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Solution to Principles of Mathematical Analysis Chapter 8 Part A


Chapter 8 Some Special Functions

Exercise 1

(By Matt Frito Lundy) First note that according to the limit definition of the derivative:
\begin{align*}
f’(0) &= \lim_{x \to 0} \frac{f(x) – f(0)}{x – 0} \\
&= \lim_{x \to 0}\frac{e^{-1/x^2}}{x} \\
& = 0
\end{align*}by theorem 8.6(f). This establishes a base case and we proceed with induction by noticing that for $x \neq 0$, $n \in \mathbf N$, we have:$$f^{(n)}(x) = e^{-1/x^2}r_n(x)$$ where $r_n(x)$ is some rational function of $x$. Then, according to the limit definition of the derivative and using the inductive hypothesis that $f^{(n)}(0) = 0$ we have:
\begin{align*}
f^{(n+1)}(0) &= \lim_{x \to 0} \frac{f^{(n)}(x) -f^{(n)}(0)}{x – 0} \\
&= \lim_{x \to 0} \frac{e^{-1/x^2}r_n(x)}{x} \\
&= 0
\end{align*} again using theorem 8.6(f). This establishes that $f(x)$ is infinitely differentiable at $x = 0$ and that $f^{(n)}(0) = 0$.

The point here is that even though $f(x)$ is infinitely differentiable at $x = 0$, it does not have a taylor series expansion about $x = 0$.


Exercise 2

(By Matt Frito Lundy) Along any row, the positive entries form a finite geometric sequence, and the sum of all the entries along a row is
$$\sum_{j = 1}^\infty a_{ij} = -\left( \frac{1}{2}\right) ^{i-1},$$ and so we have $$\sum_{i = 1}^\infty \sum_{j = 1}^\infty a_{ij} = -\sum_{i = 1}^\infty \left( \frac{1}{2}\right) ^{i-1} = -2.$$ Whereas along any column, the positive entries form an infinite geometric sequence, and the sum of all the entries along a column is $$\sum_{i = 1}^{\infty} a_{ij} = -1 + \sum_{n=1}^\infty \left( \frac{1}{2}\right) ^n = -1 + \left( \frac{1}{1 – \frac{1}{2}} – 1\right) = 0,$$ so that $$\sum_{j = 1}^\infty \sum_{i = 1}^{\infty} a_{ij} = 0.$$ Note that this double sequence $\{a_{ij}\}$ fails to meet the criteria of theorem 8.3 because if $$\sum_{j = 1}^\infty \left| a_{ij} \right| = b_i$$ then $\sum b_i$ diverges.


Exercise 3

(By analambanomenos) Let $\sum_ja_{ij}=b_i\in[0,\infty]$. If $\sum b_i$ diverges, then the conclusion follows from Theorem 8.3, so suppose $\sum b_i$ converges. Let $\sum_ia_{ij}=c_j$, we want to show that $\sum c_j=\sum b_i=\infty$. But if $\sum c_j$ converges, then we can define the transposed double sequence $\tilde{a}_{ij}=a_{ji}$, and let $\sum_j\tilde{a}_{ij}=\tilde{b_i}=c_i$. Since $\sum \tilde{b_i}=\sum c_j$ converges, we can apply Theorem 8.3 to conclude that $\sum \tilde{c}_i$ converges, where $\tilde{c}_i=\sum_j\tilde{a}_{ij}=b_i$. But this contradicts the assumption that $\sum b_i$ diverges.


Exercise 4

(By analambanomenos)

(a) By L’Hospital’s Rule, Theorem 5.13, $$\lim_{x\rightarrow 0}\frac{b^x-1}{x}=\lim_{x\rightarrow 0}\frac{(\log b)b^x}{1}=\log b.$$

(b) By L’Hospital’s Rule, $$\lim_{x\rightarrow 0}\frac{\log(1+x)}{x}=\lim_{x\rightarrow 0}\frac{1/(1+x)}{1}=1.$$

Since the exponential function is continuous, if $g(x)$ is a continuous function, and $f(x)=\log\big(g(x)\big)$, and $\lim_{x\rightarrow a}f(x)=A$, then $\lim_{x\rightarrow a}g(x)=e^A$.

