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$2\max(a,b)$ $2\min(a,b)$ $4\max(a,b,c)$

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

$\infty$(no);$-1$ (sí) $3$ (no);$-3$ (no) $\sqrt7$ (sí);$-\sqrt7$ (sí)\ $2$ (no);$-3$ (no) $1$ (no);$-1$ (no)

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

$2^n/(2n)!$ $2\cdot3^n/(2n+1)!$ $2^{-n}(2n)!/(n!)^2$

$n^n/n!$\

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

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

$[\frac{1}{2},1)$;;$(-\infty,\frac{1}{2})\cup[1,\infty)$;$(-\infty,0]\cup(\frac{3}{2},\infty)$;$(0,\frac{3}{2}]$;;$(-\infty,0]\cup(\frac{3}{2},\infty)$;$(-\infty,\frac{1}{2}]\cup[1,\infty)$ $(-3,-2) \cup (2,3)$;;$(-\infty,-3]\cup[-2,2]\cup[3,\infty)$;$\emptyset$;$(-\infty,\infty)$;;$\emptyset$;$(-\infty,-3]\cup[-2,2]\cup[3,\infty)$

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

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

$(0,3]$ $[0,2]$ $(-\infty,1)\cup(2,\infty)$

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

$\frac{1}{4}$ $\frac{1}{6}$ $6$

$1$

ni;$(-1,2)\, \cup\, (3,\infty)$;$(-\infty,-1)\,\cup\,(2,3)$;$(-\infty,-1]\,\cup\,(2,3)$;$(-\infty,-1]\,\cup\,[2,3]$ abierto;$S$;$(1,2)$;$[1,2]$ cerrado;$(-3,-2)\,\cup\,(7,8)$;$(-\infty,-3)\,\cup\,(-2,7)\,\cup(8,\infty)$;$(-\infty-3]\,\cup[-2,7]\,\cup[8,\infty)$

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

$\set{x}{x=1/n,\ n=1,2, \dots}$; $\emptyset$ , $S_1=$ racionales,$S_2=$ irracionales cualquier conjunto cuyo supremo sea un punto aislado del conjunto , los racionales

$S_1=$racionales,$S_2=$ irracionales

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

, $\set{x}{x\ne(2k+1)\pi/2\mbox{ where }k=\mbox{integer}}$ $\set{x}{x\ne0,1}$ $\set{x}{x\ne0}$

$[1,\infty)$

$4$ $12$ $-1$ $2$

$-2$

$\frac{11}{17}$ $-\frac{2}{3}$ $\frac{1}{3}$

$2$

$0,2$ $0$, ninguno $-\frac{1}{3},\frac{1}{3}$

ninguno,$0$

$0$ $0$ ninguno $0$ ninguno

$0$

$0$ $0$ ninguno ninguno ninguno

$0$

$\infty$ $-\infty$ $\infty$ $\infty$ $\infty$

$-\infty$

ninguno $\infty$ $\infty$

ninguno

$\infty$ $\infty$ $\infty$ $-\infty$ ninguno

$\infty$

$\frac{3}{2}$ $\frac{3}{2}$ $\infty$ $-\infty$ $\infty$

$\frac{1}{2}$

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

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

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

desde el derecho continuo ninguno continuo ninguno continuo

desde la izquierda

$[0,1)$,$(0,1)$,$[1,2)$,$(1,2)$,,$(1,2]$,$[1,2]$$[0,1)$,$(0,1)$,$(1,\infty)$

$\tanh x$es continuo para todos$x$,$\coth x$ para todos$x\ne0$

No $[-1,1]$,$[0,\infty)$ $\bigcup_{n=-\infty}^\infty (2n\pi,(2n+1)\pi)$,$(0,\infty)$ $\bigcup_{n=-\infty}^\infty(n\pi,(n+1)\pi)$,$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$

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

$(-1,1)$ $(-\infty,\infty)$ $x_0\ne(2k+\frac{3}{2}\pi),\ k=$entero $x\ne\frac{1}{2}$ $x\ne1$ $x\ne(k+\frac{1}{2}\pi),\ k=$ entero entero $x\ne(k+\frac{1}{2}\pi),\ k=$ entero $x\ne0$

$x\ne0$

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

$f^{(k)}(x)=n(n-1)\cdots(n-k-1)x^{n-k-1}|x|$si$1\le k\le n-1$;$f^{(n)}(x)=n!$ si$x>0$;$f^{(n)}(x)=-n!$ si$x<0$;$f^{(k)}(x)=0$ si$k>n$ y$x\ne0$;$f^{(k)}(0)$ no existe si$k\ge n$.

