Teiloro eilutė (neprofesionalams)
Išnagrinėsime vieną svarbiausių formulių matematinės analizės, turinčią daugybę pritaikymų tiek pačiame analize, tiek kitose artimose disciplinose. 1. Teiloro formulė
keisti
Teiloro teorema *. Tegu funkcija f(x) turi taške a ir kai kurioje jo aplinkoje n+1** eilės išvestines. Tegu x - bet kokia argumento reikšmė iš nurodytos aplinkos ,
x
≠
a
.
{\displaystyle x\neq a.}
Tada tarp taškų a ir x yra taškas
ξ
{\displaystyle \xi }
f
(
x
)
=
f
(
a
)
+
f
′
(
a
)
1
!
(
x
−
a
)
+
f
″
(
a
)
2
!
(
x
−
a
)
2
+
f
‴
(
a
)
3
!
(
x
−
a
)
3
+
.
.
.
+
f
(
n
)
(
a
)
n
!
(
x
−
a
)
n
+
f
(
n
+
1
)
(
ξ
)
(
n
+
1
)
!
(
x
−
a
)
n
+
1
.
(
1
)
{\displaystyle f(x)=f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+...+{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}+{\frac {f^{(n+1)}(\xi )}{(n+1)!}}(x-a)^{n+1}.\quad (1)}
Įrodymas. Tegu
ϕ
(
x
,
a
)
{\displaystyle \phi (x,\;a)}
x atžvilgiu ir laipsnio n , stovintis dešinėje pusėje formulėje (1), t. y. tarsim
ϕ
(
x
,
a
)
=
f
(
a
)
+
f
′
(
a
)
1
!
(
x
−
a
)
+
f
″
(
a
)
2
!
(
x
−
a
)
2
+
f
‴
(
a
)
3
!
(
x
−
a
)
3
+
.
.
.
+
f
(
n
)
(
a
)
n
!
(
x
−
a
)
n
.
{\displaystyle \phi (x,\;a)=f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+...+{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}.}
(Jis vadinasi Teiloro polinomu laipsnio n funkcijai f(x) .)
Toliau pažymėsime per
R
n
+
1
(
x
)
{\displaystyle R_{n+1}(x)}
R
n
+
1
(
x
)
=
f
(
x
)
−
ϕ
(
x
,
a
)
.
{\displaystyle R_{n+1}(x)=f(x)-\phi (x,\;a).}
Teorema bus įrodyta, jeigu nustatysime, kad
R
n
+
1
(
x
)
=
f
(
n
+
1
)
(
ξ
)
(
n
+
1
)
!
(
x
−
a
)
n
+
1
,
a
<
ξ
<
x
.
{\displaystyle R_{n+1}(x)={\frac {f^{(n+1)}(\xi )}{(n+1)!}}(x-a)^{n+1},\;\;a<\xi <x.}
Fiksuojame bet kokią x reikšmę iš nurodytos aplinkos (iš a aplinkos). Konkretumo dėlei laikome, kad x>a . Įvedame kintamąją reikšmė t besikeičiančia atkarpoje
a
≤
t
≤
x
,
{\displaystyle a\leq t\leq x,}
a ; x ] pagalbinę funkciją
F
(
t
)
=
f
(
x
)
−
ϕ
(
x
,
t
)
−
(
x
−
t
)
n
+
1
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
.
(
2
)
{\displaystyle F(t)=f(x)-\phi (x,\;t)-{\frac {(x-t)^{n+1}R_{n+1}(x)}{(x-a)^{n+1}}}.\quad (2)}
Funkcija F (t ) tenkina atkarpoje [a ; x ] visas sąlygas Rolio teoremos :
1) iš (2) formulės ir iš sąlygų uždėtų funkcijai f (x ), seka, kad F (t ) netrūki ir diferencijuojama atkarpoje [a ; x ], nes f(t) ir jos išvestinės iki n -tos eilės tolydžios ir diferencijuojamos atkarpoje [a ; x ];
2) parinkę formulėje (2)
t
=
a
,
{\displaystyle t=a,}
F
(
a
)
=
f
(
x
)
−
ϕ
(
x
,
a
)
−
(
x
−
a
)
n
+
1
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
=
R
n
+
1
(
x
)
−
R
n
+
1
(
x
)
=
0.
{\displaystyle F(a)=f(x)-\phi (x,\;a)-{\frac {(x-a)^{n+1}R_{n+1}(x)}{(x-a)^{n+1}}}=R_{n+1}(x)-R_{n+1}(x)=0.}
Parinkę formulėje (2)
t
=
x
,
{\displaystyle t=x,}
F
(
x
)
=
f
(
x
)
−
ϕ
(
x
,
x
)
−
(
x
−
x
)
n
+
1
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
=
{\displaystyle F(x)=f(x)-\phi (x,\;x)-{\frac {(x-x)^{n+1}R_{n+1}(x)}{(x-a)^{n+1}}}=}
=
f
(
x
)
−
f
(
x
)
−
f
′
(
x
)
1
!
(
x
−
x
)
−
f
″
(
x
)
2
!
(
x
−
x
)
2
−
f
‴
(
x
)
3
!
(
x
−
x
)
3
−
.
.
.
−
f
(
n
)
(
x
)
n
!
(
x
−
x
)
n
−
(
x
−
x
)
n
+
1
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
=
0.
{\displaystyle =f(x)-f(x)-{\frac {f'(x)}{1!}}(x-x)-{\frac {f''(x)}{2!}}(x-x)^{2}-{\frac {f'''(x)}{3!}}(x-x)^{3}-...-{\frac {f^{(n)}(x)}{n!}}(x-x)^{n}-{\frac {(x-x)^{n+1}R_{n+1}(x)}{(x-a)^{n+1}}}=0.}
Tokiu budu, sąlyga
F
(
a
)
=
F
(
x
)
{\displaystyle F(a)=F(x)}
Pagal Rolio teoremą atkarpos [a ; x ] viduje yra toks taškas
ξ
,
{\displaystyle \xi ,}
F
′
(
ξ
)
=
0.
(
3
)
{\displaystyle F'(\xi )=0.\quad (3)}
Apskaičiuosime išvestinę
F
′
(
t
)
.
{\displaystyle F'(t).}
t , turime
F
(
t
)
=
f
(
x
)
−
ϕ
(
x
,
t
)
−
(
x
−
t
)
n
+
1
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
=
f
(
x
)
−
f
(
t
)
−
f
′
(
t
)
1
!
(
x
−
t
)
−
f
″
(
t
)
2
!
(
x
−
t
)
2
−
f
‴
(
t
)
3
!
(
x
−
t
)
3
−
.
.
.
−
f
(
n
)
(
t
)
n
!
