함수 f f f a a a
f ( x ) = ∑ k = 0 ∞ f ( k ) ( a ) k ! ( x − a ) k f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k f ( x ) = k = 0 ∑ ∞ k ! f ( k ) ( a ) ( x − a ) k 이 성립하기 위한 필요충분조건은, 오차항 R n ( x ) R_n(x) R n ( x ) ∃ ε n ∈ ( a , x ) \exist \varepsilon_n \in (a, x) ∃ ε n ∈ ( a , x )
lim n → ∞ R n ( x ) = lim n → ∞ f ( n + 1 ) ( ε n ) ( n + 1 ) ! ( x − a ) ( n + 1 ) = 0 \lim_{n \to \infty} R_n(x) = \lim_{n \to \infty} \frac {f^{(n+1)}(\varepsilon_n)}{(n+1)!}(x-a)^(n+1) = 0 n → ∞ lim R n ( x ) = n → ∞ lim ( n + 1 )! f ( n + 1 ) ( ε n ) ( x − a ) ( n + 1 ) = 0 이어야 한다.
1. 유한차 테일러 공식 증명 ¶ 1.1. 보조함수 F F F ¶ n n n
P n ( t ) = ∑ k = 0 n f ( k ) ( a ) k ! ( t − a ) k P_n(t) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(t-a)^k P n ( t ) = k = 0 ∑ n k ! f ( k ) ( a ) ( t − a ) k 함수 f ( x ) f(x) f ( x ) P n ( x ) P_n(x) P n ( x )
E = f ( x ) − P n ( x ) E = f(x) - P_n(x) E = f ( x ) − P n ( x ) 다음과 같이 F ( a ) = F ( x ) = 0 F(a) = F(x) = 0 F ( a ) = F ( x ) = 0 F F F
F ( t ) = f ( t ) − P n ( t ) − E ( t − a ) ( n + 1 ) ( x − a ) ( n + 1 ) F(t) = f(t) - P_n(t) - E \frac{(t-a)^{(n+1)}}{(x-a)^{(n+1)}} F ( t ) = f ( t ) − P n ( t ) − E ( x − a ) ( n + 1 ) ( t − a ) ( n + 1 ) 위 보조함수는 a a a
P n ( t ) = f ( a ) + f ′ ( a ) ( t − a ) + f ′ ′ ( a ) 2 ! ( t − a ) 2 + ⋯ + f ( n ) n ! ( t − a ) n P_n(t) = f(a) + f'(a)(t-a) + \frac {f''(a)}{2!}(t-a)^2 + \cdots + \frac {f{(n)}}{n!}(t-a)^n P n ( t ) = f ( a ) + f ′ ( a ) ( t − a ) + 2 ! f ′′ ( a ) ( t − a ) 2 + ⋯ + n ! f ( n ) ( t − a ) n 로 표현할 수 있기 때문에 n차 미분하면 f ( n ) ( a ) f^{(n)}(a) f ( n ) ( a )
E ( t − a ) n + 1 ( x − a ) n + 1 E\frac {(t-a)^{n+1}}{(x-a)^{n+1}} E ( x − a ) n + 1 ( t − a ) n + 1 은 j ≤ n j \leq n j ≤ n t = a t=a t = a
1.2. 잔차항 중간점 ¶ F ( a ) = 0 F(a) = 0 F ( a ) = 0 F ( x ) = 0 F(x) = 0 F ( x ) = 0 ∃ c 1 ∈ H ( a , x ) , F ′ ( c 1 ) = 0 \exist c_1 \in \mathcal H (a, x), \ F'(c_1) = 0 ∃ c 1 ∈ H ( a , x ) , F ′ ( c 1 ) = 0 c 1 c_1 c 1 F ′ ( a ) = 0 F'(a) = 0 F ′ ( a ) = 0 ∃ c 2 ∈ H ( a , c 1 ) , F ′ ′ ( c 2 ) = 0 \exist c_2 \in \mathcal H (a, c_1), \ F''(c_2)=0 ∃ c 2 ∈ H ( a , c 1 ) , F ′′ ( c 2 ) = 0 c 2 c_2 c 2 n n n ∃ ξ n ∈ H ( a , x ) , F ( n + 1 ) ( ξ n ) = 0 \exist \xi_n \in \mathcal H (a, x), \ F^{(n+1)}(\xi_n) = 0 ∃ ξ n ∈ H ( a , x ) , F ( n + 1 ) ( ξ n ) = 0 ξ n \xi_n ξ n
1.3. F F F n + 1 n+1 n + 1 ¶ F ( t ) F(t) F ( t ) n + 1 n+1 n + 1
f ( t ) f(t) f ( t ) n + 1 n+1 n + 1
P n ( t ) P_n(t) P n ( t ) t t t n n n
d n + 1 d t n + 1 ( t − a ) n + 1 = ( n + 1 ) ! \frac {d^{n+1}}{dt^{n+1}}(t-a)^{n+1} = (n+1)! d t n + 1 d n + 1 ( t − a ) n + 1 = ( n + 1 )!
