Elementariųjų funkcijų reikšmių skaičiavimas
Logaritminės ir atvirkštinių trigonometrinių funkcijų reikšmių skaičiavimas pagrįstas Teiloro formulės taikymu. Čia smulkiai aptarsime, kaip skaičiuojamos logaritmo ir arktangento reikšmės. Funkcijų
arcctg
x
,
{\displaystyle \operatorname {arcctg} x,\;}
arcsin
x
{\displaystyle \operatorname {arcsin} x\;}
arccos
x
{\displaystyle \operatorname {arccos} x\;}
[ https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Relationships_among_the_inverse_trigonometric_functions ]
arcctg
x
=
π
2
−
arctg
x
,
arcsin
x
=
arctg
x
1
−
x
2
,
{\displaystyle \operatorname {arcctg} x={\frac {\pi }{2}}-\operatorname {arctg} x,\;\;\;\operatorname {arcsin} x=\operatorname {arctg} {\frac {x}{\sqrt {1-x^{2}}}},}
arccos
x
=
arcctg
x
1
−
x
2
=
arctg
1
−
x
2
x
.
{\displaystyle \operatorname {arccos} x=\operatorname {arcctg} {\frac {x}{\sqrt {1-x^{2}}}}=\operatorname {arctg} {\frac {\sqrt {1-x^{2}}}{x}}.}
[ keisti ] Skaičių a>0 išreikškime sandauga
a
=
2
p
M
.
(
8.83
)
{\displaystyle a=2^{p}M.\quad (8.83)}
kurioje p – sveikasis skaičius, o M tenkina nelygybes
1
2
≤
M
<
1.
(
8.84
)
{\displaystyle {\frac {1}{2}}\leq M<1.\quad (8.84)}
Pastebėsime, kad skaičių a išreikšti (8.83) sandauga galima vieninteliu būdu. Remdamiesi (8.83) formule, logaritmą
ln
a
{\displaystyle \ln a}
ln
a
=
ln
(
2
p
M
)
=
p
ln
2
+
ln
M
.
(
8.85
)
{\displaystyle \ln a=\ln(2^{p}M)=p\ln 2+\ln M.\quad (8.85)}
Sakykime,
M
=
1
2
1
+
x
1
−
x
,
(
8.86
)
{\displaystyle M={\frac {1}{\sqrt {2}}}\,{\frac {1+x}{1-x}},\quad (8.86)}
ir šitą M išraišką įrašykime į (8.85) lygybę. Tada logaritmo
ln
a
{\displaystyle \ln a}
ln
a
=
p
ln
2
+
ln
M
=
p
ln
2
+
ln
(
1
2
1
+
x
1
−
x
)
=
p
ln
2
+
ln
1
2
+
ln
1
+
x
1
−
x
=
{\displaystyle \ln a=p\ln 2+\ln M=p\ln 2+\ln \left({\frac {1}{\sqrt {2}}}{\frac {1+x}{1-x}}\right)=p\ln 2+\ln {\frac {1}{\sqrt {2}}}+\ln {\frac {1+x}{1-x}}=}
=
p
ln
2
−
1
2
ln
2
+
ln
1
+
x
1
−
x
=
(
p
−
1
2
)
ln
2
+
ln
1
+
x
1
−
x
.
{\displaystyle =p\ln 2-{\frac {1}{2}}\ln 2+\ln {\frac {1+x}{1-x}}=\left(p-{\frac {1}{2}}\right)\ln 2+\ln {\frac {1+x}{1-x}}.}
ln
a
=
(
p
−
1
2
)
ln
2
+
ln
1
+
x
1
−
x
.
(
8.87
)
{\displaystyle \ln a=\left(p-{\frac {1}{2}}\right)\ln 2+\ln {\frac {1+x}{1-x}}.\quad (8.87)}
Funkciją
ln
1
+
x
1
−
x
{\displaystyle \ln {\frac {1+x}{1-x}}}
g
(
x
)
=
ln
1
+
x
1
−
x
{\displaystyle g(x)=\ln {\frac {1+x}{1-x}}}
x
=
0
{\displaystyle x=0}
g
′
(
x
)
=
(
ln
1
+
x
1
−
x
)
′
=
1
−
x
1
+
x
⋅
(
1
+
x
1
−
x
)
′
=
{\displaystyle g'(x)=\left(\ln {\frac {1+x}{1-x}}\right)'={\frac {1-x}{1+x}}\cdot \left({\frac {1+x}{1-x}}\right)'=}
=
1
−
x
1
+
x
⋅
(
1
+
x
)
′
(
1
−
x
)
−
(
1
+
x
)
(
1
−
x
)
′
(
1
−
x
)
2
=
1
−
x
1
+
x
⋅
(
1
−
x
)
−
(
1
+
x
)
(
−
1
)
(
1
−
x
)
2
=
1
−
x
1
+
x
⋅
(
1
−
x
)
+
(
1
+
x
)
(
1
−
x
)
2
=
{\displaystyle ={\frac {1-x}{1+x}}\cdot {\frac {(1+x)'(1-x)-(1+x)(1-x)'}{(1-x)^{2}}}={\frac {1-x}{1+x}}\cdot {\frac {(1-x)-(1+x)(-1)}{(1-x)^{2}}}={\frac {1-x}{1+x}}\cdot {\frac {(1-x)+(1+x)}{(1-x)^{2}}}=}
=
1
−
x
1
+
x
⋅
2
(
1
−
x
)
2
=
2
(
1
+
x
)
(
1
−
x
)
=
2
1
−
x
2
.
{\displaystyle ={\frac {1-x}{1+x}}\cdot {\frac {2}{(1-x)^{2}}}={\frac {2}{(1+x)(1-x)}}={\frac {2}{1-x^{2}}}.}
g
″
(
x
)
=
(
ln
1
+
x
1
−
x
)
″
=
(
2
1
−
x
2
)
′
=
−
2
(
1
−
x
2
)
2
⋅
(
1
−
x
2
)
′
=
−
2
(
1
−
x
2
)
2
⋅
(
−
2
x
)
=
4
x
(
1
−
x
2
)
2
.
{\displaystyle g''(x)=\left(\ln {\frac {1+x}{1-x}}\right)''=\left({\frac {2}{1-x^{2}}}\right)'=-{\frac {2}{(1-x^{2})^{2}}}\cdot (1-x^{2})'=-{\frac {2}{(1-x^{2})^{2}}}\cdot (-2x)={\frac {4x}{(1-x^{2})^{2}}}.}
g
‴
(
x
)
=
(
ln
1
+
x
1
−
x
)
‴
=
(
4
x
(
1
−
x
2
)
2
)
′
=
(
4
x
)
′
(
1
−
x
2
)
2
−
4
x
[
(
1
−
x
2
)
2
]
′
(
(
1
−
x
2
)
2
)
2
=
{\displaystyle g'''(x)=\left(\ln {\frac {1+x}{1-x}}\right)'''=\left({\frac {4x}{(1-x^{2})^{2}}}\right)'={\frac {(4x)'(1-x^{2})^{2}-4x[(1-x^{2})^{2}]'}{((1-x^{2})^{2})^{2}}}=}
=
4
(
1
−
x
2
)
2
−
4
x
[
2
(
1
−
x
2
)
′
]
(
1
−
x
2
)
4
=
4
(
1
−
2
x
2
+
x
4
)
−
4
x
[
2
(
−
2
x
)
]
(
1
−
x
2
)
4
=
4
−
8
x
2
+
4
x
4
+
16
x
2
(
1
−
x
2
)
4
=
{\displaystyle ={\frac {4(1-x^{2})^{2}-4x[2(1-x^{2})']}{(1-x^{2})^{4}}}={\frac {4(1-2x^{2}+x^{4})-4x[2(-2x)]}{(1-x^{2})^{4}}}={\frac {4-8x^{2}+4x^{4}+16x^{2}}{(1-x^{2})^{4}}}=}
=
4
+
8
x
2
+
4
x
4
(
1
−
x
2
)
4
=
4
(
1
+
2
x
2
+
x
4
)
(
1
−
x
2
)
4
=
4
(
1
+
x
2
)
2
(
1
−
x
2
)
4
.
{\displaystyle ={\frac {4+8x^{2}+4x^{4}}{(1-x^{2})^{4}}}={\frac {4(1+2x^{2}+x^{4})}{(1-x^{2})^{4}}}={\frac {4(1+x^{2})^{2}}{(1-x^{2})^{4}}}.}
g
(
4
)
(
x
)
=
(
ln
1
+
x
1
−
x
)
(
4
)
=
(
4
(
1
+
x
2
)
2
(
1
−
x
2
)
4
)
′
=
{\displaystyle g^{(4)}(x)=\left(\ln {\frac {1+x}{1-x}}\right)^{(4)}=\left({\frac {4(1+x^{2})^{2}}{(1-x^{2})^{4}}}\right)'=}
=
(
8
(
1
+
x
2
)
⋅
2
x
)
(
1
−
x
2
)
4
−
4
(
1
+
x
2
)
2
⋅
4
(
1
−
x
2
)
3
⋅
(
−
2
x
)
(
1
−
x
2
)
8
=
{\displaystyle ={\frac {(8(1+x^{2})\cdot 2x)(1-x^{2})^{4}-4(1+x^{2})^{2}\cdot 4(1-x^{2})^{3}\cdot (-2x)}{(1-x^{2})^{8}}}=}
=
16
x
(
1
+
x
2
)
(
1
−
x
2
)
4
+
32
x
(
1
+
x
2
)
2
(
1
−
x
2
)
3
(
1
−
x
2
)
8
.
{\displaystyle ={\frac {16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}}{(1-x^{2})^{8}}}.}
g
(
5
)
(
x
)
=
(
ln
1
+
x
1
−
x
)
(
5
)
=
(
16
x
(
1
+
x
2
)
(
1
−
x
2
)
4
+
32
x
(
1
+
x
2
)
2
(
1
−
x
2
)
3
(
1
−
x
2
)
8
)
′
=
{\displaystyle g^{(5)}(x)=\left(\ln {\frac {1+x}{1-x}}\right)^{(5)}=\left({\frac {16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}}{(1-x^{2})^{8}}}\right)'=}
=
[
16
x
(
1
+
x
2
)
(
1
−
x
2
)
4
+
32
x
(
1
+
x
2
)
2
(
1
−
x
2
)
3
]
′
(
1
−
x
2
)
8
−
[
16
x
(
1
+
x
2
)
(
1
−
x
2
)
4
+
32
x
(
1
+
x
2
)
2
(
1
−
x
2
)
3
]
(
(
1
−
x
2
)
8
)
′
(
1
−
x
2
)
16
=
{\displaystyle ={\frac {[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}]'(1-x^{2})^{8}-[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}]((1-x^{2})^{8})'}{(1-x^{2})^{16}}}=}
=
[
16
(
1
+
x
2
)
(
1
−
x
2
)
4
+
16
x
g
1
(
x
)
+
32
(
1
+
x
2
)
2
(
1
−
x
2
)
3
+
32
x
g
2
(
x
)
]
(
1
−
x
2
)
8
−
[
16
x
(
1
+
x
2
)
(
1
−
x
2
)
4
+
32
x
(
1
+
x
2
)
2
(
1
−
x
2
)
3
]
8
(
1
−
x
2
)
7
⋅
(
−
2
x
)
(
1
−
x
2
)
16
=
{\displaystyle ={\frac {[16(1+x^{2})(1-x^{2})^{4}+16xg_{1}(x)+32(1+x^{2})^{2}(1-x^{2})^{3}+32xg_{2}(x)](1-x^{2})^{8}-[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}]8(1-x^{2})^{7}\cdot (-2x)}{(1-x^{2})^{16}}}=}
=
[
16
(
1
+
x
2
)
(
1
−
x
2
)
4
+
16
x
g
1
(
x
)
+
32
(
1
+
x
2
)
2
(
1
−
x
2
)
3
+
32
x
g
2
(
x
)
]
(
1
−
x
2
)
8
+
16
x
[
16
x
(
1
+
x
2
)
(
1
−
x
2
)
4
+
32
x
(
1
+
x
2
)
2
(
1
−
x
2
)
3
]
(
1
−
x
2
)
7
(
1
−
x
2
)
16
;
{\displaystyle ={\frac {[16(1+x^{2})(1-x^{2})^{4}+16xg_{1}(x)+32(1+x^{2})^{2}(1-x^{2})^{3}+32xg_{2}(x)](1-x^{2})^{8}+16x[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}](1-x^{2})^{7}}{(1-x^{2})^{16}}};}
čia
g
1
(
x
)
=
[
(
1
+
x
2
)
(
1
−
x
2
)
4
]
′
=
2
x
(
1
−
x
2
)
4
+
(
1
+
x
2
)
⋅
4
(
1
−
x
2
)
3
⋅
(
−
2
x
)
=
2
x
(
1
−
x
2
)
4
−
8
x
(
1
+
x
2
)
(
1
−
x
2
)
3
,
{\displaystyle g_{1}(x)=[(1+x^{2})(1-x^{2})^{4}]'=2x(1-x^{2})^{4}+(1+x^{2})\cdot 4(1-x^{2})^{3}\cdot (-2x)=2x(1-x^{2})^{4}-8x(1+x^{2})(1-x^{2})^{3},}
g
2
(
x
)
=
[
(
1
+
x
2
)
2
(
1
−
x
2
)
3
]
′
=
2
(
1
+
x
2
)
⋅
2
x
⋅
(
1
−
x
2
)
3
+
(
1
+
x
2
)
2
⋅
3
(
1
−
x
2
)
2
⋅
(
−
2
x
)
=
{\displaystyle g_{2}(x)=[(1+x^{2})^{2}(1-x^{2})^{3}]'=2(1+x^{2})\cdot 2x\cdot (1-x^{2})^{3}+(1+x^{2})^{2}\cdot 3(1-x^{2})^{2}\cdot (-2x)=}
=
4
x
(
1
+
x
2
)
(
1
−
x
2
)
3
−
6
x
(
1
+
x
2
)
2
(
1
−
x
2
)
2
.
{\displaystyle =4x(1+x^{2})(1-x^{2})^{3}-6x(1+x^{2})^{2}(1-x^{2})^{2}.}
Akivaizdu, kad
g
1
(
0
)
=
0
{\displaystyle g_{1}(0)=0}
g
2
(
0
)
=
0.
{\displaystyle g_{2}(0)=0.}
Randame funkcijos g(x) ir jos išvestinių reikšmes taške
x
=
0
{\displaystyle x=0}
g
(
0
)
=
ln
1
+
0
1
−
0
=
ln
1
=
0
,
{\displaystyle g(0)=\ln {\frac {1+0}{1-0}}=\ln 1=0,}
g
′
(
0
)
=
2
1
−
0
2
=
2
,
{\displaystyle g'(0)={\frac {2}{1-0^{2}}}=2,}
g
″
(
0
)
=
4
⋅
0
(
1
−
0
2
)
2
=
0
,
{\displaystyle g''(0)={\frac {4\cdot 0}{(1-0^{2})^{2}}}=0,}
g
‴
(
0
)
=
4
(
1
+
2
⋅
0
2
+
0
4
)
(
1
−
0
2
)
4
=
4
,
{\displaystyle g'''(0)={\frac {4(1+2\cdot 0^{2}+0^{4})}{(1-0^{2})^{4}}}=4,}
g
(
4
)
(
0
)
=
16
⋅
0
(
1
+
0
2
)
(
1
−
x
2
)
4
+
32
⋅
0
(
1
+
0
2
)
2
(
1
−
0
2
)
3
(
1
−
0
2
)
8
=
0.
{\displaystyle g^{(4)}(0)={\frac {16\cdot 0(1+0^{2})(1-x^{2})^{4}+32\cdot 0(1+0^{2})^{2}(1-0^{2})^{3}}{(1-0^{2})^{8}}}=0.}
g
(
5
)
(
0
)
=
[
16
(
1
+
0
2
)
(
1
−
0
2
)
4
+
16
⋅
0
⋅
g
1
(
0
)
+
32
(
1
+
0
2
)
2
(
1
−
0
2
)
3
+
32
⋅
0
⋅
g
2
(
0
)
]
(
1
−
0
2
)
8
+
16
⋅
0
[
16
⋅
0
(
1
+
0
2
)
(
1
−
0
2
)
4
+
32
⋅
0
(
1
+
0
2
)
2
(
1
−
0
2
)
3
]
(
1
−
0
2
)
7
(
1
−
0
2
)
16
=
{\displaystyle g^{(5)}(0)={\frac {[16(1+0^{2})(1-0^{2})^{4}+16\cdot 0\cdot g_{1}(0)+32(1+0^{2})^{2}(1-0^{2})^{3}+32\cdot 0\cdot g_{2}(0)](1-0^{2})^{8}+16\cdot 0[16\cdot 0(1+0^{2})(1-0^{2})^{4}+32\cdot 0(1+0^{2})^{2}(1-0^{2})^{3}](1-0^{2})^{7}}{(1-0^{2})^{16}}}=}
=
[
16
+
32
]
+
0
1
=
48.
