Taylor's theorem is named after Brook Taylor, who stated a version of it in 1715, although an earlier version of the result was already mentioned in 1671 by James Gregory.
Teorema ini mendapat nama dari matematikawan Brook Taylor, yang menyatakannya pada tahun 1712, meskipun hasilnya sudah ditemukan pertama kali tahun 1671 oleh James Gregory.
0. (x)(x-a),\quad \lim _{x\to a}h_{1}(x)=0.}
0 }\quad \lim _{x\to a}h_{k}(x)=0} .
f ( x ) − P 1 ( x )h 1 ( x ) ( x − a ) .
f ( a ) + x f ′ ( x )
(x)+h_{2}(x)(x-a)^{2},\quad \lim _{x\to a}h_{2}(x)=0.}
0 , (x)(x-a)^{k},\quad \lim _{x\to a}h_{k}(x)=0,} setelah itu pPk.
The precise statement of the most basic version of Taylor's theorem is as follows: The polynomial appearing in Taylor's theorem is the k {\textstyle {\boldsymbol {k}}} -th
(\xi )}{n!}}(x-\xi )^{n}\cdot {\frac {G(x)-G(a)}{G'(\xi )}}.} Pernyataan dari versi paling dasar dari teorema Taylor adalah sebagai berikut: Teorema Taylor.
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a )
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a ) k + h k ( x ) ( x − a )
k (x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}} of the function f at the point a .
k , {2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}+h_{k}(x)(x-a)^{k},} dan lim x → a h k ( x )
Taylor's theorem describes the asymptotic behavior of the remainder term R k ( x )
Teorema Taylor menggambarkan perilaku asimtotik dari istilah sisa R k ( x )
f ( x ) − P k ( x ) , (x)=f(x)-P_{k}(x),} which is the approximation error when approximating f with its Taylor polynomial.
f ( x ) − P k ( x ) , (x)=f(x)-P_{k}(x),} Salah satu kesalahan aproksimasi saat mendekati nilai f dengan teorema taylor.
Using the little-o notation, the statement in Taylor's theorem reads as R k ( x )o ( | x − a | k ) , x → a .
Menggunakan notasi o kecil, pernyataan dalam teorema Taylor dibaca sebagai berikut R k ( x )
These refinements of Taylor's theorem are usually proved using the mean value theorem, whence the name.
Perbaikan teorema Taylor tersebut terbiasa dibuktikan menggunakan teorema nilai rata-rata, dari mana namanya.
f ( k + 1 ) ( ξ )
f ( n + 1 ) ( ξ )
The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of Lebesgue integration theory for the full generality.
Pernyataan dalam bentuk integral dari sisa lebih maju dari yang sebelumnya, dan membutuhkan pemahaman tentang teori integrasi Lebesgue untuk pencarian penuh.
However, it holds also in the sense of Riemann integral provided the (k + 1)th derivative of f is continuous on the closed interval
Namun, itu berlaku juga dalam arti integral Riemann asalkan nilai (k + 1) keturunan dari f kontinu pada interval tertutup [a,x].
≤ R k ( x ) ≤ Q ( x − a ) k + 1 ( k + 1 ) ! , }{(k+1)!}}\leq R_{k}(x)\leq Q{\frac {(x-a)^{k+1}}{(k+1)!}},} if x > a, and a similar estimate if x < a. This is a simple consequence of the Lagrange form of the remainder.
Kemudian R k ( x )f ( k + 1 ) ( ξ L ) ( k + 1 ) ! ( x − a ) k + 1 (x)={\frac {f^{(k+1)}(\xi _{L})}{(k+1)!}}(x-a)^{k+1}} untuk beberapa bilangan real pada nilai ξL di antara nilai a dan x.
In particular, if | f ( k + 1 ) ( x ) | ≤ M (x)|\leq M} on an interval I(a − r,a + r) with some r > 0 , then | R k ( x ) | ≤ M | x − a | k + 1 ( k + 1 ) ! ≤ M r k + 1 ( k + 1 ) ! (x)|\leq M{\frac {|x-a|^{k+1}}{(k+1)!}}\leq M{\frac {r^{k+1}}{(k+1)!}}} for all x∈(a
Kemudian R k ( x )f ( k + 1 ) ( ξ L ) ( k + 1 ) ! ( x − a ) k + 1 (x)={\frac {f^{(k+1)}(\xi _{L})}{(k+1)!}}(x-a)^{k+1}} untuk beberapa bilangan real pada nilai ξL di antara nilai a dan x.
e ξ ( k + 1 ) !
∫ a x f ( k + 1 ) ( t )
As a conclusion, Taylor's theorem leads to the approximation e x1 + x + x 2 2 !