(
x
−
t
)
n
−
(
x
−
t
)
n
+
1
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
;
{\displaystyle F(t)=f(x)-\phi (x,\;t)-{\frac {(x-t)^{n+1}R_{n+1}(x)}{(x-a)^{n+1}}}=f(x)-f(t)-{\frac {f'(t)}{1!}}(x-t)-{\frac {f''(t)}{2!}}(x-t)^{2}-{\frac {f'''(t)}{3!}}(x-t)^{3}-...-{\frac {f^{(n)}(t)}{n!}}(x-t)^{n}-{\frac {(x-t)^{n+1}R_{n+1}(x)}{(x-a)^{n+1}}};}
F
′
(
t
)
=
−
f
′
(
t
)
+
f
′
(
t
)
1
!
−
f
″
(
t
)
1
!
(
x
−
t
)
+
f
″
(
t
)
2
!
2
(
x
−
t
)
−
f
‴
(
t
)
2
!
(
x
−
t
)
2
+
f
‴
(
t
)
3
!
3
(
x
−
t
)
2
−
f
(
4
)
(
t
)
3
!
(
x
−
t
)
3
+
.
.
.
+
f
(
n
)
(
t
)
n
!
n
(
x
−
t
)
n
−
1
−
f
(
n
+
1
)
(
t
)
n
!
(
x
−
t
)
n
+
(
n
+
1
)
(
x
−
t
)
n
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
.
{\displaystyle F'(t)=-f'(t)+{\frac {f'(t)}{1!}}-{\frac {f''(t)}{1!}}(x-t)+{\frac {f''(t)}{2!}}2(x-t)-{\frac {f'''(t)}{2!}}(x-t)^{2}+{\frac {f'''(t)}{3!}}3(x-t)^{2}-{\frac {f^{(4)}(t)}{3!}}(x-t)^{3}+...+{\frac {f^{(n)}(t)}{n!}}n(x-t)^{n-1}-{\frac {f^{(n+1)}(t)}{n!}}(x-t)^{n}+{\frac {(n+1)(x-t)^{n}R_{n+1}(x)}{(x-a)^{n+1}}}.}
Nesunku pastebėti, kad visi nariai dešinėje lygybės pusėje, išskyrus du paskutiniuosius, tarpusavyje pasinaikina. Tokiu budu,
F
′
(
t
)
=
−
f
(
n
+
1
)
(
t
)
n
!
(
x
−
t
)
n
+
(
n
+
1
)
(
x
−
t
)
n
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
.
(
4
)
{\displaystyle F'(t)=-{\frac {f^{(n+1)}(t)}{n!}}(x-t)^{n}+{\frac {(n+1)(x-t)^{n}R_{n+1}(x)}{(x-a)^{n+1}}}.\quad (4)}
Lygybėje (4) parinkę
t
=
ξ
{\displaystyle t=\xi }
F
′
(
ξ
)
=
−
f
(
n
+
1
)
(
ξ
)
n
!
(
x
−
ξ
)
n
+
(
n
+
1
)
(
x
−
ξ
)
n
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
=
0
,
{\displaystyle F'(\xi )=-{\frac {f^{(n+1)}(\xi )}{n!}}(x-\xi )^{n}+{\frac {(n+1)(x-\xi )^{n}R_{n+1}(x)}{(x-a)^{n+1}}}=0,}
iš čia
(
n
+
1
)
(
x
−
ξ
)
n
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
=
f
(
n
+
1
)
(
ξ
)
n
!
(
x
−
ξ
)
n
,
{\displaystyle {\frac {(n+1)(x-\xi )^{n}R_{n+1}(x)}{(x-a)^{n+1}}}={\frac {f^{(n+1)}(\xi )}{n!}}(x-\xi )^{n},}
(
n
+
1
)
R
n
+
1
(
x
)
(
x
−
a
)
n
+
1
=
f
(
n
+
1
)
(
ξ
)
n
!
,
{\displaystyle {\frac {(n+1)R_{n+1}(x)}{(x-a)^{n+1}}}={\frac {f^{(n+1)}(\xi )}{n!}},}
R
n
+
1
(
x
)
=
f
(
n
+
1
)
(
ξ
)
(
n
+
1
)
!
(
x
−
a
)
n
+
1
.
{\displaystyle R_{n+1}(x)={\frac {f^{(n+1)}(\xi )}{(n+1)!}}(x-a)^{n+1}.}
Teorema įrodyta.
Formulė (1) vadinasi Teiloro formule , o
R
n
+
1
(
x
)
{\displaystyle R_{n+1}(x)}
Lagranžo formos papildomuoju nariu . Jį galima perrašyti kitame pavidale. Kadangi
a
<
ξ
<
x
,
{\displaystyle a<\xi <x,}
θ
{\displaystyle \theta }
0
<
θ
<
1
,
{\displaystyle 0<\theta <1,}
ξ
=
a
+
θ
(
x
−
a
)
,
{\displaystyle \xi =a+\theta (x-a),}
R
n
+
1
(
x
)
=
f
(
n
+
1
)
[
a
+
θ
(
x
−
a
)
]
(
n
+
1
)
!
(
x
−
a
)
n
+
1
,
0
<
θ
<
1.
{\displaystyle R_{n+1}(x)={\frac {f^{(n+1)}[a+\theta (x-a)]}{(n+1)!}}(x-a)^{n+1},\;\;0<\theta <1.}
Šitą formą papildomojo nario dažniausiai naudoja taikymuose. ____________
* Teiloras Brukas (1685-1731) - anglų matematikas.** Iš čia seka, kad pati funkcija f (x ) ir jos išvestinės iki n eilės tolydžios ir diferencijuojamos šitoje aplinkoje.
2. Kitoks Teiloro formulės ir papildomojo nario užrašymas
keisti
Dažnai Teiloro formulę (1) užrašo kitokiame pavidale. Pažymėsime (1) formulėje
a
=
x
0
,
x
−
a
=
Δ
x
,
x
=
x
0
+
Δ
x
.
{\displaystyle a=x_{0},\;x-a=\Delta x,\;x=x_{0}+\Delta x.}
f
(
x
0
+
Δ
x
)
−
f
(
x
0
)
=
f
′
(
x
0
)
1
!
Δ
x
+
f
″
(
x
0
)
2
!
(
Δ
x
)
2
+
f
‴
(
x
0
)
3
!
(
Δ
x
)
3
+
.
.
.
+
f
(
n
)
(
x
0
)
n
!
(
Δ
x
)
n
+
f
(
n
+
1
)
(
x
0
+
θ
Δ
x
)
(
n
+
1
)
!
(
Δ
x
)
n
+
1
.
0
<
θ
<
1
(
5
)
.
{\displaystyle f(x_{0}+\Delta x)-f(x_{0})={\frac {f'(x_{0})}{1!}}\Delta x+{\frac {f''(x_{0})}{2!}}(\Delta x)^{2}+{\frac {f'''(x_{0})}{3!}}(\Delta x)^{3}+...+{\frac {f^{(n)}(x_{0})}{n!}}(\Delta x)^{n}+{\frac {f^{(n+1)}(x_{0}+\theta \Delta x)}{(n+1)!}}(\Delta x)^{n+1}.\;\;0<\theta <1\quad (5).}
Kai
n
=
0
{\displaystyle n=0}
Lagranžo formulė
f
(
x
0
+
Δ
x
)
−
f
(
x
0
)
=
f
(
0
+
1
)
(
x
0
+
θ
Δ
x
)
(
0
+
1
)
!
(
Δ
x
)
0
+
1
,
{\displaystyle f(x_{0}+\Delta x)-f(x_{0})={\frac {f^{(0+1)}(x_{0}+\theta \Delta x)}{(0+1)!}}(\Delta x)^{0+1},}
f
(
x
0
+
Δ
x
)
−
f
(
x
0
)
=
f
′
(
x
0
+
θ
Δ
x
)
Δ
x
.
{\displaystyle f(x_{0}+\Delta x)-f(x_{0})=f'(x_{0}+\theta \Delta x)\Delta x.}
Parodysime, kad jeigu
f
(
n
+
1
)
{\displaystyle f^{(n+1)}}
a aplinkoje, tai papildomasis narys
R
n
+
1
(
x
)
{\displaystyle R_{n+1}(x)}
(
x
−
a
)
n
,
{\displaystyle (x-a)^{n},}
x
→
a
{\displaystyle x\to a}
lim
x
→
a
R
n
+
1
(
x
)
(
x
−
a
)
n
=
lim
x
→
a
f
(
n
+
1
)
(
ξ
)
(
n
+
1
)
!
(
x
−
a
)
n
+
1
(
x
−
a
)
n
=
lim
x
→
a
f
(
n
+
1
)
(
ξ
)
(
n
+
1
)
!
(
x
−
a
)
=
0
,
{\displaystyle \lim _{x\to a}{\frac {R_{n+1}(x)}{(x-a)^{n}}}=\lim _{x\to a}{\frac {f^{(n+1)}(\xi )}{(n+1)!}}{\frac {(x-a)^{n+1}}{(x-a)^{n}}}=\lim _{x\to a}{\frac {f^{(n+1)}(\xi )}{(n+1)!}}(x-a)=0,}
kadangi funkcija
f
(
n
+
1
)
{\displaystyle f^{(n+1)}}
(
x
−
a
)
→
0
,
{\displaystyle (x-a)\to 0,}
x
→
a
.
{\displaystyle x\to a.}
R
n
+
1
(
x
)
=
o
[
(
x
−
a
)
n
]
kai
x
→
a
.
(
6
)
{\displaystyle R_{n+1}(x)=o[(x-a)^{n}]\;{\text{kai}}\;x\to a.\quad (6)}
Formulė (6) vadinama papildomuoju nariu formoje Peano *. _________
* Peano Džuzepe (1858-1932) - italų matematikas. 3. Makloreno formulė
keisti
Makloreno* formule vadina Teiloro formulę (1) kai
a
=
0
{\displaystyle a=0}
f
(
x
)
=
f
(
0
)
+
f
′
(
0
)
1
!
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
.
.
.
+
f
(
n
)
(
0
)
n
!
x
n
+
R
n
+
1
(
x
)
.
{\displaystyle f(x)=f(0)+{\frac {f'(0)}{1!}}x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+...+{\frac {f^{(n)}(0)}{n!}}x^{n}+R_{n+1}(x).}
Papildomasis narys turi pavidalą:
1) Lagranžo formoje
R
n
+
1
(
x
)
=
f
(
n
+
1
)
(
θ
x
)
(
n
+
1
)
!
x
n
+
1
,
0
<
θ
<
1
;
{\displaystyle R_{n+1}(x)={\frac {f^{(n+1)}(\theta x)}{(n+1)!}}x^{n+1},\;\;0<\theta <1;}
2) Peano formoje
R
n
+
1
(
x
)
=
o
(
x
n
)
.
{\displaystyle R_{n+1}(x)=o(x^{n}).}
Trumpai aiškinant ką reiškia
o
(
x
n
)
,
{\displaystyle o(x^{n}),}
o
(
x
n
)
≈
C
x
n
+
1
.
{\displaystyle o(x^{n})\approx Cx^{n+1}.}
C , galimai esanti reiškinyje
o
(
x
n
)
,
{\displaystyle o(x^{n}),}
o
(
x
n
)
{\displaystyle o(x^{n})}
x
n
{\displaystyle x^{n}}
x
→
0
{\displaystyle x\to 0}
x neartėja prie 0, tai
o
(
x
n
)
{\displaystyle o(x^{n})}
x
n
.
{\displaystyle x^{n}.}
___________
* Maklorenas Kolinas (1698-1746) - škotų matematikas. 4. Kai kurių elementariųjų funkcijų išdėstymas Makloreno formule
keisti
1)
f
(
x
)
=
e
x
.
{\displaystyle f(x)=e^{x}.}
f
(
x
)
=
f
′
(
x
)
=
f
″
(
x
)
=
f
‴
(
x
)
=
.
.
.
=
f
(
n
+
1
)
(
x
)
=
e
x
,
{\displaystyle f(x)=f'(x)=f''(x)=f'''(x)=...=f^{(n+1)}(x)=e^{x},}
f
(
0
)
=
f
′
(
0
)
=
f
″
(
0
)
=
f
‴
(
0
)
=
.
.
.
=
f
(
n
+
1
)
(
0
)
=
e
0
=
1
,
{\displaystyle f(0)=f'(0)=f''(0)=f'''(0)=...=f^{(n+1)}(0)=e^{0}=1,}
tai Maklorerno formulė yra tokia
e
x
=
1
+
x
1
!
+
x
2
2
!
+
x
3
3
!
+
.
.
.
+
x
n
n
!
+
o
(
x
n
)
.
(
7
)
{\displaystyle e^{x}=1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+...+{\frac {x^{n}}{n!}}+o(x^{n}).\quad (7)}
2)
f
(
x
)
=
sin
x
.
{\displaystyle f(x)=\sin x.}
f
(
n
)
(
x
)
=
sin
(
x
+
n
π
2
)
,
f
(
n
)
(
0
)
=
sin
(
n
π
2
)
=
{\displaystyle f^{(n)}(x)=\sin {\Big (}x+n{\frac {\pi }{2}}{\Big )},\;\;f^{(n)}(0)=\sin {\Big (}n{\frac {\pi }{2}}{\Big )}=}
=
{
0
,
kai
n
lyginis
,
(
−
1
)
(
n
−
1
)
/
2
,
kai
n
nelyginis
,
{\displaystyle ={\begin{cases}\quad 0,\quad \quad {\text{kai}}\;n\;{\text{lyginis}},&\\(-1)^{(n-1)/2},\quad {\text{kai}}\;n\;{\text{nelyginis}},&\end{cases}}}
tai Makloreno formulė yra tokia
sin
x
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
.
