Matematika/Teiloro eilutė
keisti
keisti
Darome prielaidą, kad funkciją
f
(
x
)
{\displaystyle f(x)\;}
f
(
x
)
=
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
a
4
x
4
+
a
5
x
5
+
⋯
.
{\displaystyle f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}+\cdots .}
Tuomet surandame visų eilių (laipsnių) išvestines funkcijos
f
(
x
)
:
{\displaystyle f(x):\;}
f
′
(
x
)
=
a
1
+
2
a
2
x
+
3
a
3
x
2
+
4
a
4
x
3
+
5
a
5
x
4
+
⋯
,
{\displaystyle f'(x)=a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}+5a_{5}x^{4}+\cdots ,}
f
″
(
x
)
=
(
f
′
(
x
)
)
′
=
2
a
2
+
3
⋅
2
a
3
x
+
4
⋅
3
a
4
x
2
+
5
⋅
4
a
5
x
3
+
⋯
,
{\displaystyle f''(x)=(f'(x))'=2a_{2}+3\cdot 2a_{3}x+4\cdot 3a_{4}x^{2}+5\cdot 4a_{5}x^{3}+\cdots ,}
f
‴
(
x
)
=
3
⋅
2
a
3
+
4
⋅
3
⋅
2
a
4
x
+
5
⋅
4
⋅
3
a
5
x
2
+
⋯
,
{\displaystyle f'''(x)=3\cdot 2a_{3}+4\cdot 3\cdot 2a_{4}x+5\cdot 4\cdot 3a_{5}x^{2}+\cdots ,}
f
(
4
)
(
x
)
=
4
⋅
3
⋅
2
a
4
+
5
⋅
4
⋅
3
⋅
2
a
5
x
+
⋯
,
{\displaystyle f^{(4)}(x)=4\cdot 3\cdot 2a_{4}+5\cdot 4\cdot 3\cdot 2a_{5}x+\cdots ,}
f
(
5
)
(
x
)
=
5
⋅
4
⋅
3
⋅
2
a
5
+
⋯
.
{\displaystyle f^{(5)}(x)=5\cdot 4\cdot 3\cdot 2a_{5}+\cdots .}
Kad surasti kam lygūs koeficientai
a
n
{\displaystyle a_{n}}
x
=
0
{\displaystyle x=0}
f
(
0
)
=
a
0
+
a
1
⋅
0
+
a
2
⋅
0
2
+
a
3
⋅
0
3
+
a
4
⋅
0
4
+
a
5
⋅
0
5
+
⋯
=
a
0
,
{\displaystyle f(0)=a_{0}+a_{1}\cdot 0+a_{2}\cdot 0^{2}+a_{3}\cdot 0^{3}+a_{4}\cdot 0^{4}+a_{5}\cdot 0^{5}+\cdots =a_{0},}
f
′
(
0
)
=
a
1
+
2
a
2
⋅
0
+
3
a
3
⋅
0
2
+
4
a
4
⋅
0
3
+
5
a
5
⋅
0
4
+
⋯
=
a
1
,
{\displaystyle f'(0)=a_{1}+2a_{2}\cdot 0+3a_{3}\cdot 0^{2}+4a_{4}\cdot 0^{3}+5a_{5}\cdot 0^{4}+\cdots =a_{1},}
f
″
(
0
)
=
2
a
2
+
3
⋅
2
a
3
⋅
0
+
4
⋅
3
a
4
⋅
0
2
+
5
⋅
4
a
5
⋅
0
3
+
⋯
=
2
a
2
=
2
!
⋅
a
2
,
{\displaystyle f''(0)=2a_{2}+3\cdot 2a_{3}\cdot 0+4\cdot 3a_{4}\cdot 0^{2}+5\cdot 4a_{5}\cdot 0^{3}+\cdots =2a_{2}=2!\cdot a_{2},}
f
‴
(
0
)
=
3
⋅
2
a
3
+
4
⋅
3
⋅
2
a
4
⋅
0
+
5
⋅
4
⋅
3
a
5
⋅
0
2
+
⋯
=
3
⋅
2
a
3
=
3
!
⋅
a
3
,
{\displaystyle f'''(0)=3\cdot 2a_{3}+4\cdot 3\cdot 2a_{4}\cdot 0+5\cdot 4\cdot 3a_{5}\cdot 0^{2}+\cdots =3\cdot 2a_{3}=3!\cdot a_{3},}
f
(
4
)
(
0
)
=
4
⋅
3
⋅
2
a
4
+
5
⋅
4
⋅
3
⋅
2
a
5
⋅
0
+
⋯
=
4
⋅
3
⋅
2
a
4
=
4
!
⋅
a
4
,
{\displaystyle f^{(4)}(0)=4\cdot 3\cdot 2a_{4}+5\cdot 4\cdot 3\cdot 2a_{5}\cdot 0+\cdots =4\cdot 3\cdot 2a_{4}=4!\cdot a_{4},}
f
(
5
)
(
0
)
=
5
⋅
4
⋅
3
⋅
2
a
5
+
⋯
=
5
⋅
4
⋅
3
⋅
2
a
5
=
5
!
⋅
a
5
.
{\displaystyle f^{(5)}(0)=5\cdot 4\cdot 3\cdot 2a_{5}+\cdots =5\cdot 4\cdot 3\cdot 2a_{5}=5!\cdot a_{5}.}
Dabar nesunku rasti koeficientus
a
n
{\displaystyle a_{n}}
a
0
=
f
(
0
)
,
{\displaystyle a_{0}=f(0),\;}
a
1
=
f
′
(
0
)
,
{\displaystyle a_{1}=f'(0),\;}
a
2
=
f
″
(
0
)
2
!
,
{\displaystyle a_{2}={\frac {f''(0)}{2!}},\;}
a
3
=
f
‴
(
0
)
3
!
,
{\displaystyle a_{3}={\frac {f'''(0)}{3!}},\;}
a
4
=
f
(
4
)
(
0
)
4
!
,
{\displaystyle a_{4}={\frac {f^{(4)}(0)}{4!}},\;}
a
5
=
f
(
5
)
(
0
)
5
!
,
{\displaystyle a_{5}={\frac {f^{(5)}(0)}{5!}},\;}
⋯
{\displaystyle \cdots }
a
n
=
f
(
n
)
(
0
)
n
!
.
{\displaystyle a_{n}={\frac {f^{(n)}(0)}{n!}}.\;}
Dabar tereikia įstatyti gautus koeficientus
a
n
{\displaystyle a_{n}}
f
(
x
)
=
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
a
4
x
4
+
a
5
x
5
+
⋯
=
f
(
0
)
+
f
′
(
0
)
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
f
(
4
)
(
0
)
4
!
x
4
+
f
(
5
)
(
0
)
5
!
x
5
+
⋯
.
{\displaystyle f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}+\cdots =f(0)+f'(0)x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+{\frac {f^{(4)}(0)}{4!}}x^{4}+{\frac {f^{(5)}(0)}{5!}}x^{5}+\cdots .}
Taigi, gavome funkciją
f
(
x
)
{\displaystyle f(x)\;}
f
(
x
)
=
f
(
0
)
+
f
′
(
0
)
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
f
(
4
)
(
0
)
4
!
x
4
+
f
(
5
)
(
0
)
5
!
x
5
+
⋯
.
{\displaystyle f(x)=f(0)+f'(0)x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+{\frac {f^{(4)}(0)}{4!}}x^{4}+{\frac {f^{(5)}(0)}{5!}}x^{5}+\cdots .}
Išreikšti funkciją
f
(
x
)
=
sin
(
x
)
{\displaystyle f(x)=\sin(x)\;}
Sprendimas .
sin
(
x
)
+
(
sin
(
x
)
)
′
x
+
(
sin
(
x
)
)
″
2
!
x
2
+
(
sin
(
x
)
)
‴
3
!
x
3
+
(
sin
(
x
)
)
(
4
)
4
!
x
4
+
(
sin
(
x
)
)
(
5
)
5
!
x
5
+
⋯
=
{\displaystyle \sin(x)+(\sin(x))'x+{\frac {(\sin(x))''}{2!}}x^{2}+{\frac {(\sin(x))'''}{3!}}x^{3}+{\frac {(\sin(x))^{(4)}}{4!}}x^{4}+{\frac {(\sin(x))^{(5)}}{5!}}x^{5}+\cdots =}
=
sin
(
x
)
+
x
cos
(
x
)
+
−
sin
(
x
)
2
!
x
2
+
−
cos
(
x
)
3
!
x
3
+
sin
(
x
)
4
!
x
4
+
cos
(
x
)
5
!
x
5
+
⋯
.
{\displaystyle =\sin(x)+x\cos(x)+{\frac {-\sin(x)}{2!}}x^{2}+{\frac {-\cos(x)}{3!}}x^{3}+{\frac {\sin(x)}{4!}}x^{4}+{\frac {\cos(x)}{5!}}x^{5}+\cdots .}
sin
(
x
)
=
sin
(
0
)
+
x
cos
(
0
)
+
−
sin
(
0
)
2
!
x
2
+
−
cos
(
0
)
3
!
x
3
+
sin
(
0
)
4
!
x
4
+
cos
(
0
)
5
!
x
5
+
⋯
=
{\displaystyle \sin(x)=\sin(0)+x\cos(0)+{\frac {-\sin(0)}{2!}}x^{2}+{\frac {-\cos(0)}{3!}}x^{3}+{\frac {\sin(0)}{4!}}x^{4}+{\frac {\cos(0)}{5!}}x^{5}+\cdots =}
=
0
+
x
⋅
1
+
0
2
!
x
2
+
−
1
3
!
x
3
+
0
4
!
x
4
+
1
5
!
x
5
+
⋯
=
{\displaystyle =0+x\cdot 1+{\frac {0}{2!}}x^{2}+{\frac {-1}{3!}}x^{3}+{\frac {0}{4!}}x^{4}+{\frac {1}{5!}}x^{5}+\cdots =}
=
x
−
x
3
3
!
+
x
5
5
!
−
⋯
.
{\displaystyle =x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots .}
Gauname, kad
sin
(
x
)
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
x
9
9
!
−
x
11
11
!
+
⋯
.
{\displaystyle \sin(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-{\frac {x^{11}}{11!}}+\cdots .}
Užrašyti funkcija
f
(
x
)
=
1
1
−
x
{\displaystyle f(x)={\frac {1}{1-x}}}
Sprendimas .
1
1
−
x
+
(
1
1
−
x
)
′
x
+
(
1
1
−
x
)
″
2
!
x
2
+
(
1
1
−
x
)
‴
3
!
x
3
+
(
1
1
−
x
)
(
4
)
4
!
x
4
+
(
1
1
−
x
)
(
5
)
5
!
x
5
+
⋯
=
{\displaystyle {\frac {1}{1-x}}+({\frac {1}{1-x}})'x+{\frac {({\frac {1}{1-x}})''}{2!}}x^{2}+{\frac {({\frac {1}{1-x}})'''}{3!}}x^{3}+{\frac {({\frac {1}{1-x}})^{(4)}}{4!}}x^{4}+{\frac {({\frac {1}{1-x}})^{(5)}}{5!}}x^{5}+\cdots =}
=
1
1
−
x
+
1
(
1
−
x
)
2
x
+
−
(
(
1
−
x
)
2
)
′
(
(
1
−
x
)
2
)
2
2
!
x
2
+
−
2
(
(
1
−
x
)
3
)
′
(
(
1
−
x
)
3
)
2
3
!
x
3
+
−
6
(
(
1
−
x
)
4
)
′
(
(
1
−
x
)
4
)
2
4
!
x
4
+
−
24
⋅
(
(
1
−
x
)
5
)
′
(
(
1
−
x
)
5
)
2
5
!
x
5
+
⋯
=
{\displaystyle ={\frac {1}{1-x}}+{\frac {1}{(1-x)^{2}}}x+{\frac {\frac {-((1-x)^{2})'}{((1-x)^{2})^{2}}}{2!}}x^{2}+{\frac {\frac {-2((1-x)^{3})'}{((1-x)^{3})^{2}}}{3!}}x^{3}+{\frac {\frac {-6((1-x)^{4})'}{((1-x)^{4})^{2}}}{4!}}x^{4}+{\frac {\frac {-24\cdot ((1-x)^{5})'}{((1-x)^{5})^{2}}}{5!}}x^{5}+\cdots =}
=
1
1
−
x
+
1
(
1
−
x
)
2
x
+
2
(
1
−
x
)
(
1
−
x
)
4
2
!
x
2
+
−
2
⋅
3
(
1
−
x
)
2
⋅
(
−
1
)
(
1
−
x
)
6
3
!
x
3
+
−
6
⋅
4
(
1
−
x
)
3
⋅
(
−
1
)
(
1
−
x
)
8
4
!
x
4
+
−
24
⋅
5
(
1
−
x
)
4
⋅
(
−
1
)
(
1
−
x
)
10
5
!
x
5
+
⋯
=
{\displaystyle ={\frac {1}{1-x}}+{\frac {1}{(1-x)^{2}}}x+{\frac {\frac {2(1-x)}{(1-x)^{4}}}{2!}}x^{2}+{\frac {\frac {-2\cdot 3(1-x)^{2}\cdot (-1)}{(1-x)^{6}}}{3!}}x^{3}+{\frac {\frac {-6\cdot 4(1-x)^{3}\cdot (-1)}{(1-x)^{8}}}{4!}}x^{4}+{\frac {\frac {-24\cdot 5(1-x)^{4}\cdot (-1)}{(1-x)^{10}}}{5!}}x^{5}+\cdots =}
=
1
1
−
x
+
1
(
1
−
x
)
2
x
+
2
(
1
−
x
)
3
2
!
x
2
+
6
(
1
−
x
)
4
3
!
x
3
+
24
(
1
−
x
)
5
4
!
x
4
+
120
(
1
−
x
)
6
5
!
x
5
+
⋯
=
{\displaystyle ={\frac {1}{1-x}}+{\frac {1}{(1-x)^{2}}}x+{\frac {\frac {2}{(1-x)^{3}}}{2!}}x^{2}+{\frac {\frac {6}{(1-x)^{4}}}{3!}}x^{3}+{\frac {\frac {24}{(1-x)^{5}}}{4!}}x^{4}+{\frac {\frac {120}{(1-x)^{6}}}{5!}}x^{5}+\cdots =}
=
1
1
−
x
+
1
(
1
−
x
)
2
x
+
1
(
1
−
x
)
3
x
2
+
1
(
1
−
x
)
4
x
3
+
1
(
1
−
x
)
5
x
4
+
1
(
1
−
x
)
6
x
5
+
⋯
.
{\displaystyle ={\frac {1}{1-x}}+{\frac {1}{(1-x)^{2}}}x+{\frac {1}{(1-x)^{3}}}x^{2}+{\frac {1}{(1-x)^{4}}}x^{3}+{\frac {1}{(1-x)^{5}}}x^{4}+{\frac {1}{(1-x)^{6}}}x^{5}+\cdots .}
f
(
x
)
=
1
1
−
x
=
f
(
0
)
+
f
′
(
0
)
1
!