(c) Let $f(x)=\log(1+x)^{1/x}=\big(\log(1+x)\big)/x$. Then, from part (b), $$\displaystyle\lim_{x\rightarrow0}(1+x)^{1/x}=e^{\lim_{x\rightarrow 0}f(x)}=e^1=e.$$

(d) Let $f(x)=\log\big(1+(x/n)\big)^n=n\log\big(1+(x/n)\big)$. By L’Hospital’s Rule $$\lim_{n\rightarrow\infty}f(x)=\lim_{n\rightarrow\infty}\frac{\log\bigg(1+\displaystyle
\frac{x}{n}\bigg)}{1/n}=\lim_{n\rightarrow\infty}\frac{-\displaystyle\frac{x}{n^2}\bigg(\frac{1}{1+x/n}\bigg)}{-1/n^2}=\lim_{n\rightarrow\infty}\frac{x}{1+x/n}=x.$$ Hence $\lim_{n\rightarrow\infty}\big(1+(x/n)\big)^n=e^x$.


Exercise 5

(By analambanomenos)

(a) Note from Exercise 3(c) that the limit of the numerator is 0, so by L’Hospital’s Rule,
\begin{align*}
\lim_{x\rightarrow 0}\frac{e-(1+x)^{1/x}}{x} &= -\lim_{x\rightarrow 0}\frac{d}{dx}\bigg(\frac{\log(1+x)}{x}\bigg)(1+x)^{1/x} \\
&= -e\lim_{x\rightarrow 0}\frac{x-(1+x)\log(1+x)}{x^2} & (\hbox{Exercise 3(c)}) \\
&= -e\lim_{x\rightarrow 0}\frac{-\log(1+x)}{2x} & (\hbox{L’Hospital’s Rule}) \\
&= -e\lim_{x\rightarrow 0}\frac{-1/(1+x)}{2} & (\hbox{L’Hospital’s Rule}) \\
&= \frac{e}{2}.
\end{align*}

(b) Applying L’Hospital’s Rule to $\log(n^{1/n})=\log n/n$, we get $\lim_{n\rightarrow\infty}\log n/n=\lim_{n\rightarrow\infty}1/n=0$, so that $\lim_{x\rightarrow\infty}
n^{1/n}=e^0=1$. Hence we can apply L’Hospital’s Rule to get $$\lim_{n\rightarrow\infty}\frac{n^{1/n}-1}{\log n/n}=\lim_{n\rightarrow\infty}\frac{n^{1/n}\displaystyle\frac{d}{dn}
\bigg(\frac{\log n}{n}\bigg)}{\displaystyle\frac{d}{dn}\bigg(\frac{\log n}{n}\bigg)}=\lim_{n\rightarrow\infty}n^{1/n}=1.$$

(c) Note that the derivatives of the numerator and denominator are
\begin{align*}
f(x) &= \tan x-x \\
f’(x) &= \cos^{-2} x-1 \\
f”(x) &= 2\cos^{-3}x\sin x \\
f”’(x) &= 6\cos^{-4}x\sin^2x+2\cos^{-2}(x) \\
g(x) &= x-x\cos x \\
g’(x) &= 1-\cos x+x\sin x \\
g”(x) &= 2\sin x+x\cos x \\
g”’(x) &= 3\cos x-x\sin x
\end{align*}
so that $f(0)=f’(0)=f”(0)=g(0)=g’(0)=g”(0)=0$, while $f”’(0)=2$ and $g”’(0)=3$. Hence applying L’Hospital’s Rule three times, we have $$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=
\lim_{x\rightarrow 0}\frac{f’(x)}{g’(x)}=\lim_{x\rightarrow 0}\frac{f”(x)}{g”(x)}=\lim_{x\rightarrow 0}\frac{f”’(x)}{g”’(x)}=\frac{2}{3}.$$

(d) Note that the derivatives of the numerator and denominator are
\begin{align*}
f(x) &= x-\sin x \\
f’(x) &= 1-\cos x \\
f”(x) &= \sin x \\
f”’(x) &= \cos x \\
g(x) &= \tan x-x \\
g’(x) &= \cos^{-2} x-1 \\
g”(x) &= 2\cos^{-3}x\sin x \\
g”’(x) &= 6\cos^{-4}x\sin^2x+2\cos^{-2}x
\end{align*}
so that $f(0)=f’(0)=f”(0)=g(0)=g’(0)=g”(0)=0$, while $f”’(0)=1$ and $g”’(0)=2$. Hence applying L’Hospital’s Rule three times, we have $$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=
\lim_{x\rightarrow 0}\frac{f’(x)}{g’(x)}=\lim_{x\rightarrow 0}\frac{f”(x)}{g”(x)}=\lim_{x\rightarrow 0}\frac{f”’(x)}{g”’(x)}=\frac{1}{2}.$$


