1 Chebyshev 展開と Clenshaw 法 定義 1.1. f ∈ C ( [ − 1 , 1 ] ) f\in C([-1,1]) f ∈ C ([ − 1 , 1 ]) k ∈ N ≥ 0 k\in\N k ∈ N ≥ 0
a k ( f ) : = 2 π ∫ 0 π f ( cos θ ) cos k θ d θ a_k(f):=\frac2\pi\int_0^\pi f(\cos\theta)\cos k\theta\,d\theta a k ( f ) := π 2 ∫ 0 π f ( cos θ ) cos k θ d θ をf f f k k k Chebyshev 係数 (Chebyshev coefficient )n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0
S n f : = a 0 ( f ) 2 + ∑ k = 1 n a k ( f ) T k S_nf:=\frac{a_0(f)}2+\sum_{k=1}^na_k(f)\,T_k S n f := 2 a 0 ( f ) + k = 1 ∑ n a k ( f ) T k をf f f n n n 打切り Chebyshev 展開 (truncated Chebyshev expansion )
命題 1.2. f ∈ C ( [ − 1 , 1 ] ) f\in C([-1,1]) f ∈ C ([ − 1 , 1 ]) ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] w T ( x ) : = ( 1 − x 2 ) − 1 / 2 w_T(x):=(1-x^2)^{-1/2} w T ( x ) := ( 1 − x 2 ) − 1/2 ( − 1 , 1 ) (-1,1) ( − 1 , 1 )
任意のn ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 S n f = Π n w T f S_nf=\Pi_n^{w_T}f S n f = Π n w T f q ∈ P n q\in\mathcal P_n q ∈ P n S n q = q S_nq=q S n q = q
∑ k = 0 ∞ ∣ a k ( f ) ∣ < ∞ \sum_{k=0}^\infty|a_k(f)|<\infty ∑ k = 0 ∞ ∣ a k ( f ) ∣ < ∞ a 0 ( f ) / 2 + ∑ k = 1 ∞ a k ( f ) T k a_0(f)/2+\sum_{k=1}^\infty a_k(f)\,T_k a 0 ( f ) /2 + ∑ k = 1 ∞ a k ( f ) T k [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f f f n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 ∥ f − S n f ∥ ∞ ≤ ∑ k = n + 1 ∞ ∣ a k ( f ) ∣ \lVert f-S_nf\rVert_\infty\le\sum_{k=n+1}^\infty|a_k(f)| ∥ f − S n f ∥ ∞ ≤ ∑ k = n + 1 ∞ ∣ a k ( f ) ∣ 証明. (1) を示す。§E20.14 命題 4.3 (4) の係数はa k ( f ) a_k(f) a k ( f ) S n f = Π n w T f S_nf=\Pi_n^{w_T}f S n f = Π n w T f q ∈ P n q\in\mathcal P_n q ∈ P n ∥ q − q ∥ w T = 0 \lVert q-q\rVert_{w_T}=0 ∥ q − q ∥ w T = 0 §E20.14 定理 1.6 (1) の一意性によりΠ n w T q = q \Pi_n^{w_T}q=q Π n w T q = q
(2) を示す。a k : = a k ( f ) a_k:=a_k(f) a k := a k ( f ) §E20.11 補題 3.2 (3) により[ − 1 , 1 ] [-1,1] [ − 1 , 1 ] ∣ a k T k ∣ ≤ ∣ a k ∣ |a_kT_k|\le|a_k| ∣ a k T k ∣ ≤ ∣ a k ∣ F : = a 0 / 2 + ∑ k ≥ 1 a k T k F:=a_0/2+\sum_{k\ge1}a_kT_k F := a 0 /2 + ∑ k ≥ 1 a k T k [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] F ∈ C ( [ − 1 , 1 ] ) F\in C([-1,1]) F ∈ C ([ − 1 , 1 ]) S n f S_nf S n f ∥ S n f − F ∥ ∞ ≤ ∑ k > n ∣ a k ∣ \lVert S_nf-F\rVert_\infty\le\sum_{k>n}|a_k| ∥ S n f − F ∥ ∞ ≤ ∑ k > n ∣ a k ∣ n → ∞ n\to\infty n → ∞ 0 0 0 §E20.14 命題 4.3 (1) により∫ − 1 1 w T ( x ) d x = ∥ T 0 ∥ w T 2 = π \int_{-1}^1w_T(x)\,dx=\lVert T_0\rVert_{w_T}^2=\pi ∫ − 1 1 w T ( x ) d x = ∥ T 0 ∥ w T 2 = π §E20.14 命題 1.5 により任意のg ∈ C ( [ − 1 , 1 ] ) g\in C([-1,1]) g ∈ C ([ − 1 , 1 ]) ∥ g ∥ w T ≤ π ∥ g ∥ ∞ \lVert g\rVert_{w_T}\le\sqrt\pi\lVert g\rVert_\infty ∥ g ∥ w T ≤ π ∥ g ∥ ∞ (1) と§E20.14 定理 5.1 (3) により∥ f − S n f ∥ w T = ∥ f − Π n w T f ∥ w T → 0 \lVert f-S_nf\rVert_{w_T}=\lVert f-\Pi_n^{w_T}f\rVert_{w_T}\to0 ∥ f − S n f ∥ w T = ∥ f − Π n w T f ∥ w T → 0
∥ f − F ∥ w T ≤ ∥ f − S n f ∥ w T + ∥ S n f − F ∥ w T ≤ ∥ f − S n f ∥ w T + π ∥ S n f − F ∥ ∞ → 0 ( n → ∞ ) \lVert f-F\rVert_{w_T}\le\lVert f-S_nf\rVert_{w_T}+\lVert S_nf-F\rVert_{w_T}\le\lVert f-S_nf\rVert_{w_T}+\sqrt\pi\lVert S_nf-F\rVert_\infty\to0\qquad(n\to\infty) ∥ f − F ∥ w T ≤ ∥ f − S n f ∥ w T + ∥ S n f − F ∥ w T ≤ ∥ f − S n f ∥ w T + π ∥ S n f − F ∥ ∞ → 0 ( n → ∞ ) であり、左辺はn n n ∥ f − F ∥ w T = 0 \lVert f-F\rVert_{w_T}=0 ∥ f − F ∥ w T = 0 §E20.14 命題 1.5 により⟨ ⋅ , ⋅ ⟩ w T \langle\cdot,\cdot\rangle_{w_T} ⟨ ⋅ , ⋅ ⟩ w T C ( [ − 1 , 1 ] ) C([-1,1]) C ([ − 1 , 1 ]) f = F f=F f = F x ∈ [ − 1 , 1 ] x\in[-1,1] x ∈ [ − 1 , 1 ] ∣ f ( x ) − S n f ( x ) ∣ = ∣ ∑ k > n a k T k ( x ) ∣ ≤ ∑ k > n ∣ a k ∣ |f(x)-S_nf(x)|=\bigl|\sum_{k>n}a_kT_k(x)\bigr|\le\sum_{k>n}|a_k| ∣ f ( x ) − S n f ( x ) ∣ = ∑ k > n a k T k ( x ) ≤ ∑ k > n ∣ a k ∣ ▨
補題 1.3. n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 c 0 , … , c n c_0,\dots,c_n c 0 , … , c n x ∈ R x\in\R x ∈ R b n + 1 : = b n + 2 : = 0 b_{n+1}:=b_{n+2}:=0 b n + 1 := b n + 2 := 0 k = n , n − 1 , … , 0 k=n,n-1,\dots,0 k = n , n − 1 , … , 0
b k : = c k + 2 x b k + 1 − b k + 2 b_k:=c_k+2xb_{k+1}-b_{k+2} b k := c k + 2 x b k + 1 − b k + 2 と定める。このとき∑ k = 0 n c k T k ( x ) = b 0 − x b 1 \sum_{k=0}^nc_kT_k(x)=b_0-xb_1 ∑ k = 0 n c k T k ( x ) = b 0 − x b 1
証明. 0 ≤ k ≤ n 0\le k\le n 0 ≤ k ≤ n c k = b k − 2 x b k + 1 + b k + 2 c_k=b_k-2xb_{k+1}+b_{k+2} c k = b k − 2 x b k + 1 + b k + 2 b n + 1 = b n + 2 = 0 b_{n+1}=b_{n+2}=0 b n + 1 = b n + 2 = 0
∑ k = 0 n c k T k ( x ) = ∑ k = 0 n b k T k ( x ) − 2 x ∑ k = 1 n b k T k − 1 ( x ) + ∑ k = 2 n b k T k − 2 ( x ) = b 0 T 0 ( x ) + b 1 ( T 1 ( x ) − 2 x T 0 ( x ) ) + ∑ k = 2 n b k ( T k ( x ) − 2 x T k − 1 ( x ) + T k − 2 ( x ) ) \sum_{k=0}^nc_kT_k(x)=\sum_{k=0}^nb_kT_k(x)-2x\sum_{k=1}^nb_kT_{k-1}(x)+\sum_{k=2}^nb_kT_{k-2}(x)=b_0T_0(x)+b_1\bigl(T_1(x)-2xT_0(x)\bigr)+\sum_{k=2}^nb_k\bigl(T_k(x)-2xT_{k-1}(x)+T_{k-2}(x)\bigr) k = 0 ∑ n c k T k ( x ) = k = 0 ∑ n b k T k ( x ) − 2 x k = 1 ∑ n b k T k − 1 ( x ) + k = 2 ∑ n b k T k − 2 ( x ) = b 0 T 0 ( x ) + b 1 ( T 1 ( x ) − 2 x T 0 ( x ) ) + k = 2 ∑ n b k ( T k ( x ) − 2 x T k − 1 ( x ) + T k − 2 ( x ) ) である。§E20.11 定義 3.1 の漸化式によりk ≥ 2 k\ge2 k ≥ 2 T k ( x ) − 2 x T k − 1 ( x ) + T k − 2 ( x ) = 0 T_k(x)-2xT_{k-1}(x)+T_{k-2}(x)=0 T k ( x ) − 2 x T k − 1 ( x ) + T k − 2 ( x ) = 0 T 0 ( x ) = 1 T_0(x)=1 T 0 ( x ) = 1 T 1 ( x ) = x T_1(x)=x T 1 ( x ) = x b 0 − x b 1 b_0-xb_1 b 0 − x b 1 ▨
2 一様最良近似と交代定理 補題 2.2. n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 t 0 < t 1 < ⋯ < t n + 1 t_0<t_1<\dots<t_{n+1} t 0 < t 1 < ⋯ < t n + 1 0 ≤ j ≤ n + 1 0\le j\le n+1 0 ≤ j ≤ n + 1 v j : = 1 / ∏ k ≠ j ( t j − t k ) v_j:=1/\prod_{k\ne j}(t_j-t_k) v j := 1/ ∏ k = j ( t j − t k ) 0 ≤ k ≤ n + 1 0\le k\le n+1 0 ≤ k ≤ n + 1 k ≠ j k\ne j k = j
任意のq ∈ P n q\in\mathcal P_n q ∈ P n ∑ j = 0 n + 1 v j q ( t j ) = 0 \sum_{j=0}^{n+1}v_jq(t_j)=0 ∑ j = 0 n + 1 v j q ( t j ) = 0
0 ≤ j ≤ n + 1 0\le j\le n+1 0 ≤ j ≤ n + 1 v j ≠ 0 v_j\ne0 v j = 0 ( − 1 ) j v j = ( − 1 ) n + 1 ∣ v j ∣ (-1)^jv_j=(-1)^{n+1}|v_j| ( − 1 ) j v j = ( − 1 ) n + 1 ∣ v j ∣ 任意の( y 0 , … , y n + 1 ) ∈ R n + 2 (y_0,\dots,y_{n+1})\in\R^{n+2} ( y 0 , … , y n + 1 ) ∈ R n + 2 y j − p ( t j ) = ( − 1 ) j h y_j-p(t_j)=(-1)^jh y j − p ( t j ) = ( − 1 ) j h 0 ≤ j ≤ n + 1 0\le j\le n+1 0 ≤ j ≤ n + 1 ( p , h ) ∈ P n × R (p,h)\in\mathcal P_n\times\R ( p , h ) ∈ P n × R h = ( − 1 ) n + 1 ∑ j = 0 n + 1 v j y j ∑ j = 0 n + 1 ∣ v j ∣ h=(-1)^{n+1}\frac{\sum_{j=0}^{n+1}v_jy_j}{\sum_{j=0}^{n+1}|v_j|} h = ( − 1 ) n + 1 ∑ j = 0 n + 1 ∣ v j ∣ ∑ j = 0 n + 1 v j y j
(3) の( p , h ) (p,h) ( p , h ) q ∈ P n q\in\mathcal P_n q ∈ P n ∣ h ∣ ≤ ∑ j = 0 n + 1 ∣ v j ∣ ∣ y j − q ( t j ) ∣ ∑ j = 0 n + 1 ∣ v j ∣ ≤ max 0 ≤ j ≤ n + 1 ∣ y j − q ( t j ) ∣ |h|\le\frac{\sum_{j=0}^{n+1}|v_j|\,|y_j-q(t_j)|}{\sum_{j=0}^{n+1}|v_j|}\le\max_{0\le j\le n+1}|y_j-q(t_j)| ∣ h ∣ ≤ ∑ j = 0 n + 1 ∣ v j ∣ ∑ j = 0 n + 1 ∣ v j ∣ ∣ y j − q ( t j ) ∣ ≤ 0 ≤ j ≤ n + 1 max ∣ y j − q ( t j ) ∣ σ ∈ { 1 , − 1 } \sigma\in\{1,-1\} σ ∈ { 1 , − 1 } j j j σ ( − 1 ) j ( y j − q ( t j ) ) ≥ 0 \sigma(-1)^j\bigl(y_j-q(t_j)\bigr)\ge0 σ ( − 1 ) j ( y j − q ( t j ) ) ≥ 0 証明. (1) を示す。q ∈ P n ⊆ P n + 1 q\in\mathcal P_n\subseteq\mathcal P_{n+1} q ∈ P n ⊆ P n + 1 §E20.12 定理 1.3 (2) によりq q q t 0 , … , t n + 1 t_0,\dots,t_{n+1} t 0 , … , t n + 1 q q q P n + 1 \mathcal P_{n+1} P n + 1 §E20.12 命題 1.4 (1) によりq q q x n + 1 x^{n+1} x n + 1 ∑ j v j q ( t j ) \sum_jv_jq(t_j) ∑ j v j q ( t j ) 0 0 0
