$P(5) = 1050\ \Rightarrow\ 8 \cdot 125 + a \cdot 25 - 11 \cdot 5 + 5 = 1050\ \Rightarrow\ \fbox{$a = 4$}$
domingo, 16 de janeiro de 2022
Qual a soma dos coeficientes de um polinômio do terceiro grau sabendo que é divisível por $x - 1$?
Seja $P(x) = ax^3 + bx^2 + cx + d$ tal polinômio.
Se ele é divisível por $x - 1$, $P(1) = 0$. Logo:
$\fbox{$a + b + c + d = 0$}$.
Calculadora: ponto simétrico a uma reta.
Entre com os argumentos, separados por barra vertical "|": primeiro: o ponto com abscissa separada da ordenada por ponto e vírgula ";"; segundo: os coeficientes $a$, $b$ e $c$ da reta $ax + by + c = 0$, separados por ponto e vírgula ";":
Exemplo:
Input: "0; 3 | -1; 1; 0". Output: "3, 0".
Ponto simétrico à reta:
Exemplo:
Input: "0; 3 | -1; 1; 0". Output: "3, 0".
Ponto simétrico à reta:
Novas coordenadas cartesianas para uma rotação do eixo $Ox$.
${\large (x_r, y_r) = \left(\dfrac{x}{\cos \theta}, y - x\tan \theta\right)}$
sábado, 15 de janeiro de 2022
Se $n$ pessoas se encontram, e cada uma aperta $1$ vez a mão de cada outra, quantos apertos de mão haverá?
$\displaystyle{n \choose 2} = \dfrac{n!}{2!(n - 2)!}\ =\ \fbox{$\dfrac{n(n - 1)}{2}$}$
Duas formas de encontrar $I = \displaystyle\int (\sin x)(\cos x)\ dx$.
Primeira:
$I = \dfrac{1}{2}\displaystyle\int \sin (2x)\ dx = -\dfrac{\cos(2x)}{4} + c$
Segunda:
Seja $u = \sin x$, $du = \cos x\ dx$.
$I = \displaystyle\int u\ du = \dfrac{u^2}{2} + C = \dfrac{\sin^2 x}{2} + C$
Observemos que $-\dfrac{\cos(2x)}{4} - \dfrac{\sin^2 x}{2} = -\dfrac{1}{4}$, que é constante. Logo as duas respostas estão corretas, pois tratam-se de integrais indefinidas.
Seja $V$ o espaço vetorial das matrizes quadradas $n\ x\ n$, $U$ o subespaço das matrizes simétricas e $W$ o subespaço das matrizes antissimétricas, $V = U \oplus W$.
$M \in U\ \Rightarrow\ M = M^t$
$M \in W\ \Rightarrow\ M = -M^t$
Seja $A \in V$, $A = \dfrac{1}{2}(A + A^t) + \dfrac{1}{2}(A - A^t)\ {\large (I)}$.
$(A + A^t)^t = A^t + A\ \Rightarrow\ (A + A^t) \in U\ {\large (II)}$
$(A - A^t)^t = -(A - A^t)\ \Rightarrow\ (A - A^t) \in W\ {\large (III)}$
${\large (I)}\ \wedge\ {\large (II)}\ \wedge\ {\large (III)}\ \Rightarrow\ V = U + W\ {\large (IV)}$
Seja $M \in U\ \wedge\ M \in W$:
$M = M^t\ \wedge\ -M = M^t\ \Rightarrow\ M = -M\ \Rightarrow\ M = O\ \Rightarrow\ U \cap W = \{O\}\ {\large (V)}$.
${\large (IV)}\ \wedge\ {\large (V)}\ \Rightarrow\ V = U \oplus W$
Quod Erat Demonstrandum.
sexta-feira, 14 de janeiro de 2022
Os lados de um triângulo medem $5\ cm$, $7\ cm$ e $8\ cm$. Qual a medida da mediana relativa ao maior lado?
Seja $\theta$ o ângulo adjacente aos lados de medida $5$ e $8$:
$49 = 25 + 64 - 80\cos \theta\ \therefore\ \theta = \arccos \dfrac{1}{2}$.
Seja $m$ a medida da mediana relativa ao maior lado:
$m^2 = 25 + 16 - 40 \cdot \dfrac{1}{2}\ \therefore\ \fbox{$m = \sqrt{21}\ cm$}$.