Contoh sederhana penerapan teorema Taylor adalah hampiran fungsi eksponensial ex di dekat x0: e x ≈ 1 + x + x 2 2 !
^{\infty }c_{k}(x-a)^{k}=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+\cdots ,\qquad |x-a|<r.}
{2!}}(x-a)^{2}+\dots {\frac {f^{(n)}(a)}{n!}}(x-a)^{n}.}
(x)=\sum _{j=k+1}^{\infty }c_{j}(x-a)^{j}=(x-a)^{k}h_{k}(x),\qquad |x-a|<r.}
(x)=\int _{a}^{x}{\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k}\,dt.}
∫ 0 1 ( 1 − t ) | β | − 1 D β f ( a + t ( x − a ) )
( x − t ) n d tf ( n + 1 ) ( ξ ) ∫ a x ( x − t ) n n !
f ( a ) + ∂ f ∂ x 1 ( a ) v 1 + ∂ f ∂ x 2 ( a ) v 2 + ∂ 2 f ∂ x 1 2 ( a ) v 1 2 2 !
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a ) k + h k ( x ) ( x − a )
{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}.}
{2!}}(x-a)^{2}+\dots {\frac {f^{(n)}(a)}{n!}}(x-a)^{n}.}
Set F {\textstyle F} and G {\textstyle G} to be F ( x )f ( x ) − ∑ k0 n − 1 f ( k ) ( a ) k ! ( x − a ) k F(x)=f(x)-\sum _{k=0}^{n-1}{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}\end{aligned}}} G ( x )
k , {2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}+h_{k}(x)(x-a)^{k},} dan lim x → a h k ( x )
− f ( n − 1 ) ( a ) ( n − 1 ) ! ( a − a ) n − 10 G ( a )
f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − a ) n + 1 .
− f ( n − 1 ) ( a ) ( n − 2 ) ! ( a − a ) n − 20 F'(a)=f'(a)-f'(a)-{\frac {f''(a)}{(2-1)!}}(a-a)^{(2-1)}-...-{\frac {f^{(n-1)}(a)}{(n-2)!}}(a-a)^{n-2}=0\end{aligned}}}
f ( a ) + f ′ ( a ) 1 ! ( x − a ) + f ( 2 ) ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( n ) ( a ) n ! ( x − a ) n + R n ( x ) , {1!}}(x-a)+{\frac {f^{(2)}(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}+R_{n}(x),} di mana fungsi sisa Rn memenuhi ketidaksamaan (dikenal sebagai estimasi Cauchy) | R n ( x ) | ≤ M n r n + 1 ( n + 1 ) !
n ( a − a ) n − 10 ⋮ G ( n − 1 ) ( a )
f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − a ) n + 1 .
F ( n − 1 ) ( a )
f ( n + 1 ) ( ξ )
f ( x ) − ∑ k0 n − 1 f ( k ) ( a )
f ( n + 1 ) ( ξ )
.1 (c)=n(n-1)...1} , so: F ( n ) ( c ) G ( n ) ( c )f ( n ) ( c ) n ( n − 1 )
(x)={\frac {f^{(n+1)}(\xi )}{(n+1)!}}(x-a)^{n+1}.}
⋯ 1f ( n ) ( c ) n ! .
∫ a x f ( k + 1 ) ( t )
(c)}{G^{(n)}(c)}}={\frac {f^{(n)}(c)}{n(n-1)\cdots 1}}={\frac {f^{(n)}(c)}{n!}}.}
(x)={\frac {f^{(n+1)}(\xi )}{n!}}(x-\xi )^{n}\cdot {\frac {G(x)-G(a)}{G'(\xi )}}.}
∑ k0 n − 1 f ( k ) ( a ) k ! ( x − a ) k + f ( n ) ( c ) n ! ( x − a ) n , f(x)=\sum _{k=0}^{n-1}{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}+{\frac {f^{(n)}(c)}{n!}}(x-a)^{n}\end{aligned}},} or because c naa} eventually: f ( x )∑ k0 n f ( k ) ( a ) k ! ( x − a ) k .
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a ) k (x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}} dari fungsi f pada titik a. Teorema Taylor adalah Teorema yang digunakan dalam arti, jika terdapat suatu fungsi hk : R → R dan k ke order teorema ``p dengan sedemikian rupa f ( x )p ( x ) + h k ( x ) ( x − a ) k , lim x → a h k ( x )
f ( t ) + f ′ ( t ) ( x − t ) + f ″ ( t ) 2 ! ( x − t ) 2 + ⋯ + f ( k ) ( t ) k ! ( x − t ) k .
