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Teiloro polinomo laipsniui didėjant, jis tampa artimesnis aproksimuojamai funkcijai. Ši iliustracija parodo
sin
x
{\displaystyle \sin x}
1 , 3 , 5 , 7 , 9 , 11 ir 13 . Teiloro eilutė – 1712 m. B. Teiloro aprašyta formulė,[1] polinomu galima aproksimuoti bet kurią tolydžią, realaus ar kompleksinio skaičiaus a aplinkoje be galo diferencijuojamą funkciją.
Formulė:
f
(
x
)
≈
∑
n
=
0
N
f
(
n
)
(
a
)
n
!
(
x
−
a
)
n
+
R
N
{\displaystyle f(x)\approx \sum _{n=0}^{N}{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}+R_{N}}
x pakankamai artimas a .Čia n! yra n faktorialas , o
f
(
n
)
(
a
)
{\displaystyle f^{(n)}(a)}
f išvestinę taške a .
Kai
a
=
0
{\displaystyle a=0}
Koliną Makloreną ).
Bendruoju atveju, Teiloro eilutės nebūtinai konverguoja į funkcijos reikšmę tame taške.
Eksponentė:
e
x
=
∑
n
=
0
∞
x
n
n
!
=
1
+
x
+
x
2
2
!
+
x
3
3
!
+
x
4
4
!
+
⋯
+
x
n
n
!
su visais
x
{\displaystyle \mathrm {e} ^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{x^{4} \over 4!}+\cdots +{x^{n} \over n!}{\text{ su visais }}x\!}
Pavyzdžiui:
e
3
=
2
,
718281828
3
=
20
,
08553692
,
{\displaystyle e^{3}=2,718281828^{3}=20,08553692,}
e
3
=
∑
n
=
0
8
3
n
n
!
=
1
+
3
+
3
2
2
⋅
1
+
3
3
3
⋅
2
⋅
1
+
3
4
4
⋅
3
⋅
2
+
3
5
5
⋅
4
⋅
3
⋅
2
+
3
6
6
!
+
3
7
7
!
+
3
8
8
!
=
{\displaystyle \mathrm {e} ^{3}=\sum _{n=0}^{8}{\frac {3^{n}}{n!}}=1+3+{\frac {3^{2}}{2\cdot 1}}+{\frac {3^{3}}{3\cdot 2\cdot 1}}+{3^{4} \over 4\cdot 3\cdot 2}+{3^{5} \over 5\cdot 4\cdot 3\cdot 2}+{3^{6} \over 6!}+{3^{7} \over 7!}+{3^{8} \over 8!}=}
=
1
+
3
+
4
,
5
+
4
,
5
+
3
,
375
+
2
,
025
+
1
,
0125
+
0
,
433928571
+
0
,
162723214
=
20
,
00915179.
{\displaystyle =1+3+4,5+4,5+3,375+2,025+1,0125+0,433928571+0,162723214=20,00915179.}
Natūrinis logaritmas:
ln
(
1
−
x
)
=
−
∑
n
=
1
∞
x
n
n
=
−
(
x
1
1
+
x
2
2
+
x
3
3
+
x
4
4
+
.
.
.
+
x
n
n
)
su
−
1
≤
x
<
1
{\displaystyle \ln(1-x)=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}=-({x^{1} \over 1}+{x^{2} \over 2}+{x^{3} \over 3}+{x^{4} \over 4}+...+{x^{n} \over n}){\text{ su }}-1\leq x<1}
Pavyzdžiui:
ln
(
1
−
0
,
8
)
=
−
1.6094379124341
;
{\displaystyle \ln(1-0,8)=-1.6094379124341;}
ln
(
1
−
0
,
8
)
=
−
0
,
8
−
0
,
8
2
2
−
0
,
8
3
3
−
0
,
8
4
4
−
0
,
8
5
5
−
0
,
8
6
6
−
0
,
8
7
7
=
{\displaystyle \ln(1-0,8)=-0,8-{0,8^{2} \over 2}-{0,8^{3} \over 3}-{0,8^{4} \over 4}-{0,8^{5} \over 5}-{0,8^{6} \over 6}-{0,8^{7} \over 7}=}
=
−
0.8
−
0.32
−
0.170
(
6
)
−
0.1024
−
0.065536
−
0.043690
(
6
)
−
0.029959314285714
=
−
1.532252647619.
{\displaystyle =-0.8-0.32-0.170(6)-0.1024-0.065536-0.043690(6)-0.029959314285714=-1.532252647619.}
ln
(
1
+
x
)
=
∑
n
=
1
∞
(
−
1
)
n
+
1
x
n
n
=
x
−
x
2
2
+
x
3
3
−
x
4
4
+
x
5
5
.
.