A sequência $\left(\dfrac{\sqrt{2} + 1}{2}, a, \dfrac{\sqrt{2} - 1}{2}\right)$ é uma PG. Qual o valor de $a$?
Resolução:
$a = \pm\sqrt{\dfrac{\sqrt{2} + 1}{2} \cdot \dfrac{\sqrt{2} - 1}{2}} = \pm\sqrt{\dfrac{1}{4}} = \fbox{$\pm\dfrac{1}{2}$}$
Encontrar o ângulo inscrito $\alpha$.
Seja $O$ o centro da circunferência.
$m(A\hat{O}B) = 180^o - 48^o = 132^o$
$\alpha = m(A\hat{Q}B) = \dfrac{360^o - m(A\hat{O}B)}{2} = \dfrac{228^o}{2}$
$\fbox{$\alpha = 114^o$}$
Resolver a equação $x + \dfrac{2x}{3} + \dfrac{4x}{9} + \dots = 18$.
$x\underbrace{\left[\displaystyle\sum_{i=0}^{+\infty} \left(\dfrac{2}{3}\right)^i\right]}_{(1 - 2/3)^{-1}} = 18\ \Rightarrow\ 3x = 18$
$S = \{6\}$
Se $A'_{k,p}$ é o número de arranjos com repetição de $k$ elementos tomados $p$ a $p$, resolver as equações $A'_{k,2} = 121$ e $A'_{5,k'} = 625$.
$k^2 = 121\ \Rightarrow\ k = 11$
$5^{k'} = 625\ \Rightarrow\ k' = \log_5 625 = 4$
$U = \{(a, b, c)\ :\ a = b = c\}$ e $W = \{(0, b, c)\}$, $\mathbb{R}^3 = U \oplus W$.
$\{(0, 0, 0)\} = U \cap W\ {\large (I)}$
Seja $(a, b, c)$ um vetor do $\mathbb{R}^3$:
$(a, b, c) = (a, a, a) + (0, b - a, c - a)\ \Rightarrow\ \mathbb{R}^3 = U + W\ {\large (II)}$.
${\large (I)}\ \wedge\ {\large (II)}\ \Rightarrow\ \mathbb{R}^3 = U \oplus W$
Quod Erat Demonstrandum.
quinta-feira, 13 de janeiro de 2022
Alguns senos e cossenos de nx.
$\sin (2x)=2\,\cos x\,\sin x$
$\cos (2x)=\cos ^2x-\sin ^2x$
$\sin (3x)=3\,\cos ^2x\,\sin x-\sin ^3x$
$\cos (3x)=\cos ^3x-3\,\cos x\,\sin ^2x$
$\sin (4x)=4\,\cos ^3x\,\sin x-4\,\cos x\,\sin ^3x$
$\cos (4x)=\sin ^4x-6\,\cos ^2x\,\sin ^2x+\cos ^4x$
$\sin (5x)=\sin ^5x-10\,\cos ^2x\,\sin ^3x+5\,\cos ^4x\,\sin x$
$\cos (5x)=5\,\cos x\,\sin ^4x-10\,\cos ^3x\,\sin ^2x+\cos ^5x$
$\sin (6x)=6\,\cos x\,\sin ^5x-20\,\cos ^3x\,\sin ^3x+6\,\cos ^5x\,\sin x$
$\cos (6x)=-\sin ^6x+15\,\cos ^2x\,\sin ^4x-15\,\cos ^4x\,\sin ^2x+\cos ^6x$
$\sin (7x)=-\sin ^7x+21\,\cos ^2x\,\sin ^5x-35\,\cos ^4x\,\sin ^3x+7\,\cos ^6x\,\sin x$
$\cos (7x)=-7\,\cos x\,\sin ^6x+35\,\cos ^3x\,\sin ^4x-21\,\cos ^5x\,\sin ^2x+\cos ^7x$
$\sin (8x)=-8\,\cos x\,\sin ^7x+56\,\cos ^3x\,\sin ^5x-56\,\cos ^5x\,\sin ^3x+8\,\cos ^7x\,\sin x$
$\cos (8x)=\sin ^8x-28\,\cos ^2x\,\sin ^6x+70\,\cos ^4x\,\sin ^4x-28\,\cos ^6x\,\sin ^2x+\cos ^8x$
$\sin (9x)=\sin ^9x-36\,\cos ^2x\,\sin ^7x+126\,\cos ^4x\,\sin ^5x-84\,\cos ^6x\,\sin ^3x+9\,\cos ^8x\,\sin x$