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
f
(
4
)
(
0
)
4
!
x
4
+
f
(
5
)
(
0
)
5
!
x
5
+
⋯
=
{\displaystyle f(x)={\frac {1}{1-x}}=f(0)+{\frac {f'(0)}{1!}}x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+{\frac {f^{(4)}(0)}{4!}}x^{4}+{\frac {f^{(5)}(0)}{5!}}x^{5}+\cdots =}
=
1
1
−
0
+
1
(
1
−
0
)
2
1
!
x
+
2
(
1
−
0
)
3
2
!
x
2
+
6
(
1
−
0
)
4
3
!
x
3
+
24
(
1
−
0
)
5
4
!
x
4
+
120
(
1
−
0
)
6
5
!
x
5
+
⋯
=
{\displaystyle ={\frac {1}{1-0}}+{\frac {\frac {1}{(1-0)^{2}}}{1!}}x+{\frac {\frac {2}{(1-0)^{3}}}{2!}}x^{2}+{\frac {\frac {6}{(1-0)^{4}}}{3!}}x^{3}+{\frac {\frac {24}{(1-0)^{5}}}{4!}}x^{4}+{\frac {\frac {120}{(1-0)^{6}}}{5!}}x^{5}+\cdots =}
=
1
1
−
0
+
1
(
1
−
0
)
2
x
+
1
(
1
−
0
)
3
x
2
+
1
(
1
−
0
)
4
x
3
+
1
(
1
−
0
)
5
x
4
+
1
(
1
−
0
)
6
x
5
+
⋯
=
{\displaystyle ={\frac {1}{1-0}}+{\frac {1}{(1-0)^{2}}}x+{\frac {1}{(1-0)^{3}}}x^{2}+{\frac {1}{(1-0)^{4}}}x^{3}+{\frac {1}{(1-0)^{5}}}x^{4}+{\frac {1}{(1-0)^{6}}}x^{5}+\cdots =}
=
1
+
x
+
x
2
+
x
3
+
x
4
+
x
5
+
⋯
.
{\displaystyle =1+x+x^{2}+x^{3}+x^{4}+x^{5}+\cdots .}
Teiloro eilute užrašomos funkcijos pavidalas yra toks:
f
(
x
)
=
f
(
x
0
)
+
f
′
(
x
0
)
1
!
(
x
−
x
0
)
+
f
″
(
x
0
)
2
!
(
x
−
x
0
)
2
+
f
‴
(
x
0
)
3
!
(
x
−
x
0
)
3
+
.
.
.
+
f
(
n
)
(
x
0
)
n
!
(
x
−
x
0
)
n
+
f
(
n
+
1
)
(
c
)
(
n
+
1
)
!
(
x
−
x
0
)
n
+
1
.
(
8
)
{\displaystyle f(x)=f(x_{0})+{\frac {f'(x_{0})}{1!}}(x-x_{0})+{\frac {f''(x_{0})}{2!}}(x-x_{0})^{2}+{\frac {f'''(x_{0})}{3!}}(x-x_{0})^{3}+...+{\frac {f^{(n)}(x_{0})}{n!}}(x-x_{0})^{n}+{\frac {f^{(n+1)}(c)}{(n+1)!}}(x-x_{0})^{n+1}.\quad (8)}
Čia liekamasis narys Lagranžo formoje yra
r
n
+
1
=
f
(
n
+
1
)
(
c
)
(
n
+
1
)
!
(
x
−
x
0
)
n
+
1
.
{\displaystyle r_{n+1}={\frac {f^{(n+1)}(c)}{(n+1)!}}(x-x_{0})^{n+1}.}
x
0
<
c
<
x
.
{\displaystyle x_{0}<c<x.}
Teiloro formulė, kurios
x
0
=
0
,
{\displaystyle x_{0}=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
+
f
(
n
+
1
)
(
c
)
(
n
+
1
)
!
x
n
+
1
.
(
9
)
{\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}+{\frac {f^{(n+1)}(c)}{(n+1)!}}x^{n+1}.\quad (9)}
Šįkart
0
<
c
<
x
.
{\displaystyle 0<c<x.}
keisti
Apskaičiuosime sin(1) su paklaida mažesne nei
10
−
6
.
{\displaystyle 10^{-6}.}
Žinoma, kad
f
(
n
)
(
x
)
=
(
sin
x
)
(
n
)
=
sin
(
n
π
2
+
x
)
.
{\displaystyle f^{(n)}(x)=(\sin x)^{(n)}=\sin(n{\frac {\pi }{2}}+x).}
Tuomet
f
(
n
)
(
0
)
=
sin
(
n
π
2
)
=
{
0
,
kai
n
=
2
k
,
(
−
1
)
k
,
kai
n
=
2
k
+
1.
{\displaystyle f^{(n)}(0)=\sin({\frac {n\pi }{2}})={\begin{cases}\quad 0,\quad \quad {\text{kai}}\;n=2k,&\\(-1)^{k},\quad {\text{kai}}\;n=2k+1.&\end{cases}}}
Čia k yra sveikasis skaičius. Todėl
sin
x
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
.
.
.
+
(
−
1
)
k
x
2
k
+
1
(
2
k
+
1
)
!
+
r
2
k
+
3
(
x
)
;
(
11
)
{\displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+...+(-1)^{k}{\frac {x^{2k+1}}{(2k+1)!}}+r_{2k+3}(x);\quad (11)}
čia
|
r
2
k
+
3
(
x
)
|
=
|
x
2
k
+
3
(
2
k
+
3
)
!
sin
(
(
2
k
+
3
)
π
2
+
c
)
|
=
|
x
2
k
+
3
(
2
k
+
3
)
!
cos
c
|
(
0
<
c
<
x
)
.
{\displaystyle |r_{2k+3}(x)|=|{\frac {x^{2k+3}}{(2k+3)!}}\sin((2k+3){\frac {\pi }{2}}+c)|=|{\frac {x^{2k+3}}{(2k+3)!}}\cos c|\;\;(0<c<x).}
Laikysime, kad cos(c)=1. Tada, tarus, kad x=1, turi būti tenkinama nelygybė
|
x
2
k
+
3
(
2
k
+
3
)
!
cos
c
|
<
10
−
6
,
{\displaystyle |{\frac {x^{2k+3}}{(2k+3)!}}\cos c|<10^{-6},}
|
1
2
k
+
3
(
2
k
+
3
)
!
⋅
1
|
<
10
−
6
,
{\displaystyle |{\frac {1^{2k+3}}{(2k+3)!}}\cdot 1|<10^{-6},}
|
1
(
2
k
+
3
)
!
|
<
10
−
6
,
{\displaystyle |{\frac {1}{(2k+3)!}}|<10^{-6},}
10
6
<
(
2
k
+
3
)
!
.
{\displaystyle 10^{6}<(2k+3)!.}
Paskutinė nelygybė tenkinama, kai
k
=
4
{\displaystyle k=4}
(
2
k
+
3
)
!
=
(
2
⋅
4
+
3
)
!
=
11
!
=
39916800.
{\displaystyle (2k+3)!=(2\cdot 4+3)!=11!=39916800.}
sin
1
=
1
−
1
3
3
!
+
1
5
5
!
−
1
7
7
!
+
1
9
9
!
=
{\displaystyle \sin 1=1-{\frac {1^{3}}{3!}}+{\frac {1^{5}}{5!}}-{\frac {1^{7}}{7!}}+{\frac {1^{9}}{9!}}=}
=
1
−
1
6
+
1
120
−
1
5040
+
1
362880
=
{\displaystyle =1-{\frac {1}{6}}+{\frac {1}{120}}-{\frac {1}{5040}}+{\frac {1}{362880}}=}
=0.84147100970017636684303350970018.
Tikroji (tiksli) sin(1) reikšmė iš skaičiuotuvo yra:
sin(1)=
=0.8414709848078965066525023216303.
Atėmus gautą mūsų skaičiuotą sin(1) reikšmę iš kalkuliatoriaus sin(1) reikšmės gauname tokią paklaidą:
0.8414709848078965066525023216303 - 0.84147100970017636684303350970018=
=-2.4892279860190531188069881000377e-8=
=
−
2.489227986019
⋅
10
−
8
;
{\displaystyle =-2.489227986019\cdot 10^{-8};}
|
−
2.489227986019
⋅
10
−
8
|
<
10
−
6
.
{\displaystyle |-2.489227986019\cdot 10^{-8}|<10^{-6}.}
Taigi, viskas apskaičiuota pagal teoriją ir paklaida surasta teisingai.
Pastebėsime, kad
11
!
=
39916800
>
10
7
,
{\displaystyle 11!=39916800>10^{7},}
|
−
2.489227986019
⋅
10
−
8
|
<
10
−
7
.
{\displaystyle |-2.489227986019\cdot 10^{-8}|<10^{-7}.}
Apskaičiuosime sin(0.1) su paklaida mažesne nei
10
−
9
.
{\displaystyle 10^{-9}.}
Pasinaudoję pirmu pavyzdžiu, išeina, kad turi būti tenkinama nelygybė
|
x
2
k
+
3
(
2
k
+
3
)
!
cos
c
|
<
10
−
9
,
{\displaystyle |{\frac {x^{2k+3}}{(2k+3)!}}\cos c|<10^{-9},}
kurioje laikysime, kad cos(c)=1, o
x
=
0.1.
{\displaystyle x=0.1.}
0.1
2
k
+
3
(
2
k
+
3
)
!
⋅
1
<
10
−
9
.
{\displaystyle {\frac {0.1^{2k+3}}{(2k+3)!}}\cdot 1<10^{-9}.}
Pabandykime atspėti, kad sveikasis skaičius k galėtų būti lygus 2. Tada
0.1
2
⋅
2
+
3
(
2
⋅
2
+
3
)
!
<
10
−
9
,
{\displaystyle {\frac {0.1^{2\cdot 2+3}}{(2\cdot 2+3)!}}<10^{-9},}
0.1
7
7
!
<
10
−
9
,
{\displaystyle {\frac {0.1^{7}}{7!}}<10^{-9},}
1
7
!
<
10
−
9
⋅
10
7
,
{\displaystyle {\frac {1}{7!}}<10^{-9}\cdot 10^{7},}
1
7
!
<
10
−
2
,
{\displaystyle {\frac {1}{7!}}<10^{-2},}
10
2
<
7
!
,
{\displaystyle 10^{2}<7!,}
100
<
7
!
=
5040.
{\displaystyle 100<7!=5040.}
Nesunku matyti, kad su skaičiumi k =2, paklaida bus mažesnė už
10
−
10
.
{\displaystyle 10^{-10}.}
x =0.1, o n =5:
sin
(
0.1
)
=
0.1
−
0.1
3
3
!
+
0.1
5
5
!
=
{\displaystyle \sin(0.1)=0.1-{\frac {0.1^{3}}{3!}}+{\frac {0.1^{5}}{5!}}=}
=
0.1
−
0.001
6
+
0.00001
120
=
{\displaystyle =0.1-{\frac {0.001}{6}}+{\frac {0.00001}{120}}=}
= 0.09983341666666666666666666666667.
Windows kalkuliatoriaus
sin
0.1
{\displaystyle \sin 0.1}
sin(0.1) =
= 0.09983341664682815230681419841062.
0.09983341664682815230681419841062 - 0.09983341666666666666666666666667 =
= -1.9838514359852468256047973010085e-11 =
=
−
1.983851
⋅
10
−
11
.
{\displaystyle =-1.983851\cdot 10^{-11}.}
|
−
1.983851
⋅
10
−
11
|
<
10
−
10
<
10
−
9
.
{\displaystyle |-1.983851\cdot 10^{-11}|<10^{-10}<10^{-9}.}
Vadinasi paklaidą radome teisingai.
Apskaičiuosime sin(1.5) su paklaida mažesne nei
10
−
6
.
{\displaystyle 10^{-6}.}
Pasinaudoję pirmu pavyzdžiu, išeina, kad turi būti tenkinama nelygybė
|
x
2
k
+
3
(
2
k
+
3
)
!
cos
c
|
<
10
−
6
,
{\displaystyle |{\frac {x^{2k+3}}{(2k+3)!}}\cos c|<10^{-6},}
kurioje laikysime, kad
cos
c
=
1
,
{\displaystyle \cos c=1,}
x
=
1.5
{\displaystyle x=1.5}
k
=
5
,
{\displaystyle k=5,}
|
1.5
2
⋅
5
+
3
(
2
⋅
5
+
3
)
!
⋅
1
|
<
10
−
6
,
{\displaystyle |{\frac {1.5^{2\cdot 5+3}}{(2\cdot 5+3)!}}\cdot 1|<10^{-6},}
|
1.5
13
13
!
⋅
1
|
<
10
−
6
,
{\displaystyle |{\frac {1.5^{13}}{13!}}\cdot 1|<10^{-6},}
1.5
13
13
!
<
10
−
6
,
{\displaystyle {\frac {1.5^{13}}{13!}}<10^{-6},}
194.6195068359375
6227020800
<
10
−
6
,
{\displaystyle {\frac {194.6195068359375}{6227020800}}<10^{-6},}
≈
3.1254031917789242
⋅
10
−
8
<
10
−
6
.
{\displaystyle \approx 3.1254031917789242\cdot 10^{-8}<10^{-6}.}
Taigi, matome, kad paklaida bus mažesnė už
10
−
7
.
{\displaystyle 10^{-7}.}
x
=
1.5
,
{\displaystyle x=1.5,}
n
=
11
{\displaystyle n=11}
sin
(
1.5
)
=
1.5
−
1.5
3
3
!
+
1.5
5
5
!
−
1.5
7
7
!
+
1.5
9
9
!
−
1.5
11
11
!
=
{\displaystyle \sin(1.5)=1.5-{\frac {1.5^{3}}{3!}}+{\frac {1.5^{5}}{5!}}-{\frac {1.5^{7}}{7!}}+{\frac {1.5^{9}}{9!}}-{\frac {1.5^{11}}{11!}}=}
=
1.5
−
3.375
6
+
7.59375
120
−
17.0859375
5040
+
38.443359375
362880
−
86.49755859375
39916800
=
{\displaystyle =1.5-{\frac {3.375}{6}}+{\frac {7.59375}{120}}-{\frac {17.0859375}{5040}}+{\frac {38.443359375}{362880}}-{\frac {86.49755859375}{39916800}}=}
= 1.5 - 3.375/6 + 7.59375/120 - 17.0859375/5040 + 38.443359375/362880 - 86.49755859375/39916800 = (šitą išraišką galima iškart įdėt (padaryt "Paste") į Windows kalkuliatorių ir jis iškart atliks visus aritmetinius veiksmus ir duos atsakymą)
= 0.99749495568213524756493506493506.
Gautą sin(1.5) reikšmę atėmę iš tikslios (kalkuliatoriaus) sin(1.5) reikšmės gauname paklaidą:
0.99749498660405443094172337114149 - 0.99749495568213524756493506493506 =
= 3.0921919183376788306206422387642e-8 =
=
3.0921919
⋅
10
−
8
<
10
−
7
<
10
−
6
.