{\displaystyle ={\frac {[16+32]+0}{1}}=48.}
Makloreno eilutės dėstinys su Lagranžo formos liekamuoju nariu bus šitoks:
ln
1
+
x
1
−
x
=
g
(
0
)
+
g
′
(
0
)
1
!
x
+
g
″
(
0
)
2
!
x
2
+
g
‴
(
0
)
3
!
x
3
+
.
.
.
+
g
(
n
)
(
0
)
n
!
x
n
+
g
(
n
+
1
)
(
θ
x
)
(
n
+
1
)
!
x
n
+
1
=
{\displaystyle \ln {\frac {1+x}{1-x}}=g(0)+{\frac {g'(0)}{1!}}x+{\frac {g''(0)}{2!}}x^{2}+{\frac {g'''(0)}{3!}}x^{3}+...+{\frac {g^{(n)}(0)}{n!}}x^{n}+{\frac {g^{(n+1)}(\theta x)}{(n+1)!}}x^{n+1}=}
=
0
+
2
1
!
x
+
0
2
!
x
2
+
4
3
!
x
3
+
0
4
!
x
4
+
48
5
!
x
5
+
.
.
.
+
R
2
n
+
2
(
x
)
=
{\displaystyle =0+{\frac {2}{1!}}x+{\frac {0}{2!}}x^{2}+{\frac {4}{3!}}x^{3}+{\frac {0}{4!}}x^{4}+{\frac {48}{5!}}x^{5}+...+R_{2n+2}(x)=}
=
2
x
+
2
3
x
3
+
6
⋅
8
5
⋅
4
⋅
3
⋅
2
x
5
+
.
.
.
+
R
2
n
+
2
(
x
)
=
{\displaystyle =2x+{\frac {2}{3}}x^{3}+{\frac {6\cdot 8}{5\cdot 4\cdot 3\cdot 2}}x^{5}+...+R_{2n+2}(x)=}
=
2
x
+
2
3
x
3
+
8
5
⋅
4
x
5
+
.
.
.
+
R
2
n
+
2
(
x
)
=
2
x
+
2
x
3
3
+
2
x
5
5
+
.
.
.
+
2
x
2
n
+
1
2
n
+
1
+
R
2
n
+
2
(
x
)
.
{\displaystyle =2x+{\frac {2}{3}}x^{3}+{\frac {8}{5\cdot 4}}x^{5}+...+R_{2n+2}(x)=2x+{\frac {2x^{3}}{3}}+{\frac {2x^{5}}{5}}+...+{\frac {2x^{2n+1}}{2n+1}}+R_{2n+2}(x).}
ln
1
+
x
1
−
x
=
2
x
+
2
x
3
3
+
2
x
5
5
+
.
.
.
+
2
x
2
n
+
1
2
n
+
1
+
R
2
n
+
2
(
x
)
,
(
8.88
)
{\displaystyle \ln {\frac {1+x}{1-x}}=2x+{\frac {2x^{3}}{3}}+{\frac {2x^{5}}{5}}+...+{\frac {2x^{2n+1}}{2n+1}}+R_{2n+2}(x),\quad (8.88)}
R
2
n
+
2
(
x
)
=
−
x
2
n
+
2
2
n
+
2
[
1
(
1
+
θ
x
)
2
n
+
2
−
1
(
1
−
θ
x
)
2
n
+
2
]
;
(
8.89
)
{\displaystyle R_{2n+2}(x)=-{\frac {x^{2n+2}}{2n+2}}\left[{\frac {1}{(1+\theta x)^{2n+2}}}-{\frac {1}{(1-\theta x)^{2n+2}}}\right];\quad (8.89)}
skaičius
θ
{\displaystyle \theta }
[Naudotojas Paraboloid neįsivaizduoja iš kur ir kaip gauta tokia
R
2
n
+
2
(
x
)
{\displaystyle R_{2n+2}(x)}
Skaičiuojant
ln
a
{\displaystyle \ln a}
ln
a
≈
(
p
−
1
2
)
ln
2
+
2
(
x
+
x
3
3
+
x
5
5
+
.
.
.
+
x
2
n
+
1
2
n
+
1
)
,
(
8.90
)
{\displaystyle \ln a\approx \left(p-{\frac {1}{2}}\right)\ln 2+2\left(x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+...+{\frac {x^{2n+1}}{2n+1}}\right),\quad (8.90)}
kuri gaunama iš (8.87) lygybės, funkciją
ln
1
+
x
1
−
x
{\displaystyle \ln {\frac {1+x}{1-x}}}
R
2
n
+
2
(
x
)
.
{\displaystyle R_{2n+2}(x).}
x apytikslėje
ln
a
{\displaystyle \ln a}
M .
Įvertinsime (8.90) formulės paklaidą. Kadangi skaičiaus
ln
a
{\displaystyle \ln a}
R
2
n
+
2
(
x
)
,
{\displaystyle R_{2n+2}(x),}
Iš pradžių išsiaiškinsime x kitimo ribas. Iš (8.86) formulės gauname
M
=
1
2
1
+
x
1
−
x
,
(
8.86
)
{\displaystyle M={\frac {1}{\sqrt {2}}}\,{\frac {1+x}{1-x}},\quad (8.86)}
M
2
=
1
+
x
1
−
x
,
{\displaystyle M{\sqrt {2}}={\frac {1+x}{1-x}},}
M
2
(
1
−
x
)
=
1
+
x
,
{\displaystyle M{\sqrt {2}}(1-x)=1+x,}
M
2
−
1
=
x
+
x
M
2
,
{\displaystyle M{\sqrt {2}}-1=x+xM{\sqrt {2}},}
x
=
M
2
−
1
M
2
+
1
.
(
8.91
)
{\displaystyle x={\frac {M{\sqrt {2}}-1}{M{\sqrt {2}}+1}}.\quad (8.91)}
Jei M reikšmės tenkina (8.84) nelygybes, tai x tenkina sąlygą*
|
x
|
<
0.172.
(
8.92
)
{\displaystyle |x|<0.172.\quad (8.92)}
[
h
′
(
M
)
=
(
M
2
−
1
M
2
+
1
)
′
=
2
(
M
2
+
1
)
−
(
M
2
−
1
)
2
(
M
2
+
1
)
2
=
(
2
M
+
2
)
−
(
2
M
−
2
)
(
M
2
+
1
)
2
=
2
2
(
M
2
+
1
)
2
.
{\displaystyle h'(M)=\left({\frac {M{\sqrt {2}}-1}{M{\sqrt {2}}+1}}\right)'={\frac {{\sqrt {2}}(M{\sqrt {2}}+1)-(M{\sqrt {2}}-1){\sqrt {2}}}{(M{\sqrt {2}}+1)^{2}}}={\frac {(2M+{\sqrt {2}})-(2M-{\sqrt {2}})}{(M{\sqrt {2}}+1)^{2}}}={\frac {2{\sqrt {2}}}{(M{\sqrt {2}}+1)^{2}}}.}
2
2
(
M
2
+
1
)
2
=
0.
{\displaystyle {\frac {2{\sqrt {2}}}{(M{\sqrt {2}}+1)^{2}}}=0.}
M reikšmių, su kuriom h'(M) būtų lygi nuliui. Todėl funkcija
h
(
M
)
=
M
2
−
1
M
2
+
1
{\displaystyle h(M)={\frac {M{\sqrt {2}}-1}{M{\sqrt {2}}+1}}}
h
(
1
/
2
)
=
1
2
2
−
1
1
2
2
+
1
=
1
2
−
1
1
2
+
1
=
{\displaystyle h(1/2)={\frac {{\frac {1}{2}}{\sqrt {2}}-1}{{\frac {1}{2}}{\sqrt {2}}+1}}={\frac {{\frac {1}{\sqrt {2}}}-1}{{\frac {1}{\sqrt {2}}}+1}}=}
=(0.70710678118654752440084436210485-1)/(0.70710678118654752440084436210485+1) =
= -0.29289321881345247559915563789515/1.7071067811865475244008443621048 = -0.17157287525380990239662255158061.
h
(
1
)
=
1
⋅
2
−
1
1
⋅
2
+
1
=
{\displaystyle h(1)={\frac {1\cdot {\sqrt {2}}-1}{1\cdot {\sqrt {2}}+1}}=}
=(1.4142135623730950488016887242097-1)/(1.4142135623730950488016887242097+1)=
= 0.4142135623730950488016887242097/2.4142135623730950488016887242097 = 0.1715728752538099023966225515806.
