If you find any mistakes, please make a comment! Thank you.

Solution to Principles of Mathematical Analysis Chapter 5 Part B

Chapter 5 Differentiation

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
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
4x & (-1<x<0) \\
\displaystyle\frac{4x}{(x^2+1)^2} & (0<x<\infty)
\end{cases} \quad\quad
4 & (-1<x<0) \\
\displaystyle\frac{4(1-3x^2)}{(x^2+1)^3} & (0<x<\infty)
\end{cases}$$ $$f^{(3)}(x)=
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
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
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
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
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
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).
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
\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:
\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} \\
\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 $$\big|\mathbf f(\beta)-\mathbf P_{\mathbf f,\alpha}(\beta)\big|=|\mathbf z|\le\frac{(\beta-\alpha)^n}{n!}\big|
\mathbf f^{(n)}(x)\big|.$$

Baby Rudin 数学分析原理完整第五章习题解答


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu