Answers to Selected Exercises

$4\min(a,b,c)$

$\sqrt7$ (no); $-\sqrt7$ (no)

$n^n/n!$\

no

$\dst{A_n=\frac{x^n}{n!}\left(\ln x-\sum_{j=1}^n\frac{1} {j}\right)}$

$f_n(x_1,x_2, \dots,x_n)=2^{n-1}\max(x_1,x_2, \dots,x_n)$, $g_n(x_1,x_2, \dots,x_n)=2^{n-1}\min(x_1,x_2, \dots,x_n)$

$\emptyset$; $(-\infty,\infty)$; $\emptyset$; $(-\infty,\infty)$; $\emptyset$; $(-\infty,\infty)$

$\emptyset$; $(-\infty,\infty)$; $[-1,1]$; $(-\infty,-1)\cup(1,\infty)$; $[-1,1]$; $(-\infty,\infty)$

$(-\infty,0]\cup(3,\infty)$

$1$

closed; $\emptyset$; $\bigcup\set{(n,n+1)}{n=\mbox{integer}}$; $(-\infty,\infty)$

$S_1=$ rationals, $S_2=$ irrationals

$D_f=[-2,1)\cup[3,\infty)$, $D_g=(-\infty,-3]\cup[3,7)\cup(7,\infty)$, $D_{f\pm g}=D_{fg}=[3,7)\cup(7,\infty)$, $D_{f/g}=(3,4)\cup(4,7)\cup(7,\infty)$

$[1,\infty)$

$-2$

$2$

none, $0$

$0$

$0$

$-\infty$

none

$\infty$

$\frac{1}{2}$

$\lim_{x\to\infty}r(x)=\infty$ if $n>m$ and $a_n/b_m>0$; $=-\infty$ if $n>m$ and $a_n/b_m<0$; $=a_n/b_m$ if $n=m$; $=0$ if $n<m$. $\lim_{x\to-\infty}r(x)=(-1)^{n-m}\lim_{x\to\infty}r(x)$

$\lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)$

$\limsup_{x\to x_0-}(f-g)(x)\le\limsup_{x\to x_0-}f(x)-\liminf_{x\to x_0-}g(x)$; $\liminf_{x\to x_0-}(f-g)(x)\ge\liminf_{x\to x_0-}f(x)-\limsup_{x\to x_0-}g(x)$

from the left

$\tanh x$ is continuous for all $x$, $\coth x$ for all $x\ne0$

$\bigcup_{n=-\infty}^\infty[n\pi,(n+\frac{1}{2})\pi]$, $[0,\infty)$

$x\ne0$

$p(c)=q(c)$ and $p'_{-}(c)=q'_{+}(c)$

$f^{(k)}(x)=n(n-1)\cdots(n-k-1)x^{n-k-1}|x|$ if $1\le k\le n-1$; $f^{(n)}(x)=n!$ if $x>0$; $f^{(n)}(x)=-n!$ if $x<0$; $f^{(k)}(x)=0$ if $k>n$ and $x\ne0$; $f^{(k)}(0)$ does not exist if $k\ge n$.

$c(x)=e^{ax}\cos bx$, $s(x)=e^{ax}\sin bx$

continuous from the right

There is no such function (Theorem~2.3.9).

Counterexample: Let $x_0=0$, $f(x)=|x|^{3/2}\sin(1/x)$ if $x\ne0$, and $f(0)~=0~$.

Counterexample: Let $x_0=0$, $f(x)=x/|x|$ if $x\ne0$, $f(0)=0$.

$-\infty$ if $\alpha\le0$, $0$ if $\alpha>0$

$\infty$ if $\alpha>0$, $-\infty$ if $\alpha\le0$

$-\infty$ $-\infty$ if $\alpha\le0$, $0$ if $\alpha>0$

Suppose that $g'$ is continuous at $x_0$ and $f(x)=g(x)$ if $x\le x_0$, $f(x)=1+g(x)$ if $x>x_0$.

$1$ $e^L$

Counterexample: Let $x_0=0$ and $f(x)=x|x|$.

Let $g(x)=1+|x-x_0|$, so $f(x)=(x-x_0)(1+|x-x_0|)$.

Let $g(x)=1+|x-x_0|$, so $f(x)=(x-x_0)^2(1+|x-x_0|)$.

$-2$, $5$, $-16$, $65$

$0$, $-1$, $0$, $5$

min

$f(x)=e^{-1/x^2}$ if $x\ne0$, $f(0)=0$ (Exercise~)

None if $b^2-4c<0$; local min at $x_1=(-b+\sqrt{b^2-4c})/2$ and local max at $x_1=(-b-\sqrt{b^2-4c})/2$ if $b^2-4c>0$; if $b^2=4c$ then $x=-b/2$ is a critical point, but not a local extreme point.

