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MATH 447: Real Variables - Homework #9

Jerich Lee

December 22, 2024

Problem.
[28.6b] Let $ f(x) = x \sin{} \frac{1}{x} $ for $ x \neq{} 0 $ and $ f(0) = 0 $. See Fig. 19.3.
  1. Is $ f $ differentiable at $ x = 0 $? Justify your answer.
Solution.
Proof. By using Definition 28.1 from Ross of the derivative, we can show that the function $f(x)$ is not differentiable at $x=0$. $$\begin{align} \frac{f(t)-f(0)}{t-0} = \frac{t\sin(\frac{1}{t})-0}{t}=\sin \left( \frac{1}{t}\right) \end{align}$$ $\sin{}(\frac{1}{t})$ does not tend to any limit as $t\to{} 0$, so the proof is done.
∎

Problem.
[29.3] Suppose $ f $ is differentiable on $ \mathbb{R} $ and $ f(0) = 0 $, $ f(1) = 1 $ and $ f(2) = 1 $.
  1. Show $ f'(x) = \frac{1}{2} $ for some $ x \in (0,2) $.
  2. Show $ f'(x) = \frac{1}{7} $ for some $ x \in (0,2) $.
Solution.
  1. Proof. By MVT, $\exists{} t \ s.t. \ \frac{f(2)-f(0)}{2-0}=f^\prime{} (t)$ $$\begin{align} f^\prime(t)=\frac{f(2)-f(0)}{2-0}=\frac{1}{2} \end{align}$$
    ∎

  2. Proof. From $(0,1)$, by MVT, $\exists{} t_1 \ s.t. \ $ $f^\prime{} (t)=1$. From $(1,2)$, by MVT, $\exists{} t_2 \ s.t. \ f^\prime{} (t)=0$. We know that $1>\frac{1}{7}>0$. By IVTD (Intermediate Value property of Derivatives), $\exists{} t_3 \in{} (0,2) \ s.t. \ f^\prime{} (t_3)=\frac{1}{7}$.
    ∎

Problem.
[29.10] Let $ f(x) = x^2 \sin{}(\frac{1}{x}) + \frac{x}{2} $ for $ x \neq{} 0 $ and $ f(0) = 0 $.
  1. Show $ f'(0) > 0 $; see Exercise 28.4.
  2. Show $ f $ is not increasing on any open interval containing 0.
  3. Compare this example with Corollary 29.7(i).
Solution. We will appeal to Corollary 29.7 and determine $f^\prime{} (x)<0$ for all $x\in{} (a,b)\cup{} \{ 0 \}$, which will prove that $f(x)$ is not increasing on any open interval containing $0$. Applying Theorem 28.3 from Ross, we get the following: $$\begin{align} f^\prime (x)=2x\sin \left( \frac{1}{x} \right) -\cos\left( \frac{1}{x} \right) +\frac{1}{2} \label{3deriv} \end{align}$$ Discussion: The idea is to find a value $x \in{} (a,b)$ containing $0$ so the equation above is negative. This involves two cases: $a<x<0$ and $0<x<b$. For the first case, we see that: $$\begin{align} \frac{1}{x} = -\frac{3\pi}{2}n \\[10pt] x>a = \frac{1}{a}>\frac{1}{x} \\[10pt] \frac{1}{a}> -\frac{3\pi}{2}n \\[10pt] n > -\frac{2}{3\pi a} \end{align}$$ Where $n\in{} \mathbb{N} $.
Proof. Choose $n> -\frac{2}{3\pi{} a} \ s.t. \ n\in{} \mathbb{N} $. Let $x = -\frac{2}{3\pi{} n}$. Then, substituting $x$ into \autoref{3deriv} gives us the following: $$\begin{align} f^\prime (x)=2x\sin \left( \frac{1}{x} \right) -\cos\left( \frac{1}{x} \right) +\frac{1}{2} \\[10pt] = 2\left( -\frac{3\pi}{2}n \right)\left( 1 \right) - 0 + \frac{1}{2} \\[10pt] = \frac{1-6\pi}{2}<0 \end{align}$$ From the above, we were able to find $a<x<0$ such that $f^\prime{} (x)<0$, which disproves that $f$ is increasing from $(a,b)$ containing $0$. The second case where $0<x<b$ is handled similarly.
∎

Problem.
[29.12]
  1. Show $ x < \tan x $ for all $ x \in \left( 0, \frac{\pi}{2} \right) $.
  2. Show $ \frac{x}{\sin x} $ is a strictly increasing function on $ \left( 0, \frac{\pi}{2} \right) $.
  3. Show $ x \leq \frac{\pi}{2} \sin x $ for $ x \in \left[ 0, \frac{\pi}{2} \right] $.
Solution.
  1. To prove that $x<\tan(x)$ such that $f(x)=\tan(x)$ for all $x \in (0, \frac{\pi}{2})$, it suffices to show that $f^\prime (x)>1$ for all $x\in(0,\frac{\pi}{2})$
    Proof. $$\begin{align} f^\prime (x)>1 \\[10pt] \sec^{2}(x)-1>0\\[10pt] \underbrace{\frac{1}{\cos^{2}(x)}}_{< 1} >1 \end{align}$$
    ∎

  2. To prove that $f(x)=\frac{x}{\sin(x)}$ is strictly increasing on $(0,\frac{\pi}{2})$, it suffices to show that $f^\prime (x)>0$ for all $x\in(0,\frac{\pi}{2})$
    Proof. $$\begin{align} f^\prime (x)=-x(\sin(x))^{-2}\cos(x)+(\sin(x))^{-1}\\[10pt] \frac{\sin(x)-x\cos(x)}{\sin^{2}(x)}>0 \\[10pt] \sin(x)-x\cos(x)>0 \label{4.2} \end{align}$$ \autoref{4.2} is always true for $x\in{} (0, \frac{\pi}{2})$.
    ∎