{\displaystyle =3.0921919\cdot 10^{-8}<10^{-7}<10^{-6}.}
Paklaidos didumas gautas pagal teoriją.
keisti
Apskaičiuosime cos(1) su paklaida mažesne nei
10
−
6
.
{\displaystyle 10^{-6}.}
f
(
x
)
=
cos
x
.
{\displaystyle f(x)=\cos x.}
f
(
n
)
(
x
)
=
(
cos
x
)
(
n
)
=
cos
(
n
π
2
+
x
)
.
{\displaystyle f^{(n)}(x)=(\cos x)^{(n)}=\cos(n{\frac {\pi }{2}}+x).}
Tuomet
f
(
n
)
(
0
)
=
cos
n
π
2
=
{
0
,
kai
n
=
2
k
+
1
,
(
−
1
)
k
,
kai
n
=
2
k
.
{\displaystyle f^{(n)}(0)=\cos {\frac {n\pi }{2}}={\begin{cases}\quad 0,\quad \quad {\text{kai}}\;n=2k+1,&\\(-1)^{k},\quad {\text{kai}}\;n=2k.&\end{cases}}}
Čia k yra sveikasis skaičius. Todėl
cos
x
=
1
−
x
2
2
!
+
x
4
4
!
−
x
6
6
!
+
.
.
.
+
(
−
1
)
k
x
2
k
(
2
k
)
!
+
r
2
k
+
2
(
x
)
;
(
12
)
{\displaystyle \cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+...+(-1)^{k}{\frac {x^{2k}}{(2k)!}}+r_{2k+2}(x);\quad (12)}
čia
|
r
2
k
+
2
(
x
)
|
=
|
x
2
k
+
2
(
2
k
+
2
)
!
cos
(
(
2
k
+
2
)
π
2
+
c
)
|
=
|
x
2
k
+
2
(
2
k
+
2
)
!
cos
c
|
(
0
<
c
<
x
)
.
{\displaystyle |r_{2k+2}(x)|=|{\frac {x^{2k+2}}{(2k+2)!}}\cos((2k+2){\frac {\pi }{2}}+c)|=|{\frac {x^{2k+2}}{(2k+2)!}}\cos c|\;\;(0<c<x).}
Tada
|
r
2
k
+
2
(
x
)
|
=
|
x
2
k
+
2
(
2
k
+
2
)
!
cos
c
|
<
x
2
k
+
2
(
2
k
+
2
)
!
{\displaystyle |r_{2k+2}(x)|=|{\frac {x^{2k+2}}{(2k+2)!}}\cos c|<{\frac {x^{2k+2}}{(2k+2)!}}}
(nes |cos(c)| negali būti daugiau už 1),
|
r
2
k
+
2
(
x
)
|
<
x
2
k
+
2
(
2
k
+
2
)
!
<
10
−
6
.
{\displaystyle |r_{2k+2}(x)|<{\frac {x^{2k+2}}{(2k+2)!}}<10^{-6}.}
Paskutinė nelygybė tenkinama, kai
k
=
4
,
{\displaystyle k=4,}
|
r
2
k
+
2
(
1
)
|
<
1
2
⋅
4
+
2
(
2
⋅
4
+
2
)
!
<
10
−
6
,
{\displaystyle |r_{2k+2}(1)|<{\frac {1^{2\cdot 4+2}}{(2\cdot 4+2)!}}<10^{-6},}
|
r
2
k
+
2
(
1
)
|
<
1
10
!
<
10
−
6
,
{\displaystyle |r_{2k+2}(1)|<{\frac {1}{10!}}<10^{-6},}
|
r
2
k
+
2
(
1
)
|
<
1
3628800
<
10
−
6
,
{\displaystyle |r_{2k+2}(1)|<{\frac {1}{3628800}}<10^{-6},}
|
r
2
k
+
2
(
1
)
|
<
2.75573192
⋅
10
−
7
<
10
−
6
.
{\displaystyle |r_{2k+2}(1)|<2.75573192\cdot 10^{-7}<10^{-6}.}
Dabar galime užrašyti
cos
1
{\displaystyle \cos 1}
cos
1
=
1
−
1
2
2
!
+
1
4
4
!
−
1
6
6
!
+
1
8
8
!
=
{\displaystyle \cos 1=1-{\frac {1^{2}}{2!}}+{\frac {1^{4}}{4!}}-{\frac {1^{6}}{6!}}+{\frac {1^{8}}{8!}}=}
=
1
−
1
2
+
1
24
−
1
720
+
1
40320
=
{\displaystyle =1-{\frac {1}{2}}+{\frac {1}{24}}-{\frac {1}{720}}+{\frac {1}{40320}}=}
= 0.54030257936507936507936507936508.
Atėmę gautą
cos
1
{\displaystyle \cos 1}
cos
1
{\displaystyle \cos 1}
cos(1) - 0.54030257936507936507936507936508 =
= 0.54030230586813971740093660744298 - 0.54030257936507936507936507936508 =
= -2.7349693964767842847192210276135e-7.
|
−
2.734969396
⋅
10
−
7
|
<
2.75573192
⋅
10
−
7
<
10
−
6
.
{\displaystyle |-2.734969396\cdot 10^{-7}|<2.75573192\cdot 10^{-7}<10^{-6}.}
Vadinasi, liekamajį narį
|
r
2
k
+
2
(
x
)
|
=
|
x
2
k
+
2
(
2
k
+
2
)
!
cos
c
|
{\displaystyle |r_{2k+2}(x)|=|{\frac {x^{2k+2}}{(2k+2)!}}\cos c|}
Apskaičiuosime cos(0.1) su paklaida mažesne nei
10
−
8
.
{\displaystyle 10^{-8}.}
Pasinaudoję pirmu pavyzdžiu, turime
|
r
2
k
+
2
(
x
)
|
=
|
x
2
k
+
2
(
2
k
+
2
)
!
cos
c
|
<
x
2
k
+
2
(
2
k
+
2
)
!
,
{\displaystyle |r_{2k+2}(x)|=|{\frac {x^{2k+2}}{(2k+2)!}}\cos c|<{\frac {x^{2k+2}}{(2k+2)!}},}
|
r
2
k
+
2
(
x
)
|
<
x
2
k
+
2
(
2
k
+
2
)
!
<
10
−
8
.
{\displaystyle |r_{2k+2}(x)|<{\frac {x^{2k+2}}{(2k+2)!}}<10^{-8}.}
Paskutinė nelygybė tenkinama, kai
k
=
2
,
{\displaystyle k=2,}
x
=
0.1
{\displaystyle x=0.1}
|
r
2
k
+
2
(
0.1
)
|
<
0.1
2
⋅
2
+
2
(
2
⋅
2
+
2
)
!
<
10
−
8
,
{\displaystyle |r_{2k+2}(0.1)|<{\frac {0.1^{2\cdot 2+2}}{(2\cdot 2+2)!}}<10^{-8},}
|
r
2
k
+
2
(
0.1
)
|
<
0.1
6
6
!
<
10
−
8
,
{\displaystyle |r_{2k+2}(0.1)|<{\frac {0.1^{6}}{6!}}<10^{-8},}
|
r
2
k
+
2
(
0.1
)
|
<
0.000001
720
<
10
−
8
,
{\displaystyle |r_{2k+2}(0.1)|<{\frac {0.000001}{720}}<10^{-8},}
|
r
2
k
+
2
(
0.1
)
|
<
1.388888889
⋅
10
−
9
<
10
−
8
.
{\displaystyle |r_{2k+2}(0.1)|<1.388888889\cdot 10^{-9}<10^{-8}.}
Dabar galime užrašyti
cos
(
0.1
)
{\displaystyle \cos(0.1)}
cos
0.1
=
1
−
0.1
2
2
!
+
0.1
4
4
!
=
1
−
0.01
2
+
0.0001
24
=
{\displaystyle \cos 0.1=1-{\frac {0.1^{2}}{2!}}+{\frac {0.1^{4}}{4!}}=1-{\frac {0.01}{2}}+{\frac {0.0001}{24}}=}
= 0.9950041(6).
Tiksli
cos
(
0.1
)
{\displaystyle \cos(0.1)}
cos(0.1) = 0.99500416527802576609556198780387.
Atėmę apskaičiuotą
cos
(
0.1
)
{\displaystyle \cos(0.1)}
cos
(
0.1
)
{\displaystyle \cos(0.1)}
0.99500416527802576609556198780387 - 0.99500416666666666666666666666667 =
= -1.3886409005711046788627997051614e-9 =
=
−
1.3886409
⋅
10
−
9
.
{\displaystyle =-1.3886409\cdot 10^{-9}.}
Taigi
|
r
2
k
+
2
(
0.1
)
|
≈
|
−
1.3886409
⋅
10
−
9
|
<
1.388888889
⋅
10
−
9
<
10
−
8
.
{\displaystyle |r_{2k+2}(0.1)|\approx |-1.3886409\cdot 10^{-9}|<1.388888889\cdot 10^{-9}<10^{-8}.}
Ką ir reikėjo parodyti.
Apskaičiuosime cos(1.5) su paklaida mažesne nei
10
−
8
.
{\displaystyle 10^{-8}.}
Pasinaudoję pirmu pavyzdžiu, turime
|
r
2
k
+
2
(
x
)
|
=
|
x
2
k
+
2
(
2
k
+
2
)
!
cos
c
|
<
x
2
k
+
2
(
2
k
+
2
)
!
,
(
0
<
c
<
x
)
{\displaystyle |r_{2k+2}(x)|=|{\frac {x^{2k+2}}{(2k+2)!}}\cos c|<{\frac {x^{2k+2}}{(2k+2)!}},\;\;(0<c<x)}
|
r
2
k
+
2
(
1.5
)
|
=
|
1.5
2
k
+
2
(
2
k
+
2
)
!
cos
c
|
<
1.5
2
k
+
2
(
2
k
+
2
)
!
,
(
0
<
c
<
1.5
)
{\displaystyle |r_{2k+2}(1.5)|=|{\frac {1.5^{2k+2}}{(2k+2)!}}\cos c|<{\frac {1.5^{2k+2}}{(2k+2)!}},\;\;(0<c<1.5)}
|
r
2
k
+
2
(
1.5
)
|
<
1.5
2
k
+
2
(
2
k
+
2
)
!
<
10
−
8
.
{\displaystyle |r_{2k+2}(1.5)|<{\frac {1.5^{2k+2}}{(2k+2)!}}<10^{-8}.}
Paskutinė nelygybė tenkinama, kai
k
=
6
,
{\displaystyle k=6,}
|
r
2
k
+
2
(
1.5
)
|
<
1.5
2
⋅
6
+
2
(
2
⋅
6
+
2
)
!
<
10
−
8
.
{\displaystyle |r_{2k+2}(1.5)|<{\frac {1.5^{2\cdot 6+2}}{(2\cdot 6+2)!}}<10^{-8}.}
|
r
14
(
1.5
)
|
<
1.5
14
14
!
<
10
−
8
.
{\displaystyle |r_{14}(1.5)|<{\frac {1.5^{14}}{14!}}<10^{-8}.}
|
r
14
(
1.5
)
|
<
291.92926025390625
87178291200
<
10
−
8
.
{\displaystyle |r_{14}(1.5)|<{\frac {291.92926025390625}{87178291200}}<10^{-8}.}
|
r
14
(
1.5
)
|
<
3.3486462769
⋅
10
−
9
<
10
−
8
.
{\displaystyle |r_{14}(1.5)|<3.3486462769\cdot 10^{-9}<10^{-8}.}
Dabar užrašysime
cos
(
1.5
)
{\displaystyle \cos(1.5)}
cos
1.5
=
1
−
1.5
2
2
!
+
1.5
4
4
!
−
1.5
6
6
!
+
1.5
8
8
!
−
1.5
10
10
!
+
1.5
12
12
!
=
{\displaystyle \cos 1.5=1-{\frac {1.5^{2}}{2!}}+{\frac {1.5^{4}}{4!}}-{\frac {1.5^{6}}{6!}}+{\frac {1.5^{8}}{8!}}-{\frac {1.5^{10}}{10!}}+{\frac {1.5^{12}}{12!}}=}
=
1
−
2.25
2
+
5.0625
24
−
11.390625
720
+
25.62890625
40320
−
57.6650390625
3628800
+
129.746337890625
479001600
=
{\displaystyle =1-{\frac {2.25}{2}}+{\frac {5.0625}{24}}-{\frac {11.390625}{720}}+{\frac {25.62890625}{40320}}-{\frac {57.6650390625}{3628800}}+{\frac {129.746337890625}{479001600}}=}
= 1 - 2.25/2 + 5.0625/24 - 11.390625/720 + 25.62890625/40320 - 57.6650390625/3628800 + 129.746337890625/479001600 =
(įstačius šią išraišką nuo "1" iki "=" į Windows'ų kalkuliatorių gaunamas iškart apskaičiuotas atsakymas)
= 0.07073720498518510298295454545455.
Tiksli
cos
(
1.5
)
{\displaystyle \cos(1.5)}
Windows kalkuliatoriaus yra
cos(1.5) = 0.07073720166770291008818985143427.
Atėmę apskaičiuotą nevisiškai tikslią cos(1.5) reikšmę iš tikslios cos(1.5) reikšmės, gauname paklaidą:
0.07073720166770291008818985143427 - 0.07073720498518510298295454545455 =
= -3.3174821928947646940202767454604e-9.
Matome, kad
|
r
14
(
1.5
)
|
≈
|
−
3.31748219
⋅
10
−
9
|
<
3.3486462769
⋅
10
−
9
<
10
−
8
.
{\displaystyle |r_{14}(1.5)|\approx |-3.31748219\cdot 10^{-9}|<3.3486462769\cdot 10^{-9}<10^{-8}.}
keisti
Natūrinio logaritmo reikšmės greitas skaičiavimas: https://en.wikipedia.org/wiki/Natural_logarithm#Efficient_computation
f
(
x
)
=
ln
(
1
+
x
)
.
{\displaystyle f(x)=\ln(1+x).}
x
=
0.5.
{\displaystyle x=0.5.}
Randame išvestines:
f
′
(
x
)
=
1
1
+
x
,
f
″
(
x
)
=
−
1
(
1
+
x
)
2
,
f
‴
(
x
)
=
(
−
1
)
2
⋅
2
(
1
+
x
)
3
,
f
(
4
)
(
x
)
=
(
−
1
)
3
⋅
2
⋅
3
(
1
+
x
)
4
,
.
.
.
,
{\displaystyle f'(x)={\frac {1}{1+x}},\;\;f''(x)=-{\frac {1}{(1+x)^{2}}},\;\;f'''(x)={\frac {(-1)^{2}\cdot 2}{(1+x)^{3}}},\;\;f^{(4)}(x)={\frac {(-1)^{3}\cdot 2\cdot 3}{(1+x)^{4}}},\;...,}
f
(
n
)
(
x
)
=
(
−
1
)
n
−
1
(
n
−
1
)
!
(
1
+
x
)
n
,
n
=
1
,
2
,
3
,
.
.
.
