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Solution to Principles of Mathematical Analysis Chapter 8 Part C

Chapter 8 Some Special Functions

Exercise 21

(By analambanomenos) Since $D_n$ is an even function, and making a substitution, we can simplify the integral to
L_n(x) &= \frac{1}{2\pi}\int_{-\pi}^\pi\big|D_n(x)\big|\,dx \\
&= \frac{1}{\pi}\int_0^\pi\frac{\big|\sin\big(n+\frac{1}{2}\big)x\big|}{\sin(x/2)}\,dx \\
&= \frac{2}{\pi}\int_0^{\pi/2}\frac{\big|\sin(2n+1)x\big|}{\sin x}\,dx
\end{align*}Letting $A=\pi/(2N+1)$, we can break up the interval of integration into the subintervals $$I_1=[0,A]\quad I_2=[A,2A]\quad\cdots\quad I_n=\big[(n-1)A,nA\big]\quad\big[nA,\pi/2\big]$$ each having length $A$ except the last, which has length $A/2$ and which we can ignore for purposes of the estimate. In the interior of the interval $I_m$, $\sin(2n+1)x$ is positive for $m$ odd, and negative for $m$ even. Also, $\sin x$, which is positive and monotonically increasing over $I_m$, has the maximum value $\sin(mA)<mA$. Hence, for $m$ odd, we have
\frac{2}{\pi}\int_{(m-1)A}^{mA}\frac{\big|\sin(2n+1)x\big|}{\sin x}\,dx &\ge \frac{2}{\pi}\cdot\frac{1}{mA}\int_{(m-1)A}^{mA}\sin(2n+1)x\,dx \\
&= \frac{2}{\pi}\cdot\frac{2n+1}{m\pi}\cdot\frac{1}{2n+1}\int_{(m-1)\pi}^{m\pi}\sin x\,dx \\
&= \frac{2}{\pi^2}\cdot\frac{1}{m}\Big(\cos\big((m-1)\pi\big)-\cos(m\pi)\Big) \\
&= \frac{4}{\pi^2}\cdot\frac{1}{m}
\end{align*}and for $m$ even, we have
\frac{2}{\pi}\int_{(m-1)A}^{mA}\frac{\big|\sin(2n+1)x\big|}{\sin x}\,dx &\ge \frac{2}{\pi}\cdot\frac{1}{mA}\int_{(m-1)A}^{mA}-\sin(2n+1)x\,dx \\
&= \frac{2}{\pi}\cdot\frac{2n+1}{m\pi}\cdot\frac{1}{2n+1}\int_{(m-1)\pi}^{m\pi}-\sin x\,dx \\
&= \frac{2}{\pi^2}\cdot\frac{1}{m}\Big(\cos(m\pi)-\cos\big((m-1)\pi\big)\Big) \\
&= \frac{4}{\pi^2}\cdot\frac{1}{m}
\end{align*}Recall that in Exercise 9 it was shown that $1+(1/2)+\cdots+(1/n)>\log n$. Hence, adding the inequalities above for $m=1,2,\ldots,n$, we get $$L_n\ge\frac{4}{\pi^2}\bigg(1+\frac{1}{2}+\cdots+
\frac{1}{n}\bigg)>\frac{4}{\pi^2}\log n$$ which shows the first part of the Exercise, with $C=4/\pi^2$. (To solve the second part, you would have to get an upper estimate for $L_n$).