$\cos (9x)=9\,\cos x\,\sin ^8x-84\,\cos ^3x\,\sin ^6x+126\,\cos ^5x\,\sin ^4x-36\,\cos ^7x\,\sin ^2x+\cos ^9x$
${\small \sin (10x)=10\,\cos x\,\sin ^9x-120\,\cos ^3x\,\sin ^7x+252\,\cos ^5x\,\sin ^5x-120\,\cos ^7x\,\sin ^3x+10\,\cos ^9x\,\sin x}$
${\small \cos (10x)=-\sin ^{10}x+45\,\cos ^2x\,\sin ^8x-210\,\cos ^4x\,\sin ^6x+210\,\cos ^6x\,\sin ^4x-45\,\cos ^8x\,\sin ^2x+\cos ^{10}x}$
${\scriptsize \sin (11x)=-\sin ^{11}x+55\,\cos ^2x\,\sin ^9x-330\,\cos ^4x\,\sin ^7x+462\,\cos ^6x\,\sin ^5x-165\,\cos ^8x\,\sin ^3x+11\,\cos ^{10}x\,\sin x}$
${\scriptsize \cos (11x)=-11\,\cos x\,\sin ^{10}x+165\,\cos ^3x\,\sin ^8x-462\,\cos ^5x\,\sin ^6x+330\,\cos ^7x\,\sin ^4x-55\,\cos ^9x\,\sin ^2x+\cos ^{11}x}$
${\tiny \sin (12x)=-12\,\cos x\,\sin ^{11}x+220\,\cos ^3x\,\sin ^9x-792\,\cos ^5x\,\sin ^7x+792\,\cos ^7x\,\sin ^5x-220\,\cos ^9x\,\sin ^3x+12\,\cos ^{11}x\,\sin x}$
${\tiny \cos (12x)=\sin ^{12}x-66\,\cos ^2x\,\sin ^{10}x+495\,\cos ^4x\,\sin ^8x-924\,\cos ^6x\,\sin ^6x+495\,\cos ^8x\,\sin ^4x-66\,\cos ^{10}x\,\sin ^2x+\cos ^{12}x}$
${\tiny \sin (13x)=\sin ^{13}x-78\,\cos ^2x\,\sin ^{11}x+715\,\cos ^4x\,\sin ^9x-1716\,\cos ^6x\,\sin ^7x+1287\,\cos ^8x\,\sin ^5x-286\,\cos ^{10}x\,\sin ^3x+13\,\cos ^{12}x\,\sin x}$
${\tiny \cos (13x)=13\,\cos x\,\sin ^{12}x-286\,\cos ^3x\,\sin ^{10}x+1287\,\cos ^5x\,\sin ^8x-1716\,\cos ^7x\,\sin ^6x+715\,\cos ^9x\,\sin ^4x-78\,\cos ^{11}x\,\sin ^2x+\cos ^{13}x}$
${\tiny \sin (14x)=14\,\cos x\,\sin ^{13}x-364\,\cos ^3x\,\sin ^{11}x+2002\,\cos ^5x\,\sin ^9x-3432\,\cos ^7x\,\sin ^7x+2002\,\cos ^9x\,\sin ^5x-364\,\cos ^{11}x\,\sin ^3x+14\,\cos ^{13}x\,\sin x}$
${\tiny \cos (14x)=-\sin ^{14}x+91\,\cos ^2x\,\sin ^{12}x-1001\,\cos ^4x\,\sin ^{10}x+3003\,\cos ^6x\,\sin ^8x-3003\,\cos ^8x\,\sin ^6x+1001\,\cos ^{10}x\,\sin ^4x-91\,\cos ^{12}x\,\sin ^2x+\cos ^{14}x}$
${\tiny \sin (15x)=-\sin ^{15}x+105\,\cos ^2x\,\sin ^{13}x-1365\,\cos ^4x\,\sin ^{11}x+5005\,\cos ^6x\,\sin ^9x-6435\,\cos ^8x\,\sin ^7x+3003\,\cos ^{10}x\,\sin ^5x-455\,\cos ^{12}x\,\sin ^3x+15\,\cos ^{14}x\,\sin x}$
${\tiny \cos (15x)=-15\,\cos x\,\sin ^{14}x+455\,\cos ^3x\,\sin ^{12}x-3003\,\cos ^5x\,\sin ^{10}x+6435\,\cos ^7x\,\sin ^8x-5005\,\cos ^9x\,\sin ^6x+1365\,\cos ^{11}x\,\sin ^4x-105\,\cos ^{13}x\,\sin ^2x+\cos ^{15}x}$
Alguns valores precisos de senos e cossenos.