Taigi, matome, kad
-0.17157287525380990239662255158061 < x < 0.1715728752538099023966225515806. Arba |x| < 0.171572875254.]
Dabar pastebėsime, kad liekamasis narys
R
2
n
+
2
(
x
)
{\displaystyle R_{2n+2}(x)}
x reikšmės teigiamos, tiek tada, kai jos neigiamos (iš (8.89) formulės matyti, kad, x pakeitus -x , liekamojo nario
R
2
n
+
2
(
x
)
{\displaystyle R_{2n+2}(x)}
R
2
n
+
2
(
x
)
,
{\displaystyle R_{2n+2}(x),}
x
>
0.
{\displaystyle x>0.}
x rašome skaičių 0.172, vietoj trupmenos
1
1
+
θ
x
{\displaystyle {\frac {1}{1+\theta x}}}
1
1
−
θ
x
{\displaystyle {\frac {1}{1-\theta x}}}
1
1
−
0.172
{\displaystyle {\frac {1}{1-0.172}}\;}
|
R
2
n
+
2
(
x
)
|
≤
(
0.172
)
2
n
+
2
2
n
+
2
[
1
+
1
(
1
−
0.172
)
2
n
+
2
]
.
{\displaystyle |R_{2n+2}(x)|\leq {\frac {(0.172)^{2n+2}}{2n+2}}\left[1+{\frac {1}{(1-0.172)^{2n+2}}}\right].}
[Taip, čia laužtiniuose skliaustuose jau pliusas pasidarė. Tai arba (8.89) formulėje klaida arba čia.]
Daugiklį
(
0.172
)
2
n
+
2
{\displaystyle (0.172)^{2n+2}}
0.172
1
−
0.172
<
0.208
{\displaystyle {\frac {0.172}{1-0.172}}<0.208\;}
|
R
2
n
+
2
(
x
)
|
≤
(
0.172
)
2
n
+
2
+
(
0.208
)
2
n
+
2
2
n
+
2
.
(
8.93
)
{\displaystyle |R_{2n+2}(x)|\leq {\frac {(0.172)^{2n+2}+(0.208)^{2n+2}}{2n+2}}.\quad (8.93)}
Skaičiuojant
ln
a
{\displaystyle \ln a}
n
=
6.
{\displaystyle n=6.}
(
0.172
)
14
+
(
0.208
)
14
14
,
{\displaystyle {\frac {(0.172)^{14}+(0.208)^{14}}{14}},\;}
1.625
⋅
10
−
10
.
{\displaystyle 1.625\cdot 10^{-10}.}
[ (0.172^14 + 0.208^14)/14 = 2.1682237054688578614343997699803e-11
≈
2.1682237
⋅
10
−
11
.
{\displaystyle \approx 2.1682237\cdot 10^{-11}.}
[O jeigu ten turi būti minusas laužtiniuose skliaustuose kaip (8.89) formulėje, tai
(0.172^14 - 0.208^14)/14 = -1.8848887113203326016424566362697e-11
≈
−
1.88488871132
⋅
10
−
11
.
{\displaystyle \approx -1.88488871132\cdot 10^{-11}.}
________________
* Kadangi x yra argumento M funkcija, tai užtenka rasti maksimalią (8.91) funkcijos modulio reikšmę segmente
[
1
2
;
1
]
.
{\displaystyle \left[{\frac {1}{2}};\;1\right].}
** Taip skaičiuojamas
ln
a
{\displaystyle \ln a}
Apskaičiuosime ln(a)=ln(143). Pagal (8.83) formulę [
a
=
2
p
M
.
(
8.83
)
{\displaystyle a=2^{p}M.\quad (8.83)}
143
=
2
8
M
,
{\displaystyle 143=2^{8}M,}
M
=
143
256
=
0.55859375.
{\displaystyle M={\frac {143}{256}}=0.55859375.}
Toliau, pagal (8.87) formulę
[
ln
a
=
(
p
−
1
2
)
ln
2
+
ln
1
+
x
1
−
x
.
(
8.87
)
{\displaystyle \ln a=\left(p-{\frac {1}{2}}\right)\ln 2+\ln {\frac {1+x}{1-x}}.\quad (8.87)}
ln
143
=
(
8
−
1
2
)
ln
2
+
ln
1
+
x
1
−
x
.
{\displaystyle \ln 143=\left(8-{\frac {1}{2}}\right)\ln 2+\ln {\frac {1+x}{1-x}}.}
(
8
−
1
2
)
ln
2
=
7.5
ln
2
=
{\displaystyle \left(8-{\frac {1}{2}}\right)\ln 2=7.5\ln 2=}
Tiksli ln(143) reikšmė yra ln(143)=4.9628446302599072801154310195304.
Iš (8.91) formulės randame x :
x
=
M
2
−
1
M
2
+
1
=
0.55859375
2
−
1
0.55859375
2
+
1
=
{\displaystyle x={\frac {M{\sqrt {2}}-1}{M{\sqrt {2}}+1}}={\frac {0.55859375{\sqrt {2}}-1}{0.55859375{\sqrt {2}}+1}}=}
= (0.55859375*1.4142135623730950488016887242097 - 1)/(0.55859375*1.4142135623730950488016887242097+1) =
= -0.11733662705136264741194733209931.
Toliau pagal (8.88) formulę apskaičiuosime
ln
1
+
x
1
−
x
,
{\displaystyle \ln {\frac {1+x}{1-x}},\,}
ln
1
+
x
1
−
x
=
2
x
+
2
x
3
3
+
2
x
5
5
+
.
.
.
+
2
x
2
n
+
1
2
n
+
1
=
{\displaystyle \ln {\frac {1+x}{1-x}}=2x+{\frac {2x^{3}}{3}}+{\frac {2x^{5}}{5}}+...+{\frac {2x^{2n+1}}{2n+1}}=}
=
2
x
+
2
x
3
3
+
2
x
5
5
+
2
x
7
7
+
2
x
9
9
+
2
x
11
11
+
2
x
13
13
=
2
(
x
+
x
3
3
+
x
5
5
+
x
7
7
+
x
9
9
+
x
11
11
+
x
13
13
)
=
{\displaystyle =2x+{\frac {2x^{3}}{3}}+{\frac {2x^{5}}{5}}+{\frac {2x^{7}}{7}}+{\frac {2x^{9}}{9}}+{\frac {2x^{11}}{11}}+{\frac {2x^{13}}{13}}=2\left(x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}+{\frac {x^{11}}{11}}+{\frac {x^{13}}{13}}\right)=}
= 2*(-0.11733662705136264741194733209931 -0.11733662705136264741194733209931^3/3 -0.11733662705136264741194733209931^5/5
-0.11733662705136264741194733209931^7/7 -0.11733662705136264741194733209931^9/9 -0.11733662705136264741194733209931^11/11
-0.11733662705136264741194733209931^13/13) =
(nukopijavus viską nuo "2" iki ženklo "=" ir įdėjus (padarius "Paste") į Windows 10 kalkuliatorių gaunamas žemiau esantis atsakymas)
= -0.23575922393968105542703851052716.
Pridėjus gautą atsakymą prie 7.5*ln(2) reikšmės, gauname ln(a)=ln(143) artinį:
ln
143
=
(
8
−
1
2
)
ln
2
+
ln
1
+
x
1
−
x
=
7.5
ln
2
+
ln
1
+
x
1
−
x
=
{\displaystyle \ln 143=\left(8-{\frac {1}{2}}\right)\ln 2+\ln {\frac {1+x}{1-x}}=7.5\ln 2+\ln {\frac {1+x}{1-x}}=}
= 5.1986038541995898206292409109363 + (-0.23575922393968105542703851052716) = 4.9628446302599087652022024004091.
Atėmę gautą ln(143) reikšmę iš tikslios ln(143) reikšmės, gauname
4.9628446302599072801154310195304 - 4.9628446302599087652022024004091 = -0.0000000000000014850867713808787 =
≈
−
1.48508677
⋅
10
−
15
.