{\displaystyle f^{(n)}(x)={\frac {(-1)^{n-1}(n-1)!}{(1+x)^{n}}},\;\;n=1,\;2,\;3,\;...}
Apskaičiuosime jų reikšmes taške
x
=
0
{\displaystyle x=0}
f
(
0
)
=
ln
(
1
+
0
)
=
0
,
f
′
(
0
)
=
1
,
f
″
(
0
)
=
−
1
,
f
‴
(
0
)
=
(
−
1
)
2
⋅
2
=
2
,
f
(
4
)
(
0
)
=
(
−
1
)
3
⋅
2
⋅
3
=
−
3
!
,
.
.
.
,
{\displaystyle f(0)=\ln(1+0)=0,\;\;f'(0)=1,\;\;f''(0)=-1,\;\;f'''(0)=(-1)^{2}\cdot 2=2,\;\;f^{(4)}(0)=(-1)^{3}\cdot 2\cdot 3=-3!,\;...,}
f
(
n
)
(
0
)
=
(
−
1
)
n
−
1
(
n
−
1
)
!
{\displaystyle f^{(n)}(0)=(-1)^{n-1}(n-1)!}
Vadinasi,
ln
(
1
+
x
)
=
f
(
0
)
+
f
′
(
0
)
1
!
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
f
(
4
)
(
0
)
4
!
x
4
+
.
.
.
+
f
(
n
)
(
0
)
n
!
x
n
+
f
(
n
+
1
)
(
c
)
(
n
+
1
)
!
x
n
+
1
=
{\displaystyle \ln(1+x)=f(0)+{\frac {f'(0)}{1!}}x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+{\frac {f^{(4)}(0)}{4!}}x^{4}+...+{\frac {f^{(n)}(0)}{n!}}x^{n}+{\frac {f^{(n+1)}(c)}{(n+1)!}}x^{n+1}=}
=
x
−
x
2
2
+
x
3
3
−
x
4
4
.
.
.
+
(
−
1
)
n
−
1
x
n
n
+
r
n
+
1
(
x
)
;
(
13
)
{\displaystyle =x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}...+(-1)^{n-1}{\frac {x^{n}}{n}}+r_{n+1}(x);\quad (13)}
čia
|
r
n
+
1
(
x
)
|
=
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
,
0
<
c
<
0.5.
{\displaystyle |r_{n+1}(x)|={\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}\;,\;\;0<c<0.5.}
Matome, kad
|
r
n
+
1
(
x
)
|
=
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
<
|
x
n
+
1
(
n
+
1
)
|
.
{\displaystyle |r_{n+1}(x)|={\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}<{\Big |}{\frac {x^{n+1}}{(n+1)}}{\Big |}.}
Kai
n
=
9
,
{\displaystyle n=9,}
x
=
0.5
,
{\displaystyle x=0.5,}
|
r
n
+
1
(
x
)
|
<
0.0001
,
{\displaystyle |r_{n+1}(x)|<0.0001,}
|
0.5
9
+
1
(
9
+
1
)
|
=
0.5
10
10
=
0.0009765625
10
=
0.00009765625
{\displaystyle {\Big |}{\frac {0.5^{9+1}}{(9+1)}}{\Big |}={\frac {0.5^{10}}{10}}={\frac {0.0009765625}{10}}=0.00009765625}
ir 0.00009765625 < 0.0001. Apskaičiuosime kam lygi ši eilutė, kai
n
=
9
,
{\displaystyle n=9,}
x
=
0.5
{\displaystyle x=0.5}
ln
(
1
+
0.5
)
=
0.5
−
0.5
2
2
+
0.5
3
3
−
0.5
4
4
+
0.5
5
5
−
0.5
6
6
+
0.5
7
7
−
0.5
8
8
+
0.5
9
9
=
{\displaystyle \ln(1+0.5)=0.5-{\frac {0.5^{2}}{2}}+{\frac {0.5^{3}}{3}}-{\frac {0.5^{4}}{4}}+{\frac {0.5^{5}}{5}}-{\frac {0.5^{6}}{6}}+{\frac {0.5^{7}}{7}}-{\frac {0.5^{8}}{8}}+{\frac {0.5^{9}}{9}}=}
=0.5-0.25/2+0.125/3-0.0625/4+0.03125/5-0.015625/6+0.0078125/7-0.00390625/8+0.001953125/9=
(įdėjus (padarius "Copy"-"Paste") šią eilutę nuo "0.5" iki "=" į Windows kalkuliatorių iškart gauname atsakymą)
=0.40553230406746031746031746031746.
Kalukiatoriaus ln(1.5) reikšmė yra tokia: ln(1.5)=
=0.40546510810816438197801311546435.
Atėmę mūsų apskaičiuotą eilutės reikšmę iš kalkuliatoriaus ln(1.5) reikšmės, gauname
0.40546510810816438197801311546435 - 0.40553230406746031746031746031746 = -6.7195959295935482304344853111181e-5.
Matome, kad
|
r
9
+
1
(
0.5
)
|
=
|
−
6.71959592959
⋅
10
−
5
|
<
0.00009765625
<
0.0001.
{\displaystyle |r_{9+1}(0.5)|=|-6.71959592959\cdot 10^{-5}|<0.00009765625<0.0001.}
f
(
x
)
=
ln
(
1
+
x
)
.
{\displaystyle f(x)=\ln(1+x).}
x
=
0.9.
{\displaystyle x=0.9.}
ln
(
1
+
x
)
=
x
−
x
2
2
+
x
3
3
−
x
4
4
.
.
.
+
(
−
1
)
n
−
1
x
n
n
+
r
n
+
1
(
x
)
;
(
13
)
{\displaystyle \ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}...+(-1)^{n-1}{\frac {x^{n}}{n}}+r_{n+1}(x);\quad (13)}
čia
|
r
n
+
1
(
x
)
|
=
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
,
0
<
c
<
0.9.
{\displaystyle |r_{n+1}(x)|={\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}\;,\;\;0<c<0.9.}
Matome, kad
|
r
n
+
1
(
x
)
|
=
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
<
|
x
n
+
1
n
+
1
|
.
{\displaystyle |r_{n+1}(x)|={\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}<{\Big |}{\frac {x^{n+1}}{n+1}}{\Big |}.}
Kai
n
=
19
,
{\displaystyle n=19,}
x
=
0.9
,
{\displaystyle x=0.9,}
|
r
n
+
1
(
x
)
|
<
0.01
,
{\displaystyle |r_{n+1}(x)|<0.01,}
0.9
19
+
1
19
+
1
=
0.9
20
20
=
0.12157665459056928801
20
=
0.0060788327295284644005
{\displaystyle {\frac {0.9^{19+1}}{19+1}}={\frac {0.9^{20}}{20}}={\frac {0.12157665459056928801}{20}}=0.0060788327295284644005}
ir 0.0060788327295284644005 < 0.01. Apskaičiuosime kam lygi ši eilutė, kai
n
=
19
,
{\displaystyle n=19,}
x
=
0.9
{\displaystyle x=0.9}
ln
(
1
+
0.9
)
=
0.9
−
0.9
2
2
+
0.9
3
3
−
0.9
4
4
+
0.9
5
5
−
0.9
6
6
+
0.9
7
7
−
0.9
8
8
+
0.9
9
9
−
0.9
10
10
+
{\displaystyle \ln(1+0.9)=0.9-{\frac {0.9^{2}}{2}}+{\frac {0.9^{3}}{3}}-{\frac {0.9^{4}}{4}}+{\frac {0.9^{5}}{5}}-{\frac {0.9^{6}}{6}}+{\frac {0.9^{7}}{7}}-{\frac {0.9^{8}}{8}}+{\frac {0.9^{9}}{9}}-{\frac {0.9^{10}}{10}}+}
+
0.9
11
11
−
0.9
12
12
+
0.9
13
13
−
0.9
14
14
+
0.9
15
15
−
0.9
16
16
+
0.9
17
17
−
0.9
18
18
+
0.9
19
19
=
{\displaystyle +{\frac {0.9^{11}}{11}}-{\frac {0.9^{12}}{12}}+{\frac {0.9^{13}}{13}}-{\frac {0.9^{14}}{14}}+{\frac {0.9^{15}}{15}}-{\frac {0.9^{16}}{16}}+{\frac {0.9^{17}}{17}}-{\frac {0.9^{18}}{18}}+{\frac {0.9^{19}}{19}}=}
= 0.62619810431142857142857142857143 + 0.01893065053370873839967787630326 = 0.64512875484513730982824930487469.
Windows kalkuliatoriaus tiksli ln(1.9) reikšmė yra tokia: ln(1.9) = 0.64185388617239477599103597720349.
Atėmę apskaičiuotą netikslią ln(1.9) reikšmę iš tikslios Windows kalkuliatoriaus ln(1.9) reikšmės, gauname:
0.64185388617239477599103597720349 - 0.64512875484513730982824930487469 = -0.0032748686727425338372133276712.
Taigi
|
r
20
(
0.9
)
|
≈
|
−
0.00327486867274
|
<
0.0060788327295284644005
<
0.01.
{\displaystyle |r_{20}(0.9)|\approx |-0.00327486867274|<0.0060788327295284644005<0.01.}
Liekamojo nario įvertinimą gavome teisingai.
Pastebėsime, kad kai x reikšmė arti 1 arba lygi 1, tai
ln
(
1
+
x
)
{\displaystyle \ln(1+x)}
ln
(
1
+
1
)
=
ln
(
0.2
⋅
10
)
=
ln
0.2
+
ln
10
=
ln
(
1
−
0.8
)
+
ln
10
{\displaystyle \ln(1+1)=\ln(0.2\cdot 10)=\ln 0.2+\ln 10=\ln(1-0.8)+\ln 10}
arba, pavyzdžiui, kai x=0.99, gautume
ln
(
1
+
0.99
)
=
ln
(
1.99
)
=
ln
(
0.199
⋅
10
)
=
ln
0.199
+
ln
10
=
ln
(
1
−
0.801
)
+
ln
10
{\displaystyle \ln(1+0.99)=\ln(1.99)=\ln(0.199\cdot 10)=\ln 0.199+\ln 10=\ln(1-0.801)+\ln 10}
ir galime apskaičiuoti
ln
(
1
+
x
)
{\displaystyle \ln(1+x)}
x
=
−
0.8
{\displaystyle x=-0.8}
x
=
−
0.801.
{\displaystyle x=-0.801.}
ln
10
{\displaystyle \ln 10}
ln
10
=
ln
(
1.25
⋅
8
)
=
ln
1.25
+
ln
8
=
ln
(
1
+
0.25
)
+
3
ln
2.
{\displaystyle \ln 10=\ln(1.25\cdot 8)=\ln 1.25+\ln 8=\ln(1+0.25)+3\ln 2.}
O
ln
2
{\displaystyle \ln 2}
ln
2
=
ln
(
2
⋅
2
)
=
ln
2
+
ln
2
≈
2
ln
(
1.414213562373
)
=
2
ln
(
1
+
0.414213562373
)
.
{\displaystyle \ln 2=\ln({\sqrt {2}}\cdot {\sqrt {2}})=\ln {\sqrt {2}}+\ln {\sqrt {2}}\approx 2\ln(1.414213562373)=2\ln(1+0.414213562373).}
Daug kartų traukiant šaknį kaip ką tik darėme su ln(2) galima gana greitai apskaičiuoti
ln
(
1
+
x
)
{\displaystyle \ln(1+x)}
−
1
<
x
≤
1
{\displaystyle -1<x\leq 1}
Pavyzdžiui, ištraukus 4 kartus kvadratinę šaknį iš 2 gausime
ln
2
=
2
⋅
2
⋅
2
⋅
2
ln
(
2
1
2
⋅
1
2
⋅
1
2
⋅
1
2
)
≈
16
ln
(
1.04427378242741384
)
=
16
ln
(
1
+
0.04427378242741384
)
.
{\displaystyle \ln 2=2\cdot 2\cdot 2\cdot 2\ln(2^{{\frac {1}{2}}\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}})\approx 16\ln(1.04427378242741384)=16\ln(1+0.04427378242741384).}
Jei užrašysime
ln
(
1
+
x
)
{\displaystyle \ln(1+x)}
x
=
0.04427378242741384
{\displaystyle x=0.04427378242741384}
n
=
9
,
{\displaystyle n=9,}
|
r
9
+
1
(
0.04427378242741384
)
|
<
|
x
n
+
1
n
+
1
|
≈
0.04427378242741384
9
+
1
9
+
1
=
{\displaystyle |r_{9+1}(0.04427378242741384)|<{\Big |}{\frac {x^{n+1}}{n+1}}{\Big |}\approx {\frac {0.04427378242741384^{9+1}}{9+1}}=}
=
2.89378502579267661242
⋅
10
−
14
10
=
2.89378502579267661242
⋅
10
−
15
<
10
−
14
.
{\displaystyle ={\frac {2.89378502579267661242\cdot 10^{-14}}{10}}=2.89378502579267661242\cdot 10^{-15}<10^{-14}.}
Gavome
ln
2
{\displaystyle \ln 2}
10
−
14
{\displaystyle 10^{-14}}
10
−
13
{\displaystyle 10^{-13}}
10
−
12
{\displaystyle 10^{-12}}
Pastebėsime, kad pavyzdžiui ln(0.00003) galima apskaičiuoti taip:
ln
(
0.00003
)
=
ln
(
3
⋅
10
−
5
)
=
ln
(
0.3
⋅
10
−
4
)
=
ln
0.3
+
ln
10
−
4
=
ln
(
1
−
0.7
)
−
4
ln
10.
{\displaystyle \ln(0.00003)=\ln(3\cdot 10^{-5})=\ln(0.3\cdot 10^{-4})=\ln 0.3+\ln 10^{-4}=\ln(1-0.7)-4\ln 10.}
Apskaičiuosime ln(0.3) su paklaida mažesne nei 0.01. Pasinaudosime pirmu pavyzdžiu. Vadinasi,
ln
(
1
+
x
)
=
f
(
0
)
+
f
′
(
0
)
1
!
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
f
(
4
)
(
0
)
4
!
x
4
+
.
.
.
+
f
(
n
)
(
0
)
n
!
x
n
+
f
(
n
+
1
)
(
c
)
(
n
+
1
)
!
x
n
+
1
=
{\displaystyle \ln(1+x)=f(0)+{\frac {f'(0)}{1!}}x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+{\frac {f^{(4)}(0)}{4!}}x^{4}+...+{\frac {f^{(n)}(0)}{n!}}x^{n}+{\frac {f^{(n+1)}(c)}{(n+1)!}}x^{n+1}=}
=
x
−
x
2
2
+
x
3
3
−
x
4
4
.
.
.
+
(
−
1
)
n
−
1
x
n
n
+
r
n
+
1
(
x
)
;
(
13
)
{\displaystyle =x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}...+(-1)^{n-1}{\frac {x^{n}}{n}}+r_{n+1}(x);\quad (13)}
čia
|
r
n
+
1
(
x
)
|
=
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
,
0
>
c
>
−
0.7.
{\displaystyle |r_{n+1}(x)|={\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}\;,\;\;0>c>-0.7.}
ln
0.3
=
ln
(
1
+
(
−
0.7
)
)
.