$c'=ac-bs$,$s'=bc+as$

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

$f(x)=-1$si$x\le0$,$f(x)=1$ si$x>0$; entonces$f'(0+)=0$, pero$f_+'(0)$ no existe.

continua desde la derecha

No existe tal función (Teoremo~2.3.9).

Contraejemplo: Dejar$x_0=0$,$f(x)=|x|^{3/2}\sin(1/x)$ si$x\ne0$, y$f(0)~=0~$.

Contraejemplo: Vamos$x_0=0$,$f(x)=x/|x|$ si$x\ne0$,$f(0)=0$.

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

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

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

$0$

Supongamos que$g'$ es continuo en$x_0$ y$f(x)=g(x)$ si$x\le x_0$,$f(x)=1+g(x)$ si$x>x_0$.

$1$ $e$

$1$$e^L$

$f^{(n+1)}(x_0)/(n+1)!$.

Contraejemplo: Vamos$x_0=0$ y$f(x)=x|x|$.

Vamos$g(x)=1+|x-x_0|$, entonces$f(x)=(x-x_0)(1+|x-x_0|)$.

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

$1$,$2$,$2$,$0$ $0$,$-\pi$,$3\pi/2$,$-4\pi+\pi^3/2$\ $-\pi^2/4$,$-2\pi$,$-6+\pi^2/4$,$4\pi$

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

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

$0$,$1$,$0$,$5$ $-1$,$0$,$6$,$-24$ $\sqrt2$,$3\sqrt2$,,$11\sqrt2$,$57\sqrt2$ $-1$,$3$,$-14$,$88$ min ni min max min ni min

min

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

Ninguno si$b^2-4c<0$; min local a$x_1=(-b+\sqrt{b^2-4c})/2$ y máximo local en$x_1=(-b-\sqrt{b^2-4c})/2$ si$b^2-4c>0$; si$b^2=4c$ entonces$x=-b/2$ es un punto crítico, pero no un punto extremo local.

$\dst\frac{1}{6}\left(\frac{\pi}{20}\right)^3$ $\dst\frac{1}{8^3}$ $\dst\frac{\pi^2}{512\sqrt2}$

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

$M_3h/3$, donde$M_3=\sup_{|x-c|\le h}|f^{(3)}(c)|$\

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

$k=-h/2$

funciones monótona

Dejar$[a,b]=[0,1]$ y$P=\{0,1\}$. Dejar$f(0)=f(1)=\frac{1}{2}$ y$f(x)=x$ si$0<x<1$. Entonces$s(P)=0$ y$S(P)=1$, pero tampoco es una suma de Riemann de$f$ más$P$.

$\frac{1}{2}$,$-\frac{1}{2}$

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

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

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

Si$g\equiv1$ y$f$ es arbitrario, entonces$\int_a^b f(x)\,dg(x)=0$.

$\overline{u}=c=\frac{2}{3}$ $\overline{u}=c=0$

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

$n!$ $\frac{1}{2}$ divergente $1$ $-1$

$0$

divergente convergente convergente convergente convergente

divergente

$p<2$ $p<1$ $p>-1$ $-1<p<2$ ninguno ninguno

$p<1$

$p-q<1$ $p,q<1$ $-1<p<2q-1$ $q>-1$,$p+q>1$ $p+q>1$

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

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

$2$ $1$ $0$ $1/2 \quad$ $1/2 \quad$ $1/2 \quad$

$1/2 \quad$

$\sqrt A$ $1$ $1$ $1$ $-\infty$

$0$

Si$s_n=1$ y$t_n=-1/n$, entonces$(\lim_{n\to\infty}s_n)/(\lim_{n\to\infty}t_n)=1/0=\infty$, pero$\lim_{n\to\infty}s_n/t_n=-\infty$.

