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

Solution to Principles of Mathematical Analysis Chapter 5 Part C


Chapter 5 Differentiation

Exercise 21

(By analambanomenos) I’m going to show this for the infinitely differentiable case (and even that won’t be complete). Cauchy’s function $F(x)=e^{-1/x^2}$ for $x\neq 0$, $F(0)=0$, is the classic counterexample of a non-constant, infinitely differentiable real function such that $f^{(n)}(0)=0$ for all orders $n$ (you can easily find a proof of this, or you can try it yourself). We can use this to define a function $\tilde{F}(x)=F(x)$ for $x>0$, $\tilde{F}(x)=0$ for $x\le0$, which is infinitely differentiable everywhere on $\mathbf R$ and whose zero set is $(-\infty,0]$. Finally, for $a<b$ let $$F_{a,b}(x)=\tilde{F}(x-a)\tilde{F}(b-x),$$ an infinitely differentiable function on $\mathbf R$ whose zero set is $(-\infty,a]\cup[b,\infty)$.

The complement of a closed set $E$ of $\mathbf R$ is an open set consisting of a (possibly infinite) collection of open intervals $(a_n,b_n)$. The function $f(x)=\sum F_{a_n,b_n}(x)$ is well-defined since at most one of the terms in the sum is non-zero for any given $x$, infinitely differentiable everywhere on $\mathbf R$, and has zero set $E$.


Exercise 22

(By analambanomenos)

(a) Suppose $f(a)=a$ and $f(b)=b$ for $a<b$. By Theorem 5.10, there is a point $t$, $a<t<b$, such that $f’(t)=\bigl(f(b)-f(a)\bigr)/(b-a)=1$, contradicting $f’(t)\neq 1$ for all real $t$.

(b) If $t=f(t)=t+(1+e^t)^{-1}$, then $(1+e^t)^{-1}=0$, which is impossible. We have $$f(t)=1-\frac{e^t}{(1+e^t)^2}\quad\quad f”(t)=\frac{e^t(e^t-1)}{(1+e^t)^3}.$$ Since $f”(t)$ has a single zero, at $t=0$, $f’(t)$ decreases from $\lim_{t\rightarrow-\infty}f’(t)=1$ to $f’(0)=3/4$, then increases to $\lim_{t\rightarrow\infty}f’(t)=1$, so that the range of $f’$ is $[3/4,1)$.

(c) Since $x_{n+1}=f(x_n)$ and $x_{n}=f(x_{n-1})$, by Theorem 5.10 there is a point $t$ between $x_{n-1}$ and $x_n$ such that $$(x_{n+1}-x_n)=(x_n-x_{n-1})f’(t).$$ Hence
$$|x_{n+1}-x_n|<A|x_n-x_{n-1}|<A^2|x_{n-1}-x_{n-2}|<\cdots<A^{n-1}|x_2-x_1|.$$ So for $m$ and $n$ greater than $N$, we have
\begin{align*}
|x_n-x_m| &\le |x_n-x_{n+1}|+\cdots+|x_{m-1}-x_m| \\
&\le (A^{n-1}+\cdots+A^{m-2})|x_2-x_1| \\
&\le |x_2-x_1|\sum_{k=N}^{\infty}A^k \\
&\le \frac{|x_2-x_1|}{1-A}A^N
\end{align*} Since $A<1$, this last term goes to 0 as $N\rightarrow\infty$, so $\{x_n\}$ is a Cauchy sequence, converging to $x$. Since $f$ is differentiable, it is continuous, hence
$$f(x)=\lim_n f(x_n)=\lim_n x_{n+1}=x$$ that is, $x$ is a fixed point of $f$.

(d) This zig-zag path is described in $\mathbf R^2$ with respect to the graph of $f$ in that space. It starts with the point $\bigl(x_1,x_2=f(x_1)\bigr)$ on the graph of $f$, goes horizonatally until it meets the diagonal $y=x$ at $(x_2,x_2)$ then goes vertically until it hits the graph of $f$ again at $\bigl(x_2,x_3=f(x_2)\bigr)$, and so forth. It will zig-zag or spiral to the point on the graph of $f$ corresponding to a fixed point of $f$, where it crosses the diagonal $y=x$.


Exercise 23

(By analambanomenos) The fixed points of $f$ are the zeros of $g(x)=f(x)-x$, of which there are at most three. Since $g(-2)=-1/3$, $g(-1)=1$, $g(0)=1/3$, $g(1)=-1/3$, and $g(2)=1$, there are three fixed points of $f$, lying in the intervals $(-2,-1)$, $(0,1)$, and $(1,2)$, as asserted.

