多元泰勒公式不能通過在每個坐標中寫出互不相關的一元展開得到。位移同時改變所有坐標，因此出現混合偏導數。多重線性導數組織這些相互作用，餘項估計則告訴我們無窮泰勒展開是否等於函數。
1 作為多重線性映射的高階導數
在實數域工作，維數 n ≥ 1 n\ge1 n ≥ 1 k k k
對 j 重線性映射 B B B j ≥ 1 j\ge1 j ≥ 1
B [ v 1 , … , v j ] = ∑ i 1 , … , i j = 1 n B [ e i 1 , … , e i j ] ( v 1 ) i 1 ⋯ ( v j ) i j . B[v_1,\ldots,v_j]=\sum_{i_1,\ldots,i_j=1}^n B[e_{i_1},\ldots,e_{i_j}](v_1)_{i_1}\cdots(v_j)_{i_j}. B [ v 1 , … , v j ] = i 1 , … , i j = 1 ∑ n B [ e i 1 , … , e i j ] ( v 1 ) i 1 ⋯ ( v j ) i j . 記 M B = max i 1 , … , i j ∣ B [ e i 1 , … , e i j ] ∣ M_B=\max_{i_1,\ldots,i_j}|B[e_{i_1},\ldots,e_{i_j}]| M B = max i 1 , … , i j ∣ B [ e i 1 , … , e i j ] ∣
∣ B [ v 1 , … , v j ] ∣ ≤ M B ∏ r = 1 j ∑ i ∣ ( v r ) i ∣ ≤ n j / 2 M B ∏ r = 1 j ∥ v r ∥ . |B[v_1,\ldots,v_j]|\le M_B\prod_{r=1}^j\sum_i|(v_r)_i|\le n^{j/2}M_B\prod_{r=1}^j\|v_r\|. ∣ B [ v 1 , … , v j ] ∣ ≤ M B r = 1 ∏ j i ∑ ∣ ( v r ) i ∣ ≤ n j /2 M B r = 1 ∏ j ∥ v r ∥. 因此多重線性算子範數
∥ B ∥ = sup { ∣ B [ v 1 , … , v j ] ∣ : ∥ v r ∥ ≤ 1 ( 1 ≤ r ≤ j ) } \|B\|=\sup\{|B[v_1,\ldots,v_j]|:\|v_r\|\le1\ (1\le r\le j)\} ∥ B ∥ = sup { ∣ B [ v 1 , … , v j ] ∣ : ∥ v r ∥ ≤ 1 ( 1 ≤ r ≤ j )} 有限，且滿足 M B ≤ ∥ B ∥ ≤ n j / 2 M B M_B\le\|B\|\le n^{j/2}M_B M B ≤ ∥ B ∥ ≤ n j /2 M B B B B k k k
具體地，對 1 ≤ j ≤ k 1\le j\le k 1 ≤ j ≤ k
D j f ( x ) [ v 1 , … , v j ] = ∑ i 1 , … , i j = 1 n ∂ i 1 ⋯ ∂ i j f ( x ) ( v 1 ) i 1 ⋯ ( v j ) i j . D^jf(x)[v_1,\ldots,v_j]=\sum_{i_1,\ldots,i_j=1}^n\partial_{i_1}\cdots\partial_{i_j}f(x)(v_1)_{i_1}\cdots(v_j)_{i_j}. D j f ( x ) [ v 1 , … , v j ] = i 1 , … , i j = 1 ∑ n ∂ i 1 ⋯ ∂ i j f ( x ) ( v 1 ) i 1 ⋯ ( v j ) i j . 將 C2 坐標矩形定理應用於相應的低階導數，連續混合偏導數便可交換一對相鄰求導次序。每個排列都是有限次相鄰交換的複合，所以這些多重線性映射對稱。對 j = 2 j=2 j = 2 D 2 f ( a ) [ u , v ] = u T H f ( a ) v D^2f(a)[u,v]=u^TH_f(a)v D 2 f ( a ) [ u , v ] = u T H f ( a ) v
固定 h h h ϕ ( t ) = f ( a + t h ) \phi(t)=f(a+th) ϕ ( t ) = f ( a + t h )
ϕ ( j ) ( t ) = D j f ( a + t h ) [ h , … , h ] . \phi^{(j)}(t)=D^jf(a+th)[h,\ldots,h]. ϕ ( j ) ( t ) = D j f ( a + t h ) [ h , … , h ] . 當 h ≠ 0 h\ne0 h = 0 ∥ h ∥ \|h\| ∥ h ∥ h = 0 h=0 h = 0 ∣ D j f ( x ) [ h , … , h ] ∣ ≤ ∥ D j f ( x ) ∥ ∥ h ∥ j |D^jf(x)[h,\ldots,h]|\le\|D^jf(x)\|\|h\|^j ∣ D j f ( x ) [ h , … , h ] ∣ ≤ ∥ D j f ( x ) ∥∥ h ∥ j
對由非負整數組成的多重指標 α = ( α 1 , … , α n ) \alpha=(\alpha_1,\ldots,\alpha_n) α = ( α 1 , … , α n ) ∣ α ∣ = ∑ i α i |\alpha|=\sum_i\alpha_i ∣ α ∣ = ∑ i α i α ! = ∏ i α i ! \alpha!=\prod_i\alpha_i! α ! = ∏ i α i ! h α = ∏ i h i α i h^\alpha=\prod_i h_i^{\alpha_i} h α = ∏ i h i α i α \alpha α j ! / α ! j!/\alpha! j ! / α !