$\dst\frac{1}{4(63)^4}$

$M_4h^2/12$ where $M_4=\sup_{|x-c|\le h}|f^{(4)}(c)|$

$k=-h/2$

Let $[a,b]=[0,1]$ and $P=\{0,1\}$. Let $f(0)=f(1)=\frac{1}{2}$ and $f(x)=x$ if $0<x<1$. Then $s(P)=0$ and $S(P)=1$, but neither is a Riemann sum of $f$ over $P$.

$\frac{1}{2}$, $1$ $e^b-e^a$ $1-\cos b$ $\sin b$

$f(a)[g_1-g(a)]+f(d)(g_2-g_1)+f(b)[g(b)-g_2]$

$f(a)[g_1-g(a)]+f(b)[g(b)-g_p]+\sum_{m=1}^{p-1}f(a_m)(g_{m+1}-g_m)$

If $g\equiv1$ and $f$ is arbitrary, then $\int_a^b f(x)\,dg(x)=0$.

$\overline{u}=(e-2)/(e-1),\ c=\sqrt{\overline{u}}$

$0$

divergent

$p<1$

$q+1<p<3q+1$

$\deg g-\deg f\ge 2$

$1/2 \quad$

$0$

If $s_n=1$ and $t_n=-1/n$, then $(\lim_{n\to\infty}s_n)/(\lim_{n\to\infty}t_n)=1/0=\infty$, but $\lim_{n\to\infty}s_n/t_n=-\infty$.

$|t|$, $-|t|$

$\sqrt{3}/2$, $-\sqrt{3}/2$

If $\{s_n\}=\{1,0,1,0, \dots\}$, then $\lim_{n\to\infty}t_n=\frac{1}{2}$

$\lim_{m\to\infty}s_{8m}=\lim_{m\to\infty}s_{8m+2}=1$, $\lim_{m\to\infty}s_{8m+1}=\sqrt2$,\ $\lim_{m\to\infty}s_{8m+3}=\lim_{m\to\infty}s_{8m+7}=0$, $\lim_{m\to\infty}s_{8m+5}=-\sqrt2$,\ $\lim_{m\to\infty}s_{8m+4}=\lim_{m\to\infty}s_{8m+6}=-1$

$\{1,2,1,2,3,1,2,3,4,1,2,3,4,5, \dots\}$

Let $\{t_n\}$ be any convergent sequence and $\{s_n\}=\{t_1,1,t_2,2, \dots,t_n,n, \dots\}$.

No; consider $\sum 1/n$

convergent

$p>1$

convergent

convergent

convergent

convergent if $\alpha<\beta-1$, divergent if $\alpha\ge\beta-1$

convergent

$\sum(-1)^n$

absolutely convergent

Let $k$ and $s$ be the degrees of the numerator and denominator, respectively. If $|r|=1$, the series converges absolutely if and only if $s\ge k+2$. The series converges conditionally if $s=k+1$ and $r=-1$, and diverges in all other cases, where $s\ge k+1$ and $|r|=1$.

$2A-a_0$

$F(x)=1,\ -\infty<x<\infty$

$F(x)=\sin x/x$

$F_n(x)=x^n$; $S_k=[-k/(k+1),k/(k+1)]$

$[-r,r],\ r>0$

Let $S=(0,1]$, $F_n(x)=\sin(x/n)$, $G_n(x)=1/x^2$; then $F=0$, $G=1/x^2$, and the convergence is uniform, but $\|F_nG_n\|_S=\infty$.

$e-1$

compact subsets of $(-\infty,0)\cup(0,\infty)$

``absolutely"

means that $\sum|f_n(x)|$ converges uniformly on $S$.

$\dst{\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)(2n+1)!}}$

$\infty$

$1$

$x(1+x)/(1-x)^3$ $e^{-x^2}$ \ $\dst{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{ n^2} }(x-1)^n;\ R=1$

$\mbox{Tan}^{-1}x=\dst{\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)}};\ f^{(2n)}(0)=0;\ f^{(2n+1)}(0)=(-1)^2(2n)!$;

$\dst{\frac{\pi}{6}=\mbox{Tan}^{-1}\frac{1}{\sqrt3}=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)3^{n+1/2}}}$

$\cosh x=\dst{\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}}$, $\sinh x=\dst{\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}}$

$(1-x)\sum_{n=0}^\infty x^n=1$ converges for all $x$

$\dst{x^2-\frac{x^3}{2}+\frac{x^4}{6}- \frac{x^5}{6}+\cdots}$

$\dst{2-x^2+\frac{x^4}{12}-\frac{x^6}{360}+\cdots}$

$\dst{F(x)=\frac{5}{(1-3x)(1+2x)}=\frac{3}{1-3x}+\frac{2}{1+2x}= \sum_{n=0}^\infty[3^{n+1}-(-2)^{n+1}]x^n}$ $1$

$(\frac{1}{6},\frac{11}{12},\frac{23}{24},\frac{5}{36})$

$\sqrt3$

$\sqrt{31}$

$27$

$\mathbf{X}=\mathbf{X}_0+t\mathbf{U}\ (-\infty<t<\infty)$ in all cases.

$\dots\mathbf{U}$ and $\mathbf{X}_1-\mathbf{X}_0$ are scalar multiples of $\mathbf{V}$.

$\mathbf{X}=(1,-3,4,2)+t(1,3,-5,3)$

$\mathbf{X}=(3,1,-2,1,4,)+t(-1,-1,1,3,-7)$

$\mathbf{X}=(1,2,-1)+t(-1,-3,0)$

$1/2\sqrt5$

$S$

$\set{(x_1,x_2,x_3,x_4)}{|x_i|>3 \mbox{ for at least one of }i=1,2,3}$

$\set{(x,y,z)}{z\ne1\mbox{ or }x^2+y^2>1}$

closed

$(1,0,e)$

$\infty$

$\set{(x,y)}{x^2+y^2=1}$

$\dots$ if for $A$ there is an integer $R$ such that $|\mathbf{X}_r|>A$ if $r\ge R$.

$0$

$a/(1+a^2)$

no

none

.

$\lim_{\mathbf{X}\to\mathbf{0}}f(\mathbf{X})=0$ if $a_1+a_2+\cdots+a_n>b$; no limit if $a_1+a_2+\cdots+a_n\le b$ and $a_1^2+a_2^2+\cdots+a_n^2\ne0$; $\lim_{\mathbf{X}\to\mathbf{0}}f(\mathbf{X})=\infty$ if $a_1=a_2=\cdots=a_n=0$ and $b>0$.

No; for example, $\lim_{x\to\infty}g(x,\sqrt x)=0$.

$\mathbb R^n$

$\mathbb R^2$

$f(x,y)=xy/(x^2+y^2)$ if $(x,y)\ne(0,0)$ and $f(0,0)=0$

$1/(1+x+y+z)$

$0$

$f_x=f_y=1/(x+y+2z)$, $f_z=2/(x+y+2z)$

$f_x=2xy\cos x^2y$, $f_y=x^2\cos x^2y$, $f_z=1$

$f_{xx}=f_{yy}=f_{xy}=f_{yx}=-1/(x+y+2z)^2$, $f_{xz}=f_{zx}=f_{yz}=f_{zy}=$ $-2/(x+y+2z)^2$, $f_{zz}=-4/(x+y+2z)^2$

$f_{xx}=0$, $f_{yy}=xz^2e^{yz}$, $f_{zz}=xy^2e^{yz}$, $f_{xy}=f_{yx}=ze^{yz}$, $f_{xz}=f_{zx}=ye^{yz}$, $f_{yz}=f_{zy}=xe^{yz}$

$f_{xx}=2y\cos x^2y-4x^2y^2\sin x^2y$, $f_{yy}=-x^4\sin x^2y$, $f_{zz}=0$, $f_{xy}=f_{yx}=2x\cos x^2y-2x^3y\sin x^2y$, $f_{xz}=f_{zx}=f_{yz}=f_{zy}=0$

$f_{xx}(0,0)=f_{yy}(0,0)=0$, $f_{xy}(0,0)=-1$, $f_{yx}(0,0)=1$

$f(x,y)=g(x,y)+h(y)$, where $g_{xy}$ exists everywhere and $h$ is nowhere differentiable.

$df=2r|\mathbf{X}|^{2r-2}\sum_{j=1}^nx_j\,dx_j$, $d_{\mathbf{X}_0}f=2rn^{r-1}\sum_{j=1}^n dx_j$, \ $(d_{\mathbf{X}_0}f)(\mathbf{X}- \mathbf{X}_0)=2rn^{r-1}\sum_{j=1}^n (x_j-1)$,

The unit vector in the direction of $(f_{x_1}(\mathbf{X}_0), f_{x_2}(\mathbf{X}_0), \dots,f_{x_n}(\mathbf{X}_0))$ provided that this is not $\mathbf{0}$; if it is $\mathbf{0}$, then $\partial f(\mathbf{X}_0)/\partial\boldsymbol{\Phi}=0$ for every $\boldsymbol{\Phi}$.

$z=x+10y+4$

$8\,du$

$h_r=f_x\cos\theta+f_y\sin\theta$, $h_\theta=r(-f_x\sin\theta+f_y\cos\theta)$, $h_z=f_z$

$h_r=f_x\sin\phi\cos\theta+f_y\sin\phi\sin\theta+f_z\cos\phi$, $h_\theta=r\sin\phi(-f_x\sin\theta+f_y\cos\theta)$, $h_\phi=r(f_x\cos\phi\cos\theta+f_y\cos\phi\sin\theta-f_z\sin\phi)$

$h_y=g_xx_y+g_y+g_ww_y$, $h_z=g_xx_z+g_z+g_ww_z$

$h_{rr}=f_{xx}\sin^2\phi\cos^2\theta+f_{yy}\sin^2\phi\sin^2\theta+ f_{zz}\cos^2\phi+f_{xy}\sin^2\phi\sin2\theta+f_{yz}\sin2\phi\sin\theta+ f_{xz}\sin2\phi\cos\theta$,

$\dst h_{r\theta} =(-f_x\sin\theta+f_y\cos\theta)\sin\phi+\frac{r}{2}(f_{yy}- f_{xx})\sin^2\phi\sin2\theta +rf_{xy}\sin^2\phi\cos2\theta+\frac{r}{2} (f_{zy}\cos\theta-f_{zx}\sin\theta)\sin2\phi$

$\dst{1+x+\frac{x^2}{2}-\frac{y^2}{2}+\frac{x^3}{6}-\frac{xy^2}{2}}$

$xyz$

$(d_{(0,0)}^2p)(x,y)=(d_{(0,0)}^2q)(x,y)=2(x-y)^2$

$\left[\begin{array}{rr}2&4\\3&-2\\7&-4\\6&1\end{array}\right]$

$\left[\begin{array}{rrr}-2&-6&0\\0&-2&-4\\-2&2&-6\end{array}\right]$

$\left[\begin{array}{rr}-1&7\\3&5\\5&14\end{array}\right]$

$\left[\begin{array}{r}29\\50\end{array}\right]$

$\mathbf{A}$ and $\mathbf{B}$ are square of the same order.

$\left[\begin{array}{rr}14&10\\6&-2\\14&2\end{array}\right]$

$\left[\begin{array}{rrr}-7&6&4\\-9&7&13\\5&0&-14\end{array}\right]$, $\left[\begin{array}{rrr}-5&6&0\\4&-12&3\\4&0&3\end{array}\right]$

$\left[\begin{array}{rrr}6xyz&3xz^2&3x^2y\end{array}\right]$;
$\left[\begin{array}{rrr}-6&3&-3\end{array}\right]$

$\cos(x+y)\left[\begin{array}{rr}1&1\end{array}\right]$; $\left[\begin{array}{rr}0&0\end{array}\right]$

$\left[\begin{array}{rrr}(1-xz)ye^{-xz}&xe^{-xz}&-x^2ye^{-xz} \end{array}\right]$; $\left[\begin{array}{rrr}2&1&-2\end{array}\right]$\

$\sec^2(x+2y+z)\left[\begin{array}{rrr}1&2&1\end{array}\right]$; $\left[\begin{array}{rrr}2&4&2\end{array}\right]$

$|\mathbf{X}|^{-1}\left[\begin{array}{rrrr}x_1&x_2&\cdots&x_n\end{array}\right]$; $\dst\frac{1}{\sqrt n}\left[\begin{array}{rrrr}1&1&\cdots&1\end{array}\right]$

$(3,1,3,2)$

$\dst\frac{1}{2}\left[\begin{array}{rrr}-1&1&2\\3&1&-4\\-1&-1&2 \end{array}\right]$

$\dst\frac{1}{2}\left[\begin{array}{rrr}1&-1&1\\-1&1&1\\1&1&-1 \end{array}\right]$

$\dst\frac{1}{10}\left[\begin{array}{rrrr}-1&-2&0&5 \\-14&-18&10&20\\21&22&-10&-25\\17&24&-10&-25\end{array}\right]$

$\mathbf{F}'(\mathbf{X})=\left[\begin{array}{ccc} 2x&1&2\\-\sin(x+y+z)&-\sin(x+y+z)&-\sin(x+y+z)\\[2\jot] yze^{xyz}&xze^{xyz}&xye^{xyz}\end{array}\right]$;\[2] $J\mathbf{F}(\mathbf{X})=e^{xyz}\sin(x+y+z) [x(1-2x)(y-z)-z(x-y)]$;\[2] $\mathbf{G}(\mathbf{X})=\left[\begin{array}{r}0\\1\\1\end{array}\right] +\left[\begin{array}{rrr}2&1&2\\0&0&0\\0&0&-1\end{array}\right] \left[\begin{array}{c}x-1\\y+1\\z\end{array}\right]$

$\mathbf{F}'(\mathbf{X})=\left[\begin{array}{rr}e^x\cos y&-e^x\sin y\\e^x\sin y&e^x\cos y\end{array}\right]$; $J\mathbf{F}(\mathbf{X})=e^{2x}$;\[2] $\mathbf{G}(\mathbf{X})=\left[\begin{array}{r}0\\1\end{array}\right] +\left[\begin{array}{rr}0&-1\\1&0\end{array}\right] \left[\begin{array}{c}x\\y-\pi/2\end{array}\right]$\[2]

$\mathbf{F}'(\mathbf{X})= \left[\begin{array}{rrr}2x&-2y&0\\0&2y&-2z\\-2x&0&2z\end{array}\right]$; $J\mathbf{F}=0$;\[2] $\mathbf{G}(\mathbf{X})= \left[\begin{array}{rrr}2&-2&0\\0&2&-2\\-2&0&2\end{array}\right] \left[\begin{array}{c}x-1\\y-1\\z-1\end{array}\right]$

$\mathbf{F}'(\mathbf{X})= \left[\begin{array}{ccc} (x+y+z+1)e^x&e^x&e^x\\(2x-x^2-y^2)e^{-x} &2ye^{-x}&0\end{array}\right]$

$\mathbf{F}'(r,\theta)= \left[\begin{array}{ccc}e^x\sin yz&ze^x\cos yz&ye^x\cos yz\\ze^y\cos xz&e^y\sin xz&xe^y\cos xz\\ye^z\cos xy&xe^z\cos xy&e^z\sin xy\end{array}\right]$

$\mathbf{F}'(r,\theta)=\left[\begin{array}{rr}\cos\theta&-r\sin\theta \\\sin\theta&r\cos\theta\end{array}\right]$; $J\mathbf{F}(r,\theta)=r$\[2]

$\mathbf{F}'(r,\theta,\phi)=\left[\begin{array}{ccc}\cos\theta\cos\phi& -r\sin\theta\cos\phi&-r\cos\theta\sin\phi\\\sin\theta\cos\phi& \phantom{-}r\cos\theta\cos\phi&-r\sin\theta\sin\phi\\\sin\phi&0&r\cos\phi \end{array}\right]$;\ $J\mathbf{F}(r,\theta,\phi)=r^2\cos\phi$\[2]

$\mathbf{F}'(r,\theta,z)=\left[\begin{array}{ccc}\cos\theta&-r\sin\theta&0 \\\sin\theta&\phantom{-}r\cos\theta&0\\0&0&1\end{array}\right]$; $J\mathbf{F}(r,\theta,z)=r$

$\left[\begin{array}{rr}9&-3\\3&-8\\1&0\end{array}\right]$

$\left[\begin{array}{rr}5&10\\9&18\\-4&-8\end{array}\right]$

$[\sqrt2,3\pi/4]$

$[\sqrt2,-5\pi/4]$

Let $f(x)=x\ (0\le x\le\frac{1}{2})$, $f(x)=x-\frac{1}{2}\ (\frac{1}{2}<x\le1)$; then $f$ is locally invertible but not invertible on $[0,1]$.

$\mathbf{F}(S)=\set{(u,v)}{-\pi+2\phi<\arg(u,v)<\pi+2\phi}$, where $\phi$ is an argument of $(a,b)$;

$\mathbf{F}_S^{-1}(u,v)=(u^2+v^2)^{1/4} \left[\begin{array}{r}\cos(\arg(u,v)/2) \\[2\jot]\sin(\arg(u,v)/2)\end{array}\right],\ 2\phi-\pi<\arg(u,v)<2\phi+\pi$

$\left[\begin{array}{c}x\\y\end{array}\right]= \dst\frac{1}{10}\left[\begin{array}{c}\phantom{3}u-2v\\3u+4v\end{array}\right]$; $(\mathbf{F}^{-1})'=\dst\frac{1}{10} \left[\begin{array}{rr}1&-2\\3&4\end{array}\right]$

$\left[\begin{array}{c}x\\y\\z\end{array}\right]= \dst\frac{1}{2}\left[\begin{array}{c}u+2v+3w\\u-w\\u+v+2w\end{array}\right]$; $(\mathbf{F}^{-1})'=\dst\frac{1}{2} \left[\begin{array}{rrr}1&2&3\\1&0&-1\\1&1&2\end{array}\right]$

$\mathbf{G}_1(u,v)=\dst\frac{1}{\sqrt2} \left[\begin{array}{r}\sqrt{u+v}\\ \sqrt{u-v}\end{array}\right]$, $\mathbf{G}_1'(u,v)=\dst\frac{1}{2\sqrt2} \left[\begin{array}{rr} 1/\sqrt{u+v}&1/\sqrt{u+v}\\ 1/\sqrt{u-v}&-1/\sqrt{u-v}\\ \end{array}\right]$

$\mathbf{G}_2(u,v)=\dst\frac{1}{\sqrt2} \left[\begin{array}{r}-\sqrt{u+v}\\ \sqrt{u-v}\end{array}\right]$, $\mathbf{G}_2'(u,v)=\dst\frac{1}{2\sqrt2} \left[\begin{array}{rr} -1/\sqrt{u+v}&-1/\sqrt{u+v}\\ 1/\sqrt{u-v}&-1/\sqrt{u-v}\\ \end{array}\right]$