Note that $\Delta x_n=x_n-x_{n-1}=f(x_{n-1})-x_{n-1}=g(x_{n-1})$. For $x<\alpha$, $g(x)<0$, and $g$ decreases monotonically to $-\infty$ as $x\rightarrow-\infty$. Hence if $x_1<\alpha$, then $g(x_{n-1})=\Delta x_n<0$, so $x_n<x_{n-1}$ and $\Delta x_{n+1}=g(x_n)<\Delta x_n$. Hence $x_n\rightarrow-\infty$ as $n\rightarrow\infty$.

Similarly, for $x>\gamma$, $g(x)>0$, and $g$ increases monotonically to $+\infty$ as $x\rightarrow+\infty$. Hence if $x_1>\gamma$, then $g(x_{n-1})=\Delta x_n>0$, so $x_n>x_{n-1}$ and $\Delta x_{n+1}=g(x_n)>\Delta x_n$. Hence $x_n\rightarrow+\infty$ as $n\rightarrow\infty$.

For $x\in(\alpha,\beta)$ we have $g(x)>0$, and for $x\in(\beta,\gamma)$ we have $g(x)<0$. I want to compare $g$ with the linear function $\beta-x$, which also has a zero at $x=\beta$ and has a slope of $-1$. Since $g’(\beta)=\beta^2-1>-1$, we have $g(x)<\beta-x$ for $x\in(\alpha,\beta)$ and $g(x)>\beta-x$ for $x\in(\beta,\gamma)$.

Putting this together, we get that if $x_n\in(\alpha,\beta)$, then $0<\Delta x_{n+1}<\beta-x_n$ so that $x_{n+1}\in(x_n,\beta)$. Similarly, if $x_n\in(\beta,\gamma)$, then $\beta-x<\Delta x_{n+1}<0$ so that $x_{n+1}\in(\beta,x_n)$.

That is, if $x_1\in(\alpha,\beta)$, then $\{x_n\}$ is a monotonically increasing sequence with upper bound $\beta$, and if $x_1\in(\beta,\gamma)$, then $\{x_n\}$ is a monotonically decreasing sequence with lower bound $\beta$. In either case they must have a limit $\beta’$. This must be a zero of $g$ (and so a fixed point of $f$) since otherwise $g(\beta’)\neq 0$, so for $x_n$ close to $\beta’$, the value of $|\Delta x_n|$ would be larger than $|\beta’-x_n|$ which would contradict $\beta’$ being a limit point of the monotonic sequence $\{x_n\}$. Hence $\beta’=\beta$.


Exercise 24

(By analambanomenos) The distance between the successive elements of the sequence $x_n=f(x_{n-1})$ is given by the function $F(x)=f(x)-x$, while the distance between the successive elements of the sequence $y_n=g(y_{n-1})$ is given by the function $G(y)=g(y)-y$. We have
$$\displaylines{
F(x)=-\frac{1}{2}\biggl(x-\frac{\alpha}{x}\biggr) \cr
F\bigl(\sqrt{\alpha})=0,\quad F’\bigl(\sqrt{\alpha}\bigr)=-1 \cr
G(y)=\frac{\alpha-y^2}{1+\alpha}\cr
G\bigl(\sqrt{\alpha})=0,\quad G’\bigl(\sqrt{\alpha}\bigr)=-\frac{2}{1+\sqrt{\alpha}}.
}$$ By Taylor’s theorem we have near $\sqrt{\alpha}$
\begin{align*}
F(x) &= -(x-\sqrt{\alpha})+K_1(x-\sqrt{\alpha})^2 \\
G(y) &= -\frac{2}{1+\sqrt{\alpha}}(y-\sqrt{\alpha})+K_2(y-\sqrt{\alpha})^2
\end{align*}Now $\sqrt{\alpha}-x$ is the distance from $x$ to $\sqrt{\alpha}$, so we see that as $x_n$ approaches $\sqrt{\alpha}$ the differences between the successive elements of the first sequence become very close to the distance to $\sqrt{\alpha}$, but not so much in the case of the second sequence, which has the additional factor of $2/\bigl(1+\sqrt{\alpha})$. In other words, the difference between the successive steps of the $x_n$ and the distance to $\sqrt{\alpha}$ is quadratic in $x-\sqrt{\alpha}$, while in the case of the second sequece the difference is only linear in $y-\sqrt{\alpha}$.


Exercise 25

(By analambanomenos)

(a) The tangent to the graph of $f$ at the point $\bigl(x_n,f(x_n)\bigr)$ has the equation $$y-f(x_n)=f’(x_n)(x-x_n).$$ Letting $y=0$ and solving for $x$, we see that this line intersects the $x$-axis at the point $(x_{n+1},0)$.