.
su
−
1
<
x
≤
1
{\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=x-{x^{2} \over 2}+{x^{3} \over 3}-{x^{4} \over 4}+{x^{5} \over 5}...{\text{ su }}-1<x\leq 1}
Pavyzdžiui:
ln
(
1
+
1
)
=
0
,
69314718
;
{\displaystyle \ln(1+1)=0,69314718;}
ln
(
1
+
x
)
=
∑
n
=
1
∞
(
−
1
)
n
+
1
x
n
n
=
1
−
1
2
2
!
+
1
3
3
!
−
1
4
4
!
+
1
5
5
!
−
1
6
!
+
1
7
!
−
1
8
!
+
1
9
!
=
{\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=1-{1^{2} \over 2!}+{1^{3} \over 3!}-{1^{4} \over 4!}+{1^{5} \over 5!}-{1 \over 6!}+{1 \over 7!}-{1 \over 8!}+{1 \over 9!}=}
=
1
−
0
,
5
+
0
,
16
(
6
)
−
0
,
0416
(
6
)
+
0
,
0083
(
3
)
−
0
,
00138
(
8
)
+
0
,
000198412
−
0
,
000024801
+
0
,
000002755
=
0
,
632120811.
{\displaystyle =1-0,5+0,16(6)-0,0416(6)+0,0083(3)-0,00138(8)+0,000198412-0,000024801+0,000002755=0,632120811.}
Teisingai taip (nors atsakymai teisingu ir neteisingu atveju beveik tiek pat skiriasi nuo ln(2)):
ln
(
1
+
1
)
=
∑
n
=
1
∞
(
−
1
)
n
+
1
x
n
n
=
1
−
1
2
2
+
1
3
3
−
1
4
4
+
1
5
5
−
1
6
+
1
7
−
1
8
+
1
9
=
{\displaystyle \ln(1+1)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=1-{1^{2} \over 2}+{1^{3} \over 3}-{1^{4} \over 4}+{1^{5} \over 5}-{1 \over 6}+{1 \over 7}-{1 \over 8}+{1 \over 9}=}
=0.74563492063492.
Ilgesnė eilutė duoda tokį atsakymą:
ln
(
1
+
1
)
=
∑
n
=
1
∞
(
−
1
)
n
+
1
x
n
n
=
1
−
1
2
2
+
1
3
3
−
1
4
4
+
1
5
5
−
1
6
+
1
7
−
1
8
+
1
9
−
1
10
+
{\displaystyle \ln(1+1)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=1-{1^{2} \over 2}+{1^{3} \over 3}-{1^{4} \over 4}+{1^{5} \over 5}-{1 \over 6}+{1 \over 7}-{1 \over 8}+{1 \over 9}-{1 \over 10}+}
+
1
11
−
1
12
+
1
13
−
1
14
+
1
15
−
1
16
+
1
17
−
1
18
+
1
19
−
1
20
=
{\displaystyle +{1 \over 11}-{1 \over 12}+{1 \over 13}-{1 \over 14}+{1 \over 15}-{1 \over 16}+{1 \over 17}-{1 \over 18}+{1 \over 19}-{1 \over 20}=}
=0.64563492063492+0.0231364825405=0.6687714031754273. Kvadratinė šaknis:
1
+
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
)
!
(
1
−
2
n
)
n
!
2
4
n
x
n
for
|
x
|
<
1
{\displaystyle {\sqrt {1+x}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)n!^{2}4^{n}}}x^{n}{\text{ for }}|x|<1\!}
Trigonometrinės funkcijos (x čia reiškiamas radianais):
sin
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
+
1
)
!
x
2
n
+
1
=
x
−
x
3
3
!
+
x
5
5
!
−
⋯
su visais
x
{\displaystyle \sin x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}\quad =x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots {\text{ su visais }}x\!}
cos
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
)
!
x
2
n
=
1
−
x
2
2
!
+
x
4
4
!
−
⋯
su visais
x
{\displaystyle \cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}\quad =1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots {\text{ su visais }}x\!}
tan
x
=
∑
n
=
1
∞
B
2
n
(
−
4
)
n
(
1
−
4
n
)
(
2
n
)
!
x
2
n
−
1
=
x
+
x
3
3
+
2
x
5
15
+
⋯
su
|
x
|
<
π
2
{\displaystyle \tan x=\sum _{n=1}^{\infty }{\frac {B_{2n}(-4)^{n}(1-4^{n})}{(2n)!}}x^{2n-1}\quad =x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+\cdots {\text{ su }}|x|<{\frac {\pi }{2}}\!}
kur
B
n
{\displaystyle B_{n}}
Atvirkštinės trigonometrinės funkcijos:
arcsin
x
=
∑
n
=
0
∞
(
2
n
)
!
4
n
(
n
!
