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$$\begin{align}z^n=&C_{n}+S_nz=-(S_{n-1})l+(S_n)z\\
&\begin{cases}
S_0&=0\\
S_1&=1\\
S_n&=-(S_{n-2})l+(S_{n-1})r\\
l&=z\cdot\bar{z}=|z|^2\\
r&=z+\bar{z}=2\mathrm{Re}(z)\\
\end{cases}\\[8pt]
&\begin{pmatrix}C_{n}&C_{n+1}\\S_{n}&S_{n+1}\end{pmatrix}=\begin{pmatrix}0&-l\\1&r\end{pmatrix}^n\\[8pt]
&\begin{pmatrix}S_{n+1}\\S_{n}\end{pmatrix}=\begin{pmatrix}r&-l\\1&0\end{pmatrix}^n\begin{pmatrix}1\\0\end{pmatrix}\\[8pt]
&C_n=-(S_{n-1})l\\
&S_n=\displaystyle\frac{\left(r+\sqrt{r^2-4l}\right)^n-\left(r-\sqrt{r^2-4l}\right)^n}{2^n\sqrt{r^2-4l}}\\
&S_n=\displaystyle\sum_{k=0}^{\lfloor (n-1)/2\rfloor}\binom{n-k-1}{k}r^{n-2k-1}l^{k}
\end{align}$$
$$\\[8pt]$$
âé¢é£èšäºã ãåæ§ã«ã¯èŠãã¡ããã¡!? ãªããã®ãããã®çµååâ¡ ã
$$\begin{align} &A+(B\times C)=A\times(B+C)\\\\ &\begin{cases} A=(a-1)b+1\\ B=a((a-1)b+1)\\ C=ab+1\\ \end{cases} \end{align}$$
$$\\[8pt]$$
âé¢é£èšäºã ãã§ã«ããŒã®æçµå®çãšãã£ããããæ°ãšãªã¥ã«æ°ãå æ°å解 ã
$n\in\mathbb{Z}^+$ ãšããŸãã$n=0$ ã®ãšã㯠$\begin{cases}x^0-y^0=0\\x^0+y^0=2\end{cases}$
$$\begin{align}
x^n-y^n=&\displaystyle\prod_{m=0}^{n-1}\sqrt{x^2+y^2+(-1)^n\cdotp2xy\cos\left(\frac{2m}n\pi\right)}\\
x^n+y^n
=&\displaystyle\prod_{m=0}^{n-1}\sqrt{x^2+y^2+(-1)^n\cdotp2xy\cos\left(\frac{2m+1}n\pi\right)}\\
\end{align}$$
$$\begin{cases}
x^{2m}-y^{2m}&=\displaystyle(x+y)(x-y)\prod_{k=1}^{m-1}\left[x^2+y^2\pm2xy\cos\left(\frac{k}{m}\pi\right)\right]\\
x^{2m+1}-y^{2m+1}
&=\displaystyle\quad\quad\quad(x-y)\prod_{k=1}^{m}\left[x^2+y^2+2xy\cos\left(\frac{2k-1}{2m+1}\pi\right)\right]\\
&=\displaystyle\quad\quad\quad(x-y)\prod_{k=1}^{m}\left[x^2+y^2-2xy\cos\left(\frac{2k}{2m+1}\pi\right)\right]\\
\end{cases}$$
$$\begin{cases}
x^{2m}+y^{2m}&=\displaystyle\quad\quad\quad\quad\quad~~~\prod_{k=1}^{m}\left[x^2+y^2\pm2xy\cos\left(\frac{2k-1}{2m}\pi\right)\right]\\
x^{2m+1}+y^{2m+1}&=\displaystyle\quad\quad~~~(x+y)\prod_{k=1}^{m}\left[x^2+y^2+2xy\cos\left(\frac{2k}{2m+1}\pi\right)\right]\\
&=\displaystyle\quad\quad~~~(x+y)\prod_{k=1}^{m}\left[x^2+y^2-2xy\cos\left(\frac{2k-1}{2m+1}\pi\right)\right]\\
\end{cases}$$
$$\\[8pt]$$
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$$\begin{align}
n=2m\text{ ã®æ}\\
\sum_{k=0}^{n-1}x^{k}
&=(x+1)\prod_{k=1}^{m-1}\left[x^2\pm2\cos\left(\frac{2k}{n}\pi\right)x+1\right]\\
&=(x+1)\prod_{k=1}^{m-1}\left[x^2\pm2\cos\left(\frac{k}{m}\pi\right)x+1\right]\\
&\begin{cases}
n=2&\Phi_n(x)&=x+1\\
n\geqq4&\Phi_n(x)
