Elementariųjų funkcijų reikšmių skaičiavimas
Šių funkcijų reikšmių skaičiavimas pagrįstas grandininėmis (arba tolydžiosiomis ) trupmenomis. Reikalingos žinios apie tas trupmenas pateiktos toliau.
Išvardytų funkcijų reikšmių skaičiavimas yra susijęs su konkrečia grandinine trupmena, gaunama išskleidus funkcją
th
x
.
{\displaystyle \operatorname {th} x.}
th
x
{\displaystyle \operatorname {th} x}
Toliau
ch
x
,
sh
x
,
th
x
{\displaystyle \operatorname {ch} x,\;\;\operatorname {sh} x,\;\;\operatorname {th} x\;}
https://en.wikipedia.org/wiki/Hyperbolic_functions
[ keisti ] Baigtine grandinine trupmena
P
n
Q
n
{\displaystyle {\frac {P_{n}}{Q_{n}}}}
[ https://en.wikipedia.org/wiki/Generalized_continued_fraction ]
P
n
Q
n
=
b
0
+
a
1
b
1
+
a
2
b
2
+
a
3
b
3
+
(
8.95
)
{\displaystyle {\frac {P_{n}}{Q_{n}}}=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+\quad \quad }}}}}}\quad \quad (8.95)}
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
⋱
{\displaystyle \quad \quad \ddots }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
+
a
n
b
n
{\displaystyle \quad \quad \quad +{\frac {a_{n}}{b_{n}}}}
Skaičiai
a
1
,
a
2
,
.
.
.
,
a
n
{\displaystyle a_{1},\;a_{2},\;...,\;a_{n}}
b
1
,
b
2
,
.
.
.
,
b
n
{\displaystyle b_{1},\;b_{2},\;...,\;b_{n}}
Grandininės trupmenos
P
0
Q
0
=
b
0
1
,
{\displaystyle {\frac {P_{0}}{Q_{0}}}={\frac {b_{0}}{1}},\quad }
P
1
Q
1
=
b
0
+
a
1
b
1
,
{\displaystyle {\frac {P_{1}}{Q_{1}}}=b_{0}+{\frac {a_{1}}{b_{1}}},\quad }
P
2
Q
2
=
b
0
+
a
1
b
1
+
a
2
b
2
,
.
.
.
(
8.96
)
{\displaystyle {\frac {P_{2}}{Q_{2}}}=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}\quad }}}},\quad ...\quad (8.96)}
vadinamos grandininės trupmenos
P
n
Q
n
{\displaystyle {\frac {P_{n}}{Q_{n}}}}
Tarę, kad
P
−
1
=
1
,
Q
−
1
=
0
,
{\displaystyle P_{-1}=1,\;\;Q_{-1}=0,}
P
k
Q
k
{\displaystyle {\frac {P_{k}}{Q_{k}}}\;}
k
=
1
,
2
,
.
.
.
,
n
{\displaystyle k=1,\;2,\;...,\;n}
P
k
{\displaystyle P_{k}}
P
k
−
1
{\displaystyle P_{k-1}}
P
k
−
2
{\displaystyle P_{k-2}}
Q
k
{\displaystyle Q_{k}}
Q
k
−
1
{\displaystyle Q_{k-1}}
Q
k
−
2
:
{\displaystyle Q_{k-2}:}
{
P
k
=
b
k
P
k
−
1
+
a
k
P
k
−
2
,
Q
k
=
b
k
Q
k
−
1
+
a
k
Q
k
−
2
.
(
8.97
)
{\displaystyle {\begin{cases}P_{k}=b_{k}P_{k-1}+a_{k}P_{k-2},&\\Q_{k}=b_{k}Q_{k-1}+a_{k}Q_{k-2}.&\end{cases}}\quad \quad (8.97)}
[Parodysime, kad jos teisingos, kai k =2.
