closed form
A closed form is an expression that allows us to understand a sequence or function without requiring summation notation or recursion. Essentially, it gives a direct formula to calculate any term. For instance, instead of iteratively adding terms of a sequence, you can use the closed form. From the exercise, \(\frac{-1}{1-x}\) is a closed form of the generating function for the sequence where \(a_{n}=-1\). Another example from the exercise is \(\frac{2x}{1-2x}\) for the sequence \(a_{n}=2^n\) from n=1 onwards and a_0=0. Obtaining a closed form makes calculations simpler and quicker.
sequence analysis
Sequence analysis involves examining the properties of a sequence to find patterns, relationships, and generating functions. It helps us derive useful characteristics like the closed form. For instance, in the provided exercise, analyzing the sequence \(a_{n} = n - 1\) involves splitting it into two parts based on their properties. One part results in \(\frac{x}{(1-x)^2}\), derived using differentiation properties of power series. The other part is simply \(\frac{1}{1-x}\) from the geometric series. Combining these insights, the generating function \(\frac{2x - 1}{(1-x)^2}\) is obtained.
integrating series
Integrating series is a technique used to derive generating functions from known series. For sequences where terms are expressed in a factorial form, integration can simplify the process. In the exercise, part (d) shows how integrating the series \(e^x = \sum_\infty^n=0 \frac{x^n}{n!}\) helps find the generating function. By integrating \(e^x\), we get: \(\int_0^x e^t \ = \ e^x - 1\). Hence, the generating function for the sequence \(a_{n} = \frac{1}{(n+1)!}\) is derived as \(\frac{e^x - 1}{x}\). This approach turns complex series into manageable forms, revealing generating functions more straightforwardly.
