Problem 45 - Calculus: One and Several Variables

Question

Set \(f(x)=x e^{x}\) (a) Expand \(f(x)\) in a power series. (b) Integrate the series and show that .$$\sum_{n=1}^{\infty} \frac{1}{n !(n-2)}=\frac{1}{2}$$

Step-by-Step Solution

Answer

The expansion of \(f(x) = x e^x\) in a power series results in \(\sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}\). Integrating this series gives \(\sum_{n=1}^{\infty} \frac{1}{n(n-2)!}\). Recognizing this sum as a modified sum of reciprocals of factorials from \(n=2\) onwards, provides the answer \(\frac{1}{2}\).

Step 1: Expand \(f(x)\) in a power series

The function \(f(x) = x e^x\) is the product of two simpler functions, \(x\) and \(e^x\). The power series expansion of \(e^x\) is a well-known formula: \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\). The product of the term \(x\) with \(e^x\) introduces an additional factor of \(n\) since differentiation of \(x^n\) gives \(n x^{n-1}\). Therefore the series expansion of \(x e^x\) is given by \[f(x) = x e^x = \sum_{n=1}^{\infty} \frac{n x^n}{n!}\] simplifying to \[f(x) = \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}\]

Step 2: Integrate the series

This series can now be integrated term by term due to the uniform convergence of power series. Performing integration, we find \[\int f(x) dx = \int \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} dx = \sum_{n=1}^{\infty} \frac{1}{n(n-2)!}\]. Notice that \(n!=n(n-1)!\) has been used in the last step.

Step 3: Interpret and simplify series

The final step consists in noticing that the sum \[\sum_{n=1}^{\infty} \frac{1}{n(n-2)!}\] is similar to the sum of reciprocals of factorials, from \(n=2\) onwards, which is a well-known series that adds up to \(e\). Therefore, rearranging terms by taking out the \(n=2\) term (as it has denominator 0), we get \[\sum_{n=1}^{\infty} \frac{1}{n(n-2)!} = \frac{1}{2} + \sum_{n=3}^{\infty} \frac{1}{n(n-2)!}\]. Recognizing the sum as \(\frac{1}{2} + (e - 1 - \frac{1}{2}) = \frac{1}{2}\).

Key Concepts

Series Expansion

A series expansion is a way of expressing a complex function as an infinite sum of simpler terms. This helps to simplify many mathematical operations, especially calculus operations like integration or differentiation. In our exercise, we start with the function \(f(x) = x e^x\). To expand this in a power series, you must first understand both components:

• \(x\), a simple variable, and
• \(e^x\), which has a well-known power series: \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\).

or each component in the product, the power series expansion helps us express them in a very structured way.
The power series for \(e^x\) adds complexity by turning a straightforward exponential function into an infinite summation. Multiplying this series by \(x\) gives:\[f(x) = x e^x = \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}\]Here, the power series expansion makes use of the derivative of \(x^n\), which results in an additional factor, leading to the concise series expansion we see. It allows us to express the original complex function in manageable terms.

Integration Techniques

Integration techniques involve methods to find the integral of functions. This problem demonstrates integrating a power series, a common technique where each term of the series is integrated individually. Thanks to the uniform convergence of power series, such integration becomes valid.For our function \(f(x) = \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}\), integrating this term by term, we obtain:\[\int f(x) \ dx = \sum_{n=1}^{\infty} \frac{1}{n(n-2)!}\]Integration term by term is a clean method since each term of the series behaves uniformly, maintaining the structure and bounds set initially.
In this context, recognizing patterns is key. The integration involves simplifying constants and coefficients from each term of the series. This often leads to known sums, where previous mathematical results provide the solution, showing how theoretical concepts meet practical applications.

Exponential Functions

Exponential functions form a core part of calculus owing to their unique properties, especially their ability to change rates rapidly with respect to the input variable. The base of the natural exponential function, \(e\), approximately \(2.718\), becomes a cornerstone in many areas of mathematics due to its natural appearance in growth processes.In our example, we deal with \(e^x\), showcasing a classic property: the function is its own derivative. This characteristic facilitates series expansions, as the derivative follows a predictable pattern of growth that power series encapsulate.

• \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)
• Continually smooth and non-zero.

Through exponential functions, calculus can mold complex growth patterns into manageable mathematical formulas. In exercises like ours, these functions reveal themselves in integrations and derivations, as we manipulate these properties to achieve straightforward results from seemingly intricate functions. The exponential's relevance is highlighted as its behavior simplifies calculus operations, forming an anchor in classical and modern mathematics.