3問すべてが必答の試験です。第1問は固有値が縮退して対角化できない3次実行列で、ジョルダン鎖 A e ⃗ 3 = e ⃗ 2 + λ 2 e ⃗ 3 A\vec{e}_3=\vec{e}_2+\lambda_2\vec{e}_3 A e 3 = e 2 + λ 2 e 3 λ \lambda λ λ → 0 \lambda\to0 λ → 0 sin ω x / ω \sin\omega x/\omega sin ω x / ω
問題 分野 主題 第1問 線形代数 縮退した固有値・ジョルダン鎖・アフィン漸化式の収束 第2問 微分方程式 定数変化法、矩形外力の応答、インパルス応答 第3問 確率統計 2連勝までの待ち時間、3項間漸化式、期待値
実パラメータ a a a
A = ( a + 1 0 0 0 a + 1 − 1 − 1 1 a − 1 ) A=\begin{pmatrix} a+1 & 0 & 0 \\ 0 & a+1 & -1 \\ -1 & 1 & a-1 \end{pmatrix} A = a + 1 0 − 1 0 a + 1 1 0 − 1 a − 1 を考えます。設問(a)は A e ⃗ 1 = λ 1 e ⃗ 1 A\vec{e}_1=\lambda_1\vec{e}_1 A e 1 = λ 1 e 1 A e ⃗ 2 = λ 2 e ⃗ 2 A\vec{e}_2=\lambda_2\vec{e}_2 A e 2 = λ 2 e 2 A e ⃗ 3 = e ⃗ 2 + λ 2 e ⃗ 3 A\vec{e}_3=\vec{e}_2+\lambda_2\vec{e}_3 A e 3 = e 2 + λ 2 e 3 λ 1 , λ 2 \lambda_1,\lambda_2 λ 1 , λ 2 R 3 \mathbb{R}^3 R 3 e ⃗ 1 , e ⃗ 2 , e ⃗ 3 \vec{e}_1,\vec{e}_2,\vec{e}_3 e 1 , e 2 , e 3 A n A^n A n r ⃗ n + 1 = A r ⃗ n + d ⃗ \vec{r}_{n+1}=A\vec{r}_n+\vec{d} r n + 1 = A r n + d e ⃗ 2 , e ⃗ 3 \vec{e}_2,\vec{e}_3 e 2 , e 3 λ 2 \lambda_2 λ 2 A A A
まず固有多項式を求めます。A − λ I A-\lambda I A − λ I ( a + 1 − λ , 0 , 0 ) (a+1-\lambda,0,0) ( a + 1 − λ , 0 , 0 )
det ( A − λ I ) = ( a + 1 − λ ) [ ( a + 1 − λ ) ( a − 1 − λ ) + 1 ] \det(A-\lambda I)=(a+1-\lambda)\bigl[(a+1-\lambda)(a-1-\lambda)+1\bigr] det ( A − λ I ) = ( a + 1 − λ ) [ ( a + 1 − λ ) ( a − 1 − λ ) + 1 ] です。μ = a − λ \mu=a-\lambda μ = a − λ ( μ + 1 ) ( μ − 1 ) + 1 = μ 2 (\mu+1)(\mu-1)+1=\mu^2 ( μ + 1 ) ( μ − 1 ) + 1 = μ 2
det ( A − λ I ) = ( μ + 1 ) μ 2 = ( a + 1 − λ ) ( a − λ ) 2 . \det(A-\lambda I)=(\mu+1)\mu^2=(a+1-\lambda)(a-\lambda)^2 . det ( A − λ I ) = ( μ + 1 ) μ 2 = ( a + 1 − λ ) ( a − λ ) 2 . 固有値は λ = a + 1 \lambda=a+1 λ = a + 1 λ = a \lambda=a λ = a a a a
λ 2 \lambda_2 λ 2 ( A − λ 2 I ) e ⃗ 3 = e ⃗ 2 ≠ 0 ⃗ (A-\lambda_2 I)\vec{e}_3=\vec{e}_2\neq\vec{0} ( A − λ 2 I ) e 3 = e 2 = 0 ( A − λ 2 I ) 2 e ⃗ 3 = ( A − λ 2 I ) e ⃗ 2 = 0 ⃗ (A-\lambda_2I)^2\vec{e}_3=(A-\lambda_2I)\vec{e}_2=\vec{0} ( A − λ 2 I ) 2 e 3 = ( A − λ 2 I ) e 2 = 0 e ⃗ 3 \vec{e}_3 e 3 e ⃗ 2 \vec{e}_2 e 2 ( A − λ 2 I ) e ⃗ 3 = 0 ⃗ (A-\lambda_2I)\vec{e}_3=\vec{0} ( A − λ 2 I ) e 3 = 0 e ⃗ 2 , e ⃗ 3 \vec{e}_2,\vec{e}_3 e 2 , e 3 ker ( A − λ 2 I ) 2 \ker(A-\lambda_2I)^2 ker ( A − λ 2 I ) 2 λ 2 \lambda_2 λ 2 λ 2 = a \lambda_2=a λ 2 = a
λ 1 \lambda_1 λ 1 a + 1 a+1 a + 1 a a a a a a λ 1 = a \lambda_1=a λ 1 = a e ⃗ 1 = e ⃗ 2 \vec{e}_1=\vec{e}_2 e 1 = e 2 R 3 \mathbb{R}^3 R 3 e ⃗ 1 , e ⃗ 2 , e ⃗ 3 \vec{e}_1,\vec{e}_2,\vec{e}_3 e 1 , e 2 , e 3 λ 1 = a + 1 \lambda_1=a+1 λ 1 = a + 1
λ 1 = a + 1 \lambda_1=a+1 λ 1 = a + 1
A − ( a + 1 ) I = ( 0 0 0 0 0 − 1 − 1 1 − 2 ) A-(a+1)I=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ -1 & 1 & -2 \end{pmatrix} A − ( a + 1 ) I = 0 0 − 1 0 0 1 0 − 1 − 2 の核です。第2行から v 3 = 0 v_3=0 v 3 = 0 − v 1 + v 2 − 2 v 3 = 0 -v_1+v_2-2v_3=0 − v 1 + v 2 − 2 v 3 = 0 v 1 = v 2 v_1=v_2 v 1 = v 2 ( 1 , 1 , 0 ) T (1,1,0)^{T} ( 1 , 1 , 0 ) T
