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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
  1. Preface:
  2. Equation of motion (EOM)
  3. Pendulum isochronism:
  4. Conservation law:
  5. Damping oscillation:
  6. Conservation law of angular momentum:
  7. Inertial frame:
  8. Coriolis force:
  9. Archimedean spiral
  10. Hyperbolic function:
  11. Inverse trigonometric function:
  12. Two-particle system motion:
  13. Rigid body dynamics:
  14. Inertia moment of rigid body:
  15. Mechanical energy of rigid body:
  16. Matrix exponential:
  17. Chaucy's functional equation:
  18. Taylor expansion:
  19. Fourier expansion:
  20. Linear algebra:
  21. Delta function:
  22. Gaussian function:
  23. Vector analysis:
  24. Integration method:
  25. Multiple integral:
  1. 序文:
  2. 運動方程式
  3. 振り子の等時性:
  4. 保存則:
  5. 減衰振動:
  6. 角運動量保存則:
  7. 慣性系:
  8. コリオリの力:
  9. アルキメデスの螺旋:
  10. 双曲線関数:
  11. 逆三角関数:
  12. 二粒子系の運動:
  13. 剛体の力学:
  14. 剛体の慣性モーメント:
  15. 剛体の力学エネルギー:
  16. 行列指数関数:
  17. コーシーの関数方程式:
  18. テイラー展開:
  19. フーリエ展開:
  20. 線形代数:
  21. デルタ関数:
  22. ガウス関数:
  23. ベクトル解析:
  24. 積分法:
  25. 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.
  • attenuation
  • Harmonic Motion:
  • harmonicmotion
  • 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:
  • harmonic oscillator
  • >Top Vector product:
    $a\times b=\\
    (a_yb_z-a_zb_y, \\
    a_zb_x-a_xb_z, \\
    a_xb_y-a_yb_x )$
  • vectorproduct

>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:
  • dampingoscillation
  • Critical damping:
  • criticaldamping
  • $\displaystyle\lim_{x\to 0}
    \frac{\sin x}{x}=1$
  • Forced oscillation:
  • forcedoscillation

 

>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: $\Delta S=v\Delta t\sin\phi$
  • areavelocity

>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:
  • coliorisforce

>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
  • $\cases{v=(x',y',z')\\e_z=(0,0,1)}$
    $→\;v\times e_z=(y'-x')$
  • inertialframe

>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:

  • polardistance

>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
  • hyperbolicsin

>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:

  • arcsin.gif

>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
  • twoparticle
  • relativevector
  • reducedmass
  • >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:
  • relativeposition
  • 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:
  • parallelaxes
  • Orthogonal axis:
    parallelplane

>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]
    1. $e^0=E$
    2. $\frac{d}{dt}e^{At}=Ae^{At}$
    3. 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:
    1. when $x=y=0→\;f(0)=f(0)+f(0)→\;f(0)=0\;$ [cross the origin]
    2. when $y=-x→\;f(0)=f(x)+f(-x)→\;f(x)=-f(x)\;$ [odd function]...(*2)
    3. 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: $f(nx)=nf(x)$
    4. from (*2), substitute: $x→nx$, then $f(-nx)=-f(nx)=-nf(x)\;[-n:$ negative number]
      $→\;f(mx)=mf(x)\;$ [m: integer]
    5. 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]
    6. $^\exists \{q_n\}$ 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:
    1. base case: show f(0) is clearly true.
    2. 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:
    1. make $^tA$
    2. make $\tilde{A}$
    3. 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:

  • gaussintegral.gif
  • surfaceintegral.gif

>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)
  • divergence.gif
  • rotation.gif
  • gradient.gif
  • solidangle.gif
  • sphericalcoordinates.gif

>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!)
  • $\int f^0g^0=f^0g^{-1}-\int f^1g^{-1}$
    $=f^0g^{-1}-f^1g^{-2}+\int f^2g^{-2}$
  • integralbyparts.gif

<*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 $J_f$
  • doubleintegral.gif
  • repeatedintegral.gif
  • doubleinteger2.gif
  • jacobian.gif

>Top 25. XXXX:

25. XXXX:

>Top 26. YYYY:

26. YYYY:

>Top 27. ZZZZ:

27. ZZZZ:

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。

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