{\displaystyle \approx -1.48508677\cdot 10^{-15}.}
1
+
x
1
−
x
=
{\displaystyle {\frac {1+x}{1-x}}=}
= (1-0.11733662705136264741194733209931)/(1+0.11733662705136264741194733209931) = 0.78997085710684606241656831078901.
Tiksli ln(0.78997085710684606241656831078901) reikšmė yra tokia:
ln(0.78997085710684606241656831078901) = -0.23575922393968254051380989140588.
ln
1
+
x
1
−
x
{\displaystyle \ln {\frac {1+x}{1-x}}\,}
-0.23575922393968254051380989140588 - (-0.23575922393968105542703851052716) =
= -0.23575922393968254051380989140588 + 0.23575922393968105542703851052716 = -0.00000000000000148508677138087872 =
≈
−
1.48508677
⋅
10
−
15
.
{\displaystyle \approx -1.48508677\cdot 10^{-15}.}
Pagal (8.89) formulę
R
2
n
+
2
(
x
)
=
−
x
2
n
+
2
2
n
+
2
[
1
(
1
+
θ
x
)
2
n
+
2
−
1
(
1
−
θ
x
)
2
n
+
2
]
.
(
8.89
)
{\displaystyle R_{2n+2}(x)=-{\frac {x^{2n+2}}{2n+2}}\left[{\frac {1}{(1+\theta x)^{2n+2}}}-{\frac {1}{(1-\theta x)^{2n+2}}}\right].\quad (8.89)}
|
R
2
n
+
2
(
x
)
|
<
(
−
0.117336627
)
2
n
+
2
2
n
+
2
[
1
(
1
+
(
−
0.117336627
)
)
2
n
+
2
−
1
]
=
(
−
0.117336627
)
2
⋅
6
+
2
2
⋅
6
+
2
[
1
(
1
−
0.117336627
)
2
⋅
6
+
2
−
1
]
=
{\displaystyle |R_{2n+2}(x)|<{\frac {(-0.117336627)^{2n+2}}{2n+2}}\left[{\frac {1}{(1+(-0.117336627))^{2n+2}}}-1\right]={\frac {(-0.117336627)^{2\cdot 6+2}}{2\cdot 6+2}}\left[{\frac {1}{(1-0.117336627)^{2\cdot 6+2}}}-1\right]=}
=
(
−
0.117336627
)
14
14
[
1
(
1
−
0.117336627
)
14
−
1
]
=
{\displaystyle ={\frac {(-0.117336627)^{14}}{14}}\left[{\frac {1}{(1-0.117336627)^{14}}}-1\right]=}
= (-0.11733662705136264741194733209931)^14 * (1/(1-0.11733662705136264741194733209931)^14 -1)/14 =
= 3.1744477658835444278622203928479e-14
≈
3.1744477658835444
⋅
10
−
14
.
{\displaystyle \approx 3.1744477658835444\cdot 10^{-14}.}
Matome, kad paklaida įvertinta teisingai, nes
1.48508677
⋅
10
−
15
<
3.1744477658835444
⋅
10
−
14
.
{\displaystyle 1.48508677\cdot 10^{-15}<3.1744477658835444\cdot 10^{-14}.}
Pastebėsime, kad
(
−
0.117336627
)
14
14
=
{\displaystyle {\frac {(-0.117336627)^{14}}{14}}=}
≈
6.6979571136742
⋅
10
−
15
{\displaystyle \approx 6.6979571136742\cdot 10^{-15}}
(
−
0.117336627
)
15
15
=
{\displaystyle {\frac {(-0.117336627)^{15}}{15}}=}
Todėl (8.88) formulėje liekamąjį narį tikriausiai galima įvertinti taip:
|
R
2
n
+
2
(
x
)
|
<
|
x
2
n
+
2
2
n
+
2
|
≈
6.6979571136742
⋅
10
−
15
.
{\displaystyle |R_{2n+2}(x)|<|{\frac {x^{2n+2}}{2n+2}}|\approx 6.6979571136742\cdot 10^{-15}.}
Arba garantuotai visais atvejais R(x) turėtų būti teisingas, kai
|
R
(
x
)
|
<
|
x
2
n
+
1
2
n
+
1
|
.
{\displaystyle |R(x)|<|{\frac {x^{2n+1}}{2n+1}}|.}
Nes
(
−
0.117336627
)
13
13
=
{\displaystyle {\frac {(-0.117336627)^{13}}{13}}=}
[ keisti ] Skaičiuojant
arctg
α
,
{\displaystyle \operatorname {arctg} \alpha ,}
|
α
|
=
x
,
{\displaystyle |\alpha |=x,}
arctg
α
=
sgn
α
⋅
arctg
x
.
{\displaystyle \operatorname {arctg} \alpha =\operatorname {sgn} \alpha \cdot \operatorname {arctg} x.}
Aprašysime standartines transformacijas, kuriomis
arctg
x
{\displaystyle \operatorname {arctg} x}
x reikšmės ne mažesnės kaip
1
8
,
{\displaystyle {\frac {1}{8}},}
1
8
.
{\displaystyle {\frac {1}{8}}.}
Iš pradžių tarkime, kad
x
≥
1.
{\displaystyle x\geq 1.}
y
=
arctg
x
,
{\displaystyle y=\operatorname {arctg} x,\;}
x
=
tg
y
,
{\displaystyle x=\operatorname {tg} y,\;}
x
1
=
tg
(
y
−
arctg
1
)
.
{\displaystyle x_{1}=\operatorname {tg} (y-\operatorname {arctg} 1).}
[pasinaudodami formule (iš https://lt.wikibooks.org/wiki/Matematika/Trigonometrinės_formulės )
tan
(
α
−
β
)
=
tan
α
−
tan
β
1
+
tan
α
tan
β
.
(
7
)
{\displaystyle \tan(\alpha -\beta )={\frac {\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }}.\quad (7)}
x
1
=
tg
(
y
−
arctg
1
)
=
tg
y
−
tg
(
arctg
1
)
1
+
tg
y
⋅
tg
(
arctg
1
)
=
tg
y
−
1
1
+
tg
y
⋅
1
=
tg
y
−
1
tg
y
+
1
=
x
−
1
x
+
1
<
1.
{\displaystyle x_{1}=\operatorname {tg} (y-\operatorname {arctg} 1)={\frac {\operatorname {tg} y-\operatorname {tg} (\operatorname {arctg} 1)}{1+\operatorname {tg} y\cdot \operatorname {tg} (\operatorname {arctg} 1)}}={\frac {\operatorname {tg} y-1}{1+\operatorname {tg} y\cdot 1}}={\frac {\operatorname {tg} y-1}{\operatorname {tg} y+1}}={\frac {x-1}{x+1}}<1.}
Kadangi
arctg
x
1
=
arctg
[
tg
(
y
−
arctg
1
)
]
=
y
−
arctg
1
=
arctg
x
−
arctg
1
,
{\displaystyle \operatorname {arctg} x_{1}=\operatorname {arctg} [\operatorname {tg} (y-\operatorname {arctg} 1)]=y-\operatorname {arctg} 1=\operatorname {arctg} x-\operatorname {arctg} 1,}
arctg
x
=
arctg
1
+
arctg
x
1
=
π
4
+
arctg
x
1
,
{\displaystyle \operatorname {arctg} x=\operatorname {arctg} 1+\operatorname {arctg} x_{1}={\frac {\pi }{4}}+\operatorname {arctg} x_{1},}
tai, norint apskaičiuoti
arctg
x
,
{\displaystyle \operatorname {arctg} x,}
x
≥
1
,
{\displaystyle x\geq 1,}
arctg
x
1
,
{\displaystyle \operatorname {arctg} x_{1},}
0
<
x
1
<
1.
{\displaystyle 0<x_{1}<1.}
Dabar aptarsime atvejį, kai argumentas tenkina nelygybes
1
8
≤
x
<
1.
{\displaystyle {\frac {1}{8}}\leq x<1.}
Sakykime,
k
1
=
1
,
k
2
=
1
2
,
k
3
=
1
4
,
k
4
=
1
8
.
{\displaystyle k_{1}=1,\;\;k_{2}={\frac {1}{2}},\;\;k_{3}={\frac {1}{4}},\;\;k_{4}={\frac {1}{8}}.}
i
=
1
,
2
,
3
,
4
{\displaystyle i=1,\;2,\;3,\;4\;}
k
i
≤
x
<
2
k
i
.