.
.
+
(
−
1
)
n
−
1
x
2
n
−
1
(
2
n
−
1
)
!
+
o
(
x
2
n
)
.
(
8
)
{\displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+...+(-1)^{n-1}{\frac {x^{2n-1}}{(2n-1)!}}+o(x^{2n}).\quad (8)}
3)
f
(
x
)
=
cos
x
.
{\displaystyle f(x)=\cos x.}
f
(
n
)
(
x
)
=
cos
(
x
+
n
π
2
)
,
f
(
n
)
(
0
)
=
cos
(
n
π
2
)
=
{\displaystyle f^{(n)}(x)=\cos {\Big (}x+n{\frac {\pi }{2}}{\Big )},\;\;f^{(n)}(0)=\cos {\Big (}n{\frac {\pi }{2}}{\Big )}=}
=
{
0
,
kai
n
nelyginis
,
(
−
1
)
n
/
2
,
kai
n
lyginis
,
{\displaystyle ={\begin{cases}\quad 0,\quad \quad {\text{kai}}\;n\;{\text{nelyginis}},&\\(-1)^{n/2},\quad {\text{kai}}\;n\;{\text{lyginis}},&\end{cases}}}
tai Makloreno formulė yra tokia
cos
x
=
1
−
x
2
2
!
+
x
4
4
!
−
x
6
6
!
+
.
.
.
+
(
−
1
)
n
x
2
n
(
2
n
)
!
+
o
(
x
2
n
+
1
)
.
(
9
)
{\displaystyle \cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+...+(-1)^{n}{\frac {x^{2n}}{(2n)!}}+o(x^{2n+1}).\quad (9)}
Formulėje (8) papildomasis narys užrašytas pavidale
o
(
x
2
n
)
,
{\displaystyle o(x^{2n}),}
o
(
x
2
n
−
1
)
,
{\displaystyle o(x^{2n-1}),}
4)
f
(
x
)
=
(
1
+
x
)
α
,
{\displaystyle f(x)=(1+x)^{\alpha },}
α
{\displaystyle \alpha }
f
(
n
)
(
x
)
=
α
(
α
−
1
)
(
α
−
2
)
(
α
−
3
)
.
.
.
(
α
−
n
+
1
)
(
1
+
x
)
α
−
n
,
{\displaystyle f^{(n)}(x)=\alpha (\alpha -1)(\alpha -2)(\alpha -3)...(\alpha -n+1)(1+x)^{\alpha -n},}
f
(
n
)
(
0
)
=
α
(
α
−
1
)
(
α
−
2
)
(
α
−
3
)
.
.
.
(
α
−
n
+
1
)
,
{\displaystyle f^{(n)}(0)=\alpha (\alpha -1)(\alpha -2)(\alpha -3)...(\alpha -n+1),}
tai Makloreno formulė yra tokia
(
1
+
x
)
α
=
1
+
α
1
!
x
+
α
(
α
−
1
)
2
!
x
2
+
α
(
α
−
1
)
(
α
−
2
)
3
!
x
3
+
.
.
.
+
α
(
α
−
1
)
(
α
−
2
)
.
.
.
(
α
−
n
+
1
)
n
!
x
n
+
R
n
+
1
(
x
)
,
{\displaystyle (1+x)^{\alpha }=1+{\frac {\alpha }{1!}}x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+{\frac {\alpha (\alpha -1)(\alpha -2)}{3!}}x^{3}+...+{\frac {\alpha (\alpha -1)(\alpha -2)...(\alpha -n+1)}{n!}}x^{n}+R_{n+1}(x),}
kur papildomasis narys Lagranžo formoje lygus
R
n
+
1
(
x
)
=
α
(
α
−
1
)
(
α
−
2
)
(
α
−
3
)
.
.
.
(
α
−
n
)
(
n
+
1
)
!
(
1
+
θ
x
)
α
−
(
n
+
1
)
x
n
+
1
,
0
<
θ
<
1.
{\displaystyle R_{n+1}(x)={\frac {\alpha (\alpha -1)(\alpha -2)(\alpha -3)...(\alpha -n)}{(n+1)!}}(1+\theta x)^{\alpha -(n+1)}x^{n+1},\;\;0<\theta <1.}
Atskiru atveju, kai
α
=
n
{\displaystyle \alpha =n}
f
(
n
+
1
)
(
x
)
=
α
(
α
−
1
)
(
α
−
2
)
(
α
−
3
)
.
.
.
(
α
−
n
)
(
1
+
x
)
α
−
(
n
+
1
)
=
0
,
{\displaystyle f^{(n+1)}(x)=\alpha (\alpha -1)(\alpha -2)(\alpha -3)...(\alpha -n)(1+x)^{\alpha -(n+1)}=0,}
todėl,
R
n
+
1
(
x
)
=
0
{\displaystyle R_{n+1}(x)=0}
Niutono binomo formulę
(
1
+
x
)
n
=
1
+
n
1
!
x
+
n
(
n
−
1
)
2
!
x
2
+
n
(
n
−
1
)
(
n
−
2
)
3
!
x
3
+
.
.
.
+
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
n
−
n
+
1
)
n
!
x
n
,
{\displaystyle (1+x)^{n}=1+{\frac {n}{1!}}x+{\frac {n(n-1)}{2!}}x^{2}+{\frac {n(n-1)(n-2)}{3!}}x^{3}+...+{\frac {n(n-1)(n-2)...(n-n+1)}{n!}}x^{n},}
(
1
+
x
)
n
=
1
+
n
1
!
x
+
n
(
n
−
1
)
2
!
x
2
+
n
(
n
−
1
)
(
n
−
2
)
3
!
x
3
+
.
.
.
+
x
n
.
(
10
)
{\displaystyle (1+x)^{n}=1+{\frac {n}{1!}}x+{\frac {n(n-1)}{2!}}x^{2}+{\frac {n(n-1)(n-2)}{3!}}x^{3}+...+x^{n}.\quad (10)}
Pateikti aukščiau dėstiniai parodo, kad su Makloreno formule funkcijas galima su nustatytu tikslumu pakeisti polinomais, esančiais paprasčiausiomis elementariosiomis funkcijomis. Su polinomais patogu daryti aritmetinius veiksmus, nesunku apskaičiuoti reikšmę bet kuriame taške ir t. t. Teiloro ir Makloreno formulės leidžia apytiksliai pakeisti polinomais ir sudėtingesnes funkcijas. Be to, šitos formulės turi platų ratą pritaikymų. Funkciją
arcsin
x
{\displaystyle \arcsin x}
(
1
+
x
)
α
=
1
+
α
1
!
x
+
α
(
α
−
1
)
2
!
x
2
+
α
(
α
−
1
)
(
α
−
2
)
3
!
x
3
+
.