Exercise 22

(By analambanomenos) We can show that the series on the right-hand side of the first equation converges by the Ratio Test, Theorem 3.34. Let $$a_n=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n.$$ Then for $n>\alpha$, $$\lim_{n\rightarrow\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg| = \lim_{n\rightarrow\infty}\bigg(\frac{n-\alpha}{n+1}\bigg)|x| = \lim_{n\rightarrow\infty}\bigg(1-\frac{\alpha+1}{n+1}\bigg)|x| = |x|<1.$$ Hence the right-hand side of the equation defines a function $f(x)$ for $|x|<1$. Following the hint, by Theorem 8.1 we can differentiate the series term-by-term to get $$f’(x) = \sum_{n=1}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{(n-1)!}x^{n-1} = \sum_{n=0}^\infty
\frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}x^{n}.$$ Hence \begin{align*}
(1+x)f’(x) &= \sum_{n=0}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}x^{n}+\sum_{n=0}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}x^{n+1} \\
&= \alpha + \sum_{n=1}^\infty\bigg(\frac{\alpha(\alpha-1)\cdots(\alpha-n)}{n!}+\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{(n-1)!}\bigg)x^n \\
&= \alpha + \sum_{n=1}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)(\alpha-n+n)}{n!}x^n \\
&= \alpha + \alpha\sum_{n=1}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n \\
&= \alpha f(x)
\end{align*}If we let $g(x)=(1+x)^\alpha$ be the left-hand side of the equation, then $$(1+x)g’(x)=(1+x)\alpha(1+x)^{n-1}=\alpha g(x),$$ so $g(x)$ also satisfies the differential equation $y’=\alpha y/(1+x)$. Also note that $g(0)=f(0)=1$. Since $$\bigg|\frac{\alpha y_1}{1+x}-\frac{\alpha y_2}{1+x}\bigg|=\bigg(\frac{\alpha}{1+x}\bigg)|y_1-y_2|$$ we can apply the uniqueness result of Exercise 5.27 to conclude that $g(x)=f(x)$ on any interval $[-1+\epsilon,1-\epsilon]$, and so on all of $(-1,1)$.

Note that by Theorem 8.18(a) we have, for $\alpha>0$,\begin{align*}
\Gamma(\alpha+n) &= (\alpha+n-1)\Gamma(\alpha+n-1) \\
&= (\alpha+n-1)(\alpha+n-2)\Gamma(\alpha+n-2) \\
&= \cdots = (\alpha+n-1)(\alpha+n-2)\cdots(\alpha+1)\alpha\Gamma(\alpha).
\end{align*} Hence, for $-1<x<1$ and $\alpha>0$,
(1-x)^{-\alpha} &= 1+\sum_{n=1}^\infty\frac{-\alpha(-\alpha-1)\cdots(-\alpha-n+1)}{n!}(-1)^nx^n \\
&= 1+\sum_{n=1}^\infty\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)}{n!}x^n \\
&= 1+\sum_{n=1}^\infty\frac{\Gamma(n+\alpha)}{n!\,\Gamma(\alpha)}x^n \\
&= \sum_{n=0}^\infty\frac{\Gamma(n+\alpha)}{n!\,\Gamma(\alpha)}x^n

Exercise 23

(By analambanomenos) Following the hint, for $x\in[a,b]$ define $$\varphi(x)=\int_a^x\frac{\gamma’(t)}{\gamma(t)}\,dt.$$ Then $\varphi(a)=0$, and by Theorem 6.20, since $\gamma’/\gamma$ is continuous on $[a,b]$, we have $\varphi’=\gamma’/\gamma$ on $[a,b]$. Let $f(x)=\gamma(x)e^{\varphi(x)}$, $x\in[a,b]$. Then $$f’(x)=\big(\gamma’(x)-\varphi’(x)\gamma(x)\big)e^{\varphi(x)}=0,$$ so $f$ is constant on $[a,b]$, equal to $f(a)=ae^{\varphi(a)}=a$. Hence $f(b)=be^{\varphi(b)}=a=b$, or $$e^{\varphi(b)}=e^{2\pi i\mathrm{Ind}(\gamma)}=1$$ so that $2\pi i\mathrm{Ind}(\gamma)=2\pi in$ for some integer $n$. Hence $\mathrm{Ind}(\gamma)=n$ is an integer.

For $\gamma(t)=e^{int}$, $a=0$, $b=2\pi$, we have $$\mathrm{Ind}(\gamma)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{ine^{int}}{e^{int}}\,dt=\frac{2\pi in}{2\pi i}=n.$$ For these $\gamma$, $\mathrm{Ind}(\gamma)$ measures the number of times $\gamma$ winds counter-clockwise around 0, with a negative value indicating that $\gamma$ winds around 0 clockwise $-\mathrm{Ind}(\gamma)$ times. This is true in a general sense for arbitrary $\gamma$, which is shown in most introductory Complex Analysis courses. Also, it can be shown that two such curves have the same index if and only if one can be “continuously deformed” to the other through an intermediate set of such curves $\gamma_t$.