$\infty$,$0$ $\infty$,$-\infty$ si$|r|>1$;$2$,$-2$ si$r=-1$;$0$,$0$ si$r=1$;$1$,$-1$ si;$|r|<1$ $\infty$,$-\infty$ si$r<-1$;$0$,$0$ si$|r|<1$; $\frac{1}{2}$,$\frac{1}{2}$ si$r=1$;$\infty$,$\infty$ si$r>1$\ $\infty$,$\infty$

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

$1$,$-1$ $2$,$-2$ $3$,$-1$

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

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

$\lim_{m\to\infty}s_{2m}=\infty$,$\lim_{m\to\infty}s_{2m+1}=-\infty$\ $\lim_{m\to\infty}s_{4m}=1$,$\lim_{m\to\infty}s_{4m+2}=-1$,$\lim_{m\to\infty}s_{2m+1}=0$\ $\lim_{m\to\infty}s_{2m}=0$,$\lim_{m\to\infty}s_{4m+1}=1$,$\lim_{m\to\infty}s_{4m+3}=-1$\ $\lim_{n\to\infty}s_{n}=0$ $\lim_{m\to\infty}s_{2m}=\infty$,$\lim_{m\to\infty}s_{2m+1}=0$\

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

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

Dejar$\{t_n\}$ ser cualquier secuencia convergente y$\{s_n\}=\{t_1,1,t_2,2, \dots,t_n,n, \dots\}$.

No; considere$\sum 1/n$

convergente convergente divergente divergente\ convergente convergente divergente

convergente

$p>1$ $p>1$

$p>1$

convergente convergente si$0<r<1$, divergente si$r\ge1$\ divergente convergente divergente

convergente

convergente convergente convergente

convergente

convergente divergente si y solo si$0<r<1$ o$r=1$ y$p<-1$ convergente convergente

convergente

convergente convergente divergente

convergente si$\alpha<\beta-1$, divergente si$\alpha\ge\beta-1$

convergente convergente divergente

convergente

$\sum(-1)^n$ $\sum(-1)^n/n$,$\sum\dst\left[\frac{(-1)^n}{ n}+\frac{1}{ n\log n}\right]$\ $\sum(-1)^n2^n$

$\sum(-1)^n$

condicionalmente convergente condicionalmente convergente absolutamente convergente

absolutamente convergente

%\ begin {exercisepatch1}

Dejar$k$ y$s$ ser los grados del numerador y denominador, respectivamente. Si$|r|=1$, la serie converge absolutamente si y solo si$s\ge k+2$. La serie converge condicionalmente si$s=k+1$ y$r=-1$, y diverge en todos los demás casos, donde$s\ge k+1$ y$|r|=1$.

$\sum(-1)^n/\sqrt n$ $0$

$2A-a_0$

$F(x)=0,\ |x|\le1$ $F(x)=0,\ |x|\le1$\ $F(x)=0,\ -1<x\le1$ $F(x)=\sin x,\ -\infty<x<\infty$\ $F(x)=1,\ -1<x\le1$;$F(x)=0,\ |x|>1$ $F(x)=x,\ -\infty<x<\infty$\ $F(x)=x^2/2,\ -\infty<x<\infty$ $F(x)=0,\ -\infty<x<\infty$\

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

$F(x)=0$ $F(x)=1,\ |x|<1$;$F(x)=0,\ |x|>1$\

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

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

$[-1,1]$ $[-r,r]\cup\{1\}\cup\{-1\},\ 0<r<1$ $[-r,r]\cup\{1\},\ 0<r<1$ $[-r,r],\ r>0$ $(-\infty,-1/r]\cup[-r,r]\cup[1/r,\infty)\cup\{1\},\ 0<r<1$\ $[-r,r],\ r>0$ $[-r,r],\ r>0$ $(-\infty,-r]\cup[r,\infty)\cup\{0\},\ r>0$\

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

Vamos$S=(0,1]$,$F_n(x)=\sin(x/n)$,$G_n(x)=1/x^2$; entonces$F=0$,$G=1/x^2$, y la convergencia es uniforme, pero$\|F_nG_n\|_S=\infty$.

