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	Memorandum of Mathematical Physics	Cat: SCI Pub: 2011 #2010
	Takumi (of Yobinori)	20625u

Title	Memorandum of Mathematical Physics	数理物理メモ
Index	Preface: Equation of motion (EOM) Pendulum isochronism: Conservation law: Damping oscillation: Conservation law of angular momentum: Inertial frame: Coriolis force: Archimedean spiral Hyperbolic function: Inverse trigonometric function: Two-particle system motion: Rigid body dynamics: Inertia moment of rigid body: Mechanical energy of rigid body: Matrix exponential: Chaucy's functional equation: Taylor expansion: Fourier expansion: Linear algebra: Delta function: Gaussian function: Vector analysis: Integration method: Multiple integral:	序文: 運動方程式 振り子の等時性: 保存則: 減衰振動: 角運動量保存則: 慣性系: コリオリの力: アルキメデスの螺旋: 双曲線関数: 逆三角関数: 二粒子系の運動: 剛体の力学: 剛体の慣性モーメント: 剛体の力学エネルギー: 行列指数関数: コーシーの関数方程式: テイラー展開: フーリエ展開: 線形代数: デルタ関数: ガウス関数: ベクトル解析: 積分法: 重積分:
Tag	; Angular momentum; Azmuthal direction; Center of gravity; Coriolis force; Commutative law; Conservation law; Constant area velocity; Critical damping; Damped oscillation; Diagonalization; Eigenvalue; EOM; Equivalence principle; Forced oscillation; Fourier expansion; Inertial frame; Integral method; Integration by parts; Inverse matrix ; Jacobian; Multiple integral; Orthogonal axes theorem; Parallel axes theorem; Pendulum isochronism; Rank of matrix; Reduced mass; Relative vector; Rigid body dynamics; Rotation; Simple harmonic motion; Solid angle; Spherical coordinates; Taylor expansion; Torque; Transposed matrix; Uniform linear motion; Vector product; ;
Why	Physical phenomena could have been more easily understood with differential equations; the world is written in DE. Dinosaurs recognized all movings are eatable. They know movement by the acceleration.

Original resume	Remarks
>Top 0. Preface: LaTex is convenient in writing of mathematical expression, though it need some kind of patience.	0. 序文:

>Top 1. Equation of motion (EOM): ¶1: Free-fall motion: $m\frac{d^2\mathbf{r}}{dt^2}=\mathbf{F}$ [2nd-order LDE] $my''=-mg$ [m=mass] $→\;y'=-gt+C→\;y=-\frac{}{}gt^2+Ct+C'$ $y'(0)=0→\;C=0;\; y(0)=y_0=C'$ [initial condition] $\therefore\; \boxed{y(t)=-\frac{1}{2}gt^2+y_0}$ ¶2: >Top Parabolic motion: $\cases{mx''=0\\my''=-mg}$ $mx''=0→\;x''=0→\;x'=C=v0\cos\theta$ $x(0)=0→\;C'=0$ $x=Ct+C'=v_0\cos\theta t\;$ [uniform linear motion] $my''=-mg→\;y''=g→\;y'=-gt+C=-gt+v_0\sin\theta$ $\therefore\;y=-\frac{1}{2}gt^2+v_0\sin\theta t+C'\;$ $y(0)=0=C'→\; \boxed{y(t)=-\frac{1}{2}gt^2+v_0\sin\theta t}$ [uniform acceleration motion]■ ¶3: Motion with air resistance: $mx''=-kx'$ let: $x'=v→\;mv'=-kv→\;v'=-\frac{k}{m}v$ $v(0)=v_0=C→\;v(t)=v_0e{-\frac{k}{m}t}$ $x'=vv_0e{-\frac{k}{m}t}$ $→\; x=-\frac{k}{m}v_0e^{-\frac{k}{m}t}+C'$ x(0)=0→\;C'=\frac{m}{k}v_0 $\therefore\;\boxed{x(t)=\frac{m}{k}v_0(1-e^{-\frac{k}{m}t})}$■ ¶4: >Top Simple harmonic motion: $mx''=-kx$...(*) [k: sprint constant; x: displacement from equilibrium position] $→\;x''=-\frac{k}{m}x=-\omega^2x;\$ here,$\omega=\sqrt{\frac{m}{k}}$ assume GS of (*): $x=C_1\sin t+C_2\cos t=C\sin\omega t_1+C_2\cos\omega t$ [linear combination] $→\;x=C\sin\omega t_1+C_2\cos\omega t$ initial condition: $\cases{x(0)=0\\x'(0)=v_0}$ $→\;\cases{C_2=0\\C_1\omega=V_0}$ $→\;C_1=\frac{v_0}{\omega},\;C_2=0$ $\therefore\;x(t)=\frac{v_0}{\omega}\sin\omega t$■ ¶5: Uniform circular motion: [$r_0=$raidus, $\;\omega_0$=angular velocity] $\mathbf{v}=r'e_r+r\phi'e_{\phi}=0･e_r+r_0\omega_0e_{\phi}=r_0\omega_0e_{\phi}$ $\mathbf{a}=(r''-r\phi'^2)e_r+(r\phi''+2r'\phi')e_{\phi}=-r_0\omega_0^2e_r =-\frac{v_0^2}{r_0}2e_r$ [using EOM] $\mathbf{F}=m\mathbf{a}=-mr_0\omega_0^2e_r$ [centripetal]

1. 運動方程式: radial direction: 動径方向 azmuthal direction: 方位角方向 <Ar. as-sumut pendulum isochronism: 振り子の等時性   $\mathbf{r}(x)$: trace of position: to know $\mathbf{r}(t)$ by integration ¶3. Harmonic Motion: Polar coordinate system: $\cases{e_r'=\phi'e_r\\ e_{\phi}'=-\phi'e_r}$ <Rec. cord.> $\cases{\mathbf{r}=xe_x+ye_y \\\mathbf{v}=x'e_x+y'e_y \\\mathbf{a}=x''e_x+y''e_y}$ <Por. cord.> $\cases{\mathbf{r}=re_r \\\mathbf{v}=r'e_r+r\phi'e_{\phi} \\\mathbf{a}=(r''-r\phi'^2)e_r\\ +(r\phi''+2r'\phi')e_{\phi}}$ Taylor expansion: $\sin x=x-\frac{x^3}{3!} +\frac{x^5}{5!}-\dots

>Top 2. Pendulum isochronism: Single pendulum: [$|\phi|<<1$] [azmuth direction of EOM] $m(r\phi''+2r'\phi')=-mg\sin\phi=ml$ [m=mass; l=length of pendulum] $ml\phi''=-mg\sin\phi$...(*) $→\;\phi''=-\frac{g}{l}\sin\phi\;$ [$\sin\phi\simeq\phi$] $→\;\phi''=-\frac{g}{l}\phi$ let: $\omega=\sqrt{\frac{g}{l}}:$ $→\;\phi''=-\omega^2\phi$ [=uniform circular motion] let: $\phi(t)=C_1\sin\omega t+C_2\cos\omega t$ here: $\phi(0)=\phi_0,\;\phi'=o\;$ [initial condition] $→\;C_2=\phi_0,\;C_1\omega=0→\;C_1=0,\; C_2=\phi_0$ [$\omega≠0$] $\therefore\;\phi(t)=\phi_0\cos\omega t$■ here: $T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{l}{g}}\;[T=$period] Exact solution of single pendulum: multiply $\phi':\; \phi''\phi'+\frac{g}{l}\phi'\sin\phi=0$ $→\;\frac{d}{dt}(\frac{1}{2}\phi^2-\frac{g}{l}\cos\phi)=0$ $→\;\frac{1}{2}\phi^2-\frac{g}{l}\cos\phi=E\;$ [E=constant; mechanical energy] when: $\theta=\theta_0,→\;\theta'=0$ $→\;E=-\frac{g}{l}\cos\theta_0$ $→\;\frac{1}{2}\phi'^2-\frac{g}{l}(\cos\phi-\cos\phi_0)$ $→\;\phi'^2=\frac{2g}{l}(\cos\phi-\cos\phi_0)$ here: $\phi'=-\sqrt{\frac{2g}{l}(\cos\phi-\cos\phi_0)}\;[0≤\phi≤\phi_0]$ $T=4\int_0^\frac{T}{4}dt=4\displaystyle\int_{\phi_0}^0 \frac{1}{-\sqrt{\frac{2g}{l}(\cos\phi-\cos\phi_0)}}d\phi$ $=2\sqrt{\frac{2l}{g}}\displaystyle\int_0^{\phi_0}\frac{1}{\sqrt{\cos\phi-\cos\phi_0}}d\phi$ change of variables: $\theta$ such that $\sin\frac{\phi}{2} =\sin\frac{\phi_0}{2}\sin\theta$ $→\;T=4\sqrt{\frac{l}{g}}\displaystyle\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-\sin^2\frac{\phi_0}{2}\sin^2\theta}}$ substitute to: $\sin (\frac{\phi_0}{2})=a$ $T=\frac{4}{\omega}\displaystyle\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-a^2\sin^2\theta}}=\frac{4}{\omega}K(a)\; [K:$complete elliptic integral 1st]$*\tiny{1}$

2. 振り子の等時性: complete elliptic integral of the 1st kind: 第1種完全楕円積分 $*\tiny1\;$第1種完全楕円積分: $F(x,k)=\displaystyle\int_0^x \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$ $F(x,0)=\displaystyle\int_0^x \frac{1}{\sqrt{1-t^2}}dt=\sin^{-1}x$ $F(x,1)=\displaystyle\int_0^x \frac{1}{1-t^2}dt=\tanh^{-1}x$