e ⃗ 1 = 1 2 ( 1 1 0 ) \vec{e}_1=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\\0\end{pmatrix} e 1 = 2 1 1 1 0 です。
λ 2 = a \lambda_2=a λ 2 = a
A − a I = ( 1 0 0 0 1 − 1 − 1 1 − 1 ) A-aI=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ -1 & 1 & -1 \end{pmatrix} A − a I = 1 0 − 1 0 1 1 0 − 1 − 1 で、第3行は（第2行）－（第1行）に等しいので階数は2、核は1次元です。第1行から v 1 = 0 v_1=0 v 1 = 0 v 2 = v 3 v_2=v_3 v 2 = v 3 ( 0 , 1 , 1 ) T (0,1,1)^{T} ( 0 , 1 , 1 ) T
e ⃗ 2 = 1 2 ( 0 1 1 ) \vec{e}_2=\frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\\1\end{pmatrix} e 2 = 2 1 0 1 1 となります。固有空間が1次元であることが確認できたので、A A A λ 1 = a \lambda_1=a λ 1 = a
e ⃗ 3 = ( x , y , z ) T \vec{e}_3=(x,y,z)^{T} e 3 = ( x , y , z ) T ( A − a I ) e ⃗ 3 = e ⃗ 2 (A-aI)\vec{e}_3=\vec{e}_2 ( A − a I ) e 3 = e 2 ( x , y − z , − x + y − z ) T (x,\;y-z,\;-x+y-z)^{T} ( x , y − z , − x + y − z ) T
x = 0 , y − z = 1 2 , − x + y − z = 1 2 x=0,\qquad y-z=\frac{1}{\sqrt{2}},\qquad -x+y-z=\frac{1}{\sqrt{2}} x = 0 , y − z = 2 1 , − x + y − z = 2 1 となり、第3式は第1・第2式から自動的に従います。よって解は e ⃗ 3 = ( 0 , y , y − 1 / 2 ) T \vec{e}_3=(0,\,y,\,y-1/\sqrt{2})^{T} e 3 = ( 0 , y , y − 1/ 2 ) T y y y
y 2 + ( y − 1 2 ) 2 = 1 ⟺ 4 y 2 − 2 2 y − 1 = 0 ⟺ y = 2 ± 6 4 y^2+\Bigl(y-\frac{1}{\sqrt{2}}\Bigr)^2=1
\quad\Longleftrightarrow\quad
4y^2-2\sqrt{2}\,y-1=0
\quad\Longleftrightarrow\quad
y=\frac{\sqrt{2}\pm\sqrt{6}}{4} y 2 + ( y − 2 1 ) 2 = 1 ⟺ 4 y 2 − 2 2 y − 1 = 0 ⟺ y = 4 2 ± 6 です。第2成分 y y y y = ( 6 + 2 ) / 4 y=(\sqrt{6}+\sqrt{2})/4 y = ( 6 + 2 ) /4 z = y − 1 / 2 = ( 6 − 2 ) / 4 z=y-1/\sqrt{2}=(\sqrt{6}-\sqrt{2})/4 z = y − 1/ 2 = ( 6 − 2 ) /4
以上より答えは
λ 1 = a + 1 , λ 2 = a , \lambda_1=a+1,\qquad \lambda_2=a, λ 1 = a + 1 , λ 2 = a , e ⃗ 1 = 1 2 ( 1 1 0 ) , e ⃗ 2 = 1 2 ( 0 1 1 ) , e ⃗ 3 = 1 4 ( 0 6 + 2 6 − 2 ) \vec{e}_1=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\\0\end{pmatrix},\qquad
\vec{e}_2=\frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\\1\end{pmatrix},\qquad
\vec{e}_3=\frac{1}{4}\begin{pmatrix}0\\ \sqrt{6}+\sqrt{2}\\ \sqrt{6}-\sqrt{2}\end{pmatrix} e 1 = 2 1 1 1 0 , e 2 = 2 1 0 1 1 , e 3 = 4 1 0 6 + 2 6 − 2 です。規格化の検算として ( 6 + 2 ) 2 + ( 6 − 2 ) 2 = ( 8 + 4 3 ) + ( 8 − 4 3 ) = 16 (\sqrt{6}+\sqrt{2})^2+(\sqrt{6}-\sqrt{2})^2=(8+4\sqrt{3})+(8-4\sqrt{3})=16 ( 6 + 2 ) 2 + ( 6 − 2 ) 2 = ( 8 + 4 3 ) + ( 8 − 4 3 ) = 16 ∥ e ⃗ 3 ∥ = 1 \lVert\vec{e}_3\rVert=1 ∥ e 3 ∥ = 1 e ⃗ 3 \vec{e}_3 e 3 ( cos 15 ∘ , sin 15 ∘ ) (\cos15^\circ,\sin15^\circ) ( cos 1 5 ∘ , sin 1 5 ∘ ) e ⃗ 2 \vec{e}_2 e 2 ( cos 45 ∘ , sin 45 ∘ ) (\cos45^\circ,\sin45^\circ) ( cos 4 5 ∘ , sin 4 5 ∘ ) y z yz y z 30 ∘ 30^\circ 3 0 ∘ a a a A = a I + ( A − a I ) A=aI+(A-aI) A = a I + ( A − a I ) a a a
e ⃗ 1 \vec{e}_1 e 1 e ⃗ 2 , e ⃗ 3 \vec{e}_2,\vec{e}_3 e 2 , e 3 y z yz y z
det ( 1 / 2 ( 6 + 2 ) / 4 1 / 2 ( 6 − 2 ) / 4 ) = 1 2 ⋅ − 2 2 4 = − 1 2 ≠ 0 \det\begin{pmatrix} 1/\sqrt{2} & (\sqrt{6}+\sqrt{2})/4 \\ 1/\sqrt{2} & (\sqrt{6}-\sqrt{2})/4 \end{pmatrix}