(2) を示す。v j v_j v j t j − t k t_j-t_k t j − t k 0 0 0 k > j k>j k > j n + 1 − j n+1-j n + 1 − j v j v_j v j ( − 1 ) n + 1 − j (-1)^{n+1-j} ( − 1 ) n + 1 − j ( − 1 ) j v j = ( − 1 ) n + 1 ( − 1 ) n + 1 − j v j = ( − 1 ) n + 1 ∣ v j ∣ (-1)^jv_j=(-1)^{n+1}(-1)^{n+1-j}v_j=(-1)^{n+1}|v_j| ( − 1 ) j v j = ( − 1 ) n + 1 ( − 1 ) n + 1 − j v j = ( − 1 ) n + 1 ∣ v j ∣
(3) を示す。( p , h ) (p,h) ( p , h ) (1) をp p p (2) を用いると
∑ j = 0 n + 1 v j y j = ∑ j = 0 n + 1 v j ( p ( t j ) + ( − 1 ) j h ) = h ∑ j = 0 n + 1 ( − 1 ) j v j = ( − 1 ) n + 1 h ∑ j = 0 n + 1 ∣ v j ∣ \sum_{j=0}^{n+1}v_jy_j=\sum_{j=0}^{n+1}v_j\bigl(p(t_j)+(-1)^jh\bigr)=h\sum_{j=0}^{n+1}(-1)^jv_j=(-1)^{n+1}h\sum_{j=0}^{n+1}|v_j| j = 0 ∑ n + 1 v j y j = j = 0 ∑ n + 1 v j ( p ( t j ) + ( − 1 ) j h ) = h j = 0 ∑ n + 1 ( − 1 ) j v j = ( − 1 ) n + 1 h j = 0 ∑ n + 1 ∣ v j ∣ であるから、h h h p ∈ P n p\in\mathcal P_n p ∈ P n n + 1 n+1 n + 1 t 0 , … , t n t_0,\dots,t_n t 0 , … , t n y j − ( − 1 ) j h y_j-(-1)^jh y j − ( − 1 ) j h §E20.12 定理 1.3 (2) によりp p p h h h §E20.12 定理 1.3 (2) によりP ( t j ) = y j − ( − 1 ) j h P(t_j)=y_j-(-1)^jh P ( t j ) = y j − ( − 1 ) j h 0 ≤ j ≤ n + 1 0\le j\le n+1 0 ≤ j ≤ n + 1 P ∈ P n + 1 P\in\mathcal P_{n+1} P ∈ P n + 1 §E20.12 命題 1.4 (1) と(2) により、P P P x n + 1 x^{n+1} x n + 1 ∑ j v j y j − ( − 1 ) n + 1 h ∑ j ∣ v j ∣ = 0 \sum_jv_jy_j-(-1)^{n+1}h\sum_j|v_j|=0 ∑ j v j y j − ( − 1 ) n + 1 h ∑ j ∣ v j ∣ = 0 P ∈ P n P\in\mathcal P_n P ∈ P n ( P , h ) (P,h) ( P , h )
(4) を示す。(1) により∑ j v j y j = ∑ j v j ( y j − q ( t j ) ) \sum_jv_jy_j=\sum_jv_j\bigl(y_j-q(t_j)\bigr) ∑ j v j y j = ∑ j v j ( y j − q ( t j ) ) (3) の式により
∣ h ∣ = ∣ ∑ j v j ( y j − q ( t j ) ) ∣ ∑ j ∣ v j ∣ ≤ ∑ j ∣ v j ∣ ∣ y j − q ( t j ) ∣ ∑ j ∣ v j ∣ |h|=\frac{\bigl|\sum_jv_j\bigl(y_j-q(t_j)\bigr)\bigr|}{\sum_j|v_j|}\le\frac{\sum_j|v_j|\,|y_j-q(t_j)|}{\sum_j|v_j|} ∣ h ∣ = ∑ j ∣ v j ∣ ∑ j v j ( y j − q ( t j ) ) ≤ ∑ j ∣ v j ∣ ∑ j ∣ v j ∣ ∣ y j − q ( t j ) ∣ である。右辺は∣ y j − q ( t j ) ∣ |y_j-q(t_j)| ∣ y j − q ( t j ) ∣ ∣ v j ∣ / ∑ k ∣ v k ∣ |v_j|/\sum_k|v_k| ∣ v j ∣/ ∑ k ∣ v k ∣ max j ∣ y j − q ( t j ) ∣ \max_j|y_j-q(t_j)| max j ∣ y j − q ( t j ) ∣ σ ( − 1 ) j ( y j − q ( t j ) ) ≥ 0 \sigma(-1)^j\bigl(y_j-q(t_j)\bigr)\ge0 σ ( − 1 ) j ( y j − q ( t j ) ) ≥ 0 (2) によりv j ( y j − q ( t j ) ) = ( − 1 ) n + 1 σ ∣ v j ∣ ∣ y j − q ( t j ) ∣ v_j\bigl(y_j-q(t_j)\bigr)=(-1)^{n+1}\sigma|v_j|\,|y_j-q(t_j)| v j ( y j − q ( t j ) ) = ( − 1 ) n + 1 σ ∣ v j ∣ ∣ y j − q ( t j ) ∣ ( − 1 ) n + 1 σ (-1)^{n+1}\sigma ( − 1 ) n + 1 σ 0 0 0 ▨
系 2.3 (de la Vallée Poussin の評価). a < b a<b a < b f ∈ C ( [ a , b ] ) f\in C([a,b]) f ∈ C ([ a , b ]) n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 E n ( f ) E_n(f) E n ( f ) [ a , b ] [a,b] [ a , b ] a ≤ t 0 < ⋯ < t n + 1 ≤ b a\le t_0<\dots<t_{n+1}\le b a ≤ t 0 < ⋯ < t n + 1 ≤ b 補題 2.2 (3) をy j : = f ( t j ) y_j:=f(t_j) y j := f ( t j ) ( p , h ) (p,h) ( p , h )
∣ h ∣ ≤ E n ( f ) |h|\le E_n(f) ∣ h ∣ ≤ E n ( f ) q ∈ P n q\in\mathcal P_n q ∈ P n σ ∈ { 1 , − 1 } \sigma\in\{1,-1\} σ ∈ { 1 , − 1 } j j j σ ( − 1 ) j ( f ( t j ) − q ( t j ) ) ≥ 0 \sigma(-1)^j\bigl(f(t_j)-q(t_j)\bigr)\ge0 σ ( − 1 ) j ( f ( t j ) − q ( t j ) ) ≥ 0 min 0 ≤ j ≤ n + 1 ∣ f ( t j ) − q ( t j ) ∣ ≤ ∣ h ∣ ≤ E n ( f ) \min_{0\le j\le n+1}|f(t_j)-q(t_j)|\le|h|\le E_n(f) min 0 ≤ j ≤ n + 1 ∣ f ( t j ) − q ( t j ) ∣ ≤ ∣ h ∣ ≤ E n ( f ) 証明. (1) を示す。q ∈ P n q\in\mathcal P_n q ∈ P n 補題 2.2 (4) により∣ h ∣ ≤ max j ∣ f ( t j ) − q ( t j ) ∣ ≤ ∥ f − q ∥ ∞ |h|\le\max_j|f(t_j)-q(t_j)|\le\lVert f-q\rVert_\infty ∣ h ∣ ≤ max j ∣ f ( t j ) − q ( t j ) ∣ ≤ ∥ f − q ∥ ∞ q q q ∣ h ∣ ≤ E n ( f ) |h|\le E_n(f) ∣ h ∣ ≤ E n ( f )
(2) を示す。符号の仮定により補題 2.2 (4) の最初の不等号は等号であり、∣ h ∣ |h| ∣ h ∣ ∣ f ( t j ) − q ( t j ) ∣ |f(t_j)-q(t_j)| ∣ f ( t j ) − q ( t j ) ∣ min j ∣ f ( t j ) − q ( t j ) ∣ \min_j|f(t_j)-q(t_j)| min j ∣ f ( t j ) − q ( t j ) ∣ (1) と合わせて主張を得る。▨
定理 2.4 (交代定理). a < b a<b a < b f ∈ C ( [ a , b ] ) f\in C([a,b]) f ∈ C ([ a , b ]) n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 p ∈ P n p\in\mathcal P_n p ∈ P n ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ a , b ] [a,b] [ a , b ] p p p f f f P n \mathcal P_n P n a ≤ t 0 < t 1 < ⋯ < t n + 1 ≤ b a\le t_0<t_1<\dots<t_{n+1}\le b a ≤ t 0 < t 1 < ⋯ < t n + 1 ≤ b t 0 , … , t n + 1 t_0,\dots,t_{n+1} t 0 , … , t n + 1 σ ∈ { 1 , − 1 } \sigma\in\{1,-1\} σ ∈ { 1 , − 1 }
f ( t j ) − p ( t j ) = σ ( − 1 ) j ∥ f − p ∥ ∞ ( 0 ≤ j ≤ n + 1 ) f(t_j)-p(t_j)=\sigma(-1)^j\lVert f-p\rVert_\infty\qquad(0\le j\le n+1) f ( t j ) − p ( t j ) = σ ( − 1 ) j ∥ f − p ∥ ∞ ( 0 ≤ j ≤ n + 1 ) が成り立つことである。
証明. 十分性を示す。δ : = ∥ f − p ∥ ∞ \delta:=\lVert f-p\rVert_\infty δ := ∥ f − p ∥ ∞ t 0 , … , t n + 1 t_0,\dots,t_{n+1} t 0 , … , t n + 1 σ \sigma σ 系 2.3 (2) をq = p q=p q = p min j ∣ f ( t j ) − p ( t j ) ∣ = δ \min_j|f(t_j)-p(t_j)|=\delta min j ∣ f ( t j ) − p ( t j ) ∣ = δ δ ≤ E n ( f ) \delta\le E_n(f) δ ≤ E n ( f ) E n ( f ) ≤ ∥ f − p ∥ ∞ = δ E_n(f)\le\lVert f-p\rVert_\infty=\delta E n ( f ) ≤ ∥ f − p ∥ ∞ = δ ∥ f − p ∥ ∞ = E n ( f ) \lVert f-p\rVert_\infty=E_n(f) ∥ f − p ∥ ∞ = E n ( f )
必要性を示す。p p p e : = f − p e:=f-p e := f − p δ : = ∥ e ∥ ∞ = E n ( f ) \delta:=\lVert e\rVert_\infty=E_n(f) δ := ∥ e ∥ ∞ = E n ( f ) δ = 0 \delta=0 δ = 0 e = 0 e=0 e = 0 a < b a<b a < b a ≤ t 0 < ⋯ < t n + 1 ≤ b a\le t_0<\dots<t_{n+1}\le b a ≤ t 0 < ⋯ < t n + 1 ≤ b σ = 1 \sigma=1 σ = 1 δ > 0 \delta>0 δ > 0 D 1 : = { x ∈ [ a , b ] ∣ e ( x ) = δ } D_1:=\{x\in[a,b]\mid e(x)=\delta\} D 1 := { x ∈ [ a , b ] ∣ e ( x ) = δ } D − 1 : = { x ∈ [ a , b ] ∣ e ( x ) = − δ } D_{-1}:=\{x\in[a,b]\mid e(x)=-\delta\} D − 1 := { x ∈ [ a , b ] ∣ e ( x ) = − δ } D : = D 1 ∪ D − 1 D:=D_1\cup D_{-1} D := D 1 ∪ D − 1 D 1 D_1 D 1 D − 1 D_{-1} D − 1 D D D ∣ e ∣ |e| ∣ e ∣ x ∈ D x\in D x ∈ D s ( x ) : = e ( x ) / δ ∈ { 1 , − 1 } s(x):=e(x)/\delta\in\{1,-1\} s ( x ) := e ( x ) / δ ∈ { 1 , − 1 }
t 0 : = min D t_0:=\min D t 0 := min D 0 ≤ i ≤ n 0\le i\le n 0 ≤ i ≤ n t 0 < ⋯ < t i t_0<\dots<t_i t 0 < ⋯ < t i F i : = D − s ( t i ) ∩ [ t i , b ] F_i:=D_{-s(t_i)}\cap[t_i,b] F i := D − s ( t i ) ∩ [ t i , b ] t i + 1 : = min F i t_{i+1}:=\min F_i t i + 1 := min F i F i F_i F i t n + 1 t_{n+1} t n + 1 t i ∉ D − s ( t i ) t_i\notin D_{-s(t_i)} t i ∈ / D − s ( t i ) t i + 1 > t i t_{i+1}>t_i t i + 1 > t i s ( t i + 1 ) = − s ( t i ) s(t_{i+1})=-s(t_i) s ( t i + 1 ) = − s ( t i ) L L L 1 ≤ L ≤ n + 2 1\le L\le n+2 1 ≤ L ≤ n + 2 L = n + 2 L=n+2 L = n + 2 σ : = s ( t 0 ) \sigma:=s(t_0) σ := s ( t 0 ) e ( t j ) = σ ( − 1 ) j δ e(t_j)=\sigma(-1)^j\delta e ( t j ) = σ ( − 1 ) j δ 0 ≤ j ≤ n + 1 0\le j\le n+1 0 ≤ j ≤ n + 1