1 j ! D j f ( a ) [ h , … , h ] = ∑ ∣ α ∣ = j ∂ α f ( a ) α ! h α . \frac1{j!}D^jf(a)[h,\ldots,h]=\sum_{|\alpha|=j}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha. j ! 1 D j f ( a ) [ h , … , h ] = ∣ α ∣ = j ∑ α ! ∂ α f ( a ) h α . 這實際計算了混合系數，而非從各坐標分別展開後猜測。特別地，二次混合項為 f x y h k f_{xy}hk f x y hk f x y h k / 2 f_{xy}hk/2 f x y hk /2
2 帶積分餘項的泰勒定理
設 f f f a a a a + h a+h a + h C k C^k C k k ≥ 1 k\ge1 k ≥ 1
f ( a + h ) = ∑ j = 0 k − 1 D j f ( a ) [ h , … , h ] j ! + 1 ( k − 1 ) ! ∫ 0 1 ( 1 − t ) k − 1 D k f ( a + t h ) [ h , … , h ] d t . f(a+h)=\sum_{j=0}^{k-1}\frac{D^jf(a)[h,\ldots,h]}{j!}
+\frac1{(k-1)!}\int_0^1(1-t)^{k-1}D^kf(a+th)[h,\ldots,h]dt. f ( a + h ) = j = 0 ∑ k − 1 j ! D j f ( a ) [ h , … , h ] + ( k − 1 )! 1 ∫ 0 1 ( 1 − t ) k − 1 D k f ( a + t h ) [ h , … , h ] d t . 其中 j = 0 j=0 j = 0 f ( a ) f(a) f ( a ) k = 1 k=1 k = 1 ( 1 − t ) (1-t) ( 1 − t )
在積分內減去常值 D k f ( a ) D^kf(a) D k f ( a )
f ( a + h ) = ∑ ∣ α ∣ ≤ k ∂ α f ( a ) α ! h α + o ( ∥ h ∥ k ) . f(a+h)=\sum_{|\alpha|\le k}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha+o(\|h\|^k). f ( a + h ) = ∣ α ∣ ≤ k ∑ α ! ∂ α f ( a ) h α + o ( ∥ h ∥ k ) . 為了給出 k k k C k + 1 C^{k+1} C k + 1 ∥ D k + 1 f ∥ ≤ M \|D^{k+1}f\|\le M ∥ D k + 1 f ∥ ≤ M k + 1 k+1 k + 1
∣ R k ( h ) ∣ ≤ M ∥ h ∥ k + 1 ( k + 1 ) ! . |R_k(h)|\le\frac{M\|h\|^{k+1}}{(k+1)!}. ∣ R k ( h ) ∣ ≤ ( k + 1 )! M ∥ h ∥ k + 1 . 該界涉及整條線段上的導數，而不只是起點處的導數。
歸納的實際步驟如下。寫 I k = ∫ 0 1 ( 1 − t ) k − 1 ϕ ( k ) ( t ) d t / ( k − 1 ) ! I_k=\int_0^1(1-t)^{k-1}\phi^{(k)}(t)dt/(k-1)! I k = ∫ 0 1 ( 1 − t ) k − 1 ϕ ( k ) ( t ) d t / ( k − 1 )! ϕ ( 1 ) = ϕ ( 0 ) + I 1 \phi(1)=\phi(0)+I_1 ϕ ( 1 ) = ϕ ( 0 ) + I 1 ϕ \phi ϕ
I k + 1 = [ ( 1 − t ) k ϕ ( k ) ( t ) k ! ] 0 1 + I k = − ϕ ( k ) ( 0 ) k ! + I k . I_{k+1}=\left[\frac{(1-t)^k\phi^{(k)}(t)}{k!}\right]_0^1+I_k=-\frac{\phi^{(k)}(0)}{k!}+I_k. I k + 1 = [ k ! ( 1 − t ) k ϕ ( k ) ( t ) ] 0 1 + I k = − k ! ϕ ( k ) ( 0 ) + I k . 所以每次替換 I k I_k I k ϕ ( k ) ( 0 ) / k ! \phi^{(k)}(0)/k! ϕ ( k ) ( 0 ) / k ! I k + 1 I_{k+1} I k + 1 a a a D k f ( a ) D^kf(a) D k f ( a ) D k f ( a ) [ h , … , h ] / k ! D^kf(a)[h,\ldots,h]/k! D k f ( a ) [ h , … , h ] / k !
∥ h ∥ k k ! sup 0 ≤ t ≤ 1 ∥ D k f ( a + t h ) − D k f ( a ) ∥ = o ( ∥ h ∥ k ) . \frac{\|h\|^k}{k!}\sup_{0\le t\le1}\|D^kf(a+th)-D^kf(a)\|=o(\|h\|^k). k ! ∥ h ∥ k 0 ≤ t ≤ 1 sup ∥ D k f ( a + t h ) − D k f ( a ) ∥ = o ( ∥ h ∥ k ) . 這是固定局部球上的算子範數連續性論證；Peano 結論不需要 C(k+1)。若另有 C(k+1)，對 ( 1 − t ) k / k ! (1-t)^k/k! ( 1 − t ) k / k ! 1 / ( k + 1 ) ! 1/(k+1)! 1/ ( k + 1 )!
1. 記號與線段條件
設d d d m m m U ⊂ R d U\subset\mathbb R^d U ⊂ R d a ∈ U a\in U a ∈ U h ∈ R d h\in\mathbb R^d h ∈ R d
[ a , a + h ] = { a + t h : 0 ≤ t ≤ 1 } ⊂ U . [a,a+h]=\{a+th:0\le t\le1\}\subset U. [ a , a + h ] = { a + t h : 0 ≤ t ≤ 1 } ⊂ U . 對於下面的積分餘項與Lagrange餘項公式，假設f : U → R f:U\to\mathbb R f : U → R V ⊂ U V\subset U V ⊂ U C m + 1 C^{m+1} C m + 1 m + 1 m+1 m + 1 V V V U U U h h h
多重指標 是α = ( α 1 , … , α d ) ∈ N 0 d \alpha=(\alpha_1,\ldots,\alpha_d)\in\mathbb N_0^d α = ( α 1 , … , α d ) ∈ N 0 d N 0 = { 0 , 1 , 2 , … } \mathbb N_0=\{0,1,2,\ldots\} N 0 = { 0 , 1 , 2 , … }