= >Top 3. Conservation law: Momentum conservation law: $m\frac{d^2\mathbf{r}}{dt^2}=\mathbf{F}$ $\mathbf{P}=m\frac{d\mathbf{r}}{dt}$\; [P=momentum] $→\;\frac{d\mathbf{p}}{dt}=\mathbf{F} →\;\mathbf{P}(t_2)-\mathbf{P}(t_1)=\int_{t_1}^{t_2}\mathbf{F}(t)dt$ $\mathbf{P}(t_2)-\mathbf{P}(t_1)=\mathbf{I}\;$ [I=impulse=change of momentum] when $\mathbf{I}=0:→\;\mathbf{P}(t_2)=\mathbf{P}(t_1)\;$ [conservation of momentum] Energy conservation law: $W=F\cos\theta･s=\mathbf{FS}\;$[Inner vector: W=work is constant] $dW=\mathbf{F}･d\mathbf{r}$ [W is changeable] $→\;W=\int_c\mathbf{F}･d\mathbf{r}$ conservative force: $W=\int_{\mathbf{r}_0}^{\mathbf{r}}\mathbf{F}･d\mathbf{r}$ potential energy: $U(\mathbf{r})= -\int_{\mathbf{r}_0}^\mathbf{r} \mathbf{F}d\mathbf{r}$ Kinetic energy: $k=\frac{1}{2}m|\mathbf{v}|^2=\frac{1}{2}m(v_x^2+v_y^2+v_z^2)$ $→\;\frac{dk}{dt}=m\mathbf{v}･\frac{d\mathbf{v}}{dt}$ $\mathbf{v}=\frac{d\mathbf{r}}{dt},\;m\frac{d\mathbf{v}}{dt}=\mathbf{F} →\;\frac{dk}{dt}=\mathbf{F}･\frac{dr}{dt}$ $→\;\int_{t_1}^{t_2}\frac{dk}{dt}dt=\int_{t_1}^{t_2} \mathbf{F}\frac{d\mathbf{r}}{dt}dt$ $k(t_2)-k(t_1)=\int_c\mathbf{F}d\mathbf{r}$ $\therefore\; \Delta k=W$ energy conservation law: $k(t_2)-k(t_1)=\int_c\mathbf{F}d\mathbf{r}$ $[F$=conservative force]: $\int_c\mathbf{F}d\mathbf{r} =\int_{\mathbf{r}_1}^{\mathbf{r}_2}\mathbf{F}d\mathbf{r} =\int_{\mathbf{r}_0}^{\mathbf{r}_2}\mathbf{F}d\mathbf{r} -\int_{\mathbf{r}_0}^{\mathbf{r}_1}\mathbf{F}d\mathbf{r} =-U(\mathbf{r}_2)+U(\mathbf{r}_1)$ mechanical energy: $k(t_2)-k(t_1)=-U(\mathbf{r}_2)+U(\mathbf{r}_1)$ $\therefore\;k(t_2)+U(\mathbf{r}_2)=k(t_1)+U(\mathbf{r}_1)$ If $\mathbf{F}$ is conservative force, the potential energy is defined as: $U(\mathbf{r})=-\int_{\mathbf{r}_0}^{\mathbf{r}}\mathbf{F}\mathbf{r}$...(*) then, $U(r')-U(r)=-\int_{r_0}^{r'}Fdr-(-\int_{r_0}^{r}Fdr = -(\int_{r}^{r_0}Fdr+\int_{r_0}^{r'}Fdr)=-\int_{r}^{r'}Fdr$ thus, conservative force: $W=-\Delta U$ ¶1: Potential energy & Force [1 dimension]: $U(x)=-\int_{x_0}^{x}F(x)dx$ $U(x+\Delta x)-U(x)=-\int_{x}^{x+\Delta x}F(x)dx\simeq -F(x)\Delta x$ $→\;F(x)=-\displaystyle\lim_{\Delta x\to 0} \frac{U(x+\Delta x)-U(x)}{\Delta x}=-\frac{U(x)}{dx}$ ¶2: Potential energy & Force [3D] $U(\mathbf{r})=-\int_{\mathbf{r}_0}^{\mathbf{r}}\mathbf{F}(\mathbf{r})d\mathbf{r}$ $U(x+\Delta x, y+\Delta y, z+\Delta z)-U(x, y, z)\simeq -(F_x\Delta x+F_y\Delta y +F_z\Delta z)$ when: $\Delta y=\Delta z=0$ $→\;U(x;\Delta x, y, z)-U(x, y, z)\simeq -F_x\Delta x$ $\therefore\; F_x=-\displaystyle\lim_{\Delta x\to 0}\frac{U(x;\Delta x, y, z)-U(x, y, z)}{\Delta x}$ thus: $\cases{F_x=-\frac{\partial U(\mathbf{r})}{\partial x}\\F_y=-\frac{\partial U(\mathbf{r})}{\partial y}\\F_z=-\frac{\partial U(\mathbf{r})}{\partial z}}$ $\therefore\; \mathbf{F}(\mathbf{r})=\bigl(-\frac{\partial U(\mathbf{r})}{\partial x},\;-\frac{\partial U(\mathbf{r})}{\partial y},\;-\frac{\partial U(\mathbf{r})}{\partial z}\bigr)=-\bigl(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\bigr)U(r)=-\nabla U(\mathbf{r})$ ¶3: Harmonic oscillation (2D): $U(x,y)=\frac{1}{2}k(x^2+y^2)$ $F_x=\frac{\partial (x,y))}{\partial x}=-kx$ $F_y=\frac{\partial (x,y))}{\partial y}=-ky$ $→\;\mathbf{F}(\mathbf{r})=-k(x,y)=-k\mathbf{r}$ $\nabla=(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z})→\;\nabla\times \mathbf{F}=\mathbf{0}$

３. 保存則: momentum: 運動量 impulse: 力積 inner product/vector product: 内積/外積 conservative force: 保存力 gradient: 勾配 harmonic oscillator: 調和振動子 =spring integration by substitution: 置換積分 Harmonic oscillator: >Top Vector product: $a\times b=\\ (a_yb_z-a_zb_y, \\ a_zb_x-a_xb_z, \\ a_xb_y-a_yb_x )$

>Top 4. Damped oscillation: Damping oscillation: $mx''=-kx-bx'\;$ [k=spring constant; b=damping coefficient] $→\; x''=-\omega_0^2x-2\gamma x'\; [\omega_0=\sqrt{\frac{k}{m}},; \gamma=\frac{b}{2m}]$...(*) assume:$x=e^{\lambda t}=-\omega_0^2e^{\lambda t}$ $→\;\lambda+2\gamma\lambda+\omega_0^2=0$ $→\;\lambda_{±}=-\omega±\sqrt{\gamma^2-\omega_0^2} [\lambda$ is -real number] 1) when: $\gamma>\omega_0:\;$ [over-damped] assume GS of (*) is $x=C_1e^{\lambda+t}+C_2e^{\lambda-t}$ here: $x(0)=a,\; x'(0)=0$ [exponentially damped] 2) when: $\gamma<\omega_0:\;$ $\lambda_{±}=-\gamma±\sqrt{\gamma^2-\omega_0^2}=-\gamma±i\sqrt{\omega_0^2-\gamma^2}=-\gamma±i\omega$ assume GS of (*) is $x=C_1e^{-\gamma+iw)t}+C_2e^{-\gamma-iw)t}$ $=e^{-\gamma t}(C_1e^{i\omega t}+C_2e^{-i\omega t})$ $=e^{-\gamma t}\bigl((C_1+C_2)\cos\omega t+i(C_1-C_2)\sin\omega t\bigr)$ $= e^{-\gamma t}(A\cos\omega t+iB\sin\omega t) =Ce^{-\omega t\cos(\omega t+\phi)};\;$ $[C=\sqrt{A^2+B^2},\; \tan\phi=-\frac{B}{A}]$ 3) when: $\gamma=\omega_0$ $\lambda_+=\lambda_-=-\lambda$ >Top assume GS of (*) is $x=C(t)e^{-\gamma t}$ then, $→\;x'=C'e^{-\gamma t}-C(\gamma e{-\gamma t})$ $x''=C''e^{-\gamma t}-2C'(\gamma e^{-\gamma t})+C(\gamma^2 e^{-\gamma t})$ substitute to (*): $C''e{-\gamma t}-2\gamma C'e{-\gamma t}+\gamma^2Ce{-\gamma t} =-\omega_0^2Ce{-\gamma t}+2\gamma^2e{-\gamma t}-2\gamma C'e{-\gamma t}$ $C''+C(\omega^2-\gamma^2)=0\; [e^{-\gamma t}≠0]$ $→\;C''=0 \; [\omega_0^2-\gamma^2=0]$ $→\;C=At+B　\therefore\; x=(At+B)e^{-\gamma t} \;$[critical damping] Energetic consideration: $mx''=-kx-bx'$ $→\;mx''x'=akx'x-bx'^2$ $→\;\frac{d}{dt}(\frac{1}{2}mv^2+\frac{1}{2}kx^2)=-bv^2<0$ [mechanical energy] $-bv^2→\;-bv\frac{dx}{dt}\;$ [work efficient] >Top Forced oscillation: $mx''=-kx+F\cos\omega t$ $x''=-w_0^2x+f\cos\omega t$...(*) [$w_0=\sqrt{\frac{k}{m}},\; f=\frac{F}{m}$] assume SS of (*): $x=a\cos\omega t$ substitute to (*): $-a\omega^2\cos\omega t =-a\omega_0^2\cos\omega t+f\cos\omega t$ $→\;a=-\frac{f}{\omega^2-\omega_0^2}$ SS of (*) is $x=-\frac{f}{\omega^2-\omega_0^2}\cos\omega t$ [resonance] GS of (*)=GS of homo.eq + SS of (*) assume GS of (*): $x=A\sin\omega_0 t+B\cos\omega_0 t -\frac{f}{\omega^2-\omega_0^2}\cos\omega t$ consider: $\omega →\omega_0; \;$initial condition: $x(0)=0,\; x'(0)=0$ $→\;a=0,\; B=\frac{f}{\omega^2-\omega_0^2}$ $→\;x=\frac{f}{\omega^2-\omega_0^2}(\cos\omega_0 t-\cos\omega t)$ $→\;x=-\frac{2f}{(\omega+\omega_0)(\omega-\omega_0)}$ $\sin\frac{\omega_0+\omega}{2}t\sin\frac{\omega_0-\omega}{2}t$ here: $\omega_0-\omega=\Delta\omega$ $x=\frac{2f}{\omega_0+\omega}\sin\frac{\omega_0+\omega}{2}t \frac{t}{2}\frac{\sin \frac{\Delta\omega}{2}t}{\frac{\Delta\omega}{2}t}$ $\Delta\omega→0\; x=\frac{ft}{2\omega_0}\sin\omega_0t$ [increasing ]

4. 減衰振動: damped oscillation/vibration: 減衰振動 damping coefficient: over-damped: 過減衰 critical damping: 臨界減衰 restoring force: 復元力 external force: 外力 forced oscillation: 強制振動 resonance: 共振･共鳴 Euler's formula: $e^{i\theta}=\cos\theta+i\sin\theta$ Damping oscillation: Critical damping: $\displaystyle\lim_{x\to 0} \frac{\sin x}{x}=1$ Forced oscillation:

>Top 5. Conservation law of angular momentum: moment of force (torque): $\mathbf{N}=\mathbf{r}\times \mathbf{F} [\mathbf{r}=$distance from the origin] conservation of angular momentum: $L=r\times p [L$=angular momentum; $p$=momentum]...(*) differentiate (*): $\frac{d\mathbf{L}}{dt}=\frac{dr}{dt}\times p+r\times \frac{dp}{dt} =v\times mv+r\times F$ here $[\frac{dr}{dt}=v,\;\frac{p}{dt}=F] $\therefore\; \frac{d\mathbf{L}}{dt}=\mathbf{N}\; [N$=torque] thus, when $F\parallel r, \; N=0$ $\frac{d\mathbf{L}}{dt}=\mathbf{0}→\;\mathbf{L}=C \;$ [conservation of angular momentum] Law of constant area velocity: $\Delta S=\frac{1}{2}rv\Delta t\sin\phi$ $→\;\frac{\Delta S}{\Delta t}＝\frac{1}{2}rv\sin\phi$...(*) while: angular momentum of planetary movement: $L=rmv\sin\phi \;[L$=angular momentum]...(**) $L=2m\frac{\Delta S}{\Delta t};\; [\frac{\Delta S}{\Delta t}$=area velocity] →\;[constant area velocity is one of law of constant angular momentum]

5. 角運動量保存則: angular momentum: 角運動量 moment of force: 力のモーメント 外積の微分: $\frac{d}{dt}(\mathbf{a}\times\mathbf{b}) =\frac{d\mathbf{a}}{dt}\times\mathbf{b} +\mathbf{a}\times\frac{d\mathbf{b}}{dt}$ Constant area velocity: $\Delta S=v\Delta t\sin\phi$

>Top 6. Inertial Frame: Inertial/Noninertial frame: $\mathbf{r}=\mathbf{R}+\mathbf{r}_0$ substitute to $mr''=F$: $m(R''+r_0'')=F→\;mr_0''=F-mR''$...(*) when, $O_0$ is $R=Vt\;$ [uniform linear motion] $→\; R'=V, \; R''=0$ substitute to (*): $mr_0''=F→\;S_0$ is inertial frame. right side of (*) $F-mR''$ looks like a kind of force: [inertial force] >Top Equivalence principle: $\mathbf{F}_g=m_g\mathbf{g}\; [m_g$=gravitational mass] while: $m_i\mathbf{a}=\mathbf{F}\; [m_i$=inertial mass] in only gravitational field: $m_i\mathbf{a}=m_g\mathbf{g} →\;\mathbf{a}=\frac{m_g}{m_i}\mathbf{g}$ regardless of any object: $\mathbf{a}=\mathbf{g}→\;m_g=m_i$ [equivalence principle]

6. 慣性系: noninertial frame: 非慣性系 uniform linear motion: 等速直線運動 equivalence principle: 等価原理 Inertial frame:

>Top 7. Coriolis Force: Rotating coordinate system: $\mathbf{r}=x_o\mathbf{e}_{ox}+y_o\mathbf{e}_{oy} =x\mathbf{e}_x+y\mathbf{e}_y$...(*) here: $\cases{e_x=\cos\omega t e_{ox}+\sin\omega t e_{oy} \\e_y=-\sin\omega t e_{ox}+\cos\omega t e_{oy}}$ $→\cases{e'_x=-\omega\sin\omega t e_{ox}+\omega\cos\omega t e_{oy} =\omega e_y \\e'_y=-\omega\cos\omega t e_{ox}-\omega\sin\omega t e_{oy} =-\omega e_x}$ $→\cases{e''_x=-\omega^2e_x\\e''_y=-\omega^2e_y}$ here EOM of (*): $F=mr''$ $→\;r''=x''e_x+y''e_y+2(x'e_x'+y'e_y')+xe_x''+ye_y''$ $→\;=x''e_x+y''e_y+2\omega(x'e_y-y'e_x)-\omega^2(xe_x+ye_y)$ substitute to (*)$mr''=F+m\omega^2r+2m\omega(y'e_x-x'e_y)$ here: $m\omega (y'e_x-x'e_y) =2m\omega \mathbf{v}\times e_z\;$ [Coriolis force]

7. コリオリの力: Coriolis force: コリオリの力 apparent force: 見かけの力 centrifugal force: 遠心力 $\cases{v=(x',y',z')\\e_z=(0,0,1)}$ $→\;v\times e_z=(y'-x')$

>Top 8. Archimedean spiral: Length in polar coordinates: $dL=\sqrt{(rd\theta)^2+(dr)^2}$...(*1) $→\;=\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta\;[r=f(\theta)]$ from (*1): $L=\displaystyle\int_0^{\pi}\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta$ $→\;=\displaystyle\int_0^{\pi}\sqrt{\theta^2+1}d\theta$ let $\theta=\sinh t→\;d\theta=\cosh tdt$ $L=\displaystyle\int_{0}^T \sqrt{\sinh^2t+1}\cosh tdt$ $=\int_{0}^{T}\cosh^2tdt=\int_{0}^{T}\frac{1+\cosh 2t}{2}dt$ $=\frac{1}{2}\left[t+\frac{1}{2}\sinh 2t\right]_0^T$ $=\frac{1}{2}T+\frac{1}{4}\sinh2T$ $=\frac{1}{2}T+\frac{1}{2}\sinh T\cosh T$ $=\frac{1}{2}\log(\pi+\sqrt{\pi^2+1})+\frac{1}{2}\pi\sqrt{\pi^2+1}\simeq 6.1099$ Re: $\theta=\frac{e^t-e^{-1}}{2}→\;(e^t)2-2\theta e^t-1=0$ $→\;e^t=\theta+\sqrt{\theta^2+1}→\;t=\log (\theta+\sqrt{\theta^2+1})$ $\begin{array}{c|l}\theta&0→\pi\\ \hline t&0→\log(\pi+\sqrt{\pi^2+1})=T\end{array}$ $\sinh T=\pi\; [←\theta~\sinh t]$ $\cosh T=\sqrt{\sinh^2T+1}=\sqrt{\pi^2+1}$

8. アルキメデスの螺旋:

>Top 9. Hyperbolic function: Hyperbolic function: $\sinh x=\frac{e^x-e^{-1}}{2}$ $\cosh x=\frac{e^x+e^{-1}}{2}$ $\tanh x=$ $(\sinh x)'=\cosh x$ $\cosh^2 x-\sinh^2 x=1$ $\cosh^2 x=\frac{1+\cosh 2x}{2}$ $\sinh 2x=2\sinh x\cosh x$ comparison: $\begin{array}{l|l}\cosh^2x-\sinh^2x=1&\cos^2x+\sin^2x=1\\ \tanh x=\frac{\sinh x}{\cosh x} &\tan x=\frac{\sin x}{\cos x}\\ 1-\tanh^2x=\frac{1}{\cosh^2x}&1+tan^2x=\frac{1}{\cos^2x}\end{array}$ $\begin{array}{l|l}(\sinh x)'=\cosh x&(\sin x)'=\cos x\\ (\cosh x)' =\sinh x&(\cos x)'=-\sin x\\ (\tanh x)'=\frac{1}{\cosh^2x}&(\tan x)'=\frac{1}{\cos^2x}\end{array}$ $\begin{array}{l|l}\sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y &\sin(x+y)=\sin x\cos y+\cos x\sin y \end{array}$ parametric: $x=\cosh\theta,\;y=\sinh\theta→\;x^2-y^2=1\;$ [hyperbola] cf: $x=\cos\theta,\;y=\sin\theta→\;x^2+y^2=1$ [circle] Euler's formula: $e^{i\theta}=\cos\theta+i\sin\theta$ $→\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$ $→\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$

9. 双曲線関数: hyperbolic function: 双曲線関数 catenary: 懸垂線

>Top 10. Inverse trigonometric function: $y=\arcsin x\; [-1≤x≤1,\;-\frac{\pi}{2}≤y≤\frac{\pi}{2}]$ $x=\sin y→\;\frac{dx}{dy}＝\cos y=\sqrt{1-\sin^2 y}$ $→\;\frac{dy}{dx}=\frac{1}{\sqrt{1-\sin^2 y}}=\frac{1}{\sqrt{1-x^2}}$ $y=\arccos x\;[-1≤x≤1,\;0≤y≤\pi]$ $x=\cos y→\;\frac{dx}{dy}=-\sin y=-\sqrt{1-\cos^2 y}$ $→\;\frac{dy}{dx}=-\frac{1}{\sqrt{1-\cos^2 y}}=-\frac{1}{\sqrt{1-x^2}}$ $y=\arctan x\; [-\infty≤x≤\infty,\;-\frac{\pi}{2}≤y≤\frac{\pi}{2}]$ $x=\tan y→\;\frac{dx}{dy}＝\frac{1}{\cos^2y}=1+\tan^2y$ $→\;\frac{dy}{dx}=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2}$

10: 逆三角関数:

>Top 11. Two-particle system motion: EOM: $m_1\mathbf{r}_1''=\mathbf{F}_1+\mathbf{F}_{12}$...(1) $m_2\mathbf{r}_2''=\mathbf{F}_2+\mathbf{F}_{21}$...(2) (1)+(2): $m_1r_1''+m_2r_2''=F_1+F_2$ here: $r_G=\frac{m_1r_1+m_2r_2}{m_1+m_2}=\frac{m_1r_1+m_2r_2}{M}$ [G=center of gravity; M=total mass] $→\;Mr_G''=F_1+F_2$ [external force only] >Top relative vector: $r=r_2-r_1$ [viewpoint from $r_1$] $r''=r_2''-r_1''=\frac{F_2+F_{21}}{m_2}-\frac{F_1+F_{12}}{m_1} =(\frac{1}{m_1}+\frac{1}{m_2})F_{21}+\frac{F_2}{m_2}-\frac{F_1}{m_1}$ $=\frac{1}{\mu}+\frac{F_2}{m_2}+\frac{F_1}{m_1}$ $[\mu:\;$ reduced mass]　 $→\;\mu r''=F_{21}+\frac{\mu}{m_2}F_2-\frac{\mu}{m_1}F_1$ $→\;\mu r''=F_{21}\;$ [unless external force] Motion of connected objects: (>Fig.) motion of center of gravity: $Mx_G''=0\;$ [constant linear motion] relative motion: $\mu x''=-k(x_2-x_1-l)$ [$\mu:$ reduced mass] here: $\omega=\sqrt{\frac{k}{\mu}}\;$ [simple oscillation] Total momentum: $P=p_1+p_2=m_1r_1'+m_2r_2'$ $P'=m_1r_1''+m_2r_2''=F_1+F_2$ $→\;P'=0→\;P=C\;$ [unless external force] Total kinetic energy: $K=k_1+k_2=\frac{1}{2}m_1r_1'^2+\frac{1}{2}m_2r_2'^2$...(*) here: $\cases{r_G=\frac{m_1r_1+m_2r_2}{M}\\r=r_2-r_1}$ $→\;\cases{r_1=r_G-\frac{m_2}{M}r\\r_2=r_G+\frac{m_1}{M}r}$ substitute to (*): $K=\frac{1}{2}m_1(r_g-\frac{m_2}{M}r)^2+\frac{1}{2}m_2(r_g+\frac{m_1}{M})^2$ $→\;=\frac{1}{2}(m_1+m_2)r_G'^2 +\frac{1}{2}\frac{m_1m_2(m_1+m_2)}{M}ｒ'^2$ $→\;=\frac{1}{2}Mr_G'^2+\frac{1}{2}\mu Mr'^2$ [momentum of COG & reduced mass] Total angular momentum: $L=L_1+L_2=r_1\times p_1+r_2\times p_2=r_1\times m_1r_1' +r_2\times m_2r_2'$...(*) $→\;L'=r_1'\times m_1r_1'+r_1\times m_1r_1''+r_2'\times m_2r_2'+r_2\times m_2r_2''$ $=r_1\times (F1+F_{12})+r_2\times (F_2+F_{21})$ $=(r_2-r_1)\times F_{21}+r_1\times F_1+r_2\times F_2 =r_1\times F_1+r_2\times F_2\;$ [each torque] consider COG and substitute to (*): $L=(r_G-\frac{m_2}{M}r)\times m_1(r_G'-\frac{m_2}{M}r') +(r_G+\frac{m_1}{M}r)\times m_2(r_G'+\frac{m_2}{M}r')$ $=m_1r_G\times r_G'-\frac{m_1m_2}{M}r_G\times r' -\frac{m_1m_2}{M}r\times r_G'+\frac{m_1m_2^2}{M^2}r\times r'$ $+m_2r_G\times r_G'+\frac{m_1m_2}{M}r_G\times r' +\frac{m_1m_2}{M}r\times r_G'+\frac{m_1^2m_2}{M^2}r\times r'$ $=r_G\times (m_1+m_2)r_G'+r\times\frac{m_1m_2(m_1+m_2)}{M^2}r'$ $\therefore\;L=r_G\times Mr_G'+r\times\mu r'\;$ [torque of COG and torque of relative vector]

11. 二粒子系の運動: relative vector: 相対位置ベクトル reduced mass: 換算質量 total momentum: 全運動量 center of gravity: COG >Top Commutative law of vector product: $\mathbf{a}\times (\mathbf{b}+\mathbf{c}) =\mathbf{a}\times \mathbf{b} +\mathbf{a}\times \mathbf{c}$ <Summary>: EOM of COG: $Mr_G''=F_1+F_2$ Relative EOM: $\mu r''=F_{21}$ [no ex-force] Total momentum: $P=p_1+p_2$ $P=$constant [without ex-force] Total kinetic energy: $K=k_1+k_2$ $\frac{1}{2}Mr_G'^2 ＋\frac{1}{2}\mu r'^2$ Total angular momentum: $L=L_1+L_2$ $=r\times Mr_G'+r\times\mu r'$ $L'=N_1+N_2$ [torque]