L ≤ n + 1 L\le n+1 L ≤ n + 1 F L − 1 = ∅ F_{L-1}=\emptyset F L − 1 = ∅ 1 ≤ i ≤ L − 1 1\le i\le L-1 1 ≤ i ≤ L − 1 x ∈ D x\in D x ∈ D t i − 1 ≤ x < t i t_{i-1}\le x<t_i t i − 1 ≤ x < t i t i = min F i − 1 t_i=\min F_{i-1} t i = min F i − 1 x ∉ F i − 1 x\notin F_{i-1} x ∈ / F i − 1 s ( x ) = s ( t i − 1 ) s(x)=s(t_{i-1}) s ( x ) = s ( t i − 1 ) x ∈ D x\in D x ∈ D x ≥ t L − 1 x\ge t_{L-1} x ≥ t L − 1 F L − 1 = ∅ F_{L-1}=\emptyset F L − 1 = ∅ s ( x ) = s ( t L − 1 ) s(x)=s(t_{L-1}) s ( x ) = s ( t L − 1 ) D ⊆ [ t 0 , b ] D\subseteq[t_0,b] D ⊆ [ t 0 , b ] 1 ≤ i ≤ L − 1 1\le i\le L-1 1 ≤ i ≤ L − 1 W i : = D s ( t i − 1 ) ∩ [ t i − 1 , t i ] W_i:=D_{s(t_{i-1})}\cap[t_{i-1},t_i] W i := D s ( t i − 1 ) ∩ [ t i − 1 , t i ] t i − 1 t_{i-1} t i − 1 t i ∉ W i t_i\notin W_i t i ∈ / W i w i : = max W i w_i:=\max W_i w i := max W i w i < t i w_i<t_i w i < t i w i < x < t i w_i<x<t_i w i < x < t i x ∈ D x\in D x ∈ D s ( x ) = s ( t i − 1 ) s(x)=s(t_{i-1}) s ( x ) = s ( t i − 1 ) x ∈ W i x\in W_i x ∈ W i x ≤ w i x\le w_i x ≤ w i x x x z i : = ( w i + t i ) / 2 z_i:=(w_i+t_i)/2 z i := ( w i + t i ) /2 t i − 1 ≤ w i < z i < t i t_{i-1}\le w_i<z_i<t_i t i − 1 ≤ w i < z i < t i
r ( x ) : = s ( t 0 ) ∏ i = 1 L − 1 ( z i − x ) r(x):=s(t_0)\prod_{i=1}^{L-1}(z_i-x) r ( x ) := s ( t 0 ) i = 1 ∏ L − 1 ( z i − x ) と置くと、r r r 0 0 0 L − 1 ≤ n L-1\le n L − 1 ≤ n r ∈ P n r\in\mathcal P_n r ∈ P n x ∈ D x\in D x ∈ D 1 ≤ i ≤ L − 1 1\le i\le L-1 1 ≤ i ≤ L − 1 t i − 1 ≤ x < t i t_{i-1}\le x<t_i t i − 1 ≤ x < t i x ∈ W i x\in W_i x ∈ W i x ≤ w i x\le w_i x ≤ w i j < i j<i j < i z j < t j ≤ t i − 1 ≤ x z_j<t_j\le t_{i-1}\le x z j < t j ≤ t i − 1 ≤ x j = i j=i j = i z i > w i ≥ x z_i>w_i\ge x z i > w i ≥ x j > i j>i j > i z j > t j − 1 ≥ t i > x z_j>t_{j-1}\ge t_i>x z j > t j − 1 ≥ t i > x r ( x ) r(x) r ( x ) s ( t 0 ) ( − 1 ) i − 1 = s ( t i − 1 ) = s ( x ) s(t_0)(-1)^{i-1}=s(t_{i-1})=s(x) s ( t 0 ) ( − 1 ) i − 1 = s ( t i − 1 ) = s ( x ) x ≥ t L − 1 x\ge t_{L-1} x ≥ t L − 1 j j j z j < t j ≤ t L − 1 ≤ x z_j<t_j\le t_{L-1}\le x z j < t j ≤ t L − 1 ≤ x r ( x ) r(x) r ( x ) s ( t 0 ) ( − 1 ) L − 1 = s ( t L − 1 ) = s ( x ) s(t_0)(-1)^{L-1}=s(t_{L-1})=s(x) s ( t 0 ) ( − 1 ) L − 1 = s ( t L − 1 ) = s ( x ) x ∈ D x\in D x ∈ D e ( x ) r ( x ) = δ s ( x ) r ( x ) > 0 e(x)r(x)=\delta s(x)r(x)>0 e ( x ) r ( x ) = δ s ( x ) r ( x ) > 0
K : = { x ∈ [ a , b ] ∣ e ( x ) r ( x ) ≤ 0 } K:=\{x\in[a,b]\mid e(x)r(x)\le0\} K := { x ∈ [ a , b ] ∣ e ( x ) r ( x ) ≤ 0 } [ a , b ] [a,b] [ a , b ] D D D K ≠ ∅ K\ne\emptyset K = ∅ δ ′ : = max x ∈ K ∣ e ( x ) ∣ \delta':=\max_{x\in K}|e(x)| δ ′ := max x ∈ K ∣ e ( x ) ∣ δ ′ < δ \delta'<\delta δ ′ < δ K = ∅ K=\emptyset K = ∅ δ ′ : = 0 \delta':=0 δ ′ := 0 λ : = ( δ − δ ′ ) / ( 2 ∥ r ∥ ∞ ) \lambda:=(\delta-\delta')/(2\lVert r\rVert_\infty) λ := ( δ − δ ′ ) / ( 2 ∥ r ∥ ∞ ) λ > 0 \lambda>0 λ > 0 x ∈ K x\in K x ∈ K ∣ e ( x ) − λ r ( x ) ∣ ≤ δ ′ + ( δ − δ ′ ) / 2 < δ |e(x)-\lambda r(x)|\le\delta'+(\delta-\delta')/2<\delta ∣ e ( x ) − λ r ( x ) ∣ ≤ δ ′ + ( δ − δ ′ ) /2 < δ x ∉ K x\notin K x ∈ / K e ( x ) e(x) e ( x ) λ r ( x ) \lambda r(x) λ r ( x ) 0 0 0 0 < λ ∣ r ( x ) ∣ ≤ ( δ − δ ′ ) / 2 < δ 0<\lambda|r(x)|\le(\delta-\delta')/2<\delta 0 < λ ∣ r ( x ) ∣ ≤ ( δ − δ ′ ) /2 < δ ∣ e ( x ) − λ r ( x ) ∣ = ∣ ∣ e ( x ) ∣ − λ ∣ r ( x ) ∣ ∣ |e(x)-\lambda r(x)|=\bigl||e(x)|-\lambda|r(x)|\bigr| ∣ e ( x ) − λ r ( x ) ∣ = ∣ e ( x ) ∣ − λ ∣ r ( x ) ∣ ∣ e ( x ) ∣ − λ ∣ r ( x ) ∣ |e(x)|-\lambda|r(x)| ∣ e ( x ) ∣ − λ ∣ r ( x ) ∣ ∣ e ( x ) ∣ ≤ δ |e(x)|\le\delta ∣ e ( x ) ∣ ≤ δ − λ ∣ r ( x ) ∣ > − δ -\lambda|r(x)|>-\delta − λ ∣ r ( x ) ∣ > − δ ∣ e − λ r ∣ |e-\lambda r| ∣ e − λ r ∣ [ a , b ] [a,b] [ a , b ] ∥ f − ( p + λ r ) ∥ ∞ = ∥ e − λ r ∥ ∞ < δ \lVert f-(p+\lambda r)\rVert_\infty=\lVert e-\lambda r\rVert_\infty<\delta ∥ f − ( p + λ r ) ∥ ∞ = ∥ e − λ r ∥ ∞ < δ p + λ r ∈ P n p+\lambda r\in\mathcal P_n p + λ r ∈ P n E n ( f ) < δ E_n(f)<\delta E n ( f ) < δ δ = E n ( f ) \delta=E_n(f) δ = E n ( f ) L = n + 2 L=n+2 L = n + 2 ▨
証明. ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ a , b ] [a,b] [ a , b ] ∣ c ∣ |c| ∣ c ∣ c ∈ R n + 1 c\in\R^{n+1} c ∈ R n + 1 c = ( c 0 , … , c n ) ∈ R n + 1 c=(c_0,\dots,c_n)\in\R^{n+1} c = ( c 0 , … , c n ) ∈ R n + 1 Φ ( c ) : = ∑ j = 0 n c j x j ∈ P n \Phi(c):=\sum_{j=0}^nc_jx^j\in\mathcal P_n Φ ( c ) := ∑ j = 0 n c j x j ∈ P n N ( c ) : = ∥ Φ ( c ) ∥ ∞ N(c):=\lVert\Phi(c)\rVert_\infty N ( c ) := ∥ Φ ( c ) ∥ ∞ G ( c ) : = ∥ f − Φ ( c ) ∥ ∞ G(c):=\lVert f-\Phi(c)\rVert_\infty G ( c ) := ∥ f − Φ ( c ) ∥ ∞ K 0 : = max ( ∣ a ∣ , ∣ b ∣ ) K_0:=\max(|a|,|b|) K 0 := max ( ∣ a ∣ , ∣ b ∣ ) c , c ′ ∈ R n + 1 c,c'\in\R^{n+1} c , c ′ ∈ R n + 1 ∣ N ( c ) − N ( c ′ ) ∣ ≤ N ( c − c ′ ) ≤ ∑ j ∣ c j − c j ′ ∣ K 0 j |N(c)-N(c')|\le N(c-c')\le\sum_j|c_j-c'_j|K_0^j ∣ N ( c ) − N ( c ′ ) ∣ ≤ N ( c − c ′ ) ≤ ∑ j ∣ c j − c j ′ ∣ K 0 j ∣ G ( c ) − G ( c ′ ) ∣ ≤ N ( c − c ′ ) |G(c)-G(c')|\le N(c-c') ∣ G ( c ) − G ( c ′ ) ∣ ≤ N ( c − c ′ ) N N N G G G Φ ( c ) \Phi(c) Φ ( c ) [ a , b ] [a,b] [ a , b ] 0 0 0 Φ ( c ) \Phi(c) Φ ( c ) [ a , b ] [a,b] [ a , b ] n + 1 n+1 n + 1 0 0 0 §E20.12 補題 1.2 (2) によりc = 0 c=0 c = 0 N N N { c ∣ ∣ c ∣ = 1 } \{c\mid|c|=1\} { c ∣ ∣ c ∣ = 1 } m m m N ( c ) = ∣ c ∣ N ( c / ∣ c ∣ ) N(c)=|c|\,N(c/|c|) N ( c ) = ∣ c ∣ N ( c /∣ c ∣ ) c ≠ 0 c\ne0 c = 0 c c c N ( c ) ≥ m ∣ c ∣ N(c)\ge m|c| N ( c ) ≥ m ∣ c ∣ B : = { c ∣ ∣ c ∣ ≤ 2 ∥ f ∥ ∞ / m } B:=\{c\mid|c|\le2\lVert f\rVert_\infty/m\} B := { c ∣ ∣ c ∣ ≤ 2 ∥ f ∥ ∞ / m } G G G B B B c ∗ ∈ B c^*\in B c ∗ ∈ B c ∉ B c\notin B c ∈ / B G ( c ) ≥ N ( c ) − ∥ f ∥ ∞ > ∥ f ∥ ∞ = G ( 0 ) ≥ G ( c ∗ ) G(c)\ge N(c)-\lVert f\rVert_\infty>\lVert f\rVert_\infty=G(0)\ge G(c^*) G ( c ) ≥ N ( c ) − ∥ f ∥ ∞ > ∥ f ∥ ∞ = G ( 0 ) ≥ G ( c ∗ ) Φ \Phi Φ ∥ f − Φ ( c ∗ ) ∥ ∞ = min c ∈ R n + 1 G ( c ) = E n ( f ) \lVert f-\Phi(c^*)\rVert_\infty=\min_{c\in\R^{n+1}}G(c)=E_n(f) ∥ f − Φ ( c ∗ ) ∥ ∞ = min c ∈ R n + 1 G ( c ) = E n ( f ) Φ ( c ∗ ) \Phi(c^*) Φ ( c ∗ )
p 1 , p 2 p_1,p_2 p 1 , p 2 p : = ( p 1 + p 2 ) / 2 p:=(p_1+p_2)/2 p := ( p 1 + p 2 ) /2 ∥ f − p ∥ ∞ ≤ ( ∥ f − p 1 ∥ ∞ + ∥ f − p 2 ∥ ∞ ) / 2 = E n ( f ) \lVert f-p\rVert_\infty\le(\lVert f-p_1\rVert_\infty+\lVert f-p_2\rVert_\infty)/2=E_n(f) ∥ f − p ∥ ∞ ≤ (∥ f − p 1 ∥ ∞ + ∥ f − p 2 ∥ ∞ ) /2 = E n ( f ) p p p 定理 2.4 により、a ≤ t 0 < ⋯ < t n + 1 ≤ b a\le t_0<\dots<t_{n+1}\le b a ≤ t 0 < ⋯ < t n + 1 ≤ b σ ∈ { 1 , − 1 } \sigma\in\{1,-1\} σ ∈ { 1 , − 1 } f ( t j ) − p ( t j ) = σ ( − 1 ) j E n ( f ) f(t_j)-p(t_j)=\sigma(-1)^jE_n(f) f ( t j ) − p ( t j ) = σ ( − 1 ) j E n ( f ) f ( t j ) − p ( t j ) f(t_j)-p(t_j) f ( t j ) − p ( t j ) E n ( f ) E_n(f) E n ( f ) f ( t j ) − p 1 ( t j ) f(t_j)-p_1(t_j) f ( t j ) − p 1 ( t j ) f ( t j ) − p 2 ( t j ) f(t_j)-p_2(t_j) f ( t j ) − p 2 ( t j ) σ ( − 1 ) j E n ( f ) \sigma(-1)^jE_n(f) σ ( − 1 ) j E n ( f ) p 1 ( t j ) = p 2 ( t j ) p_1(t_j)=p_2(t_j) p 1 ( t j ) = p 2 ( t j ) p 1 − p 2 ∈ P n p_1-p_2\in\mathcal P_n p 1 − p 2 ∈ P n n + 2 n+2 n + 2 0 0 0 §E20.12 補題 1.2 (2) によりp 1 = p 2 p_1=p_2 p 1 = p 2 ▨