P
−
1
=
1
,
Q
−
1
=
0
,
{\displaystyle P_{-1}=1,\;\;Q_{-1}=0,}
P
k
−
2
=
P
0
=
b
0
,
Q
k
−
2
=
Q
0
=
1
,
{\displaystyle P_{k-2}=P_{0}=b_{0},\;\;Q_{k-2}=Q_{0}=1,}
P
1
=
b
1
P
0
+
a
1
P
−
1
=
b
1
b
0
+
a
1
⋅
1
=
b
1
b
0
+
a
1
,
{\displaystyle P_{1}=b_{1}P_{0}+a_{1}P_{-1}=b_{1}b_{0}+a_{1}\cdot 1=b_{1}b_{0}+a_{1},}
Q
1
=
b
1
Q
0
+
a
1
Q
−
1
=
b
1
⋅
1
+
a
1
⋅
0
=
b
1
,
{\displaystyle Q_{1}=b_{1}Q_{0}+a_{1}Q_{-1}=b_{1}\cdot 1+a_{1}\cdot 0=b_{1},}
P
2
=
b
2
P
1
+
a
2
P
0
=
b
2
(
b
1
b
0
+
a
1
)
+
a
2
b
0
=
b
2
b
1
b
0
+
b
2
a
1
+
a
2
b
0
,
{\displaystyle P_{2}=b_{2}P_{1}+a_{2}P_{0}=b_{2}(b_{1}b_{0}+a_{1})+a_{2}b_{0}=b_{2}b_{1}b_{0}+b_{2}a_{1}+a_{2}b_{0},}
Q
2
=
b
2
Q
1
+
a
2
Q
0
=
b
2
b
1
+
a
2
⋅
1
=
b
2
b
1
+
a
2
.
{\displaystyle Q_{2}=b_{2}Q_{1}+a_{2}Q_{0}=b_{2}b_{1}+a_{2}\cdot 1=b_{2}b_{1}+a_{2}.}
Gauname
P
2
Q
2
=
b
2
b
1
b
0
+
b
2
a
1
+
a
2
b
0
b
2
b
1
+
a
2
.
{\displaystyle {\frac {P_{2}}{Q_{2}}}={\frac {b_{2}b_{1}b_{0}+b_{2}a_{1}+a_{2}b_{0}}{b_{2}b_{1}+a_{2}}}.}
O iš (8.96) formulės gauname
P
2
Q
2
=
b
0
+
a
1
b
1
+
a
2
b
2
=
b
0
+
a
1
b
1
b
2
+
a
2
b
2
=
b
0
+
a
1
b
2
b
1
b
2
+
a
2
=
b
0
(
b
1
b
2
+
a
2
)
+
a
1
b
2
b
1
b
2
+
a
2
=
b
0
b
1
b
2
+
b
0
a
2
+
a
1
b
2
b
1
b
2
+
a
2
.
{\displaystyle {\frac {P_{2}}{Q_{2}}}=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}\quad }}}}=b_{0}+{\cfrac {a_{1}}{\cfrac {b_{1}b_{2}+a_{2}}{b_{2}\quad }}}=b_{0}+{\cfrac {a_{1}b_{2}}{b_{1}b_{2}+a_{2}}}={\frac {b_{0}(b_{1}b_{2}+a_{2})+a_{1}b_{2}}{b_{1}b_{2}+a_{2}}}={\frac {b_{0}b_{1}b_{2}+b_{0}a_{2}+a_{1}b_{2}}{b_{1}b_{2}+a_{2}}}.}
Matome, kad trupmenos
P
2
Q
2
{\displaystyle {\frac {P_{2}}{Q_{2}}}}
Mums bus reikalinga speciali formulė, išreiškianti trupmeną
P
n
Q
n
,
{\displaystyle {\frac {P_{n}}{Q_{n}}},}
P
k
Q
k
{\displaystyle {\frac {P_{k}}{Q_{k}}}}
P
k
−
1
Q
k
−
1
.
{\displaystyle {\frac {P_{k-1}}{Q_{k-1}}}.}
P
k
Q
k
−
P
k
−
1
Q
k
−
1
=
P
k
Q
k
−
1
−
Q
k
P
k
−
1
Q
k
−
1
Q
k
.
(
8.98
)
{\displaystyle {\frac {P_{k}}{Q_{k}}}-{\frac {P_{k-1}}{Q_{k-1}}}={\frac {P_{k}Q_{k-1}-Q_{k}P_{k-1}}{Q_{k-1}Q_{k}}}.\quad (8.98)}
Paskutinės trupmenos skaitiklį, atsižvelgdami į (8.97) lygybes, galime pertvarkyti šitaip
P
k
Q
k
−
1
−
Q
k
P
k
−
1
=
(
b
k
P
k
−
1
+
a
k
P
k
−
2
)
Q
k
−
1
−
(
b
k
Q
k
−
1
+
a
k
Q
k
−
2
)
P
k
−
1
=
−
a
k
[
P
k
−
1
Q
k
−
2
−
Q
k
−
1
P
k
−
2
]
.
