Chapter 5 Differentiation
- Part A: Exercise 1 – Exercise 14
- Part B: Exercise 15 – Exercise 20
- Part C: Exercise 21 – Exercise 29
Exercise 15
(By analambanomenos) Let $g(x)=A/x+Bx$ for $x>0$ where $A$ and $B$ are positive real numbers. Then $g’(x)=-A/x^2+B$ and $g”(x)=2A/x^3>0$. Since $g’(x)=0$ for $x=\sqrt{A/B}$, $g$ has the minimum value $g\bigl(\sqrt{A/B}\bigr)=2\sqrt{AB}$.
Following the hint, by Theorem 5.15, for $h>0$ there is $\xi\in(x,x+2h)$ such that
\begin{align*}
f(x+2h) &= f(x)+2hf’(x)+2h^2f”(\xi) \\
f’(x) &= \frac{f(x+2h)-f(x)}{2h}-hf”(\xi)
\end{align*}Hence $\big|f’(x)\big|\le M_0/h+M_2h\le 2\sqrt{M_0M_2}$ by the previous result, or $M_1^2\le4M_0M_2$.
Letting $f(x)$ be the example above, we get
$$f’(x)=
\begin{cases}
4x & (-1<x<0) \\
\displaystyle\frac{4x}{(x^2+1)^2} & (0<x<\infty)
\end{cases} \quad\quad
f”(x)=
\begin{cases}
4 & (-1<x<0) \\
\displaystyle\frac{4(1-3x^2)}{(x^2+1)^3} & (0<x<\infty)
\end{cases}$$ $$f^{(3)}(x)=
\begin{cases}
0 & (-1<x<0) \\
\displaystyle\frac{48x(x^2-1)}{(x^2+1)^4} & (0<x<\infty)
\end{cases}$$ For $x<0$, $f’$ is negative, so $f(x)$ decreases from 1 to $-1$, and for $x>0$, $f’$ is positive, so $f(x)$ increases monotonically from $-1$ to a limit of $1$. Hence $M_0=1$.
For $x<0$, $f’$ increases linearly from $-4$ to 0. For $x>0$, $f”(x)=0$ has a single solution at $x=\sqrt{3}/3$, so $f’(x)$ has a maximum value of $3\sqrt{3}/4$. Hence $M_1=4$.
For $x>0$, $f^{(3)}(x)$ has the single solution $x=1$, so $f”(x)$ decreases from 4 to a minimum value of $f”(1)=-1$, then increases monotonically to a limit of 0. Hence $M_2=4$.
Hence for this example, $M_1^2=4M_0M_1=16$.
Regarding whether this theorem holds for vector-valued functions, the following proof comes from Roger Cooke’s solution manual and is reproduced here for completeness.
Consider a vector-valued function $f$ on $(a, \infty)$. As before let
\begin{align*}
M_0 &= \sup_{a < x < \infty} |f(x)| \\
M_1 &= \sup_{a < x < \infty} |f’(x)| \\
M_2 &= \sup_{a < x < \infty} |f”(x)| \,.
\end{align*}We then must show that $M_1^2 \leq 4 M_0 M_2$.
Clearly if $M_0=0$ then $f(x) = 0$ (zero vector) for all $x \in (a, \infty)$. It then follows that $f’(x) = f”(x) = 0$ so that $M_0 = M_1 = M_2 = 0$, resulting in the inequality clearly being true. It is also clearly satisfied if $M_1 = 0$ since the right side is always greater than or equal to zero.
If $M_2 = 0$ then $f”(x) = 0$ for all $x \in (a, \infty)$. It then must be that $f’$ is constant so that $f(x)$ is linear in $x$, i.e. if $f(x) = (f_1(x), \ldots, f_n(x))$ then every $f_k(x)$ is linear or constant. But since $f$ is bounded (since $M_1$ exists) it has to be that $f$ is constant. From this it follows that $M_1 = 0$ so that the inequality is again always satisfied.
So then suppose that $M_0 > 0$, $M_1 > 0$, and $M_2 > 0$. Then there is an $\alpha \in \mathbf R$ where $0 < \alpha < M_1$. Then $\alpha$ is \emph{not} an upper bound of $|f’(x)|$ so that there is an $x_0 \in (a, \infty)$ where $|f’(x_0)| > \alpha$. Let
\[ u = \frac{1}{|f’(x_0)|}f’(x_0) \]so that $|u| = 1$. Consider then the real-valued function
\[ \phi(x) = u \cdot f(x) \]and let
\[ N_0 = \sup_{a < x < \infty} |\phi(x)| \,, \]which exists since for all $x \in (a, \infty)$
\[ |\phi(x)| \leq |u||f(x)| = |f(x)| \leq M_0 \,. \]Thus $N_0 \leq M_0$ since $M_0$ is an upper bound of $|\phi(x)|$. Similarly letting
\begin{align*}
N_1 = \sup_{a < x < \infty} |\phi’(x)| \\
N_2 = \sup_{a < x < \infty} |\phi”(x)|
\end{align*}results in $N_1 \leq M_1$ and $N_2 \leq M_2$.