例 2.6. f ( x ) : = x 2 f(x):=x^2 f ( x ) := x 2 C ( [ − 1 , 1 ] ) C([-1,1]) C ([ − 1 , 1 ]) ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f − 1 / 2 f-1/2 f − 1/2 ∥ f − 1 / 2 ∥ ∞ = 1 / 2 \lVert f-1/2\rVert_\infty=1/2 ∥ f − 1/2 ∥ ∞ = 1/2 0 0 0 − 1 / 2 -1/2 − 1/2 1 1 1 1 / 2 1/2 1/2 定理 2.4 をn = 0 n=0 n = 0 t 0 = 0 t_0=0 t 0 = 0 t 1 = 1 t_1=1 t 1 = 1 σ = − 1 \sigma=-1 σ = − 1 1 / 2 1/2 1/2 f f f P 0 \mathcal P_0 P 0 E 0 ( f ) = 1 / 2 E_0(f)=1/2 E 0 ( f ) = 1/2 w P : = 1 w_P:=1 w P := 1 §E20.14 定理 1.6 (5) を基底1 1 1 Π 0 w P f = 1 2 ∫ − 1 1 x 2 d x = 1 3 \Pi_0^{w_P}f=\frac12\int_{-1}^1x^2\,dx=\frac13 Π 0 w P f = 2 1 ∫ − 1 1 x 2 d x = 3 1 ∥ f − 1 / 3 ∥ ∞ = ∣ f ( 1 ) − 1 / 3 ∣ = 2 / 3 \lVert f-1/3\rVert_\infty=|f(1)-1/3|=2/3 ∥ f − 1/3 ∥ ∞ = ∣ f ( 1 ) − 1/3∣ = 2/3 E 0 ( f ) = 1 / 2 E_0(f)=1/2 E 0 ( f ) = 1/2 a 0 ( f ) = 2 π ∫ 0 π cos 2 θ d θ = 1 a_0(f)=\frac2\pi\int_0^\pi\cos^2\theta\,d\theta=1 a 0 ( f ) = π 2 ∫ 0 π cos 2 θ d θ = 1 a 1 ( f ) = 2 π ∫ 0 π cos 3 θ d θ = 0 a_1(f)=\frac2\pi\int_0^\pi\cos^3\theta\,d\theta=0 a 1 ( f ) = π 2 ∫ 0 π cos 3 θ d θ = 0 S 0 f = S 1 f = 1 / 2 S_0f=S_1f=1/2 S 0 f = S 1 f = 1/2 f − 1 / 2 = T 2 / 2 f-1/2=T_2/2 f − 1/2 = T 2 /2 − 1 , 0 , 1 -1,0,1 − 1 , 0 , 1 1 / 2 , − 1 / 2 , 1 / 2 1/2,-1/2,1/2 1/2 , − 1/2 , 1/2 定理 2.4 をn = 1 n=1 n = 1 S 1 f = 1 / 2 S_1f=1/2 S 1 f = 1/2 P 1 \mathcal P_1 P 1 E 1 ( f ) = 1 / 2 E_1(f)=1/2 E 1 ( f ) = 1/2
3 最良近似誤差の収束 証明. (1) を示す。P n ⊆ P n + 1 \mathcal P_n\subseteq\mathcal P_{n+1} P n ⊆ P n + 1 E n + 1 ( f ) ≤ E n ( f ) E_{n+1}(f)\le E_n(f) E n + 1 ( f ) ≤ E n ( f ) ε > 0 \varepsilon>0 ε > 0 §E2.32 問題 5.2 により、実係数多項式q q q ∥ f − q ∥ ∞ < ε \lVert f-q\rVert_\infty<\varepsilon ∥ f − q ∥ ∞ < ε q q q N N N q = 0 q=0 q = 0 N : = 0 N:=0 N := 0 n ≥ N n\ge N n ≥ N q ∈ P n q\in\mathcal P_n q ∈ P n E n ( f ) ≤ ∥ f − q ∥ ∞ < ε E_n(f)\le\lVert f-q\rVert_\infty<\varepsilon E n ( f ) ≤ ∥ f − q ∥ ∞ < ε
(2) を示す。§E20.12 定理 3.2 (4) により∥ f − I n f ∥ ∞ ≤ E n ( f ) + Λ n E n ( f ) \lVert f-I_nf\rVert_\infty\le E_n(f)+\Lambda_nE_n(f) ∥ f − I n f ∥ ∞ ≤ E n ( f ) + Λ n E n ( f ) (1) と仮定により右辺は0 0 0 ▨
4 原点で値を固定した最小最大問題 定理 4.1. 0 < a < b 0<a<b 0 < a < b k ∈ N ≥ 1 k\in\NN k ∈ N ≥ 1 σ : = ( b + a ) / ( b − a ) \sigma:=(b+a)/(b-a) σ := ( b + a ) / ( b − a ) z : = ( b − a ) / ( b + a ) z:=(\sqrt b-\sqrt a)/(\sqrt b+\sqrt a) z := ( b − a ) / ( b + a ) T k ( σ ) > 0 T_k(\sigma)>0 T k ( σ ) > 0
p ∗ ( x ) : = 1 T k ( σ ) T k ( b + a − 2 x b − a ) p_*(x):=\frac1{T_k(\sigma)}\,T_k\Bigl(\frac{b+a-2x}{b-a}\Bigr) p ∗ ( x ) := T k ( σ ) 1 T k ( b − a b + a − 2 x ) はp ∗ ∈ P k p_*\in\mathcal P_k p ∗ ∈ P k p ∗ ( 0 ) = 1 p_*(0)=1 p ∗ ( 0 ) = 1 max x ∈ [ a , b ] ∣ p ∗ ( x ) ∣ = 1 / T k ( σ ) \max_{x\in[a,b]}|p_*(x)|=1/T_k(\sigma) max x ∈ [ a , b ] ∣ p ∗ ( x ) ∣ = 1/ T k ( σ )
p ∈ P k p\in\mathcal P_k p ∈ P k p ( 0 ) = 1 p(0)=1 p ( 0 ) = 1 max x ∈ [ a , b ] ∣ p ( x ) ∣ ≥ 1 / T k ( σ ) \max_{x\in[a,b]}|p(x)|\ge1/T_k(\sigma) max x ∈ [ a , b ] ∣ p ( x ) ∣ ≥ 1/ T k ( σ ) p = p ∗ p=p_* p = p ∗ 1 / T k ( σ ) ≤ 2 z k 1/T_k(\sigma)\le2z^k 1/ T k ( σ ) ≤ 2 z k 証明. b + a > b − a > 0 b+a>b-a>0 b + a > b − a > 0 σ > 1 \sigma>1 σ > 1 §E20.11 補題 3.2 (4) によりT k ( σ ) = ( ( σ + σ 2 − 1 ) k + ( σ − σ 2 − 1 ) k ) / 2 T_k(\sigma)=\bigl((\sigma+\sqrt{\sigma^2-1})^k+(\sigma-\sqrt{\sigma^2-1})^k\bigr)/2 T k ( σ ) = ( ( σ + σ 2 − 1 ) k + ( σ − σ 2 − 1 ) k ) /2 σ > σ 2 − 1 ≥ 0 \sigma>\sqrt{\sigma^2-1}\ge0 σ > σ 2 − 1 ≥ 0 T k ( σ ) > 0 T_k(\sigma)>0 T k ( σ ) > 0 φ ( x ) : = ( b + a − 2 x ) / ( b − a ) \varphi(x):=(b+a-2x)/(b-a) φ ( x ) := ( b + a − 2 x ) / ( b − a ) φ \varphi φ [ a , b ] [a,b] [ a , b ] [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] φ ( 0 ) = σ \varphi(0)=\sigma φ ( 0 ) = σ §E20.11 補題 3.2 (1) によりp ∗ = T k ∘ φ / T k ( σ ) ∈ P k p_*=T_k\circ\varphi/T_k(\sigma)\in\mathcal P_k p ∗ = T k ∘ φ / T k ( σ ) ∈ P k p ∗ ( 0 ) = 1 p_*(0)=1 p ∗ ( 0 ) = 1 0 ≤ j ≤ k 0\le j\le k 0 ≤ j ≤ k x j : = φ − 1 ( cos ( j π / k ) ) x_j:=\varphi^{-1}(\cos(j\pi/k)) x j := φ − 1 ( cos ( j π / k )) a = x 0 < x 1 < ⋯ < x k = b a=x_0<x_1<\dots<x_k=b a = x 0 < x 1 < ⋯ < x k = b §E20.12 命題 5.2 (2) により[ a , b ] [a,b] [ a , b ] ∣ p ∗ ∣ ≤ 1 / T k ( σ ) |p_*|\le1/T_k(\sigma) ∣ p ∗ ∣ ≤ 1/ T k ( σ ) p ∗ ( x j ) = ( − 1 ) j / T k ( σ ) p_*(x_j)=(-1)^j/T_k(\sigma) p ∗ ( x j ) = ( − 1 ) j / T k ( σ ) max [ a , b ] ∣ p ∗ ∣ = 1 / T k ( σ ) \max_{[a,b]}|p_*|=1/T_k(\sigma) max [ a , b ] ∣ p ∗ ∣ = 1/ T k ( σ )
(1) を示す。0 < a 0<a 0 < a 0 < x 0 < ⋯ < x k 0<x_0<\dots<x_k 0 < x 0 < ⋯ < x k 補題 2.2 をn = k n=k n = k t 0 : = 0 t_0:=0 t 0 := 0 t j + 1 : = x j t_{j+1}:=x_j t j + 1 := x j 0 ≤ j ≤ k 0\le j\le k 0 ≤ j ≤ k v 0 , … , v k + 1 v_0,\dots,v_{k+1} v 0 , … , v k + 1 補題 2.2 (2) によりv j + 1 ( − 1 ) j = ( − 1 ) k ∣ v j + 1 ∣ v_{j+1}(-1)^j=(-1)^k|v_{j+1}| v j + 1 ( − 1 ) j = ( − 1 ) k ∣ v j + 1 ∣ S : = ∑ j = 0 k ∣ v j + 1 ∣ > 0 S:=\sum_{j=0}^k|v_{j+1}|>0 S := ∑ j = 0 k ∣ v j + 1 ∣ > 0 p ∈ P k p\in\mathcal P_k p ∈ P k p ( 0 ) = 1 p(0)=1 p ( 0 ) = 1 補題 2.2 (1) により
v 0 = − ∑ j = 0 k v j + 1 p ( x j ) v_0=-\sum_{j=0}^kv_{j+1}p(x_j) v 0 = − j = 0 ∑ k v j + 1 p ( x j ) である。p = p ∗ p=p_* p = p ∗ − ( − 1 ) k S / T k ( σ ) -(-1)^kS/T_k(\sigma) − ( − 1 ) k S / T k ( σ ) ∣ v 0 ∣ = S / T k ( σ ) |v_0|=S/T_k(\sigma) ∣ v 0 ∣ = S / T k ( σ ) p p p ∣ v 0 ∣ ≤ S max j ∣ p ( x j ) ∣ ≤ S max [ a , b ] ∣ p ∣ |v_0|\le S\max_j|p(x_j)|\le S\max_{[a,b]}|p| ∣ v 0 ∣ ≤ S max j ∣ p ( x j ) ∣ ≤ S max [ a , b ] ∣ p ∣ max [ a , b ] ∣ p ∣ ≥ 1 / T k ( σ ) \max_{[a,b]}|p|\ge1/T_k(\sigma) max [ a , b ] ∣ p ∣ ≥ 1/ T k ( σ ) max [ a , b ] ∣ p ∣ = 1 / T k ( σ ) \max_{[a,b]}|p|=1/T_k(\sigma) max [ a , b ] ∣ p ∣ = 1/ T k ( σ ) d : = p ∗ − p d:=p_*-p d := p ∗ − p d ∈ P k d\in\mathcal P_k d ∈ P k d ( 0 ) = 0 d(0)=0 d ( 0 ) = 0 ( − 1 ) j d ( x j ) = 1 / T k ( σ ) − ( − 1 ) j p ( x j ) ≥ 0 (-1)^jd(x_j)=1/T_k(\sigma)-(-1)^jp(x_j)\ge0 ( − 1 ) j d ( x j ) = 1/ T k ( σ ) − ( − 1 ) j p ( x j ) ≥ 0 補題 2.2 (1) をd d d
0 = v 0 d ( 0 ) + ∑ j = 0 k v j + 1 d ( x j ) = ( − 1 ) k ∑ j = 0 k ∣ v j + 1 ∣ ( − 1 ) j d ( x j ) 0=v_0d(0)+\sum_{j=0}^kv_{j+1}d(x_j)=(-1)^k\sum_{j=0}^k|v_{j+1}|\,(-1)^jd(x_j) 0 = v 0 d ( 0 ) + j = 0 ∑ k v j + 1 d ( x j ) = ( − 1 ) k j = 0 ∑ k ∣ v j + 1 ∣ ( − 1 ) j d ( x j ) であり、各項は0 0 0 j j j d ( x j ) = 0 d(x_j)=0 d ( x j ) = 0 d ∈ P k d\in\mathcal P_k d ∈ P k k + 2 k+2 k + 2 0 , x 0 , … , x k 0,x_0,\dots,x_k 0 , x 0 , … , x k 0 0 0 §E20.12 補題 1.2 (2) によりd = 0 d=0 d = 0 p = p ∗ p=p_* p = p ∗
(2) を示す。§E20.11 補題 4.1 をλ − = a \lambda_-=a λ − = a λ + = b \lambda_+=b λ + = b q ∈ P k q\in\mathcal P_k q ∈ P k q ( 0 ) = 1 q(0)=1 q ( 0 ) = 1 max [ a , b ] ∣ q ∣ ≤ 2 z k \max_{[a,b]}|q|\le2z^k max [ a , b ] ∣ q ∣ ≤ 2 z k (1) により1 / T k ( σ ) ≤ max [ a , b ] ∣ q ∣ ≤ 2 z k 1/T_k(\sigma)\le\max_{[a,b]}|q|\le2z^k 1/ T k ( σ ) ≤ max [ a , b ] ∣ q ∣ ≤ 2 z k ▨
5 Remez 型の更新 定義 5.1. a < b a<b a < b f ∈ C ( [ a , b ] ) f\in C([a,b]) f ∈ C ([ a , b ]) n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ a , b ] [a,b] [ a , b ]