$\sin \frac{\pi}{2}=1$
$\cos \frac{\pi}{2}=0$
$\sin \frac{\pi}{4}={{1}\over{\sqrt{2}}}$
$\cos \frac{\pi}{4}={{1}\over{\sqrt{2}}}$
$\sin \frac{\pi}{8}={{\sqrt{\sqrt{2}-1}}\over{2^{{{3}\over{4}}}}}$
$\cos \frac{\pi}{8}={{\sqrt{\sqrt{2}+1}}\over{2^{{{3}\over{4}}}}}$
$\sin \frac{\pi}{16}={{\sqrt{2^{{{3}\over{4}}}-\sqrt{\sqrt{2}+1}}}\over{2^{{{7}\over{8}}}}}$
$\cos \frac{\pi}{16}={{\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}\over{2^{{{7}\over{8}}}}}$
$\sin \frac{\pi}{32}={{\sqrt{2^{{{7}\over{8}}}-\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}\over{2^{{{15}\over{16}}}}}$
$\cos \frac{\pi}{32}={{\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}\over{2^{{{15}\over{16}}}}}$
$\sin \frac{\pi}{64}={{\sqrt{2^{{{15}\over{16}}}-\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}\over{2^{{{31}\over{32}}}}}$
$\cos \frac{\pi}{64}={{\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}\over{2^{{{31}\over{32}}}}}$
$\sin \frac{\pi}{128}={{\sqrt{2^{{{31}\over{32}}}-\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}\over{2^{{{63}\over{64}}}}}$
$\cos \frac{\pi}{128}={{\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}\over{2^{{{63}\over{64}}}}}$
$\sin \frac{\pi}{256}={{\sqrt{2^{{{63}\over{64}}}-\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}\over{2^{{{127}\over{128}}}}}$
$\cos \frac{\pi}{256}={{\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}\over{2^{{{127}\over{128}}}}}$
$\sin \frac{\pi}{512}={{\sqrt{2^{{{127}\over{128}}}-\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}\over{2^{{{255}\over{256}}}}}$
$\cos \frac{\pi}{512}={{\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}\over{2^{{{255}\over{256}}}}}$
$\sin \frac{\pi}{1024}={{\sqrt{2^{{{255}\over{256}}}-\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}\over{2^{{{511}\over{512}}}}}$
$\cos \frac{\pi}{1024}={{\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}\over{2^{{{511}\over{512}}}}}$
$\sin \frac{\pi}{2048}={{\sqrt{2^{{{511}\over{512}}}-\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}\over{2^{{{1023}\over{1024}}}}}$
$\cos \frac{\pi}{2048}={{\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}\over{2^{{{1023}\over{1024}}}}}$
$\sin \frac{\pi}{4096}={{\sqrt{2^{{{1023}\over{1024}}}-\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}\over{2^{{{2047}\over{2048}}}}}$
$\cos \frac{\pi}{4096}={{\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}\over{2^{{{2047}\over{2048}}}}}$
$\sin \frac{\pi}{8192}={{\sqrt{2^{{{2047}\over{2048}}}-\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}\over{2^{{{4095}\over{4096}}}}}$
$\cos \frac{\pi}{8192}={{\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}\over{2^{{{4095}\over{4096}}}}}$
$\sin \frac{\pi}{16384}={{\sqrt{2^{{{4095}\over{4096}}}-\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}\over{2^{{{8191}\over{8192}}}}}$
$\cos \frac{\pi}{16384}={{\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}\over{2^{{{8191}\over{8192}}}}}$
$\sin \frac{\pi}{32768}={{\sqrt{2^{{{8191}\over{8192}}}-\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}\over{2^{{{16383}\over{16384}}}}}$
$\cos \frac{\pi}{32768}={{\sqrt{2^{{{8191}\over{8192}}}+\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}\over{2^{{{16383}\over{16384}}}}}$
${\small \sin \frac{\pi}{65536}={{\sqrt{2^{{{16383}\over{16384}}}-\sqrt{2^{{{8191}\over{8192}}}+\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}}\over{2^{{{32767}\over{32768}}}}}}$
${\small \cos \frac{\pi}{65536}={{\sqrt{2^{{{16383}\over{16384}}}+\sqrt{2^{{{8191}\over{8192}}}+\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}}\over{2^{{{32767}\over{32768}}}}}}$
$\sin \frac{\pi}{3}={{\sqrt{3}}\over{2}}$
$\cos \frac{\pi}{3}={{1}\over{2}}$