=\frac{1}{\sqrt{2}}\cdot\frac{-2\sqrt{2}}{4}=-\frac{1}{2}\neq 0 det ( 1/ 2 1/ 2 ( 6 + 2 ) /4 ( 6 − 2 ) /4 ) = 2 1 ⋅ 4 − 2 2 = − 2 1 = 0 より線形独立なので、{ e ⃗ 1 , e ⃗ 2 , e ⃗ 3 } \{\vec{e}_1,\vec{e}_2,\vec{e}_3\} { e 1 , e 2 , e 3 } R 3 \mathbb{R}^3 R 3
e ⃗ 1 , e ⃗ 2 \vec{e}_1,\vec{e}_2 e 1 , e 2 n n n
A n e ⃗ 1 = ( a + 1 ) n e ⃗ 1 , A n e ⃗ 2 = a n e ⃗ 2 A^n\vec{e}_1=(a+1)^n\vec{e}_1,\qquad A^n\vec{e}_2=a^n\vec{e}_2 A n e 1 = ( a + 1 ) n e 1 , A n e 2 = a n e 2 が得られます。e ⃗ 3 \vec{e}_3 e 3
A n e ⃗ 3 = n a n − 1 e ⃗ 2 + a n e ⃗ 3 ( n ≥ 1 ) A^n\vec{e}_3=n\,a^{n-1}\vec{e}_2+a^n\vec{e}_3\qquad(n\ge 1) A n e 3 = n a n − 1 e 2 + a n e 3 ( n ≥ 1 ) が成り立ちます。数学的帰納法で示します。n = 1 n=1 n = 1 A e ⃗ 3 = e ⃗ 2 + a e ⃗ 3 A\vec{e}_3=\vec{e}_2+a\vec{e}_3 A e 3 = e 2 + a e 3 n n n
A n + 1 e ⃗ 3 = A ( n a n − 1 e ⃗ 2 + a n e ⃗ 3 ) = n a n − 1 ⋅ a e ⃗ 2 + a n ( e ⃗ 2 + a e ⃗ 3 ) = ( n + 1 ) a n e ⃗ 2 + a n + 1 e ⃗ 3 \begin{aligned}
A^{n+1}\vec{e}_3
&=A\bigl(n\,a^{n-1}\vec{e}_2+a^n\vec{e}_3\bigr)\\
&=n\,a^{n-1}\cdot a\,\vec{e}_2+a^n(\vec{e}_2+a\vec{e}_3)\\
&=(n+1)a^{n}\vec{e}_2+a^{n+1}\vec{e}_3
\end{aligned} A n + 1 e 3 = A ( n a n − 1 e 2 + a n e 3 ) = n a n − 1 ⋅ a e 2 + a n ( e 2 + a e 3 ) = ( n + 1 ) a n e 2 + a n + 1 e 3 となり、n + 1 n+1 n + 1 a = 0 a=0 a = 0 n = 1 n=1 n = 1 a 0 = 1 a^{0}=1 a 0 = 1 A e ⃗ 3 = e ⃗ 2 A\vec{e}_3=\vec{e}_2 A e 3 = e 2 n ≥ 2 n\ge2 n ≥ 2 n a n − 1 n a^{n-1} n a n − 1 a n a^n a n A n e ⃗ 3 = 0 ⃗ A^n\vec{e}_3=\vec{0} A n e 3 = 0 A 2 e ⃗ 3 = A e ⃗ 2 = 0 ⋅ e ⃗ 2 = 0 ⃗ A^2\vec{e}_3=A\vec{e}_2=0\cdot\vec{e}_2=\vec{0} A 2 e 3 = A e 2 = 0 ⋅ e 2 = 0 n = 0 n=0 n = 0 A 0 e ⃗ 3 = e ⃗ 3 A^0\vec{e}_3=\vec{e}_3 A 0 e 3 = e 3
答えは
A n e ⃗ 1 = ( a + 1 ) n e ⃗ 1 , A n e ⃗ 2 = a n e ⃗ 2 , A n e ⃗ 3 = n a n − 1 e ⃗ 2 + a n e ⃗ 3 A^n\vec{e}_1=(a+1)^n\vec{e}_1,\qquad
A^n\vec{e}_2=a^n\vec{e}_2,\qquad
A^n\vec{e}_3=n\,a^{n-1}\vec{e}_2+a^n\vec{e}_3 A n e 1 = ( a + 1 ) n e 1 , A n e 2 = a n e 2 , A n e 3 = n a n − 1 e 2 + a n e 3 です。この基底で A A A λ 1 = a + 1 \lambda_1=a+1 λ 1 = a + 1 1 × 1 1\times1 1 × 1 a a a 2 × 2 2\times2 2 × 2
( a 1 0 a ) n = ( a n n a n − 1 0 a n ) \begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}^{\!n}=\begin{pmatrix} a^n & n a^{n-1} \\ 0 & a^n \end{pmatrix} ( a 0 1 a ) n = ( a n 0 n a n − 1 a n ) に対応します。
r ⃗ 0 = r ⃗ \vec{r}_0=\vec{r} r 0 = r r ⃗ n + 1 = A r ⃗ n + d ⃗ \vec{r}_{n+1}=A\vec{r}_n+\vec{d} r n + 1 = A r n + d
r ⃗ n = A n r ⃗ + ∑ k = 0 n − 1 A k d ⃗ \vec{r}_n=A^n\vec{r}+\sum_{k=0}^{n-1}A^k\vec{d} r n = A n r + k = 0 ∑ n − 1 A k d です（n = 0 n=0 n = 0 { e ⃗ i } \{\vec{e}_i\} { e i } r ⃗ = c 1 e ⃗ 1 + c 2 e ⃗ 2 + c 3 e ⃗ 3 \vec{r}=c_1\vec{e}_1+c_2\vec{e}_2+c_3\vec{e}_3 r = c 1 e 1 + c 2 e 2 + c 3 e 3 d ⃗ = d 1 e ⃗ 1 + d 2 e ⃗ 2 + d 3 e ⃗ 3 \vec{d}=d_1\vec{e}_1+d_2\vec{e}_2+d_3\vec{e}_3 d = d 1 e 1 + d 2 e 2 + d 3 e 3 r ⃗ , d ⃗ \vec{r},\vec{d} r , d R 3 \mathbb{R}^3 R 3 ( c i ) , ( d i ) (c_i),(d_i) ( c i ) , ( d i )