&=\displaystyle\prod_{k\perp m}\left[x^2-2\cos\left(\frac{k}{m}\pi\right)x+1\right]\\
&&=\displaystyle\prod_{k\perp m}\left[x^2+2\cos\left(\frac{m-k}{m}\pi\right)x+1\right]\\
\end{cases}
\\
n=2m+1\text{ ã®æ}\\
\sum_{k=0}^{n-1}x^{k}
&=\prod_{k=1}^{m}\left[x^2-2\cos\left(\frac{2k}{n}\pi\right)x+1\right]\\
&=\prod_{k=1}^{m}\left[x^2+2\cos\left(\frac{2k-1}{n}\pi\right)x+1\right]\\
\
&\begin{cases}
n=1&\Phi_n(x)&=x-1\\
n\geqq3&\Phi_n(x)
&=\displaystyle\prod_{k\perp n}\left[x^2+2\cos\left(\frac{k}{n}\pi\right)x+1\right]\\
&&=\displaystyle\prod_{k\perp n}\left[x^2-2\cos\left(\frac{n-k}{n}\pi\right)x+1\right]\\
\end{cases}\end{align}$$
$\begin{align} \Phi_1&=x-1\\ \Phi_2&=x+1\\ \Phi_3&=x^2+x+1\\ &\color{#33f}=\textstyle(x^2+2\cos(\frac13\pi)x+1)\\ &\color{#33f}=\textstyle(x^2-2\cos(\frac23\pi)x+1)\\ \Phi_4&=x^2+1\\ &\color{#33f}=\textstyle(x^2\pm2\cos(\frac12\pi)x+1)\\ \Phi_5&=x^4+x^3+x^2+x+1\\ &\color{#33f}=\textstyle(x^2+2\cos(\frac15\pi)x+1)(x^2+2\cos(\frac35\pi)x+1)\\ &\color{#33f}=\textstyle(x^2-2\cos(\frac25\pi)x+1)(x^2-2\cos(\frac45\pi)x+1)\\ \Phi_6&=x^2-x+1\\ &\color{#33f}=\textstyle(x^2-2\cos(\frac13\pi)x+1)\\ &\color{#33f}=\textstyle(x^2+2\cos(\frac23\pi)x+1)\\ \Phi_7&=x^6+x^5+x^4+x^3+x^2+x+1\\ &\color{#33f}=\textstyle(x^2+2\cos(\frac17\pi)x+1)(x^2+2\cos(\frac37\pi)x+1)(x^2+2\cos(\frac57\pi)x+1)\\ &\color{#33f}=\textstyle(x^2-2\cos(\frac27\pi)x+1)(x^2-2\cos(\frac47\pi)x+1)(x^2-2\cos(\frac67\pi)x+1)\\ \Phi_8&=x^4+1\\ &\color{#33f}=\textstyle(x^2\pm2\cos(\frac14\pi)x+1)(x^2\pm2\cos(\frac34\pi)x+1)\\ \Phi_9&=x^6+x^3+1\\ &\color{#33f}=\textstyle(x^2+2\cos(\frac19\pi)x+1)(x^2+2\cos(\frac59\pi)x+1)(x^2+2\cos(\frac79\pi)x+1)\\ &\color{#33f}=\textstyle(x^2-2\cos(\frac29\pi)x+1)(x^2-2\cos(\frac49\pi)x+1)(x^2-2\cos(\frac89\pi)x+1)\\ \Phi_{10}&=x^4-x^3+x^2-x+1\\ &\color{#33f}=\textstyle(x^2-2\cos(\frac15\pi)x+1)(x^2-2\cos(\frac35\pi)x+1)\\ &\color{#33f}=\textstyle(x^2+2\cos(\frac25\pi)x+1)(x^2+2\cos(\frac45\pi)x+1)\\ &\vdots\\ \end{align}$
$$\\[8pt]$$
âé¢é£èšäºã ãã§ã«ããŒã®æçµå®çãšãã£ããããæ°ãšãªã¥ã«æ°ãå æ°å解 ã
$n\in\mathbb{Z}^+_0$ ãšããŸãã$n=0$ ã®ãšã㯠$\quad\begin{cases} \quad F_n=\frac{\phi^0-\overline{\phi^0}}{\phi-\overline{\phi}}=\frac{1-1}{\sqrt5}=0\\ \quad L_n=\frac{\phi^0+\overline{\phi^0}}{\phi+\overline{\phi}}=\frac{1+1}1=2 \end{cases}$
$\quad\displaystyle F_n=\prod_{k=1}^{\lceil\frac{n-2}2\rceil}\left[3+2\cos\left(\frac{2k}{n}\pi\right)\right]$
$\quad\begin{cases}