Also, however,
\[ N_1 \geq |\phi’(x_0)| = |u \cdot f’(x_0)| = \left| \frac{f’(x_0)}{|f’(x_0)|} \cdot f’(x_0) \right| = \frac{|f’(x_0)|^2}{|f’(x_0)|} = |f’(x_0)| > \alpha \,. \]Then since $\phi$ is real-valued it follows from the solution above that
\[ N_1^2 \leq 4 N_0 N_2 \]so that
\[ \alpha^2 < N_1^2 \leq 4 N_0 N_2 \leq 4 M_0 M_2 \]since $0 < \alpha < N_1$, $N_0 \leq M_0$, and $N_2 \leq M_2$. Since this is true for every $\alpha$ less than $M_1$ it has to be that
\[M_1^2 \leq 4 M_0 M_2 \]so that the result also holds for vector-valued functions.
Exercise 16
(By analambanomenos) Following the hint, let $f”$ be bounded by $M$ on $(0,\infty)$. Then by Exercise 15, $$\sup_{x>a}\big|f’(x)\big|\le4M\sup_{x>a}\big|f(x)\big|$$ which tends to 0 as $a\rightarrow\infty$.
Exercise 17
(By analambanomenos) Following the hint, applying Theorem 5.15 with $\alpha=0$ and $\beta=\pm1$, we get
\begin{align*}
f(1) &= 1 = \frac{f”(0)}{2}+\frac{f^{(3)}(s)}{6} \\
f(-1) &= 0 = \frac{f”(0)}{2}-\frac{f^{(3)}(t)}{6}
\end{align*}for some $s\in(0,1)$ and $t\in(-1,0)$. Subtracting the second equation from the first, we get $6=f^{(3)}(s)+f^{(3)}(t)$, so at least one of the two terms is $\ge 3$.
Exercise 18
(By analambanomenos) We have
\begin{align*}
f’(t) &= Q(t)+(t-\beta)Q’(t) \\
f”(t) &= 2Q’(t)+(t-\beta)Q”(t) \\
f^{(3)}(t) &= 3Q”(t)+(t-\beta)Q^{(3)}(t)
\end{align*} and so forth, which can be rewritten
\begin{align*}
Q(t) &= f’(t)+Q’(t)(\beta-t) \\
Q’(t) &= \frac{1}{2}\bigl(f”(t)+Q”(t)(\beta-t)\bigr) \\
Q”(t) &= \frac{1}{3}\bigl(f^{(3)}(t)+Q^{(3)}(t)(\beta-t)\bigr).
\end{align*}Hence
\begin{align*}
f(\beta) &= f(\alpha)+Q(\alpha)(\beta-\alpha) \\
&= f(\alpha)+f’(\alpha)(\beta-\alpha)+Q’(\alpha)(\beta-\alpha)^2 \\
&= f(\alpha)+f’(\alpha)(\beta-\alpha)+\frac{1}{2}f”(\alpha)(\beta-\alpha)^2+\frac{1}{2}Q”(\alpha)(\beta-\alpha)^3 \\
&= f(\alpha)+f’(\alpha)(\beta-\alpha)+\frac{1}{2}f”(\alpha)(\beta-\alpha)^2+\frac{1}{3!}f^{(3)}(\beta-\alpha)^3+\frac{1}{3!}Q^{(3)}(\alpha)(\beta-\alpha)^4
\end{align*}which easily leads to the desired formula.
Exercise 19
(By analambanomenos) For (a) and (b), we need to find an algebraic expression that relates $D_n$ to the difference quotients found in the definition of $f’(0)$, $\bigl(f(\alpha_n)-f(0)\bigr)/
\alpha_n$ and $\bigl(f(\beta_n)-f(0)\bigr)/\beta_n$ in such a way that we can safely let $n\rightarrow\infty$. To simplify the algebra, replace $f$ with $F=f-f(0)$ so that $F(0)=0$. We start with
\begin{align*}
\frac{F(\beta_n)}{\beta_n}-\frac{F(\alpha_n)}{\alpha_n} &= \frac{\alpha_nF(\beta_n)-\beta_nF(\alpha_n)}{\alpha_n\beta_n} \\
&= \frac{\alpha_nF(\beta_n)-\alpha_nF(\alpha_n)+\alpha_nF(\alpha_n)-\beta_nF(\alpha_n)}{\alpha_n\beta_n} \\
&= \frac{F(\beta_n)-F(\alpha_n)}{\beta_n}+\frac{(\alpha_n-\beta_n)F(\alpha_n)}{\alpha_n\beta_n}.
\end{align*}To get $D_n$, we need to multiply by $\beta_n/(\beta_n-\alpha_n)$, and this gives us what we need:
\begin{align*}
\biggl(\frac{F(\beta_n)}{\beta_n}-\frac{F(\alpha_n)}{\alpha_n}\biggr)\frac{\beta_n}{\beta_n-\alpha_n} &= \biggl(\frac{F(\beta_n)-F(\alpha_n)}{\beta_n}+
\frac{(\alpha_n-\beta_n)F(\alpha_n)}{\alpha_n\beta_n}\biggr)\frac{\beta_n}{\beta_n-\alpha_n} \\
&=\frac{F(\beta_n)-F(\alpha_n)}{\beta_n-\alpha_n}-\frac{F(\alpha_n)}{\alpha_n}.