r ⃗ n = [ c 1 ( a + 1 ) n + d 1 ∑ k = 0 n − 1 ( a + 1 ) k ] e ⃗ 1 + [ c 2 a n + c 3 n a n − 1 + d 2 ∑ k = 0 n − 1 a k + d 3 ∑ k = 0 n − 1 k a k − 1 ] e ⃗ 2 + [ c 3 a n + d 3 ∑ k = 0 n − 1 a k ] e ⃗ 3 \begin{aligned}
\vec{r}_n=&\ \Bigl[c_1(a+1)^n+d_1\sum_{k=0}^{n-1}(a+1)^k\Bigr]\vec{e}_1\\
&+\Bigl[c_2a^n+c_3\,n\,a^{n-1}+d_2\sum_{k=0}^{n-1}a^k+d_3\sum_{k=0}^{n-1}k\,a^{k-1}\Bigr]\vec{e}_2\\
&+\Bigl[c_3a^n+d_3\sum_{k=0}^{n-1}a^k\Bigr]\vec{e}_3
\end{aligned} r n = [ c 1 ( a + 1 ) n + d 1 k = 0 ∑ n − 1 ( a + 1 ) k ] e 1 + [ c 2 a n + c 3 n a n − 1 + d 2 k = 0 ∑ n − 1 a k + d 3 k = 0 ∑ n − 1 k a k − 1 ] e 2 + [ c 3 a n + d 3 k = 0 ∑ n − 1 a k ] e 3 となります（k = 0 k=0 k = 0 k a k − 1 = 0 k\,a^{k-1}=0 k a k − 1 = 0 r ⃗ n \vec{r}_n r n
必要性を示します。r ⃗ = 0 ⃗ \vec{r}=\vec{0} r = 0 d ⃗ = e ⃗ 1 \vec{d}=\vec{e}_1 d = e 1 e ⃗ 1 \vec{e}_1 e 1 ∑ k = 0 n − 1 ( a + 1 ) k \sum_{k=0}^{n-1}(a+1)^k ∑ k = 0 n − 1 ( a + 1 ) k ∣ a + 1 ∣ < 1 \lvert a+1\rvert<1 ∣ a + 1 ∣ < 1 a + 1 = 1 a+1=1 a + 1 = 1 n n n ∣ a + 1 ∣ ≥ 1 \lvert a+1\rvert\ge1 ∣ a + 1 ∣ ≥ 1 a + 1 ≠ 1 a+1\neq1 a + 1 = 1 − 2 < a < 0 -2<a<0 − 2 < a < 0 r ⃗ = 0 ⃗ \vec{r}=\vec{0} r = 0 d ⃗ = e ⃗ 2 \vec{d}=\vec{e}_2 d = e 2 ∑ k = 0 n − 1 a k \sum_{k=0}^{n-1}a^k ∑ k = 0 n − 1 a k ∣ a ∣ < 1 \lvert a\rvert<1 ∣ a ∣ < 1 − 1 < a < 1 -1<a<1 − 1 < a < 1 − 1 < a < 0 -1<a<0 − 1 < a < 0
十分性を示します。− 1 < a < 0 -1<a<0 − 1 < a < 0 ∣ a ∣ < 1 \lvert a\rvert<1 ∣ a ∣ < 1 ∣ a + 1 ∣ < 1 \lvert a+1\rvert<1 ∣ a + 1 ∣ < 1 ( a + 1 ) n → 0 (a+1)^n\to0 ( a + 1 ) n → 0 a n → 0 a^n\to0 a n → 0 n a n − 1 → 0 n a^{n-1}\to0 n a n − 1 → 0
∑ k = 0 ∞ ( a + 1 ) k = 1 1 − ( a + 1 ) = − 1 a , ∑ k = 0 ∞ a k = 1 1 − a , ∑ k = 0 ∞ k a k − 1 = 1 ( 1 − a ) 2 \sum_{k=0}^{\infty}(a+1)^k=\frac{1}{1-(a+1)}=-\frac{1}{a},\qquad
\sum_{k=0}^{\infty}a^k=\frac{1}{1-a},\qquad
\sum_{k=0}^{\infty}k\,a^{k-1}=\frac{1}{(1-a)^2} k = 0 ∑ ∞ ( a + 1 ) k = 1 − ( a + 1 ) 1 = − a 1 , k = 0 ∑ ∞ a k = 1 − a 1 , k = 0 ∑ ∞ k a k − 1 = ( 1 − a ) 2 1 がいずれも収束します（項別微分は収束半径の内部で正当）。よって3つの係数列はすべて収束し、
lim n → ∞ r ⃗ n = − d 1 a e ⃗ 1 + [ d 2 1 − a + d 3 ( 1 − a ) 2 ] e ⃗ 2 + d 3 1 − a e ⃗ 3 \lim_{n\to\infty}\vec{r}_n=-\frac{d_1}{a}\vec{e}_1+\Bigl[\frac{d_2}{1-a}+\frac{d_3}{(1-a)^2}\Bigr]\vec{e}_2+\frac{d_3}{1-a}\vec{e}_3 n → ∞ lim r n = − a d 1 e 1 + [ 1 − a d 2 + ( 1 − a ) 2 d 3 ] e 2 + 1 − a d 3 e 3 となります。極限は r ⃗ \vec{r} r
− 1 < a < 0 -1<a<0 − 1 < a < 0 です。
得られた条件は max ( ∣ a + 1 ∣ , ∣ a ∣ ) < 1 \max(\lvert a+1\rvert,\lvert a\rvert)<1 max (∣ a + 1 ∣ , ∣ a ∣) < 1 A A A I − A I-A I − A r ⃗ ∞ = ( I − A ) − 1 d ⃗ \vec{r}_\infty=(I-A)^{-1}\vec{d} r ∞ = ( I − A ) − 1 d ( I − A ) − 1 d ⃗ (I-A)^{-1}\vec{d} ( I − A ) − 1 d a = − 0.4 a=-0.4 a = − 0.4 a = 0 a=0 a = 0 λ 1 = 1 \lambda_1=1 λ 1 = 1 e ⃗ 1 \vec{e}_1 e 1 n n n a = − 1 a=-1 a = − 1 λ 2 = − 1 \lambda_2=-1 λ 2 = − 1 n a n − 1 n a^{n-1} n a n − 1