∣ α ∣ = ∑ j = 1 d α j , α ! = ∏ j = 1 d α j ! , h α = ∏ j = 1 d h j α j , ∂ α f = ∂ 1 α 1 ⋯ ∂ d α d f . |\alpha|=\sum_{j=1}^d\alpha_j,\qquad \alpha!=\prod_{j=1}^d\alpha_j!,\qquad h^\alpha=\prod_{j=1}^dh_j^{\alpha_j},\qquad \partial^\alpha f=\partial_1^{\alpha_1}\cdots\partial_d^{\alpha_d}f. ∣ α ∣ = j = 1 ∑ d α j , α ! = j = 1 ∏ d α j ! , h α = j = 1 ∏ d h j α j , ∂ α f = ∂ 1 α 1 ⋯ ∂ d α d f . 指數為零時，相應因子按1計算，即使該座標為零亦如此；0 ! = 1 0!=1 0 ! = 1 h 0 = 1 h^0=1 h 0 = 1 ∂ 0 f = f \partial^0f=f ∂ 0 f = f e j e_j e j j j j m m m
T m , a ( h ) = ∑ ∣ α ∣ ≤ m ∂ α f ( a ) α ! h α . T_{m,a}(h)=\sum_{|\alpha|\le m}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha. T m , a ( h ) = ∣ α ∣ ≤ m ∑ α ! ∂ α f ( a ) h α . 它的次數可以小於m m m
2. 一元工具與閉區間端點
在一元積分餘項定理中，「導數在閉區間上連續」是指區間內部的各階導數連續延伸至端點，端點處取相應的單側導數。Taylor多項式中的導數值就是這些端點值。能延伸至開區間上是一個充分但更強的條件。在當前多元情形中，這一條件自動滿足：開集{ t ∈ R : a + t h ∈ V } \{t\in\mathbb R:a+th\in V\} { t ∈ R : a + t h ∈ V } [ 0 , 1 ] [0,1] [ 0 , 1 ]
費馬必要條件。 在可微的內點局部最大值或最小值處，f ′ ( a ) = 0 f'(a)=0 f ′ ( a ) = 0
證明。 在局部最大值處，充分小的正增量給出非正差商，負增量給出非負差商。雙側導數的極限要同時滿足兩種符號，只能為零。對函數取負處理最小值。□ \square □
內點條件很重要：f ( x ) = x f(x)=x f ( x ) = x [ 0 , 1 ] [0,1] [ 0 , 1 ] x 3 x^3 x 3
Rolle 定理。 若 a < b a<b a < b f f f [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) f ( a ) = f ( b ) f(a)=f(b) f ( a ) = f ( b ) c ∈ ( a , b ) c\in(a,b) c ∈ ( a , b ) f ′ ( c ) = 0 f'(c)=0 f ′ ( c ) = 0
證明。 極值定理保證最大值和最小值存在。若函數為常數，任意內點均符合要求。否則某值不同於相等的端點值，因此最大值或最小值之一出現在內部。可在該處應用費馬條件。□ \square □
中值定理。 在相同的連續與可微條件下，不要求端點值相等，則存在 c ∈ ( a , b ) c\in(a,b) c ∈ ( a , b ) f ( b ) − f ( a ) = f ′ ( c ) ( b − a ) f(b)-f(a)=f'(c)(b-a) f ( b ) − f ( a ) = f ′ ( c ) ( b − a )
從 f f f f ( a ) + [ f ( b ) − f ( a ) ] ( x − a ) / ( b − a ) f(a)+[f(b)-f(a)](x-a)/(b-a) f ( a ) + [ f ( b ) − f ( a )] ( x − a ) / ( b − a ) f ′ f' f ′ ∣ f ′ ∣ ≤ M |f'|\le M ∣ f ′ ∣ ≤ M ∣ f ( y ) − f ( x ) ∣ ≤ M ∣ y − x ∣ |f(y)-f(x)|\le M|y-x| ∣ f ( y ) − f ( x ) ∣ ≤ M ∣ y − x ∣
在 a a a n n n T n ( x ) = ∑ k = 0 n f ( k ) ( a ) ( x − a ) k / k ! T_n(x)=\sum_{k=0}^nf^{(k)}(a)(x-a)^k/k! T n ( x ) = ∑ k = 0 n f ( k ) ( a ) ( x − a ) k / k ! n n n
定理：Taylor 積分餘項
設 n ≥ 0 n\ge0 n ≥ 0 f f f a a a x x x n + 1 n+1 n + 1
f ( x ) = T n ( x ) + 1 n ! ∫ a x f ( n + 1 ) ( t ) ( x − t ) n d t . f(x)=T_n(x)+\frac1{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt. f ( x ) = T n ( x ) + n ! 1 ∫ a x f ( n + 1 ) ( t ) ( x − t ) n d t . 因此，若該區間上 ∣ f ( n + 1 ) ( t ) ∣ ≤ M |f^{(n+1)}(t)|\le M ∣ f ( n + 1 ) ( t ) ∣ ≤ M ∣ f ( x ) − T n ( x ) ∣ ≤ M ∣ x − a ∣ n + 1 / ( n + 1 ) ! |f(x)-T_n(x)|\le M|x-a|^{n+1}/(n+1)! ∣ f ( x ) − T n ( x ) ∣ ≤ M ∣ x − a ∣ n + 1 / ( n + 1 )!
證明。 當 n = 0 n=0 n = 0 f ( x ) − f ( a ) = ∫ a x f ′ ( t ) d t f(x)-f(a)=\int_a^xf'(t)dt f ( x ) − f ( a ) = ∫ a x f ′ ( t ) d t n ≥ 1 n\ge1 n ≥ 1 u = ( x − t ) n / n ! u=(x-t)^n/n! u = ( x − t ) n / n ! d v = f ( n + 1 ) ( t ) d t dv=f^{(n+1)}(t)dt d v = f ( n + 1 ) ( t ) d t x x x a a a − f ( n ) ( a ) ( x − a ) n / n ! -f^{(n)}(a)(x-a)^n/n! − f ( n ) ( a ) ( x − a ) n / n ! ( n − 1 ) (n-1) ( n − 1 ) x < a x<a x < a ∣ x − t ∣ n |x-t|^n ∣ x − t ∣ n
此定理給出充分的正則性條件，並非最弱假設。Taylor 級數要在固定的 x x x f f f n → ∞ n\to\infty n → ∞ n n n
3. 重複方向求導與係數的完整計數
固定h h h D h q ( x ) = d d s q ( x + s h ) ∣ s = 0 D_hq(x)=\left.\frac{d}{ds}q(x+sh)\right|_{s=0} D h q ( x ) = d s d q ( x + s h ) s = 0 D h 0 f = f D_h^0f=f D h 0 f = f D h k + 1 f = D h ( D h k f ) D_h^{k+1}f=D_h(D_h^kf) D h k + 1 f = D h ( D h k f )