Exercise 24

(By analambanomenos) Following the hint, for $0\le c<\infty$ let $\gamma_c(t)=\gamma(t)+c$. Then $\gamma_c$ is also a continuous differentiable closed curve in $\mathbf C$ such that $\gamma_c(t)\ne 0$ for all $t\in[a,b]$, so we can define $\mathrm{Ind}(\gamma_c)$. Let $\{c_n\}$ be a sequence of nonnegative real numbers converging to $c$. Then $$\frac{\gamma_{c_n}’(t)}{\gamma_{c_n}}=\frac{\gamma’(t)}{\gamma(t)+c_n}\rightarrow\frac{\gamma’(t)}{\gamma(t)}\hbox{ as }n\rightarrow\infty,$$ and the convergence is uniform for $t\in[a,b]$. Hence by Theorem 7.16 $$\mathrm{Ind}(\gamma_{c_n})=\frac{1}{2\pi i}\int_a^b\frac{\gamma’(t)}{\gamma(t)+c_n}\,dt\rightarrow\frac{1}{2\pi i}\int_a^b\frac{\gamma’(t)}{\gamma(t)}\,dt=\mathrm{Ind}(\gamma),$$ so that $\mathrm{Ind}(\gamma_{c})$ is a continuous function of $c$. Since $\mathrm{Ind}(\gamma_c)$ has only integer values, this forces $\mathrm{Ind}(\gamma_c)=\mathrm{Ind}(\gamma)$ for all $0\le c<\infty$.

Since $[a,b]$ is compact, $\min\gamma(t)>-\infty$ for $t\in[a,b]$. Hence, for large enough $c$ we have $\min\big|\gamma(t)+c\big|$ as large as we like. In that case, we have $$\big|\mathrm{Ind}(\gamma_c)\big|\le\frac{1}{2\pi}\int_a^b\bigg|\frac{\gamma’(t)}{\gamma(t)+c}\bigg|\,dt\le\frac{1}{2\pi}\Bigg(\frac{\max\big|\gamma’(t)\big|}{\min\big|\gamma(t)+c\big|}\Bigg)(b-a)
\rightarrow 0$$ as $c\rightarrow\infty$. Hence $\mathrm{Ind}(\gamma)=\mathrm{Ind}(\gamma_c)=0$ for large enough $c$.

Exercise 25

(By analambanomenos) Following the hint, let $\gamma=\gamma_2/\gamma_1$. This is well-defined since $\gamma_1$ is nowhere 0, and so $\gamma$ is also a continuous differentiable closed curve defined on $[a,b]$ which is nowhere 0, so that we can define $\mathrm{Ind}(\gamma)$. By the condition on $\gamma_1$ and $\gamma_2$ we have for all $t\in[a,b]$ $$\big|1-\gamma(t)\big|=\frac{\big|\gamma_1(t)-\gamma_2(t)\big|}{\big|\gamma_1(t)\big|}<\frac{\big|\gamma_1(t)\big|}{\big|\gamma_1(t)\big|}=1.$$ Hence the range of $\gamma$ does not intersect the negative real axis, so by Exercise 24 we have
0 &= \mathrm{Ind}(\gamma) \\
&= \frac{1}{2\pi i}\int_a^b\frac{\gamma’(t)}{\gamma(t)}\,dt \\
&= \frac{1}{2\pi i}\int_a^b\frac{\gamma_1(t)\gamma_2′(t)-\gamma_2(t)\gamma_1′(t)}{\gamma_1^2(t)}\cdot\frac{\gamma_1(t)}{\gamma_2(t)}\,dt \\
&= \frac{1}{2\pi i}\int_a^b\bigg(\frac{\gamma_2′(t)}{\gamma_2(t)}-\frac{\gamma_1′(t)}{\gamma_1)}\bigg)\,dt \\
&= \mathrm{Ind}(\gamma_2)-\mathrm{Ind}(\gamma_1)
\end{align*}so that $\mathrm{Ind}(\gamma_1)=\mathrm{Ind}(\gamma_2)$.