>Top 12. Rigid body dynamics: Rigid body: deformation is zero or negligible; the distance between any two point remains constant regardless of external forces exerted; continuous distribution of mass. Rigid body of EOM: $P_G'=F$ [EOM of COG] $L'=N$ [rotation around the axis passing COG] EOM of COG: $m_ir_i''=f_i+\displaystyle\sum_jFf_{ij}\; [ex-power+in-power;\;i≠j]$ here; $Mr_G''=\frac{\sum_im_ir_i}{\sum_im_i}\;$ [COG] feature: 1) degree of freedom: 3n→6; 2) offset of internal-power $Mr_G''=\sum_if_i$ here: $P_G=Mr_G',\; F=\sum_if_i$ $P_G'=F$ Rotary motion: $P_i'=f_i+\sum_jf_{ij}\; [momentum=exforce + inforce]$→\;r_i\times P_i'=r_i\times (f_i+\sum_jf_{ij})\; $r_i\times p_i'=\frac{d}{dt}(r_i\times p_i)$...(*1) $\because\; \frac{d}{dt}(r_i\times p_i)=r_i'\times p_i+r_i\times p_i'$...(*2) here: $r_i'\times p_i=0$ [parallel vector] here: $\sum_i\sum_j(r_i\times f_{ij})=\sum_{i>j}(r_i-r_j)\times f_{ij}=0$ $\because\;$if$\; n=3→\;\sum_i\sum_j(r_i\times f_{ij})$ $=r_1\times f_{12}+r_1\times f_{13}+r_1\times f_{21} +r_1\times f_{23}+r_1\times f_{31}+r_1\times f_{32} =0$ from (*2): $\frac{d}{dt}\sum_i(r_i\times p_i)=\sum_i(r_i\times f_i)$...(*3) from (*3): $L'=\sum_i(r_i\times p_i);\;N=\sum_i(r_i\times f_i)=\;$ [L= angular momentum; N=moment of force] $\therefore\; \boxed{L'=N}$ [around the origin] Relative position from COG: here;$r_G=\frac{\sum_im_ir_i}{M};\;r_G'=\frac{\sum_im_ir_i}{M} =frac{\sum_ip_i}{M}$ substitute to (*3): $r_i=r_o+r_G$ $→\;\frac{d}{dt}\sum_i((r_{oi}+r_G)\times p_i)=\sum_i((r_{oi}+r_G)\times f_i)$ $→\;\frac{d}{dt}\sum_i(r_{oi}\times p_i)+\frac{d}{dt}\sum_i(r_G\times p_i) =\sum_i(r_{oi}\times f_i)+\sum_i(r_G\times f_i)$...(*4) here: $\frac{d}{dt}\sum_i(r_G\times p_i)=\sum_i(r_G'\times p_i+r_G\times p_i') =\sum_i(r_G\times p_i')$ here: $p_i'=f_i+\sum_jf_{ij}=f_i$ (*4) is: $\frac{d}{dt}\sum_i(r_{oi}\times p_i)=\sum_i(r_{oi}\times f_i)$ $→\;p_{oi}'=p_i-m_ir_G'$ $→\;\frac{d}{dt}\sum_i(r_i'\times (p_i'+m_ir_G'))=\sum_i(r_{oi}\times f_i)$ $→\;\frac{d}{dt}\sum_i(r_{oi}\times p_{oi}') +\frac{d}{dt}((\sum_im_ir_{oi})\times r_G')=\sum_i(r_{oi}\times f_i)$ here: $\sum_im_ir_{oi}=\sum_i(r_i-r_G)=\sum_im_ir_i-r_G\sum_im_i=0$ here: $L_o=\sum_ir_{oi}\times p_{oi};\; N_o=\sum_i(r_{oi}\times f_i)$ $\therefore\; \boxed{L_o'=N_o}$ [rotary motion around COG]

12. 剛体の力学: rigid body: 剛体 point mass: 質点 point of application: 作用点 line of action: 作用線 degree of freedom: 自由度 manifold: 多様体 deformation: 変形 nonlocal action; action at a distance: 遠隔作用 roll/yaw/pitch: position of a rigid body Rotation around the axis passing COG: $\frac{d}{dt}(r_i\times p_i) =r_i'\times p_i +r_i\times p_i'$ here; $r_i'\times p_i=0$ Relative position: 原点: $L'=N$ 重心: $L_o'=N_o$

>Top 13. Inertia moment of rigid body: Inertia moment: Motion with fixed axis: →degree of freedom is 1. $L_z'=N_z$ here: $L_z=\sum_i(r_i\times p_i)_z=\sum_i(x_iy_i'-y_ix_i') =\int \rho (r)(xy'-yx')d^3r$..(*) here: $x=r\cos\phi,\; y=r\sin\phi$ $x'=-r\phi'\sin\phi,\; y'=r\phi'\cos\phi$ $→\;xy'-yx'=r^2\phi'\cos^2\phi+r^2\phi'\sin^2\phi=r^2\phi'$ substiture to (*): $L_z=(\int r^2\rho (r)d^3r)\phi'$ $=I_z\phi'\;$[inertia moment around z-axis] $\therefore\;\boxed{I_z\phi''=N_z}\;$ [I_z: difficulty of rotation] Inertia moment of COG: $I=\int r^2\rho(r)d^3r$...(*2) >Top Parallel axes theorem: $I=I_G+Mh^2\;$ [M=total mass; h=distance between axes] here: $r_o=r-r_G$ substitute to (*2): $I=\int r^2\rho(r)d^3r$ $=\int (x^2+y^2)\rho(r)d^3r$ $=((x_o+x_g)^2+(y_o+y_G)^2)\rho(r)d^3r$ $=(x_g^2+y_g^2)\int \rho(r)d^3r+2x_g\int x_o\rho(r)d^3r$ $++2y_g\int y_o\rho(r)d^3r+\int (x_o^2+y_o^2)\rho(r)d^3r$ here: $\int x_o\rho(r)d^3r=\int(x-x_G)\rho(r)d^3r$ $=\int x\rho(r)d^3r-x_g\int\rho(r)d^3r =Mx_g-x_gM=0$ $\therefore\; \boxed{I=I_g+Mh^2}$ >Top Orthogonal axes theorem: $I_z=\in r^2\sigma(r)d^3r=\int (x^2+y^2)\sigma(r)d^2r$ $=\int x^2\sigma(r)d^2r+\int y^2\sigma(r)d^2r=I_y+I_x$ $\therefore\; \boxed{I_z=I_x+I_y}$

13. 剛体の慣性モーメント: inertia moment: 慣性モーメント orthogonal axis: 直交軸 Parallel axis theorem: Orthogonal axis:

>Top 14. Mechanical energy of rigid body: Kinetic energy of rigid body: $\sum_i\frac{1}{2}m_i|r_i'|^2$ $=\sum_i\frac{1}{2}m_i|r_{oi}'+r_G'|^2$ $=\frac{1}{2}\sum_im_i|r_G'|^2+\sum_im_ir_{oi}'r_G' +\frac{1}{2}\sum_im_i|r_{oi}|^2$ $=\frac{1}{2}|r_g'|^2+r_g(\sum_im_ir_{oi}')+\frac{1}{2}\sum_im_i|r_{oi}|^2$ =translational motion + kinetic energy around COG [$\because \sum_im_ir_{oi}=0$] here [motion around fixed axis passing COG] $\frac{1}{2}\sum_im_i|r_{oi}'|^2=\frac{1}{2}\sum_im_i(r_i\omega)^2$ $=\frac{1}{2}\omega^2\sum_im_ir_i^2=\frac{1}{2}\omega^2\int r^2\rho(r)d^3r$ $=\frac{1}{2}I\omega^2$ $\therefore\; K=\frac{1}{2}Mv_G^2+\frac{1}{2}I\omega^2$ [translational.e+rotational.e] Potential energy of rigid body: $\sum_im_igz_i=g\sum_im_iz_i=Mgz_G$ $\therefore\; \boxed{U=Mgh}$ [h=height of COG from the origin]

14. 剛体の力学的エネルギー: translational motion: 並進運動

>Top 15. Matrix exponential: Proof: $e^0=E+0+\frac{1}{2}0^2+\cdots=E$ $\frac{d}{dt}e^{At}=\frac{d}{dt}(E+At+\frac{1}{2}(At)^2+\cdots)$ $=0+A+A^2t+\frac{1}{2!}A^3t^2+cdots$ $=A(E+At+\frac{1}{2!}A^2+cdots)=Ae^{At}$ ¶ $\frac{d}{dt}\mathbf{y}(t)=A\mathbf{y}(t)$ $→\;y(t)=e^{At}y_0\; [y_0=y(0)]$

15. 行列指数関数: matrix exponential: 行列乗 commutative: 可換 Maclaurin expansion: $e^x=1+x+\frac{1}{2!}x^2 +\frac{1}{3!}x^3+\cdots$ $=\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}x^k$ $e^A=\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}A^k\;$ [A=square matrix] $e^0=E$ $\frac{d}{dt}e^{At}=Ae^{At}$ if $AB=BA →e^Ae^B=e^{A+B}$

>Top 16. Chauchy's functional equation: definition: $f(x+y)=f(x)+f(y)\;$...(*) [real continuous function] is $f(x)=ax\;$ the only function to satisfy (*)?] proof: when $x=y=0→\;f(0)=f(0)+f(0)→\;f(0)=0\;$ [cross the origin] when $y=-x→\;f(0)=f(x)+f(-x)→\;f(x)=-f(x)\;$ [odd function]...(*2) when $n$ is natural number, and $x$ is real number, then $f(nx)=nf(x)$ when $n=1$ is established. if n=k is established, then replace $x→kx$ and $y→x$ then, $f(kx+x)=f(kx)+f(x)=kf(x)+f(x)$ $\therefore\;f((k+1)x)=(k+1)f(x)$ is established. then, any $n$ of natural number: $f(nx)=nf(x)$ from (*2), substitute: $x→nx$, then $f(-nx)=-f(nx)=-nf(x)\;[-n:$ negative number] $→\;f(mx)=mf(x)\;$ [m: integer] when rational number $r=\frac{m}{n}:$ $→nf(rx)=f(nrx)=f(mx)=mf(x)$ $→f(rx)=\frac{m}{n}f(x)=rf(x)$...(*3) substitute to (*3): $x=1→\;f(r)=rf(1)$ let $f(1)=a$ then: $f(r)=ar\;[x:$rational number] $^\exists \{q_n\}$ such that $^\forall x=\displaystyle\lim_{n\to\infty}q_n=x\; [\{q_n\}:$ rational series] and $f(x)$ is continuous function. $→f(x)=\displaystyle\lim_{n\to\infty}f(q_n) =\displaystyle\lim_{x\to\infty}aq_n=ax$ $\therefore\;f(x)=ax$■ when: $f(x)$ is continuous function, then consider 'rational number' only: [density] when $x$ is real number, which is expressed as follows: $x=m+\displaystyle\sum_{i=1}^{\infty}C_i10^{-i}\; [m, C_i:$ integer, $0≤C_i≤9$] rational series: $q_n=m+\displaystyle\sum_{i=1}^nC_i10^{-i}→x (n→\infty)$

16. コーシーの関数方程式: real function: 実関数 mathematical induction: 数学的帰納法 density: 稠密性   数学的帰納法, mathematical induction: base case: show f(0) is clearly true. inductive step: show for any k≥0, if f(k) holds, then f(k+1) also holds.