F_{2m}&\displaystyle=\prod_{k=1}^{m-1}\left[3\pm2\cos\left(\frac{k}{m}\pi\right)\right]&\cdots~~n=2m\gt0\\
F_{2m+1}&\displaystyle=\prod_{k=1}^{m}\left[3+2\cos\left(\frac{2k}{2m+1}\pi\right)\right]&\cdots~~n=2m+1\gt0\\
\end{cases}$
$\quad\displaystyle L_n=\prod_{k=0}^{\lfloor\frac{n-2}2\rfloor}\left[3+2\cos\left(\frac{2k+1}{n}\pi\right)\right]$
$\quad\begin{cases}
L_{2m}&\displaystyle=\prod_{k=1}^{m}\left[3\pm2\cos\left(\frac{2k-1}{2m}\pi\right)\right]&\cdots~~n=2m\gt0\\
L_{2m+1}&\displaystyle=\prod_{k=1}^{m}\left[3+2\cos\left(\frac{2k-1}{2m+1}\pi\right)\right]&\cdots~~n=2m+1\gt0\\
\end{cases}$
$$\\[8pt]$$
âé¢é£èšäºã äžè§é¢æ°ã®ïœåè§ã®å ¬åŒãšåæ²ç·é¢æ°ãå æ°å解 ã
$m,~n\in\mathbb{Z}_0^+$ ãšããŸãã$n=0$ ã®ãšã㯠$\begin{cases}\sin(n\theta)=0\\\cos(n\theta)=1\\\tan(n\theta)=0\end{cases}$
$\quad\begin{cases} \sin(2m\theta)&=\displaystyle2^{2m-1}\cos\theta\sin\theta\displaystyle\prod_{k=1}^{m-1}\left[\sin\left(\frac{k}{2m}\pi-\theta\right)\sin\left(\frac{k}{2m}\pi+\theta\right)\right]\\ \cos(2m\theta)&=\displaystyle2^{2m-1}\displaystyle\prod_{k=1}^{m}\left[\cos\left(\frac{2k-1}{4m}\pi-\theta\right)\cos\left(\frac{2k-1}{4m}\pi+\theta\right)\right]\\ \end{cases}\quad\cdots~~n=2m\gt0$
$\quad\begin{cases} \sin((2m+1)\theta)&=\displaystyle2^{2m}\sin\theta\prod_{k=1}^{m}\left[\sin\left(\frac{k}{2m+1}\pi-\theta\right)\sin\left(\frac{k}{2m+1}\pi+\theta\right)\right]\\ \cos((2m+1)\theta)&=\displaystyle2^{2m}\cos\theta\prod_{k=1}^{m}\left[\cos\left(\frac{k}{2m+1}\pi-\theta\right)\cos\left(\frac{k}{2m+1}\pi+\theta\right)\right]\\ \tan((2m+1)\theta)&=\displaystyle\quad~\tan\theta\prod_{k=1}^{m}\left[\tan\left(\frac{k}{2m+1}\pi-\theta\right)\tan\left(\frac{k}{2m+1}\pi+\theta\right)\right]\\ \end{cases}\quad\cdots~~n=2m+1\gt0$
$$\\[8pt]$$
âé¢é£èšäºã äžè§é¢æ°ã®ïœåè§ã®å ¬åŒãšåæ²ç·é¢æ°ãå æ°å解 ã
$m,~n\in\mathbb{Z}_0^+$ ãšããŸãã$n=0$ ã®ãšã㯠$\begin{cases}\sinh(nx)=0\\\cosh(nx)=1\end{cases}$
$\quad\begin{cases}
\sinh(2mx)&=\displaystyle2^m\cosh x\sinh x\prod_{k=1}^{m-1}\left[2\sinh^2x+1\pm\cos\left(\frac{k}{m}\pi\right)\right]\\
\cosh(2mx)&=\displaystyle2^{m-1}\prod_{k=1}^{m}\left[2\cosh^2x-1-\cos\left(\frac{2k-1}{2m}\pi\right)\right]\\
\end{cases}$
$\quad\begin{cases}
\sinh((2m+1)x)&=\displaystyle\quad2^m\sinh x\prod_{k=1}^{m}\left[2\sinh^2x+1-\cos\left(\frac{2k}{2m+1}\pi\right)\right]\\
\cosh((2m+1)x)&=\displaystyle\quad2^m\cosh x\prod_{k=1}^{m}\left[2\cosh^2x-1-\cos\left(\frac{2k-1}{2m+1}\pi\right)\right]\\
\end{cases}$
$$\\[8pt]$$