Exercise 6

(By analambanomenos)

(a) By assumption, there is a real $x$ such that $f(x)\ne 0$. Hence $f(x)=f(x+0)=f(x)f(0)$, showing that $f(0)=1$. Since $f$ is differentiable, we have for all real $x$ $$f’(x)=
\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0}\frac{f(x)f(h)-f(x)}{h}=f(x)\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=f(x)f’(0).$$ So, letting $c=f’(0)$, we have $f’(x)=cf’(x)$ for all real $x$. Since this is also true for the function $g(x)=e^{cx}$, we have $$\frac{d}{dx}\bigg(\frac{f}{g}\bigg)=\frac{cfg-cfg}{g^2}=0,$$ so that $f=Kg$ for some constant $K$. Since $f(0)=1=g(0)$, we have $K=1$, or $$f(x)=g(x)=e^{cx}.$$(b) For all rational numbers $p=m/n$, we have $f(p)=f(1)^{m/n}$. Let $c=\log f(1)$, and let $g(x)=e^{cx}$. Since we also have $g(p)=g(1)^{m/n}$ and $g(1)=e^{\log f(1)}=f(1)$, we have $f(x)=g(x)$ on all rational numbers. Hence, since the continuous functions $f$ and $g$ are equal on a dense subset of the real numbers, they are equal for all real numbers.


Exercise 7

(By analambanomenos) Note that by L’Hospital’s Rule, $$\lim_{x\rightarrow 0}\frac{\sin x}{x}=\lim_{x\rightarrow 0}\frac{\cos x}{1}=1.$$ If $f(x)=\sin x/x$, then $$f’(x)=\frac{x\cos x-\sin x}{x^2}.$$ Letting $g(x)=x\cos x-\sin x$, the numerator in the expression above for $f’(x)$, we have $g’(x)=-x\sin x\le 0$ for $0\le x\le\pi/2$, so that $g(x)$ decreases in this interval from $g(0)=0$ to $g(\pi/2)=-1$. Hence $f’(x)<0$ for $0<x<\pi/2$, so that it decreases steadily in this interval, and its values lie between $\lim_{x\rightarrow 0}f(x)=1$ and $f(\pi/2)=2/\pi$.


Exercise 8

(By analambanomenos) Since the functions on both sides of the inequality are periodic functions with period $\pi$, and since they are both even functions, we only have to show this for $x\in[0,\pi/2]$.

Let $f(x)=n\sin x-\sin nx$, $f(0)=0$, $f’(x)=n(\cos x-\cos nx)$. Since $\cos x$ is monotonically decreasing on $[0,\pi/2]$, we have $f’(x)\ge 0$ for $x\in\big[0,\pi/(2n)\big]$, hence $f(x)\ge 0$ in this interval, that is, $n\sin x\ge\sin nx$.

For $x\in[\pi/(2n),\pi/2]$, since $\sin x$ is monotonically increasing, we have $$n\sin x\ge n\sin\big(\pi/(2n)\big)\ge\sin\big(n\pi/(2n)\big)=1,$$ while $|\sin nx|\le 1$. Hence $n\sin x\ge|\sin nx|$ on all of $[0,\pi/2]$.


Exercise 9

(By analambanomenos)

(a) Since the minimum and maximum values of $1/x$ in the interval $[n,n+1]$ are $1/(n+1)$ and $1/n$, respectively, we have $1/(n+1)\le\int_n^{n+1}(1/x)\,dx=\log(n+1)-\log(1)\le 1/n$. Hence,
$$\big(s_{N+1}-\log(N+1)\big)-(s_N-\log N)=\frac{1}{N}-\int_N^{N+1}\frac{1}{x}\,dx\ge 0,$$ so that the sequence $s_N-\log N$ is nondecreasing. Also, since $\log(1)=0$, we have
\begin{align*}
\frac{1}{2}+\cdots+\frac{1}{N} &\le \sum_{n=0}^{N-1}\big(\log(n+1)-\log(n)\big) = \log N \\
-\log N &\le -\frac{1}{2}-\cdots-\frac{1}{N} \\
s_N-\log N &\le 1.
\end{align*}so that the sequence is also bounded above. Hence by Theorem 3.14 the limit of the sequence exists.