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a ) k + h k ( x ) ( x − a )
{2!}}(x-t)^{2}+\cdots +{\frac {f^{(k)}(t)}{k!}}(x-t)^{k}.}
{2!}}(x-a)^{2}+\dots {\frac {f^{(n)}(a)}{n!}}(x-a)^{n}.}
+ ( f ″ ( t ) ( x − t ) − f ′ ( t ) ) + ( f ( 3 ) ( t ) 2 ! ( x − t ) 2 − f ( 2 ) ( t ) 1 ! ( x − t ) ) + ⋯ ⋯ + ( f ( k + 1 ) ( t )
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a ) k + h k ( x ) ( x − a )
k ! ( x − t ) k − f ( k ) ( t ) ( k − 1 ) ! ( x − t ) k − 1 )f ( k + 1 ) ( t ) k ! ( x − t ) k , F'(t)={}&f'(t)+{\big (}f''(t)(x-t)-f'(t){\big )}+\left({\frac {f^{(3)}(t)}{2!}}(x-t)^{2}-{\frac {f^{(2)}(t)}{1!}}(x-t)\right)+\cdots \\&\cdots +\left({\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k}-{\frac {f^{(k)}(t)}{(k-1)!}}(x-t)^{k-1}\right)={\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k},\end{aligned}}} plug it into (★★★) and rearrange terms to find that R k ( x )f ( k + 1 ) ( ξ ) k ! ( x − ξ ) k G ( x ) − G ( a )
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a ) k (x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}} dari fungsi f pada titik a. Teorema Taylor adalah Teorema yang digunakan dalam arti, jika terdapat suatu fungsi hk : R → R dan k ke order teorema ``p dengan sedemikian rupa f ( x )p ( x ) + h k ( x ) ( x − a ) k , lim x → a h k ( x )
∫ a t f ( k + 1 ) ( s )
∫ a x f ( k + 1 ) ( t )
f ( a ) + ( x f ′ ( x )
f ( a ) + x f ′ ( x )
− a f ′ ( a ) ) − ∫ a x t f ″ ( t )
− a f ′ ( a ) − ∫ a x t f ″ ( t )
d tf ( a ) + x ( f ′ ( a ) + ∫ a x f ″ ( t )
d tf ( a ) + ∫ a x x f ″ ( t )
d t ) − a f ′ ( a ) − ∫ a x t f ″ ( t ) d tf ( a ) + ( x − a ) f ′ ( a ) + ∫ a x ( x − t ) f ″ ( t )
d t + x f ′ ( a ) − a f ′ ( a ) − ∫ a x t f ″ ( t )
d t , f(x)&=f(a)+{\Big (}xf'(x)-af'(a){\Big )}-\int _{a}^{x}tf''(t)\,dt\\&=f(a)+x\left(f'(a)+\int _{a}^{x}f''(t)\,dt\right)-af'(a)-\int _{a}^{x}tf''(t)\,dt\\&=f(a)+(x-a)f'(a)+\int _{a}^{x}\,(x-t)f''(t)\,dt,\end{aligned}}} which is exactly Taylor's theorem with remainder in the integral form in the case k1 .
f(x)&=f(a)+xf'(x)-af'(a)-\int _{a}^{x}\,tf''(t)\,dt\\&=f(a)+\int _{a}^{x}\,xf''(t)\,dt+xf'(a)-af'(a)-\int _{a}^{x}\,tf''(t)\,dt\\&=f(a)+(x-a)f'(a)+\int _{a}^{x}\,(x-t)f''(t)\,dt.\end{aligned}}} Persamaan pertama diperoleh dengan memisalkan uf ′ ( t ) dandvdt; persamaan kedua didapatkan dengan mencatat bahwa ∫ a x x f ″ ( t )
( x − a ) k + 1 + ∫ a x f ( k + 2 ) ( t ) ( k + 1 ) ! ( x − t ) k + 1 d t .
( x − t ) n d tf ( n + 1 ) ( ξ ) ∫ a x ( x − t ) n n !
d j d t j f ( u ( t ) )d j d t j f ( a + t ( x − a ) )
d tf ( a ) + ( x − a ) f ′ ( a ) + ∫ a x ( x − t ) f ″ ( t )
f ( a ) + ∑ 1 ≤ | α | ≤ k 1 α ! ( D α f ) ( a ) ( x − a ) α + ∑ | α |k + 1 k + 1 α ! ( x − a ) α ∫ 0 1 ( 1 − t ) k ( D α f ) ( a + t ( x − a ) )
f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( k ) ( a ) k ! ( x − a ) k + h k ( x ) ( x − a )