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

Short Answer

Expert verified
a) \ e^{-2x} b) \-e^x c) \ x e^x d) \ x^2 e^x e) \ \frac{x}{1- x} - \frac{x^2}{2(1-x)^2}

Step by step solution

01

Understanding Exponential Generating Function

The exponential generating function for a sequence \(a_n\) is given by \[A(x) = \sum_{n=0}^{\infty} a_n \frac{x^n}{n!}.\]
02

Generating Function for \(a_n = (-2)^n\)

The exponential generating function is \[A(x) = \sum_{n=0}^{\infty} (-2)^n \frac{x^n}{n!} = e^{-2x}.\]
03

Generating Function for \(a_n = -1\)

For \(a_n = -1\), the sum converges to \[A(x) = \sum_{n=0}^{\infty} -1 \frac{x^n}{n!} = -e^x.\]
04

Generating Function for \(a_n = n\)

For \(a_n = n\), multiply the generating function by the derivative of \(e^x\) to get \[A(x) = \sum_{n=1}^{\infty} n \frac{x^n}{n!} = x e^x.\]
05

Generating Function for \(a_n = n(n-1)\)

For \(a_n = n(n-1)\), the generating function uses the second derivative: \[A(x) = \sum_{n=2}^{\infty} n(n-1) \frac{x^n}{n!} = x^2 e^x.\]
06

Generating Function for \(a_n = \frac{1}{(n+1)(n+2)}\)

Rewriting \(a_n\) and finding the generating function's closed form: \[A(x) = \int_0^x \frac{t}{(1-t)^2} dt = \frac{x}{1- x} - \frac{x^2}{2(1-x)^2}.\]

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Key Concepts

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

Understanding Exponential Generating Functions
An exponential generating function is a way to represent a sequence of numbers using an infinite series. This series has a unique twist: each term is divided by the factorial of its index. The exponential generating function for a sequence \(a_n\) is written as \[A(x) = \sum_{n=0}^{\infty} a_n \frac{x^n}{n!}.\] This representation helps in finding closed-form solutions for different sequences. It transforms problems into manageable mathematical expressions, allowing us to understand the properties of sequences better.
Discrete Mathematics and Series Representation
In discrete mathematics, we often deal with sequences and their properties. Representing these sequences as series is a common technique because it simplifies complex expressions.
For instance, to represent a sequence \(a_n = (-2)^n\), we can use:
\[A(x) = \sum_{n=0}^{\infty} (-2)^n \frac{x^n}{n!} = e^{-2x}.\]
This shows that the sequence's exponential generating function simplifies to \( e^{-2x} \). This method is helpful for solving recurrence relations and understanding series behavior.
By converting sequences into series representations, such as for \(a_n = n\), we get:
\[A(x) = \sum_{n=1}^{\infty} n \frac{x^n}{n!} = x e^x.\] These examples illustrate how powerful series representation can be in discrete mathematics.
Closed Form Solutions and Their Importance
A closed-form solution is an explicit formula for a sequence or function, which doesn't require summation or recurrence. It's like a shortcut that reveals the big picture instantly. For example, finding a closed-form for \(a_n = n(n-1)\), we derive:
\[A(x) = \sum_{n=2}^{\infty} n(n-1) \frac{x^n}{n!} = x^2 e^x.\]
This closed-form allows us to quickly understand the sequence's nature without computing every term. Another example is \(a_n = -1\), which gives:
\[A(x) = \sum_{n=0}^{\infty} -1 \frac{x^n}{n!} = -e^x.\]
Closed-form solutions are essential for mathematical analysis, simplifying complex problems, and providing clear insights into the behavior of sequences and functions. They are invaluable tools both in theory and practical applications.

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