Exercise 26

(By analambanomenos) Since $$\Big|\big|P_j(t)\big|-\big|\gamma(t)\big|\Big|\le\big|P_j(t)-\gamma(t)\big|<\frac{\delta}{4}$$ we have $$\big|P_j(t)\big|\ge\big|\gamma(t)\big|-\frac{\delta}{4}>\frac{3\delta}{4}.$$ Hence $P_1(t)$ and $P_2(t)$, $0\le g\le2\pi$, define continuous closed differentiable curves in the complex plane which are nowhere 0, so we can define $\mathrm{Ind}(P_j)$. We have $$\big|P_1(t)-P_2(t)\big|\le\big|P_1(t)-\gamma(t)\big|+\big|P_2(t)-\gamma(t)\big|<\frac{\delta}{2}<\frac{3\delta}{4}<\big|P_1(t)\big|$$ so by Exercise 25 we have $\mathrm{Ind}(P_1)=\mathrm{Ind}(P_2)$.

Before showing Exercise 24 for the continuous case, I want to show a general theorem which is pretty standard and may have been done in a previous exercise, or even the text, although I couldn’t find it. Let $X$ be a metric space with disjoint subsets $F$ and $K$ such that $F$ is closed and $K$ is compact. Then there is a positive minimum distance between them, that is, there is a $\delta>0$ such that $d(x,y)>\delta$ for all $x\in F$ and $y\in K$. If not, then there are subsequences $\{x_n\}\subset F$ and $\{y_n\}\subset K$ such that $\lim d(x_n,y_n)=0$. Since $K$ is compact, $\{y_n\}$ has a subsequence $\{y_{n_m}\}$ converging to $y\in K$. Then $$d(x_{n_m},y)\le d(x_{n_m},y_{n_m})+d(y_{n_m},y)\rightarrow 0\hbox{ as }m\rightarrow\infty$$ which, since $F$ is closed, implies that $y\in F$, contradicting the fact that $F$ and $K$ are disjoint. (If you prefer a direct proof, for each $y\in K$ let $N_y$ be an open neighborhood of $y$ disjoint from $F$. Then $\{N_y\}$ is a cover of $K$, so you can pass to a finite subcover whose sets have a positive minimum radius and go from there.)

Suppose $\gamma$ is a closed, continuous curve in the complex plane with domain $[0,2\pi]$, whose range does not intersect the negative real axis. Since the negative real axis is a closed set and the range of $\gamma$ is compact, by the preceding paragraph there is a $\delta>0$ such that if $P$ is a trigonometric polynomial satisfying $\big|P(t)-\gamma(t)\big|<\delta$ for all $t\in[0,2\pi]$, so that the range of $P$ also does not intersect the negative real axis. Since $P$ is differentiable, $\mathrm{Ind}(P)=0$ by Exercise 25, hence $\mathrm{Ind}(\gamma)=0$.