>Top 17. Taylor expansion: $f(x)=f(a)+f'(a)(x-a)+\frac{1}{2!}f''(a)(x-a)^2+\frac{1}{3!}f'''(a)(x-a)^3+\cdots$...(*) where: $h=x-a$, then (*) is: $f(a+h)=f(a)+f'(a)h+\frac{1}{2!}f''(a)h^2+\frac{1}{3!}f'''(a)h^3+\cdots\;$ [Taylor expansion around $a$] where: $a=0=x→\;f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+\cdots$ $=\displaystyle\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$ [Maclaurin expansion] ¶1: $f(x)=e^x\; x=0$...(*) to find linear function in the vicinity of (*): $f(0)=1,\; f'(0)=1→\;g(x)=1+x$ to find quadratic function in the vicinity of (*): $f(0)=1,\; f'(0)=1,\;f''(0)=1→\; g(x)=1+x+\frac{1}{2}x^2 $e^x=e^0+e^0x+\frac{1}{2!}e^0x^2+\frac{1}{3!}e^0x^3+\frac{1}{4!}e^0x^4+\cdots$ $=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+\frac{1}{5!}x^5+\cdots$ $\therefore\;\boxed{e^x=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}x^n}$ $\log (1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\cdots =\displaystyle\sum_{n=0}^{infty}\frac{(-1)^n}{n+1}x^{n+1}$ $f(x)=\log(1+x)→\;f^{(n)}(x)=$ $\log 2=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\cdots =\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\log 2\;$[Mercator series] $(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$ $\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x7+\cdots$ $f(0)=\sin 0=0,\;f'(0)=\cos 0=1,\;f''=-\sin 0=0,\;f'''=-\cos 0=-1, \cdots$ $\sin x=0+1x+\frac{0}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+\cdots$ $=0+1x+\frac{0}{2!}x^3+\frac{-1}{3!}x^5-\frac{0}{4!}x^6+\cdots$ $=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots$ $\therefore\;\boxed{\sin x=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}}$ $\cos x=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\cdots$ $\therefore\;\boxed{\cos x=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}}$ $\tan x=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\cdots$ Euler's formula: $i\sin x=ix-i\frac{1}{3!}x^3+i\frac{1}{5!}x^5-i\frac{1}{7!}x^7+\cdots$ $e^{ix}=1+ix-\frac{1}{2!}x^2-i\frac{1}{3!}x^3+\frac{1}{4!}x^4+i\frac{1}{5!}x^5-\cdots$ $→e^{ix}=\cos x+i\sin x\;$ [Euler' formula] $→\cos x=\frac{e^{ix}+e^{-ix}}{2},\;\sin x=\frac{e^{ix}-e^{-ix}}{2}$ $e^{i(a+b)}=e^{ia}e^{ib}$ $→\;\cos (a+b)+i\sin (a+b)=(\cos a+i\sin a)(\cos b+i\sin b)$ $=(\cos a\cos b-\sin a\sin b)+i(\cos a\sin b+\sin a\cos b)\;$ [addition theorem] $e^{in\theta}=(e^{i\theta})^n$ $\cos(2\theta)+i\sin(2\theta)=(\cos\theta+i\sin\theta)^2$ $(\cos^2\theta-\sin^2\theta)+i2\cos\theta\sin\theta\;$ [double-angle formula] $\cos(3\theta)+i\sin(3\theta)=(\cos\theta+i\sin\theta)^3$ $=(\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)$ $=(4\cos^3\theta-3\cos\theta)+i(3\sin\theta-4\sin^3\theta)\;$ [tripple-angle formula] $\sin^2(\frac{x}{2})=\bigl(\frac{e^{i\frac{x}{2}}-e^{-i\frac{x}{2}}}{2i}\bigr)^2=\frac{e^{ix}+e^{-ix}-2}{-4}=\frac{1-\cos x}{2\;}$ [half-angle formula] $\cos^2(\frac{x}{2})=\bigl(\frac{e^{i\frac{x}{2}}+e^{-i\frac{x}{2}}}{2}\bigr)^2=\frac{e^{ix}+e^{-ix}+2}{4}=\frac{1+\cos x}{2\;}$ $\cos a\sin b=\frac{e^{ia}+e^{-ia}}{2}･\frac{e^{ib}-e^{-ib}}{2i}$ $=\frac{e^{i(a+b)}-e^{-i(a+b)}-e^{i(a-b)}+e^{-i(a-b)}}{4i}$ $=\frac{1}{2}\bigl(\frac{e^{i(a+b)}-e^{-i(a+b)}}{2i}-\frac{e^{i(a-b)}-e^{-i(a-b)}}{2i}\bigr)=\frac{1}{2}(\sin (a+b)-\sin (a-b))\;$[sum of products] $\cos a\cos b=\frac{1}{2}(\cos (a+b)+\cos (a-b))$ $\sin a\sin b=-\frac{1}{2}(\cos (a+b)-\cos (a-b))$ $\sin a\cos b=\frac{1}{2}(\sin (a+b)+\sin (a-b))$ $e^{ia}+e^{ib}=e^{\frac{i(a+b)}{2}}e^{\frac{i(a-b)}{2}} +e^{\frac{i(a+b)}{2}}e^{\frac{-i(a-b)}{2}}$ $=e^{\frac{i(a+b}{2}}\left(e^{\frac{i(a-b)}{2}}+e^{\frac{-i(a-b)}{2}}\right)$ $=\left(\cos (\frac{a+b}{2})+i\sin (\frac{a+b}{2})\right) ･\left(\cos (\frac{a-b}{2})+i\sin (\frac{a-b}{2})+\cos (\frac{a-b}{2})-i\sin (\frac{a-b}{2})\right)$ $=2\cos (\frac{a-b}{2})\left(\cos (\frac{a+b}{2})+i\sin (\frac{a+b}{2})\right)$ $=2\cos (\frac{a-b}{2})\cos (\frac{a+b}{2})+i2 \cos (\frac{a-b}{2})\sin (\frac{a+b}{2})$ while $e^{ia}+e^{ib}=(\cos a+i\sin a)+(cos b+\sin b)$ $=(\cos a+\cos b)+i(\sin a+\sin b)→\;$ $\cos a+\cos b=2\cos (\frac{a+b}{2})\cos (\frac{a-b}{2})\;$ [product of sum] $\cos a-\cos b=-2\sin (\frac{a+b}{2})\sin (\frac{a-b}{2})$ $\sin a+\sin b=2\sin (\frac{a+b}{2})\cos (\frac{a-b}{2})$ $\sin a-\sin b=2\cos (\frac{a+b}{2})\sin (\frac{a-b}{2})$

17. テイラー展開: radius of convergence: 収束半径 : ダランベールの収束判定法 d'Alembert's ratio test: $\displaystyle\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}|=r$ $r<1:\;$ convergence $r>1:\;$ divergence

>Top 18. Fourier expansion: continuous $f(x)$ defined in $0≤x≤2\pi$ can be expressed by the combination of trigonometric funtions, i.e.: $f(x)=c+\displaystyle\sum_{n=1}^{\infty}\bigl(a_n\cos (nx)+b_n\sin (nx)\bigr)$...(*1) where $c$ is the average of (*1): $c=\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(x)dx$...(*2) coefficient $a_n, \;b_n$ can be: $\cases{a_n=\frac{1}{\pi}\displaystyle\int_0^{2\pi}f(x)\cos(nx)dx \\b_n=\frac{1}{\pi}\displaystyle\int_0^{2\pi}f(x)\sin(nx)dx}$ where, $n=0$ of (*2): $→\;c=\frac{1}{2}a_0 so, $\boxed{f(x)=\frac{a_0}{2}+\displaystyle\sum_{n=1}^{\infty} \bigl(a_n\cos (nx)+b_n\sin (nx)\bigr)}$...(*3) [Fourier series] here: $\displaystyle\int_0^{2\pi}\cos(mx)\cos(nx)dx$ $=\frac{1}{2}\displaystyle\int_0^{2\pi}\bigl(\cos(m+n)x+\cos(m-n)x\bigr)dx$ $=\frac{1}{2}\left[\frac{1}{m+n}\sin (m+n)x+\frac{1}{m-n}\sin (m-n)x\right]_0^{2\pi}=\pi \delta_{mn}\;$ [Kronecker delta: $\delta_{mn}=1\;(m=n),\;0\;(m\ne n)]$ similarly: $\displaystyle\int_0^{2\pi}\sin (mx)\sin (nx)dx=\pi\delta_{mn}$ $\displaystyle\int_0^{2\pi}\sin (mx)\cos (nx)dx=0$

18. フーリエ展開:

>Top 19. Linear algebra: Gaussian elimination method: multiply 1st row: $-\frac{d}{a}$ then add to 2nd row: $\pmatrix{a&b&c\\d&e&f\\g&h&i}\;$ multiply 1st row: $-\frac{g}{a}$ then add to 3rd row: $\pmatrix{a&b&c\\0&e'&f'\\g&h&i}\;$ multiply 1st row: $-\frac{g}{a}$ then add to 3rd row: $\pmatrix{a&b&c\\0&e'&f'\\0&h'&i'}\;$ multiply 2nd row: $-\frac{h'}{e'}$ then add to 3rd row: $\pmatrix{a&b&c\\0&e'&f'\\0&0&i''}\;$ Cofactor expansion: $\pmatrix{a&b&c&d&\\0&e&f&g\\0&0&h&-i&\\0&0&0&j}=a･e･h･j$ $|A|=a_{i1}\tilde{a}_{i1}+a_{i2}\tilde{a}_{i2}+\cdots+a_{in}\tilde{a}_{in} =\displaystyle\sum_{j=1}^na_{ij}\tilde{a}_{ij}$ $=a_{1j}\tilde{a}_{1j}+a_{2j}\tilde{a}_{2j}+\cdots+a_{nj}\tilde{a}_{nj} =\displaystyle\sum_{i=1}^na_{ij}\tilde{a}_{ij}$ Properties of matrix (elementary row operations): Row switching: $R_i \leftrightarrow R_j$ Scalar times: row multiplication: $kR_i→R_i, \;\mathrm{where}\;k≠0$ row additon: $R_i+kR_j\leftrightarrow R_i, \;\mathrm{where}\; i≠j$ invere of matrix: $T_{ij}^{-1}=T_{ij}$ same two rows makes the matrix is $0$. $k$ times of a row and then add it to other row makes no change of the matrix. Commutative law: $AB\ne BA$ $(A+B)+C=A+(B+C)$ Associative law: $A(BC)=(AB)C$ Indenty matrix: $E=\pmatrix{1&0&0\\0&1&0\\0&0&1}$ $AE=A,\; EA=A$ >Top Inverse matrix: $BA=E→\;A^{-1}A=E→\;AA^{-1}=E\;$ [regular matrix] from Cramer's rule: $\pmatrix{a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}} \pmatrix{x_{11}&x_{12}&x_{13}\\x_{21}&x_{22}&x_{23}\\ x_{31}&x_{32}&x_{33}} =\pmatrix{1&0&0\\0&1&0\\0&0&1}$ $x_{11}=\frac{\tilde{a}_{11}}{|A|},\; x_{21}=\frac{\tilde{a}_{12}}{|A|},\; x_{31}=\frac{\tilde{a}_{13}}{|A|},\;$ $X=\frac{1}{|A|}\pmatrix{\tilde{a}_{11}&\tilde{a}_{21}&\tilde{a}_{31}\\ \tilde{a}_{12}&\tilde{a}_{22}&\tilde{a}_{32}\\ \tilde{a}_{13}&\tilde{a}_{23}&\tilde{a}_{33}} =\frac{1}{|A|}^tA^{ij}$ $A=\pmatrix{a&b\\c&d},\; A^{-1}=\frac{1}{ad-bc}\pmatrix{d&-b\\-c&a}$ Matrix multiplication : $|AB|=|A||B|$ $|AA^{-1}|=|A||A^{-1}|=1→\;|A^{-1}|=\frac{1}{|A|}$ >Top Linearly dependent: if there exist scalars $a_1,a_2,\cdots,a_n$ , not all zero, such that: $a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+a_n\mathbf{v}_n=\mathbf{0}$...(*) $\mathbf{v}_1=\frac{-a_2}{a_1}\mathbf{v}_2+\cdots+\frac{-a_n}{a_1}\mathbf{v}_n\;$ [if $a_1≠0$] thus $\mathbf{v}_1$ is shown to be a linear combination of the remaining vectors. a sequence of vectors ($\mathbf{v}_1,\mathbf{v}_2,\cdots,\mathbf{v}_n$) is linearly independent if (*) can only be satisfied by $a_i=0$ for $i=1,\cdots,n.$ this implies that no vector in the sequence can be represented as a linear combination of the remaining vectors. $\displaystyle\sum_{i=1}^na_i\mathbf{v}_i=\mathbf{0} \Rightarrow a_1=\cdots=a_n=0$ a certain vector $\mathbf{x}$ can be uniquely expressed by linear combination of linearly independent vectors ($\mathbf{v}_1,\mathbf{v}_2,\cdots,\mathbf{v}_n).$ >Top Rank of a matrix $A$: is the maximum number of linearly independent row of vectors of the matrix: rank $A=n$ is the dimension of the image of the linear map represented by $A$; which is converted to number of the rank dimension. by simplification of a square matrix, if $^\exists EA$, then $^\exists A^{-1}$ eg.: $B=\pmatrix{1&0&1\\-2&-3&1\\3&3&0}=\pmatrix{1&0&1\\0&-3&3\\0&0&0}$: the first two columns linearly independent, but the third is a linear combination of the first two; thus the rank is 2. when the upper triangle matrix B has a column (or row) of all 0 elements (=rank down); the rank is $n-1$, here 3. Trace: sum of its diagonal entries. tr$(AB)=\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^na_{ij}b_{ij} =\mathrm{tr}(BA)$ tr$(A)=\mathrm{tr}(^tA)$ >Top Transposed matrix: $A=\pmatrix{a_{11}&\cdots&a_{1n}\\ \vdots&&\vdots\\a_{m1}&\cdots&a_{mn}}$ $^tA=\pmatrix{a_{11}&\cdots&a_{m1}\\ \vdots&&\vdots\\a_{1n}&\cdots&a_{mn}}$ $^{tt}A=A,\;^t(A+B)=^tA+^tB,\;^t(kA)=k^tA,\; ^t(kA+lB)=k^tA+l^tB,\;^t(AB)=^tB^tA$ (square matrix): $^t(A^{-1})=(^tA)^{-1},\; \mathrm{tr}A=\mathrm{tr}^tA$ $\mathrm{det} A=\mathrm{det}^tA$ $\langle Ax,y\rangle=\langle x,^tAy\rangle$ Coordinates transformation: $\pmatrix{p_x&q_x&r_x\\p_y&q_y&r_y\\p_z&q_z&r_z}\pmatrix{x'\\y'\\z'} =\pmatrix{x\\y\\z}$ $→\;\pmatrix{x'\\y'\\z'} =\pmatrix{p_x&q_x&r_x\\p_y&q_y&r_y\\p_z&q_z&r_z}^{-1}\pmatrix{x\\y\\z}$ $\pmatrix{p_x&q_x&r_x\\p_y&q_y&r_y\\p_z&q_z&r_z}\pmatrix{x'\\y'\\z'} =\pmatrix{s_x&t_x&u_x\\s_y&t_y&u_y\\s_z&t_z&u_z}\pmatrix{x\\y\\z}$ $→\;\pmatrix{x'\\y'\\z'} =\pmatrix{p_x&q_x&r_x\\p_y&q_y&r_y\\p_z&q_z&r_z}^{-1} \pmatrix{s_x&t_x&u_x\\s_y&t_y&u_y\\s_z&t_z&u_z}\pmatrix{x\\y\\z}$ >Top Eigenvalue and eigenvector: $\bf{Ax}=\lambda \bf{x}\;[x:$ eigenvector] $\bf{Ax}-\lambda \bf{Ex}=0→\;(\bf{A}-\lambda \bf{E})\bf{x}=0$ $(\bf{A}-\lambda \bf{E})=0\; [x\ne 0$; Eigen equation] $\lambda=\lambda_i,\;(i=1,2,\cdots,n)\;$ [eigenvalue] solve: $(A-\lambda_iE=0)x=0\;$ [n-unknown simul.linear.eq] $→\;x=x_\;(i=1,2,\cdots,n)\;$ [eigenvector] ¶1:$A=\pmatrix{2&3\\4&1}$ $\pmatrix{2-\lambda&3\\4&1-\lambda}=(2-\lambda)(1-\lambda)-12$ $→\;=(\lambda-5)(\lambda+2)=0\;\therefore\;\lambda=5, -2$ i) when: $\lambda=5:→\;\pmatrix{-3&3\\4&-4}\pmatrix{x\\y} =\pmatrix{0\\0} \Leftrightarrow \\cases{-3x+3y=0\\4x-4y=0} \Leftrightarrow　x-y=0$ $x=s_1:, y=s_1→\;x_1=s_1\pmatrix{1\\1} \; (s_1≠0)$ ii) when: $\lambda=-2:→\;\pmatrix{4&3\\4&3}\pmatrix{x\\y} =\pmatrix{0\\0} \Leftrightarrow 4x+3y=0$ $x=-3s_2:, y=4s_2→\;x_2=s_2\pmatrix{-3\\4} \; (s_2≠0)$ ¶2:$A=\pmatrix{2&1$1\\1&2$1\\1&1&2}$ $\pmatrix{2-\lambda&1$1\\1&2-\lambda&1\\1&1&2-\lambda} =-(\lambda-1)^2(\lambda-4)$ $→\therefore\;\lambda=1, 4$ i) when: $\lambda=1:→\;\pmatrix{1&1&1\\1&1&1\\1&1&1}\pmatrix{x\\y\\z} =\pmatrix{0\\0\\0} \Leftrightarrow x+y+z=0$ $x=s_1:, y=t_1→\;z=-s_1-t_1$ $→\;x_1=\pmatrix{s_1\\t_1\\-s_1-t_1} \; (s_1, t_1≠0)$ $=s_1\pmatrix{1\\0\\-1}+t_1\pmatrix{0\\1\\-1}$ ii) when: $\lambda=4:→\;\pmatrix{-2&1&1\\1&-2&1\\1&1&-2}\pmatrix{x\\y\\z} =\pmatrix{0\\0\\0}$ $→\left(\begin{array}{ccc|c}1&0&-1&0\\0&1&-1&0\\0&0&0&0\\\end{array}\right)$ $\cases{x-z=0\\y-z=0}→\;x=y=z=s_2→\;s_2\pmatrix{1\\1\\1}$ >Top Diagonalization-I (no-double solution): diagonalizable square matrix $A$: Diagonal matrix of $A$ exists in the form of: $P^{-1}AP=\pmatrix{\lambda_1&&&0\\&\lambda_2&&\\ &&\ddots&\\0&&&\lambda_n\\}\; [P\;$ is transformation matrix; $\lambda_n\;$eigenvalue] $(P^{-1}AP)^n=\underbrace{(P^{-1}AP)(P^{-1}AP)\cdots(P^{-1}AP)}_{n}=P^{-1}A^nP\therefore\;A^n=P(P^{-1}A^nP)P^{-1}$ Formula: where, linearly independent eigenvector ($x_1,x_2,\cdots,x_n$) of a square matrix $A$: let $P=(x_1,x_2,\cdots,x_n)$ then, $P^{-1}AP=\pmatrix{\lambda_1&&&0\\&\lambda_2&&\\ &&\ddots&\\0&&&\lambda_n\\}\; [P\;$ is transformation matrix; $\lambda_n$ Proof: $P$ has inverse matrix: $C_1x_1+C_2x_2+\cdots+C_nx_n=(x_1,x_2,\cdots,x_n) \pmatrix{C_1\\C_2\\\vdots\\C_n}=0\;$ [$n$-dimensional simultaneous linear eq.] ...(*) if rank$(P<0)$, then (*) has the solution other than trivial one $(C_1=C_2=\cdots=C_n=0)$, which contracts linearly independence of $x_1,x_2,\cdots,x_n$; thus rank$(P)=n→\;P$ is regular matrix. Proof-2: being diagonal matrix: $P^{-1}AP^=P^{-1}A(x_1,x_2,\cdots,x_n)=P^{-1}(Ax_1,Ax_2,\cdots,Ax_n)$ $=P^{-1}(\lambda_1x_1,\lambda_2x_2,\cdots,\lambda_nx_n)$ $=P^{-1}(x_1,x_2,\cdots,x_n)\pmatrix{\lambda_1&&&0\\&\lambda_2&&\\ &&\ddots&\\0&&&\lambda_n\\}\; [P=(x_1,x_2,\cdots,x_n)]$ here cf.: $(\lambda_1x_1,\lambda_2x_2) =\pmatrix{\lambda_1x_1&\lambda_2x_2\\\lambda_1y_1&\lambda_2y_2} =\pmatrix{x_1&x_2\\y_1&y_2}\pmatrix{\lambda_1&0\\0&\lambda_2}$ Formula: where, eigenvector $x_1,x_2,\cdots,x_k$ corresponding different eigenvalue of $\lambda_1,\lambda_2,\cdots,\lambda_k$ of $n$-sized square matrix $A$ is linearly independent. $(1≤k≤n)$...(*0) Proof: when $k=1$, (*0) is obvious. [$c_1x_1=0,\; x_1≠\mathbf{0}→\;c_1=0$] assume $k=m$ (*) is true, and consider: $C_1x_1+C_2x_2+\cdots+C_mx_m+C_{m+1}x_{m+1}=0$...(*1) multiply $A$ to (*1) from left: $→\;C_1\lambda_1x_1+C_2\lambda_2x_2+\cdots+ C_m\lambda_mx_m+C_{m+1}\lambda_{m+1}x_{m+1}=0$...(a) multiply $\lambda_{m+1}$ to (*1) from left: $→\;C_1\lambda_{m+1}x_1+C_2\lambda_{m+1}x_2+\cdots+ C_m\lambda_{m+1}x_m+C_{m+1}\lambda_{m+1}x_{m+1}=0$...(b) (a)-(b):$C_1(\lambda_1-\lambda_{m+1})x_1 +C_2(\lambda_2-\lambda_{m+1})x_2+\cdots +C_m(\lambda_m-\lambda_{m+1})x_m=0$ here: $x_1,x_2,\cdots,x_m$ is linearly independent, then: $C_i(\lambda_i-\lambda_{m+1})=0,\;(i=1,2,\cdots,m)$ here: $\lambda_i-\lambda_{m+1}≠0,\; (i=1,2,\cdots,m)$ $→\;C_1=C_2=\cdots=C_m=0$ substitute this to (*1): $→\;C_{m+1}x_{m+1}=0→\;C_{m+1}=0$ thus, $x_1,x_2,\cdots,x_{m+1}$ is linearly independent. ¶1: diagonalize: $A=\pmatrix{-2&1\\5&2}$ $\pmatrix{-2-\lambda&1\\5&2-\lambda}=(-2-\lambda)(2-\lambda)-5 =(\lambda-3)(\lambda+3)=0→\;\lambda03, -3$ i) when $\lambda=3:$ $\pmatrix{-5&1\\5&-1}\pmatrix{x\\y}=\pmatrix{0\\0} \Leftrightarrow -5x+y=0$ $→\;x=x_1,\;y=5s_1→\;x_1=s_1\pmatrix{1\\5}$ ii) when $\lambda=-3$ $\pmatrix{1&1\\5&5}\pmatrix{x\\y}=\pmatrix{0\\0}$ $\Leftrightarrow x+y=0→\;x=s_2, \;y=-s_2→\;x_1=s_2\pmatrix{1\\-1}$ thus: $P=\pmatrix{1&1\\5&-1}$ then $P^{-1}AP=\pmatrix{3&0\\0&-3}$ check: $P^{-1}=\frac{1}{-6}\pmatrix{-1&-1\\-5&1} =\frac{1}{6}\pmatrix{1&1\\5&-1}$ $→\;P^{-1}AP=\frac{1}{6}\pmatrix{1&1\\5&-1} \pmatrix{-2&1\\5&2}\pmatrix{1&1\\5&-1}$ here: $\frac{1}{6}\pmatrix{1&1\\5&-1}\pmatrix{-2&1\\5&2}\bigl(\bigr)$ $=\frac{1}{6}\pmatrix{3&3\\-15&3}\pmatrix{1&1\\5&-1} =\frac{1}{6}\pmatrix{18&0\\0&18}=\pmatrix{3&0\\0&-3}$ Diagonalization-II (double solution): diagonalize square matrix $A=\pmatrix{-2&2&4\\-2&3&2\\-2&1&4}$ $\pmatrix{-2-\lambda&2&4\\-2&3-\lambda&2\\12&1&4-\lambda}$ $=-(\lambda-1)(\lambda-2)^2=0→\;\lambda=1,\;2$(double)$ i) $\lambda=1:$ $\pmatrix{-3&2&4\\-2&2&2\\-2&1&3}\pmatrix{x\\y\\z} =\pmatrix{0\\0\\0}→\;x_1=s_1\pmatrix{2\\1\\1}$ ii) $\lambda=2$ $\pmatrix{-4&2&4\\-2&1&2\\-2&1&2}\pmatrix{x\\y\\z} =\pmatrix{0\\0\\0}\;\Leftrightarrow -2x+y+2z=0$→\;$x=s_2,\;z=t_2, y=2s_2-2t_2$ $→\;x_2=s_2\pmatrix{1\\2\\0}+t_2\pmatrix{0\\-2\\1}$ here: $\pmatrix{2\\1\\1},\;\pmatrix{1\\2\\0},\;\pmatrix{0\\-2\\1}$ are linealy independent, and eigenvector. $p^{-1}AP=\pmatrix{1&0&0\\0&2&0\\0&0&2}$ Diagonalization-III (no diagnalizable case) $A=\pmatrix{-3&-1\\1&-1}$ $\pmatrix{-3-\lambda&-1\\1&-1-\lambda}=(-3-\lambda)(-1-\lambda)+1 =\lambda^2+4\lambda+4=(\lambda+2)^2=0→\;\lambda=-2\;$ (double) $\Leftrightarrow x+y=0$→\;$x=s_1,\;y=-s_1 →\; x_1=s_1\pmatrix{1\\-1}$ there is no other independent eigenvector than this: thus 'no diagonal'