一階方向導數。 若q q q x x x C 1 C^1 C 1 x x x x + s h x+sh x + s h d d d s s s h j ≠ 0 h_j\ne0 h j = 0 j j j z j , s z_{j,s} z j , s
q ( x + s h ) − q ( x ) = ∑ j : h j ≠ 0 s h j ∂ j q ( z j , s ) . q(x+sh)-q(x)=\sum_{j:h_j\ne0}s h_j\partial_jq(z_{j,s}). q ( x + s h ) − q ( x ) = j : h j = 0 ∑ s h j ∂ j q ( z j , s ) . 當s → 0 s\to0 s → 0 z j , s → x z_{j,s}\to x z j , s → x h j = 0 h_j=0 h j = 0 s ≠ 0 s\ne0 s = 0
D h q ( x ) = ∑ j = 1 d h j ∂ j q ( x ) . (1) D_hq(x)=\sum_{j=1}^dh_j\partial_jq(x). \tag{1} D h q ( x ) = j = 1 ∑ d h j ∂ j q ( x ) . ( 1 ) 這個論證亦適用於負的s s s
為甚麼混合偏導數可以交換。 對C 2 C^2 C 2 q q q i , j i,j i , j u , v u,v u , v x x x
Δ = q ( x + u e i + v e j ) − q ( x + u e i ) − q ( x + v e j ) + q ( x ) . \Delta=q(x+ue_i+ve_j)-q(x+ue_i)-q(x+ve_j)+q(x). Δ = q ( x + u e i + v e j ) − q ( x + u e i ) − q ( x + v e j ) + q ( x ) . 先對[ 0 , u ] [0,u] [ 0 , u ] s ↦ q ( x + s e i + v e j ) − q ( x + s e i ) s\mapsto q(x+se_i+ve_j)-q(x+se_i) s ↦ q ( x + s e i + v e j ) − q ( x + s e i ) ∂ i q \partial_iq ∂ i q j j j
Δ u v = ∂ j ∂ i q ( x + ξ e i + η e j ) , 0 < ξ < u , 0 < η < v . \frac{\Delta}{uv}=\partial_j\partial_iq(x+\xi e_i+\eta e_j),\qquad 0<\xi<u,\quad0<\eta<v. uv Δ = ∂ j ∂ i q ( x + ξ e i + η e j ) , 0 < ξ < u , 0 < η < v . 交換兩個座標的角色，則在同一矩形的另一個點上有Δ / ( u v ) = ∂ i ∂ j q ( x + ξ ~ e i + η ~ e j ) \Delta/(uv)=\partial_i\partial_jq(x+\widetilde\xi e_i+\widetilde\eta e_j) Δ/ ( uv ) = ∂ i ∂ j q ( x + ξ e i + η e j ) u , v ↓ 0 u,v\downarrow0 u , v ↓ 0 ∂ j ∂ i q ( x ) = ∂ i ∂ j q ( x ) \partial_j\partial_iq(x)=\partial_i\partial_jq(x) ∂ j ∂ i q ( x ) = ∂ i ∂ j q ( x ) m + 1 m+1 m + 1
斷言。 對每個0 ≤ k ≤ m + 1 0\le k\le m+1 0 ≤ k ≤ m + 1 x ∈ V x\in V x ∈ V
D h k f ( x ) = ∑ ∣ α ∣ = k k ! α ! h α ∂ α f ( x ) . (2) D_h^kf(x)=\sum_{|\alpha|=k}\frac{k!}{\alpha!}h^\alpha\partial^\alpha f(x). \tag{2} D h k f ( x ) = ∣ α ∣ = k ∑ α ! k ! h α ∂ α f ( x ) . ( 2 ) 完整歸納證明。 當k = 0 k=0 k = 0 0 0 0 f ( x ) f(x) f ( x ) 0 ≤ k ≤ m 0\le k\le m 0 ≤ k ≤ m ∂ α f \partial^\alpha f ∂ α f C 1 C^1 C 1
D h k + 1 f ( x ) = ∑ ∣ α ∣ = k ∑ j = 1 d k ! α ! h α h j ∂ α + e j f ( x ) = ∑ ∣ β ∣ = k + 1 ( ∑ j : β j ≥ 1 k ! ( β − e j ) ! ) h β ∂ β f ( x ) . \begin{aligned}
D_h^{k+1}f(x)
&=\sum_{|\alpha|=k}\sum_{j=1}^d\frac{k!}{\alpha!}h^\alpha h_j\partial^{\alpha+e_j}f(x)\\
&=\sum_{|\beta|=k+1}\left(\sum_{j:\beta_j\ge1}\frac{k!}{(\beta-e_j)!}\right)h^\beta\partial^\beta f(x).
\end{aligned} D h k + 1 f ( x ) = ∣ α ∣ = k ∑ j = 1 ∑ d α ! k ! h α h j ∂ α + e j f ( x ) = ∣ β ∣ = k + 1 ∑ j : β j ≥ 1 ∑ ( β − e j )! k ! h β ∂ β f ( x ) . 固定β \beta β ( α , j ) = ( β − e j , j ) (\alpha,j)=(\beta-e_j,j) ( α , j ) = ( β − e j , j ) β j ≥ 1 \beta_j\ge1 β j ≥ 1
( β − e j ) ! = β ! β j , ∑ j : β j ≥ 1 k ! ( β − e j ) ! = k ! β ! ∑ j : β j ≥ 1 β j = k ! ( k + 1 ) β ! = ( k + 1 ) ! β ! . (\beta-e_j)!=\frac{\beta!}{\beta_j},\qquad
\sum_{j:\beta_j\ge1}\frac{k!}{(\beta-e_j)!}
=\frac{k!}{\beta!}\sum_{j:\beta_j\ge1}\beta_j
=\frac{k!(k+1)}{\beta!}
=\frac{(k+1)!}{\beta!}. ( β − e j )! = β j β ! , j : β j ≥ 1 ∑ ( β − e j )! k ! = β ! k ! j : β j ≥ 1 ∑ β j = β ! k ! ( k + 1 ) = β ! ( k + 1 )! . 未列入內層求和的座標滿足β j = 0 \beta_j=0 β j = 0 k + 1 k+1 k + 1 k ! / α ! k!/\alpha! k ! / α ! k k k j j j α j \alpha_j α j
現在令g ( t ) = f ( a + t h ) g(t)=f(a+th) g ( t ) = f ( a + t h ) t t t t t t a + t h a+th a + t h h h h f f f g ( 0 ) ( t ) = f ( a + t h ) g^{(0)}(t)=f(a+th) g ( 0 ) ( t ) = f ( a + t h )
