ポアンカレ計量による距離ではX軸に沿った移動では線分が最短にならない

計量による距離

$$\begin{eqnarray} g_{(x,y)}\left(\boldsymbol{v},\boldsymbol{w}\right)&=& \frac{\left\langle \boldsymbol{v},\boldsymbol{w}\right\rangle}{y^2} \;\cdots\;ポアンカレ計量,\;\left\langle \boldsymbol{x},\boldsymbol{y} \right\rangle=\sum_{i=1}^n{x_i y_i}は標準内積 \\\gamma\left(t\right)&=&\left(x(t),y(t)\right)\;\cdots\;曲線\gamma,\;tによる媒介表示,\;始点\gamma(a),\;終点\gamma(b) \\\gamma^\prime\left(t\right)&=&\left(x^\prime(t),y^\prime(t)\right)\;\cdots\;(x(t),y(t))での接ベクトル \\L{\left(\gamma\right)}&=&\int_a^b \sqrt{g\left(\gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right)} \;\cdots\;\gammaの計量による距離 \\&=&\int_a^b \sqrt{\frac{\left\langle \gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right\rangle}{y^2}}\mathrm{d}t \\&=&\int_a^b \frac{\sqrt{\left\langle \gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right\rangle}}{\sqrt{y^2}}\mathrm{d}t \\&=&\int_a^b \frac{\left||\gamma^\prime\left(t\right)\right||}{y}\mathrm{d}t \;\cdots\;\left||x\right||=\sqrt{\left\langle \boldsymbol{x},\boldsymbol{x} \right\rangle} \end{eqnarray}$$

\((\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)から \((-\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)までの線分で移動(ただし\(\theta\in \left(0, \frac{\pi}{2}\right)\))

$$\begin{eqnarray} \gamma_3(t)&=&\left((x_1-x_0)t+x_0,(y_1-y_0)t+y_0 \right) \\&=&\left( (-\cos{\left(\theta\right)}-\cos{\left(\theta\right)})t+\cos{\left(\theta\right)}, (\sin{\left(\theta\right)}-\sin{\left(\theta\right)})t+\sin{\left(\theta\right)} \right) \\&=&\left((1-2t)\cos{\left(\theta\right)},\sin{\left(\theta\right)}\right) \\\gamma_3^\prime(t)&=&\left(-2\cos{\left(\theta\right)}, 0\right) \\L_3=L\left(\gamma_3\right)&=&\int_0^1 \frac{\left||\gamma_3^\prime\left(t\right)\right||}{y(t)}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{ \left(-2\cos{\left(\theta\right)}\right)^2 +0^2 }}{\sin{\left(\theta\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{2\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{2}{\tan{\left(\theta\right)}}\mathrm{d}t \\&=&\frac{2}{\tan{\left(\theta\right)}}\int_0^1\mathrm{d}t \\&=&\frac{2}{\tan{\left(\theta\right)}}[t]_0^1 \\&=&\frac{2}{\tan{\left(\theta\right)}}[1-0] \\&=&\frac{2}{\tan{\left(\theta\right)}}\cdot 1 \\&=&\frac{2}{\tan{\left(\theta\right)}} \end{eqnarray}$$

\((\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)から \((-\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)まで，原点を中心とした半径1の円弧で移動(ただし \( \theta\in \left(0,\frac{\pi}{2}\right) \))

$$\begin{eqnarray} \gamma_4(t)&=&\left(\cos{\left(\theta+(\pi-2\theta)t\right)}, \sin{\left(\theta+(\pi-2\theta)t\right)} \right) \\\gamma_4^\prime(t)&=&\left( -(\pi-2\theta)\sin{\left(\theta+(\pi-2\theta)t\right)}, (\pi-2\theta)\cos{\left(\theta+(\pi-2\theta)t\right)} \right) \\L_4=L\left(\gamma_4\right)&=&\int_0^1 \frac{\left||\gamma_4^\prime\left(t\right)\right||}{y(t)}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{ \left( -(\pi-2\theta)\sin{\left(\theta+(\pi-2\theta)t\right)} \right)^2 +\left( (\pi-2\theta)\cos{\left(\theta+(\pi-2\theta)t\right)} \right)^2 }} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{(\pi-2\theta)^2\left( \sin^2{\left(\theta+(\pi-2\theta)t\right)} +\cos^2{\left(\theta+(\pi-2\theta)t\right)} \right) }} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{(\pi-2\theta)^2\cdot 1}} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\pi-2\theta} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\alpha} {\sin{\left(\beta+\alpha t\right)}}\mathrm{d}t \\&=&\int_\beta^{\alpha+\beta} \frac{\cancel{\alpha}} {\sin{\left(u\right)}}\frac{1}{\cancel{\alpha}}\mathrm{d}u \\&=&\left[\ln{\left|\tan{\left(\frac{u}{2}\right)}\right|}\right]_\beta^{\alpha+\beta} \;\cdots\;\int \frac{1}{\sin{\left(x\right)}} \mathrm{d}x= \ln{\left|\tan{\left(\frac{x}{2}\right)}\right|}+C \\&=&\left[\ln{\left|\tan{\left(\frac{u}{2}\right)}\right|}\right]_\theta^{\pi-2\theta+\theta} \\&=&\left[\ln{\left|\tan{\left(\frac{u}{2}\right)}\right|}\right]_\theta^{\pi-\theta} \\&=&\ln{\left|\tan{\left(\frac{\pi-\theta}{2}\right)}\right|} -\ln{\left|\tan{\left(\frac{\theta}{2}\right)}\right|} \\&=&\ln{\left| \frac{\tan{\left(\frac{\pi-\theta}{2}\right)}}{\tan{\left( \frac{\theta}{2}\right)}}\right|} \\&=&\ln{\left| \frac{\cot{\left(\frac{\theta}{2}\right)}}{\tan{\left( \frac{\theta}{2}\right)}}\right|} \\&=&\ln{\left| \frac{1}{\tan^2{\left( \frac{\theta}{2}\right)}} \right|} \\&=&\ln{\left| \tan^{-2}{\left( \frac{\theta}{2}\right)} \right|} \\&=&-2\ln{\left| \tan{\left( \frac{\theta}{2}\right)} \right|} \end{eqnarray}$$