(b) From part (a), $s_N\approx\gamma+\log N$. Using the estimate $\gamma\approx 0.577$ and solving for $$100\approx\gamma+\log 10^m,$$ we get the approximate solution $m=43$.


Exercise 10

(By analambanomenos) Following the hint, let $p_1,\ldots,p_k$ be distinct primes. Then the product of the convergent geometric series of $p_k^{-1}$ yields $$\prod_{j=1}^k\bigg(1+\frac{1}{p_j}+\frac{1}{p_j^2}+\cdots \bigg)=\sum\frac{1}{p_1^{n_1}\cdots p_k^{n_k}}$$ where the sum is over all ordered $k$-tuples $(n_1,\ldots,n_k)$ of non-negative integers.

Hence if $N$ is a positive integer, and $p_1,\ldots,p_k$ are the prime numbers that divide at least one integer $\le N$, we have, using Theorem 3.26, $$\sum_{n=1}^N\frac{1}{n}\le\prod_{j=1}^k \bigg(1+\frac{1}{p_j}+\frac{1}{p_j^2}+\cdots\bigg)=\prod_{j=1}^k\bigg(1-\frac{1}{p_j}\bigg)^{-1}.$$Let $f(x)=2x+\log(1-x)$ for $x\in[0,1/2]$. Then $f’(x)=(1-2x)/(1-x)\ge 0$ in this interval. Since $f(0)=0$, we have $f(x)\ge 0$ for $0\le x\le 1/2$, or
\begin{align*}
0 &\le 2x+\log(1-x) \\
\log(1-x)^{-1} &\le 2x \\
(1-x)^{-1} &\le e^{2x}.
\end{align*}Applying this to the above, we get
\begin{align*}
\sum_{n=1}^N\frac{1}{n} &\le \prod_{j=1}^k\bigg(1-\frac{1}{p_j}\bigg)^{-1} \\
&\le \prod_{j=1}e^{2/p_j} \\
&= \exp\Bigg(2\sum_{j=1}^k\frac{1}{p_j}\Bigg).
\end{align*}Since the harmonic series diverges, so must the series in the argument of $\exp$ above.


Exercise 11

(By analambanomenos) Since $|e^{-tx}|\le 1$ for $x\ge 0$ and $t>0$, we have, for $A>0$ and $t>0$ \begin{equation}\label{8.11.1}\lim_{t\rightarrow0}\Bigg|t\int_0^Ae^{-tx}f(x)\,dx\Bigg|\le\lim_{t\rightarrow0}t\int_0^A\big|f(x)\big|\,dx=0.\end{equation} Let $\varepsilon>0$ and let $A$ be large enough so that $1-\varepsilon<f(x)<1+\varepsilon$ for $x\ge A$. For any constant $K$ we have, for $t>0$, $$t\int_A^BKe^{-tx}\,dx=
K(e^{-tA}-e^{-tB})\rightarrow Ke^{-tA}\hbox{ as }B\rightarrow\infty.$$ Hence $$(1-\varepsilon)e^{-tA}=t\int_A^\infty(1-\varepsilon)e^{-tx}\,dx\le t\int_A^\infty e^{-tx}f(x)\,dx\le
t\int_A^\infty(1+\varepsilon)e^{-tx}\,dx=(1+\varepsilon)e^{-tA}.$$ Letting $t\rightarrow 0+$, we get $$(1-\varepsilon)\le\lim_{t\rightarrow0}t\int_A^\infty e^{-tx}f(x)\,dx\le(1+\varepsilon).$$
Combining this with \eqref{8.11.1}, we get $$(1-\varepsilon)\le\lim_{t\rightarrow0}t\int_0^\infty e^{-tx}f(x)\,dx\le(1+\varepsilon).$$ Since $\varepsilon$ was arbitrary, we finally get
$$\lim_{t\rightarrow0}t\int_0^\infty e^{-tx}f(x)\,dx=1.$$


