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Chapter 3: Problem 31

Let \(F(x)=x^{2} \sin \left(x^{-1}\right)\) and \(G(x)=x^{2} \sin \left(x^{-2}\right)\) for \(x \neq 0\), and \(F(0)=\) \(G(0)=0\). a. \(F\) and \(G\) are differentiable everywhere (including \(x=0\) ). b. \(F \in B V([-1,1])\), but \(G \notin B V([-1,1])\).

Short Answer

Expert verified
a. Both are differentiable everywhere. b. \( F \) is of bounded variation, \( G \) is not.

Step by step solution

01

Check Differentiability at x 鈮� 0 for F(x)

To check differentiability of \( F(x) = x^2 \sin(x^{-1}) \) at \( x eq 0 \), apply the product rule: differentiate \( x^2 \) and \( \sin(x^{-1}) \). The derivative of \( x^2 \) is \( 2x \), and the derivative of \( \sin(x^{-1}) \) using the chain rule is \( \cos(x^{-1})(-x^{-2})\). Thus, the derivative \( F'(x) = 2x \sin(x^{-1}) + x \cos(x^{-1})(-x^{-2}) = 2x \sin(x^{-1}) - \cos(x^{-1})/x \). This is differentiable for \( x eq 0 \).
02

Check Differentiability at x = 0 for F(x)

We need to compute \( F'(0) \). Since \( F(0) = 0 \), compute the limit \( \lim_{x \to 0} \frac{F(x) - F(0)}{x - 0} \), which yields \( \lim_{x \to 0} x \sin(x^{-1}) \). By using the fact that \( |\sin(x^{-1})| \leq 1 \), the limit becomes \( \lim_{x \to 0} x = 0\), making \( F \) differentiable at 0 with \( F'(0) = 0 \). Thus, \( F \) is differentiable everywhere.
03

Check Differentiability at x 鈮� 0 for G(x)

For \( G(x) = x^2 \sin(x^{-2}) \), differentiate using the product rule similarly: The derivative is \( 2x \sin(x^{-2}) + x^2 (-2x^{-3}) \cos(x^{-2}) = 2x \sin(x^{-2}) - 2x^{-1} \cos(x^{-2}) \), which is differentiable at \( x eq 0 \).
04

Check Differentiability at x = 0 for G(x)

We need to evaluate \( G'(0) \). Calculate \( \lim_{x \to 0} \frac{G(x) - G(0)}{x - 0} = \lim_{x \to 0} x \sin(x^{-2}) \). Again, since \( |\sin(x^{-2})| \leq 1 \), the limit is \( \lim_{x \to 0} x = 0 \). Hence, \( G \) is differentiable at 0 with \( G'(0) = 0 \). Thus, \( G \) is differentiable everywhere.
05

Analyze Bounded Variation for F(x)

For bounded variation, we need the function to have a bounded total variation over \([-1, 1]\). For \( F(x) \), the primary behavior is determined by \( x^2 \) which smoothens the oscillations of \( \sin(x^{-1}) \) as \( x \to 0 \). This causes \( F(x) \) to have bounded variation on \([-1, 1]\).
06

Analyze Bounded Variation for G(x)

For \( G(x) \), the function \( \sin(x^{-2}) \) oscillates wildly as \( x \to 0 \) and is only multiplied by \( x^2 \), which isn't enough to smooth out the oscillations unlike in \( F(x) \). Therefore, the variation of \( G(x) \) diverges as \( x \to 0 \), proving it does not have bounded variation on \([-1, 1]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bounded Variation
The concept of **bounded variation** refers to how much a function can "wiggle" or oscillate. It's a measure of the total amount that a function changes over a given interval. For a function to have bounded variation on an interval like \([-1, 1]\), the total variation must be a finite number.

In the context of the functions \(F(x)\) and \(G(x)\):
  • For \(F(x) = x^{2} \sin\left(x^{-1}\right)\), the behavior is dominated by \(x^2\). This term helps "calm" down the oscillations of \(\sin\left(x^{-1}\right)\) as \(x\) approaches zero. Thus, \(F(x)\) has bounded variation over \([-1, 1]\).
  • On the other hand, \(G(x) = x^{2} \sin\left(x^{-2}\right)\) does not benefit as much from the stabilizing factor of \(x^2\) with \(\sin(x^{-2})\), causing it to oscillate too wildly near zero. Hence, it does not have bounded variation over \([-1, 1]\).
Understanding bounded variation is important as it gives insights into the function's behavior across its domain and helps determine its differentiability and integrability characteristics.
Product Rule
The product rule is a fundamental concept for differentiating functions that are products of two functions. If you have a function defined as a product \(h(x) = f(x)g(x)\), the product rule states that its derivative is:
\[h'(x) = f'(x)g(x) + f(x)g'(x)\]

Looking at the functions we are considering:
  • For \(F(x) = x^2 \sin(x^{-1})\), the product rule is used as follows: differentiate \(x^2\) to get \(2x\), leaving \(\sin(x^{-1})\) unchanged, and vice verso. The result: \( F'(x) = 2x \sin(x^{-1}) - \cos(x^{-1})/x \).
  • For \(G(x) = x^2 \sin(x^{-2})\), the process is similar with its result: \( G'(x) = 2x \sin(x^{-2}) - 2x^{-1} \cos(x^{-2}) \).
The product rule is essential whenever you are working with functions that are combinations of two or more differentiable functions. It allows us to understand and compute the derivative efficiently.
Chain Rule
The chain rule is another essential rule in calculus for differentiating compositions of functions, such as \(f(g(x))\). It states that the derivative is:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

In our case:
  • For \( F(x) = x^2 \sin(x^{-1}) \), consider \( \sin(x^{-1}) \) as the composition function. Here, \( u = x^{-1} \) leads to \( \sin(u) \), and \( u' = -x^{-2} \). Therefore, the derivative becomes \( \cos(x^{-1})(-x^{-2}) \), part of the differentiation steps for \( F'(x) \).
  • Similarly, for \( G(x) = x^2 \sin(x^{-2}) \), \( \sin(x^{-2}) \) involves a composition where \( u = x^{-2} \) with \( u' = -2x^{-3} \), obtaining \( \cos(x^{-2})(-2x^{-3}) \).
Using the chain rule correctly ensures precise differentiation of nested functions, which frequently occur in mathematical problems.
Derivatives
Derivatives are a fundamental concept in calculus representing the rate of change of a function with respect to its variable. It essentially tells how a function changes at any given point.

In this exercise, derivatives help identify differentiability everywhere, meaning at every point of their respective domains:
  • For \( F(x) = x^{2} \sin\left(x^{-1}\right) \), its derivatives both at \( x eq 0 \) and \( x = 0 \) confirm it is smooth across its domain.
  • The same logic applies for \( G(x) = x^{2} \sin\left(x^{-2}\right) \), where computations show it too is differentiable at all points.
Understanding how to compute and interpret derivatives is crucial, as it paves the way for understanding the behavior and characteristics of functions in calculus.

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