正の定数 ω \omega ω
d 2 y d x 2 + ω 2 y = f ( x ) ( x ≥ 0 ) \frac{d^2y}{dx^2}+\omega^2 y=f(x)\qquad(x\ge 0) d x 2 d 2 y + ω 2 y = f ( x ) ( x ≥ 0 ) を扱います。設問(a)では y = a ( x ) cos ω x + b ( x ) sin ω x y=a(x)\cos\omega x+b(x)\sin\omega x y = a ( x ) cos ω x + b ( x ) sin ω x
d a d x cos ω x + d b d x sin ω x = 0 \frac{da}{dx}\cos\omega x+\frac{db}{dx}\sin\omega x=0 d x d a cos ω x + d x d b sin ω x = 0 のもとで d a / d x da/dx d a / d x d b / d x db/dx d b / d x a ( x ) , b ( x ) a(x),b(x) a ( x ) , b ( x ) f ( x ) = 1 / λ f(x)=1/\lambda f ( x ) = 1/ λ 0 ≤ x ≤ λ 0\le x\le\lambda 0 ≤ x ≤ λ f ( x ) = 0 f(x)=0 f ( x ) = 0 x > λ x>\lambda x > λ y ( 0 ) = y ′ ( 0 ) = 0 y(0)=y'(0)=0 y ( 0 ) = y ′ ( 0 ) = 0 y ( x ) y(x) y ( x ) λ → 0 \lambda\to0 λ → 0
補助条件があるので、y y y a , b a,b a , b
d y d x = d a d x cos ω x + d b d x sin ω x − a ω sin ω x + b ω cos ω x = − a ω sin ω x + b ω cos ω x . \frac{dy}{dx}=\frac{da}{dx}\cos\omega x+\frac{db}{dx}\sin\omega x-a\omega\sin\omega x+b\omega\cos\omega x
=-a\omega\sin\omega x+b\omega\cos\omega x . d x d y = d x d a cos ω x + d x d b sin ω x − aω sin ω x + bω cos ω x = − aω sin ω x + bω cos ω x . もう一度微分すると
d 2 y d x 2 = − d a d x ω sin ω x + d b d x ω cos ω x − ω 2 ( a cos ω x + b sin ω x ) \frac{d^2y}{dx^2}=-\frac{da}{dx}\,\omega\sin\omega x+\frac{db}{dx}\,\omega\cos\omega x-\omega^2\bigl(a\cos\omega x+b\sin\omega x\bigr) d x 2 d 2 y = − d x d a ω sin ω x + d x d b ω cos ω x − ω 2 ( a cos ω x + b sin ω x ) なので、微分方程式に代入すると − ω 2 y -\omega^2 y − ω 2 y
− d a d x sin ω x + d b d x cos ω x = f ( x ) ω -\frac{da}{dx}\sin\omega x+\frac{db}{dx}\cos\omega x=\frac{f(x)}{\omega} − d x d a sin ω x + d x d b cos ω x = ω f ( x ) が残ります。これが補助条件と対になる関係式です。まとめると、d a / d x da/dx d a / d x d b / d x db/dx d b / d x
d a d x cos ω x + d b d x sin ω x = 0 , − d a d x sin ω x + d b d x cos ω x = f ( x ) ω \frac{da}{dx}\cos\omega x+\frac{db}{dx}\sin\omega x=0,\qquad
-\frac{da}{dx}\sin\omega x+\frac{db}{dx}\cos\omega x=\frac{f(x)}{\omega} d x d a cos ω x + d x d b sin ω x = 0 , − d x d a sin ω x + d x d b cos ω x = ω f ( x ) を満たします。係数行列は回転行列で行列式は cos 2 ω x + sin 2 ω x = 1 \cos^2\omega x+\sin^2\omega x=1 cos 2 ω x + sin 2 ω x = 1 x x x
d a d x = − f ( x ) sin ω x ω , d b d x = f ( x ) cos ω x ω \frac{da}{dx}=-\frac{f(x)\sin\omega x}{\omega},\qquad
\frac{db}{dx}=\frac{f(x)\cos\omega x}{\omega} d x d a = − ω f ( x ) sin ω x , d x d b = ω f ( x ) cos ω x となります。
設問(a)の式をそのまま 0 0 0 x x x
a ( x ) = a 0 − 1 ω ∫ 0 x f ( t ) sin ω t d t , b ( x ) = b 0 + 1 ω ∫ 0 x f ( t ) cos ω t d t a(x)=a_0-\frac{1}{\omega}\int_0^x f(t)\sin\omega t\,dt,\qquad
b(x)=b_0+\frac{1}{\omega}\int_0^x f(t)\cos\omega t\,dt a ( x ) = a 0 − ω 1 ∫ 0 x f ( t ) sin ω t d t , b ( x ) = b 0 + ω 1 ∫ 0 x f ( t ) cos ω t d t です。a 0 = a ( 0 ) a_0=a(0) a 0 = a ( 0 ) b 0 = b ( 0 ) b_0=b(0) b 0 = b ( 0 ) y y y