Now suppose $\gamma_1$ and $\gamma_2$ are closed, continuous curves in the complex plane with domain $[0,2\pi]$ such that $\big|\gamma_1(t)-\gamma_2(t)\big|<\big|\gamma_1(t)\big|$ for all $t\in[0,2\pi]$. Then by Theorem 4.16 there is a $\delta>0$ such that $$\big|\gamma_1(t)\big|-\big|\gamma_1(t)-\gamma_2(t)\big|>\delta$$ on the compact set $[0,2\pi]$. Let $P_1$, $P_2$ be trigonometric polynomials such that $\big|P_j(t)-\gamma_j(t)\big|<\delta/4$ for all $t\in[0,2\pi]$, $j=1,2$. Then, for all $t\in[0,2\pi]$, \begin{align*}
\Big|\big|P_1(t)-P_2(t)\big|-\big|\gamma_1(t)-\gamma_2(t)\big|\Big| &\le \Big|\big(P_1(t)-P_2(t)\big)-\big(\gamma_1(t)-\gamma_2(t)\big)\Big| \\
&= \Big|\big(P_1(t)-\gamma_1(t)\big)-\big(P_2(t)-\gamma_2(t)\big)\Big| \\
&\le \big|P_1(t)-\gamma_1(t)\big|+\big|P_2(t)-\gamma_2(t)\big| \\
&< \frac{\delta}{2} \\
\Big|\big|P_1(t)\big|-\big|\gamma_1(t)\big|\Big| &\le \big|P_1(t)-\gamma_1(t)\big| \\
&< \frac{\delta}{4}
\end{align*}Hence $\big|P_1(t)\big|-\big|P_1(t)-P_2(t)\big|>\delta/4$ for all $t\in[0,2\pi]$. Hence, by Exercise 25, $$\mathrm{Ind}(\gamma_1)=\mathrm{Ind}(P_1)=\mathrm{Ind}(P_2)=\mathrm{Ind}(\gamma_2).$$

Exercise 27

(By analambanomenos) Following the hint, suppose that $f(z)\ne 0$ for all $z$. Then the continuous closed curves $\gamma_r(t)=f(re^{it})$, $r\in[0,\infty)$, do not intersect the origin, so we can define the function $\mathrm{Ind}(\gamma_r)$ for $r\in[0,\infty)$. We now show that this function would have the following three properties, which leads to a contradiction.

(a) $\mathrm{Ind}(\gamma_0)=0$.

Since $\gamma_0(t)=f(0)$ is a differentiable function, we have $\mathrm{Ind}(\gamma_0)=(2\pi)^{-1}\int \gamma_0′(t)/\gamma_0(t)\,dt = 0$.

(b) $\mathrm{Ind}(\gamma_r)=n$ for all sufficiently large $r$.

By the assumption, we have $$\lim_{r\rightarrow\infty}\frac{f(re^{it})}{r^ne^{int}}=c$$ That is, for sufficiently large $r$ there is a $\delta>0$ such that
\bigg|\frac{\gamma_r(t)}{r^ne^{int}}-c\bigg| < \delta \\
\big|\gamma_r(t)-cr^ne^{int}\big| < r^n\delta < r^n|c|=|cr^ne^{int}|.
\end{gather*} Let $\gamma(t)=cr^ne^{int}$, a differentiable closed curve which is nowhere 0. By the extension of Exercise 25 shown in Exercise 26, we have $$\mathrm{Ind}(\gamma_r)=\mathrm{Ind}(\gamma)=\frac{1}{2\pi}
\int_0^{2\pi}\frac{\gamma’(t)}{\gamma(t)}\,dt=\frac{1}{2\pi i}\int_0^{2\pi i}\frac{in\gamma(t)}{\gamma(t)}\,dt=n.$$

(c) $\mathrm{Ind}(\gamma_r)$ is a continuous function of $r$ on $[0,\infty)$.

Fix $r_0\in[0,\infty)$. Then the positive continuous function $\big|\gamma_{r_0}(t)\big|$ has a minimum value $\epsilon>0$ on the compact set $[0,2\pi]$ by Theorem 4.16. Since $f$ is continuous, by Theorem 4.19 it is uniformly continuous a compact neighborhood of the circle $\{r_0e^{it}\}$, $0\le t\le 2\pi$. Hence there is a $\delta>0$ such that if $|r-r_0|<\delta$ (limited to small positive values of $r$ in case $r_0=0$), we have $$\big|\gamma_r(t)-\gamma_{r_0}(t)\big|<\epsilon<\big|\gamma_{r_0}(t)\big|.$$ Hence, by the extension of Exercise 25 given in Exercise 26, we have $$\mathrm{Ind}(\gamma_r)=\mathrm{Ind}(\gamma_{r_0})\hbox{, for }|r-r_0|<\delta.$$

Note that a continuous, real-valued function on a connected set $X$ with only integer values must be constant. For then the inverse images of the open sets $(z-0.5,z+0.5)$ for the integers $z$ define a set of disjoint open sets which cover $X$, so by the connectedness of $X$ they must all be empty except one. Hence, since $[0,\infty)$ is a connected set, the properties (a), (b), and (c) of the function $\mathrm{Ind}(\gamma_r)$ on $[0,\infty)$ lead to a contradiction.