g ( k ) ( t ) = D h k f ( a + t h ) = ∑ ∣ α ∣ = k k ! α ! h α ∂ α f ( a + t h ) , 0 ≤ k ≤ m + 1. (3) g^{(k)}(t)=D_h^kf(a+th)=\sum_{|\alpha|=k}\frac{k!}{\alpha!}h^\alpha\partial^\alpha f(a+th),\qquad0\le k\le m+1. \tag{3} g ( k ) ( t ) = D h k f ( a + t h ) = ∣ α ∣ = k ∑ α ! k ! h α ∂ α f ( a + t h ) , 0 ≤ k ≤ m + 1. ( 3 ) 這些導數都在包含[ 0 , 1 ] [0,1] [ 0 , 1 ]
4. Taylor定理與兩種精確餘項
定理。 在第1節的假設下，
f ( a + h ) = T m , a ( h ) + R m ( a , h ) , f(a+h)=T_{m,a}(h)+R_m(a,h), f ( a + h ) = T m , a ( h ) + R m ( a , h ) , 其中
R m ( a , h ) = 1 m ! ∫ 0 1 ( 1 − t ) m D h m + 1 f ( a + t h ) d t = ∑ ∣ α ∣ = m + 1 m + 1 α ! h α ∫ 0 1 ( 1 − t ) m ∂ α f ( a + t h ) d t . (4) \begin{aligned}
R_m(a,h)
&=\frac1{m!}\int_0^1(1-t)^mD_h^{m+1}f(a+th)\,dt\\
&=\sum_{|\alpha|=m+1}\frac{m+1}{\alpha!}h^\alpha\int_0^1(1-t)^m\partial^\alpha f(a+th)\,dt. \tag{4}
\end{aligned} R m ( a , h ) = m ! 1 ∫ 0 1 ( 1 − t ) m D h m + 1 f ( a + t h ) d t = ∣ α ∣ = m + 1 ∑ α ! m + 1 h α ∫ 0 1 ( 1 − t ) m ∂ α f ( a + t h ) d t . ( 4 ) 若h ≠ 0 h\ne0 h = 0 θ ∈ ( 0 , 1 ) \theta\in(0,1) θ ∈ ( 0 , 1 )
R m ( a , h ) = D h m + 1 f ( a + θ h ) ( m + 1 ) ! = ∑ ∣ α ∣ = m + 1 ∂ α f ( a + θ h ) α ! h α . (5) \begin{aligned}
R_m(a,h)
&=\frac{D_h^{m+1}f(a+\theta h)}{(m+1)!}\\
&=\sum_{|\alpha|=m+1}\frac{\partial^\alpha f(a+\theta h)}{\alpha!}h^\alpha. \tag{5}
\end{aligned} R m ( a , h ) = ( m + 1 )! D h m + 1 f ( a + θ h ) = ∣ α ∣ = m + 1 ∑ α ! ∂ α f ( a + θ h ) h α . ( 5 ) 積分形式的證明。 上面的一元定理可用於g g g
I r = 1 r ! ∫ 0 1 ( 1 − t ) r g ( r + 1 ) ( t ) d t ( 0 ≤ r ≤ m ) . I_r=\frac1{r!}\int_0^1(1-t)^rg^{(r+1)}(t)\,dt\qquad(0\le r\le m). I r = r ! 1 ∫ 0 1 ( 1 − t ) r g ( r + 1 ) ( t ) d t ( 0 ≤ r ≤ m ) . 微積分基本定理給出I 0 = g ( 1 ) − g ( 0 ) I_0=g(1)-g(0) I 0 = g ( 1 ) − g ( 0 ) 1 ≤ r ≤ m 1\le r\le m 1 ≤ r ≤ m
I r = [ ( 1 − t ) r r ! g ( r ) ( t ) ] 0 1 + 1 ( r − 1 ) ! ∫ 0 1 ( 1 − t ) r − 1 g ( r ) ( t ) d t = − g ( r ) ( 0 ) r ! + I r − 1 . \begin{aligned}
I_r&=\left[\frac{(1-t)^r}{r!}g^{(r)}(t)\right]_0^1+\frac1{(r-1)!}\int_0^1(1-t)^{r-1}g^{(r)}(t)\,dt\\
&=-\frac{g^{(r)}(0)}{r!}+I_{r-1}.
\end{aligned} I r = [ r ! ( 1 − t ) r g ( r ) ( t ) ] 0 1 + ( r − 1 )! 1 ∫ 0 1 ( 1 − t ) r − 1 g ( r ) ( t ) d t = − r ! g ( r ) ( 0 ) + I r − 1 . 所以I 0 = g ( 1 ) − ∑ k = 0 0 g ( k ) ( 0 ) / k ! I_0=g(1)-\sum_{k=0}^0g^{(k)}(0)/k! I 0 = g ( 1 ) − ∑ k = 0 0 g ( k ) ( 0 ) / k ! I r − 1 = g ( 1 ) − ∑ k = 0 r − 1 g ( k ) ( 0 ) / k ! I_{r-1}=g(1)-\sum_{k=0}^{r-1}g^{(k)}(0)/k! I r − 1 = g ( 1 ) − ∑ k = 0 r − 1 g ( k ) ( 0 ) / k ! I r = g ( 1 ) − ∑ k = 0 r g ( k ) ( 0 ) / k ! I_r=g(1)-\sum_{k=0}^rg^{(k)}(0)/k! I r = g ( 1 ) − ∑ k = 0 r g ( k ) ( 0 ) / k ! r = m r=m r = m m = 0 m=0 m = 0
∑ k = 0 m g ( k ) ( 0 ) k ! = ∑ k = 0 m ∑ ∣ α ∣ = k ∂ α f ( a ) α ! h α = T m , a ( h ) . \sum_{k=0}^m\frac{g^{(k)}(0)}{k!}=\sum_{k=0}^m\sum_{|\alpha|=k}\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha=T_{m,a}(h). k = 0 ∑ m k ! g ( k ) ( 0 ) = k = 0 ∑ m ∣ α ∣ = k ∑ α ! ∂ α f ( a ) h α = T m , a ( h ) . 由(3)的m + 1 m+1 m + 1 ( m + 1 ) ! / m ! = m + 1 (m+1)!/m!=m+1 ( m + 1 )! / m ! = m + 1 m m m m + 1 m+1 m + 1
嚴格中間點形式的證明。 令R = g ( 1 ) − ∑ k = 0 m g ( k ) ( 0 ) / k ! R=g(1)-\sum_{k=0}^mg^{(k)}(0)/k! R = g ( 1 ) − ∑ k = 0 m g ( k ) ( 0 ) / k !