\(L_3-L_4\)

$$\begin{eqnarray} L_3-L_4 &=&\left\{\frac{2}{\tan{\left(\theta\right)}}\right\} -\left\{-2\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}\right\} \\&=&2\left\{ \frac{1}{\tan{\left(\theta\right)}}+\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}\right\} \\\frac{\mathrm{d}}{\mathrm{d}\theta}\left(L_3-L_4\right) &=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{1}{2\cos{\left(\frac{\theta}{2}\right)}\sin{\left(\frac{\theta}{2}\right)}} \right\} \\&=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{1}{\sin{\left(\theta\right)}} \right\} \\&=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{1}{\sin{\left(\theta\right)}}\frac{\sin{\left(\theta\right)}}{\sin{\left(\theta\right)}} \right\} \\&=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{\sin{\left(\theta\right)}}{\sin^2{\left(\theta\right)}} \right\} \\&=&2\left\{ \frac{\sin{\left(\theta\right)}-1}{\sin^2{\left(\theta\right)}} \right\}\leq0\;\cdots\; 分母は常に正，分子は\theta\in \left(0,\frac{\pi}{2}\right)なので(-1,0)であり，全体では常に負となる． \\\lim_{\theta \rightarrow \frac{\pi}{2}-0}L_3-L_4&=&0 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2024/11/lim-x2-cotx.html}{ \lim_{\theta \rightarrow \frac{\pi}{2}-0}{\frac{1}{\tan{\left(\theta\right)}}}=0} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2024/12/lim-x2-lntanx2.html}{\lim_{\theta \rightarrow \frac{\pi}{2}-0}{\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}}=0} \\よって \\L_3-L_4&\geq&0\;\cdots\;，\theta\in \left(0,\frac{\pi}{2}\right)では微分が常に負(減少関数)であり，\theta=\frac{\pi}{2}で0なので，範囲内では常に正． \end{eqnarray}$$

まとめ

ポアンカレ計量では水平の線分より，\(x\)軸から離れる側となる円弧に沿った経路の方が短いことになる．
これは境界\(x\)軸\(y=0\)に近いほど計量中にある\(\frac{1}{y^2}\)の効果で値が大きくなり，距離としては長くなることが影響した結果．

ポアンカレ計量による距離でもY軸に沿った移動なら線分が最短になる

計量による距離

$$\begin{eqnarray} g_{(x,y)}\left(\boldsymbol{v},\boldsymbol{w}\right)&=& \frac{\left\langle \boldsymbol{v},\boldsymbol{w}\right\rangle}{y^2} \;\cdots\;ポアンカレ計量,\;\left\langle \boldsymbol{x},\boldsymbol{y} \right\rangle=\sum_{i=1}^n{x_i y_i}は標準内積 \\\gamma\left(t\right)&=&\left(x(t),y(t)\right)\;\cdots\;曲線\gamma,\;tによる媒介表示,\;始点\gamma(a),\;終点\gamma(b) \\\gamma^\prime\left(t\right)&=&\left(x^\prime(t),y^\prime(t)\right)\;\cdots\;(x(t),y(t))での接ベクトル \\L{\left(\gamma\right)}&=&\int_a^b \sqrt{g\left(\gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right)} \;\cdots\;\gammaの計量による距離 \\&=&\int_a^b \sqrt{\frac{\left\langle \gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right\rangle}{y^2}}\mathrm{d}t \\&=&\int_a^b \frac{\sqrt{\left\langle \gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right\rangle}}{\sqrt{y^2}}\mathrm{d}t \\&=&\int_a^b \frac{\left||\gamma^\prime\left(t\right)\right||}{y}\mathrm{d}t \;\cdots\;\left||x\right||=\sqrt{\left\langle \boldsymbol{x},\boldsymbol{x} \right\rangle} \end{eqnarray}$$

