Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let $$ f(x)= \begin{cases}e^{-1 / x^{2}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases} $$ (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

The Maclaurin series is 0 and does not represent \( f(x) \).

Step 1: Definition of Derivative at Zero

The derivative of a function at a point is defined by:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]For \( f'(0) \), substituting the given function:\[ f'(0) = \lim_{h \to 0} \frac{e^{-1/h^2} - 0}{h} \]As \( h \to 0 \), \( e^{-1/h^2} \to 0 \). Thus, the limit is \( \lim_{h \to 0} \frac{0}{h} = 0 \) confirming \( f'(0) = 0 \).

Step 2: Second Derivative at Zero

To find \( f''(0) \), we use the definition:\[ f''(0) = \lim_{h \to 0} \frac{f'(h) - f'(0)}{h} \]Since we showed \( f'(0) = 0 \) and \( f'(h) \approx 0 \) for small \( h \):\[ \lim_{h \to 0} \frac{0 - 0}{h} = 0 \]Thus, \( f''(0) = 0 \).

Step 3: Maclaurin Series Using Known Derivatives

The Maclaurin series for a function is given by:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]Since \( f^{(n)}(0) = 0 \) for all \( n \), all the terms in the series are zero:\[ f(x) = 0 \]

Step 4: Checking Series Representation

The derived Maclaurin series \( f(x) = 0 \) does not match the given function for \( x eq 0 \), since \( f(x) = e^{-1/x^2} \) when \( x eq 0 \). Thus, the series does not represent the actual function.

Step 5: Maclaurin Series Remainder Analysis

The remainder \( R_n(x) \) in the Maclaurin series is:\[ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1} \]for some \( \xi \) between 0 and \( x \). Since \( f^{(n)}(0) = 0 \) and \( f(x) eq 0 \) for \( x eq 0 \), the remainder \( R_n(x) \to e^{-1/x^2} \) showing the series does not converge to \( f(x) \).

Key Concepts

Definition of Derivative

In calculus, the derivative of a function at a point provides the rate at which the function's value changes as its input changes. This concept is foundational for understanding how functions behave. The derivative is formally defined as the limit of the difference quotient as the step size approaches zero. Formally, if we have a function \(f(x)\), its derivative at a point \(a\) is given by:

• \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\)

This definition tells us how to calculate the slope of the tangent line to the curve of \(f(x)\) at the point \(x = a\). For our function \(f(x)\), especially at \(x = 0\), substituting the given piecewise function in the definition:

• \(f'(0) = \lim_{h \to 0} \frac{e^{-1/h^2} - 0}{h}\)

As \(h\) approaches zero, \(e^{-1/h^2}\) quickly approaches zero as well, leading the limit to be 0. Therefore, this confirms that the derivative \(f'(0) = 0\). This approach is critical for delving deeper into how the function behaves at a pivotal point like zero.

Calculus Function Analysis

Analyzing functions using calculus involves understanding how derivatives affect the function's graph and behavior. Knowing both the first and second derivatives can provide significant insights. For a given function like \(f(x)\), we ascertain the following:

• \(f'(x)\) tells us about the slope or rate of change.
• \(f''(x)\) typically provides information on the concavity and inflection points.

In our problem's context, finding the second derivative \(f''(0)\) helps confirm the behavior at zero. Using:

• \(f''(0) = \lim_{h \to 0} \frac{f'(h) - f'(0)}{h}\)

Given that \(f'(0)=0\) and that \(f'(h)\) is approximated by zero for very small \(h\), it shows a consistent rate of non-increasing values at zero, thus resulting in \(f''(0)=0\). These derivatives, therefore, support examining whether our function changes directions or how it behaves at specific intervals.

Series Representation

The Maclaurin series is a method of expressing a function as an infinite series of terms calculated from the derivatives at a single point, often zero. This series helps in analyzing complex functions by breaking them into simpler polynomial components.

• The formula for the Maclaurin series is \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\).
• This allows us to represent a function using its derivatives at 0 (or another specified point).

In the given exercise, since \(f^{(n)}(0) = 0\) for all \(n\), the entire series simplifies to zero:

• \(f(x) = 0\)

This result suggests that the Maclaurin series in this instance doesn't accurately depict the function beyond zero. Specifically, for \(x eq 0\), \(f(x) = e^{-1/x^2}\), showing a significant deviation. Such differences highlight scenarios wherein a function's series representation might not seamlessly align or converge with the original function, especially regarding the concept of remainder terms in the series expansion.