{\displaystyle \ln 0.3=\ln(1+(-0.7)).}
Turime
|
r
n
+
1
(
x
)
|
=
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
<
|
x
n
+
1
(
1
−
0.7
)
n
+
1
(
n
+
1
)
|
=
|
x
n
+
1
0.3
n
+
1
(
n
+
1
)
|
.
{\displaystyle |r_{n+1}(x)|={\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}<{\Big |}{\frac {x^{n+1}}{(1-0.7)^{n+1}(n+1)}}{\Big |}={\Big |}{\frac {x^{n+1}}{0.3^{n+1}(n+1)}}{\Big |}.}
Kai
n
=
19
,
{\displaystyle n=19,}
x
=
−
0.7
,
{\displaystyle x=-0.7,}
|
(
−
0.7
)
19
+
1
0.3
19
+
1
(
19
+
1
)
|
=
(
−
0.7
)
20
0.3
20
⋅
20
=
{\displaystyle {\Big |}{\frac {(-0.7)^{19+1}}{0.3^{19+1}(19+1)}}{\Big |}={\frac {(-0.7)^{20}}{0.3^{20}\cdot 20}}=}
=
0.00079792266297612001
0.00000000003486784401
⋅
20
=
{\displaystyle ={\frac {0.00079792266297612001}{0.00000000003486784401\cdot 20}}=}
=1144209.9241170145552684546382425.
Matome, kad tokiu budu ln(0.3) apskaičiuoti negalima (tiksliau, negalima sužinoti paklaidos).
Galima iš 0.3 ištraukti, pavyzdžiui, 4 kartus kvadratinę šaknį. Tada turėsime:
ln
(
0.3
)
=
16
ln
(
0.3
1
16
)
≈
16
ln
(
0.9275131559667915
)
=
16
ln
(
1
−
0.07248684403320849
)
.
{\displaystyle \ln(0.3)=16\ln(0.3^{1 \over 16})\approx 16\ln(0.9275131559667915)=16\ln(1-0.07248684403320849).}
Tada
|
r
n
+
1
(
x
)
|
=
16
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
,
0
>
c
>
−
0.07248684403320849.
{\displaystyle |r_{n+1}(x)|=16{\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}\;,\;\;0>c>-0.07248684403320849.}
Turime
|
r
n
+
1
(
x
)
|
=
16
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
<
16
|
x
n
+
1
(
1
−
0.07248684403320849
)
n
+
1
(
n
+
1
)
|
=
16
|
x
n
+
1
0.9275131559667915
n
+
1
(
n
+
1
)
|
.
{\displaystyle |r_{n+1}(x)|=16{\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}<16{\Big |}{\frac {x^{n+1}}{(1-0.07248684403320849)^{n+1}(n+1)}}{\Big |}=16{\Big |}{\frac {x^{n+1}}{0.9275131559667915^{n+1}(n+1)}}{\Big |}.}
Kai
n
=
19
,
{\displaystyle n=19,}
x
=
−
0.07248684403320849
,
{\displaystyle x=-0.07248684403320849,}
|
(
−
0.07248684403320849
)
19
+
1
0.9275131559667915
19
+
1
(
19
+
1
)
|
=
(
−
0.07248684403320849
)
20
0.9275131559667915
20
⋅
20
=
{\displaystyle {\Big |}{\frac {(-0.07248684403320849)^{19+1}}{0.9275131559667915^{19+1}(19+1)}}{\Big |}={\frac {(-0.07248684403320849)^{20}}{0.9275131559667915^{20}\cdot 20}}=}
=
1.603924173409687
⋅
10
−
23
0.222024841347685575
⋅
20
=
{\displaystyle ={\frac {1.603924173409687\cdot 10^{-23}}{0.222024841347685575\cdot 20}}=}
=3.6120376523498566762096049279603e-24=
≈
3.6120376523498566762
⋅
10
−
24
.
{\displaystyle \approx 3.6120376523498566762\cdot 10^{-24}.}
|
r
n
+
1
(
x
)
|
=
|
r
20
(
−
0.07248684403320849
)
|
<
16
⋅
3.6120376523498566762
⋅
10
−
24
=
5.77926024375977
⋅
10
−
23
.
{\displaystyle |r_{n+1}(x)|=|r_{20}(-0.07248684403320849)|<16\cdot 3.6120376523498566762\cdot 10^{-24}=5.77926024375977\cdot 10^{-23}.}
ln
(
1
−
0.07248684403320849
)
=
{\displaystyle \ln(1-0.07248684403320849)=}
=
−
0.0724
−
(
−
0.0724
)
2
2
+
(
−
0.0724
)
3
3
−
(
−
0.0724
)
4
4
+
(
−
0.0724
)
5
5
−
(
−
0.0724
)
6
6
+
(
−
0.0724
)
7
7
−
(
−
0.0724
)
8
8
+
(
−
0.0724
)
9
9
−
(
−
0.0724
)
10
10
+
{\displaystyle =-0.0724-{\frac {(-0.0724)^{2}}{2}}+{\frac {(-0.0724)^{3}}{3}}-{\frac {(-0.0724)^{4}}{4}}+{\frac {(-0.0724)^{5}}{5}}-{\frac {(-0.0724)^{6}}{6}}+{\frac {(-0.0724)^{7}}{7}}-{\frac {(-0.0724)^{8}}{8}}+{\frac {(-0.0724)^{9}}{9}}-{\frac {(-0.0724)^{10}}{10}}+}
+
(
−
0.0724
)
11
11
−
(
−
0.0724
)
12
12
+
(
−
0.0724
)
13
13
−
(
−
0.0724
)
14
14
+
(
−
0.0724
)
15
15
−
(
−
0.0724
)
16
16
+
(
−
0.0724
)
17
17
−
(
−
0.0724
)
18
18
+
(
−
0.0724
)
19
19
=
{\displaystyle +{\frac {(-0.0724)^{11}}{11}}-{\frac {(-0.0724)^{12}}{12}}+{\frac {(-0.0724)^{13}}{13}}-{\frac {(-0.0724)^{14}}{14}}+{\frac {(-0.0724)^{15}}{15}}-{\frac {(-0.0724)^{16}}{16}}+{\frac {(-0.0724)^{17}}{17}}-{\frac {(-0.0724)^{18}}{18}}+{\frac {(-0.0724)^{19}}{19}}=}
[vietoje (-0.0724) imama -0.07248684403320849089929261472692 reikšmė]
= -0.07524830027034272898276362268037 + (-2.8270556157154487988345737336471e-14) =
= -0.07524830027037099953892077716836.
-0.07524830027037099953892077716836 * 16 = -1.2039728043259359926227324346937.
Atėmę gautą ln(0.3) reikšmę iš tikslios kalkuliatoriaus reikšmės (-1.2039728043259359926227462177618), gauname:
-1.2039728043259359926227462177618 - (-1.2039728043259359926227324346937) = -1.3783068138502953610930772512816e-23.
Taigi gavome paklaidą |-1.3783068138502953610930772512816e-23| (tai reiškia
≈
|
−
1.37830681385
⋅
10
−
23
|
{\displaystyle \approx |-1.37830681385\cdot 10^{-23}|}
|
r
n
+
1
(
x
)
|
=
|
r
20
(
−
0.07248684403320849
)
|
≈
|
−
1.37830681385
⋅
10
−
23
|
<
5.77926024375977
⋅
10
−
23
.
{\displaystyle |r_{n+1}(x)|=|r_{20}(-0.07248684403320849)|\approx |-1.37830681385\cdot 10^{-23}|<5.77926024375977\cdot 10^{-23}.}
Kas atitinka teoriją.
Pastebėsime, kad jeigu vietoje c=-0.07248684403320849, parinksime c=0, tai gausime, kad paklaida turėtų būti mažesnė už
5.77926024375977
⋅
10
−
23
⋅
0.222024841347685575
=
1.2831393387277
⋅
10
−
23
.
{\displaystyle 5.77926024375977\cdot 10^{-23}\cdot 0.222024841347685575=1.2831393387277\cdot 10^{-23}.}
Bet
1.2831393387277
⋅
10
−
23
<
|
−
1.37830681385
⋅
10
−
23
|
.
{\displaystyle 1.2831393387277\cdot 10^{-23}<|-1.37830681385\cdot 10^{-23}|.}
Ir gaunasi, kad su c=0, gautume truputi melagingą paklaidą.
Apskaičiuosime ln(0.11) ir įvertinsime paklaidą. Iš 0.11 ištrauksime 4 kartus kvadratinę šaknį. Tada turėsime:
ln
(
0.11
)
=
2
⋅
2
⋅
2
⋅
2
ln
(
0.11
1
2
⋅
1
2
⋅
1
2
⋅
1
2
)
=
16
ln
(
0.11
1
16
)
≈
16
ln
(
0.871138169026
)
=
16
ln
(
1
−
0.12886183
)
.
{\displaystyle \ln(0.11)=2\cdot 2\cdot 2\cdot 2\ln(0.11^{{\frac {1}{2}}\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}})=16\ln(0.11^{1 \over 16})\approx 16\ln(0.871138169026)=16\ln(1-0.12886183).}
Tai yra, x=-0.12886183097395441374956757206472.
Tada
|
r
n
+
1
(
x
)
|
=
16
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
,
0
>
c
>
−
0.12886183.
{\displaystyle |r_{n+1}(x)|=16{\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}\;,\;\;0>c>-0.12886183.}
Turime
|
r
n
+
1
(
x
)
|
=
16
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
<
16
|
x
n
+
1
(
1
−
0.1288618309739544
)
n
+
1
(
n
+
1
)
|
=
16
|
x
n
+
1
0.871138169026
n
+
1
(
n
+
1
)
|
.
{\displaystyle |r_{n+1}(x)|=16{\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |}<16{\Big |}{\frac {x^{n+1}}{(1-0.1288618309739544)^{n+1}(n+1)}}{\Big |}=16{\Big |}{\frac {x^{n+1}}{0.871138169026^{n+1}(n+1)}}{\Big |}.}
Kai
n
=
19
,
{\displaystyle n=19,}
x
=
−
0.12886183097395
,
{\displaystyle x=-0.12886183097395,}
|
(
−
0.12886183097395
)
19
+
1
0.871138169026
19
+
1
(
19
+
1
)
|
=
(
−
0.12886183097395
)
20
0.871138169026
20
⋅
20
=
1.5939913388918
⋅
10
−
18
0.063349159397
⋅
20
=
{\displaystyle {\Big |}{\frac {(-0.12886183097395)^{19+1}}{0.871138169026^{19+1}(19+1)}}{\Big |}={\frac {(-0.12886183097395)^{20}}{0.871138169026^{20}\cdot 20}}={\frac {1.5939913388918\cdot 10^{-18}}{0.063349159397\cdot 20}}=}
= 1.2580998343623026293204200398239e-18 =
≈
1.2580998343623
⋅
10
−
18
.
{\displaystyle \approx 1.2580998343623\cdot 10^{-18}.}
|
r
n
+
1
(
x
)
|
=
|
r
20
(
−
0.12886183
)
|
<
16
⋅
1.2580998343623
⋅
10
−
18
=
2.0129597349796842
⋅
10
−
17
.
{\displaystyle |r_{n+1}(x)|=|r_{20}(-0.12886183)|<16\cdot 1.2580998343623\cdot 10^{-18}=2.0129597349796842\cdot 10^{-17}.}
ln
(
1
−
0.12886183097395
)
=
{\displaystyle \ln(1-0.12886183097395)=}
=
−
0.1288
−
(
−
0.1288
)
2
2
+
(
−
0.1288
)
3
3
−
(
−
0.1288
)
4
4
+
(
−
0.1288
)
5
5
−
(
−
0.1288
)
6
6
+
(
−
0.1288
)
7
7
−
(
−
0.1288
)
8
8
+
(
−
0.1288
)
9
9
−
(
−
0.1288
)
10
10
+
{\displaystyle =-0.1288-{\frac {(-0.1288)^{2}}{2}}+{\frac {(-0.1288)^{3}}{3}}-{\frac {(-0.1288)^{4}}{4}}+{\frac {(-0.1288)^{5}}{5}}-{\frac {(-0.1288)^{6}}{6}}+{\frac {(-0.1288)^{7}}{7}}-{\frac {(-0.1288)^{8}}{8}}+{\frac {(-0.1288)^{9}}{9}}-{\frac {(-0.1288)^{10}}{10}}+}
+
(
−
0.1288
)
11
11
−
(
−
0.1288
)
12
12
+
(
−
0.1288
)
13
13
−
(
−
0.1288
)
14
14
+
(
−
0.1288
)
15
15
−
(
−
0.1288
)
16
16
+
(
−
0.1288
)
17
17
−
(
−
0.1288
)
18
18
+
(
−
0.1288
)
19
19
=
{\displaystyle +{\frac {(-0.1288)^{11}}{11}}-{\frac {(-0.1288)^{12}}{12}}+{\frac {(-0.1288)^{13}}{13}}-{\frac {(-0.1288)^{14}}{14}}+{\frac {(-0.1288)^{15}}{15}}-{\frac {(-0.1288)^{16}}{16}}+{\frac {(-0.1288)^{17}}{17}}-{\frac {(-0.1288)^{18}}{18}}+{\frac {(-0.1288)^{19}}{19}}=}
[vietoje (-0.1288) imama -0.12886183097395441374956757206472 reikšmė]
= -0.13795468205758418170298735700763 -1.6773369704537036636530891506237e-11 =
= -0.13795468207435755140752439364416.
ln(0.11) = -0.13795468207435755140752439364416 * 16 = -2.2072749131897208225203902983066.
Atėmę šią gautą ln(0.11) reikšmę iš tikslios (ln(0.11)=-2.2072749131897208239740393314036) kalkuliatoriaus reikšmės, gauname:
-2.2072749131897208239740393314036 - (-2.2072749131897208225203902983066) =
= -2.2072749131897208239740393314036 + 2.2072749131897208225203902983066 =
= -1.4536490330969991153804961232531e-18.
Tada
|
r
n
+
1
(
x
)
|
=
|
r
20
(
−
0.12886183
)
|
≈
|
−
1.453649033096999
⋅
10
−
18
|
<
2.0129597349796842
⋅
10
−
17
.
{\displaystyle |r_{n+1}(x)|=|r_{20}(-0.12886183)|\approx |-1.453649033096999\cdot 10^{-18}|<2.0129597349796842\cdot 10^{-17}.}
Paklaidos įvertinimas gavosi pagal teoriją. Apskaičiavome ln(0.11) su paklaida mažesne nei
10
−
16
{\displaystyle 10^{-16}}
10
−
17
,
{\displaystyle 10^{-17},}
Pastebėsime, kad
2.0129597349796842
⋅
10
−
17
⋅
0.871138169026
20
=
2.0129597349796842
⋅
10
−
17
⋅
0.063349159397
=
1.2751930711097
⋅
10
−
18
.
{\displaystyle 2.0129597349796842\cdot 10^{-17}\cdot 0.871138169026^{20}=2.0129597349796842\cdot 10^{-17}\cdot 0.063349159397=1.2751930711097\cdot 10^{-18}.}
1.2751930711097
⋅
10
−
18
<
|
−
1.453649033096999
⋅
10
−
18
|
.