\((0, y_0)\)から\((0, y_1)\)までの線分\(\gamma_1\)

$$\begin{eqnarray} \gamma_1\left(t\right)&=&\left(0,\;(y_1 - y_0)t+y_0 \right)\;\cdots\;t\in[0, 1] \\\gamma_1^\prime\left(t\right)&=&\left(0,\;y_1 - y_0 \right) \\L_1=L\left(\gamma_1\right)&=&\int_0^1 \frac{\left||\gamma_1^\prime\left(t\right)\right||}{y(t)}\mathrm{d}t \\&=&\int_0^1\frac{\sqrt{0^2+(y_1 - y_0)^2}}{(y_1 - y_0)t+y_0}\mathrm{d}t \\&=&\int_0^1\frac{y_1 - y_0}{(y_1 - y_0)t+y_0}\mathrm{d}t \\&=&\int_{y_0}^{y_1}\frac{\cancel{y_1 - y_0}}{u}\frac{1}{\cancel{y_1-y_0}}\mathrm{d}u \\&=&\int_{y_0}^{y_1}\frac{1}{u}\mathrm{d}u \\&=&\left[\log{\left(u\right)}\right]_{y_0}^{y_1} \\&=&\log{\left(y_1\right)}-\log{\left(y_0\right)} \\&=&\log{\left(\frac{y_1}{y_0}\right)} \end{eqnarray}$$

\((0, y_0)\)から\((0, y_1)\)までの任意の曲線\(\gamma_2\)

$$\begin{eqnarray} \gamma_2\left(t\right)&=&\left(x(t),y(t)\right)\;\cdots\;t\in[0, 1] \\\gamma_2^\prime\left(t\right)&=&\left(x^\prime(t),y^\prime(t)\right) \\L_2=L\left(\gamma_2\right)&=&\int_0^1 \frac{\left||\gamma_2^\prime\left(t\right)\right||}{y(t)}\mathrm{d}t \\&=&\int_0^1\frac{\sqrt{x^\prime(t)^2+y^\prime(t)^2}}{y(t)}\mathrm{d}t \\&\geq&\int_0^1\frac{\sqrt{y^\prime(t)^2}}{y(t)}\mathrm{d}t\;\cdots\;x^\prime(t)^2(\geq0)を取り除くので等しいか小さくなる \\&=&\int_0^1\frac{|y^\prime(t)|}{y(t)}\mathrm{d}t \\&=&\left[\log{\left(y(t)\right)}\right]_{0}^{1} \\&=&\log{\left(y(1)\right)}-\log{\left(y(0)\right)} \\&=&\log{\left(y_1\right)}-\log{\left(y_0\right)} \\&=&\log{\left(\frac{y_1}{y_0}\right)}=L_1 \\L_2&\geq&L_1\;\cdots\;y軸方向では線分が一番短い \end{eqnarray}$$

独立なガウス分布に従う誤差を含む回帰モデルがn個の同時確率を考える

独立なガウス分布\(N(\mu_i,\sigma_i^2)\)に従う誤差\(\epsilon_i\)を含む回帰モデル\(y_i=\beta x_i+\alpha+\epsilon_i\)が\(n\)個の同時確率を考える

準備

\begin{eqnarray} \bar{x}&=&\frac{1}{n}\sum_{i=1}^n x_i \\\bar{y}&=&\frac{1}{n}\sum_{i=1}^n y_i \\\sum_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right) &=&\sum_{i=1}^n\left(x_iy_i-x_i\bar{y}-\bar{x}y_i+\bar{x}\bar{y}\right) \\&=&\sum_{i=1}^nx_iy_i-\bar{y}\sum_{i=1}^nx_i-\bar{x}\sum_{i=1}^ny_i+\bar{x}\bar{y}\sum_{i=1}^n1 \\&=&\sum_{i=1}^nx_iy_i-\bar{y}n\bar{x}-\bar{x}n\bar{y}+\bar{x}\bar{y}n \\&=&\sum_{i=1}^nx_iy_i-2n\bar{x}\bar{y}+n\bar{x}\bar{y} \\&=&\sum_{i=1}^nx_iy_i-n\bar{x}\bar{y}\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/10/sxy.html}{S_{xy}の式展開} \\\sum_{i=1}^n(x_i-\bar{x})^2 &=&\sum_{i=1}^n(x_i^2-2x_i\bar{x}+\bar{x}^2) \\&=&\sum_{i=1}^nx_i^2-2\bar{x}\sum_{i=1}^nx_i+\bar{x}^2\sum_{i=1}^n1 \\&=&\sum_{i=1}^nx_i^2-2\bar{x}n\bar{x}+\bar{x}^2n \\&=&\sum_{i=1}^nx_i^2-2n\bar{x}^2+n\bar{x}^2 \\&=&\sum_{i=1}^nx_i^2-n\bar{x}^2\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/10/sxx.html}{S_{xx}の式展開} \end{eqnarray}