H ( t ) = g ( t ) − ∑ k = 0 m g ( k ) ( 0 ) k ! t k − R t m + 1 . H(t)=g(t)-\sum_{k=0}^m\frac{g^{(k)}(0)}{k!}t^k-Rt^{m+1}. H ( t ) = g ( t ) − k = 0 ∑ m k ! g ( k ) ( 0 ) t k − R t m + 1 . 於是H ( 1 ) = 0 H(1)=0 H ( 1 ) = 0 0 ≤ r ≤ m 0\le r\le m 0 ≤ r ≤ m H ( r ) ( 0 ) = 0 H^{(r)}(0)=0 H ( r ) ( 0 ) = 0 [ 0 , 1 ] [0,1] [ 0 , 1 ] c 1 ∈ ( 0 , 1 ) c_1\in(0,1) c 1 ∈ ( 0 , 1 ) H ′ ( c 1 ) = 0 H'(c_1)=0 H ′ ( c 1 ) = 0 m ≥ 1 m\ge1 m ≥ 1 H ′ ( 0 ) = 0 H'(0)=0 H ′ ( 0 ) = 0 [ 0 , c 1 ] [0,c_1] [ 0 , c 1 ] c 2 ∈ ( 0 , c 1 ) c_2\in(0,c_1) c 2 ∈ ( 0 , c 1 ) H ′ ′ ( c 2 ) = 0 H''(c_2)=0 H ′′ ( c 2 ) = 0 1 ≤ r ≤ m 1\le r\le m 1 ≤ r ≤ m H ( r ) ( c r ) = 0 H^{(r)}(c_r)=0 H ( r ) ( c r ) = 0 c r > 0 c_r>0 c r > 0 H ( r ) ( 0 ) = 0 H^{(r)}(0)=0 H ( r ) ( 0 ) = 0 [ 0 , c r ] [0,c_r] [ 0 , c r ] c r + 1 ∈ ( 0 , c r ) c_{r+1}\in(0,c_r) c r + 1 ∈ ( 0 , c r ) H ( r + 1 ) ( c r + 1 ) = 0 H^{(r+1)}(c_{r+1})=0 H ( r + 1 ) ( c r + 1 ) = 0 m + 1 m+1 m + 1 θ = c m + 1 ∈ ( 0 , 1 ) \theta=c_{m+1}\in(0,1) θ = c m + 1 ∈ ( 0 , 1 )
0 = H ( m + 1 ) ( θ ) = g ( m + 1 ) ( θ ) − ( m + 1 ) ! R . 0=H^{(m+1)}(\theta)=g^{(m+1)}(\theta)-(m+1)!R. 0 = H ( m + 1 ) ( θ ) = g ( m + 1 ) ( θ ) − ( m + 1 )! R . 當m = 0 m=0 m = 0
(5)中每一項使用同一個θ \theta θ g g g
若h = 0 h=0 h = 0 g g g T m , a ( 0 ) = f ( a ) T_{m,a}(0)=f(a) T m , a ( 0 ) = f ( a ) R m ( a , 0 ) = 0 R_m(a,0)=0 R m ( a , 0 ) = 0 h ≠ 0 h\ne0 h = 0 m = 0 m=0 m = 0 f ( a + h ) − f ( a ) = ∫ 0 1 ∇ f ( a + t h ) ⋅ h d t f(a+h)-f(a)=\int_0^1\nabla f(a+th)\cdot h\,dt f ( a + h ) − f ( a ) = ∫ 0 1 ∇ f ( a + t h ) ⋅ h d t
5. 誤差界與較弱的局部Taylor結論
若對所有∣ α ∣ = m + 1 |\alpha|=m+1 ∣ α ∣ = m + 1 x x x ∣ ∂ α f ( x ) ∣ ≤ M |\partial^\alpha f(x)|\le M ∣ ∂ α f ( x ) ∣ ≤ M
∣ R m ( a , h ) ∣ ≤ M ∑ ∣ α ∣ = m + 1 ∣ h α ∣ α ! = M ( m + 1 ) ! ( ∑ j = 1 d ∣ h j ∣ ) m + 1 . (6) |R_m(a,h)|\le M\sum_{|\alpha|=m+1}\frac{|h^\alpha|}{\alpha!}
=\frac{M}{(m+1)!}\left(\sum_{j=1}^d|h_j|\right)^{m+1}. \tag{6} ∣ R m ( a , h ) ∣ ≤ M ∣ α ∣ = m + 1 ∑ α ! ∣ h α ∣ = ( m + 1 )! M ( j = 1 ∑ d ∣ h j ∣ ) m + 1 . ( 6 ) 這是因為∫ 0 1 ( 1 − t ) m d t = 1 / ( m + 1 ) \int_0^1(1-t)^m\,dt=1/(m+1) ∫ 0 1 ( 1 − t ) m d t = 1/ ( m + 1 ) m + 1 m+1 m + 1 ∑ j ∣ h j ∣ \sum_j|h_j| ∑ j ∣ h j ∣ a a a ∣ D h m + 1 f ( a + t h ) ∣ ≤ K ∥ h ∥ m + 1 |D_h^{m+1}f(a+th)|\le K\|h\|^{m+1} ∣ D h m + 1 f ( a + t h ) ∣ ≤ K ∥ h ∥ m + 1 ∣ R m ∣ ≤ K ∥ h ∥ m + 1 / ( m + 1 ) ! |R_m|\le K\|h\|^{m+1}/(m+1)! ∣ R m ∣ ≤ K ∥ h ∥ m + 1 / ( m + 1 )!
C m + 1 C^{m+1} C m + 1 m ≥ 1 m\ge1 m ≥ 1 f f f a a a C m C^m C m
f ( a + h ) = T m , a ( h ) + o ( ∥ h ∥ m ) ( h → 0 ) . (7) f(a+h)=T_{m,a}(h)+o(\|h\|^m)\qquad(h\to0). \tag{7} f ( a + h ) = T m , a ( h ) + o ( ∥ h ∥ m ) ( h → 0 ) . ( 7 ) 證明這一較弱結論不需要多加一階導數：對充分短的線段應用階數為m − 1 m-1 m − 1 D h m f ( a ) D_h^mf(a) D h m f ( a ) D h m f ( a ) / m ! D_h^mf(a)/m! D h m f ( a ) / m ! m m m
ε ( h ) m ! ( ∑ j = 1 d ∣ h j ∣ ) m , ε ( h ) = max ∣ α ∣ = m sup 0 ≤ t ≤ 1 ∣ ∂ α f ( a + t h ) − ∂ α f ( a ) ∣ ⟶ 0. \frac{\varepsilon(h)}{m!}\left(\sum_{j=1}^d|h_j|\right)^m,\qquad
\varepsilon(h)=\max_{|\alpha|=m}\sup_{0\le t\le1}|\partial^\alpha f(a+th)-\partial^\alpha f(a)|\longrightarrow0. m ! ε ( h ) ( j = 1 ∑ d ∣ h j ∣ ) m , ε ( h ) = ∣ α ∣ = m max 0 ≤ t ≤ 1 sup ∣ ∂ α f ( a + t h ) − ∂ α f ( a ) ∣ ⟶ 0. 這樣的α \alpha α ∥ t h ∥ ≤ ∥ h ∥ \|th\|\le\|h\| ∥ t h ∥ ≤ ∥ h ∥ a a a ∑ j ∣ h j ∣ ≤ d ∥ h ∥ \sum_j|h_j|\le\sqrt d\|h\| ∑ j ∣ h j ∣ ≤ d ∥ h ∥ m = 0 m=0 m = 0 f ( a + h ) = f ( a ) + o ( 1 ) f(a+h)=f(a)+o(1) f ( a + h ) = f ( a ) + o ( 1 ) m + 1 m+1 m + 1
