Problem 21 - Essential Calculus

Question

In Exercises \(21-30,\) find the Maclaurin series for the function. (Use the table of power series for elementary functions.) $$ f(x)=e^{x^{2} / 2} $$

Step-by-Step Solution

Answer

The Maclaurin series for the function \(f(x)=e^{x^{2}/2}\) is \(\sum_{n=0}^{\infty} \frac{x^{2n}}{2^{n} \cdot n!}\).

Step 1: Recognize the Elementary Function

The given function \(e^{x^{2}/2}\) can be seen as an elementary function \(e^{x}\) with a substituted variable. This become evident when you consider \(t = x^{2}/2\), thus transforming the function into \(e^{t}\). This is an important first step in order to access the power series table and find the correct base series. In this case, the power series for \(e^{x}\).

Step 2: Establish the Maclaurin Series of the Elementary Function

The Maclaurin series of \(e^{x}\) is found by summing the terms \(x^{n}/n!\) from \(n = 0\) to infinity. Hence, it is represented as: \(\sum_{n=0}^{\infty} \frac{x^{n}}{n!}\). This series represents the expansion of \(e^{x}\) around the point 0.

Step 3: Apply Substitutions to the Maclaurin Series

To find the Maclaurin series of the given function, we need to substitute \(x^{2}/2\) for \(x\) in the power series of \(e^{x}\). As a result, the Maclaurin series becomes: \(\sum_{n=0}^{\infty} \frac{(x^{2}/2)^{n}}{n!}\). This is the Maclaurin series for the function \(f(x)=e^{x^{2}/2}\).

Step 4: Simplify the Maclaurin Series

The last step is to simplify the obtained series: \(\sum_{n=0}^{\infty} \frac{(x^{2}/2)^{n}}{n!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{2^{n} \cdot n!}\). Here, in each term of the series, the exponent of \(x\) is \(2n\), and in the denominator we have \(2^{n} \cdot n!\).

Key Concepts

Power Series

A power series is a way to express functions as an infinite sum of terms, using the powers of a variable. It is essentially like a polynomial, but it can have infinitely many terms. The general form of a power series around a point (often zero, which is called a Maclaurin series) is:

• \[ a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_n x^n \ldots \]

Here, the coefficients \(a_0, a_1, a_2, \ldots\) determine the specific function represented by the series.
Power series are particularly useful because they can simplify complex functions into an easily computable sum, especially around a point.
In the case of the function \( e^{x^2/2} \), we see a variation of the exponential function being expressed as a power series. This allows us to study and understand the behavior of complicated functions using basic algebra.

Exponential Functions

Exponential functions are foundational in calculus and mathematical analysis. They are functions of the form \( f(x) = a^x \) where \( a \) is a constant, typically \( e \), Euler's number, which is approximately 2.718. This special number leads to the function \( f(x) = e^x \), known for its unique mathematical properties:

• \( f'(x) = e^x \), meaning that the derivative of the function is the function itself.
• This property also makes it straightforward to expand into a power series.

When dealing with exponential functions, the Maclaurin series provides a tool to express them in terms of simpler polynomial forms.
For instance, the Maclaurin series for \( e^x \) is:

• \[ \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

This reveals how exponential functions can be expanded into an infinite sum of terms, which makes them easier to evaluate.

Series Expansion

Series expansion is the process of expressing a complex function as an infinite sum of simpler components. In calculus, we often expand functions using different kinds of series, like Taylor or Maclaurin, depending on the function and desired expansion point.
The main idea is to approximate functions with a sum of polynomials that becomes an exact representation as the number of terms approaches infinity. This allows us to approximate values and gain insights into a function's behavior.
For the Maclaurin series of \( e^{x^2/2} \), we expand the function by substituting \( x^2 / 2 \) into the series of \( e^x \):

• \[ \sum_{n=0}^{\infty} \frac{\left(x^2/2\right)^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{2^n \cdot n!} \]

This conversion of a function into a series helps in both theoretical analysis and practical computation, providing a way to handle functions that might otherwise be difficult to work with.