Problem 57 - Essential Calculus Early Transcendentals

Question

$$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}}} & {\text { if } x \neq 0} \\\ {0} & {\text { if } x=0}\end{array}\right.$$ (a) Use the definition of derivative to compute \(f^{\prime}(0)\) (b) Show that \(f\) has derivatives of all orders that are defined on \(\mathbb{R} .[\)Hint. . First show by induction that there is a polynomial \(p_{n}(x)\) and a nonnegative integer \(k_{n}\) such that \(f^{(n)}(x)=p_{n}(x) f(x) / x^{k_{n}}\) for \(x \neq 0 . ]\)

Step-by-Step Solution

Answer

(a) \(f'(0) = 0\). (b) \(f\) has all order derivatives on \(\mathbb{R}\).

Step 1: Analyze and Define the Derivative

To find the derivative of the function at the point where it is defined piecewise, we use the limit definition of the derivative: \[f'(0) = \lim_{{h \to 0}} \frac{f(h) - f(0)}{h}.\]Given that \(f(0) = 0\), this simplifies to:\[f'(0) = \lim_{{h \to 0}} \frac{f(h)}{h}.\]

Step 2: Substitute the Piecewise Function in Limit

For \(x eq 0\), \(f(x) = e^{-1/x^2}\). Therefore, substitute \(f(h) = e^{-1/h^2}\) in the limit:\[f'(0) = \lim_{{h \to 0}} \frac{e^{-1/h^2}}{h}.\]

Step 3: Evaluate the Limit

To evaluate the limit \( \lim_{{h \to 0}} \frac{e^{-1/h^2}}{h}\), note that as \(h \to 0\), \(-1/h^2\) approaches \(-\infty\). Therefore, \(e^{-1/h^2}\) approaches \(0\) faster than any polynomial \(h\). The numerator approaches 0 significantly faster than the denominator, leading to:\[f'(0) = 0.\]

Step 4: Define the Induction Hypothesis

We need to find a formula for derivatives of all orders at any point \(x eq 0\). The induction hypothesis is that for some \(n\) there exists a polynomial \(p_n(x)\) and integer \(k_n\) such that:\[f^{(n)}(x) = \frac{p_n(x) f(x)}{x^{k_n}}.\]

Step 5: Base Case for Induction

For the base case where \(n = 0\):\[f^{(0)}(x) = f(x).\]Clearly, it can be expressed by:\[p_0(x) = 1, \quad k_0 = 0.\]

Step 6: Inductive Step

Assume the induction hypothesis holds for \(n\). We need to show it holds for \(n+1\). Using the product rule:\[f^{(n+1)}(x) = \left(\frac{p_n(x) f(x)}{x^{k_n}}\right)'.\]Apply the quotient and product rule, and simplify, assuming each term can be broken into polynomial factors with appropriate \(e^{-1/x^2}\) and power of \(x\). Continue this pattern.

Step 7: Conclude from Induction

Because each differentiation introduces terms of polynomial \(p_n(x)\) multiplied by \(\frac{1}{x^k}\), and because each differentiation only multiplies existing polynomial factors without creating non-polynomial terms, every derivative \(f^{(n)}(x)\) can be expressed in the desired form.Given that this is true for any \(x eq 0\), and \(f\) is smooth at \(x=0\), the derivatives of all orders exist on \(\mathbb{R}\).

Key Concepts

Limit Definition of Derivative

To compute the derivative of a function at a specific point, particularly when dealing with piecewise functions, we leverage the limit definition of the derivative. This foundational concept in calculus states that the derivative of a function at a point \(c\) is defined as:\[f'(c) = \lim_{{h \to 0}} \frac{f(c+h) - f(c)}{h}.\]
In our exercise, the objective is to find \(f'(0)\) for the given piecewise function. Substituting \(x = 0\) and using \(f(x)\) for \(x eq 0\), we simplify the limit to:

• \[f'(0) = \lim_{{h \to 0}} \frac{f(h)}{h}.\]

This form arises because the function \(f(x)\) is defined such that \(f(0) = 0\). Evaluating such limits often requires understanding how rapidly terms approach zero or infinity. In this case, as \(h \to 0\), \(-1/h^2\) goes to \(-\infty\), hence \(e^{-1/h^2}\) approaches zero very quickly, yielding:

• \[f'(0) = 0.\]

This demonstrates the importance of understanding exponential functions within the limit framework when dealing with derivatives at specific points.

Induction and Recurrence

Induction is a powerful mathematical technique used to prove assertions for an infinite sequence of statements. In our exercise, it helps establish the form of derivatives of any order by demonstrating a principle holds at each step.The base case, which we initially identify as \(f^{(0)}(x) = f(x)\), connects the function's starting point to its derived forms. Using the induction hypothesis, we assume that there exists a polynomial \(p_n(x)\) and integer \(k_n\) such that:

• \[f^{(n)}(x) = \frac{p_n(x) f(x)}{x^{k_n}}.\]

To prove the next order derivative, \(f^{(n+1)}(x)\), we use the product and quotient rules. These allow us to differentiate the expression, following a pattern where each step introduces polynomial factors without disrupting the polynomial nature of the equation. By maintaining polynomials, the induction process facilitates that every derivative can be expressed this way for any \(x eq 0\). This structured approach shows that all derivatives exist and are defined on \(\mathbb{R}\).

Piecewise Functions

Piecewise functions are unique because they define separate expressions over different portions of their domain. This characteristic offers flexibility in modeling real-world scenarios where behaviors change. In our exercise, the function \(f(x)=\begin{cases} e^{-1/x^2} & \text{if } x eq 0 \ 0 & \text{if } x=0 \end{cases}\) is piecewise, altering its formula based on whether \(x\) equals zero or not.This duality in definition demands particular attention especially when calculating limits, derivatives, or integrals, as continuity might not be apparent without a deeper dive. When solving for derivatives, this function illustrates the sharp contrast between its rapid decline toward zero as \(x\) approaches zero from either side, compared to its constant state at zero.

• Evaluating piecewise functions often involves matching conditions at boundaries, ensuring smooth transitions where definitions change.
• The derivative at \(x=0\) highlights the critical role the exponential function plays, significantly influencing behavior in the surrounding areas.

Understanding these nuances can illuminate how piecewise setups efficiently model variable phenomena, underlying the importance of comprehensive analysis in calculus.