\end{align*}Rearranging and substituting back $f-f(0)$ for $F$, we get $$D_n=\frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}=\biggl(\frac{f(\beta_n)-f(0)}{\beta_n}-
\frac{f(\alpha_n)-f(0)}{\alpha_n}\biggr)\frac{\beta_n}{\beta_n-\alpha_n}+ \frac{f(\alpha_n)-f(0)}{\alpha_n}.$$ For (b), the $\beta_n/(\beta_n-\alpha_n)$ factor is assumed to be bounded, and for part (a) it is bounded by 1. In either case, we can pass to a subsequence where the factor converges to a finite limit, so letting $n\rightarrow\infty$ we get $\lim D_n=f’(0)$.
For part (c), we can apply Theorem 5.10 to get $D_n=f’(\gamma_n)$ for some $\gamma_n$ between $\alpha_n$ and $\beta_n$. Since $\lim\gamma_n=0$ and $f’$ is continuous, we get $\lim D_n=f’(0)$.
Let $f$ be the function in Example 5.6(b), $f(x)=x^2\sin(1/x)$ for $x\neq0$ and $f(0)=0$. It was shown that $f’(0)=0$, although $f’$ is not continuous at 0. Let $$\alpha_n=
\frac{2}{(4n-1)\pi}\quad\quad\beta_n=\frac{2}{(4n-3)\pi}.$$ Then $\sin(1/\alpha_n)=-1$ and $\sin(1/\beta_n)=1$, so that $f(\alpha_n)=-\alpha_n^2$ and $f(\beta_n)=\beta_n^2$. Hence $$D_n=\frac{\beta_n^2+\alpha_n^2}{\beta_n-\alpha_n}=\cdots=\frac{2}{\pi}\biggl(\frac{16n^2-16n+5}{16n^2-16n+3}\biggr)$$ which tends to $2/\pi$ as $n\rightarrow\infty$.
Exercise 20
(By analambanomenos) As in Theorem 5.15, let $f$ be a real function on $[a,b]$ and $n$ a positive integer such that $f^{(n-1)}$ is continuous on $[a,b]$ and $f^{(n)}(t)$ exists for $t\in(a,b)$. Let $\alpha$ and $\beta$ be distinct points of $[a,b]$ and define $$P_{f,\alpha}(t)=\sum_{k=0}^{n-1}\frac{f^{(k)}(\alpha)}{k!}(t-\alpha)^k.$$ Then by Theorem 5.15 there is a point $x$ between $\alpha$ and $\beta$ such that $$\big|f(\beta)-P_{f,\alpha}(\beta)\big|\le\frac{(\beta-\alpha)^n}{n!}\big|f^{(n)}(x)\big|.$$ We can extend this result to vector-valued functions just as Theorem 5.19 extended the Mean-Value theorem. Let $\mathbf f$ be a continuous mapping of $[a,b]$ into $\mathbf R^k$ such that $\mathbf f^{(n-1)}$ is continuous on $[a,b]$ and $\mathbf f^{(n)}$ exists for $t\in(a,b)$. Let $\alpha$ and $\beta$ be distinct points in $[a,b]$ and define
$$\mathbf P_{\mathbf f,\alpha}(t)=\sum_{k=0}^{n-1}\frac{\mathbf f^{(k)}(\alpha)}{k!}(t-\alpha)^k.$$ Put $\mathbf z=\mathbf f(\beta)-\mathbf P_{\mathbf f,\alpha}(\beta)$, and let $\varphi(t)=\mathbf z\cdot\mathbf f(t)$. Then by Theorem 5.15 there is a point $x$ between $\alpha$ and $\beta$ such that $$\big|\varphi(\beta)-P_{\varphi,\alpha}(\beta)\big|=
\frac{(\beta-\alpha)^n}{n!}\big|\varphi^{(n)}(x)\big|=\frac{(\beta-\alpha)^n}{n!}\bigl|\mathbf z\cdot\mathbf f^{(n)}(x)\bigr|.$$ We also have $$\big|\varphi(\beta)-
P_{\varphi,\alpha}(\beta)\big|=\big|\mathbf z\cdot\mathbf f(\beta)-\mathbf z\cdot\mathbf P_{\mathbf f,\alpha}(\beta)\big|=\big|\mathbf z\cdot\mathbf z\big|=
|\mathbf z|^2.$$ By the Schwartz inequality, we have $$|\mathbf z|^2=\frac{(\beta-\alpha)^n}{n!}\big|\mathbf z\cdot\mathbf f^{(n)}(x)\big|\le\frac{(\beta-\alpha)^n}{n!}
|\mathbf z|\big|\mathbf f^{(n)}(x)\big|$$ so that $$\big|\mathbf f(\beta)-\mathbf P_{\mathbf f,\alpha}(\beta)\big|=|\mathbf z|\le\frac{(\beta-\alpha)^n}{n!}\big|
\mathbf f^{(n)}(x)\big|.$$