$+1$ ãš $e^{(\pi-\theta)i}$ ãåºåºã®å
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$$\begin{align} &(a+be^{(\pi-\theta)i})(a+be^{-(\pi-\theta)i}))=a^2+b^2-2ab\cos\theta\\ \\ &\textstyleç¹ã«\theta=\frac\pi2~ã®æã(a+bi)(a-bi)=a^2+b^2\\ \end{align}$$
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±åœ¹è€çŽ æ°å士ã®åãšç©ã¯å¹ŸäœåŠçã«$\begin{cases}
z+\overline{z}=2\operatorname{Re}z\\
z\cdotp\overline{z}=z^2\\
\end{cases}$ïŒ
$$\\[8pt]$$
âé¢é£èšäºã
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解説] 13ã®8ä¹ãå¹³æ¹æ°2ã€ã®åã§4éãã«è¡šãã
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ä»»æã®å¶æ° $2ab$ ãïŒèŸºã«ãã€ãã¿ãŽã©ã¹äžè§åœ¢ã®äžèŸºæ¯
$$
|(a+bi)^2|=|(a^2-b^2)+(2ab)i|=a^2+b^2\\[4pt]
\quad\rightarrow~2ab~:~a^2-b^2~:~a^2+b^2\\
$$
ä»»æã®å¥æ° $(2a+1)(2b+1)$ ãïŒèŸºã«ãã€ãã¿ãŽã©ã¹äžè§åœ¢ã®äžèŸºæ¯
$$\begin{align}
&|(2a+1)(2b+1)+[2a(a+1)-2b(b+1)]i|=2a(a+1)+2b(b+1)+1\\[4pt]
&\rightarrow~(2a+1)(2b+1)~:~2a(a+1)-2b(b+1)~:~2a(a+1)+2b(b+1)+1\\
&\rightarrow~(a+b+1)^2-(a-b)^2~:~2(a-b)(a+b+1)~:~(a+b+1)^2+(a-b)^2
\end{align}\\
$$
çŽåŸ $R=2mn$ ã®åã«å€æ¥ãããã¿ãŽã©ã¹äžè§åœ¢ã®äžèŸºæ¯
$R+m^2:R+2n^2:R+m^2+2n^2$
$$\\[8pt]$$
$$
|(1+i)(a+bi)(a+bi)|^2=|(1+i)(a+bi)(a-bi)|^2\\
\quad\rightarrow~2(a^2+b^2)^2=(a^2-b^2+2ab)^2+(a^2-b^2-2ab)^2\\
$$
$$\begin{cases}
s&=\operatorname{Re}(a+bi)^2&=a^2-b^2\\
t&=\operatorname{Im}(a+bi)^2&=2ab\\
\end{cases}\quad
\begin{cases}
x&=\operatorname{Re}(s+ti)^2&=s^2-t^2\\
y&=\operatorname{Im}(s+ti)^2&=2st\\
\end{cases}\\
\rightarrow~2(a^2+b^2)^4=(x+y)^2+(x-y)^2
$$
$$\begin{align} &(a+bi)^2(c+di)^2\\ =&|(a-bi)(c+di)(a+bi)(c-di)|=|[(ac+bd)+(ad-bc)i][(ac+bd)-(ad-bc)i]|=(ac+bd)^2+(ad-bc)^2\\ =&|(a+bi)(a+bi)(c+di)(c+di)|=|[(ac-bd)+(ad+bc)i][(ac-bd)+(ad+bc)i]|=(ac-bd)^2+(ad+bc)^2\\ &\rightarrow~(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2\\ \end{align}$$
$$[(a^2+b^2)a]^2+[(a^2+b^2)b]^2=[(a^2-3b^2)a]^2+[(3a^2-b^2)b]^2$$
$$(a^2-b^2+ac)^2+(2ab+bc)^2=(a^2+b^2+ac)^2+(bc)^2$$
$$\begin{align} &(2a+1)(2b+1)(2c+1)(2d+1)\\ &=[(a+b+1)^2-(a-b)^2][(c+d+1)^2-(c-d)^2]\\ &=[(a+b+1)+(2a+1)c+(2b+1)d]^2-[(a-b)+(2a+1)c-(2b+1)d]^2\\ &=[2(ac+bd)+(a+b)+(c+d)+1]^2-[2(ac-bd)+(a-b)+(c-d)]^2 \end{align}$$
$$\\[8pt]$$
âé¢é£èšäºã å¹³æ¹åãšç«æ¹åã®æçåŒ ã
$$\begin{align} &a^2+c^2=b^2+d^2\\ &(am+bn)^2+(cm+dn)^2\\ =&(an+bm)^2+(cn+dm)^2 \end{align}$$
$$\\[8pt]$$
âé¢é£èšäºã å¹³æ¹åãšç«æ¹åã®æçåŒ ã