.
.
+
α
(
α
−
1
)
(
α
−
2
)
.
.
.
(
α
−
n
+
1
)
n
!
x
n
+
R
n
+
1
(
x
)
(
9.5
)
{\displaystyle (1+x)^{\alpha }=1+{\frac {\alpha }{1!}}x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+{\frac {\alpha (\alpha -1)(\alpha -2)}{3!}}x^{3}+...+{\frac {\alpha (\alpha -1)(\alpha -2)...(\alpha -n+1)}{n!}}x^{n}+R_{n+1}(x)\quad (9.5)}
vietoje
α
{\displaystyle \alpha }
−
1
2
,
{\displaystyle -{\frac {1}{2}},}
x įrašę
−
x
2
{\displaystyle -x^{2}}
R
n
+
1
(
x
)
{\displaystyle R_{n+1}(x)}
1
1
−
x
2
=
(
1
−
x
2
)
−
1
2
=
1
+
−
1
2
1
!
(
−
x
2
)
+
−
1
2
(
−
1
2
−
1
)
2
!
x
4
+
−
1
2
(
−
1
2
−
1
)
(
−
1
2
−
2
)
3
!
(
−
x
6
)
+
.
.
.
=
{\displaystyle {\frac {1}{\sqrt {1-x^{2}}}}=(1-x^{2})^{-{1 \over 2}}=1+{\frac {-{1 \over 2}}{1!}}(-x^{2})+{\frac {-{1 \over 2}(-{1 \over 2}-1)}{2!}}x^{4}+{\frac {-{1 \over 2}(-{1 \over 2}-1)(-{1 \over 2}-2)}{3!}}(-x^{6})+...=}
=
1
+
1
2
x
2
+
−
1
2
(
−
3
2
)
2
x
4
+
−
1
2
(
−
3
2
)
(
−
5
2
)
2
⋅
3
(
−
x
6
)
+
.
.
.
=
{\displaystyle =1+{\frac {1}{2}}x^{2}+{\frac {-{1 \over 2}(-{3 \over 2})}{2}}x^{4}+{\frac {-{1 \over 2}(-{3 \over 2})(-{5 \over 2})}{2\cdot 3}}(-x^{6})+...=}
=
1
+
1
2
x
2
+
3
2
⋅
4
x
4
−
3
⋅
5
2
⋅
3
⋅
8
(
−
x
6
)
+
.
.
.
=
{\displaystyle =1+{\frac {1}{2}}x^{2}+{\frac {3}{2\cdot 4}}x^{4}-{\frac {3\cdot 5}{2\cdot 3\cdot 8}}(-x^{6})+...=}
=
1
+
1
2
x
2
+
1
⋅
3
2
⋅
4
x
4
+
1
⋅
3
⋅
5
2
⋅
4
⋅
6
x
6
+
.
.
.
=
{\displaystyle =1+{\frac {1}{2}}x^{2}+{\frac {1\cdot 3}{2\cdot 4}}x^{4}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}x^{6}+...=}
=
1
+
∑
n
=
1
∞
(
2
n
−
1
)
!
!
(
2
n
)
!
!
x
2
n
.
{\displaystyle =1+\sum _{n=1}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}x^{2n}.}
Ši eilutė konverguoja intervale (-1; 1). Integruodami ją panariui, atkarpoje [0; x ], kai |x | < 1, gauname eilutę
arcsin
x
=
∫
0
x
d
x
1
−
x
2
=
∫
0
x
(
1
+
1
2
x
2
+
1
⋅
3
2
⋅
4
x
4
+
1
⋅
3
⋅
5
2
⋅
4
⋅
6
x
6
+
.
.
.
)
d
x
=
{\displaystyle \arcsin x=\int _{0}^{x}{\frac {dx}{\sqrt {1-x^{2}}}}=\int _{0}^{x}{\Big (}1+{\frac {1}{2}}x^{2}+{\frac {1\cdot 3}{2\cdot 4}}x^{4}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}x^{6}+...{\Big )}dx=}
=
x
+
1
2
x
3
3
+
1
⋅
3
2
⋅
4
x
5
5
+
1
⋅
3
⋅
5
2
⋅
4
⋅
6
x
7
7
+
.
.
.
=
{\displaystyle =x+{\frac {1}{2}}{\frac {x^{3}}{3}}+{\frac {1\cdot 3}{2\cdot 4}}{\frac {x^{5}}{5}}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}{\frac {x^{7}}{7}}+...=}
=
x
+
∑
n
=
1
∞
(
2
n
−
1
)
!
!
x
2
n
+
1
(
2
n
)
!
!
(
2
n
+
1
)
,
{\displaystyle =x+\sum _{n=1}^{\infty }{\frac {(2n-1)!!\;x^{2n+1}}{(2n)!!\;(2n+1)}},}
kuri taip pat konverguoja intervale (-1; 1). Funkciją
f
(
x
)
=
ln
(
1
+
x
)
{\displaystyle f(x)=\ln(1+x)}
(9.5) formulėje vietoj
α
{\displaystyle \alpha }
−
1
{\displaystyle -1}
(
1
+
x
)
−
1
=
1
1
+
x
=
1
+
−
1
1
!
x
+
−
1
(
−
1
−
1
)
2
!
x
2
+
−
1
(
−
1
−
1
)
(
−
1
−
2
)
3
!
x
3
+
−
1
(
−
1
−
1
)
(
−
1
−
2
)
(
−
1
−
3
)
4
!
x
4
.
.
.
=
{\displaystyle (1+x)^{-1}={\frac {1}{1+x}}=1+{\frac {-1}{1!}}x+{\frac {-1(-1-1)}{2!}}x^{2}+{\frac {-1(-1-1)(-1-2)}{3!}}x^{3}+{\frac {-1(-1-1)(-1-2)(-1-3)}{4!}}x^{4}...=}
=
1
−
x
+
−
1
(
−
2
)
2
!
x
2
+
−
1
(
−
2
)
(
−
3
)
3
!
x
3
+
−
1
(
−
2
)
(
−
3
)
(
−
4
)
4
!
x
4
.
.
.
=
{\displaystyle =1-x+{\frac {-1(-2)}{2!}}x^{2}+{\frac {-1(-2)(-3)}{3!}}x^{3}+{\frac {-1(-2)(-3)(-4)}{4!}}x^{4}...=}
=
1
−
x
+
x
2
−
x
3
+
x
4
−
.
.
.
+
(
−
1
)
n
x
n
+
.
.
.
,
{\displaystyle =1-x+x^{2}-x^{3}+x^{4}-...+(-1)^{n}x^{n}+...,}
kuri konverguoja intervale (-1; 1). Ją galima integruoti bet kurioje atkarpoje [0; x ] iš intervalo (-1; 1). Integruojant gauname:
ln
(
1
+
x
)
=
∫
0
x
d
x
1
+
x
=
∫
0
x
(
1
−
x
+
x
2
−
x
3
+
x
4
−
.