19. 線形代数: Gaussian elimination: ガウスの消去法, 掃き出し法 identity (unit) matrix: 単位行列 upper triangle matrix: 上三角行列 cofactor expansion: 余因子展開 rank: 階数 trace: 跡, 対角線成分 linear combination: 線形結合 transposed matrix: 転置行列 coordinate transformation: 座標変換 eigenvalue: 固有値 eigenvector: 固有ベクトル diagonalization: 対角化 diagonalizable: 対角化可能 regular matrix =reversible matrix: 正則行列 (=可逆行列) transformation (=diagonalizable) matrix: 変換(=対角化)行列 掃き出し法: 余因子展開: 特徴: スカラー倍 可換法則 結合法則 単位行列 乗算 線形独立(=一次独立) ランク, 階数 転置行列 逆行列, Inverse matrix: make $^tA$ make $\tilde{A}$ consider sgn $\tilde{A}$ 座標変換: 固有値と固有ベクトル: 対角化:

>Top 20. Delta function: Paul Dirac's delta function: definition-1: $\delta(x-a)=\cases{\infty,\;(x=a)\\0,\;(x≠0)}$ definition-2: $\int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a)$ when $f(x)=1→\;\int_{-\infty}^{\infty}\delta(x-a)dx=1$ [size of infinity]...(*) point mass of a mass $m$ on $x=a:\;\int_{-\infty}^{\infty}m\delta(x-a)dx=m$ but, mathematically (*) normal function will be zero; which is called distribution (by Schwarz) or hyperfunction (by Sato).

20. デルタ関数: point mass: 質点

>Top 21. Gaussian integration: Definition: $\boxed{\displaystyle\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}}$ proof: $I=\displaystyle\int_{-\infty}^{\infty}e^{-x^2}dx$ $I^2=(\displaystyle\int_{-\infty}^{\infty}e^{-x^2}dx) (\displaystyle\int_{-\infty}^{\infty}e^{-y^2}dy)\;[x↔y]$ $=\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}e^{x^2+y^2}dxdy$ let: $x=r\cos\theta,\;y=r\sin\theta$, then $I^2=\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\infty}e^{-r^2}rdrd\theta$ surface integral: $ds=\pi(r+dr)^2\frac{d\theta}{2\pi}-\pi r^2\frac{d\theta}{2\pi} =rdrd\theta+\frac{1}{2}(dr)^2d\theta \simeq rdrd\theta$ $I^2=\displaystyle\int_0^{2\pi}d\theta\displaystyle\int_0^{\infty}re^{-r^2}rd$ $=2\pi\left[-\frac{1}{2}e^{-r^2}\right]_0^{\infty}=\pi\;→I=\sqrt{\pi},\;(I>0)$ Net outflow:$=(\frac{\partial V_x}{\partial x}+\frac{\partial V_y}{\partial y} +\frac{\partial V_z}{\partial z} )\Delta x\Delta y\Delta z$ thus outflow per volume: $=div \mathbf{V}$

21. ガウス積分:

>Top 22. Vector analysis: Divergence: outflow-inflow $V_x(x+\frac{\Delta x}{2},y,z)\Delta y\Delta z-V_x(x-\frac{\Delta x}{2},y,z)\Delta y\Delta z$ $=\bigl(V_x(x+\frac{\Delta x}{2},y,z)-V_x(x-\frac{\Delta x}{2},y,z)\bigr)\Delta y\Delta z$...(*) Taylor expansion: $f(a+h)=f(a)+f'(a)h+\frac{1}{2!}f''(a)h^2+\cdots$ $Vx(x±\frac{\Delta x}{2},y,z)\simeq Vx(x,y,z) ±\frac{\partial V_x}{\partial x}\frac{\Delta x}{2}$ (*)$\simeq\bigl(V_x(x,y,z)+\frac{\partial V_x}{\partial x}\frac{\Delta x}{2} -V_x(x,y,z) +\frac{\partial V_x}{\partial x}\frac{\Delta x}{2}\bigr)\Delta y\Delta z =\frac{\partial V_x}{\partial x} \Delta x\Delta y\Delta z$ >Top Rotation: Definition: $rot \mathbf{V} =\bigl(\frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z}, \frac{\partial V_x}{\partial z}-\frac{\partial V_z}{\partial x}, \frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y}\bigr)$ $=\nabla \times \mathbf{V}\; [\nabla=(\frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z})]$ $=\bigl(V_y(x+\frac{\Delta x}{2},y)-V_y(x-\frac{\Delta x}{2},y)\bigr)\Delta y +\bigl(V_x(x,y-\frac{\Delta y}{2})-V_x(x,y+\frac{\Delta x}{2})\bigr) \Delta x$ $\simeq \bigl(V_y(x,y)+\frac{\partial V_y}{\partial x}\frac{\Delta x}{2} -V_y(x,y)+\frac{\partial V_y}{\partial x}\frac{\Delta x}{2}\bigr)\Delta y +\bigl(V_x(x,y)-\frac{\partial V_x}{\partial y}\frac{\Delta y}{2} -V_x(x,y)-\frac{\partial V_x}{\partial y}\frac{\Delta y}{2}\bigr)\Delta x$ $=(\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y})\Delta x\Delta y$ per unit space: $=\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y} \; [z$-component of $rot\;\mathbf{V}$] Gradient: Definition: grad $f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})\;$[grad $f$ means mostly† increasing direction] Proof (†):$\Delta f=f(r+\Delta r)-f(r)$ $=f(x+\Delta x, y+\Delta y, z+\Delta z)-f(x,y,z)$ $\simeq f(x,y,z)+\frac{\partial f}{\partial x}\Delta x +\frac{\partial f}{\partial y}\Delta y+\frac{\partial f}{\partial z}\Delta z$ $=(\mathrm{grad}\;f)･\Delta r=|\mathrm{grad}\;f||\Delta r|\cos\theta ≤|\mathrm{grad}\;f||\Delta \mathbf{r}|\; [\theta=0\;$maximum] >Top Solid angle: (unit: Steradian $d\Omega$) $dS=r^2\sin\theta d\theta d\phi$ $d\Omega\equiv \frac{dS}{r^2}=\sin\theta d\theta d\phi$ ¶1: [global surface] $=\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi}\sin\theta d\theta d\phi=4\pi$ ¶2: $[\theta_1→\theta_2,\;\phi_1→\phi_2]$ $=(\phi_2-\phi_1)(\cos\theta_1-\cos\theta_2)$ ¶3: [half-apex angle=$\alpha$] $=\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\alpha}\sin\theta d\theta d\phi =2\pi(1-\cos\alpha)$ >Top Spherical coordinates: $\cases{x=r\sin\theta\cos\phi\\y=r\sin\theta\sin\phi\\z=r\cos\theta}$ $\cases{r=\sqrt{x^2+y^2+z^2}\;(0≤r)\\\theta=\cos^{-1}(\frac{z}{\sqrt{s^2+y^2+z^2}}) \;(0≤\theta ≤\pi) \\\phi=\mathrm{sgn}(y)\cos^{-1}(\frac{x}{\sqrt{x^2+y^2}})\; (0≤\phi<2\pi) }$ here: sgn$(y)=\cases{1\;(0≤y)\\-1\;(y<0)}$

22. ベクトル解析: vector field: ベクトル場 position vector: 位置ベクトル solid angle: 立体角 steradian: 立体角の単位 (sr)

>Top 23. Integration method: Some Calculus formula: $y=fg\\ \log y=\log(fg)=\log f+\log g\\ \frac{y'}{y}=\frac{f'}{f} +\frac{g'}{g}\\ y'=f'g+fg'$ $y=(\frac{f}{g})$ $\log y=\log\frac{f}{g} =\log f-\log g\\ \frac{y'}{y}=\frac{f'}{f} -\frac{g'}{g}\\ y'=\frac{f'}{g}-\frac{fg'}{g^2} =\frac{f'g-fg'}{g^2}$ $(e^x)'=e^x$ $(a^x)'=a^x\log a$ $y=(x^x)'$ $\log y=x\log x$ $\frac{y'}{y}=\log x+1\\ y'=y(\log x+1)=x^x(\log x+1)$ $(\log x)'=\frac{1}{x}$ $(\log y)'=\frac{y'}{y}$ $(\log_ax)'\\ =(\frac{\log x}{\log a})'\\ =\frac{1}{\log a}(\log x)'\\ =\frac{1}{x\log a}$ Some Integral formula: $\int\frac{1}{x}=\log|x|+C$ $\int\frac{1}{x^2}=-\frac{1}{x}+C$ $\int \frac{f'}{f}=\log|f|+C$ $\int e^xdx=e^x+C$ $\int a^x=\frac{a^x}{\log a}+C$ $\int a^x\log adx=a^x+C$ $\int\log xdx=\int x'\log xdx=x\log x-x(\log x)'dx=x\log x-\int dx$ $=x\log x-x+C\;$ [\log = 1･\log] $\int \frac{1}{x\log a}dx=\log_ax$ $\int \sin xdx=-\cos x+C$ $\int \cos xdx=\sin x+C$ $\int \tan xdx=\int \frac{\sin x}{\cos x}dx=-\int\frac{(\cos x)'}{\cos x}dx$ $=-log|\cos x|+C$ $\int \sec^2xdx=\tan x+C$ $\int \csc^2xdx=-cotx+C$ $\int \frac{1}{\cos^2x}dx=\tan x+C$ $\int \frac{1}{\sin^2x}dx=-\frac{1}{\tan x}+C$ $\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+C,\;(a≠0)$ $\int\frac{1}{\sqrt{x^2-a^2}}dx=\sin^{-1}\frac{x}{a}+C,\;(a>0)$ $\int\frac{1}{\sqrt{x^2+a}}dx=\log |x+\sqrt{x^2+a}|+C,\;(a≠0)$ Integration by parts: $\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$ ¶$\int\log|x|dx=\int x'\log|x|dx=x\log|x|-\int x(\log|x|)'dx$ $=x\log|x|-\int x\frac{1}{x}dx=x\log|x|-x+C$ $\int f(g(x))g'(x)dx=$ [differential contact type] let:$g(x)=t$ $\int \frac{f'(x)}{f(x)}dx=\log|f(x)|+C$ $\int(f(x))^af'(x)dx=\frac{(f(x)^{a+1})}{a+1}+C$ ¶$\int x^ne^xdx=x^ne^x-\int (x^n)'e^x$ ¶$\int \frac{1}{\cos^3xdx}$ $=\int \frac{1}{\cos^2x}\frac{1}{\cos x}dx =\tan x\frac{1}{\cos x}-\int \tan x\frac{\sin x}{\cos^2x}dx$ $=\frac{\sin x}{\cos^2x}-\int \frac{\sin^2 x}{\cos^3 x}$ $=\frac{\sin x}{\cos^2x}-\int \frac{1}{\cos^3x}dx+\int \frac{1}{\cos x}dx$ $2\int \frac{1}{\cos^3x}dx =\frac{\sin x}{\cos^2x}+\frac{1}{2}\log\big|\frac{1+\sin x}{1-\sin x}\big|+C$ ¶$\int (\log x)^3dx)$ $t=\log x→\;dt=\frac{1}{x}dx→dx=e^tdt$ $=\int t^3e^tdt=t^3e^t-3t^2e^t+6te^t-6e^t+C=(t^3-et^2+6t-6)e^t+C$ $=\{(\log x)^3-(\log x)^2+6\log x-6\}x+C$ Integration by substitution: $\int f(x)dx=\int f(g(t))g'(t)dt$ ¶$\int_0^1x^2dx =\int_0^2\frac{1}{4}t^2 \frac{1}{2}dt$ $x=\frac{1}{2}t$ $dx=\frac{1}{2}dt$ ¶$\int_0^1\sqrt{1-x^2}dx$...(*) let: $x=\sin(u),\;dx=\cos udu$ $→\sqrt{1-x^2}=\cos(u)$ $=\int_0^{\frac{\pi}{2}} \cos^2(u) du =(\frac{u}{2}+ \frac{\sin(2u)}{4}) \big|_0^{\frac{\pi}{2}} =\frac{\pi}{4}$ ¶$\int_a^b(x-a)(x-b)dx =\int_a^b\{(x-a)^2-(b-a)(x-a)\}dx$ $=\frac{1}{3}(x-a)^3-(b-a)\frac{1}{2}(x-a)^2\big|_a^b=-\frac{1}{6}(b-a)^3$ ¶$\int \frac{1}{\cos x}dx\;$ [→*tips] $=\int \frac{1+t^2}{1-t^2}\frac{2}{1+t^2}dt=\int \frac{2}{1-t^2}dt$ $=\int (\frac{1}{1+t}+\frac{1}{1-t})dt=\log |1+t|-\log |1-t|+C =\log |\frac{1+t}{1-t}|+C$ $=\log \big|\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\big|+C$ ¶$\int \frac{1}{\sin x}dx\;$...(*) [→*tips] $\frac{1}{\sin x}=\frac{\sin x}{1-\cos^2x},\; t=\cos x→dt=-sin xdx$ $(*)=\int \frac{\sin x}{1-t^2}\frac{1}{-\sin x}dt=\int \frac{1}{t^2-1} =\frac{1}{2}\int (\frac{1}{t-1}-\frac{1}{t+1})dt$ $=\frac{1}{2}(\log|t-1|-log|t+1|)+C=\frac{1}{2}\log |\frac{t-1}{t+1}|+C$ $=\frac{1}{2}\log \big|\frac{1-\cos x}{1+\cos x}\big|+C$ Rotational body integral: around x-axis: $V=\pi\int_a^bf(x)^2dx$ Baumkuchen type integral (shell integration): around y-axis: $f(x)=g(x)-h(x)$ $V=2\pi\int_a^bxf(x)dx$ Pappus-Guildinus theorem: volume of a solid revolution by roatating about an external axis ($y$): $V=2\pi rS\;$ [r=distance from axis to the gravity, S=space of the graph] $x_G=\frac{m_1x_1+m_2x_2+\cdots+m_nx_n}{m_1+m_2+\cdots+m_n}$ $=\frac{\int_a^b x\rho f(x)dx}{\int_a^b\rho f(x)dx}\; [\rho=$specific density] $=\frac{\int_a^b xf(x)dx}{S}$ $S=\displaystyle\int_a^b f(x)dx$ $V=\int_a^b 2\pi xf(x)dx\;$ [Baumkuchen] $x_G=\frac{\frac{V}{2\pi}}{S}\therefore\;\boxed{V=2\pi Sx_G}$ King property: $\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$ effective when $f(x)+f(a+b-x)$ is easily calculated. ¶1:$\int_{-1}^1\frac{\sin^2(\pi x)}{1+e^x}dx$...I $2I=\{\int_{-1}^1\frac{\sin^2(\pi x)}{1+e^x} +\frac{\sin^2(\pi x)}{1+e^{-x}}\}dx$ $=\int_{-1}^1\sin^2(\pi x)dx=\int_{-1}^1\frac{1-\cos (2\pi x)}{2}dx$ $=\frac{1}{2}x-\frac{1}{4\pi}\sin(2\pi x)\big|_{-1}^1=1$ $\therefore\;I=\frac{1}{2}$