同時確率を考える

\begin{eqnarray} p(\epsilon_1,\epsilon_2,\cdots,\epsilon_n) &=&\frac{1}{\sqrt{2\pi{\sigma_1}^2}}e^{-\frac{\left(\epsilon_1-\mu_1\right)^2}{2{\sigma_1}^2}} \times\frac{1}{\sqrt{2\pi{\sigma_2}^2}}e^{-\frac{\left(\epsilon_2-\mu_2\right)^2}{2{\sigma_2}^2}} \times\cdots\times\frac{1}{\sqrt{2\pi{\sigma_n}^2}}e^{-\frac{\left(\epsilon_n-\mu_n\right)^2}{2{\sigma_n}^2}} \\&=&\left(\prod_{i=1}^n{\frac{1}{\sqrt{2\pi{\sigma_i}^2}}}\right) \left(\prod_{i=1}^n{e^{-\frac{\left(\epsilon_i-\mu_i\right)^2}{2{\sigma_i}^2}}}\right) \\&=&\left(\prod_{i=1}^n{\left(2\pi{\sigma_i}^2\right)^{-\frac{1}{2}}}\right) \left(\prod_{i=1}^n{e^{-\frac{\left(y_i-\beta x_i-\alpha -\mu_i\right)^2}{2{\sigma_i}^2}}}\right) \;\cdots\;y_i=\beta x_i+\alpha+\epsilon_i,\;\epsilon_i=y_i-\beta x_i-\alpha \end{eqnarray}

負の対数によって情報量を求める

\begin{eqnarray} \\-\ln(p)&=&-\ln\left( \left(\prod_{i=1}^n{\left(2\pi{\sigma_i}^2\right)^{-\frac{1}{2}}}\right) \left(\prod_{i=1}^n{e^{-\frac{1}{2{\sigma_i}^2}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2}}\right) \right) \\&=&-\ln\left( \prod_{i=1}^n{\left(2\pi{\sigma_i}^2\right)^{-\frac{1}{2}}} \right) -\ln\left( \prod_{i=1}^n{e^{-\frac{1}{2{\sigma_i}^2}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2}} \right) \\&=&-\sum_{i=1}^n\ln\left( {\left(2\pi{\sigma_i}^2\right)^{-\frac{1}{2}}} \right) -\sum_{i=1}^n\ln\left( {e^{-\frac{1}{2{\sigma_i}^2}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2}} \right) \\&=&-\sum_{i=1}^n \left\{-\frac{1}{2}\ln\left( 2\pi{\sigma_i}^2 \right) \right\} -\sum_{i=1}^n\left(-\frac{1}{2{\sigma_i}^2}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2 \right) \\&=&\sum_{i=1}^n \left\{\frac{1}{2}\ln\left( 2\pi{\sigma_i}^2 \right) \right\} +\sum_{i=1}^n\frac{1}{2{\sigma_i}^2}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2 \\&=&\sum_{i=1}^n \left\{\frac{1}{2}\ln\left( 2\pi{\sigma_i}^2 \right) \right\} +\sum_{i=1}^n\frac{1}{2{\sigma_i}^2}\left( y_i^2 +\beta^2x_i^2 +\alpha^2 +\mu_i^2 -2\beta x_i y_i -2\alpha y_i -2y_i \mu_i +2\alpha \beta x_i +2\beta x_i\mu_i +2\alpha \mu_i \right) \end{eqnarray}

回帰モデルの係数で微分する

\(\beta\)について． \begin{eqnarray} \frac{\partial}{\partial \beta}\left(-\ln(p)\right)&=&0 +\sum_{i=1}^n\frac{1}{\cancel{2}{\sigma_i}^2}\left(0+\cancel{2}{\beta}x_i^2+0+0-\cancel{2}x_i y_i-0-0 +\cancel{2}\alpha x_i+\cancel{2}x_i \mu_i+0\right) \\&=&\sum_{i=1}^n\frac{1}{{\sigma_i}^2}\left({\beta}x_i^2-x_iy_i+\alpha x_i+x_i\mu_i\right) \\&=&\beta\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2} -\sum_{i=1}^n\frac{x_i\left(y_i-\alpha-\mu_i\right)}{{\sigma_i}^2} \end{eqnarray} \(\alpha\)について． \begin{eqnarray} \frac{\partial}{\partial \alpha}\left(-\ln(p)\right)&=&0 +\sum_{i=1}^n\frac{1}{\cancel{2}{\sigma_i}^2}\left(0+0+\cancel{2}\alpha+0-0-\cancel{2}y_i -0 +\cancel{2}\beta x_i+0+\cancel{2}\mu_i\right) \\&=&\sum_{i=1}^n\frac{1}{{\sigma_i}^2}\left(\alpha -y_i+\beta x_i+\mu_i\right) \\&=&\alpha \sum_{i=1}^n\frac{1}{{\sigma_i}^2}-\sum_{i=1}^n\frac{y_i-\beta x_i-\mu_i}{{\sigma_i}^2} \end{eqnarray}