$\mathbf{G}_3(u,v)=\dst\frac{1}{\sqrt2} \left[\begin{array}{r}\sqrt{u+v}\\ -\sqrt{u-v}\end{array}\right]$, $\mathbf{G}_3'(u,v)=\dst\frac{1}{2\sqrt2} \left[\begin{array}{rr} 1/\sqrt{u+v}&1/\sqrt{u+v}\\ -1/\sqrt{u-v}&1/\sqrt{u-v}\\ \end{array}\right]$

$\mathbf{G}_4(u,v)=\dst\frac{1}{\sqrt2} \left[\begin{array}{r}-\sqrt{u+v}\\ -\sqrt{u-v}\end{array}\right]$, $\mathbf{G}_4'(u,v)=\dst\frac{1}{2\sqrt2} \left[\begin{array}{rr} -1/\sqrt{u+v}&-1/\sqrt{u+v}\\ -1/\sqrt{u-v}&1/\sqrt{u-v}\\ \end{array}\right]$

does not hold if $x=0$.

$\left[\begin{array}{c}x\\y\end{array}\right]= \mathbf{G}(u,v)=(u^2+v^2)^{1/4} \left[\begin{array}{r}\cos[\frac{1}{2}\arg(u,v)] \\[2\jot] \sin(\arg(u,v)/2)\end{array}\right]$,

where $\beta-\pi/2<\arg(u,v)<\beta+\pi/2$ and $\beta$ is an argument of $(a,b)$;

$\mathbf{G}'(u,v)=\dst\frac{1}{2(x^2+y^2)} \left[\begin{array}{rr}x&y\\-y&x\end{array}\right]$

If $\mathbf{F}(x_1,x_2, \dots,x_n)=(x_1^3,x_2^3, \dots,x_n^3)$, then $\mathbf{F}$ is invertible, but\ $J\mathbf{F}(\mathbf{0})~=0$.

$\mathbf{A}(\mathbf{U})= \left[\begin{array}{r}1\\-1\end{array}\right] -\dst\frac{1}{25}\left[\begin{array}{rr}5&5\\3&8\end{array}\right] \left[\begin{array}{c}u+5\\v-4\end{array}\right]$

$\mathbf{A}(\mathbf{U})= \left[\begin{array}{r}1\\1\end{array}\right] +\dst\frac{1}{6}\left[\begin{array}{rr}4&-2\\-3&3\end{array}\right] \left[\begin{array}{c}u-2\\v-3\end{array}\right]$

$\mathbf{A}(\mathbf{U})= \left[\begin{array}{r}0\\1\\1\end{array}\right]+ \left[\begin{array}{rrr}0&-1&1\\-1&1&0\\1&0&0\end{array}\right] \left[\begin{array}{c}u-1\\v-1\\w-2\end{array}\right]$

$\mathbf{A}(\mathbf{U})= \left[\begin{array}{c}1\\\pi/2\\\pi\end{array}\right]+ \left[\begin{array}{rrr}0&-1&0\\1&0&0\\0&0&-1\end{array}\right] \left[\begin{array}{c}u\\v+1\\w\end{array}\right]$

$\mathbf{G}'(x,y,z)= \left[\begin{array}{ccc} \cos\theta\cos\phi&\sin\theta\cos\phi&\sin\phi\\[2\jot] -\dst\frac{\sin\theta}{ r\cos\phi}&\dst\frac{\cos\theta}{ r\cos\phi}&0\\[2\jot] -\dst\frac{1}{ r}\cos\theta\sin\phi&-\dst\frac{1}{ r}\sin\theta\sin\phi&\dst\frac{1}{ r}\cos\phi\end{array}\right]$

$\mathbf{G}'(x,y,z)= \left[\begin{array}{ccc}\cos\theta&\sin\theta&0\\[2\jot] -\dst\frac{1}{ r}\sin\theta&\dst\frac{1}{ r}\cos\theta&0\\[2\jot] 0&0&1\end{array}\right]$

$\left[\begin{array}{c}u\\v\end{array}\right]= \dst\frac{1}{2}\left[\begin{array}{rr}-3&4\\1&-2\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]$

$\left[\begin{array}{c}u\\v\end{array}\right]= \dst\frac{1}{5}\left[\begin{array}{rr}2&-1\\-1&3\end{array}\right] \left[\begin{array}{c}-y+\sin x\\-x+\sin y\end{array}\right]$