(
8.94
)
{\displaystyle k_{i}\leq x<2k_{i}.\quad (8.94)}
[Kai x daugiau už 1, tai x gali būti daugiau už
2
k
i
.
{\displaystyle 2k_{i}.}
x >2, tai
x
>
2
k
1
=
2.
{\displaystyle x>2k_{1}=2.}
Tarkime, kad
y
=
arctg
x
,
{\displaystyle y=\operatorname {arctg} x,\;}
x
=
tg
y
,
{\displaystyle x=\operatorname {tg} y,\;}
x
i
=
tg
(
y
−
arctg
k
i
)
.
{\displaystyle x_{i}=\operatorname {tg} (y-\operatorname {arctg} k_{i}).}
[pasinaudodami formule
tan
(
α
−
β
)
=
tan
α
−
tan
β
1
+
tan
α
tan
β
{\displaystyle \tan(\alpha -\beta )={\frac {\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }}}
x
i
=
tg
(
y
−
arctg
k
i
)
=
tg
y
−
tg
(
arctg
k
i
)
1
+
tg
y
⋅
tg
(
arctg
k
i
)
=
tg
y
−
k
i
1
+
tg
y
⋅
k
i
=
x
−
k
i
1
+
k
i
x
.
{\displaystyle x_{i}=\operatorname {tg} (y-\operatorname {arctg} k_{i})={\frac {\operatorname {tg} y-\operatorname {tg} (\operatorname {arctg} k_{i})}{1+\operatorname {tg} y\cdot \operatorname {tg} (\operatorname {arctg} k_{i})}}={\frac {\operatorname {tg} y-k_{i}}{1+\operatorname {tg} y\cdot k_{i}}}={\frac {x-k_{i}}{1+k_{i}x}}.}
Kadangi
x
>
0
,
{\displaystyle x>0,}
1
+
k
i
x
>
1.
{\displaystyle 1+k_{i}x>1.}
x
−
k
i
<
2
k
i
−
k
1
=
k
i
.
{\displaystyle x-k_{i}<2k_{i}-k_{1}=k_{i}.}
x
i
{\displaystyle x_{i}}
x
i
<
k
i
.
{\displaystyle x_{i}<k_{i}.}
arctg
x
i
=
arctg
[
tg
(
y
−
arctg
k
i
)
]
=
y
−
arctg
k
i
=
arctg
x
−
arctg
k
i
,
{\displaystyle \operatorname {arctg} x_{i}=\operatorname {arctg} [\operatorname {tg} (y-\operatorname {arctg} k_{i})]=y-\operatorname {arctg} k_{i}=\operatorname {arctg} x-\operatorname {arctg} k_{i},}
arctg
x
=
arctg
k
i
+
arctg
x
i
,
{\displaystyle \operatorname {arctg} x=\operatorname {arctg} k_{i}+\operatorname {arctg} x_{i},}
tai
arctg
x
{\displaystyle \operatorname {arctg} x}
x tenkina (8.94) nelygybes, pakeičiamas
arctg
x
i
{\displaystyle \operatorname {arctg} x_{i}}
0
<
x
i
<
k
i
.
{\displaystyle 0<x_{i}<k_{i}.}
Jei aprašytąsias argumento x transformacijas pakartosime daugių daugiausia keturis kartus, tai
arctg
x
{\displaystyle \operatorname {arctg} x}
x reikšmės priklauso pusintervaliui
1
8
≤
x
<
1
,
{\displaystyle {\frac {1}{8}}\leq x<1,}
1
8
.
{\displaystyle {\frac {1}{8}}.}
[Ten klaidelė. Maksimum keturių transformacijų reikia, kai x reikšmės priklauso pusintervaliui
1
8
≤
x
,
{\displaystyle {\frac {1}{8}}\leq x,}
x reikšmės priklauso pusintervaliui
1
8
≤
x
<
1
,
{\displaystyle {\frac {1}{8}}\leq x<1,}
Skaičiuojant
arctg
x
,
{\displaystyle \operatorname {arctg} x,}
x
<
1
8
,
{\displaystyle x<{\frac {1}{8}},}
https://lt.wikibooks.org/wiki/Matematika/Matematinės_eilutės )
arctg
x
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
.
.
.
+
(
−
1
)
n
x
2
n
+
1
2
n
+
1
+
R
2
n
+
2
(
x
)
.
{\displaystyle \operatorname {arctg} x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+...+(-1)^{n}{\frac {x^{2n+1}}{2n+1}}+R_{2n+2}(x).}
Dažniausiai šioje formulėje imama
n
=
6
{\displaystyle n=6}
x
2
n
+
1
2
n
+
1
{\displaystyle {\frac {x^{2n+1}}{2n+1}}}
Apskaičiuosime
arctg
5.
{\displaystyle \operatorname {arctg} 5.}
x
=
5.
{\displaystyle x=5.}
x
i
=
x
−
k
i
1
+
k
i
x
.
{\displaystyle x_{i}={\frac {x-k_{i}}{1+k_{i}x}}.}
Pradėsime nuo to, kad
k
1
=
1.
{\displaystyle k_{1}=1.}
x
1
=
x
−
k
1
1
+
k
1
x
=
x
−
1
1
+
x
=
5
−
1
1
+
5
=
4
6
=
2
3
.
{\displaystyle x_{1}={\frac {x-k_{1}}{1+k_{1}x}}={\frac {x-1}{1+x}}={\frac {5-1}{1+5}}={\frac {4}{6}}={\frac {2}{3}}.}
Toliau
arctg
x
=
arctg
k
i
+
arctg
x
i
=
arctg
k
1
+
arctg
x
1
=
arctg
1
+
arctg
2
3
=
π
4
+
arctg
2
3
.
{\displaystyle \operatorname {arctg} x=\operatorname {arctg} k_{i}+\operatorname {arctg} x_{i}=\operatorname {arctg} k_{1}+\operatorname {arctg} x_{1}=\operatorname {arctg} 1+\operatorname {arctg} {\frac {2}{3}}={\frac {\pi }{4}}+\operatorname {arctg} {\frac {2}{3}}.}
Toliau imsime
k
2
=
1
2
.
{\displaystyle k_{2}={\frac {1}{2}}.}
x
2
=
x
1
−
k
2
1
+
k
2
x
1
=
x
1
−
1
2
1
+
1
2
x
1
=
2
3
−
1
2
1
+
1
2
⋅
2
3
=
4
−
3
6
1
+
1
3
=
1
6
3
+
1
3
=
1
6
4
3
=
1
6
⋅
3
4
=
1
8
.
{\displaystyle x_{2}={\frac {x_{1}-k_{2}}{1+k_{2}x_{1}}}={\frac {x_{1}-{\frac {1}{2}}}{1+{\frac {1}{2}}x_{1}}}={\frac {{\frac {2}{3}}-{\frac {1}{2}}}{1+{\frac {1}{2}}\cdot {\frac {2}{3}}}}={\frac {\frac {4-3}{6}}{1+{\frac {1}{3}}}}={\frac {\frac {1}{6}}{\frac {3+1}{3}}}={\frac {\frac {1}{6}}{\frac {4}{3}}}={\frac {1}{6}}\cdot {\frac {3}{4}}={\frac {1}{8}}.}
Toliau
arctg
2
3
=
arctg
x
1
=
arctg
k
2
+
arctg
x
2
=
arctg
1
2
+
arctg
1
8
.
{\displaystyle \operatorname {arctg} {\frac {2}{3}}=\operatorname {arctg} x_{1}=\operatorname {arctg} k_{2}+\operatorname {arctg} x_{2}=\operatorname {arctg} {\frac {1}{2}}+\operatorname {arctg} {\frac {1}{8}}.}
Toliau imsime
k
4
=
1
8
.
{\displaystyle k_{4}={\frac {1}{8}}.}
x
4
=
x
2
−
k
4
1
+
k
4
x
2
=
x
2
−
1
8
1
+
1
8
x
2
=
1
8
−
1
8
1
+
1
8
⋅
1
8
=
0
1
+
1
64
=
0
64
+
1
64
=
0
65
64
=
0.