y ( x ) = a 0 cos ω x + b 0 sin ω x + 1 ω ∫ 0 x f ( t ) ( sin ω x cos ω t − cos ω x sin ω t ) d t = a 0 cos ω x + b 0 sin ω x + 1 ω ∫ 0 x f ( t ) sin ω ( x − t ) d t y(x)=a_0\cos\omega x+b_0\sin\omega x+\frac{1}{\omega}\int_0^x f(t)\bigl(\sin\omega x\cos\omega t-\cos\omega x\sin\omega t\bigr)dt
=a_0\cos\omega x+b_0\sin\omega x+\frac{1}{\omega}\int_0^x f(t)\sin\omega(x-t)\,dt y ( x ) = a 0 cos ω x + b 0 sin ω x + ω 1 ∫ 0 x f ( t ) ( sin ω x cos ω t − cos ω x sin ω t ) d t = a 0 cos ω x + b 0 sin ω x + ω 1 ∫ 0 x f ( t ) sin ω ( x − t ) d t となり、斉次解と特解（デュアメル積分）の和という見慣れた形が再現されます。
初期条件を積分定数に翻訳します。y ( 0 ) = a ( 0 ) = a 0 y(0)=a(0)=a_0 y ( 0 ) = a ( 0 ) = a 0 y ′ ( 0 ) = ω b ( 0 ) = ω b 0 y'(0)=\omega b(0)=\omega b_0 y ′ ( 0 ) = ω b ( 0 ) = ω b 0 y ( 0 ) = y ′ ( 0 ) = 0 y(0)=y'(0)=0 y ( 0 ) = y ′ ( 0 ) = 0 a 0 = b 0 = 0 a_0=b_0=0 a 0 = b 0 = 0
0 ≤ x ≤ λ 0\le x\le\lambda 0 ≤ x ≤ λ f = 1 / λ f=1/\lambda f = 1/ λ
a ( x ) = − 1 ω λ ∫ 0 x sin ω t d t = cos ω x − 1 ω 2 λ , b ( x ) = 1 ω λ ∫ 0 x cos ω t d t = sin ω x ω 2 λ a(x)=-\frac{1}{\omega\lambda}\int_0^x\sin\omega t\,dt=\frac{\cos\omega x-1}{\omega^2\lambda},\qquad
b(x)=\frac{1}{\omega\lambda}\int_0^x\cos\omega t\,dt=\frac{\sin\omega x}{\omega^2\lambda} a ( x ) = − ω λ 1 ∫ 0 x sin ω t d t = ω 2 λ cos ω x − 1 , b ( x ) = ω λ 1 ∫ 0 x cos ω t d t = ω 2 λ sin ω x です。よって
y ( x ) = ( cos ω x − 1 ) cos ω x + sin 2 ω x ω 2 λ = 1 − cos ω x ω 2 λ ( 0 ≤ x ≤ λ ) . y(x)=\frac{(\cos\omega x-1)\cos\omega x+\sin^2\omega x}{\omega^2\lambda}
=\frac{1-\cos\omega x}{\omega^2\lambda}\qquad(0\le x\le\lambda). y ( x ) = ω 2 λ ( cos ω x − 1 ) cos ω x + sin 2 ω x = ω 2 λ 1 − cos ω x ( 0 ≤ x ≤ λ ) . 検算すると y ′ ′ = cos ω x / λ y''=\cos\omega x/\lambda y ′′ = cos ω x / λ ω 2 y = ( 1 − cos ω x ) / λ \omega^2 y=(1-\cos\omega x)/\lambda ω 2 y = ( 1 − cos ω x ) / λ 1 / λ = f 1/\lambda=f 1/ λ = f y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 y ′ ( 0 ) = sin 0 / ( ω λ ) = 0 y'(0)=\sin 0/(\omega\lambda)=0 y ′ ( 0 ) = sin 0/ ( ω λ ) = 0
x > λ x>\lambda x > λ f = 0 f=0 f = 0 a , b a,b a , b x = λ x=\lambda x = λ
a ( x ) = cos ω λ − 1 ω 2 λ , b ( x ) = sin ω λ ω 2 λ ( x > λ ) a(x)=\frac{\cos\omega\lambda-1}{\omega^2\lambda},\qquad
b(x)=\frac{\sin\omega\lambda}{\omega^2\lambda}\qquad(x>\lambda) a ( x ) = ω 2 λ cos ω λ − 1 , b ( x ) = ω 2 λ sin ω λ ( x > λ ) です。加法定理でまとめると
y ( x ) = ( cos ω λ − 1 ) cos ω x + sin ω λ sin ω x ω 2 λ = cos ω ( x − λ ) − cos ω x ω 2 λ ( x > λ ) . y(x)=\frac{(\cos\omega\lambda-1)\cos\omega x+\sin\omega\lambda\sin\omega x}{\omega^2\lambda}
=\frac{\cos\omega(x-\lambda)-\cos\omega x}{\omega^2\lambda}\qquad(x>\lambda). y ( x ) = ω 2 λ ( cos ω λ − 1 ) cos ω x + sin ω λ sin ω x = ω 2 λ cos ω ( x − λ ) − cos ω x ( x > λ ) . x = λ x=\lambda x = λ ( 1 − cos ω λ ) / ( ω 2 λ ) (1-\cos\omega\lambda)/(\omega^2\lambda) ( 1 − cos ω λ ) / ( ω 2 λ ) y y y x = λ x=\lambda x = λ a , b a,b a , b y ′ y' y ′