.
.
+
(
−
1
)
n
x
n
+
.
.
.
)
d
x
=
{\displaystyle \ln(1+x)=\int _{0}^{x}{\frac {dx}{1+x}}=\int _{0}^{x}(1-x+x^{2}-x^{3}+x^{4}-...+(-1)^{n}x^{n}+...)dx=}
=
x
−
x
2
2
+
x
3
3
−
x
4
4
+
x
5
5
−
.
.
.
+
(
−
1
)
n
x
n
+
1
n
+
1
+
.
.
.
.
(
9.6
)
{\displaystyle =x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-...+(-1)^{n}{\frac {x^{n+1}}{n+1}}+...\;.\quad (9.6)}
Nesunku patikrinti, kad eilutė
x
−
x
2
2
+
x
3
3
−
x
4
4
+
x
5
5
−
.
.
.
+
(
−
1
)
n
x
n
+
1
n
+
1
+
.
.
.
{\displaystyle x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-...+(-1)^{n}{\frac {x^{n+1}}{n+1}}+...}
konverguoja intervale (-1; 1]. Tuomet iš (9.6) formulės, įrašę įrašę į ją x = 1, gauname eilutę
ln
(
1
+
1
)
=
ln
2
=
1
−
1
2
2
+
1
3
3
−
1
4
4
+
1
5
5
−
.
.
.
+
(
−
1
)
n
1
n
+
1
n
+
1
+
.
.
.
=
{\displaystyle \ln(1+1)=\ln 2=1-{\frac {1^{2}}{2}}+{\frac {1^{3}}{3}}-{\frac {1^{4}}{4}}+{\frac {1^{5}}{5}}-...+(-1)^{n}{\frac {1^{n+1}}{n+1}}+...=}
=
1
−
1
2
+
1
3
−
1
4
+
1
5
−
.
.
.
+
(
−
1
)
n
1
n
+
1
+
.
.
.
,
{\displaystyle =1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-...+(-1)^{n}{\frac {1}{n+1}}+...,}
kuri vadinama Leibnico eilute . Atvirkštinį hiperbolinį sinusą išskleisime eilute. Atvirkštinis hiperbolinis sinusas yra lygus
arsinh
x
=
sinh
−
1
x
=
ln
(
x
+
1
+
x
2
)
.
{\displaystyle \operatorname {arsinh} x=\sinh ^{-1}x=\ln(x+{\sqrt {1+x^{2}}}).}
Apie atvirkštines hiperbolines funkcijas žiūrėti čia: https://en.wikipedia.org/wiki/Inverse_hyperbolic_functions
arsinh(x) išvestinės įrodymas (kai nežinoma kam lygus arsinh(x)): https://proofwiki.org/wiki/Derivative_of_Inverse_Hyperbolic_Sine
Jo išvestinė yra
(
ln
(
x
+
1
+
x
2
)
)
′
=
(
x
+
1
+
x
2
)
′
x
+
1
+
x
2
=
1
+
1
2
2
x
1
+
x
2
x
+
1
+
x
2
=
{\displaystyle {\big (}\ln(x+{\sqrt {1+x^{2}}}){\big )}'={\frac {(x+{\sqrt {1+x^{2}}})'}{x+{\sqrt {1+x^{2}}}}}={\frac {1+{\frac {1}{2}}{\frac {2x}{\sqrt {1+x^{2}}}}}{x+{\sqrt {1+x^{2}}}}}=}
=
1
+
x
1
+
x
2
x
+
1
+
x
2
=
x
+
1
+
x
2
1
+
x
2
x
+
1
+
x
2
=
1
1
+
x
2
.
{\displaystyle ={\frac {1+{\frac {x}{\sqrt {1+x^{2}}}}}{x+{\sqrt {1+x^{2}}}}}={\frac {\frac {x+{\sqrt {1+x^{2}}}}{\sqrt {1+x^{2}}}}{x+{\sqrt {1+x^{2}}}}}={\frac {1}{\sqrt {1+x^{2}}}}.}
Taikysime (9.5) formulę
(
1
+
x
)
α
=
1
+
α
1
!
x
+
α
(
α
−
1
)
2
!
x
2
+
α
(
α
−
1
)
(
α
−
2
)
3
!
x
3
+
.
.
.
+
α
(
α
−
1
)
(
α
−
2
)
.
.
.
(
α
−
n
+
1
)
n
!
x
n
+
R
n
+
1
(
x
)
.
(
9.5
)
{\displaystyle (1+x)^{\alpha }=1+{\frac {\alpha }{1!}}x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+{\frac {\alpha (\alpha -1)(\alpha -2)}{3!}}x^{3}+...+{\frac {\alpha (\alpha -1)(\alpha -2)...(\alpha -n+1)}{n!}}x^{n}+R_{n+1}(x).\quad (9.5)}
Vietoje
α
{\displaystyle \alpha }
−
1
2
,
{\displaystyle -{\frac {1}{2}},}
x įrašę
x
2
{\displaystyle x^{2}}
R
n
+
1
(
x
)
{\displaystyle R_{n+1}(x)}
1
1
+
x
2
=
(
1
+
x
2
)
−
1
2
=
1
+
−
1
2
1
!
x
2
+
−
1
2
(
−
1
2
−
1
)
2
!
x
4
+
−
1
2
(
−
1
2
−
1
)
(
−
1
2
−
2
)
3
!
x
6
+
.
.
.
=
{\displaystyle {\frac {1}{\sqrt {1+x^{2}}}}=(1+x^{2})^{-{1 \over 2}}=1+{\frac {-{1 \over 2}}{1!}}x^{2}+{\frac {-{1 \over 2}(-{1 \over 2}-1)}{2!}}x^{4}+{\frac {-{1 \over 2}(-{1 \over 2}-1)(-{1 \over 2}-2)}{3!}}x^{6}+...=}
=
1
−
1
2
x
2
+
−
1
2
(
−
3
2
)
2
x
4
+
−
1
2
(
−
3
2
)
(
−
5
2
)
2
⋅
3
x
6
+
.
.
.
=
{\displaystyle =1-{\frac {1}{2}}x^{2}+{\frac {-{1 \over 2}(-{3 \over 2})}{2}}x^{4}+{\frac {-{1 \over 2}(-{3 \over 2})(-{5 \over 2})}{2\cdot 3}}x^{6}+...=}
=
1
−
1
2
x
2
+
3
2
⋅
4
x
4
−
3
⋅
5
2
⋅
3
⋅
8
x
6
+
.
.
.