$3$ $1$ $\frac{1}{2}$

$e-1$

subconjuntos compactos de subconjuntos$(-\frac{1}{2},\infty)$ $[-\frac{1}{2},\infty)$ cerrados de$\dst\left(\frac{1-\sqrt5}{2},\frac{1+\sqrt5}{2}\right)$ $(-\infty,\infty)$ $[r,\infty),\ r>1$

subconjuntos compactos de$(-\infty,0)\cup(0,\infty)$

Dejar$S=(-\infty,\infty)$,$f_n=a_n$ (constante), donde$\sum a_n$ converge condicionalmente, y$g_n=|a_n|$.

``absolutamente”

significa que$\sum|f_n(x)|$ converge puntualmente y$\sum f_n(x)$ converge uniformemente en$S$, mientras

significa que$\sum|f_n(x)|$ converge uniformemente en$S$.

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

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

$1/3e$ $1$ $\frac{1}{3}$ $1$

$\infty$

$1$ $\frac{1}{2}$ $\frac{1}{4}$ $4$ $1/e$

$1$

$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)! \);

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

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

$(1-x)\sum_{n=0}^\infty x^n=1$converge para todos$x$

$\dst{x+x^2+\frac{x^3}{3}-\frac{3x^5}{40}+\cdots}$ $\dst{1-x-\frac{x^2}{2}+\frac{5x^3}{6}+\cdots}$ $\dst{1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{721x^6}{720}+\cdots}$

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

$\dst{1+x+\frac{2x^2}{3}+\frac{x^3}{3}+\cdots}$ $\dst{1-x-\frac{x^2}{2}+\frac{3x^3}{2}+\cdots}$ $\dst{1+\frac{x^2}{2}+\frac{5x^4}{24}+\frac{61x^6}{720}+\cdots}$ $\dst{1+\frac{x^2}{6}+\frac{7x^4}{360}+\frac{31x^6}{15120} +\cdots}$

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

$\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$

$(3,0,3,3)$ $(-1,-1,4)$

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

$\sqrt{15}$ $\sqrt{65}/12$ $\sqrt{31}$

$\sqrt3$

$\sqrt{89}$ $\sqrt{166}/12$ $3$

$\sqrt{31}$

$12$ $\frac{1}{32}$

$27$

$\mathbf{X}=\mathbf{X}_0+t\mathbf{U}\ (-\infty<t<\infty)$en todos los casos.

$\dots\mathbf{U}$y$\mathbf{X}_1-\mathbf{X}_0$ son múltiplos escalares de$\mathbf{V}$.

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

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

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

$5$ $2$

$1/2\sqrt5$

$\set{(x_1,x_2,x_3,x_4)}{|x_i|\le3\ (i=1,2,3) \mbox{ with at least one equality}}$ \ (\ set {(x_1, x_2, x_3, x_4)} {|x_i|\ le3\ (i=1,2,3)}\)

$S$

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

$S$ $S$ $\emptyset$

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

abrir ni

cerrado

$(\pi,1,0)$

$(1,0,e)$

$6$ $6$ $2\sqrt5$ $2L\sqrt n$

$\infty$

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

$\dots$si para$A$ hay un entero$R$ tal que$|\mathbf{X}_r|>A$ si$r\ge R$.

$10$ $3$ $1$ $0$ $0$

$0$

$a/(1+a^2)$

$\infty$ $\infty$ no $-\infty$

$0$ $0$ ninguno $0$

ninguno

si no$D_f$ es acotado y para cada uno$M$ hay$R$ tal que$f(\mathbf{X})>M$ si$\mathbf{X}\in D_f$ y$|\mathbf{X}|>R$. Reemplazar $>M$” por$<M$” en

$\lim_{\mathbf{X}\to\mathbf{0}}f(\mathbf{X})=0$si$a_1+a_2+\cdots+a_n>b$; sin límite si$a_1+a_2+\cdots+a_n\le b$ y$a_1^2+a_2^2+\cdots+a_n^2\ne0$;$\lim_{\mathbf{X}\to\mathbf{0}}f(\mathbf{X})=\infty$ si$a_1=a_2=\cdots=a_n=0$ y$b>0$.