微分が0となる\(\hat{\alpha},\hat{\beta}\),つまり極値となるパラメータを求める

\(\hat{\beta}\)について． \begin{eqnarray} \\0&=&\frac{\partial}{\partial \beta}\left(-\ln(p)\right) \\&=&\hat{\beta}\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2} -\sum_{i=1}^n\frac{x_i\left(y_i-\hat{\alpha}-\mu_i\right)}{{\sigma_i}^2} \\\hat{\beta}\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2}&=& \sum_{i=1}^n\frac{x_i\left(y_i-\hat{\alpha}-\mu_i\right)}{{\sigma_i}^2} \\\hat{\beta}&=&\frac{ \sum_{i=1}^n\frac{x_i\left(y_i-\hat{\alpha}-\mu_i\right)}{{\sigma_i}^2} }{\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2}} \end{eqnarray} \(\hat{\alpha}\)について． \begin{eqnarray} \\0&=&\frac{\partial}{\partial \alpha}\left(-\ln(p)\right) \\&=&\hat{\alpha}\sum_{i=1}^n\frac{1}{{\sigma_i}^2}-\sum_{i=1}^n\frac{y_i-\hat{\beta}x_i-\mu_i}{{\sigma_i}^2} \\\hat{\alpha}\sum_{i=1}^n\frac{1}{{\sigma_i}^2}&=&\sum_{i=1}^n\frac{y_i-\hat{\beta}x_i-\mu_i}{{\sigma_i}^2} \\\hat{\alpha}&=&\frac{\sum_{i=1}^n\frac{y_i-\hat{\beta}x_i-\mu_i}{{\sigma_i}^2}} {\sum_{i=1}^n\frac{1}{{\sigma_i}^2}} \end{eqnarray}

誤差の発生が全て同一モデル\(\epsilon_i \sim N(0,\sigma^2)\)から発生しているとする

\(\hat{\alpha}\)について． \begin{eqnarray} \hat{\alpha}&=&\frac{\frac{1}{{\sigma}^2} \sum_{i=1}^n \left(y_i-\hat{\beta}x_i-0\right) }{\frac{n}{{\sigma}^2}} \;\cdots\;\epsilon_i \sim N(0,\sigma^2) \\&=&\frac{1}{n}\sum_{i=1}^n \left(y_i-\hat{\beta}x_i\right) \\&=&\frac{\sum_{i=1}^n y_i}{n}-\hat{\beta}\frac{\sum_{i=1}^n x_i}{n} \\&=&\bar{y}-\hat{\beta}\bar{x} \end{eqnarray} \(\hat{\beta}\)について． \begin{eqnarray} \\\hat{\beta}&=&\frac{ \frac{1}{{\sigma}^2} \sum_{i=1}^nx_i\left(y_i-\hat{\alpha}-0\right) }{ \frac{1}{{\sigma}^2}\sum_{i=1}^nx_i^2 } \;\cdots\;\epsilon_i \sim N(0,\sigma^2) \\&=&\frac{ \sum_{i=1}^nx_i y_i-\hat{\alpha}\sum_{i=1}^nx_i }{ \sum_{i=1}^nx_i^2 } \\&=&\frac{\sum_{i=1}^nx_i y_i-\hat{\alpha}n\bar{x}}{\sum_{i=1}^nx_i^2} \\&=&\frac{\sum_{i=1}^nx_i y_i-\left(\bar{y}-\hat{\beta}\bar{x}\right)n\bar{x}}{\sum_{i=1}^nx_i^2} \;\cdots\;\hat{\alpha}=\bar{y}-\hat{\beta}\bar{x},\;\hat{\alpha}での結果を代入する \\&=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}+\hat{\beta}n\bar{x}^2}{\sum_{i=1}^nx_i^2} \\\hat{\beta}-\frac{\hat{\beta}n\bar{x}^2}{\sum_{i=1}^nx_i^2} &=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}}{\sum_{i=1}^nx_i^2} \\\hat{\beta}\left(\frac{\sum_{i=1}^nx_i^2-n\bar{x}^2}{\cancel{\sum_{i=1}^nx_i^2}}\right) &=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}}{\cancel{\sum_{i=1}^nx_i^2}} \\\hat{\beta}&=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}}{\sum_{i=1}^nx_i^2-n\bar{x}^2} \\&=&\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)}{\sum_{i=1}^n(x_i-\bar{x})^2}\;\cdots\;準備 より \end{eqnarray}