(
8.99
)
{\displaystyle P_{k}Q_{k-1}-Q_{k}P_{k-1}=(b_{k}P_{k-1}+a_{k}P_{k-2})Q_{k-1}-(b_{k}Q_{k-1}+a_{k}Q_{k-2})P_{k-1}=-a_{k}[P_{k-1}Q_{k-2}-Q_{k-1}P_{k-2}].\quad (8.99)}
Paeiliui remdamiesi (8.99) sąryšiu reikšmėms
k
,
k
−
1
,
k
−
2
,
.
.
.
,
1
{\displaystyle k,\;k-1,\;k-2,\;...,\;1\;}
P
−
1
=
1
,
Q
−
1
=
0
,
Q
0
=
1
,
{\displaystyle P_{-1}=1,\;\;Q_{-1}=0,\;\;Q_{0}=1,\;}
P
k
Q
k
−
P
k
−
1
Q
k
−
1
=
−
a
k
[
P
k
−
1
Q
k
−
2
−
Q
k
−
1
P
k
−
2
]
Q
k
−
1
Q
k
=
−
a
k
(
−
a
k
−
1
)
[
P
k
−
2
Q
k
−
3
−
Q
k
−
2
P
k
−
3
]
Q
k
−
1
Q
k
=
{\displaystyle {\frac {P_{k}}{Q_{k}}}-{\frac {P_{k-1}}{Q_{k-1}}}={\frac {-a_{k}[P_{k-1}Q_{k-2}-Q_{k-1}P_{k-2}]}{Q_{k-1}Q_{k}}}={\frac {-a_{k}(-a_{k-1})[P_{k-2}Q_{k-3}-Q_{k-2}P_{k-3}]}{Q_{k-1}Q_{k}}}=}
=
−
a
k
(
−
a
k
−
1
)
(
−
a
k
−
2
)
[
P
k
−
3
Q
k
−
4
−
Q
k
−
3
P
k
−
4
]
Q
k
−
1
Q
k
=
(
−
1
)
k
+
1
a
k
a
k
−
1
a
k
−
2
.
.
.
a
1
1
Q
k
−
1
Q
k
.
(
8.100
)
{\displaystyle ={\frac {-a_{k}(-a_{k-1})(-a_{k-2})[P_{k-3}Q_{k-4}-Q_{k-3}P_{k-4}]}{Q_{k-1}Q_{k}}}=(-1)^{k+1}a_{k}a_{k-1}a_{k-2}\;...\;a_{1}{\frac {1}{Q_{k-1}Q_{k}}}.\quad (8.100)}
[
P
k
Q
k
−
1
−
Q
k
P
k
−
1
=
−
a
k
[
P
k
−
1
Q
k
−
2
−
Q
k
−
1
P
k
−
2
]
.
{\displaystyle P_{k}Q_{k-1}-Q_{k}P_{k-1}=-a_{k}[P_{k-1}Q_{k-2}-Q_{k-1}P_{k-2}].}
Kai k=1, gauname:
P
1
Q
0
−
Q
1
P
0
=
−
a
1
[
P
0
Q
−
1
−
Q
0
P
−
1
]
=
−
a
1
[
b
0
⋅
0
−
1
⋅
1
]
=
a
1
.
{\displaystyle P_{1}Q_{0}-Q_{1}P_{0}=-a_{1}[P_{0}Q_{-1}-Q_{0}P_{-1}]=-a_{1}[b_{0}\cdot 0-1\cdot 1]=a_{1}.}
Kadangi
P
n
Q
n
=
P
0
Q
0
+
(
P
1
Q
1
−
P
0
Q
0
)
+
(
P
2
Q
2
−
P
1
Q
1
)
+
.
.
.