$\sin \frac{\pi}{6}={{1}\over{2}}$
$\cos \frac{\pi}{6}={{\sqrt{3}}\over{2}}$
$\sin \frac{\pi}{12}={{\sqrt{2-\sqrt{3}}}\over{2}}$
$\cos \frac{\pi}{12}={{\sqrt{\sqrt{3}+2}}\over{2}}$
$\sin \frac{\pi}{24}={{\sqrt{2-\sqrt{\sqrt{3}+2}}}\over{2}}$
$\cos \frac{\pi}{24}={{\sqrt{\sqrt{\sqrt{3}+2}+2}}\over{2}}$
$\sin \frac{\pi}{48}={{\sqrt{2-\sqrt{\sqrt{\sqrt{3}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{48}={{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{96}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{96}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{192}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{192}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{384}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{384}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{768}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{768}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{1536}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{1536}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{3072}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{3072}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{6144}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{6144}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{12288}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{12288}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{24576}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{24576}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{49152}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{49152}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\sin \frac{\pi}{98304}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$
$\cos \frac{\pi}{98304}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$
$\cos \frac{\pi}{5}={{\sqrt{5}+1}\over{4}}$
$\sin \frac{\pi}{10}={{\sqrt{3-\sqrt{5}}}\over{2^{{{3}\over{2}}}}}$
$\cos \frac{\pi}{10}={{\sqrt{\sqrt{5}+5}}\over{2^{{{3}\over{2}}}}}$
$\sin \frac{\pi}{20}={{\sqrt{2^{{{3}\over{2}}}-\sqrt{\sqrt{5}+5}}}\over{2^{{{5}\over{4}}}}}$
$\cos \frac{\pi}{20}={{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}}\over{2^{{{5}\over{4}}}}}$
$\sin \frac{\pi}{40}={{\sqrt{2^{{{5}\over{4}}}-\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}}}\over{2^{{{9}\over{8}}}}}$
$\cos \frac{\pi}{40}={{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}}\over{2^{{{9}\over{8}}}}}$
$\sin \frac{\pi}{80}={{\sqrt{2^{{{9}\over{8}}}-\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}}}\over{2^{{{17}\over{16}}}}}$
$\cos \frac{\pi}{80}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}}\over{2^{{{17}\over{16}}}}}$
$\sin \frac{\pi}{160}={{\sqrt{2^{{{17}\over{16}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}}}\over{2^{{{33}\over{32}}}}}$
$\cos \frac{\pi}{160}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}}\over{2^{{{33}\over{32}}}}}$
$\sin \frac{\pi}{320}={{\sqrt{2^{{{33}\over{32}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}}}\over{2^{{{65}\over{64}}}}}$
$\cos \frac{\pi}{320}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}}\over{2^{{{65}\over{64}}}}}$
$\sin \frac{\pi}{640}={{\sqrt{2^{{{65}\over{64}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}}}\over{2^{{{129}\over{128}}}}}$
$\cos \frac{\pi}{640}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}}\over{2^{{{129}\over{128}}}}}$
$\sin \frac{\pi}{1280}={{\sqrt{2^{{{129}\over{128}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}}}\over{2^{{{257}\over{256}}}}}$
$\cos \frac{\pi}{1280}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}}\over{2^{{{257}\over{256}}}}}$
$\sin \frac{\pi}{2560}={{\sqrt{2^{{{257}\over{256}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}}}\over{2^{{{513}\over{512}}}}}$