Exercise 28

(By analambanomenos) Following the hint, for $0\le r\le 1$, $0\le t\le 2\pi$, put $\gamma_r(t)=g(re^{it})$, a closed, continuous curve with values on $T$ so that we can define $\mathrm{Ind}(\gamma_r)$. Note that since $|z|>1/2$ for $z\in T$, if a trigonometric polynomial $P$ satisfies $|P-\gamma|<1/8$ for any curve with values on $T$, then $\mathrm{Ind}(\gamma)=\mathrm{Ind}(P)$. Since $g$ is uniformly continuous on the compact set $\bar D$, this implies that for every $r\in[0,1]$ there is a $\delta>0$ such that if $|r_0-r|<\delta$, then $\mathrm{Ind}(\gamma_{r_0})=\mathrm{Ind}(\gamma_r)$, that is, $\mathrm{Ind}(\gamma_r)$ is a continuous function on the connected set $[0,1]$. As shown at the end of the solution to Exercise 27, this indicates that $\mathrm{Ind}(\gamma_r)$ is constant on $[0,1]$. Since $\gamma_0$ is the curve with the constant value $f(0)$, this constant must be $\mathrm{Ind}(\gamma_0)=0$.

Put $\psi(t)=e^{it}\gamma_1(t)$, which is also a closed continuous curve with values on $T$. If $\psi(t)=-1$ for some $t$, then $g(e^{it})=\gamma_1(t)=-e^{it}$. So if we assume that $g(z)\ne-z$ for all $z\in T$, then $\psi(z)\ne-1$, which is the intersection of $T$ with the negative real axis. Hence by the extension of Exercise 24 shown in Exercise 26, we have $\mathrm{Ind}(\psi)=0$. Let $P_1$ be a trigonometric polynomial such that $|P_1-\psi|<1/8$ on $T$ so that $\mathrm{Ind}(P_1)=\mathrm{Ind}(\psi)=0$. If $P_2(t)=e^{it}P_1(t)$, then $$\big|P_2(t)-\gamma_1(t)\big|=
\big|e^{it}P_1(t)-e^{it}\psi(t)\big|=\big|P_1(t)-\psi(t)\big|<\frac{1}{8}$$ so that
\mathrm{Ind}(\gamma_1) &= \mathrm{Ind}(P_2) \\
&= \frac{1}{2\pi i}\int_0^{2\pi}\frac{P_2′(t)}{P_2(t)}\,dt \\
&= \frac{1}{2\pi i}\int_0^{2\pi}\frac{ie^{it}P_1(t)+e^{it}P_1′(t)}{e^{it}P_1(t)}\,dt \\
&= \frac{1}{2\pi i}\int_0^{2\pi}i\,dt+\frac{1}{2\pi i}\int_0^{2\pi}\frac{P_1′(t)}{P_1(t)}\,dt \\
&= 1+\mathrm{Ind}(P_1) \\
&= 1
\end{align*}which contradicts $\mathrm{Ind}(\gamma_1)=0$ shown in the first paragraph.

Exercise 29

(By analambanomenos) Following the hint, assume $f(z)\ne z$ for all $z\in\bar D$. For $z\in\bar D$ let $g(z)\in T$ be the point which lies on the ray starting at $f(z)$ (but not including $f(z)$) and passes through $z$. Note that if $z\in T$, then $g(z)=z$. The points on that ray are the points $$f(z)+r\big(z-f(z)\big),\quad r\in(0,\infty),$$ so the $r$ defining $g(z)$ satisfies
1 &= \big|f(z)+r\big(z-f(z)\big)\big|^2 \\
1 &= \Big(f(z)+r\big(z-f(z)\big)\Big)\Big(\overline{f(z)}+r\big(\overline{z}-\overline{f(z)}\big)\Big) \\
0 &= \big|z-f(z)|^2r^2+2\Re\big(f(z)\overline{z}-|f(z)|^2\big)r+\big(|f(z)|^2-1\big)
\end{align*} This is a quadratic equation in $r$ with coefficients which are continuous functions of $z$ and $f(z)$. From the geometry of the ray, for all $z\in\bar{D}$ there will always be only one positive solution $r(z)$, given by the familar quadratic formula, and so $r(z)$ and $g(z)$ are continuous functions of $z$. Then $g(z)$, a continuous function which maps $\bar{D}$ into $T$, satisfies the conditions of Exercise 28, so there must be a $z\in T$ such that $g(z)=-z$, contradicting $g(z)=z$ for all $z\in T$.