+
(
P
n
Q
n
−
P
n
−
1
Q
n
−
1
)
,
{\displaystyle {\frac {P_{n}}{Q_{n}}}={\frac {P_{0}}{Q_{0}}}+\left({\frac {P_{1}}{Q_{1}}}-{\frac {P_{0}}{Q_{0}}}\right)+\left({\frac {P_{2}}{Q_{2}}}-{\frac {P_{1}}{Q_{1}}}\right)+...+\left({\frac {P_{n}}{Q_{n}}}-{\frac {P_{n-1}}{Q_{n-1}}}\right),}
tai, pasinaudoję (8.100) lygybe, gausime reikalingą konkrečią formulę trupmenai
P
n
Q
n
{\displaystyle {\frac {P_{n}}{Q_{n}}}}
P
n
Q
n
=
b
0
+
a
1
Q
0
Q
1
−
a
1
a
2
Q
1
Q
2
+
a
1
a
2
a
3
Q
2
Q
3
−
.
.
.
+
(
−
1
)
n
+
1
a
1
a
2
.
.
.
a
n
Q
n
−
1
Q
n
.
(
8.101
)
{\displaystyle {\frac {P_{n}}{Q_{n}}}=b_{0}+{\frac {a_{1}}{Q_{0}Q_{1}}}-{\frac {a_{1}a_{2}}{Q_{1}Q_{2}}}+{\frac {a_{1}a_{2}a_{3}}{Q_{2}Q_{3}}}-...+(-1)^{n+1}{\frac {a_{1}a_{2}\;...\;a_{n}}{Q_{n-1}Q_{n}}}.\quad (8.101)}
[ keisti ] Šiame skirsnyje aprašomą funkcijos
th
x
{\displaystyle \operatorname {th} x}
tg
x
.
{\displaystyle \operatorname {tg} x.}
Imkime funkciją
y
=
ch
x
,
{\displaystyle y=\operatorname {ch} {\sqrt {x}},}
x
>
0.
{\displaystyle x>0.}
2
x
y
′
=
sh
x
,
2
x
y
″
+
y
′
x
−
y
2
x
=
0.
{\displaystyle 2{\sqrt {x}}y'=\operatorname {sh} {\sqrt {x}},\quad 2{\sqrt {x}}y''+{\frac {y'}{\sqrt {x}}}-{\frac {y}{2{\sqrt {x}}}}=0.}
[
y
′
=
(
ch
x
)
′
=
sh
x
⋅
1
2
x
.
{\displaystyle y'=(\operatorname {ch} {\sqrt {x}})'=\operatorname {sh} {\sqrt {x}}\cdot {\frac {1}{2{\sqrt {x}}}}.}
y
″
=
(
sh
x
⋅
1
2
x
)
′
=
1
2
x
ch
x
⋅
1
2
x
+
sh
x
(
1
2
x
)
′
=
1
4
x
ch
x
+
sh
x
⋅
−
1
⋅
1
2
x
2
(
x
)
2
=
ch
x
4
x
−
1
4
x
3
sh
x
.
{\displaystyle y''=(\operatorname {sh} {\sqrt {x}}\cdot {\frac {1}{2{\sqrt {x}}}})'={\frac {1}{2{\sqrt {x}}}}\operatorname {ch} {\sqrt {x}}\cdot {\frac {1}{2{\sqrt {x}}}}+\operatorname {sh} {\sqrt {x}}({\frac {1}{2{\sqrt {x}}}})'={\frac {1}{4x}}\operatorname {ch} {\sqrt {x}}+\operatorname {sh} {\sqrt {x}}\cdot {\frac {-1\cdot {\frac {1}{2{\sqrt {x}}}}}{2({\sqrt {x}})^{2}}}={\frac {\operatorname {ch} {\sqrt {x}}}{4x}}-{\frac {1}{4{\sqrt {x^{3}}}}}\operatorname {sh} {\sqrt {x}}.}
Bet antra y išvestinė nieko nepasako. Tiesiog reikia žinot, kad (sh(x))'=ch(x). ]
Iš lygybės
2
x
y
″
+
y
′
x
−
y
2
x
=
0
{\displaystyle 2{\sqrt {x}}y''+{\frac {y'}{\sqrt {x}}}-{\frac {y}{2{\sqrt {x}}}}=0}
išplaukia tapatybė, kuri teisinga, kai
x
>
0
{\displaystyle x>0}
2
x
{\displaystyle 2{\sqrt {x}}}
4
x
y
″
+
2
y
′
−
y
=
0.