まとめ

\begin{cases}\begin{eqnarray} \hat{\beta}&=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}}{\sum_{i=1}^nx_i^2-n\bar{x}^2}=\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)}{\sum_{i=1}^n(x_i-\bar{x})^2} \\\hat{\alpha}&=&\bar{y}-\hat{\beta}\bar{x} \end{eqnarray}\end{cases}

lim x→π/2 ln(tan(x/2)) を求める

\(\lim_{x\rightarrow \frac{\pi}{2}}\ln{\left(\tan{\left(\frac{x}{2}\right)}\right)}\)を求める

高階の微分を求めておく

一階から順に求めておく． $$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} &=&\frac{1}{\tan{\left(\frac{x}{2}\right)}}\left(\frac{\mathrm{d}}{\mathrm{d}x}\tan{\left(\frac{x}{2}\right)}\right) \;\cdots\;u=\tan{\left(\frac{x}{2}\right)},f=\ln{\left(u\right)},\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}x} \\&=&\frac{1}{ \frac{\sin{\left(\frac{x}{2}\right)}}{\cos{\left(\frac{x}{2}\right)}} } \left(\frac{\mathrm{d}}{\mathrm{d}x} \frac{\sin{\left(\frac{x}{2}\right)}}{\cos{\left(\frac{x}{2}\right)}} \right) \\&=&\frac{\cos{\left(\frac{x}{2}\right)}}{\sin{\left(\frac{x}{2}\right)}} \left(\frac{\mathrm{d}}{\mathrm{d}x} \sin{\left(\frac{x}{2}\right)}\cos^{-1}{\left(\frac{x}{2}\right)} \right) \\&=&\frac{\cos{\left(\frac{x}{2}\right)}}{\sin{\left(\frac{x}{2}\right)}} \left\{ \left(\frac{\mathrm{d}}{\mathrm{d}x}\sin{\left(\frac{x}{2}\right)}\right)\cos^{-1}{\left(\frac{x}{2}\right)} +\sin{\left(\frac{x}{2}\right)}\left(\frac{\mathrm{d}}{\mathrm{d}x}\cos^{-1}{\left(\frac{x}{2}\right)}\right) \right\} \\&=&\frac{\cos{\left(\frac{x}{2}\right)}}{\sin{\left(\frac{x}{2}\right)}} \left[ \left(\cancel{\cos{\left(\frac{x}{2}\right)}}\cdot\frac{1}{2}\right)\cancel{\cos^{-1}{\left(\frac{x}{2}\right)}} +\sin{\left(\frac{x}{2}\right)}\left\{-\cos^{-2}{\left(\frac{x}{2}\right)}\left(-\sin{\left(\frac{x}{2}\right)}\cdot\frac{1}{2}\right)\right\} \right] \\&=&\frac{\cos{\left(\frac{x}{2}\right)}}{\sin{\left(\frac{x}{2}\right)}} \cdot\frac{1}{2}\left\{ 1+\sin^2{\left(\frac{x}{2}\right)}\cos^{-2}{\left(\frac{x}{2}\right)} \right\} \\&=&\frac{1}{2}\left( \frac{\cos{\left(\frac{x}{2}\right)}}{\sin{\left(\frac{x}{2}\right)}} +\frac{\cancel{\cos{\left(\frac{x}{2}\right)}}}{\cancel{\sin{\left(\frac{x}{2}\right)}}}\sin^{\cancel{2}1}{\left(\frac{x}{2}\right)}\cos^{\cancel{-2}-1}{\left(\frac{x}{2}\right)} \right) \\&=&\frac{1}{2}\left( \frac{\cos{\left(\frac{x}{2}\right)}}{\sin{\left(\frac{x}{2}\right)}} +\frac{\sin{\left(\frac{x}{2}\right)}}{\cos{\left(\frac{x}{2}\right)}} \right) \\&=&\frac{1}{2} \frac{ \cos^2{\left(\frac{x}{2}\right)}+\sin^2{\left(\frac{x}{2}\right)} }{\sin{\left(\frac{x}{2}\right)}\cos{\left(\frac{x}{2}\right)}} \\&=& \frac{1}{2\sin{\left(\frac{x}{2}\right)}\cos{\left(\frac{x}{2}\right)}} \\&=&\frac{1}{\sin{(x)}}\;\cdots\;\sin{(x)}=2\sin{\left(\frac{x}{2}\right)}\cos{\left(\frac{x}{2}\right)} \\\; \\\frac{\mathrm{d}^2}{\mathrm{d}x^2} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{\sin{(x)}}\right) \\&=&-\frac{1}{\sin^2{(x)}}\left\{\frac{\mathrm{d}}{\mathrm{d}x}\sin{(x)}\right\} \\&=&-\frac{1}{\sin^2{(x)}}\cos{(x)} \\&=&-\frac{\cos{(x)}}{\sin^2{(x)}} \\\; \\\frac{\mathrm{d}^3}{\mathrm{d}x^3} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left(-\frac{\cos{(x)}}{\sin^2{(x)}}\right) \\&=&-\frac{\mathrm{d}}{\mathrm{d}x}\cos{(x)}\sin^{-2}{(x)} \\&=&-\left\{ \left( \frac{\mathrm{d}}{\mathrm{d}x}\cos{(x)}\right) \sin^{-2}{(x)} + \cos{(x)}\left(\frac{\mathrm{d}}{\mathrm{d}x}\sin^{-2}{(x)}\right) \right\} \\&=&-\left\{ \left( -\sin{(x)}\right) \sin^{-2}{(x)} + \cos{(x)}\left(-2\sin^{-3}{(x)}\cos{(x)}\right) \right\} \\&=&-\left\{ -\sin^{-1}{(x)}-2\sin^{-3}{(x)}\cos^2{(x)} \right\} \\&=&\sin^{-1}{(x)}+2\sin^{-3}{(x)}\cos^2{(x)} \\&=&\frac{1}{\sin{(x)}}+\frac{2\cos^2{(x)}}{\sin^3{(x)}} \\\; \\\frac{\mathrm{d}^4}{\mathrm{d}x^4} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{1}{\sin{(x)}}+\frac{2\cos^2{(x)}}{\sin^3{(x)}} \right) \\&=&\frac{\mathrm{d}}{\mathrm{d}x}\sin^{-1}{(x)} +2\frac{\mathrm{d}}{\mathrm{d}x}\cos^2{(x)}\sin^{-3}{(x)} \\&=& -\sin^{-2}{(x)}\cos{(x)} +2\left\{ \left(\frac{\mathrm{d}}{\mathrm{d}x}\cos^2{(x)}\right)\sin^{-3}{(x)} +\cos^2{(x)}\left(\frac{\mathrm{d}}{\mathrm{d}x}\sin^{-3}{(x)}\right) \right\} \\&=& -\sin^{-2}{(x)}\cos{(x)} +2\left[ \left\{2\cos{(x)}\left(-\sin{(x)}\right)\right\}\sin^{-3}{(x)} +\cos^2{(x)}\left(-3\sin^{-4}{(x)}\cos{(x)}\right) \right] \\&=&-\frac{\cos{(x)}}{\sin^{2}{(x)}} +2\left[ -2\frac{\cos{(x)}}{\sin^{2}{(x)}} -3\frac{\cos^3{(x)}}{\sin^{4}{(x)}} \right] \\&=&-\frac{\cos{(x)}}{\sin^{2}{(x)}} -4\frac{\cos{(x)}}{\sin^{2}{(x)}} -6\frac{\cos^3{(x)}}{\sin^{4}{(x)}} \\&=&-5\frac{\cos{(x)}}{\sin^{2}{(x)}} -6\frac{\cos^3{(x)}}{\sin^{4}{(x)}} \\\; \\\frac{\mathrm{d}^5}{\mathrm{d}x^5} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left( -5\frac{\cos{(x)}}{\sin^{2}{(x)}} -6\frac{\cos^3{(x)}}{\sin^{4}{(x)}} \right) \\&=&-5\frac{\mathrm{d}}{\mathrm{d}x}\cos{(x)}\sin^{-2}{(x)} -6\frac{\mathrm{d}}{\mathrm{d}x}\cos^3{(x)}\sin^{-4}{(x)} \\&=&-5\left\{ \left(\frac{\mathrm{d}}{\mathrm{d}x}\cos{(x)}\right)\sin^{-2}{(x)} +\cos{(x)}\frac{\mathrm{d}}{\mathrm{d}x}\sin^{-2}{(x)} \right\} -6\left\{ \left(\frac{\mathrm{d}}{\mathrm{d}x}\cos^3{(x)}\right)\sin^{-4}{(x)} +\cos^3{(x)}\frac{\mathrm{d}}{\mathrm{d}x}\sin^{-4}{(x)} \right\} \\&=&-5\left\{ \left(-\sin{(x)}\right)\sin^{-2}{(x)} +\cos{(x)}\left(-2\sin^{-3}{(x)}\cos{(x)}\right) \right\} -6\left\{ \left(-3\cos^2{(x)}\sin{(x)}\right)\sin^{-4}{(x)} +\cos^3{(x)}\left(-4\sin^{-5}{(x)}\cos{(x)}\right) \right\} \\&=&5\frac{1}{\sin{(x)}} +28\frac{\cos^2{(x)}}{\sin^{3}{(x)}} +24\frac{\cos^4{(x)}}{\sin^{5}{(x)}} \end{eqnarray}$$