(b) Note that if $x_n>\xi$ then $f(x_n)>0$ so that $x_{n+1}=x_n-f(x_n)/f’(x_n)<x_n$. By Theorem 5.10, there is a $t\in(\xi,x_n)$ such that $$f(x_n)=f(x_n)-f(\xi)=f’(t)(x_n-\xi)\quad\hbox{or}\quad \xi=x_n-\frac{f(x_n)}{f’(t)}\le x_n-\frac{f(x_n)}{f’(x_n)}=x_{n+1}$$ since $f”\ge0$ implies $f’(t)\le f’(x_n)$. If for some $m$ $x_{m}=\xi$ then $x_{m+1}=x_{m+2}=\cdots=\xi$ and so the infinite sequence converges to $\xi$. Otherwise, it is a monotonically decreasing sequence with lower bound $\xi$ and so by Theorem 3.14 converges to a limit $y\ge\xi$.

To see that $y=\xi$, let $g$ be the continuous function defined in part (e), $g(x)=x-f(x)/f’(x)$. Assume that $y>\xi$. Then $f(y)>0$, so $g(y)<y$. Let $\varepsilon>0$ be small enough so that $g(y)+\varepsilon<y$. Then there is a $\delta>0$ such that $\big|g(x)-g(y)\big|<\varepsilon$ if $|x-y|<\delta$. Let $N$ be large enough so that $|x_N-y|<\delta$. Then $g(x_N)<g(y)+\varepsilon<y$, but this contradicts $g(x_N)=x_{N+1}\ge y$. Hence $y=\xi$.

(c) Letting $\alpha=x_n$ and $\beta=\xi$ in Theorem 5.15, and using $f(x_n)=f’(x_n)(x_n-x_{n+1})$, there is some $t_n\in(\xi,x_n)$ such that
\begin{align*}
f(\xi) &= f(x_n)+f’(x_n)(\xi-x_n)+\frac{f”(t_n)}{2}(\xi-x_n)^2 \\
0 &= f’(x_n)(x_n-x_{n+1})+f’(x_n)(\xi-x_n)+\frac{f”(t_n)}{2}(\xi-x_n)^2 \\
0 &= f’(x_n)(\xi-x_{n+1})+\frac{f”(t_n)}{2}(\xi-x_n)^2 \\
x_{n+1}-\xi &= \frac{f”(t_n)}{2f’(x_n)}(x_n-\xi)^2
\end{align*}(d) Part (c) gives us
\begin{align*}
(x_{n+1}-\xi) &\le A(x_n-\xi)^2 \\
&\le A\cdot A^2(x_{n-1}-\xi)^{2^2} \\
&\le A\cdot A^2\cdot A^{2^2}(x_{n-2}-\xi)^{2^3} \\
&\le\cdots \\
&\le A^{1+2+2^2+\cdots+2^{n-1}}(x_1-\xi)^{2^n} = A^{2^n-1}(x_1-\xi)^{2^n} = \frac{1}{A}\bigl(A(x_1-\xi)\bigr)^{2^n}.
\end{align*}The algorithm described in Exercises 3.16 and 3.18 is Newton’s method applied to the functions $x^2-\alpha$ and $x^p-\alpha$, respectively.

(e) We have $g(\xi)=\xi$ if and only if $f(\xi)=0$. Since $$g’(x)=1-\frac{f’(x)^2-f(x)f”(x)}{f’(x)^2}=\frac{f(x)f”(x)}{f’(x)^2}\le f(x)\frac{M}{\delta^2},$$ $g’(x)$ tends to 0 as $x$ approaches $\xi$.

(f) For $f(x)=x^{1/3}$, note that $$x_{n+1}=x_1-\frac{x_n^{1/3}}{(1/3)x_n^{-2/3}}=-2x_n$$ so that $x_n=(-2)^{n-1}x_1$. That is, $\{x_n\}$ is an alternating sequence such that $|x_n|\rightarrow\infty$ as $x\rightarrow\infty$.


Exercise 26

(By analambanomenos) I’m going to show this for vector-valued functions $\mathbf f=(f_1,\ldots,f_k)$ mapping $[a,b]$ into $\mathbf R^k$ such that $\mathbf f(a)=\mathbf 0$ and such that $\big|\mathbf f’(x)\big|\le A|\mathbf f(x)|$, since this is needed for Exercise 28.