{\displaystyle 1.2751930711097\cdot 10^{-18}<|-1.453649033096999\cdot 10^{-18}|.}
|
r
n
+
1
(
x
)
|
=
16
|
x
n
+
1
(
1
+
c
)
n
+
1
(
n
+
1
)
|
,
{\displaystyle |r_{n+1}(x)|=16{\Big |}{\frac {x^{n+1}}{(1+c)^{n+1}(n+1)}}{\Big |},}
vietoje c=-0.12886183, parinkti c=0.
Pastebėsime, kad ln(0.11) galima užrašyti ir apskaičiuoti taip:
ln
0.11
=
ln
(
1.1
⋅
10
−
1
)
=
ln
(
1.1
)
+
ln
10
−
1
=
ln
(
1
+
0.1
)
−
ln
10.
{\displaystyle \ln 0.11=\ln(1.1\cdot 10^{-1})=\ln(1.1)+\ln 10^{-1}=\ln(1+0.1)-\ln 10.}
O ln(0.3) galima užrašyti ir apskaičiuoti taip:
ln
0.3
=
ln
(
3
⋅
10
−
1
)
=
ln
3
+
ln
10
−
1
=
ln
(
1.5
⋅
2
)
−
ln
10
=
ln
(
1
+
0.5
)
+
ln
2
−
ln
10.
{\displaystyle \ln 0.3=\ln(3\cdot 10^{-1})=\ln 3+\ln 10^{-1}=\ln(1.5\cdot 2)-\ln 10=\ln(1+0.5)+\ln 2-\ln 10.}
Kadangi skaičiuojant ln(x) gali prireikti traukti kvadratinę šaknį, tai pateiksime paprastą algoritmą kaip ištraukti kvadratinę šaknį iš bet kokio skaičiaus.
Kad gauti
x
:
{\displaystyle {\sqrt {x}}:}
Žingsnis 1: Spėjimas G = 1.
Zingsnis 2: Naujas Spėjimas = (G + x/G)/2
Pakartoti Žingsnį 2 kiek norima kartų, kad gauti kiek norima tikslų resultatą. Pavyzdžiui,
2
{\displaystyle {\sqrt {2}}}
G = 1
G = (1+2/1)/2 = 3/2 = 1.5
G = (3/2 + 4/3)/2 = 17/12 = 1.4166666667
G = (17/12 + 24/17)/2 = 577/408 = 1.414215686
G = (577/408 + 816/577)/2 = 665857/470832 = 1.4142135623746899106262955788901.
Atėmę gautą
2
{\displaystyle {\sqrt {2}}}
2
{\displaystyle {\sqrt {2}}}
1.4142135623730950488016887242097 - 1.4142135623746899106262955788901 = -1.5948618246068546804368315468877e-12 =
≈
−
1.5948618246
⋅
10
−
12
.
{\displaystyle \approx -1.5948618246\cdot 10^{-12}.}
Tačiau yra daug greitesnis šaknies traukimo būdas apie kurį galima paskaityti čia https://lt.wikibooks.org/wiki/Sekos_riba#Greitas_šaknies_iš_skaičiaus_a_traukimo_būdas Dar pastebėsime, kad jeigu skaičiuojamas ln(x) ir x>1.4, tai yra protinga užrašyti, pvz. ln(1.8) taip:
ln
1.8
=
ln
(
2
⋅
0.9
)
=
ln
2
+
ln
0.9
=
ln
2
+
ln
(
1
−
0.1
)
.
{\displaystyle \ln 1.8=\ln(2\cdot 0.9)=\ln 2+\ln 0.9=\ln 2+\ln(1-0.1).}
[
ln
1.4
=
ln
(
2
⋅
0.7
)
=
ln
2
+
ln
0.7
=
ln
2
+
ln
(
1
−
0.3
)
.
{\displaystyle \ln 1.4=\ln(2\cdot 0.7)=\ln 2+\ln 0.7=\ln 2+\ln(1-0.3).}
Matome, kad tokiu budu ln(0.9)=ln(1-0.1) galima labai greitai apskaičiuoti.
keisti
Apskaičiuosime
e
1
2
{\displaystyle e^{1 \over 2}}
0.000001
=
10
−
6
.
{\displaystyle 0.000001=10^{-6}.}
f
(
x
)
=
e
x
.
{\displaystyle f(x)=e^{x}.}
f
(
n
)
(
x
)
=
e
x
,
f
(
n
)
(
0
)
=
e
0
=
1
{\displaystyle f^{(n)}(x)=e^{x},\,f^{(n)}(0)=e^{0}=1\;}
n , priklausantiems natūraliesiems skaičiams, tai
e
x
=
1
+
x
+
x
2
2
!
+
x
3
3
!
+
x
4
4
!
+
.
.
.
+
x
n
n
!
+
r
n
+
1
(
x
)
;
(
10
)
{\displaystyle e^{x}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+...+{\frac {x^{n}}{n!}}+r_{n+1}(x);\quad (10)}
čia
r
n
+
1
(
x
)
=
f
(
n
+
1
)
(
c
)
(
n
+
1
)
!
x
n
+
1
=
e
c
(
n
+
1
)
!
x
n
+
1
,
0
<
c
<
x
.
{\displaystyle r_{n+1}(x)={\frac {f^{(n+1)}(c)}{(n+1)!}}x^{n+1}={\frac {e^{c}}{(n+1)!}}x^{n+1},\;\;0<c<x.}
Įrašę į (10) formulę vietoje x skaičių
1
2
,
{\displaystyle {1 \over 2},}
e
=
e
1
2
=
1
+
1
2
+
(
1
2
)
2
2
!
+
(
1
2
)
3
3
!
+
(
1
2
)
4
4
!
+
.
.
.
+
(
1
2
)
n
n
!
+
r
n
+
1
(
0.5
)
;
(
15
)
{\displaystyle {\sqrt {e}}=e^{1 \over 2}=1+{1 \over 2}+{\frac {({1 \over 2})^{2}}{2!}}+{\frac {({1 \over 2})^{3}}{3!}}+{\frac {({1 \over 2})^{4}}{4!}}+...+{\frac {({1 \over 2})^{n}}{n!}}+r_{n+1}(0.5);\quad (15)}
čia
r
n
+
1
(
0.5
)
=
e
c
(
n
+
1
)
!
1
2
n
+
1
,
0
<
c
<
1
2
.
{\displaystyle r_{n+1}(0.5)={\frac {e^{c}}{(n+1)!}}{1 \over 2^{n+1}},\;\;0<c<{1 \over 2}.}
Kadangi
e
c
<
e
1
2
<
2
{\displaystyle e^{c}<e^{1 \over 2}<2}
3
<
1.8
{\displaystyle {\sqrt {3}}<1.8}
r
n
+
1
(
0.5
)
<
e
1
2
(
n
+
1
)
!
1
2
n
+
1
<
2
(
n
+
1
)
!
1
2
n
+
1
=
1
(
n
+
1
)
!
1
2
n
.
{\displaystyle r_{n+1}(0.5)<{\frac {e^{1 \over 2}}{(n+1)!}}{1 \over 2^{n+1}}<{\frac {2}{(n+1)!}}{1 \over 2^{n+1}}={\frac {1}{(n+1)!}}{1 \over 2^{n}}.}
Mums reikia, kad
1
(
n
+
1
)
!
1
2
n
<
10
−
6
.
{\displaystyle {\frac {1}{(n+1)!}}{1 \over 2^{n}}<10^{-6}.}
Ši nelygybė jau teisinga, kai
n
=
7
,
{\displaystyle n=7,}
1
(
7
+
1
)
!
1
2
7
=
1
8
!
⋅
2
7
=
1
40320
⋅
128
≈
1.9376240079365
⋅
10
−
7
<
10
−
6
.
{\displaystyle {\frac {1}{(7+1)!}}{1 \over 2^{7}}={\frac {1}{8!\cdot 2^{7}}}={\frac {1}{40320\cdot 128}}\approx 1.9376240079365\cdot 10^{-7}<10^{-6}.}
Taigi
e
≈
1
+
1
2
+
1
2
2
⋅
2
!
+
1
2
3
⋅
3
!
+
1
2
4
⋅
4
!
+
1
2
5
⋅
5
!
+
1
2
6
⋅
6
!
+
1
2
7
⋅
7
!
{\displaystyle {\sqrt {e}}\approx 1+{1 \over 2}+{\frac {1}{2^{2}\cdot 2!}}+{\frac {1}{2^{3}\cdot 3!}}+{\frac {1}{2^{4}\cdot 4!}}+{\frac {1}{2^{5}\cdot 5!}}+{\frac {1}{2^{6}\cdot 6!}}+{\frac {1}{2^{7}\cdot 7!}}}
= 1 + 1/2 + 1/8 + 1/48 + 1/384 + 1/3840 + 1/46080 + 1/645120 =
(galima įdėt šitą eilutę nuo "1" iki "1/645120" į kalkuliatorių ir paspaust "=" ir tada bus iškart apskaičiuota šitos eilutės reikšmė)
=1.6487211681547619047619047619048=
≈
1.648721.
{\displaystyle \approx 1.648721.}
Tiksli
e
{\displaystyle {\sqrt {e}}}
1.6487212707001281468486507878142.
Atėmę netikslią (gautą)
e
{\displaystyle {\sqrt {e}}}
e
{\displaystyle {\sqrt {e}}}
1.6487212707001281468486507878142 - 1.6487211681547619047619047619048 = 1.0254536624208674602590940166689e-7 =
≈
1.02545366
⋅
10
−
7
.
{\displaystyle \approx 1.02545366\cdot 10^{-7}.}
Apskaičiuosime
e
5
=
e
1
5
=
e
0.2
{\displaystyle {\sqrt[{5}]{e}}=e^{1 \over 5}=e^{0.2}}
10
−
10
.
{\displaystyle 10^{-10}.}
Įrašę į (10) formulę vietoje x skaičių
1
5
,
{\displaystyle {1 \over 5},}
e
5
=
e
1
5
=
1
+
1
5
+
(
1
5
)
2
2
!
+
(
1
5
)
3
3
!
+
(
1
5
)
4
4
!
+
.
.
.
+
(
1
5
)
n
n
!
+
r
n
+
1
(
0.2
)
;
{\displaystyle {\sqrt[{5}]{e}}=e^{1 \over 5}=1+{1 \over 5}+{\frac {({1 \over 5})^{2}}{2!}}+{\frac {({1 \over 5})^{3}}{3!}}+{\frac {({1 \over 5})^{4}}{4!}}+...+{\frac {({1 \over 5})^{n}}{n!}}+r_{n+1}(0.2);}
čia
r
n
+
1
(
0.2
)
=
e
c
(
n
+
1
)
!
1
5
n
+
1
,
0
<
c
<
1
5
.
{\displaystyle r_{n+1}(0.2)={\frac {e^{c}}{(n+1)!}}{1 \over 5^{n+1}},\;\;0<c<{1 \over 5}.}
Kadangi
e
c
<
e
1
5
<
1.3
{\displaystyle e^{c}<e^{1 \over 5}<1.3}
3
0.2
<
1.3
{\displaystyle 3^{0.2}<1.3}
r
n
+
1
(
0.2
)
<
e
1
5
(
n
+
1
)
!
1
5
n
+
1
<
1.3
(
n
+
1
)
!
1
5
n
+
1
=
1.3
(
n
+
1
)
!
⋅
5
n
+
1
.
{\displaystyle r_{n+1}(0.2)<{\frac {e^{1 \over 5}}{(n+1)!}}{1 \over 5^{n+1}}<{\frac {1.3}{(n+1)!}}{1 \over 5^{n+1}}={\frac {1.3}{(n+1)!\cdot 5^{n+1}}}.}
Mums reikia, kad
1.3
(
n
+
1
)
!
⋅
5
n
+
1
<
10
−
10
.
{\displaystyle {\frac {1.3}{(n+1)!\cdot 5^{n+1}}}<10^{-10}.}
Ši nelygybė jau teisinga, kai
n
=
7
,
{\displaystyle n=7,}
1.3
(
7
+
1
)
!
1
5
7
+
1
=
1.3
8
!
⋅
5
8
=
1.3
40320
⋅
390625
=
1.3
15750000000
≈
8.253968253968
⋅
10
−
11
<
10
−
10
.
{\displaystyle {\frac {1.3}{(7+1)!}}{1 \over 5^{7+1}}={\frac {1.3}{8!\cdot 5^{8}}}={\frac {1.3}{40320\cdot 390625}}={\frac {1.3}{15750000000}}\approx 8.253968253968\cdot 10^{-11}<10^{-10}.}
Taigi
e
1
/
5
≈
1
+
1
5
+
1
5
2
⋅
2
!
+
1
5
3
⋅
3
!
+
1
5
4
⋅
4
!
+
1
5
5
⋅
5
!
+
1
5
6
⋅
6
!
+
1
5
7
⋅
7
!
=
{\displaystyle e^{1/5}\approx 1+{1 \over 5}+{\frac {1}{5^{2}\cdot 2!}}+{\frac {1}{5^{3}\cdot 3!}}+{\frac {1}{5^{4}\cdot 4!}}+{\frac {1}{5^{5}\cdot 5!}}+{\frac {1}{5^{6}\cdot 6!}}+{\frac {1}{5^{7}\cdot 7!}}=}
= 1 + 1/5 + 1/50 + 1/750 + 1/15000 + 1/375000 + 1/11250000 + 1/393750000 =
(galima nukopijuot šitą eilutę nuo "1" iki "1/393750000" ir padaryt "Paste" į "Windows" kalkuliatorių ir paspaust "=" ir tada bus iškart apskaičiuota šitos eilutės reikšmė)
=1.2214027580952380952380952380952=
≈
1.2214027581.
{\displaystyle \approx 1.2214027581.}
Tiksli
e
5
=
e
0.2
{\displaystyle {\sqrt[{5}]{e}}=e^{0.2}}
1.2214027581601698339210719946397.
Atėmę netikslią (gautą)
e
5
{\displaystyle {\sqrt[{5}]{e}}}
e
5
{\displaystyle {\sqrt[{5}]{e}}}
1.2214027581601698339210719946397 - 1.2214027580952380952380952380952 = 0.0000000000649317386829767565445 =
≈
6.49317386829767565445
⋅
10
−
11
.
{\displaystyle \approx 6.49317386829767565445\cdot 10^{-11}.}
Pastebėsime, kad
1
8
!
⋅
5
8
=
1
15750000000
≈
6.3492063492
⋅
10
−
11
{\displaystyle {\frac {1}{8!\cdot 5^{8}}}={\frac {1}{15750000000}}\approx 6.3492063492\cdot 10^{-11}}
6.3492063492
⋅
10
−
11
<
6.49317386829767565445
⋅
10
−
11
.
{\displaystyle 6.3492063492\cdot 10^{-11}<6.49317386829767565445\cdot 10^{-11}.}
Apskaičiuosime
e
=
e
1
{\displaystyle e=e^{1}}
0.0001
=
10
−
4
.