a ≤ t 0 < t 1 < ⋯ < t n + 1 ≤ b a\le t_0<t_1<\dots<t_{n+1}\le b a ≤ t 0 < t 1 < ⋯ < t n + 1 ≤ b t = ( t 0 , … , t n + 1 ) t=(t_0,\dots,t_{n+1}) t = ( t 0 , … , t n + 1 ) [ a , b ] [a,b] [ a , b ] n + 2 n+2 n + 2 参照点列 (reference )補題 2.2 (3) をy j : = f ( t j ) y_j:=f(t_j) y j := f ( t j ) ( p , h ) (p,h) ( p , h ) t t t f f f 水準付き近似 (levelled approximation )h h h 水準 (level )t t t ( p , h ) (p,h) ( p , h ) t t t f f f h ≠ 0 h\ne0 h = 0 e : = f − p e:=f-p e := f − p s ∈ [ a , b ] s\in[a,b] s ∈ [ a , b ] ∣ e ( s ) ∣ = ∥ e ∥ ∞ > ∣ h ∣ |e(s)|=\lVert e\rVert_\infty>|h| ∣ e ( s ) ∣ = ∥ e ∥ ∞ > ∣ h ∣ ∣ e ( t j ) ∣ = ∣ h ∣ < ∣ e ( s ) ∣ |e(t_j)|=|h|<|e(s)| ∣ e ( t j ) ∣ = ∣ h ∣ < ∣ e ( s ) ∣ s s s t 0 , … , t n + 1 t_0,\dots,t_{n+1} t 0 , … , t n + 1 t ′ t' t ′ 0 ≤ i ≤ n 0\le i\le n 0 ≤ i ≤ n t i < s < t i + 1 t_i<s<t_{i+1} t i < s < t i + 1 e ( s ) e ( t i ) > 0 e(s)e(t_i)>0 e ( s ) e ( t i ) > 0 t i t_i t i s s s e ( s ) e ( t i ) < 0 e(s)e(t_i)<0 e ( s ) e ( t i ) < 0 t i + 1 t_{i+1} t i + 1 s s s s < t 0 s<t_0 s < t 0 e ( s ) e ( t 0 ) > 0 e(s)e(t_0)>0 e ( s ) e ( t 0 ) > 0 t 0 t_0 t 0 s s s e ( s ) e ( t 0 ) < 0 e(s)e(t_0)<0 e ( s ) e ( t 0 ) < 0 t ′ : = ( s , t 0 , … , t n ) t':=(s,t_0,\dots,t_n) t ′ := ( s , t 0 , … , t n ) s > t n + 1 s>t_{n+1} s > t n + 1 e ( s ) e ( t n + 1 ) > 0 e(s)e(t_{n+1})>0 e ( s ) e ( t n + 1 ) > 0 t n + 1 t_{n+1} t n + 1 s s s e ( s ) e ( t n + 1 ) < 0 e(s)e(t_{n+1})<0 e ( s ) e ( t n + 1 ) < 0 t ′ : = ( t 1 , … , t n + 1 , s ) t':=(t_1,\dots,t_{n+1},s) t ′ := ( t 1 , … , t n + 1 , s ) t ′ t' t ′ t ′ t' t ′ f f f ( p ′ , h ′ ) (p',h') ( p ′ , h ′ ) Remez 型の更新 (Remez exchange step ) 例 5.2. f ( x ) : = x 3 f(x):=x^3 f ( x ) := x 3 C ( [ − 1 , 1 ] ) C([-1,1]) C ([ − 1 , 1 ]) n = 1 n=1 n = 1 t = ( − 1 , 0 , 1 ) t=(-1,0,1) t = ( − 1 , 0 , 1 ) ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] p ( x ) = c 0 + c 1 x p(x)=c_0+c_1x p ( x ) = c 0 + c 1 x h h h t j 3 − c 0 − c 1 t j = ( − 1 ) j h t_j^3-c_0-c_1t_j=(-1)^jh t j 3 − c 0 − c 1 t j = ( − 1 ) j h j = 0 , 1 , 2 j=0,1,2 j = 0 , 1 , 2 j = 1 j=1 j = 1 c 0 = h c_0=h c 0 = h j = 0 j=0 j = 0 j = 2 j=2 j = 2 − 2 c 0 = 2 h -2c_0=2h − 2 c 0 = 2 h c 0 = h = 0 c_0=h=0 c 0 = h = 0 j = 0 j=0 j = 0 j = 2 j=2 j = 2 c 1 = 1 c_1=1 c 1 = 1 p ( x ) = x p(x)=x p ( x ) = x h = 0 h=0 h = 0 定義 5.1 の Remez 型の更新は定義されない。
g ( x ) : = x 3 − x g(x):=x^3-x g ( x ) := x 3 − x g ′ ( x ) = 3 x 2 − 1 g'(x)=3x^2-1 g ′ ( x ) = 3 x 2 − 1 ± 1 / 3 \pm1/\sqrt3 ± 1/ 3 0 0 0 g ( ± 1 ) = 0 g(\pm1)=0 g ( ± 1 ) = 0 g ( ± 1 / 3 ) = ∓ 2 / ( 3 3 ) g(\pm1/\sqrt3)=\mp2/(3\sqrt3) g ( ± 1/ 3 ) = ∓ 2/ ( 3 3 ) ∥ f − p ∥ ∞ = 2 / ( 3 3 ) ≈ 0.3849 \lVert f-p\rVert_\infty=2/(3\sqrt3)\approx0.3849 ∥ f − p ∥ ∞ = 2/ ( 3 3 ) ≈ 0.3849 r ( x ) : = x 3 − 3 4 x r(x):=x^3-\tfrac34x r ( x ) := x 3 − 4 3 x r ′ ( x ) = 3 x 2 − 3 4 r'(x)=3x^2-\tfrac34 r ′ ( x ) = 3 x 2 − 4 3 ± 1 / 2 \pm1/2 ± 1/2 0 0 0 r ( ± 1 ) = ± 1 / 4 r(\pm1)=\pm1/4 r ( ± 1 ) = ± 1/4 r ( ± 1 / 2 ) = ∓ 1 / 4 r(\pm1/2)=\mp1/4 r ( ± 1/2 ) = ∓ 1/4 ∥ r ∥ ∞ = 1 / 4 \lVert r\rVert_\infty=1/4 ∥ r ∥ ∞ = 1/4 r r r − 1 , − 1 / 2 , 1 / 2 -1,-1/2,1/2 − 1 , − 1/2 , 1/2 − 1 / 4 , 1 / 4 , − 1 / 4 -1/4,1/4,-1/4 − 1/4 , 1/4 , − 1/4 定理 2.4 をt 0 = − 1 t_0=-1 t 0 = − 1 t 1 = − 1 / 2 t_1=-1/2 t 1 = − 1/2 t 2 = 1 / 2 t_2=1/2 t 2 = 1/2 σ = − 1 \sigma=-1 σ = − 1 3 4 x \tfrac34x 4 3 x f f f P 1 \mathcal P_1 P 1 E 1 ( f ) = 1 / 4 E_1(f)=1/4 E 1 ( f ) = 1/4 ( 2 / ( 3 3 ) ) 2 = 4 / 27 > 1 / 16 = ( 1 / 4 ) 2 \bigl(2/(3\sqrt3)\bigr)^2=4/27>1/16=(1/4)^2 ( 2/ ( 3 3 ) ) 2 = 4/27 > 1/16 = ( 1/4 ) 2 ∥ f − p ∥ ∞ > E 1 ( f ) \lVert f-p\rVert_\infty>E_1(f) ∥ f − p ∥ ∞ > E 1 ( f ) p p p f f f
命題 5.3. a < b a<b a < b f ∈ C ( [ a , b ] ) f\in C([a,b]) f ∈ C ([ a , b ]) n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ E n ( f ) E_n(f) E n ( f ) [ a , b ] [a,b] [ a , b ] t t t [ a , b ] [a,b] [ a , b ] n + 2 n+2 n + 2 ( p , h ) (p,h) ( p , h ) t t t f f f
∣ h ∣ ≤ E n ( f ) ≤ ∥ f − p ∥ ∞ |h|\le E_n(f)\le\lVert f-p\rVert_\infty ∣ h ∣ ≤ E n ( f ) ≤ ∥ f − p ∥ ∞ ∥ f − p ∥ ∞ = ∣ h ∣ \lVert f-p\rVert_\infty=|h| ∥ f − p ∥ ∞ = ∣ h ∣ p p p f f f P n \mathcal P_n P n E n ( f ) = ∣ h ∣ E_n(f)=|h| E n ( f ) = ∣ h ∣ h ≠ 0 h\ne0 h = 0 s s s t ′ t' t ′ ( p ′ , h ′ ) (p',h') ( p ′ , h ′ ) 定義 5.1 の Remez 型の更新で得るものとする。このときt ′ t' t ′ s s s [ a , b ] [a,b] [ a , b ] n + 2 n+2 n + 2 ∣ h ∣ < ∣ h ′ ∣ ≤ E n ( f ) ≤ min ( ∥ f − p ∥ ∞ , ∥ f − p ′ ∥ ∞ ) |h|<|h'|\le E_n(f)\le\min\bigl(\lVert f-p\rVert_\infty,\lVert f-p'\rVert_\infty\bigr) ∣ h ∣ < ∣ h ′ ∣ ≤ E n ( f ) ≤ min ( ∥ f − p ∥ ∞ , ∥ f − p ′ ∥ ∞ ) 証明. (1) を示す。系 2.3 (1) により∣ h ∣ ≤ E n ( f ) |h|\le E_n(f) ∣ h ∣ ≤ E n ( f ) p ∈ P n p\in\mathcal P_n p ∈ P n E n ( f ) ≤ ∥ f − p ∥ ∞ E_n(f)\le\lVert f-p\rVert_\infty E n ( f ) ≤ ∥ f − p ∥ ∞
(2) を示す。(1) と仮定により∣ h ∣ ≤ E n ( f ) ≤ ∥ f − p ∥ ∞ = ∣ h ∣ |h|\le E_n(f)\le\lVert f-p\rVert_\infty=|h| ∣ h ∣ ≤ E n ( f ) ≤ ∥ f − p ∥ ∞ = ∣ h ∣ E n ( f ) = ∥ f − p ∥ ∞ = ∣ h ∣ E_n(f)=\lVert f-p\rVert_\infty=|h| E n ( f ) = ∥ f − p ∥ ∞ = ∣ h ∣ p p p
(3) を示す。e : = f − p e:=f-p e := f − p e ( t j ) = ( − 1 ) j h e(t_j)=(-1)^jh e ( t j ) = ( − 1 ) j h h ≠ 0 h\ne0 h = 0 e ( t 0 ) , … , e ( t n + 1 ) e(t_0),\dots,e(t_{n+1}) e ( t 0 ) , … , e ( t n + 1 ) 0 0 0 t i < s < t i + 1 t_i<s<t_{i+1} t i < s < t i + 1 e ( s ) e ( t i ) > 0 e(s)e(t_i)>0 e ( s ) e ( t i ) > 0 t i t_i t i s s s e e e t i < s < t i + 1 t_i<s<t_{i+1} t i < s < t i + 1 e ( s ) e ( t i ) < 0 e(s)e(t_i)<0 e ( s ) e ( t i ) < 0 e ( s ) e ( t i + 1 ) > 0 e(s)e(t_{i+1})>0 e ( s ) e ( t i + 1 ) > 0 t i + 1 t_{i+1} t i + 1 s s s e e e s < t 0 s<t_0 s < t 0 s > t n + 1 s>t_{n+1} s > t n + 1 t ′ = ( s , t 0 , … , t n ) t'=(s,t_0,\dots,t_n) t ′ = ( s , t 0 , … , t n ) e ( s ) e(s) e ( s ) e ( t 0 ) e(t_0) e ( t 0 ) t ′ = ( t 1 , … , t n + 1 , s ) t'=(t_1,\dots,t_{n+1},s) t ′ = ( t 1 , … , t n + 1 , s ) e ( t n + 1 ) e(t_{n+1}) e ( t n + 1 ) e ( s ) e(s) e ( s ) t ′ t' t ′ s s s [ a , b ] [a,b] [ a , b ] n + 2 n+2 n + 2 σ ′ ∈ { 1 , − 1 } \sigma'\in\{1,-1\} σ ′ ∈ { 1 , − 1 } j j j σ ′ ( − 1 ) j e ( t j ′ ) > 0 \sigma'(-1)^je(t'_j)>0 σ ′ ( − 1 ) j e ( t j ′ ) > 0 t ′ t' t ′ s s s t t t ∣ e ∣ = ∣ h ∣ |e|=|h| ∣ e ∣ = ∣ h ∣ ∣ e ( s ) ∣ > ∣ h ∣ |e(s)|>|h| ∣ e ( s ) ∣ > ∣ h ∣ 補題 2.2 をt ′ t' t ′ v 0 ′ , … , v n + 1 ′ v'_0,\dots,v'_{n+1} v 0 ′ , … , v n + 1 ′ 補題 2.2 (4) をy j : = f ( t j ′ ) y_j:=f(t'_j) y j := f ( t j ′ ) q : = p q:=p q := p
∣ h ′ ∣ = ∑ j = 0 n + 1 ∣ v j ′ ∣ ∣ e ( t j ′ ) ∣ ∑ j = 0 n + 1 ∣ v j ′ ∣ |h'|=\frac{\sum_{j=0}^{n+1}|v'_j|\,|e(t'_j)|}{\sum_{j=0}^{n+1}|v'_j|} ∣ h ′ ∣ = ∑ j = 0 n + 1 ∣ v j ′ ∣ ∑ j = 0 n + 1 ∣ v j ′ ∣ ∣ e ( t j ′ ) ∣ である。補題 2.2 (2) により各∣ v j ′ ∣ |v'_j| ∣ v j ′ ∣ ∣ e ( t j ′ ) ∣ |e(t'_j)| ∣ e ( t j ′ ) ∣ ∣ h ∣ |h| ∣ h ∣ s s s ∣ h ∣ |h| ∣ h ∣ ∣ h ′ ∣ > ∣ h ∣ |h'|>|h| ∣ h ′ ∣ > ∣ h ∣ (1) をt t t t ′ t' t ′ ▨
例 5.4. f ( x ) : = e x f(x):=e^x f ( x ) := e x C ( [ − 1 , 1 ] ) C([-1,1]) C ([ − 1 , 1 ]) n = 1 n=1 n = 1 t = ( − 1 , 0 , 1 ) t=(-1,0,1) t = ( − 1 , 0 , 1 ) ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] p ( x ) = c 0 + c 1 x p(x)=c_0+c_1x p ( x ) = c 0 + c 1 x h h h e t j − c 0 − c 1 t j = ( − 1 ) j h e^{t_j}-c_0-c_1t_j=(-1)^jh e t j − c 0 − c 1 t j = ( − 1 ) j h j = 0 , 1 , 2 j=0,1,2 j = 0 , 1 , 2 j = 0 j=0 j = 0 j = 2 j=2 j = 2 c 1 = sinh 1 c_1=\sinh1 c 1 = sinh 1 j = 1 j=1 j = 1 c 0 = 1 + h c_0=1+h c 0 = 1 + h j = 0 j=0 j = 0 j = 2 j=2 j = 2 h = ( cosh 1 − 1 ) / 2 ≈ 0.271540 h=(\cosh1-1)/2\approx0.271540 h = ( cosh 1 − 1 ) /2 ≈ 0.271540 g 1 : = f − p g_1:=f-p g 1 := f − p g 1 ′ ′ ( x ) = e x > 0 g_1''(x)=e^x>0 g 1 ′′ ( x ) = e x > 0 g 1 ′ ( x ) = e x − sinh 1 g_1'(x)=e^x-\sinh1 g 1 ′ ( x ) = e x − sinh 1 s : = log sinh 1 ≈ 0.161439 s:=\log\sinh1\approx0.161439 s := log sinh 1 ≈ 0.161439 0 0 0 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] g 1 g_1 g 1 g 1 ( ± 1 ) = h g_1(\pm1)=h g 1 ( ± 1 ) = h g 1 ( s ) = sinh 1 − c 0 − s sinh 1 ≈ − 0.286063 g_1(s)=\sinh1-c_0-s\sinh1\approx-0.286063 g 1 ( s ) = sinh 1 − c 0 − s sinh 1 ≈ − 0.286063 ∥ g 1 ∥ ∞ = ∣ g 1 ( s ) ∣ \lVert g_1\rVert_\infty=|g_1(s)| ∥ g 1 ∥ ∞ = ∣ g 1 ( s ) ∣ 命題 5.3 (1) により0.271540 < E 1 ( f ) < 0.286063 0.271540<E_1(f)<0.286063 0.271540 < E 1 ( f ) < 0.286063