Exercise 30

(By analambanomenos) From Stirling’s formula we get the limits $$\lim_{x\rightarrow\infty}\frac{x^c\Gamma(x)}{x^c\big(\frac{x-1}{e}\big)^{x-1}\sqrt{2\pi(x-1)}}=1\quad\quad\lim_{x\rightarrow\infty}\frac{\Gamma(x+c)}
{\big(\frac{x+c-1}{e}\big)^{x+c-1}\sqrt{2\pi(x+c-1)}}=1.$$ Hence
1 &= \lim_{x\rightarrow\infty}\frac{\Gamma(x+c)}{x^c\Gamma(x)}\cdot\frac{x^c\big(\frac{x-1}{e}\big)^{x-1}\sqrt{2\pi(x-1)}}{\big(\frac{x+c-1}{e}\big)^{x+c-1}\sqrt{2\pi(x+c-1)}} \\
&= \lim_{x\rightarrow\infty}\frac{\Gamma(x+c)}{x^c\Gamma(x)}\cdot\frac{x^ce^c(x-1)^{x-1}}{(x+c-1)^{x+c-1}}\sqrt{1-\frac{c}{x+c-1}} \\
&= \lim_{x\rightarrow\infty}\frac{\Gamma(x+c)}{x^c\Gamma(x)}\cdot\frac{x^ce^c(x-1)^{x-1}}{(x+c-1)^{x+c-1}}.
\end{align*}So to show that the limit of the first factor is 1, it suffices to show that the limit of the second factor is also 1.

By substituting $y$ for $x-1$, we have
\lim_{x\rightarrow\infty}\frac{x^ce^c(x-1)^{x-1}}{(x+c-1)^{x+c-1}} &= \lim_{y\rightarrow\infty}\frac{(y+1)^ce^cy^y}{(y+c)^{y+c}} \\
&= \lim_{y\rightarrow\infty}\bigg(1+\frac{1-c}{y+c}\bigg)^c\cdot\frac{e^cy^y}{(y+c)^y} \\
&= \lim_{y\rightarrow\infty}\frac{e^c}{(1+c/y)^y} \\
&= 1
\end{align*}The last equality comes from Exercise 4(d).

Exercise 31

(By analambanomenos) We have\begin{align*}
\lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^1(1-x^2)^n\,dx &= \lim_{n\rightarrow\infty}2\sqrt{n}\int_0^1(1-x^2)^n\,dx & \hbox{by symmetry} \\
&= \lim_{n\rightarrow\infty}\sqrt{n}\int_0^1t^{-1/2}(1-t)^n\,dt & \hbox{substituting $t$ for $x^2$} \\
&= \lim_{n\rightarrow\infty}\frac{\sqrt{n}\,\Gamma\big(\frac{1}{2}\big)\Gamma(n+1)}{\Gamma\big(n+\frac{3}{2}\big)} & \hbox{Theorem 8.20} \\
&= \Gamma\big(\textstyle\frac{1}{2}\big)\displaystyle\lim_{n\rightarrow\infty}\frac{n^{3/2}\Gamma(n)}{\Gamma\big(n+\frac{3}{2}\big)} & \hbox{Theorem 8.18(a)} \\
&= \Gamma\big(\textstyle\frac{1}{2}\big) & \hbox{Exercise 30} \\
&= \sqrt{\pi} & \hbox{8.21 (99)}

Baby Rudin 数学分析原理完整第八章习题解答


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