{\displaystyle x_{4}={\frac {x_{2}-k_{4}}{1+k_{4}x_{2}}}={\frac {x_{2}-{\frac {1}{8}}}{1+{\frac {1}{8}}x_{2}}}={\frac {{\frac {1}{8}}-{\frac {1}{8}}}{1+{\frac {1}{8}}\cdot {\frac {1}{8}}}}={\frac {0}{1+{\frac {1}{64}}}}={\frac {0}{\frac {64+1}{64}}}={\frac {0}{\frac {65}{64}}}=0.}
Ir gauname
arctg
1
8
=
arctg
x
2
=
arctg
k
4
+
arctg
x
4
=
arctg
1
8
+
arctg
0
=
arctg
1
8
+
0
=
arctg
1
8
.
{\displaystyle \operatorname {arctg} {\frac {1}{8}}=\operatorname {arctg} x_{2}=\operatorname {arctg} k_{4}+\operatorname {arctg} x_{4}=\operatorname {arctg} {\frac {1}{8}}+\operatorname {arctg} 0=\operatorname {arctg} {\frac {1}{8}}+0=\operatorname {arctg} {\frac {1}{8}}.}
Tokiu budu
arctg
5
=
arctg
x
=
π
4
+
arctg
2
3
=
π
4
+
arctg
1
2
+
arctg
1
8
.
{\displaystyle \operatorname {arctg} 5=\operatorname {arctg} x={\frac {\pi }{4}}+\operatorname {arctg} {\frac {2}{3}}={\frac {\pi }{4}}+\operatorname {arctg} {\frac {1}{2}}+\operatorname {arctg} {\frac {1}{8}}.}
π
4
=
{\displaystyle {\frac {\pi }{4}}=}
arctg
1
2
=
{\displaystyle \operatorname {arctg} {\frac {1}{2}}=}
arctg
1
8
=
{\displaystyle \operatorname {arctg} {\frac {1}{8}}=}
Šiame pavyzdyje skaičiuoti Makloreno eilutės nereikia su x reikšme mažesne už 1/8, nes
x
4
=
0.
{\displaystyle x_{4}=0.}
Taigi,
arctg(5) = 0.78539816339744830961566084581988 + 0.46364760900080611621425623146121 + 0.12435499454676143503135484916387 =
= 1.373400766945015860861271926445.
Tiksli arctg(5) reikšmė iš kalkuliatoriaus yra tokia:
arctg(5) = 1.373400766945015860861271926445.
Gavome tą patį atsakymą.
Apskaičiuosime
arctg
7.
{\displaystyle \operatorname {arctg} 7.}
x
=
7.
{\displaystyle x=7.}
Pradėsime nuo to, kad
k
1
=
1.
{\displaystyle k_{1}=1.}
x
1
=
x
−
k
1
1
+
k
1
x
=
x
−
1
1
+
x
=
7
−
1
1
+
7
=
6
8
=
3
4
.
{\displaystyle x_{1}={\frac {x-k_{1}}{1+k_{1}x}}={\frac {x-1}{1+x}}={\frac {7-1}{1+7}}={\frac {6}{8}}={\frac {3}{4}}.}
Toliau
arctg
x
=
arctg
k
1
+
arctg
x
1
=
arctg
1
+
arctg
3
4
=
π
4
+
arctg
3
4
.
{\displaystyle \operatorname {arctg} x=\operatorname {arctg} k_{1}+\operatorname {arctg} x_{1}=\operatorname {arctg} 1+\operatorname {arctg} {\frac {3}{4}}={\frac {\pi }{4}}+\operatorname {arctg} {\frac {3}{4}}.}
Toliau imsime
k
2
=
1
2
.
{\displaystyle k_{2}={\frac {1}{2}}.}
x
2
=
x
1
−
k
2
1
+
k
2
x
1
=
x
1
−
1
2
1
+
1
2
x
1
=
3
4
−
1
2
1
+
1
2
⋅
3
4
=
3
−
2
4
1
+
3
8
=
1
4
8
+
3
8
=
1
4
11
8
=
1
4
⋅
8
11
=
2
11
=
0.
(
18
)
.
{\displaystyle x_{2}={\frac {x_{1}-k_{2}}{1+k_{2}x_{1}}}={\frac {x_{1}-{\frac {1}{2}}}{1+{\frac {1}{2}}x_{1}}}={\frac {{\frac {3}{4}}-{\frac {1}{2}}}{1+{\frac {1}{2}}\cdot {\frac {3}{4}}}}={\frac {\frac {3-2}{4}}{1+{\frac {3}{8}}}}={\frac {\frac {1}{4}}{\frac {8+3}{8}}}={\frac {\frac {1}{4}}{\frac {11}{8}}}={\frac {1}{4}}\cdot {\frac {8}{11}}={\frac {2}{11}}=0.(18).}
Tuo tarpu
k
3
=
1
4
=
0.25
{\displaystyle k_{3}={\frac {1}{4}}=0.25}
k
3
=
1
4
{\displaystyle k_{3}={\frac {1}{4}}}
k
4
=
1
8
=
0.125
{\displaystyle k_{4}={\frac {1}{8}}=0.125}
x
4
=
x
2
−
k
4
1
+
k
4
x
2
=
x
2
−
1
8
1
+
1
8
x
2
=
2
11
−
1
8
1
+
1
8
⋅
2
11
=
16
−
11
88
1
+
2
88
=
5
88
88
+
2
88
=
5
88
90
88
=
5
88
⋅
88
90
=
5
90
=
1
18
=
0.0
(
5
)
.
{\displaystyle x_{4}={\frac {x_{2}-k_{4}}{1+k_{4}x_{2}}}={\frac {x_{2}-{\frac {1}{8}}}{1+{\frac {1}{8}}x_{2}}}={\frac {{\frac {2}{11}}-{\frac {1}{8}}}{1+{\frac {1}{8}}\cdot {\frac {2}{11}}}}={\frac {\frac {16-11}{88}}{1+{\frac {2}{88}}}}={\frac {\frac {5}{88}}{\frac {88+2}{88}}}={\frac {\frac {5}{88}}{\frac {90}{88}}}={\frac {5}{88}}\cdot {\frac {88}{90}}={\frac {5}{90}}={\frac {1}{18}}=0.0(5).}
[1/18 = 0.05555555555555555555555555555556.]
Toliau galime skaičiuoti pagal Makloreno formulę,
arctg
x
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
.
.
.
+
(
−
1
)
n
x
2
n
+
1
2
n
+
1
+
R
2
n
+
2
(
x
)
,
{\displaystyle \operatorname {arctg} x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+...+(-1)^{n}{\frac {x^{2n+1}}{2n+1}}+R_{2n+2}(x),}
imdami n=6 ir vietoje x įrašę
x
4
.
{\displaystyle x_{4}.}
arctg
1
18
=
arctg
x
4
=
x
4
−
x
4
3
3
+
x
4
5
5
−
x
4
7
7
+
x
4
9
9
−
x
4
11
11
+
x
4
13
13
=
{\displaystyle \operatorname {arctg} {\frac {1}{18}}=\operatorname {arctg} x_{4}=x_{4}-{\frac {x_{4}^{3}}{3}}+{\frac {x_{4}^{5}}{5}}-{\frac {x_{4}^{7}}{7}}+{\frac {x_{4}^{9}}{9}}-{\frac {x_{4}^{11}}{11}}+{\frac {x_{4}^{13}}{13}}=}
= 1/18 - 1/18^3/3 + 1/18^5/5 - 1/18^7/7 + 1/18^9/9 - 1/18^11/11 + 1/18^13/13 = 0.05549850524571683556705277298107.
(Kad gauti tokį atsakymą greitai, tereikia aukščiau esančią eilutę nukopijuoti ir įdėti į Windows 10 kalkuliaorių, padarius "Paste").
Atėmę gautą arctg(1/18) reikšmę iš tikslios arctg(1/18) reikšmės, gauname paklaidą:
0.05549850524571683555719814809224 - 0.05549850524571683556705277298107 = -0.00000000000000000000985462488883 =
=
−
9.85462488883
⋅
10
−
21
.
{\displaystyle =-9.85462488883\cdot 10^{-21}.}
Kaip ir su natūrinio logaritmo skaičiavimu, liekamąjį narį galima įvertinti taip:
|
R
2
n
+
2
(
x
4
)
|
<
|
x
4
2
n
+
2
2
n
+
2
|
=
{\displaystyle |R_{2n+2}(x_{4})|<|{\frac {x_{4}^{2n+2}}{2n+2}}|=}
≈
1.90571035
⋅
10
−
19
.
