DFDUDU+++shrrrrlk*K

46 :kjmglakhngv:2015/06/16(火) 19:53:17.470 ID:V7C3nsa2
$b = ln (2 \pm \sqrt {3})$
47 :kjmglakhngv:2015/06/16(火) 19:55:09.956 ID:V7C3nsa2
$\cos x = 2$
は$x = bi = i ln (2 \pm \sqrt {3})$
という根を持つ
48 :kjmglakhngv:2015/06/16(火) 19:55:28.335 ID:V7C3nsa2
*つづく*
49 :kjmglakhngv:2015/06/18(木) 19:54:31.281 ID:K1KdSR2l
$\begin{eqnarray}
\sin (bi) &=& bi - \frac {(bi)^3} {1 \cdot 2 \cdot 3 } + \frac { (bi)^5} {1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 } - \cdots
&=& i [ b + \frac {b^3} {1 \cdot 2 \cdot 3 } + \frac { b^5} {1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 } + \cdots ] = i \frac {e^b - e^{-b}} {2}\end{eqnarray}$
50 :kjmglakhngv:2015/06/18(木) 19:55:07.185 ID:K1KdSR2l
やっぱり改行できないのかな?
51 :kjmglakhngv:2015/06/18(木) 19:56:25.947 ID:K1KdSR2l
$\sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
52 :kjmglakhngv:2015/06/18(木) 19:56:40.801 ID:K1KdSR2l
markdownのほうでやればいいのか
53 :kjmglakhngv:2015/06/18(木) 19:58:33.923 ID:K1KdSR2l
$\sin (a + bi) = \sin a \cos (bi) + \cos a \sin (bi) = \sin a \frac {e^b - e^{-b}} {2} + i \cos a \frac {e^b - e^{-b}} {2}, $

$\cos (a + bi) = \cos a\frac {e^b + e^{-b}} {2} - i \sin a \frac {e^b - e^{-b}} {2}$
54 :kjmglakhngv:2015/06/18(木) 19:59:17.281 ID:K1KdSR2l
位置あわせは
55 :kjmglakhngv:2015/06/18(木) 20:00:29.471 ID:K1KdSR2l
実数$a$を複素数$a + 0 \cdot i$
56 :kjmglakhngv:2015/06/18(木) 20:02:38.871 ID:K1KdSR2l
$\sin (a + 0 \cdot i ) = \sin a (1) + i \cos a (0) = \sin a,$

$\cos (a + 0 \cdot i ) = \cos a (1) - i \sin a (0) = \cos a$
57 :kjmglakhngv:2015/06/18(木) 20:03:20.690 ID:K1KdSR2l
*つづく*
58 :kjmglakhngv:2015/06/24(水) 09:59:12.523 ID:VTLp01YQ
$\sin^2 (a + bi) + \cos^2 (a + bi)
= [ \sin a \frac {e^b + e^{-b}} {2} +
i \cos a \frac {e^b - e^{-b}} {2} ]^2 + [ \cos a \frac {e^b + e^{-b}} {2} - i \sin a \frac {e^b - e^{-b}} {2} ]^2$
59 :kjmglakhngv:2015/06/24(水) 10:07:29.187 ID:VTLp01YQ
$= \sin^2 a \frac {e^{2b} + 2 + e^{-2b}} {4} + 2i \sin a \cos a \frac {e^{2b} - e^{-2b}} {4} - \cos^2 a \frac {e^{2b} - 2 + e^{-2b}} {4}
+ \cos^2 a \frac {e^{2b} + 2 + e^{-2b}} {4} - 2i \sin a \cos a \frac {e^{2b} - e^{-2b}} {4} - \sin^2 a \frac {e^{2b} - 2 + e^{-2b}} {4}$

$= \sin^2 a + \cos^2 a = 1$
60 :kjmglakhngv:2015/06/26(金) 00:06:11.719 ID:mlF3Kn1F
$1 = \cos^2 \theta + \sin^2 \theta = (\cos \theta + i \sin \theta) (\cos \theta - i \sin \theta)$
61 :kjmglakhngv:2015/06/26(金) 00:12:33.252 ID:mlF3Kn1F

$ (\cos \theta \pm i \sin \theta)(\cos \phi \pm i \sin \phi)
= (\cos \theta cos \phi - \sin \theta \sin \phi) \pm i (\sin \theta \cos \phi + \cos \theta \sin \phi)
= \cos ( \theta + \phi) \pm i \sin (\theta + \phi)$
62 :kjmglakhngv:2015/06/26(金) 00:16:09.193 ID:mlF3Kn1F
$\theta = \phi$

