Problem 70 - Essential Calculus Early Transcendentals

Question

(a) Show that the function defined by $$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}}} & {\text { if } x \neq 0} \\\ {0} & {\text { if } x=0}\end{array}\right.$$ is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin.

Step-by-Step Solution

Answer

The function is rapidly decreasing near zero but is not represented by its zero Maclaurin series.

Step 1: Define the Maclaurin Series

The Maclaurin series for a function \(f(x)\) is the Taylor series expansion around \(x = 0\). It is represented as \(f(x) = \sum_{n=0}^{fty} \frac{f^{(n)}(0)}{n!}x^{n}\). To find the Maclaurin series of the given function \(f(x)\), we need to determine the derivatives \(f^{(n)}(0)\) for all \(n\).

Step 2: Calculate Derivatives at Zero

Given that \(f(0) = 0\), and for \(x eq 0\), \(f(x) = e^{-1/x^2}\), we need to calculate the derivatives at \(x = 0\): \(f^{(n)}(0)\). Given the rapid decay of \(e^{-1/x^2}\) as \(x \to 0\), all derivatives \(f^{(n)}(0)\) will be zero. Therefore, the Maclaurin series is \(0 + 0x + 0x^2 + \cdots = 0\).

Step 3: Compare Function and Series

The given function \(f(x)\) has different values for non-zero \(x\), specifically \(e^{-1/x^2}\), which approaches zero as \(x\) approaches zero. However, the Maclaurin series remains zero for all \(x\). Hence, the series and the function are not equal since the series cannot capture the behavior of \(f(x)\) around zero except at \(x=0\).

Step 4: Graph the Function

Graph \(f(x) = e^{-1/x^2}\) for \(x eq 0\) and \(f(x) = 0\) for \(x = 0\). You will see that as \(x\) approaches zero from either side, \(f(x)\) rapidly approaches zero, showing a sharp dip. This implies that while near-zero values of \(x\) make \(f(x)\) very small, it is never zero except exactly at \(x = 0\).

Step 5: Analyze Behavior Near the Origin

The function \(f(x)\) is sharply decreasing toward the origin, and is "flat" at \(x = 0\) due to the rapid decay of \(e^{-1/x^2}\). This "flatness" causes all its derivatives at \(x = 0\) to be zero, demonstrating why the Maclaurin series is just zero.

Key Concepts

Functions and Limits

Functions are mathematical expressions that describe the relationship between inputs and outputs. The function defined in this problem, \( f(x) = e^{-1/x^2} \) if \( x eq 0 \) and \( f(x) = 0 \) if \( x = 0 \), is a piecewise function. This means it has distinct expressions based on different intervals of its domain.

Understanding limits is essential here because they assist in describing the behavior of functions as the input (\( x \)) approaches a particular value. For the given function, examining the limit as \( x \) approaches zero is insightful. As \( x \) gets closer to zero from either the positive or negative direction, the expression \( e^{-1/x^2} \) decreases rapidly, tending toward zero. Notably, this means:

• While \( f(x) \) approaches zero as \( x \to 0 \), it is technically only exactly zero when \( x = 0 \).
• The limit of \( f(x) \) as \( x \to 0 \) is 0, aligning with the piecewise definition at \( x = 0 \).

Understanding this limit behavior helps clarify why the function is defined the way it is, especially at the dividing line where the behavior changes at \( x = 0 \).

Derivatives

Derivatives are tools used in mathematics to determine the rate at which a function changes at any given point. They play a crucial role in constructing series expansions like the Maclaurin series. To create the Maclaurin series, we need the derivatives of \( f(x) \) evaluated at zero.

For the function given in this exercise, although \( f(x) \) exhibits different values when \( x eq 0 \), at \( x = 0 \), the function value and all its derivatives are zero:

• Since \( f(0) = 0 \), it follows logically that the first derivative \( f'(0) = 0 \), and continuing this pattern reveals that \( f^{(n)}(0) = 0 \) for all \( n \).
• This indicates that the Maclaurin series, which is supposed to approximate \( f(x) \) near \( x = 0 \), is itself zero: \( 0 + 0x + 0x^2 + \ldots \).

Thus, while derivatives offer essential insights for building the series, in this context, they underscore the peculiar nature of \( f(x) \); despite \( e^{-1/x^2} \) being nonzero for \( x eq 0 \), its tangent characteristics around zero make all derivatives at zero vanish.

Graphical Analysis

Graphical analysis provides a visual interpretation of mathematical concepts, allowing us to "see" the behavior of functions. In this task, graphing the function \( f(x) = e^{-1/x^2} \) for \( x eq 0 \) and \( f(x) = 0 \) for \( x = 0 \) is crucial.

When we plot the graph, two key observations arise:

• As \( x \) nears zero from either side, \( f(x) \) rapidly approaches zero, depicted as steep slopes collapsing towards the origin.
• At \( x = 0 \), there's a flatness in the graph — the function is itself zero, and all its derivatives there are zero, creating an extremely flat region.

This graphical flatness at the origin translates to why the Maclaurin series fails to depict \( f(x) \) accurately beyond worldly zero. The series cannot "capture" the non-zero behavior of \( f(x) \) for \( x eq 0 \).

Thus, graphing reveals not just the difference in \( f(x) \) and its Maclaurin series but emphasizes why detailed insight is necessary, confirmations of which visual analysis provides fixedly.