3 混合展開例題
令 f ( x , y ) = e x cos y f(x,y)=e^x\cos y f ( x , y ) = e x cos y
T 3 = 1 + x + x 2 − y 2 2 + x 3 6 − x y 2 2 . T_3=1+x+\frac{x^2-y^2}{2}+\frac{x^3}{6}-\frac{xy^2}{2}. T 3 = 1 + x + 2 x 2 − y 2 + 6 x 3 − 2 x y 2 . 一種求法是將 1 + x + x 2 / 2 + x 3 / 6 + ⋯ 1+x+x^2/2+x^3/6+\cdots 1 + x + x 2 /2 + x 3 /6 + ⋯ 1 − y 2 / 2 + ⋯ 1-y^2/2+\cdots 1 − y 2 /2 + ⋯ f x y y ( 0 , 0 ) = − 1 f_{xyy}(0,0)=-1 f x y y ( 0 , 0 ) = − 1 − 1 / ( 1 ! 2 ! ) = − 1 / 2 -1/(1!2!)=-1/2 − 1/ ( 1 ! 2 !) = − 1/2
為了得到具體餘項界，限制到 ( t x , t y ) (tx,ty) ( t x , t y ) 0 ≤ t ≤ 1 0\le t\le1 0 ≤ t ≤ 1 e ∣ x ∣ ( ∣ x ∣ + ∣ y ∣ ) 4 e^{|x|}(|x|+|y|)^4 e ∣ x ∣ ( ∣ x ∣ + ∣ y ∣ ) 4
∣ f ( x , y ) − T 3 ( x , y ) ∣ ≤ e ∣ x ∣ ( ∣ x ∣ + ∣ y ∣ ) 4 24 . |f(x,y)-T_3(x,y)|\le\frac{e^{|x|}(|x|+|y|)^4}{24}. ∣ f ( x , y ) − T 3 ( x , y ) ∣ ≤ 24 e ∣ x ∣ ( ∣ x ∣ + ∣ y ∣ ) 4 . 此界刻意求簡單而非尖銳。它適用於坐標同時變化，並顯示四階誤差的量級。
4 解析性的充分判據
導數增長判據。 設 n ≥ 1 n\ge1 n ≥ 1 r 0 , C , R > 0 r_0,C,R>0 r 0 , C , R > 0 f ∈ C ∞ ( B ( a , r 0 ) ) f\in C^\infty(B(a,r_0)) f ∈ C ∞ ( B ( a , r 0 )) j = 0 j=0 j = 0 ∥ D 0 f ( x ) ∥ = ∣ f ( x ) ∣ \|D^0f(x)\|=|f(x)| ∥ D 0 f ( x ) ∥ = ∣ f ( x ) ∣ j ≥ 0 j\ge0 j ≥ 0 x x x ∥ D j f ( x ) ∥ ≤ C j ! R − j \|D^jf(x)\|\le Cj!R^{-j} ∥ D j f ( x ) ∥ ≤ C j ! R − j ∥ h ∥ < min ( r 0 , R ) \|h\|<\min(r_0,R) ∥ h ∥ < min ( r 0 , R )
f ( a + h ) = ∑ j = 0 ∞ D j f ( a ) [ h , … , h ] j ! . f(a+h)=\sum_{j=0}^\infty\frac{D^jf(a)[h,\ldots,h]}{j!}. f ( a + h ) = j = 0 ∑ ∞ j ! D j f ( a ) [ h , … , h ] . 證明與精確的收斂區域。 閉線段位於球內。k 次積分餘項的界為 C ( ∥ h ∥ / R ) k + 1 C(\|h\|/R)^{k+1} C ( ∥ h ∥/ R ) k + 1 C ( ∥ h ∥ / R ) j C(\|h\|/R)^j C ( ∥ h ∥/ R ) j 0 < ρ < min ( r 0 , R ) 0<\rho<\min(r_0,R) 0 < ρ < min ( r 0 , R ) ∥ h ∥ ≤ ρ \|h\|\le\rho ∥ h ∥ ≤ ρ
這尚未說明所有單個單項式的絕對值之和在同一個歐氏球範圍內收斂。把坐標基向量代入 D j f ( a ) D^jf(a) D j f ( a ) ∣ α ∣ = j |\alpha|=j ∣ α ∣ = j ∣ ∂ α f ( a ) ∣ ≤ C j ! R − j |\partial^\alpha f(a)|\le Cj!R^{-j} ∣ ∂ α f ( a ) ∣ ≤ C j ! R − j
∑ ∣ α ∣ = j ∣ ∂ α f ( a ) α ! h α ∣ ≤ C R − j ∑ ∣ α ∣ = j j ! α ! ∏ i ∣ h i ∣ α i = C ( ∑ i ∣ h i ∣ R ) j . \sum_{|\alpha|=j}\left|\frac{\partial^\alpha f(a)}{\alpha!}h^\alpha\right|\le C R^{-j}\sum_{|\alpha|=j}\frac{j!}{\alpha!}\prod_i|h_i|^{\alpha_i}=C\left(\frac{\sum_i|h_i|}{R}\right)^j. ∣ α ∣ = j ∑ α ! ∂ α f ( a ) h α ≤ C R − j ∣ α ∣ = j ∑ α ! j ! i ∏ ∣ h i ∣ α i = C ( R ∑ i ∣ h i ∣ ) j . 因此未分組的多重指標級數至少在 ∥ h ∥ < r 0 \|h\|<r_0 ∥ h ∥ < r 0 ∑ i ∣ h i ∣ < R \sum_i|h_i|<R ∑ i ∣ h i ∣ < R 0 < r < min ( r 0 / n , R / n ) 0<r<\min(r_0/\sqrt n,R/n) 0 < r < min ( r 0 / n , R / n ) ∣ h i ∣ ≤ r |h_i|\le r ∣ h i ∣ ≤ r f f f a a a
僅在 a a a
以下逆向結論中，取正整數 n n n a ∈ R n a\in\mathbb R^n a ∈ R n s > 0 s>0 s > 0 c α c_\alpha c α ∣ h i ∣ < s |h_i|<s ∣ h i ∣ < s f ( a + h ) = ∑ α c α h α f(a+h)=\sum_\alpha c_\alpha h^\alpha f ( a + h ) = ∑ α c α h α C ∞ C^\infty C ∞
逆向結論：絕對係數控制允許逐項求導。 假設存在 s > 0 s>0 s > 0
A = ∑ α ∈ N 0 n ∣ c α ∣ s ∣ α ∣ < ∞ . A=\sum_{\alpha\in\mathbb N_0^n}|c_\alpha|s^{|\alpha|}<\infty. A = α ∈ N 0 n ∑ ∣ c α ∣ s ∣ α ∣ < ∞. 把這個絕對和解釋為有限子和的上確界，或按總次數遞增枚舉多重指標。對每個閉立方體 ∣ h i ∣ ≤ r < s |h_i|\le r<s ∣ h i ∣ ≤ r < s m = ∣ α ∣ ≥ 1 m=|\alpha|\ge1 m = ∣ α ∣ ≥ 1 α j = 0 \alpha_j=0 α j = 0 0 < r < s 0<r<s 0 < r < s