$ (\cos \theta + i \sin \theta)^2 = \cos (2 \theta) + i \sin (2 \theta),$

$(\cos \theta - i \sin \theta)^2 = \cos (2 \theta) - i \sin (2 \theta)$
63 :kjmglakhngv:2015/06/26(金) 00:24:20.663 ID:mlF3Kn1F
$ (\cos \theta \pm i \sin \theta)^n = \cos (n \theta) \pm i \sin (n \theta)$ $\forall n \geq 1$
64 :kjmglakhngv:2015/06/28(日) 10:34:55.040 ID:dD7S0+6t
$r_1 = \sqrt[ 3 ]{ 2 + 2 \sqrt {-1} } + \sqrt[3]{2 - 2 \sqrt {-1}} \\\\
= \sqrt 2 [ \cos \frac {\pi}{12} + i \sin \frac {\pi}{12}] + \sqrt 2 [\cos (- \frac {\pi}{12}) + i \sin ( - \frac {\pi}{12})]\\\\
= \sqrt 2 [ \cos \frac {\pi}{12} + i \sin \frac {\pi}{12} + \cos \frac {\pi}{12} - i \sin \frac {\pi}{12}] \\\\
= \sqrt 2 [2 \cos \frac {\pi}{12}] = 2 \sqrt 2 \sqrt {\frac {1}{2} + \frac {\sqrt 3}{4}} = \sqrt {4 + 2 \sqrt3}$
65 :kjmglakhngv:2015/07/03(金) 19:54:33.564 ID:F52mUsoT
$r_2 = \sqrt[ 3 ]{ 2 + 2 \sqrt{-1} } + \sqrt[ 3 ]{ 2 - 2 \sqrt {-1} }\\\\
= \sqrt 2 [ \cos \frac {3 \pi }{4} + i \sin \frac {3 \pi }{4} ] + \sqrt 2 [ \cos (- \frac{3 \pi }{4}) + i \sin ( - \frac {3 \pi}{4})]\\\\
= \sqrt 2 [ 2 \cos\frac {3 \pi}{4} ] = -2$
66 :kjmglakhngv:2015/07/03(金) 20:00:49.162 ID:F52mUsoT
$r_3 = \sqrt[ 3 ]{ 2 + 2 \sqrt {-1} } + \sqrt[ 3 ]{ 2 - 2 \sqrt {-1} }\\\\
= \sqrt2 [\cos \frac {17 \pi}{12} + i \sin \frac {17 \pi}{12}] + \sqrt2 [ \cos (\frac {17 \pi}{12}) + i \sin ( \frac {17 \pi}{12})]\\\\
= \sqrt 2 [2 \cos \frac {17 \pi}{12}] = - \sqrt {4 - 2 \sqrt3}$
67 :kjmglakhngv:2015/07/05(日) 13:51:31.369 ID:CuXQBslB
$\displaystyle \sum_{ k = 1 }^{ \infty } 1/k$は発散している
68 :kjmglakhngv:2015/07/05(日) 13:53:41.193 ID:CuXQBslB
$\displaystyle \prod_{ p } \frac {1}{1 - \frac {1}{p}}$も発散している
69 :kjmglakhngv:2015/07/05(日) 13:59:23.242 ID:CuXQBslB
$M = \frac {2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot \cdots}{1 \cdot 2 \cdot 4 \cdot 6 \cdot 10 \cdot 12 \cdot \cdots}
= \frac {1}{\frac {1}{2} \times \frac {2}{3} \times \frac {4}{5} \times \frac {6}{7} \times \frac {10}{11} \times \cdots}$
70 :kjmglakhngv:2015/07/05(日) 14:01:19.057 ID:CuXQBslB
>>69の前に入れ忘れ