{\displaystyle 0.0001=10^{-4}.}
Įrašę į (10) formulę vietoje x skaičių
1
{\displaystyle 1}
e
1
=
1
+
1
+
1
2
2
!
+
1
3
3
!
+
1
4
4
!
+
.
.
.
+
1
n
n
!
+
r
n
+
1
(
1
)
;
{\displaystyle e^{1}=1+1+{\frac {1^{2}}{2!}}+{\frac {1^{3}}{3!}}+{\frac {1^{4}}{4!}}+...+{\frac {1^{n}}{n!}}+r_{n+1}(1);}
čia
r
n
+
1
(
1
)
=
e
c
(
n
+
1
)
!
⋅
1
n
+
1
,
0
<
c
<
1.
{\displaystyle r_{n+1}(1)={\frac {e^{c}}{(n+1)!}}\cdot 1^{n+1},\;\;0<c<1.}
Kadangi
e
c
<
e
1
<
3
,
{\displaystyle e^{c}<e^{1}<3,}
r
n
+
1
(
1
)
<
e
1
(
n
+
1
)
!
⋅
1
n
+
1
<
3
(
n
+
1
)
!
.
{\displaystyle r_{n+1}(1)<{\frac {e^{1}}{(n+1)!}}\cdot 1^{n+1}<{\frac {3}{(n+1)!}}.}
Mums reikia, kad
3
(
n
+
1
)
!
<
10
−
4
.
{\displaystyle {\frac {3}{(n+1)!}}<10^{-4}.}
Ši nelygybė jau teisinga, kai
n
=
7
,
{\displaystyle n=7,}
3
(
7
+
1
)
!
=
3
8
!
=
3
40320
≈
7.44047619
⋅
10
−
5
<
10
−
4
.
{\displaystyle {\frac {3}{(7+1)!}}={\frac {3}{8!}}={\frac {3}{40320}}\approx 7.44047619\cdot 10^{-5}<10^{-4}.}
Taigi
e
1
≈
1
+
1
+
1
2
!
+
1
3
!
+
1
4
!
+
1
5
!
+
1
6
!
+
1
7
!
=
{\displaystyle e^{1}\approx 1+1+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+{\frac {1}{5!}}+{\frac {1}{6!}}+{\frac {1}{7!}}=}
= 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 + 1/5040 =
(galima nukopijuot šitą eilutę nuo "1" iki "1/5040" ir padaryt "Paste" į "Windows" kalkuliatorių ir paspaust "=" ir tada bus iškart apskaičiuota šitos eilutės reikšmė)
=2.718253968253968253968253968254=
≈
2.7183.
{\displaystyle \approx 2.7183.}
Atėmę netikslią (gautą) e reikšmę iš tikslios e reikšmės, gauname paklaidą:
2.7182818284590452353602874713527 - 2.718253968253968253968253968254 = 2.7860205076981392033503098694244e-5 =
≈
2.786020507698
⋅
10
−
5
.
{\displaystyle \approx 2.786020507698\cdot 10^{-5}.}
Pastebėsime, kad
1
8
!
=
1
40320
≈
2.48015873015873
⋅
10
−
5
.
{\displaystyle {\frac {1}{8!}}={\frac {1}{40320}}\approx 2.48015873015873\cdot 10^{-5}.}
Parodysime kaip ištraukti bet kokio laipsnio šaknį iš bet kokio skaičiaus a. Kadangi
a
y
=
e
ln
a
y
=
e
y
ln
a
,
{\displaystyle a^{y}=e^{\ln a^{y}}=e^{y\ln a},}
3
1
5
=
e
ln
3
1
/
5
=
e
1
5
ln
3
=
e
x
;
{\displaystyle 3^{1 \over 5}=e^{\ln 3^{1/5}}=e^{{\frac {1}{5}}\ln 3}=e^{x};}
x
=
1
5
ln
3
≈
0.2
⋅
1.0986122886681
=
{\displaystyle x={\frac {1}{5}}\ln 3\approx 0.2\cdot 1.0986122886681=}
=0.21972245773362193827904904738451.
Tada
e
x
=
e
0.21972245773362
=
{\displaystyle e^{x}=e^{0.21972245773362}=}
O iš kalkuliatoriaus
3
0.2
=
{\displaystyle 3^{0.2}=}
Kai skaičius x didesnis tai Makloreno eilutė, skaičiuojant tokiu budu, darosi ilgesnė. Tai yra, jeigu, pavyzdžiui, a=300, o y=1/5, tai
300
1
5
=
a
y
=
e
ln
a
y
=
e
1
5
ln
300
=
e
x
;
{\displaystyle 300^{1 \over 5}=a^{y}=e^{\ln a^{y}}=e^{{1 \over 5}\ln 300}=e^{x};}
x
=
1
5
ln
300
≈
0.2
⋅
5.7037824746562
=
{\displaystyle x={\frac {1}{5}}\ln 300\approx 0.2\cdot 5.7037824746562=}
Tada
e
x
=
e
1.14075649493124
=
{\displaystyle e^{x}=e^{1.14075649493124}=}
O iš kalkuliatoriaus
300
0.2
=
{\displaystyle 300^{0.2}=}
Kai skaičius a iš kurio traukiame šaknį yra labai didelis, o y nėra labai mažas, tai tokiu budu skaičiuojant Makloreno eilutė gaunasi labai ilga, jei reikia didelio tikslumo. Pavyzdžiui, kai a=50000, o y=1/3, tai
50000
1
3
=
a
y
=
e
y
ln
a
=
e
1
3
ln
50000
=
e
x
;
{\displaystyle 50000^{1 \over 3}=a^{y}=e^{y\ln a}=e^{{1 \over 3}\ln 50000}=e^{x};}
x
=
1
3
ln
50000
≈
10.81977828441
/
3
=
{\displaystyle x={\frac {1}{3}}\ln 50000\approx 10.81977828441/3=}
Tada
e
x
=
e
3.60659276147
=
{\displaystyle e^{x}=e^{3.60659276147}=}
O iš kalkuliatoriaus
50000
1
/
3
=
{\displaystyle 50000^{1/3}=}
Parodysime kaip greitai ištraukti bet kokio laipsnio šaknį iš bet kokio didelio skaičiaus a. Kaip žinome,
a
y
=
e
ln
a
y
=
e
y
ln
a
.
{\displaystyle a^{y}=e^{\ln a^{y}}=e^{y\ln a}.}
50000
1
3
=
a
y
=
e
y
ln
a
=
e
1
3
ln
50000
=
e
x
;
(
10.1
)
{\displaystyle 50000^{1 \over 3}=a^{y}=e^{y\ln a}=e^{{1 \over 3}\ln 50000}=e^{x};\quad (10.1)}
x
=
1
3
ln
50000
=
1
3
ln
(
5
⋅
10
4
)
=
1
3
(
ln
5
+
ln
10
4
)
=
1
3
(
ln
5
+
4
ln
10
)
.
{\displaystyle x={\frac {1}{3}}\ln 50000={\frac {1}{3}}\ln(5\cdot 10^{4})={\frac {1}{3}}(\ln 5+\ln 10^{4})={\frac {1}{3}}(\ln 5+4\ln 10).}
Įstatome gautą x reikšmę į (10.1) eilutę:
50000
1
3
=
e
x
=
e
1
3
(
ln
5
+
4
ln
10
)
=
e
1
3
ln
5
⋅
e
4
3
ln
10
=
(
e
ln
5
)
1
/
3
⋅
(
e
ln
10
)
4
/
3
=
5
1
/
3
⋅
10
4
/
3
.
{\displaystyle 50000^{1 \over 3}=e^{x}=e^{{\frac {1}{3}}(\ln 5+4\ln 10)}=e^{{\frac {1}{3}}\ln 5}\cdot e^{{\frac {4}{3}}\ln 10}={\Big (}e^{\ln 5}{\Big )}^{1/3}\cdot {\Big (}e^{\ln 10}{\Big )}^{4/3}=5^{1/3}\cdot 10^{4/3}.}
Matome, kad visgi šituo budu negalima pagreitinti Makloreno eilutės skaičiavimų. Pastebėsime tik, kad
1
3
ln
5
≈
1.6094379124341
/
3
=
{\displaystyle {\frac {1}{3}}\ln 5\approx 1.6094379124341/3=}
4
3
ln
10
≈
4
⋅
2.302585092994
/
3
=
{\displaystyle {\frac {4}{3}}\ln 10\approx 4\cdot 2.302585092994/3=}
Bet
4
3
ln
10
{\displaystyle {\frac {4}{3}}\ln 10}
4
3
ln
10
=
4
3
(
ln
1.25
+
ln
2
+
ln
2
+
ln
2
)
.
{\displaystyle {\frac {4}{3}}\ln 10={\frac {4}{3}}(\ln 1.25+\ln 2+\ln 2+\ln 2).}
Ir tada
e
4
3
ln
10
=
e
4
3
ln
1.25
⋅
(
e
4
3
ln
2
)
3
=
(
e
1
3
ln
1.25
)
4
⋅
(
e
1
3
ln
2
)
12
.
{\displaystyle e^{{\frac {4}{3}}\ln 10}=e^{{\frac {4}{3}}\ln 1.25}\cdot (e^{{\frac {4}{3}}\ln 2})^{3}=(e^{{\frac {1}{3}}\ln 1.25})^{4}\cdot (e^{{\frac {1}{3}}\ln 2})^{12}.}
Mums tereikia apskaičiuoti
x
1
=
1
3
ln
1.25
,
{\displaystyle x_{1}={\frac {1}{3}}\ln 1.25,}
e
x
1
{\displaystyle e^{x_{1}}}
e
x
1
{\displaystyle e^{x_{1}}}
x
2
=
1
3
ln
2
,
{\displaystyle x_{2}={\frac {1}{3}}\ln 2,}
e
x
2
{\displaystyle e^{x_{2}}}
e
x
2
{\displaystyle e^{x_{2}}}
e
x
1
{\displaystyle e^{x_{1}}}
e
x
2
{\displaystyle e^{x_{2}}}
Grįžkime, be didesnių išsidirbinėjimų, prie (10.1) eilutės. Tada
x
=
1
3
ln
50000
=
1
3
ln
(
5
⋅
10
4
)
=
1
3
(
ln
5
+
ln
10
4
)
=
1
3
(
ln
5
+
4
ln
10
)
;
{\displaystyle x={\frac {1}{3}}\ln 50000={\frac {1}{3}}\ln(5\cdot 10^{4})={\frac {1}{3}}(\ln 5+\ln 10^{4})={\frac {1}{3}}(\ln 5+4\ln 10);}
50000
1
3
=
e
x
=
e
1
3
(
ln
5
+
4
ln
10
)
=
e
1
3
ln
5
⋅
e
4
3
ln
10
=
e
1
3
ln
5
⋅
(
e
1
3
ln
10
)
4
.
{\displaystyle 50000^{1 \over 3}=e^{x}=e^{{\frac {1}{3}}(\ln 5+4\ln 10)}=e^{{\frac {1}{3}}\ln 5}\cdot e^{{\frac {4}{3}}\ln 10}=e^{{\frac {1}{3}}\ln 5}\cdot (e^{{\frac {1}{3}}\ln 10})^{4}.}
Toliau mums reikia apskaičiuoti
x
1
=
1
3
ln
5
≈
1.6094379124341
/
3
=
{\displaystyle x_{1}={\frac {1}{3}}\ln 5\approx 1.6094379124341/3=}
x
2
=
1
3
ln
10
≈
2.302585092994
/
3
=
{\displaystyle x_{2}={\frac {1}{3}}\ln 10\approx 2.302585092994/3=}
Dar toliau reikia apskaičiuoti
e
x
1
{\displaystyle e^{x_{1}}}
e
x
2
{\displaystyle e^{x_{2}}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
e
x
1
=
e
0.5364793041447
=
{\displaystyle e^{x_{1}}=e^{0.5364793041447}=}
e
x
2
=
e
0.76752836433
=
{\displaystyle e^{x_{2}}=e^{0.76752836433}=}
Gautą
e
x
2
{\displaystyle e^{x_{2}}}
p
=
(
e
x
2
)
4
=
2.15443469
4
=
{\displaystyle p=(e^{x_{2}})^{4}=2.15443469^{4}=}
Taigi,
50000
1
3
=
e
x
=
e
x
1
⋅
p
=
{\displaystyle 50000^{1 \over 3}=e^{x}=e^{x_{1}}\cdot p=}
=36.8403149864038660577982283358.
Iš "Windows" kalkuliatoriaus
50000
1
3
=
{\displaystyle 50000^{1 \over 3}=}
Taigi, suradome būdą kaip labai pagreitinti šaknies traukimą iš labai didelių skaičių.
Pateiksime dar vieną pavyzdį kaip greitai ištrauktį šaknį iš didelio skaičiaus. Tarkime, reikia ištraukti penkto laipsnio šaknį iš skaičiaus a=7358000. Kaip žinome,
a
y
=
e
ln
a
y
=
e
y
ln
a
.
{\displaystyle a^{y}=e^{\ln a^{y}}=e^{y\ln a}.}
7358000
1
5
=
a
y
=
e
y
ln
a
=
e
1
5
ln
7358000
=
e
x
;
(
10.2
)
{\displaystyle 7358000^{1 \over 5}=a^{y}=e^{y\ln a}=e^{{1 \over 5}\ln 7358000}=e^{x};\quad (10.2)}
x
=
1
5
ln
7358000
=
1
5
ln
(
7.358
⋅
10
6
)
=
1
5
(
ln
7.358
+
ln
10
6
)
=
1
5
(
ln
7.358
+
6
ln
10
)
;
{\displaystyle x={\frac {1}{5}}\ln 7358000={\frac {1}{5}}\ln(7.358\cdot 10^{6})={\frac {1}{5}}(\ln 7.358+\ln 10^{6})={\frac {1}{5}}(\ln 7.358+6\ln 10);}
7358000
1
5
=
e
x
=
e
1
5
(
ln
7.358
+
6
ln
10
)
=
e
1
5
ln
7.358
⋅
e
6
5
ln
10
=
e
1
5
ln
7.358
⋅
(
e
1
5
ln
10
)
6
.
{\displaystyle 7358000^{1 \over 5}=e^{x}=e^{{\frac {1}{5}}(\ln 7.358+6\ln 10)}=e^{{\frac {1}{5}}\ln 7.358}\cdot e^{{\frac {6}{5}}\ln 10}=e^{{\frac {1}{5}}\ln 7.358}\cdot (e^{{\frac {1}{5}}\ln 10})^{6}.}
Toliau mums reikia apskaičiuoti
x
1
=
1
5
ln
7.358
≈
1.99578815668
/
5
=
{\displaystyle x_{1}={\frac {1}{5}}\ln 7.358\approx 1.99578815668/5=}
x
2
=
1
5
ln
10
≈
2.302585092994
/
5
=
{\displaystyle x_{2}={\frac {1}{5}}\ln 10\approx 2.302585092994/5=}
Dar toliau reikia apskaičiuoti
e
x
1
{\displaystyle e^{x_{1}}}
e
x
2
{\displaystyle e^{x_{2}}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
e
x
1
=
e
0.3991576313365
=
{\displaystyle e^{x_{1}}=e^{0.3991576313365}=}
e
x
2
=
e
0.4605170185988
=
{\displaystyle e^{x_{2}}=e^{0.4605170185988}=}
Gautą
e
x
2
{\displaystyle e^{x_{2}}}
p
=
(
e
x
2
)
6
=
1.58489319246
6
=
{\displaystyle p=(e^{x_{2}})^{6}=1.58489319246^{6}=}
Taigi,
7358000
1
/
5
=
e
x
=
e
x
1
⋅
p
=
{\displaystyle 7358000^{1/5}=e^{x}=e^{x_{1}}\cdot p=}
=23.623919642811269319097605866468.