F ( n + 1 ) ( t ) = f ( n + 1 ) ( t ) − E ( n + 1 ) ! ( x − a ) n + 1 F^{(n+1)}(t) = f^{(n+1)}(t) - E\frac{(n+1)!}{(x-a)^{n+1}} F ( n + 1 ) ( t ) = f ( n + 1 ) ( t ) − E ( x − a ) n + 1 ( n + 1 )! 그런데 ξ n ∈ H ( a , x ) \xi_n \in \mathcal H(a,x) ξ n ∈ H ( a , x ) F ( n + 1 ) ( ξ n ) = 0 F^{(n+1)}(\xi_n) = 0 F ( n + 1 ) ( ξ n ) = 0
F ( n + 1 ) ( ξ n ) = f ( n + 1 ) ( ξ n ) − E ( n + 1 ) ! ( x − a ) n + 1 = 0 ⇒ E = f ( n + 1 ) ( ξ n ) ( n + 1 ) ! ( x − a ) n + 1 = f ( x ) − P n ( x ) f ( x ) = P n ( x ) + f ( n + 1 ) ( ξ n ) ( n + 1 ) ! ( x − a ) n + 1 ⇒ f ( x ) = ∑ k = 0 n f ( k ) ( a ) k ! ( t − a ) k + f ( n + 1 ) ( ξ n ) ( n + 1 ) ! ( x − a ) n + 1 유한차 테일러 공식 \begin{gathered}
F^{(n+1)}(\xi_n) = f^{(n+1)}(\xi_n) - E\frac{(n+1)!}{(x-a)^{n+1}} = 0 \\
\\
\Rightarrow E = \frac{f^{(n+1)}(\xi_n)}{(n+1)!}(x-a)^{n+1} = f(x) - P_n(x) \\
\\
f(x) = P_n(x) + \frac{f^{(n+1)}(\xi_n)}{(n+1)!}(x-a)^{n+1}\\
\\
\Rightarrow \underset{\text{유한차 테일러 공식}}{\boxed{f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(t-a)^k + \frac{f^{(n+1)}(\xi_n)}{(n+1)!}(x-a)^{n+1}}}
\end{gathered} F ( n + 1 ) ( ξ n ) = f ( n + 1 ) ( ξ n ) − E ( x − a ) n + 1 ( n + 1 )! = 0 ⇒ E = ( n + 1 )! f ( n + 1 ) ( ξ n ) ( x − a ) n + 1 = f ( x ) − P n ( x ) f ( x ) = P n ( x ) + ( n + 1 )! f ( n + 1 ) ( ξ n ) ( x − a ) n + 1 ⇒ 유한차 테일러 공식 f ( x ) = k = 0 ∑ n k ! f ( k ) ( a ) ( t − a ) k + ( n + 1 )! f ( n + 1 ) ( ξ n ) ( x − a ) n + 1 2. 충분성 증명 ¶ R n ( x ) = f ( n + 1 ) ( ξ n ) ( n + 1 ) ! ( x − a ) n + 1 R_n(x) = \frac{f^{(n+1)}(\xi_n)}{(n+1)!}(x-a)^{n+1} R n ( x ) = ( n + 1 )! f ( n + 1 ) ( ξ n ) ( x − a ) n + 1 라고 표기하고,
lim n → ∞ f ( n + 1 ) ( ξ n ) ( n + 1 ) ! ( x − a ) n + 1 = 0 \lim_{n \to \infty}\frac{f^{(n+1)}(\xi_n)}{(n+1)!}(x-a)^{n+1} = 0 n → ∞ lim ( n + 1 )! f ( n + 1 ) ( ξ n ) ( x − a ) n + 1 = 0 이라고 가정하자. 유한차 테일러 공식의 극한을 구하면,
lim n → ∞ f ( x ) = lim n → ∞ ∑ k = 0 n f ( k ) ( a ) k ! ( x − a ) k + lim n → ∞ f ( n + 1 ) ( ξ n ) ( n + 1 ) ! ( x − a ) n + 1 = ∑ k = 0 ∞ f ( k ) ( a ) k ! ( x − a ) k \begin{aligned}