Exercise 12

(By analambanomenos) (a) For $n\ne 0$
\begin{align*}
c_n &= \frac{1}{2\pi}\int_{-\delta}^\delta e^{-inx}\,dx \\
&= \frac{1}{2\pi}\cdot\frac{1}{in}(e^{in\delta}-e^{-in\delta}) \\
&= \frac{\sin n\delta}{\pi n} \\
c_0 &= \frac{1}{2\pi}\int_{-\delta}^\delta 1\,dx \\
&= \frac{\delta}{\pi}
\end{align*}(b) By Theorem 8.14, the Fourier series for $f(x)$ converges for $x=0$ to $f(0)=1$, hence (noting that $(\sin x)/x$ is an even function)
\begin{align*}
1 &= \frac{\delta}{\pi}+\sum_{|n|\ne 0}\frac{\sin n\delta}{\pi n} \\
1-\frac{\delta}{\pi} &= \frac{2}{\pi}\sum_{n=1}^\infty\frac{\sin n\delta}{n} \\
\frac{\pi-\delta}{2} & = \sum_{n=1}^\infty\frac{\sin n\delta}{n}
\end{align*}(c) We have by Parseval’s theorem, using $c_{-n}=c_{n}$ for $n\ne 0$,
\begin{align*}
\frac{1}{2\pi}\int_{-\delta}^\delta\big|f(x)\big|^2\,dx &= \sum_{-\infty}^\infty|c_n|^2 \\
\frac{1}{2\pi}\int_{-\delta}^\delta 1\,dx &= \frac{\delta^2}{\pi^2}+2\sum_{n=1}^\infty\frac{\sin^2n\delta}{\pi^2n^2} \\
\frac{\delta}{\pi}-\frac{\delta^2}{\pi^2} &= \frac{2\delta}{\pi^2}\sum_{n=1}^\infty\frac{\sin^2n\delta}{n^2\delta} \\
\frac{\pi^2}{2\delta}\cdot\frac{\delta(\pi-\delta)}{\pi^2} &= \sum_{n=1}^\infty\frac{\sin^2n\delta}{n^2\delta} \\
\frac{\pi-\delta}{2} &= \sum_{n=1}^\infty\frac{\sin^2n\delta}{n^2\delta}
\end{align*}(d) First note that by L’Hospital’s Rule, $$\lim_{x\rightarrow 0}\frac{\sin x}{x}=\lim_{x\rightarrow 0}\frac{\cos x}{x}=1,$$ so that the integral $$\int_0^A\bigg(\frac{\sin x}{x}\bigg)^2\,dx$$ is well-defined. Also, for $\delta>0$,
\begin{align*}
\lim_{A\rightarrow\infty}\int_\delta^A\bigg|\frac{\sin x}{x}\bigg|^2\,dx &\le \lim_{A\rightarrow\infty}\int_\delta^A\frac{1}{x^2}\,dx \\
&= \lim_{A\rightarrow\infty}\bigg(\frac{1}{\delta}-\frac{1}{A}\bigg) \\
&= \frac{1}{\delta}
\end{align*} so that the improper integral converges. That is, if $\varepsilon>0$, there is an $A>0$ large enough so that $$\Bigg|\int_0^\infty\bigg(\frac{\sin x}{x}\bigg)^2\,dx-
\int_0^A\bigg(\frac{\sin x}{x}\bigg)^2\,dx\Bigg|\le\frac{\varepsilon}{4}.$$Let $\delta=A/M$ for some large integer $M$, and let $P=\{0,\delta,\ldots,M\delta=A\}$ be a partition of $[0,A]$. Then $$\sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta^2}\cdot\delta=
\sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta}$$ is a Riemann sum of the finite integral. Hence there is an $M$ large enough so that $$\Bigg|\int_0^A\bigg(\frac{\sin x}{x}\bigg)^2\,dx-
\sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta}\Bigg|\le\frac{\varepsilon}{4}.$$From part (c), we can make $M$ large enough so that $$\Bigg|\sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta}-\frac{\pi-\delta}{2}\Bigg|\le\frac{\varepsilon}{4}$$ and we can make $M$ large enough so that $\delta/2\le\varepsilon/4$, so that $$\bigg|\frac{\pi-\delta}{2}-\frac{\pi}{2}\bigg|\le\frac{\varepsilon}{4}.$$Putting together these four inequalities, we get that, for all $\varepsilon>0$, $$\Bigg|\int_0^\infty\bigg(\frac{\sin x}{x}\bigg)^2\,dx-\frac{\pi}{2}\Bigg|\le\varepsilon,$$ so that the equality follows.

(e) Since $\sin^2(n\pi/2)=1$ if $n$ is odd, and 0 otherwise, we get from part (c) that $$\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi}{2}\cdot\frac{\pi-(\pi/2)}{2}=\frac{\pi^2}{8}.$$

Baby Rudin 数学分析原理完整第八章习题解答

Linearity

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