$M = \displaystyle \sum_{ k = 1 }^{ \infty } 1/k$
71 :kjmglakhngv:2015/07/05(日) 14:05:14.873 ID:CuXQBslB
$\ln M = - \ln (1/2) - \ln (2/3) - \ln (4/5) - \ln (6/7) - \ln (10/11) - \cdots\\\\
= - \ln (1 - 1/2) - \ln (1-1/3) - \ln (1 - 1/5) - \ln (1 - 1/7) - \ln (1 - 1/11) - \cdots$
72 :kjmglakhngv:2015/07/05(日) 14:07:07.592 ID:CuXQBslB
$\ln (1 - x ) = -x - \frac {x^2}{2} - \frac {x3}{3} - \frac {x^4}{4} - \cdots$
73 :kjmglakhngv:2015/07/05(日) 14:07:48.510 ID:CuXQBslB
*つづく*
74 :kjmglakhngv:2015/07/09(木) 01:35:55.829 ID:4ALEAqmq
このすれなんた
75 :kjmglakhngv:2015/07/10(金) 19:41:28.199 ID:qAqQNxrf
$\ln M = \frac {1}{2} + \frac {1}{2} (\frac {1}{2})^2 + \frac {1}{3} (\frac {1}{2})^3 + \frac {1}{4} ( \frac {1}{2})^4 + \frac {1}{5} ( \frac {1}{2})^5 + \cdots \\\\
+ \frac {1}{3} + \frac {1}{2} ( \frac {1}{3})^2 + \frac {1}{3} (\frac {1}{3})^3 + \frac {1}{4} ( \frac {1}{3})^4 + \frac {1}{5} (\frac {1}{3})^5 + \cdots \\\\
+ \frac {1}{5} + \frac {1}{2} ( \frac {1}{5})^2 + \frac {1}{3} ( \frac {1}{5})^3 + \frac {1}{4} ( \frac {1}{5})^4 + \frac {1}{5} ( \frac {1}{5})^5 + \cdots \\\\
+ \frac {1}{7} + \frac {1}{2} ( \frac {1}{7})^2 + \frac {1}{3} ( \frac {1}{7})^3 + \frac {1}4( \frac {1}{7} )^4 + \frac {1}{5} ( \frac {1}{7})^5 + \cdots \\\\
\vdots \qquad \vdots \qquad \vdots \qquad \vdots \qquad \vdots \qquad \vdots \qquad$
76 :kjmglakhngv:2015/07/10(金) 19:41:50.600 ID:qAqQNxrf
>>74
MathJaxスレ
77 :kjmglakhngv:2015/07/10(金) 19:51:17.502 ID:qAqQNxrf
$ \ln M = [ \frac {1}{2} + \frac {1}{3} + \frac {1}{5} + \frac {1}{7} + \frac {1}{11} + \frac {1}{13} + \cdots ]\\\\
+ \frac {1}{2} [ ( \frac {1}{2})^2 + ( \frac {1}{3})^2 + ( \frac {1}{5})^2 + ( \frac {1}{7})^2 + ( \frac {1}{11})^2 + \cdots]\\\\
+ \frac {1}{3} [ ( \frac {1}{2})^3 + ( \frac {1}{3})^3 + ( \frac {1}{5})^3 + ( \frac {1}{7})^3 + ( \frac {1}{11})^3 + \cdots]\\\\
+ \frac {1}{4} [ ( \frac {1}{2})^4 + ( \frac {1}{3})^4 + ( \frac {1}{5})^4 + ( \frac {1}{7})^4 + ( \frac {1}{11})^4 + \cdots]\\\\
+ \frac {1}{5} [ ( \frac {1}{2})^5 + ( \frac {1}{3})^5 + ( \frac {1}{5})^5 + ( \frac {1}{7})^5 + ( \frac {1}{11})^5 + \cdots]\\\\
\vdots \qquad \vdots \qquad \vdots \qquad \vdots \qquad \vdots \qquad \vdots \qquad$
78 :kjmglakhngv:2015/07/10(金) 19:52:52.026 ID:qAqQNxrf
$ \ln M = A + \frac {1}{2} B + \frac {1}{3} C + \frac {1}{4} D + \frac {1}{5} E + \cdots$
79 :kjmglakhngv:2015/07/10(金) 19:56:49.490 ID:qAqQNxrf
$ A = \displaystyle \sum_{ p } \frac {1}{p}, \quad B = \displaystyle \sum_{ p} \frac {1}{p^2}, \quad C = \displaystyle \sum_{p} \frac {1}{p^3}, \cdots$
80 :kjmglakhngv:2015/07/19(日) 19:45:23.020 ID:kM3Cz2Ko
$n \geq 2$に対して$\displaystyle \sum_{ k = 2 }^{ \infty } \frac {1}{k^n} \geq\frac {1}{n-1}$が成り立つ
81 :kjmglakhngv:2015/07/19(日) 19:53:55.776 ID:kM3Cz2Ko
http://i.imgur.com/vvgSBEO.png
82 :kjmglakhngv:2015/07/19(日) 19:59:28.727 ID:kM3Cz2Ko