\(x=\frac{\pi}{2}\)でのテーラー展開を求めておく

$$\begin{eqnarray} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} &=& \frac{1}{0!}\left[\left. \frac{\mathrm{d}^0}{\mathrm{d}x^0} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} \right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^0 \\&&+\frac{1}{1!}\left[\left. \frac{\mathrm{d}^1}{\mathrm{d}x^1} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} \right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^1 \\&&+\frac{1}{2!}\left[\left. \frac{\mathrm{d}^2}{\mathrm{d}x^2} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} \right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^2 \\&&+\frac{1}{3!}\left[\left. \frac{\mathrm{d}^3}{\mathrm{d}x^3} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} \right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^3 \\&&+\frac{1}{4!}\left[\left. \frac{\mathrm{d}^4}{\mathrm{d}x^4} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} \right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^4 \\&&+\frac{1}{5!}\left[\left. \frac{\mathrm{d}^5}{\mathrm{d}x^5} \ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} \right|_{x=\frac{\pi}{2}}\right]\left(x-\frac{\pi}{2}\right)^5 \\&&+\cdots \\&=& \frac{1}{1} \left[ \ln{\left(\tan{\left(\frac{\pi}{4}\right)}\right)} \right]\left(x-\frac{\pi}{2}\right)^0 \\&&+\frac{1}{1}\left[ \frac{1}{\sin{\left(\frac{\pi}{2}\right)}} \right]\left(x-\frac{\pi}{2}\right)^1 \\&&+\frac{1}{2}\left[ -\frac{\cos{\left(\frac{\pi}{2}\right)}}{\sin^2{\left(\frac{\pi}{2}\right)}} \right]\left(x-\frac{\pi}{2}\right)^2 \\&&+\frac{1}{6}\left[ \frac{1}{\sin{\left(\frac{\pi}{2}\right)}} +\frac{2\cos^2{\left(\frac{\pi}{2}\right)}}{\sin^3{\left(\frac{\pi}{2}\right)}} \right]\left(x-\frac{\pi}{2}\right)^3 \\&&+\frac{1}{24}\left[ -5\frac{\cos{\left(\frac{\pi}{2}\right)}}{\sin^{2}{\left(\frac{\pi}{2}\right)}} -6\frac{\cos^3{\left(\frac{\pi}{2}\right)}}{\sin^{4}{\left(\frac{\pi}{2}\right)}} \right]\left(x-\frac{\pi}{2}\right)^4 \\&&+\frac{1}{120}\left[ 5\frac{1}{\sin{\left(\frac{\pi}{2}\right)}} +28\frac{\cos^2{\left(\frac{\pi}{2}\right)}}{\sin^{3}{\left(\frac{\pi}{2}\right)}} +24\frac{\cos^4{\left(\frac{\pi}{2}\right)}}{\sin^{5}{\left(\frac{\pi}{2}\right)}} \right]\left(x-\frac{\pi}{2}\right)^5 \\&&+\cdots \\&=& \left[0\right]\cdot 1 \\&&+\left[ \frac{1}{1} \right]\left(x-\frac{\pi}{2}\right) \\&&+\frac{1}{2}\left[ -\frac{0}{1} \right]\left(x-\frac{\pi}{2}\right)^2 \\&&+\frac{1}{6}\left[ \frac{1}{1} +\frac{2\cdot0}{1} \right]\left(x-\frac{\pi}{2}\right)^3 \\&&+\frac{1}{24}\left[ -5\frac{0}{1} -6\frac{0}{1} \right]\left(x-\frac{\pi}{2}\right)^4 \\&&+\frac{1}{120}\left[ 5\frac{1}{1} +28\frac{0}{1} +24\frac{0}{1} \right]\left(x-\frac{\pi}{2}\right)^5 \\&&+\cdots \\&=& \left(x-\frac{\pi}{2}\right) +\frac{1}{6}\left(x-\frac{\pi}{2}\right)^3 +\frac{1}{24}\left(x-\frac{\pi}{2}\right)^5+\cdots \end{eqnarray}$$

\(\lim_{x\rightarrow \frac{\pi}{2}}\ln{\left(\tan{\left(\frac{x}{2}\right)}\right)}\)を求める

$$\begin{eqnarray} \\\lim_{x\rightarrow \frac{\pi}{2}}\ln{\left(\tan{\left(\frac{x}{2}\right)}\right)} &=& \lim_{x\rightarrow \frac{\pi}{2}}\left[ \left(x-\frac{\pi}{2}\right)+\frac{1}{6}\left(x-\frac{\pi}{2}\right)^3+\frac{1}{24}\left(x-\frac{\pi}{2}\right)^5+\cdots \right] \\&=&0 \end{eqnarray}$$