$u=-x$, $v=-y$, $z=-w$

$r=s=2$

$u_x(1,1)=-\frac{5}{8}$, $u_y(1,1)=-\frac{1}{2}$

$u_x(1,1,1)=\frac{5}{8}$, $u_y(1,1,1)=-\frac{9}{8}$, $u_z(1,1,1)=\frac{1}{2}$

$u(1,2)=0$, $u_x(1,2)=u_y(1,2)=-4$

$u(-1,-2)=2$, $u_x(-1,-2)=1$, $u_y(-1,-2)=-\frac{1}{2}$

$u(\pi/2,\pi/2)=u_x(\pi/2,\pi/2)=u_y(\pi/2,\pi/2)=0$

$u(1,1)=1$, $u_x(1,1)=u_y(1,1)=-1$

$u_1(1,1)=1$,$\dst\frac{\partial u_1(1,1)}{\partial x}=5$, $\dst\frac{\partial u_1(1,1)}{\partial y}=2$

$u_2(1,1)=2$, $\dst\frac{\partial u_2(1,1)}{\partial x}=-14$; $\dst\frac{\partial u_2(1,1)}{\partial y}=-2$

$u_k(0,\pi)=(2k+1)\pi/2$, $\dst\frac{\partial u_k(0,\pi)}{\partial x}=0$, $\dst\frac{\partial u_k(0,\pi)}{\partial y}=-1$,$k=$ integer

$\dst\frac{1}{5}\left[\begin{array}{rrr}-1&-2&1\\-1&-2&1\end{array}\right]$ $u'(0)=3$, $v'(0)=-1$

$\dst\frac{1}{6}\left[\begin{array}{rr}5&5\\-5&-5\\6&6\end{array}\right]$

$\mathbf{U}_1(1,1)=\left[\begin{array}{r}3\\1\end{array}\right]$, $\mathbf{U}_1'(1,1)=\left[\begin{array}{rr}1&3\\-1&2\end{array}\right]$;

$\mathbf{U}_2(1,1)=-\left[\begin{array}{r}3\\1\end{array}\right]$, $\mathbf{U}_2'(1,1)=-\left[\begin{array}{rr}1&3\\-1&2\end{array}\right]$

$u_x(0,0,0)=2$, $v_x(0,0,0)= w_x(0,0,0)=-2$

$y_x=-\dst\frac{\dst\frac{\partial(f,g,h)}{\partial(x,z,u)}}{ \dst\frac{\partial(f,g,h)}{\partial(y,z,u)}}$, $y_v=-\dst\frac{\dst\frac{\partial(f,g,h)}{\partial(v,z,u)}}{ \dst\frac{\partial(f,g,h)}{\partial(y,z,u)}}$, $z_x=-\dst\frac{\dst\frac{\partial(f,g,h)}{\partial(y,x,u)}}{ \dst\frac{\partial(f,g,h)}{\partial(y,z,u)}}$,

$z_v= -\dst\frac{\dst\frac{\partial(f,g,h)}{\partial(y,v,u)}}{ \dst\frac{\partial(f,g,h)}{\partial(y,z,u)}}$, $u_x=-\dst\frac{\dst\frac{\partial(f,g,h)}{\partial(y,z,x)}}{ \dst\frac{\partial(f,g,h)}{\partial(y,z,u)}}$, $u_v=-\dst\frac{\dst\frac{\partial(f,g,h)}{\partial(y,z,v)}}{ \dst\frac{\partial(f,g,h)}{\partial(y,z,u)}}$

$x=-2y-u$, $z=-2v$; $x=-2y-u$, $v=-\dst\frac{z}{2}$; $y=-\dst\frac{x}{2}-\dst\frac{u}{2}$, $z=-2v$; $y=-\dst\frac{x}{2}-\dst\frac{u}{2}$, $v=-\dst\frac{z}{2}$; $z=-2v$, $u=-x-2y$; $u=-x-2y$, $v=-\dst\frac{z}{2}$

$y_x(1,-1,-2)=-\frac{1}{2}$, $v_u(1,-1,-2)=1$

$u_w(0,-1)=\frac{5}{6}$, $u_y(0,-1)=0$, $v_w(0,-1)=-\frac{5}{6}$, $v_y(0,-1)=0$,\ $x_w(0,-1)=1$, $x_y(0,-1)=-1$

$u_x(1,1)=0$, $u_y(1,1)=0$, $v_x(1,1)=-1$, $v_y(1,1)=-1$, $u_{xx}(1,1)=2$,\ $u_{xy}(1,1)=1$, $u_{yy}(1,1)=2$, $v_{xx}(1,1)=-2$, $v_{xy}(1,1)=-1$, $v_{yy}(1,1)=-2$

$u_x(1,-1)=0$, $u_y(1,-1)=\dst\frac{1}{2}$, $v_x(1,-1)=-\dst\frac{1}{2}$, $v_y(1,-1)=0$,\ $u_{xx}(1,-1)=-\dst\frac{1}{8}$, $u_{xy}(1,-1)=\dst\frac{1}{8}$, $u_{yy}(1,-1)=\dst\frac{1}{8}$, $v_{xx}(1,-1)=-\dst\frac{1}{8}$,\ $v_{xy}(1,-1)=-\dst\frac{1}{8}$, $v_{yy}(1,-1)=\dst\frac{1}{8}$