$\displaystyle \sum_{ k = 2 }^{ \infty } \frac {1}{k^n}$網目の面積$\geq\displaystyle \int_{ 1 }^{ \infty } \frac {1}{x^n} dx = \frac {1}{n-1}$
83 :kjmglakhngv:2015/07/19(日) 20:01:25.279 ID:kM3Cz2Ko
$\frac {1}{2} B + \frac {1}{3} C + \frac {1}{4} D + \frac {1}{5} E + \cdots$は有限である
84 :kjmglakhngv:2015/07/19(日) 20:09:49.292 ID:kM3Cz2Ko
うわわ
>>80
$n \geq 2$に対して$\displaystyle \sum_{ k = 2 }^{ \infty } \frac {1}{k^n} \leq\frac {1}{n-1}$が成り立つ
>>82
$\displaystyle \sum_{ k = 2 }^{ \infty } \frac {1}{k^n}$網目の面積$\leq\displaystyle \int_{ 1 }^{ \infty } \frac {1}{x^n} dx = \frac {1}{n-1}$
85 :kjmglakhngv:2015/07/19(日) 20:10:23.408 ID:kM3Cz2Ko
コピペってこわいな
86 :kjmglakhngv:2015/07/19(日) 20:21:16.029 ID:kM3Cz2Ko
$\frac {1}{2} B + \frac {1}{3} C + \frac {1}{4} D + \cdots = \frac {1}{2} \displaystyle \sum_{ p } \frac {1} {p^2} + \frac {1}{3} \displaystyle \sum_{ p } \frac {1}{p^3} + \frac {1}{4} \displaystyle \sum_{ p } \frac {1}{p^4} + \cdots\\\\
\leq \frac {1}{2} \displaystyle \sum_{ k = 2 }^{ \infty } \frac {1}{k^2} + \frac {1}{3} \displaystyle \sum_{ k = 2 }^{ \infty } \frac {1}{k^3} + \frac {1}{4} \displaystyle \sum_{ k = 2 }^{ \infty } \frac {1}{k^4} + \cdots\\\\
\leq \frac {1}{2} (1) + \frac {1}{3} (\frac {1}{2}) + \frac {1}{4} (\frac {1}{3}) + \frac {1}{5} (\frac {1}{4}) + \cdots\\\\
\leq 1 + \frac {1}{2} ( \frac {1}{2}) + \frac {1}{3} ( \frac {1}{3}) + \frac {1}{4} (\frac {1}{4}) + \cdots = \displaystyle \sum_{ k = 1 }^{ \infty } \frac {1}{k^2} = \frac {\pi^2} {6} < \infty$
87 :kjmglakhngv:2015/07/19(日) 20:24:16.999 ID:kM3Cz2Ko
$\sum_p 1/p$は発散する
88 :kjmglakhngv:2015/07/19(日) 20:26:28.094 ID:kM3Cz2Ko
$\ln M =A + \frac {1}{2} B + \frac {1}{3} C + \frac {1}{4} D + \frac {1}{5} E \cdots$
89 :kjmglakhngv:2015/07/19(日) 20:32:53.147 ID:kM3Cz2Ko
$M = e^{\ln M} = e^{A + \frac {1}{2} B + \frac {1}{3} C + \frac {1}{4} D + \cdots} = e^A \times e^{\frac {1}{2} B + \frac {1}{3} C + \frac {1}{4} D + \cdots}$
90 :kjmglakhngv:2015/07/19(日) 20:34:21.085 ID:kM3Cz2Ko
$A = \ln (e^A) = \ln (\infty) = \infty$
91 :kjmglakhngv:2015/07/19(日) 20:35:55.664 ID:kM3Cz2Ko
$\frac {1}{2} + \frac {1}{3} + \frac {1}{5} + \frac {1}{7} + \frac {1}{11} + \frac {1}{13} + \cdots = A = \infty$
92 :test:2015/10/10(土) 01:51:15.151 ID:bfD8KX9A
x &=& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\newline
% This is a comment, it is not shown in the final output.
% The following shows a little of the typesetting power of LaTeX
\begin{eqnarray}
E &=& mc^2 \\
m &=& \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}
\end{eqnarray}
93 :kjmglakhngv:2015/10/10(土) 08:02:17.985 ID:bfD8KX9A
$$E=mc^2$$
94 :kjmglakhngv:2015/10/13(火) 17:39:27.323 ID:nCWzMCHd

\begin{eqnarray}
(x+2)(x+3) &=& x^2+(2+3)x+6 \nonumber \\
&=& x^2+5x+6
\end{eqnarray}
95 :kjmglakhngv:2015/10/13(火) 17:41:34.094 ID:nCWzMCHd
$$
(x+2)(x+3)=x2+(2+3)x+6\\
=x2+5x+6
$$
96 :kjmglakhngv:2015/10/25(日) 20:00:20.929 ID:ziV9egxT
$\sqrt{11}$
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