  3. To prove that $x \leq \frac{\pi}{2}\sin(x)$ such that $f(x) = \frac{\pi}{2}\sin(x)$ for all $x \in [0, \frac{\pi}{2}]$, it suffices to show that $g(x) = \frac{\pi}{2}\sin(x) - x \geq 0$ on the interval.
    Proof. Let $g(x) = \frac{\pi}{2}\sin{}(x) - x$. To analyze $g(x)$, we compute its derivative: $$\begin{align} g'(x) &= \frac{\pi}{2}\cos(x) - 1 \\[10pt] \frac{\pi}{2}\cos(x) &\geq 1 \\[10pt] \cos(x) &\geq \frac{2}{\pi}. \end{align}$$ The inequality $\cos{}(x) \geq{} \frac{2}{\pi{}}$ implies that $x \leq{} \arccos{}(\frac{2}{\pi{}})$, since $\cos{}(x)$ is decreasing on $[0, \frac{\pi}{2}]$. Thus, $g'(x) \geq{} 0$ for $x \in{} [0, \arccos{}(\frac{2}{\pi{}})]$, and $g'(x) \leq{} 0$ for $x \in{} [\arccos{}(\frac{2}{\pi{}}), \frac{\pi}{2}]$. Therefore, $g(x)$ increases on $[0, \arccos{}(\frac{2}{\pi{}})]$ and decreases on $[\arccos{}(\frac{2}{\pi{}}), \frac{\pi}{2}]$, reaching its maximum at $x = \arccos{}(\frac{2}{\pi{}})$. Now, compute $g(x)$ at the boundaries: $$\begin{align} g(0) &= \frac{\pi}{2}\sin(0) - 0 = 0, \\[10pt] g\left(\frac{\pi}{2}\right) &= \frac{\pi}{2}\sin\left(\frac{\pi}{2}\right) - \frac{\pi}{2} = 0. \end{align}$$ Since $g(x) \geq{} 0$ on $[0, \frac{\pi}{2}]$, it follows that $x \leq{} \frac{\pi}{2}\sin{}(x)$ for all $x \in{} [0, \frac{\pi}{2}]$.
    ∎

Problem.
[29.16] Use Theorem 29.9 to obtain the derivative of the inverse $ g = \tan{}^{-1} = \arctan{} $ of $ f $ where $ f(x) = \tan{} x $ for $ x \in{} ( -\frac{\pi}{2}, \frac{\pi}{2} ) $. \bigskip{} 29.9 Theorem. \\ Let $ f $ be a one-to-one continuous function on an open interval $ I $, and let $ J = f(I) $. If $ f $ is differentiable at $ x_0 \in{} I $ and if $ f'(x_0) \neq{} 0 $, then $ f^{-1} $ is differentiable at $ y_0 = f(x_0) $ and $$\begin{align} (f^{-1})'(y_0) = \frac{1}{f'(x_0)}. \end{align}$$
Solution. $$\begin{align} f(x)=\tan(x) \\[10pt] f^\prime (x)=\sec^{2}(x) \\[10pt] x=\arctan(y) \\[10pt] \left( f^{-1} \right) \left( y \right) =\frac{1}{\sec^{2}\left( \arctan(y) \right) } \end{align}$$
Problem.
[32.7] Let $ f $ be integrable on $[a, b]$, and suppose $ g $ is a function on $[a, b]$ such that $ g(x) = f(x) $ except for finitely many $ x $ in $[a, b]$. Show $ g $ is integrable and $$\begin{align} \int_a^b f = \int_a^b g. \end{align}$$ Hint: First reduce to the case where $ f $ is the function identically equal to 0.
Solution.
Proof. Let $g$ be a bounded function from $[a,b]$ (we can deduce the boundedness of $g$ from the integrability of $f$). Let $M=\sup{}\vert g(x) \vert $, and let $E$ be the finite set of points such that $g(x)\neq{} f(x)$. As $E$ is finite, we are able to cover $E$ with disjoint intervals $[u_j, v_j]\subset{} [a,b]$, and also make the sum of all disjoint intervals $[u_j, v_j]$ less than $\varepsilon{}$ (arbitrarily small). By removing these intervals from $[a,b]$, we obtain a new set $K$ (this set is compact, as it is bounded and closed). Using Theorem 21.4 in Ross, we can say that $g$ is uniformly continuous on $K$. This implies the following: $$\begin{align} s \in K, t \in K, \left\vert s-t \right\vert < \delta \implies \left\vert g(s)-g(t) \right\vert \end{align}$$ We can then create a partition $P$ of $[a,b]$ such that $u_j, v_j \in{} P$, but $(u_j,v_j)\notin{} P$. If $x_{i-1}$ is not $u_j$, then $\Delta{} x_i < \delta{}$. We know that $g$ is bounded for all $x\in{} [a,b]$, so $M_{i}-m_i \leq{} 2M $ for all $i$ (this includes the points $u_j$ in the finite set $E$), and if $x_{i-1}$ is not one of the finite $u_j$, then $M_{i}-m_i <\varepsilon{}$. This implies the following: $$\begin{align} U(P,f,x)-L(P,f,x)\leq [b-a]\varepsilon + 2M\varepsilon \end{align}$$ As $\varepsilon{}$ is arbitrary, this proves that $g$ is integrable.
∎