Iš "Windows" kalkuliatoriaus
7358000
0.2
=
{\displaystyle 7358000^{0.2}=}
Atsakymai sutampa.
Pastebėsime, kad jeigu, pavyzdžiui, reikia apskaičiuoti
7358000
1234.567
,
{\displaystyle 7358000^{1234.567},}
7358000
1234.567
=
7358000
1234567
1000
=
7358000
1234567
/
1000
=
(
7358000
1
/
1000
)
1234567
.
{\displaystyle 7358000^{1234.567}=7358000^{1234567 \over 1000}=7358000^{1234567/1000}=(7358000^{1/1000})^{1234567}.}
Tada toliau tereikia apskaičiuoti
7358000
1
/
1000
{\displaystyle 7358000^{1/1000}}
Dar
7358000
1234.567
{\displaystyle 7358000^{1234.567}}
7358000
1234.567
=
7358000
0.1234567
⋅
10000
=
(
7358000
0.1234567
)
10000
.
{\displaystyle 7358000^{1234.567}=7358000^{0.1234567\cdot 10000}=(7358000^{0.1234567})^{10000}.}
Tada tereikia apskaičiuoti
7358000
0.1234567
{\displaystyle 7358000^{0.1234567}}
Dar pastebėsime, kad jeigu skaičiuosime
7358000
0.9876
,
{\displaystyle 7358000^{0.9876},}
e
x
1
{\displaystyle e^{x_{1}}}
e
x
2
{\displaystyle e^{x_{2}}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
1
=
0.9876
ln
7.358
≈
0.9876
⋅
1.99578815668
=
{\displaystyle x_{1}=0.9876\ln 7.358\approx 0.9876\cdot 1.99578815668=}
x
2
=
0.9876
ln
10
≈
0.9876
⋅
2.302585092994
=
{\displaystyle x_{2}=0.9876\ln 10\approx 0.9876\cdot 2.302585092994=}
Su tokiom
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
e
x
1
{\displaystyle e^{x_{1}}}
e
x
2
{\displaystyle e^{x_{2}}}
7358000
0.9876
{\displaystyle 7358000^{0.9876}}
7358000
0.9876
=
(
7358000
0.09876
)
10
.
{\displaystyle 7358000^{0.9876}=(7358000^{0.09876})^{10}.}
Toliau
7358000
0.09876
{\displaystyle 7358000^{0.09876}}
a
y
=
e
y
ln
a
{\displaystyle a^{y}=e^{y\ln a}}
x
1
=
0.09876
ln
7.358
≈
0.09876
⋅
0.199578815668
=
{\displaystyle x_{1}=0.09876\ln 7.358\approx 0.09876\cdot 0.199578815668=}
x
2
=
0.09876
ln
10
≈
0.09876
⋅
0.2302585092994
=
{\displaystyle x_{2}=0.09876\ln 10\approx 0.09876\cdot 0.2302585092994=}
Su tokiom
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
e
x
1
{\displaystyle e^{x_{1}}}
e
x
2
{\displaystyle e^{x_{2}}}
e
x
2
{\displaystyle e^{x_{2}}}
p
=
(
e
x
2
)
6
,
{\displaystyle p=(e^{x_{2}})^{6},}
e
x
1
{\displaystyle e^{x_{1}}}
p ir rezultatą pakelti 10 laipsniu. Ir tada gausime:
7358000
0.9876
=
(
e
x
1
⋅
p
)
10
=
{\displaystyle 7358000^{0.9876}=(e^{x_{1}}\cdot p)^{10}=}
Parodysime dar gudresnį būdą apskaičiuoti
a
y
,
{\displaystyle a^{y},}
a yra labai didelis skaičius. "Windows 10" kalkuliatoriaus didžiausias įmanomas skaičius yra 9.9999999999999999999999999999999e+9999 (tai reiškia
9.999999...
⋅
10
9999
{\displaystyle 9.999999...\cdot 10^{9999}}
9.999999...
⋅
10
9999
{\displaystyle 9.999999...\cdot 10^{9999}}
Vikipedijoj :
"The 80-bit floating-point format has a range (including subnormals) from approximately 3.65×10−4951 to 1.18×104932 . Although log10 (264 ) ≅ 19.266, this format is usually described as giving approximately eighteen significant digits of precision (the floor of log10 (263 ), the minimum guaranteed precision)."
Taigi, Extended Precision (80 bit) turi apie 18 ar 19 skaitmenų (log10 (263 ) = log10 (9,223,372,036,854,775,808) = 18.96488972683) ir apie 4 skaitmenis eksponentės. Tuo tarpu Double Precision (64 bit) turi apie 3 eksponentės skaitmenis (t. y. 999): "Exponents range from −1022 to +1023 because exponents of −1023 (all 0s) and +1024 (all 1s) are reserved for special numbers."
Taigi, iš kalkuliatoriaus
ln(9.9999999999999999999999999999999e+9999) = 23025.850929940456840179914546844.
Ne toks jau didelis skaičius. Tarkim, norime apskaičiuoti
(
9.99
⋅
10
9999
)
0.9876
=
a
y
;
y
=
0.9876.
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=a^{y};\;\;y=0.9876.}
(
9.99
⋅
10
9999
)
0.9876
=
a
y
=
e
y
ln
a
=
e
x
;
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=a^{y}=e^{y\ln a}=e^{x};}
x
=
y
ln
a
=
0.9876
ln
(
9.99
⋅
10
9999
)
=
0.9876
(
ln
(
9.99
)
+
9999
ln
10
)
;
{\displaystyle x=y\ln a=0.9876\ln(9.99\cdot 10^{9999})=0.9876(\ln(9.99)+9999\ln 10);}
x
1
=
0.9876
ln
(
9.99
)
=
{\displaystyle x_{1}=0.9876\ln(9.99)=}
x
2
=
0.9876
⋅
9999
ln
10
=
{\displaystyle x_{2}=0.9876\cdot 9999\ln 10=}
Toliau darome taip. Padauginame x iš
0.00001
=
10
−
5
{\displaystyle 0.00001=10^{-5}}
10
−
7
{\displaystyle 10^{-7}}
x
0
=
x
⋅
10
−
5
=
(
x
1
+
x
2
)
⋅
10
−
5
=
{\displaystyle x_{0}=x\cdot 10^{-5}=(x_{1}+x_{2})\cdot 10^{-5}=}
x
01
=
x
1
⋅
10
−
5
≈
2.2730449437
⋅
10
−
5
=
{\displaystyle x_{01}=x_{1}\cdot 10^{-5}\approx 2.2730449437\cdot 10^{-5}=}
x
02
=
x
2
⋅
10
−
5
≈
22738.05634537
⋅
10
−
5
=
{\displaystyle x_{02}=x_{2}\cdot 10^{-5}\approx 22738.05634537\cdot 10^{-5}=}
Toliau, pagal Makloreno eilutę, apskaičiuojame
e
x
01
=
e
0.000022730449437
=
{\displaystyle e^{x_{01}}=e^{0.000022730449437}=}
e
x
02
=
e
0.2273805634537
=
{\displaystyle e^{x_{02}}=e^{0.2273805634537}=}
Ir galiausiai
e
x
0
=
e
x
1
+
x
2
=
e
x
1
⋅
e
x
2
{\displaystyle e^{x_{0}}=e^{x_{1}+x_{2}}=e^{x_{1}}\cdot e^{x_{2}}}
10
5
=
100000
{\displaystyle 10^{5}=100000}
x dauginome iš
10
−
7
{\displaystyle 10^{-7}}
(
9.99
⋅
10
9999
)
0.9876
=
(
e
x
0
)
100000
=
(
e
x
01
⋅
e
x
02
)
100000
=
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=(e^{x_{0}})^{100000}=(e^{x_{01}}\cdot e^{x_{02}})^{100000}=}
=
(
1.0000227307
⋅
1.2553075012
)
100000
=
{\displaystyle =(1.0000227307\cdot 1.2553075012)^{100000}=}
= 9.990123938748126108971250386732e+9875.
Kalkuliatoriaus reikšmė yra tokia
(
9.99
⋅
10
9999
)
0.9876
=
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=}
= 9.9901239387481261089712504055752e+9875 =
≈
9.990123938748
⋅
10
9875
.
{\displaystyle \approx 9.990123938748\cdot 10^{9875}.}
Iš tikro, galima buvo padaryt greičiau (tiesiog truputi susipainiojęs buvau (kėliau 5 laipsniu, o ne
10
5
,
{\displaystyle 10^{5},}
x
=
y
ln
a
=
0.9876
ln
(
9.99
⋅
10
9999
)
=
{\displaystyle x=y\ln a=0.9876\ln(9.99\cdot 10^{9999})=}
x
0
=
x
⋅
10
−
5
=
{\displaystyle x_{0}=x\cdot 10^{-5}=}
Toliau, pagal Makloreno eilutę, apskaičiuojame
e
x
0
=
e
0.227403293
=
{\displaystyle e^{x_{0}}=e^{0.227403293}=}
Ir toliau tiesiog
e
x
0
{\displaystyle e^{x_{0}}}
10
5
{\displaystyle 10^{5}}
(
9.99
⋅
10
9999
)
0.9876
=
(
e
x
0
)
100000
=
(
e
0.22740329390315
)
100000
=
1.25533603523574
100000
=
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=(e^{x_{0}})^{100000}=(e^{0.22740329390315})^{100000}=1.25533603523574^{100000}=}
= 1.2553360352357425652203277670022^100000 =
= 9.9901239387481261089712504105024e+9875.
Bet kokio skaičiaus pakelimas bet kokiu laipsniu (universalus algoritmas) Jei pakelsime
2
{\displaystyle 2}
2
2
=
4.
{\displaystyle 2^{2}=4.}
4
{\displaystyle 4}
(
2
2
)
2
=
2
2
⋅
2
=
4
2
=
16.
{\displaystyle (2^{2})^{2}=2^{2\cdot 2}=4^{2}=16.}
2
2
⋅
2
⋅
2
=
16
2
=
256.
{\displaystyle 2^{2\cdot 2\cdot 2}=16^{2}=256.}
2
16
=
2
2
⋅
2
⋅
2
⋅
2
=
256
2
=
65536.
{\displaystyle 2^{16}=2^{2\cdot 2\cdot 2\cdot 2}=256^{2}=65536.}
Kad gauti skaičių 65536, reikia ant "Windows 10" kalkuliatoriaus paspasuti 2, o paskui 4 kartus paspausti mygtuką x2 .
Parodysime kaip galima pakelti bet kokį skaičių bet kokiu laipsniu, naudojant tuos pačius veiksmus (nauduojant tą patį algoritmą). Bet šitas algoritmas labiau akcentuotas į pakelimą y laipsniu, skaičių a , kuris yra didesnis už 1.
Tarkim, norime apskaičiuoti
(
9.99
⋅
10
9999
)
0.9876
=
a
y
;
y
=
0.9876.
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=a^{y};\;\;y=0.9876.}
(
9.99
⋅
10
9999
)
0.9876
=
a
y
=
e
y
ln
a
=
e
x
;
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=a^{y}=e^{y\ln a}=e^{x};}
x
=
y
ln
a
=
0.9876
ln
(
9.99
⋅
10
9999
)
=
{\displaystyle x=y\ln a=0.9876\ln(9.99\cdot 10^{9999})=}
Žinome, kad
e
x
=
(
e
x
/
65536
)
65536
=
(
e
0.0000152587890625
⋅
x
)
65536
.
{\displaystyle e^{x}=(e^{x/65536})^{65536}=(e^{0.0000152587890625\cdot x})^{65536}.}
Tada pažymėkime
x
0
=
0.0000152587890625
⋅
x
=
{\displaystyle x_{0}=0.0000152587890625\cdot x=}
= 0.34698988937858681836340330299764.
Toliau apskaičiuojame Makloreno eilutę
e
x
0
,
{\displaystyle e^{x_{0}},}
x
0
<
1.
{\displaystyle x_{0}<1.}
e
x
0
=
e
0.3469898893785868
=
{\displaystyle e^{x_{0}}=e^{0.3469898893785868}=}
Toliau
e
x
0
{\displaystyle e^{x_{0}}}
65536
=
2
16
.
{\displaystyle 65536=2^{16}.}
e
x
0
{\displaystyle e^{x_{0}}}
2 ant kalkuliatoriaus 16 kartų). Tada gausime
(
9.99
⋅
10
9999
)
0.9876
=
a
y
=
e
y
ln
a
=
e
x
=
(
e
x
0
)
65536
=
(
e
x
0
)
2
16
=
1.41480242
65536
=
{\displaystyle (9.99\cdot 10^{9999})^{0.9876}=a^{y}=e^{y\ln a}=e^{x}=(e^{x_{0}})^{65536}=(e^{x_{0}})^{2^{16}}=1.41480242^{65536}=}
= 1.414802420766463739969945373195^65536 = 9.9901239387481261089712504055752e+9875.
Atėmę gautą ay reikšmę iš tikslios "Windows 10" kalkuliatoriaus reikšmės, gauname:
9.9901239387481261089712504055752e+9875 - 9.9901239387481261089712504055752e+9875 =
= 3.8064298062079638161811295770571e+9833.
Kaip sakyta anksčiau, "Windows 10" kalkuliatoriaus tikslumas yra daugiau nei 32 skaitmenys (tie skaitmenys nerodomi). Apytiksli paklaida yra 1.0e+9875 / 1.0e+9833 = 1.e+42, t. y. apie
10
−
42
.
{\displaystyle 10^{-42}.}
10
−
46
{\displaystyle 10^{-46}}
![{\displaystyle {\sqrt[{5}]{e}}=e^{1 \over 5}=e^{0.2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a561113239f480d0488cabdf918426f5859e1f1d)
![{\displaystyle {\sqrt[{5}]{e}}=e^{1 \over 5}=1+{1 \over 5}+{\frac {({1 \over 5})^{2}}{2!}}+{\frac {({1 \over 5})^{3}}{3!}}+{\frac {({1 \over 5})^{4}}{4!}}+...+{\frac {({1 \over 5})^{n}}{n!}}+r_{n+1}(0.2);}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1877458ddbea7060f8f9d67084c471c2cce1a113)
![{\displaystyle {\sqrt[{5}]{e}}=e^{0.2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31910644bf15ee8cd0e7278d8ba66255b95be3b0)
![{\displaystyle {\sqrt[{5}]{e}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c67b5756526145b3e2e4b80df6e6fca3e2e0fd73)