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

$\mathbb R^3$ $\mathbb R^2$ $\mathbb R^3$ $\mathbb R^2$ $\set{(x,y)}{x\ge y}$

$\mathbb R^n$

$\mathbb R^3-\{(0,0,0)\}$ $\mathbb R^2$ $\mathbb R^2$ $\mathbb R^2$

$\mathbb R^2$

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

$\dst{\frac{2}{\sqrt3}(x+y\cos x-xy\sin x)-2\sqrt\frac{2}{3}(x\cos x)}$ $\dst{\frac{1-2y}{\sqrt3}e^{-x+y^2+2z}}$ $\dst{\frac{2}{\sqrt n}(x_1+x_2+\cdots+x_n)}$

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

$\phi_1^2\phi_2$ $-5\pi/\sqrt6$ $-2e$ $0$

$0$

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

$f_x=2x+3yz+2y$,$f_y=3xz+2x$,$f_z=3xy$ $f_x=e^{yz}$,$f_y=xze^{yz}$,$f_z=xye^{yz}$

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

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

$f_{xx}=2$,$f_{yy}=f_{zz}=0$,$f_{xy}=f_{yx}=3z+2$,$f_{xz}=f_{zx}=3y$,$f_{yz}=f_{zy}=3x$

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

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

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

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

$f(x,y)=g(x,y)+h(y)$, donde$g_{xy}$ existe en todas partes y no$h$ es en ninguna parte diferenciable.

$df=(3x^2+4y^2+2y\sin x+2xy\cos x)\,dx+(8xy+2x\sin x)\, dy$,\$d_{\mathbf{X}_0}f=16\,dx$,$(d_{\mathbf{X}_0}f)(\mathbf{X}-\mathbf{X}_0)=16x$\ $df=-e^{-x-y-z}\,(dx+dy+dz)$,$d_{\mathbf{X}_0}f=-dx-dy-dz$,$(d_{\mathbf{X}_0}f)(\mathbf{X}- \mathbf{X}_0)=-x-y-z$\ $df=(1+x_1+2x_2+\cdots+nx_n)^{-1}\sum_{j=1}^nj\,dx_j$,,$d_{\mathbf{X}_0}f=\sum_{j=1}^n j\,dx_j$,\$(d_{\mathbf{X}_0}f)(\mathbf{X}-\mathbf{X}_0)=\sum_{j=1}^n jx_j$,\

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

El vector unitario en la dirección de$(f_{x_1}(\mathbf{X}_0), f_{x_2}(\mathbf{X}_0), \dots,f_{x_n}(\mathbf{X}_0))$ siempre que esto no sea$\mathbf{0}$; si lo es$\mathbf{0}$, entonces$\partial f(\mathbf{X}_0)/\partial\boldsymbol{\Phi}=0$ para cada$\boldsymbol{\Phi}$.

$z=2x+4y-6$ $z=2x+3y+1$ $z=(\pi x)/2+y-\pi/2$

$z=x+10y+4$

$5\,du+34\,dv$ $0$ $6\,du-18\,dv$

$8\,du$

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

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

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

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

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

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

$\dst{1-x-y+\frac{x^2}{2}+xy+\frac{y^2}{2}-\frac{x^3}{6}-\frac{x^2y}{2} -\frac{xy^2}{2}-\frac{y^3}{6}}$\ $0$

$xyz$

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

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

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

$\left[\begin{array}{rrrr}8&8&16&24\\0&0&4&12\\ 12&16&28&44\end{array}\right]$

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

$\left[\begin{array}{rrr}-2&2&6\\6&7&-3\\0&-2&6\end{array}\right]$

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

$\left[\begin{array}{rr}13&25\\16&31\\16&25\end{array}\right]$

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

$\mathbf{A}$y$\mathbf{B}$ son cuadrados del mismo orden.