=
{\displaystyle =1-{\frac {1}{2}}x^{2}+{\frac {3}{2\cdot 4}}x^{4}-{\frac {3\cdot 5}{2\cdot 3\cdot 8}}x^{6}+...=}
=
1
−
1
2
x
2
+
1
⋅
3
2
⋅
4
x
4
−
1
⋅
3
⋅
5
2
⋅
4
⋅
6
x
6
+
.
.
.
=
{\displaystyle =1-{\frac {1}{2}}x^{2}+{\frac {1\cdot 3}{2\cdot 4}}x^{4}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}x^{6}+...=}
=
1
+
∑
n
=
1
∞
(
−
1
)
n
(
2
n
−
1
)
!
!
(
2
n
)
!
!
x
2
n
.
{\displaystyle =1+\sum _{n=1}^{\infty }(-1)^{n}{\frac {(2n-1)!!}{(2n)!!}}x^{2n}.}
Kuri konverguoja intervale (-1; 1). Integruodami panariui atkarpoje [0; x ] iš intervalo (-1; 1), turime:
arsinh
x
=
∫
0
x
d
x
1
+
x
2
=
∫
0
x
(
1
−
1
2
x
2
+
1
⋅
3
2
⋅
4
x
4
−
1
⋅
3
⋅
5
2
⋅
4
⋅
6
x
6
+
.
.
.
)
d
x
=
{\displaystyle \operatorname {arsinh} x=\int _{0}^{x}{\frac {dx}{\sqrt {1+x^{2}}}}=\int _{0}^{x}{\Big (}1-{\frac {1}{2}}x^{2}+{\frac {1\cdot 3}{2\cdot 4}}x^{4}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}x^{6}+...{\Big )}dx=}
=
x
−
1
2
x
3
3
+
1
⋅
3
2
⋅
4
x
5
5
−
1
⋅
3
⋅
5
2
⋅
4
⋅
6
x
7
7
+
.
.
.
=
{\displaystyle =x-{\frac {1}{2}}{\frac {x^{3}}{3}}+{\frac {1\cdot 3}{2\cdot 4}}{\frac {x^{5}}{5}}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}{\frac {x^{7}}{7}}+...=}
=
x
+
∑
n
=
1
∞
(
−
1
)
n
(
2
n
−
1
)
!
!
x
2
n
+
1
(
2
n
)
!
!
(
2
n
+
1
)
.
{\displaystyle =x+\sum _{n=1}^{\infty }(-1)^{n}{\frac {(2n-1)!!\;x^{2n+1}}{(2n)!!\;(2n+1)}}.}
5. Makloreno formulės naudojimas ribų skaičiavimui
keisti
6. Skaičiaus e apskaičiavimas
keisti
Matematikos knygose įvestas skaičius e kaip sekos
{
(
1
+
1
/
n
)
n
}
{\displaystyle \{(1+1/n)^{n}\}}
gautas grubus jo įvertinimas
2
≤
e
≤
3.
{\displaystyle 2\leq e\leq 3.}
Parodysime, kaip apskaičiuoti skaičių e bet kokiu reikalingu tikslumu. Tam užrašysime (7) formulę su Lagranžo formos papildomuoju nariu:
e
x
=
1
+
x
1
!
+
x
2
2
!
+
x
3
3
!
+
.
.
.
+
x
n
n
!
+
e
θ
x
(
n
+
1
)
!
x
n
+
1
,
0
<
θ
<
1.
(
11
)
{\displaystyle e^{x}=1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+...+{\frac {x^{n}}{n!}}+{\frac {e^{\theta x}}{(n+1)!}}x^{n+1},\;\;0<\theta <1.\quad (11)}
Jeigu funkciją
e
x
{\displaystyle e^{x}}
n , tai gausime apytikslę lygybę
e
x
≈
1
+
x
1
!
+
x
2
2
!
+
x
3
3
!
+
.
.
.
+
x
n
n
!
,
(
12
)
{\displaystyle e^{x}\approx 1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+...+{\frac {x^{n}}{n!}},\quad (12)}
absoliuti paklaida kurio
|
R
n
+
1
(
x
)
|
=
e
θ
x
(
n
+
1
)
!
|
x
|
n
+
1
,
0
<
θ
<
1.
{\displaystyle |R_{n+1}(x)|={\frac {e^{\theta x}}{(n+1)!}}|x|^{n+1},\;\;0<\theta <1.}
Jeigu nagrinėti funkciją
e
x
{\displaystyle e^{x}}
−
1
≤
x
≤
1
,
{\displaystyle -1\leq x\leq 1,}
|
R
n
+
1
(
x
)
|
≤
e
θ
x
(
n
+
1
)
!
<
3
(
n
+
1
)
!
.
(
13
)
{\displaystyle |R_{n+1}(x)|\leq {\frac {e^{\theta x}}{(n+1)!}}<{\frac {3}{(n+1)!}}.\quad (13)}
Tarę (12) formulėje, kad
x
=
1
,
{\displaystyle x=1,}
e reikšmę:
e
≈
1
+
1
+
1
2
!
+
1
3
!
+
.
.
.
+
1
n
!
.
{\displaystyle e\approx 1+1+{\frac {1}{2!}}+{\frac {1}{3!}}+...+{\frac {1}{n!}}.}
Be to, absoliuti paklaida mažesnė už
3
(
n
+
1
)
!
.
{\displaystyle {\frac {3}{(n+1)!}}.}
e reikšmę tikslumu iki 0,001, tai skaičius n nustatomas iš nelygybės
3
(
n
+
1
)
!
<
0.001
,
{\displaystyle {\frac {3}{(n+1)!}}<0.001,}
(
n
+
1
)
!
>
3000.
{\displaystyle (n+1)!>3000.}
n
=
6
,
{\displaystyle n=6,}
(
n
+
1
)
!
=
7
!
=
5040
{\displaystyle (n+1)!=7!=5040}
Tokiu budu, panaudojant Makloreno formulę, galimą apskaičiuoti skaičių e bet kokiu tikslumu, be to skaičiavimo algoritmas skaičiaus e , pagrįstas formulėmis (11) ir (13), lengvai realizuojamas su ESM (elektronine skaičiavimo mašina).
Apskaičiuosime e , kai
n
=
6
,
{\displaystyle n=6,}
e
≈
1
+
1
+
1
2
!
+
1
3
!
+
1
4
!
+
1
5
!
+
1
6
!
=
2
+
1
2
+
1
6
+
1
24
+
1
120
+
1
720
=
2.71805555556.
{\displaystyle e\approx 1+1+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+{\frac {1}{5!}}+{\frac {1}{6!}}=2+{\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{24}}+{\frac {1}{120}}+{\frac {1}{720}}=2.71805555556.}
Tuo tarpu, tikroji e reikšmė lygi e = 2.7182818284590452353602874713527.
2.718281828459 - 2.71805555556 = 0.000226272899 < 0.001.