\lim_{n \to \infty} f(x) &= \lim_{n \to \infty} \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k + \lim_{n \to \infty} \frac{f^{(n+1)}(\xi_n)}{(n+1)!}(x-a)^{n+1} \\
&= \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k
\end{aligned} n → ∞ lim f ( x ) = n → ∞ lim k = 0 ∑ n k ! f ( k ) ( a ) ( x − a ) k + n → ∞ lim ( n + 1 )! f ( n + 1 ) ( ξ n ) ( x − a ) n + 1 = k = 0 ∑ ∞ k ! f ( k ) ( a ) ( x − a ) k 따라서,
f ( x ) = ∑ k = 0 ∞ f ( k ) ( a ) k ! ( x − a ) k 테일러 급수 f(x) = \underset{\text{테일러 급수}}{\boxed{\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k}} f ( x ) = 테일러 급수 k = 0 ∑ ∞ k ! f ( k ) ( a ) ( x − a ) k 정리하면, R n ( x ) → 0 ⇒ R_n(x) \to 0 \Rightarrow R n ( x ) → 0 ⇒
3. 필요성 증명 ¶ 유한차 테일러 급수를 S n ( x ) S_n(x) S n ( x )
S n ( x ) = ∑ k = 0 n f ( k ) ( a ) k ! ( x − a ) k S_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k S n ( x ) = k = 0 ∑ n k ! f ( k ) ( a ) ( x − a ) k 이제,
f ( x ) = lim n → ∞ S n ( x ) f(x) = \lim_{n \to \infty} S_n(x) f ( x ) = n → ∞ lim S n ( x ) 가 성립한다고 가정하자. 유한차 테일러 공식에서 잔차항의 극한을 구하면,
R n ( x ) = f ( x ) − S n ( x ) → lim n → ∞ R n ( x ) = f ( x ) − lim n → ∞ S n ( x ) = f ( x ) − f ( x ) = 0 \begin{aligned}
& R_n(x) = f(x) - S_n(x) \\
&\rightarrow \lim_{n \to \infty} R_n(x) = f(x) - \lim_{n \to \infty} S_n(x) = f(x) - f(x) = 0
\end{aligned} R n ( x ) = f ( x ) − S n ( x ) → n → ∞ lim R n ( x ) = f ( x ) − n → ∞ lim S n ( x ) = f ( x ) − f ( x ) = 0 정리하면, S n ( x ) → f ( x ) ⇒ R n ( x ) → 0 S_n(x) \to f(x) \Rightarrow R_n(x) \to 0 S n ( x ) → f ( x ) ⇒ R n ( x ) → 0
충분성과 필요성의 증명을 정리하면 아래와 같은 정리가 완성된다.
lim n → ∞ f ( n + 1 ) ( ξ n ) ( n + 1 ) ! ( x − a ) n + 1 = 0 ⇔ f ( x ) = ∑ k = 0 ∞ f ( k ) ( a ) k ! ( x − a ) k \lim_{n \to \infty} \frac{f^{(n+1)}(\xi_n)}{(n+1)!}(x-a)^{n+1} = 0 \\
\Leftrightarrow f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k n → ∞ lim ( n + 1 )! f ( n + 1 ) ( ξ n ) ( x − a ) n + 1 = 0 ⇔ f ( x ) = k = 0 ∑ ∞ k ! f ( k ) ( a ) ( x − a ) k ■ \blacksquare ■