y ( x ) = { 1 − cos ω x ω 2 λ ( 0 ≤ x ≤ λ ) cos ω ( x − λ ) − cos ω x ω 2 λ ( x > λ ) y(x)=
\begin{cases}
\dfrac{1-\cos\omega x}{\omega^2\lambda} & (0\le x\le\lambda)\\[2ex]
\dfrac{\cos\omega(x-\lambda)-\cos\omega x}{\omega^2\lambda} & (x>\lambda)
\end{cases} y ( x ) = ⎩ ⎨ ⎧ ω 2 λ 1 − cos ω x ω 2 λ cos ω ( x − λ ) − cos ω x ( 0 ≤ x ≤ λ ) ( x > λ ) です。x > λ x>\lambda x > λ
y ( x ) = 2 ω 2 λ sin ω λ 2 sin ω ( x − λ 2 ) y(x)=\frac{2}{\omega^2\lambda}\sin\frac{\omega\lambda}{2}\,\sin\omega\Bigl(x-\frac{\lambda}{2}\Bigr) y ( x ) = ω 2 λ 2 sin 2 ω λ sin ω ( x − 2 λ ) とも書け、外力が切れたあとは振幅 2 sin ( ω λ / 2 ) ω 2 λ \dfrac{2\sin(\omega\lambda/2)}{\omega^2\lambda} ω 2 λ 2 sin ( ω λ /2 )
x > 0 x>0 x > 0 λ → 0 \lambda\to0 λ → 0 λ < x \lambda<x λ < x x > λ x>\lambda x > λ
2 sin ( ω λ / 2 ) ω 2 λ ⟶ 2 ω 2 ⋅ ω 2 = 1 ω , sin ω ( x − λ 2 ) ⟶ sin ω x \frac{2\sin(\omega\lambda/2)}{\omega^2\lambda}\longrightarrow\frac{2}{\omega^2}\cdot\frac{\omega}{2}=\frac{1}{\omega},\qquad
\sin\omega\Bigl(x-\frac{\lambda}{2}\Bigr)\longrightarrow\sin\omega x ω 2 λ 2 sin ( ω λ /2 ) ⟶ ω 2 2 ⋅ 2 ω = ω 1 , sin ω ( x − 2 λ ) ⟶ sin ω x なので
lim λ → 0 y ( x ) = sin ω x ω \lim_{\lambda\to 0}y(x)=\frac{\sin\omega x}{\omega} λ → 0 lim y ( x ) = ω sin ω x です（x = 0 x=0 x = 0 y = 0 y=0 y = 0 sin 0 / ω = 0 \sin 0/\omega=0 sin 0/ ω = 0 1 / λ 1/\lambda 1/ λ λ \lambda λ λ → 0 \lambda\to0 λ → 0 f ( x ) = δ ( x ) f(x)=\delta(x) f ( x ) = δ ( x ) sin ω x / ω \sin\omega x/\omega sin ω x / ω y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 y ′ ( 0 ) = 1 y'(0)=1 y ′ ( 0 ) = 1 G ( x ) = sin ω x / ω G(x)=\sin\omega x/\omega G ( x ) = sin ω x / ω
AとBがゲームを繰り返し、Aが2回続けて勝った時点で終了します。各ゲームは独立で、Aが勝つ確率は 2 / 3 2/3 2/3 1 / 3 1/3 1/3 N N N x N x_N x N N N N x N x_N x N
最初の N N N N N N N ≥ 2 N\ge2 N ≥ 2
N N N N N N N − 1 N-1 N − 1 1 3 x N − 1 \dfrac{1}{3}x_{N-1} 3 1 x N − 1
N N N N − 1 N-1 N − 1 N N N N − 2 N-2 N − 2 N N N 1 3 ⋅ 2 3 x N − 2 = 2 9 x N − 2 \dfrac{1}{3}\cdot\dfrac{2}{3}\,x_{N-2}=\dfrac{2}{9}x_{N-2} 3 1 ⋅ 3 2 x N − 2 = 9 2 x N − 2
両者は排反なので、答えは
x N = 1 3 x N − 1 + 2 9 x N − 2 ( N ≥ 2 ) x_N=\frac{1}{3}x_{N-1}+\frac{2}{9}x_{N-2}\qquad(N\ge 2) x N = 3 1 x N − 1 + 9 2 x N − 2 ( N ≥ 2 ) です。初期値は x 0 = 1 x_0=1 x 0 = 1 x 1 = 1 x_1=1 x 1 = 1 N = 2 N=2 N = 2 x 2 = 1 − ( 2 / 3 ) 2 = 5 / 9 x_2=1-(2/3)^2=5/9 x 2 = 1 − ( 2/3 ) 2 = 5/9 x 2 = 1 3 + 2 9 = 5 9 x_2=\frac13+\frac29=\frac59 x 2 = 3 1 + 9 2 = 9 5
定数係数の3項間漸化式なので特性方程式を解きます。x N = t N x_N=t^N x N = t N
t 2 = 1 3 t + 2 9 ⟺ 9 t 2 − 3 t − 2 = 0 ⟺ t = 3 ± 9 + 72 18 = 3 ± 9 18 t^2=\frac{1}{3}t+\frac{2}{9}
\quad\Longleftrightarrow\quad
9t^2-3t-2=0
\quad\Longleftrightarrow\quad
t=\frac{3\pm\sqrt{9+72}}{18}=\frac{3\pm 9}{18} t 2 = 3 1 t + 9 2 ⟺ 9 t 2 − 3 t − 2 = 0 ⟺ t = 18 3 ± 9 + 72 = 18 3 ± 9 より t = 2 3 , − 1 3 t=\dfrac{2}{3},\ -\dfrac{1}{3} t = 3 2 , − 3 1 x N = α ( 2 3 ) N + β ( − 1 3 ) N x_N=\alpha\left(\frac23\right)^N+\beta\left(-\frac13\right)^N x N = α ( 3 2 ) N + β ( − 3 1 ) N