$$\begin{align} &a^3+c^3=b^3+d^3\\ &\left(am^2+bn^2\pm(c+d)\sqrt{\frac{c-d}{b-a}}mn\right)^3+\left(cm^2+dn^2\pm(a+b)\sqrt{\frac{a-b}{d-c}}mn\right)^3\\ =&\left(an^2+bm^2\pm(c+d)\sqrt{\frac{c-d}{b-a}}nm\right)^3+\left(cn^2+dm^2\pm(a+b)\sqrt{\frac{a-b}{d-c}}nm\right)^3 \end{align}$$
$$\\[8pt]$$
$$\begin{align}
&(2n+1)^2+[2n(n+1)]^2=[2n(n+1)+1]^2\\[8pt]
&(2m-1)(2n+1)^2+[\underbrace{2n(n+1)-(m-1)}_{n(2n+1)+(n-m)+1}]^2=[\underbrace{2n(n+1)+m}_{n(2n+1)+(n+m)}]^2\\
&[n(2n+1)]^2+\underbrace{\sum_{m=1}^n\left[n(2n+1)+(n-m)+1\right]^2}_{n(2n+1)+1ïœn(2n+1)+n~ãŸã§ã®å¹³æ¹å}=\underbrace{\sum_{m=1}^n\left[n(2n+1)+(n+m)\right]^2}_{n(2n+1)+n+1ïœn(2n+1)+2n~ãŸã§ã®å¹³æ¹å}
\end{align}$$
$$\\[8pt]$$
å¹³æ¹åã®éæ¹é£ãæ¢çŽ¢ããã®ã«äœ¿ãããã§ãã
$$\begin{align}
&A^2+B^2=C^2+D^2=E^2+F^2=G^2+H^2=I^2+I^2\\
&A^2+I^2+B^2=C^2+I^2+D^2=E^2+I^2+F^2=G^2+I^2+H^2=3I^2\\\\
&\begin{cases}
A=(a^2+b^2)(c^2-d^2-2cd)\\
B=(a^2+b^2)(c^2-d^2+2cd)\\
C=(a^2-b^2-2ab)(c^2+d^2)\\
D=(a^2-b^2+2ab)(c^2+d^2)\\
E=(ac+bd)^2-(ad-bc)^2-2(ac+bd)(ad-bc)\\
F=(ac+bd)^2-(ad-bc)^2+2(ac+bd)(ad-bc)\\
G=(ac-bd)^2-(ad+bc)^2-2(ac-bd)(ad+bc)\\
H=(ac-bd)^2-(ad+bc)^2+2(ac-bd)(ad+bc)\\
I=(a^2+b^2)(c^2+d^2)
\end{cases}
\end{align}$$
$$\\[8pt]$$
$$\begin{align} &(a^2+b^2-2ab\cos\theta)(c^2+d^2-2cd\cos\theta)\\ =&(ac-bd)^2+(ad+bc-2bd\cos\theta)^2-2(ac-bd)(ad+bc-2bd\cos\theta)\cos\theta\\ =&(ad-bc)^2+(ac+bd-2bc\cos\theta)^2-2(ad-bc)(ac+bd-2bc\cos\theta)\cos\theta\\ \\ &(a^2+b^2-2ab\cos\theta)^2\\ =&(a^2-b^2)^2+(2ab-2b^2\cos\theta)^2-2(a^2-b^2)(2ab-2b^2\cos\theta)\cos\theta\\ \\ &(a^2+b^2+abx)(c^2+d^2+cdx)\\ =&(ac-bd)^2+(ad+bc+bdx)^2+(ac-bd)(ad+bc+bdx)x\\ =&(ad-bc)^2+(ac+bd+bcx)^2+(ad-bc)(ac+bd+bcx)x\\ \\ &(a^2+b^2+abx)^2\\ =&(a^2-b^2)^2+(2ab+b^2x)^2+(a^2-b^2)(2ab+b^2x)x\\ \\ &(a^2+b^2)^2\\ =&(a^2-b^2)^2+(2ab)^2\\ =&|(a^2-b^2)+(2ab)i|^2\\ =&|(a+bi)^2|^2\\ =&|(a+bi)(a-bi)|^2 \end{align}$$
$$\\[8pt]$$
$$\begin{align} &(a^2-b^2+ac)^2+(2ab+bc-2b^2\cos\theta)^2-2(a^2-b^2+ac)(2ab+bc-2b^2\cos\theta)\cos\theta\\ =&(a^2+b^2+ac-2ab\cos\theta)^2+(bc)^2-2(a^2+b^2+ac-2ab\cos\theta)(bc)\cos\theta \end{align}$$
$$\\[8pt]$$
âé¢é£èšäºã ïœä¹åã®å ¬åŒãäœãè£ã¯ã¶ ã
$m=1$ ã«åºå®ãããšäžåŒãåžžã« $1^n=1$ ãšããæçåŒã«ãªãããã$n=0$ ããç®çã®æ¬¡æ°ãŸã§é 次代å
¥ããŠãããšæŽå²çãã«ããŒã€æ°ïŒé¢ã»ãã«ããŒã€æ°ïŒ $\hat B_k$ ãæ±ãŸããŸãã
$$\displaystyle\sum_{k=1}^mk^n=\frac1{n+1}\sum_{k=0}^n\pmatrix{n+1\\k}\hat B_km^{n+1-k}$$
$$\\[8pt]$$
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$$\begin{align}
\arctan\frac1p=\arctan\frac{q}{pq\pm r}\pm\arctan\frac{r}{(pq\pm r)p+q}\\[8pt]
\arctan\frac{q}{pq\pm r}=\arctan\frac1p\mp\arctan\frac{r}{(pq\pm r)p+q}