s ∈ ( t 1 , t 2 ) s\in(t_1,t_2) s ∈ ( t 1 , t 2 ) g 1 ( s ) < 0 g_1(s)<0 g 1 ( s ) < 0 g 1 ( t 1 ) = − h < 0 g_1(t_1)=-h<0 g 1 ( t 1 ) = − h < 0 t 1 t_1 t 1 s s s t ′ = ( − 1 , s , 1 ) t'=(-1,s,1) t ′ = ( − 1 , s , 1 ) t ′ t' t ′ p ′ ( x ) = c 0 ′ + c 1 ′ x p'(x)=c'_0+c'_1x p ′ ( x ) = c 0 ′ + c 1 ′ x h ′ h' h ′ j = 0 j=0 j = 0 j = 2 j=2 j = 2 c 1 ′ = sinh 1 c'_1=\sinh1 c 1 ′ = sinh 1 e s = sinh 1 e^s=\sinh1 e s = sinh 1 j = 0 j=0 j = 0 j = 1 j=1 j = 1 h ′ = ( e − 1 + s sinh 1 ) / 2 ≈ 0.278802 h'=(e^{-1}+s\sinh1)/2\approx0.278802 h ′ = ( e − 1 + s sinh 1 ) /2 ≈ 0.278802 c 0 ′ = ( e − 1 + 2 sinh 1 − s sinh 1 ) / 2 ≈ 1.264279 c'_0=(e^{-1}+2\sinh1-s\sinh1)/2\approx1.264279 c 0 ′ = ( e − 1 + 2 sinh 1 − s sinh 1 ) /2 ≈ 1.264279 g 2 : = f − p ′ g_2:=f-p' g 2 := f − p ′ g 2 ′ ′ > 0 g_2''>0 g 2 ′′ > 0 g 2 ′ g_2' g 2 ′ s s s 0 0 0 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] g 2 g_2 g 2 g 2 ( ± 1 ) = h ′ g_2(\pm1)=h' g 2 ( ± 1 ) = h ′ g 2 ( s ) = − h ′ g_2(s)=-h' g 2 ( s ) = − h ′ ∥ g 2 ∥ ∞ = h ′ \lVert g_2\rVert_\infty=h' ∥ g 2 ∥ ∞ = h ′ 命題 5.3 (2) によりp ′ p' p ′ f f f P 1 \mathcal P_1 P 1 E 1 ( f ) = h ′ ≈ 0.278802 E_1(f)=h'\approx0.278802 E 1 ( f ) = h ′ ≈ 0.278802
6 正則関数の Chebyshev 係数と幾何的な収束 定義 6.1. ρ > 1 \rho>1 ρ > 1 α ρ : = ( ρ + ρ − 1 ) / 2 \alpha_\rho:=(\rho+\rho^{-1})/2 α ρ := ( ρ + ρ − 1 ) /2 β ρ : = ( ρ − ρ − 1 ) / 2 \beta_\rho:=(\rho-\rho^{-1})/2 β ρ := ( ρ − ρ − 1 ) /2
E ρ : = { x + i y ∈ C ∣ x , y ∈ R , x 2 α ρ 2 + y 2 β ρ 2 ≤ 1 } \mathcal E_\rho:=\Bigl\{x+iy\in\C\Bigm|x,y\in\R,\ \frac{x^2}{\alpha_\rho^2}+\frac{y^2}{\beta_\rho^2}\le1\Bigr\} E ρ := { x + i y ∈ C x , y ∈ R , α ρ 2 x 2 + β ρ 2 y 2 ≤ 1 } を Bernstein 楕円 (Bernstein ellipse ){ α ρ cos θ + i β ρ sin θ ∣ θ ∈ R } \{\alpha_\rho\cos\theta+i\beta_\rho\sin\theta\mid\theta\in\R\} { α ρ cos θ + i β ρ sin θ ∣ θ ∈ R } ∂ E ρ \partial\mathcal E_\rho ∂ E ρ
補題 6.2. ρ > 1 \rho>1 ρ > 1 z ∈ C ∖ { 0 } z\in\C\setminus\{0\} z ∈ C ∖ { 0 } J ( z ) : = ( z + z − 1 ) / 2 J(z):=(z+z^{-1})/2 J ( z ) := ( z + z − 1 ) /2
r > 0 r>0 r > 0 θ ∈ R \theta\in\R θ ∈ R J ( r e i θ ) = r + r − 1 2 cos θ + i r − r − 1 2 sin θ J(re^{i\theta})=\frac{r+r^{-1}}2\cos\theta+i\,\frac{r-r^{-1}}2\sin\theta J ( r e i θ ) = 2 r + r − 1 cos θ + i 2 r − r − 1 sin θ J ( e i θ ) = cos θ J(e^{i\theta})=\cos\theta J ( e i θ ) = cos θ ∣ z ∣ = ρ |z|=\rho ∣ z ∣ = ρ J ( z ) ∈ ∂ E ρ J(z)\in\partial\mathcal E_\rho J ( z ) ∈ ∂ E ρ ρ − 1 ≤ ∣ z ∣ ≤ ρ \rho^{-1}\le|z|\le\rho ρ − 1 ≤ ∣ z ∣ ≤ ρ J ( z ) ∈ E ρ J(z)\in\mathcal E_\rho J ( z ) ∈ E ρ 命題 6.3. U ⊆ C U\subseteq\C U ⊆ C [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f f f U U U f ( [ − 1 , 1 ] ) ⊆ R f([-1,1])\subseteq\R f ([ − 1 , 1 ]) ⊆ R J ( z ) : = ( z + z − 1 ) / 2 J(z):=(z+z^{-1})/2 J ( z ) := ( z + z − 1 ) /2 V : = { z ∈ C ∖ { 0 } ∣ J ( z ) ∈ U } V:=\{z\in\C\setminus\{0\}\mid J(z)\in U\} V := { z ∈ C ∖ { 0 } ∣ J ( z ) ∈ U } g : = f ∘ J : V → C g:=f\circ J\colon V\to\C g := f ∘ J : V → C V V V g g g V V V 0 < r < 1 < R 0<r<1<R 0 < r < 1 < R A ( 0 ; r , R ) = { z ∈ C ∣ r < ∣ z ∣ < R } ⊆ V A(0;r,R)=\{z\in\C\mid r<|z|<R\}\subseteq V A ( 0 ; r , R ) = { z ∈ C ∣ r < ∣ z ∣ < R } ⊆ V ( c k ) k ∈ Z (c_k)_{k\in\Z} ( c k ) k ∈ Z g g g A ( 0 ; r , R ) A(0;r,R) A ( 0 ; r , R ) f f f [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f f f k ∈ N ≥ 0 k\in\N k ∈ N ≥ 0 c − k = c k c_{-k}=c_k c − k = c k a k ( f ) = 2 c k a_k(f)=2c_k a k ( f ) = 2 c k
証明. J J J C ∖ { 0 } \C\setminus\{0\} C ∖ { 0 } V V V g g g V V V §E5.11 定理 2.1 の係数を半径1 ∈ ( r , R ) 1\in(r,R) 1 ∈ ( r , R ) w = e i t w=e^{it} w = e i t 0 ≤ t ≤ 2 π 0\le t\le2\pi 0 ≤ t ≤ 2 π 補題 6.2 (1) のJ ( e i t ) = cos t J(e^{it})=\cos t J ( e i t ) = cos t k ∈ Z k\in\Z k ∈ Z
c k = 1 2 π i ∫ 0 2 π g ( e i t ) e i ( k + 1 ) t i e i t d t = 1 2 π ∫ 0 2 π f ( cos t ) e − i k t d t c_k=\frac1{2\pi i}\int_0^{2\pi}\frac{g(e^{it})}{e^{i(k+1)t}}\,ie^{it}\,dt=\frac1{2\pi}\int_0^{2\pi}f(\cos t)e^{-ikt}\,dt c k = 2 π i 1 ∫ 0 2 π e i ( k + 1 ) t g ( e i t ) i e i t d t = 2 π 1 ∫ 0 2 π f ( cos t ) e − ik t d t である。右辺に置換t = 2 π − u t=2\pi-u t = 2 π − u 1 2 π ∫ 0 2 π f ( cos u ) e i k u d u = c − k \frac1{2\pi}\int_0^{2\pi}f(\cos u)e^{iku}\,du=c_{-k} 2 π 1 ∫ 0 2 π f ( cos u ) e ik u d u = c − k c k = ( c k + c − k ) / 2 = 1 2 π ∫ 0 2 π f ( cos t ) cos k t d t c_k=(c_k+c_{-k})/2=\frac1{2\pi}\int_0^{2\pi}f(\cos t)\cos kt\,dt c k = ( c k + c − k ) /2 = 2 π 1 ∫ 0 2 π f ( cos t ) cos k t d t [ π , 2 π ] [\pi,2\pi] [ π , 2 π ] t = 2 π − u t=2\pi-u t = 2 π − u c k = 1 π ∫ 0 π f ( cos t ) cos k t d t = a k ( f ) / 2 c_k=\frac1\pi\int_0^\pi f(\cos t)\cos kt\,dt=a_k(f)/2 c k = π 1 ∫ 0 π f ( cos t ) cos k t d t = a k ( f ) /2 ▨
定理 6.4. ρ > 1 \rho>1 ρ > 1 U ⊆ C U\subseteq\C U ⊆ C E ρ \mathcal E_\rho E ρ f f f U U U f ( [ − 1 , 1 ] ) ⊆ R f([-1,1])\subseteq\R f ([ − 1 , 1 ]) ⊆ R M : = max w ∈ ∂ E ρ ∣ f ( w ) ∣ M:=\max_{w\in\partial\mathcal E_\rho}|f(w)| M := max w ∈ ∂ E ρ ∣ f ( w ) ∣ f f f [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f f f ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ E n ( f ) E_n(f) E n ( f ) [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
∣ a 0 ( f ) ∣ ≤ 2 M |a_0(f)|\le2M ∣ a 0 ( f ) ∣ ≤ 2 M k ∈ N ≥ 1 k\in\NN k ∈ N ≥ 1 ∣ a k ( f ) ∣ ≤ 2 M ρ − k |a_k(f)|\le2M\rho^{-k} ∣ a k ( f ) ∣ ≤ 2 M ρ − k 任意のn ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 ∥ f − S n f ∥ ∞ ≤ 2 M ρ − n / ( ρ − 1 ) \lVert f-S_nf\rVert_\infty\le2M\rho^{-n}/(\rho-1) ∥ f − S n f ∥ ∞ ≤ 2 M ρ − n / ( ρ − 1 ) E n ( f ) ≤ 2 M ρ − n / ( ρ − 1 ) E_n(f)\le2M\rho^{-n}/(\rho-1) E n ( f ) ≤ 2 M ρ − n / ( ρ − 1 )
証明. (1) を示す。α ρ > 1 \alpha_\rho>1 α ρ > 1 [ − 1 , 1 ] ⊆ E ρ ⊆ U [-1,1]\subseteq\mathcal E_\rho\subseteq U [ − 1 , 1 ] ⊆ E ρ ⊆ U ∂ E ρ ⊆ E ρ \partial\mathcal E_\rho\subseteq\mathcal E_\rho ∂ E ρ ⊆ E ρ [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] M M M J ( z ) : = ( z + z − 1 ) / 2 J(z):=(z+z^{-1})/2 J ( z ) := ( z + z − 1 ) /2 V : = { z ∈ C ∖ { 0 } ∣ J ( z ) ∈ U } V:=\{z\in\C\setminus\{0\}\mid J(z)\in U\} V := { z ∈ C ∖ { 0 } ∣ J ( z ) ∈ U } g : = f ∘ J g:=f\circ J g := f ∘ J 命題 6.3 によりV V V 補題 6.2 (3) によりコンパクト集合K : = { z ∣ ρ − 1 ≤ ∣ z ∣ ≤ ρ } K:=\{z\mid\rho^{-1}\le|z|\le\rho\} K := { z ∣ ρ − 1 ≤ ∣ z ∣ ≤ ρ } V V V 0 ∉ V 0\notin V 0 ∈ / V C ∖ V \C\setminus V C ∖ V K K K η : = inf { ∣ z − w ∣ ∣ z ∈ K , w ∈ C ∖ V } > 0 \eta:=\inf\{|z-w|\mid z\in K,\ w\in\C\setminus V\}>0 η := inf { ∣ z − w ∣ ∣ z ∈ K , w ∈ C ∖ V } > 0 R > ρ R>\rho R > ρ R − ρ < η R-\rho<\eta R − ρ < η ρ − 1 − R − 1 < η \rho^{-1}-R^{-1}<\eta ρ − 1 − R − 1 < η R − 1 < ∣ z ∣ < R R^{-1}<|z|<R R − 1 < ∣ z ∣ < R z z z K K K η \eta η A ( 0 ; R − 1 , R ) ⊆ V A(0;R^{-1},R)\subseteq V A ( 0 ; R − 1 , R ) ⊆ V g g g A ( 0 ; R − 1 , R ) A(0;R^{-1},R) A ( 0 ; R − 1 , R ) ( c k ) k ∈ Z (c_k)_{k\in\Z} ( c k ) k ∈ Z 命題 6.3 によりk ∈ N ≥ 0 k\in\N k ∈ N ≥ 0 a k ( f ) = 2 c k a_k(f)=2c_k a k ( f ) = 2 c k §E5.11 定理 2.1 により、w = ρ e i t w=\rho e^{it} w = ρ e i t 0 ≤ t ≤ 2 π 0\le t\le2\pi 0 ≤ t ≤ 2 π C ρ C_\rho C ρ c k = 1 2 π i ∮ C ρ g ( w ) w − k − 1 d w c_k=\frac1{2\pi i}\oint_{C_\rho}g(w)w^{-k-1}\,dw c k = 2 π i 1 ∮ C ρ g ( w ) w − k − 1 d w 補題 6.2 (2) によりC ρ C_\rho C ρ J ( w ) ∈ ∂ E ρ J(w)\in\partial\mathcal E_\rho J ( w ) ∈ ∂ E ρ ∣ g ( w ) w − k − 1 ∣ ≤ M ρ − k − 1 |g(w)w^{-k-1}|\le M\rho^{-k-1} ∣ g ( w ) w − k − 1 ∣ ≤ M ρ − k − 1 C ρ C_\rho C ρ 2 π ρ 2\pi\rho 2 π ρ §E5.4 命題 3.2 により∣ c k ∣ ≤ 1 2 π M ρ − k − 1 ⋅ 2 π ρ = M ρ − k |c_k|\le\frac1{2\pi}M\rho^{-k-1}\cdot2\pi\rho=M\rho^{-k} ∣ c k ∣ ≤ 2 π 1 M ρ − k − 1 ⋅ 2 π ρ = M ρ − k ∣ a 0 ( f ) ∣ ≤ 2 M |a_0(f)|\le2M ∣ a 0 ( f ) ∣ ≤ 2 M ∣ a k ( f ) ∣ ≤ 2 M ρ − k |a_k(f)|\le2M\rho^{-k} ∣ a k ( f ) ∣ ≤ 2 M ρ − k