Naudodamiesi Windows 10 kalkuliatoriumi, apskaičiuosime ribą
e
=
lim
n
→
∞
(
1
+
1
n
)
n
,
{\displaystyle e=\lim _{n\to \infty }{\Big (}1+{\frac {1}{n}}{\Big )}^{n},}
a) n = 1000; b) n = 1000000 ; c)
n
=
10
9
.
{\displaystyle n=10^{9}.}
Sprendimas .a)
e
≈
(
1
+
1
n
)
n
=
(
1
+
1
1000
)
1000
=
1.001
1000
=
2.71692393223589
;
{\displaystyle e\approx {\Big (}1+{\frac {1}{n}}{\Big )}^{n}={\Big (}1+{\frac {1}{1000}}{\Big )}^{1000}=1.001^{1000}=2.71692393223589;}
b)
e
≈
(
1
+
1
n
)
n
=
(
1
+
1
10
6
)
1000000
=
1.000001
1000000
=
2.7182804693193768838
;
{\displaystyle e\approx {\Big (}1+{\frac {1}{n}}{\Big )}^{n}={\Big (}1+{\frac {1}{10^{6}}}{\Big )}^{1000000}=1.000001^{1000000}=2.7182804693193768838;}
c)
e
≈
(
1
+
1
n
)
n
=
(
1
+
1
10
9
)
10
9
=
1.000000001
10
9
=
2.718281827099904322.
{\displaystyle e\approx {\Big (}1+{\frac {1}{n}}{\Big )}^{n}={\Big (}1+{\frac {1}{10^{9}}}{\Big )}^{10^{9}}=1.000000001^{10^{9}}=2.718281827099904322.}
c) atveju gaunami 8 teisingi [skaičiaus e ] skaitmenys po kablelio. Jeigu n parinkti
n
=
10
10
,
{\displaystyle n=10^{10},}
Windows 10 kalkuliatorius išduoda klaidą ("Invalid input").
Pagal Makloreno eilutę apskaičiuosime skaičiaus e reikšmę su paklaida mažesne nei
1
10
6
=
1
1000000
=
0.000001.
{\displaystyle {\frac {1}{10^{6}}}={\frac {1}{1000000}}=0.000001.}
R
n
+
1
(
x
)
=
e
θ
x
(
n
+
1
)
!
x
n
+
1
=
e
θ
(
n
+
1
)
!
<
3
(
n
+
1
)
!
,
{\displaystyle R_{n+1}(x)={\frac {e^{\theta x}}{(n+1)!}}x^{n+1}={\frac {e^{\theta }}{(n+1)!}}<{\frac {3}{(n+1)!}},}
x = 1 (čia
0
<
θ
<
1
{\displaystyle 0<\theta <1}
3
(
n
+
1
)
!
{\displaystyle {\frac {3}{(n+1)!}}}
3
(
n
+
1
)
!
<
0.000001
;
{\displaystyle {\frac {3}{(n+1)!}}<0.000001;}
3000000
<
(
n
+
1
)
!
;
{\displaystyle 3000000<(n+1)!;}
n = 9, nes
(
n
+
1
)
!
=
(
9
+
1
)
!
=
10
!
=
3628800.
{\displaystyle (n+1)!=(9+1)!=10!=3628800.}
e
≈
1
+
1
+
1
2
!
+
1
3
!
+
1
4
!
+
1
5
!
+
1
6
!
+
1
7
!
+
1
8
!
+
1
9
!
=
{\displaystyle e\approx 1+1+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+{\frac {1}{5!}}+{\frac {1}{6!}}+{\frac {1}{7!}}+{\frac {1}{8!}}+{\frac {1}{9!}}=}
=
2
+
1
2
+
1
6
+
1
24
+
1
120
+
1
720
+
1
5040
+
1
40320
+
1
362880
=
2.7182815255731922398589.
{\displaystyle =2+{\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{24}}+{\frac {1}{120}}+{\frac {1}{720}}+{\frac {1}{5040}}+{\frac {1}{40320}}+{\frac {1}{362880}}=2.7182815255731922398589.}
Atsakymą gavome teisingą, nes
2.718281828459
−
2.71828152557319
=
3.028858529955
/
10
7
=
0.00000030288585
<
0.000001.
{\displaystyle 2.718281828459-2.71828152557319=3.028858529955/10^{7}=0.00000030288585<0.000001.}
Pagal Makloreno eilutę apskaičiuosime skaičiaus e reikšmę su paklaida mažesne nei
1
10
9
.
{\displaystyle {\frac {1}{10^{9}}}.}
3
(
n
+
1
)
!
{\displaystyle {\frac {3}{(n+1)!}}}
1
10
9
.
{\displaystyle {\frac {1}{10^{9}}}.}
3
(
n
+
1
)
!
<
1
10
9
,
3
⋅
10
9
<
(
n
+
1
)
!
.
{\displaystyle {\frac {3}{(n+1)!}}<{\frac {1}{10^{9}}},\;\;3\cdot 10^{9}<(n+1)!.}
n = 12, paskutinė nelygybė tenkinama, nes
(
n
+
1
)
!
=
(
12
+
1
)
!
=
13
!
=
6227020800
=
6.2270208
⋅
10
9
.
{\displaystyle (n+1)!=(12+1)!=13!=6227020800=6.2270208\cdot 10^{9}.}
Dabar galime apskaičiuoti skaičiaus e reikšmę, kai n = 12 su paklaida mažesne nei
1
/
10
9
:
{\displaystyle 1/10^{9}:}
e
≈
1
+
1
+
1
2
!
+
1
3
!
+
1
4
!
+
1
5
!
+
1
6
!
+
1
7
!
+
1
8
!
+
1
9
!
+
1
10
!
+
1
11
!
+
1
12
!
=
{\displaystyle e\approx 1+1+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+{\frac {1}{5!}}+{\frac {1}{6!}}+{\frac {1}{7!}}+{\frac {1}{8!}}+{\frac {1}{9!}}+{\frac {1}{10!}}+{\frac {1}{11!}}+{\frac {1}{12!}}=}
=
2.71828182828616856.
{\displaystyle =2.71828182828616856.}
Atėmę gautą e reikšmę iš tikslios e reikšmės gauname skirtumą (paklaidą):
2.718281828459
−
2.71828182828616856
=
1.7287667141394574723316
/
10
10
<
1
/
10
9
.
{\displaystyle 2.718281828459-2.71828182828616856=1.7287667141394574723316/10^{10}<1/10^{9}.}
![{\displaystyle R_{n+1}(x)={\frac {f^{(n+1)}[a+\theta (x-a)]}{(n+1)!}}(x-a)^{n+1},\;\;0<\theta <1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/215fc50f2ba09d1a46f55e7e2b88fcf83a147fce)
![{\displaystyle R_{n+1}(x)=o[(x-a)^{n}]\;{\text{kai}}\;x\to a.\quad (6)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71415632a8fc49de6aea3b16e322e39591f5a4ab)