∣ α j c α h α − e j ∣ ≤ ∣ c α ∣ s m s m ( r / s ) m − 1 . |\alpha_jc_\alpha h^{\alpha-e_j}|\le\frac{|c_\alpha|s^m}{s}\,m(r/s)^{m-1}. ∣ α j c α h α − e j ∣ ≤ s ∣ c α ∣ s m m ( r / s ) m − 1 . 當 q = r / s < 1 q=r/s<1 q = r / s < 1 m q m − 1 m q^{m-1} m q m − 1 ∑ m ≥ 1 m q m − 1 = 1 / ( 1 − q ) 2 \sum_{m\ge1}m q^{m-1}=1/(1-q)^2 ∑ m ≥ 1 m q m − 1 = 1/ ( 1 − q ) 2
為甚麼這種收斂確實給出導數？令 P N P_N P N N N N [ − r , r ] [-r,r] [ − r , r ] P N P_N P N ∂ j P N \partial_jP_N ∂ j P N
對給定的更高求導階數 ℓ \ell ℓ m ℓ m^\ell m ℓ ∣ h i ∣ ≤ r |h_i|\le r ∣ h i ∣ ≤ r ∣ c α ∣ s m s − ℓ m ℓ ( r / s ) m − ℓ |c_\alpha|s^m s^{-\ell}m^\ell(r/s)^{m-\ell} ∣ c α ∣ s m s − ℓ m ℓ ( r / s ) m − ℓ r r r s s s h = 0 h=0 h = 0 ∂ α f ( a ) = α ! c α \partial^\alpha f(a)=\alpha!c_\alpha ∂ α f ( a ) = α ! c α
定義
f ( x ) = { e − 1 / x 2 , x ≠ 0 , 0 , x = 0. f(x)=\begin{cases}e^{-1/x^2},&x\ne0,\\0,&x=0.\end{cases} f ( x ) = { e − 1/ x 2 , 0 , x = 0 , x = 0. 下面給出光滑但不解析的完整論證。對 x ≠ 0 x\ne0 x = 0 t = 1 / x t=1/x t = 1/ x
P 0 ( t ) = 1 , P j + 1 ( t ) = 2 t 3 P j ( t ) − t 2 P j ′ ( t ) . P_0(t)=1,\qquad P_{j+1}(t)=2t^3P_j(t)-t^2P_j'(t). P 0 ( t ) = 1 , P j + 1 ( t ) = 2 t 3 P j ( t ) − t 2 P j ′ ( t ) . 由 d t / d x = − t 2 dt/dx=-t^2 d t / d x = − t 2 P j ( 1 / x ) e − 1 / x 2 P_j(1/x)e^{-1/x^2} P j ( 1/ x ) e − 1/ x 2 P j + 1 ( 1 / x ) e − 1 / x 2 P_{j+1}(1/x)e^{-1/x^2} P j + 1 ( 1/ x ) e − 1/ x 2 p ≥ 0 p\ge0 p ≥ 0 N > p N>p N > p u > 0 u>0 u > 0 e u ≥ u N / N ! e^u\ge u^N/N! e u ≥ u N / N ! u → ∞ u\to\infty u → ∞ u p e − u ≤ N ! u p − N → 0 u^pe^{-u}\le N!u^{p-N}\to0 u p e − u ≤ N ! u p − N → 0 1 / x 1/x 1/ x e − 1 / x 2 e^{-1/x^2} e − 1/ x 2 1 / x 1/x 1/ x
對非零 x x x g j ( x ) = P j ( 1 / x ) e − 1 / x 2 g_j(x)=P_j(1/x)e^{-1/x^2} g j ( x ) = P j ( 1/ x ) e − 1/ x 2 g j ( 0 ) = 0 g_j(0)=0 g j ( 0 ) = 0 g j g_j g j
g j ( x ) − g j ( 0 ) x = 1 x P j ( 1 / x ) e − 1 / x 2 ⟶ 0. \frac{g_j(x)-g_j(0)}x=\frac1xP_j(1/x)e^{-1/x^2}\longrightarrow0. x g j ( x ) − g j ( 0 ) = x 1 P j ( 1/ x ) e − 1/ x 2 ⟶ 0. 在零點以外，其導數是 g j + 1 g_{j+1} g j + 1 g j + 1 ( 0 ) = 0 g_{j+1}(0)=0 g j + 1 ( 0 ) = 0 g 0 = f g_0=f g 0 = f f ∈ C ∞ ( R ) f\in C^\infty(\mathbb R) f ∈ C ∞ ( R ) j j j f ( j ) ( 0 ) = 0 f^{(j)}(0)=0 f ( j ) ( 0 ) = 0 x x x f ( x ) > 0 f(x)>0 f ( x ) > 0
練習與完整解答
E1 基礎。 求 log ( 1 + x + y ) \log(1+x+y) log ( 1 + x + y )
S1. 令 u = x + y u=x+y u = x + y u − u 2 / 2 = x + y − x 2 / 2 − x y − y 2 / 2 u-u^2/2=x+y-x^2/2-xy-y^2/2 u − u 2 /2 = x + y − x 2 /2 − x y − y 2 /2 1 + x + y > 0 1+x+y>0 1 + x + y > 0 ∣ x ∣ + ∣ y ∣ < 1 |x|+|y|<1 ∣ x ∣ + ∣ y ∣ < 1 ∣ u ∣ < 1 |u|<1 ∣ u ∣ < 1
E2 混合項。 展開 ( x + 2 y ) 3 (x+2y)^3 ( x + 2 y ) 3
S2. 展開為 x 3 + 6 x 2 y + 12 x y 2 + 8 y 3 x^3+6x^2y+12xy^2+8y^3 x 3 + 6 x 2 y + 12 x y 2 + 8 y 3 3 ! 3! 3 ! 2 ! 1 ! 2!1! 2 ! 1 ! 1 ! 2 ! 1!2! 1 ! 2 ! 3 ! 3! 3 !
E3 餘項。 若從 a a a a + h a+h a + h ∥ D 3 f ∥ ≤ 12 \|D^3f\|\le12 ∥ D 3 f ∥ ≤ 12 ∥ h ∥ = 0.1 \|h\|=0.1 ∥ h ∥ = 0.1
S3. 界為 12 ( 0.1 ) 3 / 3 ! = 0.002 12(0.1)^3/3!=0.002 12 ( 0.1 ) 3 /3 ! = 0.002 a a a
E4 解析性。 解釋為甚麼 f ( x , y ) = e − 1 / x 2 f(x,y)=e^{-1/x^2} f ( x , y ) = e − 1/ x 2 x ≠ 0 x\ne0 x = 0 x = 0 x=0 x = 0 ( 0 , 0 ) (0,0) ( 0 , 0 )
S4. 含 y y y x x x ( 0 , 0 ) (0,0) ( 0 , 0 ) x ≠ 0 x\ne0 x = 0 y = 0 y=0 y = 0
總結
多重線性導數組織同時發生的方向變化，多重指標正確計算混合項。積分型泰勒餘項提供小 o 及定量界。解析性需要無窮階的鄰域控制或收斂級數論證，而非僅在一點存在各階導數。