\end{align}$$
$p=F_{2m}$ã$q=1$ã$r=F_{2m-1}$ ã®ãšã
$$\quad\arctan\frac1{F_{2m}}=\arctan\frac1{F_{2m+1}}+\arctan\frac1{F_{2m+2}}$$
$q=1$ã$r=1$ ã®ãšã
$$\quad\arctan\frac1p=\arctan\frac{1}{p+1}+\arctan\frac{1}{p^2+p+1}$$
$p=1$ ã®ãšã
$$\quad\frac{\pi}4=\arctan\frac q{q\pm r}\pm\arctan\frac{r}{2q\pm r}$$
$$\\[8pt]$$
âé¢é£èšäºã ãåæã âçtanå解ã§å°ãååšçã®èå³æ·±ãçåŒâ ãå¯èŠåã ã
$$\begin{align}\frac\pi2 =&\underbrace{\sum_{n=0}^\infty\arctan\frac1{L_{2n}}}_{\arctan\phi}+\underbrace{\sum_{n=0} ^\infty\arctan\frac1{F_{4n+3}}}_{\arctan\frac1\phi}\\ =&\underbrace{\sum_{n=0}^\infty\arctan\frac1{F_{4n+1}}}_{\arctan\phi}+\underbrace{\sum_{n=0}^\infty\arctan\frac1{F_{4n+3}}}_{\arctan\frac1\phi}\\ =&\sum_{n=0}^\infty\arctan\frac1{F_{2n+1}}\\ =&\sum_{n=0}^\infty\arctan\frac1{n^2+n+1}\\ \end{align}$$
ã$\begin{cases} \pi~ã¯ååšç\\ \phi~ã¯é»éæ°~\frac{1+\sqrt5}2\\ F_m~ã¯~m~çªç®ã®ãã£ããããæ°\\ L_m~ã¯~m~çªç®ã®ãªã¥ã«æ° \end{cases}$
$$\\[8pt]$$
äŸãã° $Z_\theta=1+i$ã$Z=5+i$ã$N=4$ ãšããã°ããã³ã®å
¬åŒãå°åºã§ããŸãã
$$\begin{align}
&\mathrm{Arg}(Z_\theta)&&=N\arctan\frac{\mathrm{Im}(Z)}{\mathrm{Re}(Z)}+\arctan\frac{\mathrm{Re}(Z^N)\sin\theta-\mathrm{Im}(Z^N)\cos\theta}{\mathrm{Re}(Z^N)\cos\theta+\mathrm{Im}(Z^N)\sin\theta}\\
&\mathrm{Arg}(1+i)&=\frac{\pi}4&=N\arctan\frac{\mathrm{Im}(Z)}{\mathrm{Re}(Z)}+\arctan\frac{\mathrm{Re}(Z^N)-\mathrm{Im}(Z^N)}{\mathrm{Re}(Z^N)+\mathrm{Im}(Z^N)}\\
&\mathrm{Arg}(i)&=\frac{\pi}2&=N\arctan\frac{\mathrm{Im}(Z)}{\mathrm{Re}(Z)}+\arctan\frac{\mathrm{Re}(Z^N)}{\mathrm{Im}(Z^N)}\\\\
\end{align}$$
$\quad\quad\begin{cases}
Z,\ Z_\theta\in\mathbb{C}\\
N\in\mathbb{Z}\\
-\frac\pi2\lt\mathrm{Arg}(Z_\theta)=\theta\leqq\frac\pi2\\
0\leqq\mathrm{Arg}(Z^N)\lt\frac\pi2
\end{cases}$
$$\\[8pt]$$
$$8\lim_{n\to\infty}\sum_{k=1}^n\frac{n\|n(2k-1)+(n^2-k(k-1))i\|_p}{(n^2+k^2)(n^2+(k-1)^2)}$$
$p=2$ ã®ãšã
\begin{align}
&8\lim_{n\to\infty}\sum_{k=1}^n\frac{n\sqrt{n^2(2k-1)^2+(n^2-k(k-1))^2}}{(n^2+k^2)(n^2+(k-1)^2)}\\
=&8\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{\sqrt{(n^2+k^2)(n^2+(k-1)^2)}}\\
~~~~\pi=&4\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{\sqrt{(n^2+k^2)(n^2+(k-1)^2)}}\\
=&4\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{n^2+(k+\lambda)^2}~\cdots~\lambda~\text{ã¯ä»»æã®å®æ°(è€çŽ æ°ã¯æªç¢ºèª)}\\
\LARGE(\normalsize=&4\int_0^1\frac1{1+x^2}dx=4\left[\arctan x\right]_0^1\LARGE)\normalsize
\end{align}
$$\\[8pt]$$
$$\phi^n=\displaystyle\sum_{k=2}^\infty\phi^{n-k}$$