$\left[\begin{array}{rrr}7&3&3\\4&7&7\\6&-9&1\end{array}\right]$

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

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

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

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

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

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

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

$(2,3,-2)$ $(2,3,0)$ $(-2,0,-1)$

$(3,1,3,2)$

$\dst\frac{1}{10}\left[\begin{array}{rr}4&2\\-3&1\end{array}\right]$

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

$\dst\frac{1}{25}\left[\begin{array}{rrr}4&3&-5\\6&-8&5\\-3&4&10 \end{array}\right]$

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

$\dst\frac{1}{7}\left[\begin{array}{rrrr}3&-2&0&0\\2&1&0&0\\0&0&2&-3 \\0&0&1&2\end{array}\right]$

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

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

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

$\mathbf{F}'(\mathbf{X})=\left[\begin{array}{c}g_1'(x)\\g_2'(x)\\\vdots\\ g_n'(x)\end{array}\right]$\

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

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

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

$\left[\begin{array}{rrr}0&0&4\\0&-\frac{1}{2}&0\end{array}\right]$ $\left[\begin{array}{rr}-18&0\\2&0\end{array}\right]$

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

$\left[\begin{array}{rrr}4&-3&1\\0&1&1\end{array}\right]$ $\left[\begin{array}{rr}2&0\\2&0\end{array}\right]$

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

$[1,\pi/2]$ $[1,2\pi]$ $[1,\pi]$ $[2\sqrt2,9\pi/4]$

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

$[1,-3\pi/2]$ $[1,-2\pi]$ $[1,-\pi]$ $[2\sqrt2,-7\pi/4]$

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

Dejar$f(x)=x\ (0\le x\le\frac{1}{2})$,\ (f (x) =x-\ frac {1} {2}\ (\ frac {1} {2} <x\ le1)\); entonces$f$ es localmente invertible pero no invertible on$[0,1]$.

$\mathbf{F}(S)=\set{(u,v)}{-\pi+2\phi<\arg(u,v)<\pi+2\phi}$, donde$\phi$ es un argumento de$(a,b)$;

\ (\ 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}\ derecha],\ 2\ phi-\ pi<\ arg (u, v) <2\ phi+\ pi\)

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

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

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

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

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

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

De resolver$x=r\cos\theta$,$y=r\sin\theta$ para$\theta=\arg(x,y)$. Cada ecuación es satisfecha por ángulos que no son argumentos de$(x,y)$, ya que ninguna de las fórmulas identifica el cuadrante de$(x,y)$ manera única. Además,

no se sostiene si$x=0$.

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

donde$\beta-\pi/2<\arg(u,v)<\beta+\pi/2$ y$\beta$ es un argumento de$(a,b)$;

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

Si$\mathbf{F}(x_1,x_2, \dots,x_n)=(x_1^3,x_2^3, \dots,x_n^3)$, entonces$\mathbf{F}$ es invertible, pero\$J\mathbf{F}(\mathbf{0})~=0$.

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

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

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

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

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

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

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

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

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

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

$f_i(\mathbf{X},\mathbf{U})=\dst\left(\sum_{j=1}^n a_{ij}(x_j-x_{j0})\right)^r-(u_i-u_{i0})^s$,$1\le i\le m$, donde$r$ y$s$ son enteros positivos y no todos$a_{ij}=0$. $r=s=3$; $r=1$,$s=3$;

$r=s=2$

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

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

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

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

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

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

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

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

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

$\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}$,\$v_{xy}(1,-1)=-\dst\frac{1}{8}$,$v_{yy}(1,-1)=\dst\frac{1}{8}$

$28$

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

$12$ $\frac{79}{20}$ $-1$

$(1-\log2)/2$

$\frac{7}{4}$ $17$ $\frac{2}{3}(\sqrt2-1)$

$1/4\pi$

$\frac{3}{8}$,$\frac{5}{8}$ $\frac{3}{8}$,$\frac{5}{8}$ $\frac{3}{4}$,$\frac{5}{4}$ $\frac{3}{4}\left(z+\frac{1}{2}\right)$,$\frac{5}{4}\left(z+\frac{1}{2}\right)$

