Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let $$ f(x)=\left\\{\begin{array}{ll} e^{-1 / x^{2}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ (a) Show that \(f^{\prime}(0)=0\) by using the definition of the derivative. (b) Show that \(f^{\prime \prime}(0)=0\). (c) Assuming the known fact that \(f^{(n)}(0)=0\) for all \(n\), find the Maclaurin series for \(f(x)\). (d) Does the Maclaurin series represent \(f(x) ?\) (e) When \(a=0\), the formula in Theorem \(\mathrm{B}\) is called Maclaurin's Formula. What is the remainder in Maclaurin's Formula for \(f(x)\) ? This shows that a Maclaurin series may exist and yet not represent the given function (the remainder does not tend to 0 as \(n \rightarrow \infty)\)

Step-by-Step Solution

Answer

(a) and (b) show 0; (c) Series is 0; (d) No, (e) Remainder does not vanish.

Step 1: Find the Derivative at Zero

To show that \(f'(0) = 0\), apply the definition of the derivative: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}. \]Since \(f(0)=0\) and for \(x eq 0\), \(f(x) = e^{-1 / x^2}\), we have:\[ f'(0) = \lim_{h \to 0} \frac{e^{-1 / h^2} - 0}{h} = \lim_{h \to 0} \frac{e^{-1 / h^2}}{h}. \]Consider the exponential expression as \(h\) approaches 0. The term \(e^{-1/h^2}\) approaches 0 faster than any polynomial, thus the whole term approaches 0. Thus \(f'(0) = 0\).

Step 2: Find the Second Derivative at Zero

To find \(f''(0)\), again use the definition of the derivative on \(f'(x)\):\[ f''(0) = \lim_{h \to 0} \frac{f'(h) - f'(0)}{h}. \]Following a similar logic from Step 1, and the exponential decay nature of \(e^{-1/h^2}\) yielding \(f'(h)\), the limit as \(h\) approaches 0 results in:\[ f''(0) = \lim_{h \to 0} \frac{g(h)}{h} \] where \(g(h) = f'(h) \text{ as a function approaching } 0\). Thus, \(f''(0) = 0\).

Step 3: Construct the Maclaurin Series for f(x)

A Maclaurin series expansion for a function \(f(x)\) around 0 is given by:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n. \]Since all derivatives at zero, \(f^{(n)}(0) = 0\) for all \(n\), each term in the series is 0. Hence, the Maclaurin series for \(f(x)\) is simply 0.

Step 4: Determine if Maclaurin Series Represents f(x)

Since the series consists entirely of zero terms, the Maclaurin series for \(f(x)\) is the constant function 0. However, this does not represent \(f(x)\) completely since for \(x eq 0\), \(f(x) = e^{-1/x^2}\), which is not zero.

Step 5: Explain Remainder in Maclaurin's Formula

Maclaurin's Formula includes a remainder term modeled by:\[ R_n(x) = \frac{f^{(n+1)}(c)x^{n+1}}{(n+1)!} \text{ for some } c \in (0, x). \]For \(f(x)\), this remainder does not tend to zero as \(n \to \infty\) because \(f(x)\) itself approaches zero more quickly than any polynomial, demonstrating the series cannot converge to \(f(x)\) on any interval around 0 where \(x eq 0\).

Key Concepts

Understanding Maclaurin Series

A Maclaurin series is a type of Taylor series expansion of a function about zero. It provides a way to represent a function as an infinite sum of its derivatives evaluated at zero, multiplied by powers of the variable divided by factorials. The formula for a Maclaurin series is given by:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \)

To construct a Maclaurin series, calculate the derivatives of the function at zero. The interesting case here is when all derivatives at zero are zero, as it was with the function \( f(x) = e^{-1/x^2} \) around zero. Despite each derivative at zero being zero, indicating a Maclaurin series composed solely of zero terms, this series doesn’t always represent the original function fully.

• In some cases, including this exercise, the series may fail to represent \( f(x) \) because of certain function behaviors, like rapid decay or oscillation, that the series cannot capture effectively.

Understanding when a series represents a function is crucial in calculus, linking back to convergence and remainder terms.

Exploring Derivatives

Derivatives are fundamental in calculus, providing insights into how functions change. They describe the rate of change of a function with respect to its variable. The first derivative \( f'(x) \) gives the slope of the tangent line to the function at any point. Subsequent derivatives, like the second derivative \( f''(x) \), reveal information about the function's curvature and acceleration.In this exercise, finding \( f'(0) \) and \( f''(0) \) for the function \( f(x) = e^{-1/x^2} \) involved applying the definition of derivatives:

• First derivative: \( f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \)
• Second derivative: \( f''(0) = \lim_{h \to 0} \frac{f'(h) - f'(0)}{h} \)

Understanding these limits can seem tricky, but central to the process is how quickly \( e^{-1/h^2} \) approaches zero. The function approaches zero faster than any polynomial function as \( x \to 0 \), hence the derivatives at zero are all zero. This behavior is significant as it forms the basis for Maclaurin expansion conclusions.

The Concept of Exponential Decay

Exponential decay describes a process where quantities reduce at a rate proportional to their current value. In calculus, this concept is represented by functions like \( e^{-x} \), which decay rapidly. The function \( f(x) = e^{-1/x^2} \) presents an even more dramatic form of exponential decay.As \( x \to 0 \), \( e^{-1/x^2} \) approaches zero extremely fast. This principle was crucial in analyzing the behavior of derivatives in this exercise:

• The function decayed so quickly as \( x \) approached zero that it rendered all derivatives at zero to vanish, \( f^{(n)}(0) = 0 \).
• This rapid decay made the function's Maclaurin series tricky - while it exists, it fails to represent \( f(x) \) for values other than zero.

In calculus and real-world situations, understanding exponential decay is valuable for predicting how processes behave over time, whether it's radioactive decay, depreciation of assets, or cooling under certain conditions. Here, it highlights the limitations of polynomial approximations in capturing swiftly vanishing behaviors.