(2) を示す。(1) により∑ k ∣ a k ( f ) ∣ < ∞ \sum_k|a_k(f)|<\infty ∑ k ∣ a k ( f ) ∣ < ∞ 命題 1.2 (2) により∥ f − S n f ∥ ∞ ≤ ∑ k > n 2 M ρ − k = 2 M ρ − n / ( ρ − 1 ) \lVert f-S_nf\rVert_\infty\le\sum_{k>n}2M\rho^{-k}=2M\rho^{-n}/(\rho-1) ∥ f − S n f ∥ ∞ ≤ ∑ k > n 2 M ρ − k = 2 M ρ − n / ( ρ − 1 ) S n f ∈ P n S_nf\in\mathcal P_n S n f ∈ P n E n ( f ) ≤ ∥ f − S n f ∥ ∞ E_n(f)\le\lVert f-S_nf\rVert_\infty E n ( f ) ≤ ∥ f − S n f ∥ ∞ ▨
補題 6.5. n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 N : = n + 1 N:=n+1 N := n + 1 c j : = cos ( ( 2 j + 1 ) π / ( 2 N ) ) c_j:=\cos\bigl((2j+1)\pi/(2N)\bigr) c j := cos ( ( 2 j + 1 ) π / ( 2 N ) ) 0 ≤ j ≤ n 0\le j\le n 0 ≤ j ≤ n [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] n + 1 n+1 n + 1 I n I_n I n k ∈ N ≥ 0 k\in\N k ∈ N ≥ 0 k = 2 l N + r k=2lN+r k = 2 l N + r l ∈ N ≥ 0 l\in\N l ∈ N ≥ 0 0 ≤ r ≤ 2 N − 1 0\le r\le2N-1 0 ≤ r ≤ 2 N − 1 0 ≤ j ≤ n 0\le j\le n 0 ≤ j ≤ n r ≤ n r\le n r ≤ n T k ( c j ) = ( − 1 ) l T r ( c j ) T_k(c_j)=(-1)^lT_r(c_j) T k ( c j ) = ( − 1 ) l T r ( c j ) r = N r=N r = N T k ( c j ) = 0 T_k(c_j)=0 T k ( c j ) = 0 N < r ≤ 2 N − 1 N<r\le2N-1 N < r ≤ 2 N − 1 T k ( c j ) = ( − 1 ) l + 1 T 2 N − r ( c j ) T_k(c_j)=(-1)^{l+1}T_{2N-r}(c_j) T k ( c j ) = ( − 1 ) l + 1 T 2 N − r ( c j ) I n T k I_nT_k I n T k ( − 1 ) l T r (-1)^lT_r ( − 1 ) l T r 0 0 0 ( − 1 ) l + 1 T 2 N − r (-1)^{l+1}T_{2N-r} ( − 1 ) l + 1 T 2 N − r [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] ∣ I n T k ∣ ≤ 1 |I_nT_k|\le1 ∣ I n T k ∣ ≤ 1
証明. θ j : = ( 2 j + 1 ) π / ( 2 N ) \theta_j:=(2j+1)\pi/(2N) θ j := ( 2 j + 1 ) π / ( 2 N ) §E20.11 補題 3.2 (2) によりT k ( c j ) = cos k θ j T_k(c_j)=\cos k\theta_j T k ( c j ) = cos k θ j k θ j = l ( 2 j + 1 ) π + r θ j k\theta_j=l(2j+1)\pi+r\theta_j k θ j = l ( 2 j + 1 ) π + r θ j cos k θ j = ( − 1 ) l cos r θ j \cos k\theta_j=(-1)^l\cos r\theta_j cos k θ j = ( − 1 ) l cos r θ j r ≤ n r\le n r ≤ n §E20.11 補題 3.2 (2) によりcos r θ j = T r ( c j ) \cos r\theta_j=T_r(c_j) cos r θ j = T r ( c j ) r = N r=N r = N cos r θ j = cos ( ( 2 j + 1 ) π / 2 ) = 0 \cos r\theta_j=\cos((2j+1)\pi/2)=0 cos r θ j = cos (( 2 j + 1 ) π /2 ) = 0 N < r ≤ 2 N − 1 N<r\le2N-1 N < r ≤ 2 N − 1 r ′ : = 2 N − r r':=2N-r r ′ := 2 N − r 1 ≤ r ′ ≤ n 1\le r'\le n 1 ≤ r ′ ≤ n r θ j = ( 2 j + 1 ) π − r ′ θ j r\theta_j=(2j+1)\pi-r'\theta_j r θ j = ( 2 j + 1 ) π − r ′ θ j cos r θ j = − cos r ′ θ j = − T r ′ ( c j ) \cos r\theta_j=-\cos r'\theta_j=-T_{r'}(c_j) cos r θ j = − cos r ′ θ j = − T r ′ ( c j ) §E20.12 命題 5.2 (3) によりc 0 , … , c n c_0,\dots,c_n c 0 , … , c n §E20.11 補題 3.2 (1) によりP n \mathcal P_n P n T k T_k T k §E20.12 定理 1.3 (2) によりI n T k I_nT_k I n T k §E20.11 補題 3.2 (3) により[ − 1 , 1 ] [-1,1] [ − 1 , 1 ] ∣ I n T k ∣ ≤ 1 |I_nT_k|\le1 ∣ I n T k ∣ ≤ 1 ▨
定理 6.6. n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 I n I_n I n [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] n + 1 n+1 n + 1 ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
f ∈ C ( [ − 1 , 1 ] ) f\in C([-1,1]) f ∈ C ([ − 1 , 1 ]) ∑ k = 0 ∞ ∣ a k ( f ) ∣ < ∞ \sum_{k=0}^\infty|a_k(f)|<\infty ∑ k = 0 ∞ ∣ a k ( f ) ∣ < ∞ ∥ f − I n f ∥ ∞ ≤ 2 ∑ k = n + 1 ∞ ∣ a k ( f ) ∣ \lVert f-I_nf\rVert_\infty\le2\sum_{k=n+1}^\infty|a_k(f)| ∥ f − I n f ∥ ∞ ≤ 2 ∑ k = n + 1 ∞ ∣ a k ( f ) ∣ ρ > 1 \rho>1 ρ > 1 U U U f f f M M M 定理 6.4 の仮定を満たすならば、f f f [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] ∥ f − I n f ∥ ∞ ≤ 4 M ρ − n / ( ρ − 1 ) \lVert f-I_nf\rVert_\infty\le4M\rho^{-n}/(\rho-1) ∥ f − I n f ∥ ∞ ≤ 4 M ρ − n / ( ρ − 1 ) 証明. (1) を示す。a 0 ′ : = a 0 ( f ) / 2 a'_0:=a_0(f)/2 a 0 ′ := a 0 ( f ) /2 k ≥ 1 k\ge1 k ≥ 1 a k ′ : = a k ( f ) a'_k:=a_k(f) a k ′ := a k ( f ) c 0 , … , c n c_0,\dots,c_n c 0 , … , c n ℓ 0 , … , ℓ n \ell_0,\dots,\ell_n ℓ 0 , … , ℓ n 命題 1.2 (2) により、任意のx ∈ [ − 1 , 1 ] x\in[-1,1] x ∈ [ − 1 , 1 ] f ( x ) = ∑ k ≥ 0 a k ′ T k ( x ) f(x)=\sum_{k\ge0}a'_kT_k(x) f ( x ) = ∑ k ≥ 0 a k ′ T k ( x ) j j j x ∈ [ − 1 , 1 ] x\in[-1,1] x ∈ [ − 1 , 1 ]
I n f ( x ) = ∑ j = 0 n ∑ k ≥ 0 a k ′ T k ( c j ) ℓ j ( x ) = ∑ k ≥ 0 a k ′ I n T k ( x ) I_nf(x)=\sum_{j=0}^n\sum_{k\ge0}a'_kT_k(c_j)\ell_j(x)=\sum_{k\ge0}a'_k\,I_nT_k(x) I n f ( x ) = j = 0 ∑ n k ≥ 0 ∑ a k ′ T k ( c j ) ℓ j ( x ) = k ≥ 0 ∑ a k ′ I n T k ( x ) である。§E20.12 定理 3.2 (1) によりk ≤ n k\le n k ≤ n I n T k = T k I_nT_k=T_k I n T k = T k f ( x ) − I n f ( x ) = ∑ k > n a k ( f ) ( T k ( x ) − I n T k ( x ) ) f(x)-I_nf(x)=\sum_{k>n}a_k(f)\bigl(T_k(x)-I_nT_k(x)\bigr) f ( x ) − I n f ( x ) = ∑ k > n a k ( f ) ( T k ( x ) − I n T k ( x ) ) §E20.11 補題 3.2 (3) と補題 6.5 により∣ T k ( x ) − I n T k ( x ) ∣ ≤ 2 |T_k(x)-I_nT_k(x)|\le2 ∣ T k ( x ) − I n T k ( x ) ∣ ≤ 2
(2) を示す。定理 6.4 (1) により∑ k ∣ a k ( f ) ∣ < ∞ \sum_k|a_k(f)|<\infty ∑ k ∣ a k ( f ) ∣ < ∞ (1) により∥ f − I n f ∥ ∞ ≤ 2 ∑ k > n 2 M ρ − k = 4 M ρ − n / ( ρ − 1 ) \lVert f-I_nf\rVert_\infty\le2\sum_{k>n}2M\rho^{-k}=4M\rho^{-n}/(\rho-1) ∥ f − I n f ∥ ∞ ≤ 2 ∑ k > n 2 M ρ − k = 4 M ρ − n / ( ρ − 1 ) ▨
例 6.7. f ( z ) : = 1 / ( 1 + 25 z 2 ) f(z):=1/(1+25z^2) f ( z ) := 1/ ( 1 + 25 z 2 ) U : = C ∖ { i / 5 , − i / 5 } U:=\C\setminus\{i/5,-i/5\} U := C ∖ { i /5 , − i /5 } f ( [ − 1 , 1 ] ) ⊆ R f([-1,1])\subseteq\R f ([ − 1 , 1 ]) ⊆ R ρ 0 : = ( 1 + 26 ) / 5 ≈ 1.21980 \rho_0:=(1+\sqrt{26})/5\approx1.21980 ρ 0 := ( 1 + 26 ) /5 ≈ 1.21980 ρ 0 − 1 = ( 26 − 1 ) / 5 \rho_0^{-1}=(\sqrt{26}-1)/5 ρ 0 − 1 = ( 26 − 1 ) /5 β ρ 0 = 1 / 5 \beta_{\rho_0}=1/5 β ρ 0 = 1/5 ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