{\displaystyle \approx 1.90571035\cdot 10^{-19}.}
Tada
arctg
3
4
=
arctg
x
1
=
arctg
k
2
+
arctg
x
2
=
arctg
1
2
+
arctg
2
11
;
{\displaystyle \operatorname {arctg} {\frac {3}{4}}=\operatorname {arctg} x_{1}=\operatorname {arctg} k_{2}+\operatorname {arctg} x_{2}=\operatorname {arctg} {\frac {1}{2}}+\operatorname {arctg} {\frac {2}{11}};}
arctg
2
11
=
arctg
x
2
=
arctg
k
4
+
arctg
x
4
=
arctg
1
8
+
arctg
1
18
.
{\displaystyle \operatorname {arctg} {\frac {2}{11}}=\operatorname {arctg} x_{2}=\operatorname {arctg} k_{4}+\operatorname {arctg} x_{4}=\operatorname {arctg} {\frac {1}{8}}+\operatorname {arctg} {\frac {1}{18}}.}
Todėl
arctg
7
=
arctg
x
=
π
4
+
arctg
3
4
=
π
4
+
arctg
1
2
+
arctg
2
11
=
{\displaystyle \operatorname {arctg} 7=\operatorname {arctg} x={\frac {\pi }{4}}+\operatorname {arctg} {\frac {3}{4}}={\frac {\pi }{4}}+\operatorname {arctg} {\frac {1}{2}}+\operatorname {arctg} {\frac {2}{11}}=}
=
π
4
+
arctg
1
2
+
arctg
1
8
+
arctg
1
18
=
{\displaystyle ={\frac {\pi }{4}}+\operatorname {arctg} {\frac {1}{2}}+\operatorname {arctg} {\frac {1}{8}}+\operatorname {arctg} {\frac {1}{18}}=}
= 0.78539816339744830961566084581988 + 0.46364760900080611621425623146121 +0.12435499454676143503135484916387 + 0.05549850524571683556705277298107 = 1.428899272190732696428324699426.
Atėmę ką tik gautą arctg(7) reikšmę iš kalkuliatoriaus tikslios arctg(7) reikšmės, gauname paklaidą:
1.4288992721907326964184700745372 - 1.428899272190732696428324699426 = -9.8546248888316409091970590727231e-21 =
= -0.0000000000000000000098546248888
≈
−
9.8546248888
⋅
10
−
21
.
{\displaystyle \approx -9.8546248888\cdot 10^{-21}.}
Pastabos. Čia visos arkatangento reikšmės skaičiuojamos radianais. Kad kalkuliatorius skaičiuotų radianais, reikia paspausti mygtuką, kuris keičia tarp laipsnių, radianų ir gradianų (iš "DEG", "RAD" ir "GRAD" reikia išrinkti, kad rodytų "RAD"). O tam, kad kalkuliatorius apskaičiuotų arktangento reikšmę, reikia iš pradžių įrašyti skaičių į kalkuliatorių ir paskui Windows 10 kalkuliatoriuje (kuris turi būti nustatytas [vienu paspaudimu] per patį kalkuliatorių, kad būtų "Scientific") paspausti "Trigonometry", vėliau paspausti
2
n
d
{\displaystyle 2^{nd}}
tan
−
1
{\displaystyle \tan ^{-1}}
![{\displaystyle g'''(x)=\left(\ln {\frac {1+x}{1-x}}\right)'''=\left({\frac {4x}{(1-x^{2})^{2}}}\right)'={\frac {(4x)'(1-x^{2})^{2}-4x[(1-x^{2})^{2}]'}{((1-x^{2})^{2})^{2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02d19a21a71fb25e559f62bd3993648e2986d54a)
![{\displaystyle ={\frac {4(1-x^{2})^{2}-4x[2(1-x^{2})']}{(1-x^{2})^{4}}}={\frac {4(1-2x^{2}+x^{4})-4x[2(-2x)]}{(1-x^{2})^{4}}}={\frac {4-8x^{2}+4x^{4}+16x^{2}}{(1-x^{2})^{4}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c691c02903d76c97e206b3d94d19cd22c2bdcd11)
![{\displaystyle ={\frac {[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}]'(1-x^{2})^{8}-[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}]((1-x^{2})^{8})'}{(1-x^{2})^{16}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c841307ef6a81df3c09662facac1d392af14b64b)
^{8}-[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}]8(1-x^{2})^{7}\cdot (-2x)}{(1-x^{2})^{16}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d276b3b64f8124191ff07d20984b185f5cc97863)
^{8}+16x[16x(1+x^{2})(1-x^{2})^{4}+32x(1+x^{2})^{2}(1-x^{2})^{3}](1-x^{2})^{7}}{(1-x^{2})^{16}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93f79802ae996b6c8995793f138b103a52dc2daa)
![{\displaystyle g_{1}(x)=[(1+x^{2})(1-x^{2})^{4}]'=2x(1-x^{2})^{4}+(1+x^{2})\cdot 4(1-x^{2})^{3}\cdot (-2x)=2x(1-x^{2})^{4}-8x(1+x^{2})(1-x^{2})^{3},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb50ae57c1e6d738bb2164378b30c00da0f4f0a5)
![{\displaystyle g_{2}(x)=[(1+x^{2})^{2}(1-x^{2})^{3}]'=2(1+x^{2})\cdot 2x\cdot (1-x^{2})^{3}+(1+x^{2})^{2}\cdot 3(1-x^{2})^{2}\cdot (-2x)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc8671bbb4a6d796948bf79a798c28ee02b074b1)
^{8}+16\cdot 0[16\cdot 0(1+0^{2})(1-0^{2})^{4}+32\cdot 0(1+0^{2})^{2}(1-0^{2})^{3}](1-0^{2})^{7}}{(1-0^{2})^{16}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/427e85749573c263cb73ab65ac818894d133ade1)
![{\displaystyle ={\frac {[16+32]+0}{1}}=48.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d568e06e69922ad512b1db13b89ec50787a4a3ce)
![{\displaystyle R_{2n+2}(x)=-{\frac {x^{2n+2}}{2n+2}}\left[{\frac {1}{(1+\theta x)^{2n+2}}}-{\frac {1}{(1-\theta x)^{2n+2}}}\right];\quad (8.89)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58514dc649fcc79d6eaca3ebb248ad095cff411e)
![{\displaystyle |R_{2n+2}(x)|\leq {\frac {(0.172)^{2n+2}}{2n+2}}\left[1+{\frac {1}{(1-0.172)^{2n+2}}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/441e5d61a43c7a6a41b90641e1a5303a3ee49ee6)
![{\displaystyle \left[{\frac {1}{2}};\;1\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfdbab664324673b6d5efb783a5688ca243b13f3)
![{\displaystyle R_{2n+2}(x)=-{\frac {x^{2n+2}}{2n+2}}\left[{\frac {1}{(1+\theta x)^{2n+2}}}-{\frac {1}{(1-\theta x)^{2n+2}}}\right].\quad (8.89)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3aa001f37775d6379e9dc17648b86b255953b981)
![{\displaystyle |R_{2n+2}(x)|<{\frac {(-0.117336627)^{2n+2}}{2n+2}}\left[{\frac {1}{(1+(-0.117336627))^{2n+2}}}-1\right]={\frac {(-0.117336627)^{2\cdot 6+2}}{2\cdot 6+2}}\left[{\frac {1}{(1-0.117336627)^{2\cdot 6+2}}}-1\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8412a5a0a8c3fca5892f8f8838f16d4ea7526f8e)
![{\displaystyle ={\frac {(-0.117336627)^{14}}{14}}\left[{\frac {1}{(1-0.117336627)^{14}}}-1\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ed725416a8e5526fe2cffe9430650804ea64a33)
![{\displaystyle \operatorname {arctg} x_{1}=\operatorname {arctg} [\operatorname {tg} (y-\operatorname {arctg} 1)]=y-\operatorname {arctg} 1=\operatorname {arctg} x-\operatorname {arctg} 1,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10936fdf3829fad08f9ca3bec4ccbab91a6ab660)
![{\displaystyle \operatorname {arctg} x_{i}=\operatorname {arctg} [\operatorname {tg} (y-\operatorname {arctg} k_{i})]=y-\operatorname {arctg} k_{i}=\operatorname {arctg} x-\operatorname {arctg} k_{i},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0323cd30beb8643c90d93db3442da1b853086b39)