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

$-285$ $0$ $0$

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

$324$ $\frac{1}{6}$

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

$36$ $1$ $\frac{64}{3}$

$(e^6+17)/2$

$\frac{2}{27}$ $\frac{1}{2}(e-\frac{5}{2})$ $\frac{1}{24}$

$\frac{1}{36}$

$16\pi$ $\frac{1}{6}$ $\frac{128}{21}$

$\frac{\pi}{2}$

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

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

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

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

Dejar$S_1$ y$S_2$ ser densos subconjuntos de$\mathbb R$ tal manera que$S_1\cup S_2=\mathbb R$.

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

$\frac{5}{6}$ $\frac{4}{9}$

$\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$

$(\pi/8)\log5$ $(\pi/4)(e^4-1)$

$2\pi/15$

$\pi^2a^4/2$

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

\ end {documento}

{}

Supongamos$f\in C[a,b]$. Acordando a la, si$\epsilon>0$, hay un polinomio$p$ tal que$|f(x)-p(x)|<\epsilon$,$a\le x\le b$. Ahora supongamos que\ [ 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. \] y\ [ \ lim_ {n\ a\ infty}\ frac {1} {n}\ suma_ {i=1} ^ {n} |x_ {in} -y_ {in} |=0. \] Mostrar que si$f\in C[a,b]$, entonces\ [ \ lim_ {n\ a\ infty}\ frac {1} {n}\ sum_ {i=1} ^ {n} |f (x_ {in}) -f (y_ {in}) |=0. \]

Por la desigualdad del triángulo,\ [ |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}) |, \] entonces\ [\ begin {ecuación}\ tag {A} |f (x_ {in}) -f (y_ {in}) |<|p (x_ {in}) -p (y_ {in}) |+2\ épsilon. \ end {ecuación}\] Vamos$M=\max_{a\le x\le b}|p'(x)|$. Por el teorema del valor medio,\ [ |p (x_ {in}) -p (y_ {in}) |\ le M |x_ {in} -y_ {in} |. \] Esto y (A) implican que\ [ \ 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} |. \] De esto y (A),\ [ \ limsup_ {n\ a\ infty} \ frac {1} {n}\ sum_ {i=1} ^n|f (x_ {in}) -f (y_ {in}) |\ le 2\ épsilon. \] Dado que$\epsilon$ es arbitrario, esto implica la conclusión.

{} En el ajuste de Ejercicio~1, vamos\ [ y_ {in} =a+\ frac {i} {n}\ sum_ {i=1} ^ {n} (b-a),\ quad 1\ le i\ le n,\ quad n\ ge 1. \] Mostrar que\ [ \ lim_ {n\ a\ infty}\ frac {1} {n}\ sum_ {i=1} ^ {n} f (x_ {in}) =\ frac {1} {b-a}\ int_ {a} ^ {b} f (x)\, dx. \]

{Solución 2.} Ya que$f\in C[a,b]$,$\int_{a}^{b}f(x)\,dx$ existe (Teoremo~3.2.8, p.~133).

Por lo tanto, de Definición~3.1.1 (p. _{114),\ [
\ lim_ {n\ a\ infty}\ sum_ {i=1} ^ {n} f (y_ {in}) =\ frac {1} {b-a}\ int_ {a} ^ {b} f (x)\, dx.
\] Desde\ [
\ frac {1} {n}\ izquierda|
\ suma_ {i=1} ^ {n} f (x_ {in}) -\ frac {1} {b-a}\ int_ {a} ^ {b} f (x)\, dx\ derecha|
\ le\ frac {1} {n}\ sum_ {i=1} ^ {n} |f (_ {in}) -f (y_ {in}) |+
\ izquierda|\ frac {1} {n}
\ sum_ {i=1} ^ {n} f (y_ {in}) -\ frac {1} {b-a}\ int_ {a} ^ {b} f (x)\, dx\ derecha|,
\] El ejercicio} 1 implica la conclusión.

Supongamos que$g$ es continuo y no decreciente en$[c,d]$. Para

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