$\cos \frac{\pi}{2560}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}}\over{2^{{{513}\over{512}}}}}$
$\sin \frac{\pi}{5120}={{\sqrt{2^{{{513}\over{512}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}}}\over{2^{{{1025}\over{1024}}}}}$
$\cos \frac{\pi}{5120}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}}\over{2^{{{1025}\over{1024}}}}}$
$\sin \frac{\pi}{10240}={{\sqrt{2^{{{1025}\over{1024}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}}}\over{2^{{{2049}\over{2048}}}}}$
$\cos \frac{\pi}{10240}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}}\over{2^{{{2049}\over{2048}}}}}$
$\sin \frac{\pi}{20480}={{\sqrt{2^{{{2049}\over{2048}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}}}\over{2^{{{4097}\over{4096}}}}}$
$\cos \frac{\pi}{20480}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}}\over{2^{{{4097}\over{4096}}}}}$
$\sin \frac{\pi}{40960}={{\sqrt{2^{{{4097}\over{4096}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}}}\over{2^{{{8193}\over{8192}}}}}$
$\cos \frac{\pi}{40960}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}}\over{2^{{{8193}\over{8192}}}}}$
${\small \sin \frac{\pi}{81920}={{\sqrt{2^{{{8193}\over{8192}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}}}\over{2^{{{16385}\over{16384}}}}}}$
${\small \cos \frac{\pi}{81920}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}+2^{{{8193}\over{8192}}}}}\over{2^{{{16385}\over{16384}}}}}}$
${\small \sin \frac{\pi}{163840}={{\sqrt{2^{{{16385}\over{16384}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}+2^{{{8193}\over{8192}}}}}}\over{2^{{{32769}\over{32768}}}}}}$
${\small \cos \frac{\pi}{163840}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}+2^{{{8193}\over{8192}}}}+2^{{{16385}\over{16384}}}}}\over{2^{{{32769}\over{32768}}}}}}$
Os termos $8$, $3$, $10$, $7$, $6$, $5$ e $15$ são $7$ dos $9$ de uma sequência. Qual a maior mediana possível da sequência?
Para que a mediana seja máxima, devemos adicionar termos maiores que o maior. Assim o maior termo central possível (a sequência tem um número ímpar de termos) é $8$.
Um produto teve um aumento total de preço de $61\%$ por meio de dois aumentos sucessivos. Se o primeiro aumento foi de $15\%$, de quanto foi o segundo aumento?
Suponha $p$ o preço original do produto, e $x - 1$ o percentual de aumento:
$1,61 \cdot p = x \cdot 1,15 \cdot p\ \Rightarrow\ x = \dfrac{1,61}{1,15} = 1,4\ \Rightarrow\ x - 1 = \fbox{$40\%$}$.
Em uma sequência de $n$ termos, $n > 1$, um termo é $1 - \dfrac{1}{n}$ e os restantes são iguais a $1$. Qual a média aritmética dos $n$ termos?
A soma dos $n - 1$ termos iguais a $1$ será $n - 1$.
$m = \dfrac{n - 1 + 1 - \dfrac{1}{n}}{n} = \dfrac{n - \dfrac{1}{n}}{n} = \fbox{$1 - \dfrac{1}{n^2}$}$
domingo, 9 de janeiro de 2022
quinta-feira, 6 de janeiro de 2022
Soma dos termos de uma PG infinita convergente.
De $S_n = \dfrac{a_1(q^n - 1)}{q - 1}$, com $|q| < 1$:
$\displaystyle\lim_{n \rightarrow +\infty} S_n = \fbox{$\dfrac{a_1}{1 - q}$}$.
Soma dos termos de uma PG finita.
$S_n = \displaystyle\sum_{i=1}^n a_i$
Se $q = 1$, $S_n = na_1$. Se não:
$qS_n = \left(\displaystyle\sum_{i=2}^n a_i\right) + qa_n$
$qS_n - S_n = \cancelto{a_1 q^n}{qa_n} - a_1$
$\fbox{$S_n = \dfrac{a_1(q^n - 1)}{q - 1}$}$.
Soma dos termos de uma PA finita.
Observemos que os termos equidistantes dos extremos de uma PA finita tem soma constante:
$a_1 + a_n = a_{1 + p} + a_{n - p},\ p \in \mathbb{N},\ p < n$.
$2S_n = \displaystyle\sum_{i=0}^{n - 1} \left(a_{1 + i} + a_{n - i}\right) = \displaystyle\sum_{i=0}^{n - 1} \left(a_1 + a_n\right) = n\left(a_1 + a_n\right)$
Logo, $\fbox{$S_n = \dfrac{\left(a_1 + a_n\right)n}{2}$}$.