$$\\[8pt]$$
$\omega=\frac{-1\pm\sqrt3i}2$
ã»$ax^3+bx^2+cx+d=0$ ã®è§£ã®å
¬åŒ
$\quad\begin{cases}
B=&2b^3+9a(3ad-bc)\\
C=&(b^2-3ac)^3\\
\end{cases}$
$$\quad\quad\quad\begin{align}
x=&\frac{-b+\omega\sqrt[3]{\frac{-B+\sqrt{B^2-4C}}2}+\omega^{-1}\sqrt[3]{\frac{-B-\sqrt{B^2-4C}}2}}{3a}\\
\end{align}$$
$\quad\begin{cases}
B=&b^3+\frac{9a(3ad-bc)}2\\
C=&(b^2-3ac)^3\\
\end{cases}$
$$\quad\quad\begin{align}
\quad x=&\frac{-b+\omega\sqrt[3]{-B+\sqrt{B^2-C}}+\omega^{-1}\sqrt[3]{-B-\sqrt{B^2-C}}}{3a}\\
\end{align}$$
$ $
ã»$x^3+bx^2+cx+d=0$ ã®æŽæ°è§£ã®å
¬åŒ
$$\quad\quad\{a,b,c,d,\alpha,\beta\}\in\mathbb{Z}$$
$\quad\begin{cases}
\alpha\beta=2b^3+9(3d-bc)\\
\beta^2-\alpha=3(b^2-3c)\\
\end{cases}$
$$\quad\quad\quad\begin{align}
x=\begin{cases}
-\frac{b+\beta}3\\
\frac{-2b+\beta\pm\sqrt{\beta^2-4\alpha}}6
\end{cases}
\end{align}$$
$ $
ã»$x^3+3bx^2+cx+d=0$ ã®è§£ã®å
¬åŒ
$\quad\begin{cases}
B=2b^3-bc+d\\
C=\left(b^2-\frac c3\right)^3\\
\end{cases}$
$$\quad\quad\quad\begin{align}
x=&-b+\omega\sqrt[3]{\frac{-B+\sqrt{B^2-4C}}2}+\omega^{-1}\sqrt[3]{\frac{-B-\sqrt{B^2-4C}}2}\\
\end{align}$$
$\quad\begin{cases}
B&=b^3-\frac{bc-d}2\\
C&=\left(b^2-\frac c3\right)^3\\
B'C'&=2b^3-bc+d~(=2B)\\
B'^2-C'&=3b^2-c\\
\end{cases}$
$$\quad\quad\quad\begin{align}
x=&-b+\omega\sqrt[3]{-B+\sqrt{B^2-C}}+\omega^{-1}\sqrt[3]{-B-\sqrt{B^2-C}}\\
=&\begin{cases}
-b-B'\\
\displaystyle-b+\underbrace{\frac{B'\pm\sqrt{B'-4C'}}2}_{X^2-B'X+C'=0~ã®è§£}=-b+\underbrace{\frac{B'}2\pm\sqrt{\left(\frac{B'}2\right)^2-C'}}_{X^2-B'X+C'=0~ã®è§£}
\end{cases}
\end{align}$$
$ $
ã»$x^3+ax+b=0$ ã®è§£ã®å
¬åŒ
$$\quad\quad\begin{align}
x=&\omega\sqrt[3]{\frac{-b+\sqrt{b^2-4\left(-\frac a3\right)^3}}2}+\omega^{-1}\sqrt[3]{\frac{-b-\sqrt{b^2-4\left(-\frac a3\right)^3}}2}\\
=&\omega\sqrt[3]{-\left(\frac{b}2\right)+\sqrt{\left(\frac{b}2\right)^2+\left(\frac a3\right)^3}}+\omega^{-1}\sqrt[3]{-\left(\frac{b}2\right)-\sqrt{\left(\frac{b}2\right)^2+\left(\frac a3\right)^3}}\\
\end{align}$$
$$\\[8pt]$$
âé¢é£èšäºã ééã£ãèšç®ã§æç«ããäžä¹æ ¹ã®çåŒ ã
ããããäžä¹æ ¹ã®äºéæ ¹å·åé¡ã¯ãã»ãŒããã§è§£ããŸãã
$$\begin{align} \begin{cases} ~~~~~d&=\sqrt[3]{\left(\textcolor{#f7c}{T}\sqrt{\textcolor{#f00}{P}}\right)^2-\left(\textcolor{#07f}{S}\sqrt{R}\right)^2}\\ +3d&=2\textcolor{#07f}{S}-\textcolor{#00f}{R}~\rightarrow~\textcolor{#f00}{P}=2\textcolor{#07f}{S}+d\\ -3d&=2\textcolor{#f7c}{T}-\textcolor{#f00}{P}~\rightarrow~\textcolor{#00f}{R}=2\textcolor{#f7c}{T}-d\\ \end{cases}\quad\quad\quad\quad\quad\\[8pt] \Rightarrow\begin{cases}\displaystyle\sqrt[3]{\textcolor{#f00}{\pm}\textcolor{#f7c}{T}\sqrt{\textcolor{#f00}{P}}\textcolor{#00f}{\pm}\textcolor{#07f}{S}\sqrt{\textcolor{#00f}{R}}}=\frac{\textcolor{#f00}{\pm}\sqrt{\textcolor{#f00}{P}}\textcolor{#00f}{\pm}\sqrt{\textcolor{#00f}{R}}}2\\ ~d=\textcolor{#07f}{S}-\textcolor{#f7c}{T},~4d=\textcolor{#f00}{P}-\textcolor{#00f}{R}\\ \textcolor{#f00}{P}=3\textcolor{#07f}{S}-\textcolor{#f7c}{T}\\ \textcolor{#00f}{R}=3\textcolor{#f7c}{T}-\textcolor{#07f}{S}\\ \end{cases} \quad\quad \end{align}$$
$ $
$$\begin{align}n^3=\frac{\textcolor{#f7c}{T}}{\textcolor{#f00}{P}+3\textcolor{#00f}{R}}=\frac{\textcolor{#07f}{S}}{\textcolor{#00f}{R}+3\textcolor{#f00}{P}}~\left(=\frac{\textcolor{#07f}{S}-\textcolor{#f7c}{T}}{2(\textcolor{#f00}{P}-\textcolor{#00f}{R})}\right)\\[8pt] \Rightarrow\displaystyle\sqrt[3]{\textcolor{#f00}{\pm}\textcolor{#f7c}{T}\sqrt{\textcolor{#f00}{P}}\textcolor{#00f}{\pm}\textcolor{#07f}{S}\sqrt{\textcolor{#00f}{R}}}=n\left(\textcolor{#f00}{\pm}\sqrt{\textcolor{#f00}{P}}\textcolor{#00f}{\pm}\sqrt{\textcolor{#00f}{R}}\right)\end{align}$$
$ $
$$\begin{align} y^3=\textcolor{#f7c}{\left(\textcolor{#000}{\frac{\textcolor{#f7c}{x-y}}{2}}\right)}^2\textcolor{#f00}{(x+2y)}-\textcolor{#07f}{\left(\textcolor{#000}{\frac{\textcolor{#07f}{x+y}}2}\right)}^2\textcolor{#00f}{(x-2y)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\\[8pt] \Rightarrow\displaystyle\sqrt[3]{\frac{\textcolor{#f7c}{\pm(x-y)}\textcolor{#f00}{\sqrt{x+2y}}\textcolor{#07f}{\pm(x+y)}\textcolor{#00f}{\sqrt{x-2y}}}2}=\frac{\textcolor{#f7c}{\pm}\textcolor{#f00}{\sqrt{x+2y}}\textcolor{#07f}{\pm}\textcolor{#00f}{\sqrt{x-2y}}}2 \end{align}$$
$ $
$$\begin{align} q^3=\textcolor{#f7c}{(q+4r)}^2\textcolor{#f00}{(q+r)}-\textcolor{#07f}{(3q+4r)}^2\textcolor{#00f}{r}\quad\quad\quad\quad\quad\quad\quad\quad\quad\\[8pt] \Rightarrow\displaystyle\sqrt[3]{\textcolor{#f7c}{\pm(q+4r)}\textcolor{#f00}{\sqrt{q+r}}\textcolor{#07f}{\pm(3q+4r)}\textcolor{#00f}{\sqrt{r}}}=\textcolor{#f7c}{\pm}\textcolor{#f00}{\sqrt{q+r}}\textcolor{#07f}{\pm}\textcolor{#00f}{\sqrt{r}} \end{align}$$
$ $
$$\sqrt[3]{\textcolor{#f00}\pm\frac{s+8r}3\sqrt{\frac{s-r}3}\textcolor{#00f}{\pm}\sqrt{s^2r}}=\textcolor{#f00}\pm\sqrt{\frac{s-r}3}\textcolor{#00f}\pm\sqrt{r}$$
$$\\[8pt]$$
$$\sqrt{\frac{n^{2a-b-1}+n^b}2\sqrt{n}\pm n^a}=\sqrt[4]{\frac{n}4}\left(\sqrt{n^{2a-b-1}}\pm\sqrt{n^b}\right)$$
$$\\[8pt]$$