)
2
(
2
n
+
1
)
x
2
n
+
1
su
|
x
|
<
1
{\displaystyle \arcsin x=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}{\text{ su }}|x|<1\!}
arcsin
x
=
x
+
x
3
2
⋅
3
+
1
⋅
3
x
5
2
⋅
4
⋅
5
+
1
⋅
3
⋅
5
x
7
2
⋅
4
⋅
6
⋅
7
+
.
.
.
+
1
⋅
3
⋅
5...
(
2
n
−
1
)
x
2
n
+
1
2
⋅
4
⋅
6...
(
2
n
)
(
2
n
+
1
)
+
.
.
.
,
|
x
|
<
1
{\displaystyle \arcsin x=x+{\frac {x^{3}}{2\cdot 3}}+{\frac {1\cdot 3x^{5}}{2\cdot 4\cdot 5}}+{\frac {1\cdot 3\cdot 5x^{7}}{2\cdot 4\cdot 6\cdot 7}}+...+{\frac {1\cdot 3\cdot 5...(2n-1)x^{2n+1}}{2\cdot 4\cdot 6...(2n)(2n+1)}}+...,\;\;|x|<1}
arccos
x
=
π
2
−
[
x
+
x
3
2
⋅
3
+
1
⋅
3
x
5
2
⋅
4
⋅
5
+
1
⋅
3
⋅
5
x
7
2
⋅
4
⋅
6
⋅
7
+
.
.
.
+
1
⋅
3
⋅
5...
(
2
n
−
1
)
x
2
n
+
1
2
⋅
4
⋅
6...
(
2
n
)
(
2
n
+
1
)
+
.
.
.
]
,
|
x
|
<
1
{\displaystyle \arccos x={\frac {\pi }{2}}-{\Big [}x+{\frac {x^{3}}{2\cdot 3}}+{\frac {1\cdot 3x^{5}}{2\cdot 4\cdot 5}}+{\frac {1\cdot 3\cdot 5x^{7}}{2\cdot 4\cdot 6\cdot 7}}+...+{\frac {1\cdot 3\cdot 5...(2n-1)x^{2n+1}}{2\cdot 4\cdot 6...(2n)(2n+1)}}+...{\Big ]},\;\;|x|<1}
arctan
x
=
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
x
2
n
+
1
su
|
x
|
≤
1
{\displaystyle \arctan x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}{\text{ su }}|x|\leq 1\!}
arctan
x
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
.
.
.
+
(
−
1
)
n
x
2
n
+
1
2
n
+
1
±
.
.
.
,
|
x
|
<
1
{\displaystyle \arctan x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+...+(-1)^{n}{\frac {x^{2n+1}}{2n+1}}\pm ...,\;\;|x|<1}
arctan
x
=
±
π
2
−
1
x
+
1
3
x
3
−
1
5
x
5
+
1
7
x
7
−
.
.
.
+
(
−
1
)
n
+
1
1
(
2
n
+
1
)
x
2
n
+
1
±
.
.
.
∗
,
|
x
|
>
1
{\displaystyle \arctan x=\pm {\frac {\pi }{2}}-{\frac {1}{x}}+{\frac {1}{3x^{3}}}-{\frac {1}{5x^{5}}}+{\frac {1}{7x^{7}}}-...+(-1)^{n+1}{\frac {1}{(2n+1)x^{2n+1}}}\pm ...*,\;\;|x|>1}
* Pirmas narys
π
2
{\displaystyle {\frac {\pi }{2}}}
x >1 ir su ženklu "-" kai
x
<
−
1.
{\displaystyle x<-1.}
arccot
x
=
π
2
−
[
x
−
x
3
3
+
x
5
5
−
x
7
7
+
.
.
.
+
(
−
1
)
n
x
2
n
+
1
2
n
+
1
±
.
.
.
]
,
|
x
|
<
1
{\displaystyle \operatorname {arccot} x={\frac {\pi }{2}}-{\Big [}x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+...+(-1)^{n}{\frac {x^{2n+1}}{2n+1}}\pm ...{\Big ]},\;\;|x|<1}
![{\displaystyle \arccos x={\frac {\pi }{2}}-{\Big [}x+{\frac {x^{3}}{2\cdot 3}}+{\frac {1\cdot 3x^{5}}{2\cdot 4\cdot 5}}+{\frac {1\cdot 3\cdot 5x^{7}}{2\cdot 4\cdot 6\cdot 7}}+...+{\frac {1\cdot 3\cdot 5...(2n-1)x^{2n+1}}{2\cdot 4\cdot 6...(2n)(2n+1)}}+...{\Big ]},\;\;|x|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25926cee19f619374d00bc79e9c0d739c3517d13)
![{\displaystyle \operatorname {arccot} x={\frac {\pi }{2}}-{\Big [}x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+...+(-1)^{n}{\frac {x^{2n+1}}{2n+1}}\pm ...{\Big ]},\;\;|x|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfb0b8be466b22c4eec7a807486025da5976eea1)