If $A=0$, then $\mathbf f’=0$, so $\mathbf f=\mathbf f(a)=\mathbf 0$ by Theorem 5.11(b). So assume that $A>0$. Following the hint, let $\delta>0$ be small enough so that $\delta<1/A$ and $x_0=a+\delta\le b$. For $i=1,\ldots,k$, let $$M_{i,0}=\sup\big|f_i(x)\big|, \quad M_{i,1}=\sup\big|f_i’(x)\big|$$ for $a\le x\le x_0$. By Theorem 5.19, for any $x\in(a,x_0)$ and for every $i=1,\ldots,k$ there is a $t_i\in(a,x)$ such that $$\big|f_i(x)\big|=\big|f_i(x)-f_i(a)\big|=\big|f_i’(t)\big|(x-a)\le M_{i,1}(x-a)\le AM_{i,0}\delta.$$ Since $A\delta<1$, this can only happen if $M_{i,0}=0$, so $f_i(x)=0$ in $[a,a+\delta]$ for each $i=1,\ldots,k$, that is, $\mathbf f=\mathbf 0$ in $[a,a+\delta]$. Repeating this argument with $a+\delta$ replacing $a$, we get $\mathbf f(x)=\mathbf 0$ in $[a,a+2\delta]$. After about $(b-a)/\delta$ steps, we’ve shown that $\mathbf f(x)=\mathbf 0$ in $[a,b]$.


Exercise 27

(By analambanomenos) Following the hint, let $f_1$, $f_2$ be two solutions of the initial value problem and let $f=f_1-f_2$. Then $f$ is differentiable on $[a,b]$, $f(a)=f_1(a)-f_2(a)=0$, and
$$\big|f’(x)\big|=\big|f_1′(x)-f_2′(x)\big|=\big|\phi(x,f_1(x))-\phi(x,f_2(x))\big|\le A\big|f_1(x)-f_2(x)\big|=A\big|f(x)\big|.$$ Applying Exercise 27, we get $f(x)=0$ for all $x\in[a,b]$.


Exercise 28

(By analambanomenos) If there is a constant $A$ such that $$\big|\boldsymbol\phi(x,\mathbf y_2)-\boldsymbol\phi(x,\mathbf y_1)\big|\le A|\mathbf y_2-\mathbf y_1|,$$ then $\mathbf y’=
\boldsymbol\phi(x,\mathbf y)$, $\mathbf y(a)=\mathbf c$ has at most one solution. For suppose $\mathbf f_1$ and $\mathbf f_2$ are two such solutions, and let $\mathbf f=
\mathbf f_1-\mathbf f_2$. Then $\mathbf f$ is differentiable on $[a,b]$, $\mathbf f(a)=\mathbf 0$, and $$\big|\mathbf f’(x)\big|= \big|\boldsymbol\phi(x,\mathbf f_1(x))-
\boldsymbol\phi(x,\mathbf f_2(x))\big|\le A|\mathbf f_1-\mathbf f_2|=A|\mathbf f|.$$ Hence by the vector-valued version of Exercise 26 shown above, $\mathbf f(x)=\mathbf 0$ for all $x\in[a,b]$.


Exercise 29

(By analambanomenos) Using the notation of Exercise 28, $\boldsymbol\phi(x,\mathbf y)=\bigl(\phi_1(x,\mathbf y),\ldots,\phi_k(x,\mathbf y)\bigr)$ where
$$\mathbf \phi_j(x,\mathbf y)=
\begin{cases}
y_{j+1} & j=1,\ldots,k-1 \\
f(x)+\sum_{i=1}^k g_i(x)y_i = f(x)+\mathbf g(x)\cdot\mathbf y & j=k
\end{cases}$$ where $\mathbf g(x)=\bigl(g_1(x),\ldots,g_k(x)\bigr)$. Then
\begin{align*}
\big|\boldsymbol\phi(x,\mathbf y_2)-\boldsymbol\phi(x,\mathbf y_2)\big|^2 &= \sum_{j=1}^k\big|\phi_j(x,\mathbf y_2)-\phi_j(x,\mathbf y_1)\big|^2 \\
&= \sum_{j=1}^{k-1}|y_{2,j+1}-y_{1,j+1}|^2 + \big|\mathbf g(x)\cdot(\mathbf y_2-\mathbf y_1)\big|^2 \\
&\le |\mathbf y_2-\mathbf y_1|^2+|\mathbf g(x)|^2|\mathbf y_2-\mathbf y_1|^2 \\
&=\bigl(1+|\mathbf g(x)|^2\bigr)|\mathbf y_2-\mathbf y_1|^2 \\ \\
\big|\boldsymbol\phi(x,\mathbf y_2)-\boldsymbol\phi(x,\mathbf y_2)\big| &\le \sqrt{\bigl(1+|\mathbf g(x)|^2\bigr)}\;|\mathbf y_2-\mathbf y_1|
\end{align*}Since the $g_i(x)$ are continuous on $[a,b]$, they are bounded, so that there is a constant $A$ such that $\sqrt{\bigl(1+|\mathbf g(x)|^2\bigr)}\le A$ for all $x\in[a,b]$. Hence by Exercise 28, the equation has at most one solution.

Baby Rudin 数学分析原理完整第五章习题解答

Linearity

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