α + β = 1 , 2 3 α − 1 3 β = 1 \alpha+\beta=1,\qquad \frac{2}{3}\alpha-\frac{1}{3}\beta=1 α + β = 1 , 3 2 α − 3 1 β = 1 です。第2式を3倍して第1式と加えると 3 α = 4 3\alpha=4 3 α = 4 α = 4 3 \alpha=\dfrac43 α = 3 4 β = − 1 3 \beta=-\dfrac13 β = − 3 1
x N = 4 3 ( 2 3 ) N − 1 3 ( − 1 3 ) N = 2 N + 2 − ( − 1 ) N 3 N + 1 ( N = 0 , 1 , 2 , … ) x_N=\frac{4}{3}\Bigl(\frac{2}{3}\Bigr)^{N}-\frac{1}{3}\Bigl(-\frac{1}{3}\Bigr)^{N}
=\frac{2^{N+2}-(-1)^N}{3^{N+1}}\qquad(N=0,1,2,\dots) x N = 3 4 ( 3 2 ) N − 3 1 ( − 3 1 ) N = 3 N + 1 2 N + 2 − ( − 1 ) N ( N = 0 , 1 , 2 , … ) が答えです。検算すると N = 0 N=0 N = 0 ( 4 − 1 ) / 3 = 1 (4-1)/3=1 ( 4 − 1 ) /3 = 1 N = 1 N=1 N = 1 ( 8 + 1 ) / 9 = 1 (8+1)/9=1 ( 8 + 1 ) /9 = 1 N = 2 N=2 N = 2 ( 16 − 1 ) / 27 = 5 / 9 (16-1)/27=5/9 ( 16 − 1 ) /27 = 5/9 N = 3 N=3 N = 3 ( 32 + 1 ) / 81 = 11 / 27 (32+1)/81=11/27 ( 32 + 1 ) /81 = 11/27 N = 3 N=3 N = 3 1 3 ⋅ 5 9 + 2 9 ⋅ 1 = 5 + 6 27 = 11 27 \frac13\cdot\frac59+\frac29\cdot1=\frac{5+6}{27}=\frac{11}{27} 3 1 ⋅ 9 5 + 9 2 ⋅ 1 = 27 5 + 6 = 27 11 N → ∞ N\to\infty N → ∞ x N → 0 x_N\to0 x N → 0
行われるゲームの総回数を T T T T > N T>N T > N N N N P ( T > N ) = x N P(T>N)=x_N P ( T > N ) = x N
E [ T ] = ∑ N = 0 ∞ P ( T > N ) = ∑ N = 0 ∞ x N E[T]=\sum_{N=0}^{\infty}P(T>N)=\sum_{N=0}^{\infty}x_N E [ T ] = N = 0 ∑ ∞ P ( T > N ) = N = 0 ∑ ∞ x N を使います（x N x_N x N 2 / 3 2/3 2/3
E [ T ] = 4 3 ∑ N = 0 ∞ ( 2 3 ) N − 1 3 ∑ N = 0 ∞ ( − 1 3 ) N = 4 3 ⋅ 1 1 − 2 3 − 1 3 ⋅ 1 1 + 1 3 = 4 3 ⋅ 3 − 1 3 ⋅ 3 4 = 4 − 1 4 E[T]=\frac{4}{3}\sum_{N=0}^{\infty}\Bigl(\frac{2}{3}\Bigr)^{N}-\frac{1}{3}\sum_{N=0}^{\infty}\Bigl(-\frac{1}{3}\Bigr)^{N}
=\frac{4}{3}\cdot\frac{1}{1-\frac23}-\frac{1}{3}\cdot\frac{1}{1+\frac13}
=\frac{4}{3}\cdot 3-\frac{1}{3}\cdot\frac{3}{4}=4-\frac{1}{4} E [ T ] = 3 4 N = 0 ∑ ∞ ( 3 2 ) N − 3 1 N = 0 ∑ ∞ ( − 3 1 ) N = 3 4 ⋅ 1 − 3 2 1 − 3 1 ⋅ 1 + 3 1 1 = 3 4 ⋅ 3 − 3 1 ⋅ 4 3 = 4 − 4 1 となり、答えは
E [ T ] = 15 4 = 3.75 E[T]=\frac{15}{4}=3.75 E [ T ] = 4 15 = 3.75 です。
別解による検算をします。状態を「直前のゲームがAの勝ちでない」（開始状態を含む）と「直前がAの勝ち」の2つに分け、それぞれからの終了までの期待回数を E 0 , E 1 E_0,E_1 E 0 , E 1
E 0 = 1 + 2 3 E 1 + 1 3 E 0 , E 1 = 1 + 2 3 ⋅ 0 + 1 3 E 0 E_0=1+\frac{2}{3}E_1+\frac{1}{3}E_0,\qquad
E_1=1+\frac{2}{3}\cdot 0+\frac{1}{3}E_0 E 0 = 1 + 3 2 E 1 + 3 1 E 0 , E 1 = 1 + 3 2 ⋅ 0 + 3 1 E 0 です。第1式から E 0 = 3 2 + E 1 E_0=\frac{3}{2}+E_1 E 0 = 2 3 + E 1 E 1 = 1 + 1 3 ( 3 2 + E 1 ) = 3 2 + 1 3 E 1 E_1=1+\frac13\bigl(\frac32+E_1\bigr)=\frac32+\frac13E_1 E 1 = 1 + 3 1 ( 2 3 + E 1 ) = 2 3 + 3 1 E 1 2 3 E 1 = 3 2 \frac23E_1=\frac32 3 2 E 1 = 2 3 E 1 = 9 4 E_1=\frac{9}{4} E 1 = 4 9 E 0 = 3 2 + 9 4 = 15 4 E_0=\frac{3}{2}+\frac{9}{4}=\frac{15}{4} E 0 = 2 3 + 4 9 = 4 15 2 / ( 2 / 3 ) 2 = 4.5 2/(2/3)^2=4.5 2/ ( 2/3 ) 2 = 4.5
出典: 東京大学大学院理学系研究科 物理学専攻 平成10年度 修士課程 入学試験問題 数学。問題文は要約して引用しています。