β ρ \beta_\rho β ρ ρ > 1 \rho>1 ρ > 1 1 < ρ < ρ 0 1<\rho<\rho_0 1 < ρ < ρ 0 β ρ < 1 / 5 \beta_\rho<1/5 β ρ < 1/5 ( ± 1 / 5 ) 2 / β ρ 2 > 1 (\pm1/5)^2/\beta_\rho^2>1 ( ± 1/5 ) 2 / β ρ 2 > 1 ± i / 5 ∉ E ρ \pm i/5\notin\mathcal E_\rho ± i /5 ∈ / E ρ E ρ ⊆ U \mathcal E_\rho\subseteq U E ρ ⊆ U 定理 6.4 と定理 6.6 は任意のρ ∈ ( 1 , ρ 0 ) \rho\in(1,\rho_0) ρ ∈ ( 1 , ρ 0 ) ρ > ρ 0 \rho>\rho_0 ρ > ρ 0 β ρ > 1 / 5 \beta_\rho>1/5 β ρ > 1/5 i / 5 ∈ E ρ i/5\in\mathcal E_\rho i /5 ∈ E ρ J ( z ) : = ( z + z − 1 ) / 2 J(z):=(z+z^{-1})/2 J ( z ) := ( z + z − 1 ) /2 q : = ρ 0 − 2 q:=\rho_0^{-2} q := ρ 0 − 2 q + q − 1 = ( ρ 0 + ρ 0 − 1 ) 2 − 2 = 54 / 25 q+q^{-1}=(\rho_0+\rho_0^{-1})^2-2=54/25 q + q − 1 = ( ρ 0 + ρ 0 − 1 ) 2 − 2 = 54/25 z ≠ 0 z\ne0 z = 0 1 + 25 J ( z ) 2 = ( 25 z 4 + 54 z 2 + 25 ) / ( 4 z 2 ) = 25 ( z 2 + q ) ( z 2 + q − 1 ) / ( 4 z 2 ) 1+25J(z)^2=(25z^4+54z^2+25)/(4z^2)=25(z^2+q)(z^2+q^{-1})/(4z^2) 1 + 25 J ( z ) 2 = ( 25 z 4 + 54 z 2 + 25 ) / ( 4 z 2 ) = 25 ( z 2 + q ) ( z 2 + q − 1 ) / ( 4 z 2 ) 1 + 25 J ( z ) 2 = 0 1+25J(z)^2=0 1 + 25 J ( z ) 2 = 0 z z z ∣ z ∣ ∈ { ρ 0 − 1 , ρ 0 } |z|\in\{\rho_0^{-1},\rho_0\} ∣ z ∣ ∈ { ρ 0 − 1 , ρ 0 } g : = f ∘ J g:=f\circ J g := f ∘ J A ( 0 ; ρ 0 − 1 , ρ 0 ) A(0;\rho_0^{-1},\rho_0) A ( 0 ; ρ 0 − 1 , ρ 0 ) u : = z 2 u:=z^2 u := z 2 25 ( q − 1 − q ) = 4 26 25(q^{-1}-q)=4\sqrt{26} 25 ( q − 1 − q ) = 4 26 g ( z ) = 4 u 25 ( u + q ) ( u + q − 1 ) = 1 26 ( 1 1 + q u − q / u 1 + q / u ) = 1 26 ( ∑ m = 0 ∞ ( − 1 ) m q m z 2 m + ∑ m = 1 ∞ ( − 1 ) m q m z − 2 m ) g(z)=\frac{4u}{25(u+q)(u+q^{-1})}=\frac1{\sqrt{26}}\Bigl(\frac{1}{1+qu}-\frac{q/u}{1+q/u}\Bigr)=\frac1{\sqrt{26}}\Bigl(\sum_{m=0}^\infty(-1)^mq^mz^{2m}+\sum_{m=1}^\infty(-1)^mq^mz^{-2m}\Bigr) g ( z ) = 25 ( u + q ) ( u + q − 1 ) 4 u = 26 1 ( 1 + q u 1 − 1 + q / u q / u ) = 26 1 ( m = 0 ∑ ∞ ( − 1 ) m q m z 2 m + m = 1 ∑ ∞ ( − 1 ) m q m z − 2 m ) ρ 0 − 1 < ρ 1 ≤ ρ 2 < ρ 0 \rho_0^{-1}<\rho_1\le\rho_2<\rho_0 ρ 0 − 1 < ρ 1 ≤ ρ 2 < ρ 0 ρ 1 ≤ ∣ z ∣ ≤ ρ 2 \rho_1\le|z|\le\rho_2 ρ 1 ≤ ∣ z ∣ ≤ ρ 2 ∣ q z 2 ∣ ≤ q ρ 2 2 < 1 |qz^2|\le q\rho_2^2<1 ∣ q z 2 ∣ ≤ q ρ 2 2 < 1 ∣ q / z 2 ∣ ≤ q ρ 1 − 2 < 1 |q/z^2|\le q\rho_1^{-2}<1 ∣ q / z 2 ∣ ≤ q ρ 1 − 2 < 1 §E5.11 命題 2.2 によりg g g A ( 0 ; ρ 0 − 1 , ρ 0 ) A(0;\rho_0^{-1},\rho_0) A ( 0 ; ρ 0 − 1 , ρ 0 ) c 2 m = c − 2 m = ( − 1 ) m q m / 26 c_{2m}=c_{-2m}=(-1)^mq^m/\sqrt{26} c 2 m = c − 2 m = ( − 1 ) m q m / 26 m ∈ N ≥ 0 m\in\N m ∈ N ≥ 0 c 2 m + 1 = 0 c_{2m+1}=0 c 2 m + 1 = 0 命題 6.3 によりk ∈ N ≥ 0 k\in\N k ∈ N ≥ 0 a k ( f ) = { 2 ( − 1 ) k / 2 26 ρ 0 − k ( k が偶数 ) , 0 ( k が奇数 ) a_k(f)=\begin{cases}\dfrac{2(-1)^{k/2}}{\sqrt{26}}\,\rho_0^{-k}&(k\text{ が偶数}),\\[1ex]0&(k\text{ が奇数})\end{cases} a k ( f ) = ⎩ ⎨ ⎧ 26 2 ( − 1 ) k /2 ρ 0 − k 0 ( k が偶数 ) , ( k が奇数 ) ρ > ρ 0 \rho>\rho_0 ρ > ρ 0 ∣ a 2 m ( f ) ∣ ρ 2 m = ( 2 / 26 ) ( ρ / ρ 0 ) 2 m → ∞ |a_{2m}(f)|\rho^{2m}=(2/\sqrt{26})(\rho/\rho_0)^{2m}\to\infty ∣ a 2 m ( f ) ∣ ρ 2 m = ( 2/ 26 ) ( ρ / ρ 0 ) 2 m → ∞ m → ∞ m\to\infty m → ∞ C C C k k k ∣ a k ( f ) ∣ ≤ C ρ − k |a_k(f)|\le C\rho^{-k} ∣ a k ( f ) ∣ ≤ C ρ − k n n n T 2 m ( 0 ) = cos ( m π ) = ( − 1 ) m T_{2m}(0)=\cos(m\pi)=(-1)^m T 2 m ( 0 ) = cos ( mπ ) = ( − 1 ) m 命題 1.2 (2) の級数によりf ( 0 ) − S n f ( 0 ) = ∑ 2 m > n a 2 m ( f ) ( − 1 ) m = ∑ k > n ∣ a k ( f ) ∣ f(0)-S_nf(0)=\sum_{2m>n}a_{2m}(f)(-1)^m=\sum_{k>n}|a_k(f)| f ( 0 ) − S n f ( 0 ) = ∑ 2 m > n a 2 m ( f ) ( − 1 ) m = ∑ k > n ∣ a k ( f ) ∣ 命題 1.2 (2) の評価と合わせて
∥ f − S n f ∥ ∞ = ∑ k > n ∣ a k ( f ) ∣ = 2 26 ⋅ ρ 0 − ( n + 1 ) 1 − ρ 0 − 2 \lVert f-S_nf\rVert_\infty=\sum_{k>n}|a_k(f)|=\frac2{\sqrt{26}}\cdot\frac{\rho_0^{-(n+1)}}{1-\rho_0^{-2}} ∥ f − S n f ∥ ∞ = k > n ∑ ∣ a k ( f ) ∣ = 26 2 ⋅ 1 − ρ 0 − 2 ρ 0 − ( n + 1 ) n = 21 n=21 n = 21 1.511 × 10 − 2 1.511\times10^{-2} 1.511 × 1 0 − 2 定理 6.6 (1) により∥ f − I 21 f ∥ ∞ ≤ 3.023 × 10 − 2 \lVert f-I_{21}f\rVert_\infty\le3.023\times10^{-2} ∥ f − I 21 f ∥ ∞ ≤ 3.023 × 1 0 − 2 §E20.12 例 6.3 で観察した Chebyshev 節点での誤差2.527 × 10 − 2 2.527\times10^{-2} 2.527 × 1 0 − 2 7 演習 解答. 補題 6.2 (1) を示す。z = r e i θ z=re^{i\theta} z = r e i θ z − 1 = r − 1 e − i θ z^{-1}=r^{-1}e^{-i\theta} z − 1 = r − 1 e − i θ z + z − 1 = ( r + r − 1 ) cos θ + i ( r − r − 1 ) sin θ z+z^{-1}=(r+r^{-1})\cos\theta+i(r-r^{-1})\sin\theta z + z − 1 = ( r + r − 1 ) cos θ + i ( r − r − 1 ) sin θ 2 2 2 r = 1 r=1 r = 1 J ( e i θ ) = cos θ J(e^{i\theta})=\cos\theta J ( e i θ ) = cos θ
補題 6.2 (2) を示す。z = ρ e i θ z=\rho e^{i\theta} z = ρ e i θ 補題 6.2 (1) によりJ ( z ) = α ρ cos θ + i β ρ sin θ ∈ ∂ E ρ J(z)=\alpha_\rho\cos\theta+i\beta_\rho\sin\theta\in\partial\mathcal E_\rho J ( z ) = α ρ cos θ + i β ρ sin θ ∈ ∂ E ρ
補題 6.2 (3) を示す。ϕ ( r ) : = r + r − 1 \phi(r):=r+r^{-1} ϕ ( r ) := r + r − 1 ψ ( r ) : = ∣ r − r − 1 ∣ \psi(r):=|r-r^{-1}| ψ ( r ) := ∣ r − r − 1 ∣ r > 0 r>0 r > 0 ϕ ( r ) = ϕ ( r − 1 ) \phi(r)=\phi(r^{-1}) ϕ ( r ) = ϕ ( r − 1 ) ψ ( r ) = ψ ( r − 1 ) \psi(r)=\psi(r^{-1}) ψ ( r ) = ψ ( r − 1 ) r ≥ 1 r\ge1 r ≥ 1 ϕ ′ ( r ) = 1 − r − 2 ≥ 0 \phi'(r)=1-r^{-2}\ge0 ϕ ′ ( r ) = 1 − r − 2 ≥ 0 ψ ( r ) = r − r − 1 \psi(r)=r-r^{-1} ψ ( r ) = r − r − 1 ψ ′ ( r ) = 1 + r − 2 > 0 \psi'(r)=1+r^{-2}>0 ψ ′ ( r ) = 1 + r − 2 > 0 ϕ \phi ϕ ψ \psi ψ [ 1 , ∞ ) [1,\infty) [ 1 , ∞ ) ρ − 1 ≤ r ≤ ρ \rho^{-1}\le r\le\rho ρ − 1 ≤ r ≤ ρ max ( r , r − 1 ) ∈ [ 1 , ρ ] \max(r,r^{-1})\in[1,\rho] max ( r , r − 1 ) ∈ [ 1 , ρ ] ϕ ( r ) ≤ ϕ ( ρ ) = 2 α ρ \phi(r)\le\phi(\rho)=2\alpha_\rho ϕ ( r ) ≤ ϕ ( ρ ) = 2 α ρ ψ ( r ) ≤ ψ ( ρ ) = 2 β ρ \psi(r)\le\psi(\rho)=2\beta_\rho ψ ( r ) ≤ ψ ( ρ ) = 2 β ρ z = r e i θ z=re^{i\theta} z = r e i θ J ( z ) = x + i y J(z)=x+iy J ( z ) = x + i y x , y ∈ R x,y\in\R x , y ∈ R 補題 6.2 (1) により∣ x ∣ = ϕ ( r ) ∣ cos θ ∣ / 2 ≤ α ρ ∣ cos θ ∣ |x|=\phi(r)|\cos\theta|/2\le\alpha_\rho|\cos\theta| ∣ x ∣ = ϕ ( r ) ∣ cos θ ∣/2 ≤ α ρ ∣ cos θ ∣ ∣ y ∣ = ψ ( r ) ∣ sin θ ∣ / 2 ≤ β ρ ∣ sin θ ∣ |y|=\psi(r)|\sin\theta|/2\le\beta_\rho|\sin\theta| ∣ y ∣ = ψ ( r ) ∣ sin θ ∣/2 ≤ β ρ ∣ sin θ ∣ x 2 / α ρ 2 + y 2 / β ρ 2 ≤ cos 2 θ + sin 2 θ = 1 x^2/\alpha_\rho^2+y^2/\beta_\rho^2\le\cos^2\theta+\sin^2\theta=1 x 2 / α ρ 2 + y 2 / β ρ 2 ≤ cos 2 θ + sin 2 θ = 1 J ( z ) ∈ E ρ J(z)\in\mathcal E_\rho J ( z ) ∈ E ρ ▨
問題 7.2. n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 f ∈ P n + 1 f\in\mathcal P_{n+1} f ∈ P n + 1 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] C ( [ − 1 , 1 ] ) C([-1,1]) C ([ − 1 , 1 ]) S n f S_nf S n f f f f P n \mathcal P_n P n [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] E n ( f ) = ∣ a n + 1 ( f ) ∣ E_n(f)=|a_{n+1}(f)| E n ( f ) = ∣ a n + 1 ( f ) ∣
解答. 命題 1.2 (1) によりS n + 1 f = f S_{n+1}f=f S n + 1 f = f f − S n f = a n + 1 ( f ) T n + 1 f-S_nf=a_{n+1}(f)\,T_{n+1} f − S n f = a n + 1 ( f ) T n + 1 0 ≤ j ≤ n + 1 0\le j\le n+1 0 ≤ j ≤ n + 1 η j : = cos ( ( n + 1 − j ) π / ( n + 1 ) ) \eta_j:=\cos\bigl((n+1-j)\pi/(n+1)\bigr) η j := cos ( ( n + 1 − j ) π / ( n + 1 ) ) cos \cos cos [ 0 , π ] [0,\pi] [ 0 , π ] − 1 = η 0 < η 1 < ⋯ < η n + 1 = 1 -1=\eta_0<\eta_1<\dots<\eta_{n+1}=1 − 1 = η 0 < η 1 < ⋯ < η n + 1 = 1 §E20.12 命題 5.2 (2) によりT n + 1 ( η j ) = ( − 1 ) n + 1 − j T_{n+1}(\eta_j)=(-1)^{n+1-j} T n + 1 ( η j ) = ( − 1 ) n + 1 − j max [ − 1 , 1 ] ∣ T n + 1 ∣ = 1 \max_{[-1,1]}|T_{n+1}|=1 max [ − 1 , 1 ] ∣ T n + 1 ∣ = 1 ∥ f − S n f ∥ ∞ = ∣ a n + 1 ( f ) ∣ \lVert f-S_nf\rVert_\infty=|a_{n+1}(f)| ∥ f − S n f ∥ ∞ = ∣ a n + 1 ( f ) ∣ σ \sigma σ ( − 1 ) n + 1 a n + 1 ( f ) ≥ 0 (-1)^{n+1}a_{n+1}(f)\ge0 ( − 1 ) n + 1 a n + 1 ( f ) ≥ 0 1 1 1 − 1 -1 − 1
f ( η j ) − S n f ( η j ) = ( − 1 ) n + 1 a n + 1 ( f ) ( − 1 ) j = σ ( − 1 ) j ∣ a n + 1 ( f ) ∣ ( 0 ≤ j ≤ n + 1 ) f(\eta_j)-S_nf(\eta_j)=(-1)^{n+1}a_{n+1}(f)(-1)^j=\sigma(-1)^j|a_{n+1}(f)|\qquad(0\le j\le n+1) f ( η j ) − S n f ( η j ) = ( − 1 ) n + 1 a n + 1 ( f ) ( − 1 ) j = σ ( − 1 ) j ∣ a n + 1 ( f ) ∣ ( 0 ≤ j ≤ n + 1 ) である。定理 2.4 によりS n f S_nf S n f E n ( f ) = ∥ f − S n f ∥ ∞ = ∣ a n + 1 ( f ) ∣ E_n(f)=\lVert f-S_nf\rVert_\infty=|a_{n+1}(f)| E n ( f ) = ∥ f − S n f ∥ ∞ = ∣ a n + 1 ( f ) ∣ ▨