Posições dos elementos distinguidos em matrizes com o mesmo espaço de linhas.
Se duas matrizes escalonadas tem o mesmo espaço de linhas, os elementos distinguidos estão nas mesmas posições. Ou seja, se $A = (a_{ij})$ e $B = (b_{k\ell})$ tem o mesmo espaço de linhas, se $a_{ij_m}$ e $b_{k\ell_n}$ são os elementos distinguidos das linhas $i$ de $A$ e $B$, $j_m = \ell_n$ para $i = k$.
Tomemos a linha $R_1$ de $A$. $R_1$ é uma combinação linear das linhas de $B$. Como, $a_{1j_1} = \displaystyle\sum_{o = 1}^s c_o b_{o \ell_1} = c_1 b_{1 \ell_1}$ e $a_{1j_1} \neq 0$ e $b_{1 \ell_1} \neq 0$, $c_1 \neq 0$, logo $j_1 = \ell_1$.
Provemos agora que a matriz $A'$, resultante da remoção da primeira linha de $A$ tem o mesmo espaço de linhas da matriz $B'$, resultante da remoção da primeira linha da matriz $B$.
Sejam $R_i,\ i \neq 1$, uma linha de $A$ e $R'_k$ uma linha de $B$, $R_i$ é uma combinação linear das linhas de $B$. Como $a_{ij_1} = 0,\ \forall i \neq 1$, $R_i = \displaystyle\sum_{o=2}^s c_o R'_o$, logo $A'$ e $B'$ tem o mesmo espaço de linhas.
Procedendo recursivamente estas duas etapas até que se tenha chegado à última linha não nula de $A$, repetindo todo o procedimento permutando-se $A$ e $B$, o teorema está demonstrado.
Quod Erat Demonstrandum.
Se uma matriz quadrada $A$ tem uma linha nula, não tem inversa.
Seja $R_i$ a linha nula de $A$. O produto de $A$ e qualquer outra matriz quadrada de mesma ordem terá a linha $i$ nula.
Como $I$ não tem linhas nulas, $A$ não é invertível.
Quod Erat Demonstrandum.
Seja $A$ uma matriz invertível, $(A^{-1})^t = (A^t)^{-1}$.
Seja $B = A^{-1}$, $I = I^t = (AB)^t = B^t A^t$.
Logo $B^t$ é a inversa de $A^t$, ou seja, $(A^{-1})^t = (A^t)^{-1}$.
Quod Erat Demonstrandum.
Seja $V$ o espaço vetorial das matrizes quadradas $n\ x\ n$, o subconjunto $W$ das matrizes que comutam com $T$ formam um subespaço.
$W$ não é vazio, pois $0T = T0 = 0$.
Sejam $a$ e $b$ escalares e $M_1$ e $M_2$ elementos de $W$:
$(aM_1 + bM_2)T = aM_1 T + bM_2 T = aTM_1 + bTM_2 = TaM_1 + TbM_2 = T(aM_1 + bM_2)$.
Logo $aM_1 + bM_2 \in W$.
Quod Erat Demonstrandum.
quarta-feira, 5 de janeiro de 2022
Calcular $I\ =\ \displaystyle\int \sin (\sqrt{x})\ dx$.
$I\ =\ x\sin (\sqrt{x}) - \displaystyle\int \dfrac{x\cos (\sqrt{x})}{2\sqrt{x}}\ dx$
Seja $u = \sqrt{x}$, $du = \dfrac{dx}{2\sqrt{x}}$.
$I\ =\ x\sin (\sqrt{x}) - \displaystyle\int u^2\cos u\ du$
$I\ =\ x\sin (\sqrt{x}) - u^2\sin u + 2\displaystyle\int u\sin u\ du$
$I\ =\ x\sin (\sqrt{x}) - u^2\sin u - 2u\cos u + 2\displaystyle\int \cos u\ du$
$I\ =\ \cancel{x\sin (\sqrt{x})} - \cancel{x\sin (\sqrt{x})} - 2\sqrt{x}\cos (\sqrt{x}) + 2\sin (\sqrt{x}) + c$
$\fbox{$I = -2\sqrt{x}\cos (\sqrt{x}) + 2\sin (\sqrt{x}) + c$}$
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