23. 積分法: integration by substitution: 置換積分法 integration by parts: 部分積分法 Basic: $\sin(\frac{\pi}{2}-x)=\cos x$ $\cos(\frac{\pi}{2}-x)=\sin x$ $\sin(x+\frac{\pi}{2})=\cos x$ $\cos(x+\frac{\pi}{2})=-\sin x$ $\sin(x+\pi)=-\sin x$ $\cos(x+\pi)=-\cos x$ Addition formula: $\sin(a\pm b)=\sin a\cos b\pm \cos a\sin b$ $\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b$ $\tan(a\pm b)=\frac{\tan a\pm\tan b} {1\mp\tan a\tan b}$ Double angle formula: $\sin 2x=2\sin x\cos x$ $\cos 2x=\cos^2x-\sin^2x =2\cos^2x-1=1-2\sin^2x$ $\tan 2x=\frac{2\tan x}{1-2\tan^2 x}$ Half angle formula: $\sin^2x=\frac{1}{2}(1-\cos 2x)$ $\cos^2x=\frac{1}{2}(x+\cos 2x)$ $\tan^2x=\frac{1-\cos 2x}{1+\cos 2x}$ $\tan x=\pm\frac{1-\cos 2x}{\sin 2x}$ Tripple angle formula: $\sin 3x=3\sin x-4\sin^3x$ $\cos 3x=-3\cos x+4\cos^3x$ Product-Sum formula: $\sin a\cos b=\frac{1}{2}\bigl(\sin(a+b) +\sin(a-b)\bigr)$ $\sin a\sin b=\frac{1}{2}\bigl(-\cos(a+b) +\cos(a-b)\bigr)$ $\cos a\cos b=\frac{1}{2}\bigl(\cos(a+b) +\cos(a-b)\bigr)$ Sum-Product formula: $\sin a+\sin b=2\sin\frac{a+b}{2} \cos\frac{a-b}{s}$ $\sin a-\sin b=2\cos\frac{a+b}{2} \sin\frac{a-b}{s}$ $\cos a+\cos b=2\cos\frac{a+b}{2} \cos\frac{a-b}{2}$ $\cos a-\cos b=-2\sin\frac{a+b}{2} \sin\frac{a-b}{2}$ $(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}$ $(\cos^{-1}x)'=-\frac{1}{\sqrt{1-x^2}}$ $(\tan^{-1}x)'=\frac{1}{1+x^2}$ >Top Integration by parts: (integral-first!) $\int f^0g^0=f^0g^{-1}-\int f^1g^{-1}$ $=f^0g^{-1}-f^1g^{-2}+\int f^2g^{-2}$ <*Tips> $f(\sin x)\cos x→t=\sin x$ →dt=\cos xdx $f(\cos x)\sin x→t=\cos x$ $f(\tan x)\frac{1}{\cos^2x} →t=\tan x$ $→dt=\frac{1}{\cos^2x}dx$ $f(\sin x,\;\cos x)$...(*) $→t=\tan\frac{x}{2}$ $→\sin x=\frac{2t}{1+t^2}$ $→\cos x=\frac{1-t^2}{1+t^2}$ $→dt=\frac{1+t^2}{2}dx$ $\int(*)dx =f(\frac{2t}{1+t^2},\; \frac{1-t^2}{1+t^2}) \frac{2}{1+t^2}dt$ $\int\log xdx→\int 1･\log xdx$ $\int_0^{\pi}xf(\sin x)dx$ $=\frac{\pi}{2}\int_0^{\pi}f(\sin x)dx$ make square of trigonometry!

>Top 24. Multiple integral: Repeated integral: $\displaystyle\iint_D f(x,y)dxdy=\displaystyle\int_c^d\displaystyle\int_a^bf(x,y)dxdy$ $=\displaystyle\int_a^b\displaystyle\int_c^df(x,y)dydx$ ¶1: $\displaystyle\iint_D(3x-y)dxdy,\;D=\{(xy)|0≤x≤1,\;-1≤y≤2|\}$ $=\displaystyle\int_{-1}^2\displaystyle\int_0^1(3x-y)dxdy$ $=\displaystyle\int_{-1}^2\bigl[\frac{3}{2}x^2-yx\bigr]_0^1dy$ $=\displaystyle\int_{-1}^2(\frac{3}{2}-y)dy =\bigl[\frac{3}{2}y-\frac{1}{2}y^2\bigr]_{-1}^2=3$ $\displaystyle\int_0^1\displaystyle\int_{-1}^2(3x-y)dydx$ $=\displaystyle\int_0^1\bigl[3xy-\frac{1}{2}y^2\bigr]_{-1}^2dx$ $=\displaystyle\int_0^1(9x-\frac{3}{2})dx$ $=\bigl[\frac{9}{2}x^2-\frac{3}{2}x\bigr]_0^1=3$ ¶2: $\displaystyle\iint_De^{x^2}dxdy,\;D=\{(x,y)|0≤y≤1,\;y≤x≤1\}$ $=\displaystyle\int_0^1\displaystyle\int_y^1e^{x^2}dxdy$ [antiderivative?] $=\displaystyle\int_0^1\displaystyle\int_0^xe^{x^2}dydx$ $=\displaystyle\int_0^1\bigl[ye^{x^2}\bigr]_0^xdx$ $=\displaystyle\int_0^1xe^{x^2}dx=\bigl[\frac{1}{2}e^{x^2}\bigr]_0^1=\frac{1}{2}(e-1)$ Substitutive integral: ¶1: $\displaystyle\iint_D\frac{1}{x^2+y^2}dxdy,\;D=\{(x,y)|1≤x^2+y^2≤4,\;y≥0\}$ $=\displaystyle\int_0^{\pi}\displaystyle\int_1^2\frac{1}{r^2}rdrd\theta$ $=\displaystyle\int_0^{\pi}\bigl[\log r\bigr]_1^2d\theta$ $=\displaystyle\int_0^{\pi}\log 2\theta=\pi\log 2$ >Top Jacobian: $0\bigl(x(u,v),\;y(u,v)\bigr)$ $A\bigl(x(u+\Delta u,v),\;y(u+\Delta u,v)\bigr)$ $B\bigl(x(u,v+\Delta v),\;y(u,v+\Delta v)\bigr)$ $x(u+\Delta u,v)\simeq x(u,v)+\frac{\partial x}{\partial u}\Delta u$ $vec{OA}\simeq \bigl(\frac{\partial x}{\partial u}\Delta u,\;, \frac{\partial y}{\partial u}\Delta u\bigr)$ $vec{OB}\simeq \bigl(\frac{\partial x}{\partial v}\Delta v,\;, \frac{\partial y}{\partial v}\Delta v\bigr)$ $\boxed{S\simeq \left|det\pmatrix{\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}}\right|\Delta u\Delta v =J(u,v)\Delta u\Delta v}$ thus: $\displaystyle\iint_D f(x,y)dxdy =\displaystyle\iint_E f\left(x(u,v),\;y(u,v)\right)|J(u,v)|dudv$ ¶1: J(u,v) on polar coordinates: $\cases{x=r\cos\theta\\y=r\sin\theta},\;\cases{u↔r\\v↔\theta}$ $J(u,v)=\|det\pmatrix{\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta}\|=r$ ¶2: $\displaystyle\iint_D (x-y)e^{x+y}dxdy,\;D=\{(x,y)|0≤x+y≤2,\;0≤x-y≤2\}$...(*) let: $x+y=u,\;v=x-y;\;\cases{x=\frac{1}{2}(u+v)\\y=\frac{1}{2}(u-v)}$ $J(u,v)=\left|det\pmatrix{\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}}\right| =\frac{1}{2}$ (*)=$\displaystyle\int_0^2\displaystyle\int_0^2 ve^{u}\frac{1}{2}dudv$ $=\frac{1}{2}\displaystyle\int_0^2\bigl[ve^u\bigr]_0^2dv$ $=\displaystyle\int_0^2(ve^2-v)dv=\frac{1}{2}\bigl[\frac{1}{2}e^2v^2-\frac{1}{2}v^2\bigr]_0^2=e^2-1$

24. 重積分: iterated/repeated integral: 累次積分 substitutive integral: 置換積分 antiderivative=primitive function: 原子関数 Jacobian matrix: ヤコビ行列 $J_f$

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Comment

In studying effectively, it's important to accumulate in the manner of dicrete, heurisitic, empirical, and then formalized knowledge.

効率的に学ぶには、離散的、発見的、実証的、そして形式化した知識を集めることが肝要である。