$\frac{1}{4}$ $3(b-a)(d-c)$, $0$ $\set{(m,n)}{m,n=\mbox{integers}}$

$(1-\log2)/2$

$1/4\pi$

$z+\frac{1}{2}$, $1$

$\frac{1}{4}(e-\frac{5}{2})$

$1$ $\frac{52}{15}$

$(e^6+17)/2$

$\frac{1}{36}$

$\frac{\pi}{2}$

$\frac{1}{2}(b_1-a_1)\cdots(b_n-a_n)\sum_{j=1}^n(a_j+b_j)$

$\frac{1}{3}(b_1-a_1)\cdots(b_n-a_n)\sum_{j=1}^n(a_j^2+a_jb_j+b_j^2)$

$2^{-n}(b_1^2-a_1^2)\cdots(b_n^2-a_n^2)$

$\int_{-\sqrt3/2}^{\sqrt3/2}dx\int_{1/2}^{\sqrt{1-x^2}}f(x,y)\,dy$ $\frac{1}{2}$

Let $S_1$ and $S_2$ be dense subsets of $\mathbb R$ such that $S_1\cup S_2=\mathbb R$.

$-1$; $c$ (constant); $1$ $(u_2-u_1)(v_2-v_1)/|ad-bc|$

$\log\frac{5}{2}$ $3$ $\frac{1}{2}$ $\frac{5}{4}e(e-1)$

$\frac{4}{3}\pi abc$ $2\pi(e^{25}-e^9)$ $16\pi/3$ $21/64$

$2\pi/15$

$\pi^2a^4/2$

$(\beta_1-\alpha_1)\cdots(\beta_n-\alpha_n)/|\det(\mathbf{A})|$ $|a_1a_2\cdots a_n|V_n$

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{}

Suppose $f\in C[a,b]$. Acording to the , if $\epsilon>0$, there is a polynonmial $p$ such that $|f(x)-p(x)|<\epsilon$, $a\le x\le b$. Now suppose that $$a\le x_{1n} < x_{2n}<\cdots<x_{nn}\le b,\quad a\le y_{1n} < y_{2n}<\cdots<y_{nn}\le b,\quad n\ge 1. \nonumber$$ and $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}|x_{in}-y_{in}|=0. \nonumber$$ Show that if $f\in C[a,b]$, then $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}|f(x_{in})-f(y_{in})|=0. \nonumber$$

By the triangle inequality, $$|f(x_{in})-f(y_{in})|\le |f(x_{in})-p(x_{in})| +|p(x_{in})-p(y_{in})|+|p(y_{in})-f(y_{in})|, \nonumber$$ so $$\begin{equation} \tag{A} |f(x_{in})-f(y_{in})|<|p(x_{in})-p(y_{in})|+2\epsilon. \end{equation} \nonumber$$ Let $M=\max_{a\le x\le b}|p'(x)|$. By the mean value theorem, $$|p(x_{in})-p(y_{in})|\le M |x_{in}-y_{in}|. \nonumber$$ This and (A) imply that $$\frac{1}{n}\sum_{i=1}^n|f(x_{in})-f(y_{in})|< 2\epsilon+\frac{M}{n}\sum_{i=1}^n|x_{in}-y_{in}|. \nonumber$$ From this and (A), $$\limsup_{n\to\infty} \frac{1}{n}\sum_{i=1}^n|f(x_{in})-f(y_{in})|\le 2\epsilon. \nonumber$$ Since $\epsilon$ is arbitrary, this implies the conclusion.

{} In the setting of Exercise~1, let $$y_{in}=a+\frac{i}{n}\sum_{i=1}^{n}(b-a), \quad 1\le i\le n,\quad n\ge 1. \nonumber$$ Show that $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} f(x_{in})=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx. \nonumber$$

{Solution 2.} Since $f\in C[a,b]$, $\int_{a}^{b}f(x)\,dx$ exists (Theorem~3.2.8, p.~133).

Therefore, from Definition~3.1.1 (p._{114), $$\lim_{n\to\infty}\sum_{i=1}^{n} f(y_{in})=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx. \nonumber$$ Since $$\frac{1}{n}\left| \sum_{i=1}^{n}f(x_{in})-\frac{1}{b-a}\int_{a}^{b}f(x)\,dx\right|\le \frac{1}{n}\sum_{i=1}^{n}|f(x_{in})-f(y_{in})|+ \left|\frac{1}{n} \sum_{i=1}^{n}f(y_{in})-\frac{1}{b-a}\int_{a}^{b}f(x)\,dx\right|, \nonumber$$ Exercise}1 implies the conclusion.

Suppose $g$ is continuous and nondecreasing on $[c,d]$. For

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