(
8.102
)
{\displaystyle 4xy''+2y'-y=0.\quad (8.102)}
Šią tapatybę diferencijuodami toliau, gausime
{
4
y
″
+
4
x
y
‴
+
2
y
″
−
y
′
=
4
x
y
‴
+
(
4
+
2
)
y
″
−
y
′
=
0
,
4
y
‴
+
4
x
y
(
4
)
+
6
y
‴
−
y
″
=
4
x
y
(
4
)
+
(
4
⋅
2
+
2
)
y
‴
−
y
″
=
0
,
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4
x
y
(
n
+
2
)
+
(
4
n
+
2
)
y
(
n
+
1
)
−
y
(
n
)
=
0.
(
8.103
)
{\displaystyle {\begin{cases}4y''+4xy'''+2y''-y'=4xy'''+(4+2)y''-y'=0,&\\4y'''+4xy^{(4)}+6y'''-y''=4xy^{(4)}+(4\cdot 2+2)y'''-y''=0,&\\........................................&\\4xy^{(n+2)}+(4n+2)y^{(n+1)}-y^{(n)}=0.&\end{cases}}\quad \quad (8.103)}
Santykį
y
(
n
+
1
)
y
(
n
)
{\displaystyle {\frac {y^{(n+1)}}{y^{(n)}}}}
u
n
+
1
.
{\displaystyle u_{n+1}.}
y
(
n
+
1
)
{\displaystyle y^{(n+1)}}
4
x
u
n
+
2
+
4
n
+
2
=
1
u
n
+
1
,
{\displaystyle 4xu_{n+2}+4n+2={\frac {1}{u_{n+1}}},}
iš kurios
u
n
+
1
=
1
4
x
u
n
+
2
+
4
n
+
2
=
1
2
2
n
+
1
+
2
x
u
n
+
2
.
(
8.104
)
{\displaystyle u_{n+1}={\frac {1}{4xu_{n+2}+4n+2}}={\frac {\frac {1}{2}}{2n+1+2xu_{n+2}}}.\quad (8.104)}
Kadangi
u
1
=
y
′
y
=
sh
x
⋅
1
2
x
ch
x
=
th
x
2
x
,
{\displaystyle u_{1}={\frac {y'}{y}}={\frac {\operatorname {sh} {\sqrt {x}}\cdot {\frac {1}{2{\sqrt {x}}}}}{\operatorname {ch} {\sqrt {x}}}}={\frac {\operatorname {th} {\sqrt {x}}}{2{\sqrt {x}}}},}
tai (8.104) sąryšį, kai
n
=
0
,
{\displaystyle n=0,}
th
x
=
u
1
⋅
2
x
=
1
2
⋅
2
x
2
⋅
0
+
1
+
2
x
u
0
+
2
=
x
1
+
2
x
u
2
.
{\displaystyle \operatorname {th} {\sqrt {x}}=u_{1}\cdot 2{\sqrt {x}}={\frac {{\frac {1}{2}}\cdot 2{\sqrt {x}}}{2\cdot 0+1+2xu_{0+2}}}={\frac {\sqrt {x}}{1+2xu_{2}}}.}
Dešinėje paskutinės formulės pusėje vietoj
u
2
{\displaystyle u_{2}}
n
=
1.
{\displaystyle n=1.}
th
x
=
x
1
+
2
x
⋅
1
2
2
⋅
1
+
1
+
2
x
u
1
+
2
=
x
1
+
x
3
+
2
x
u
3
.
{\displaystyle \operatorname {th} {\sqrt {x}}={\cfrac {\sqrt {x}}{1+2x\cdot {\cfrac {\frac {1}{2}}{2\cdot 1+1+2xu_{1+2}}}}}={\cfrac {\sqrt {x}}{1+{\cfrac {x}{3+2xu_{3}}}}}.}
Šiame reiškinyje vietoje
u
3
{\displaystyle u_{3}}
n
=
2.
{\displaystyle n=2.}
th
x
{\displaystyle \operatorname {th} {\sqrt {x}}}
x
{\displaystyle {\sqrt {x}}}
x
,
{\displaystyle x,}
th
x
{\displaystyle \operatorname {th} x}
th
x
=
x
1
+
x
2
3
+
x
2
5
+
(
8.105
)
{\displaystyle \operatorname {th} x={\cfrac {x}{1+{\cfrac {x^{2}}{3+{\cfrac {x^{2}}{5+\quad \quad }}}}}}\quad \quad (8.105)}
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
⋱
{\displaystyle \quad \quad \ddots }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
+
x
2
2
n
+
1
+
2
x
2
u
n
+
2
{\displaystyle \quad \quad \quad +{\frac {x^{2}}{2n+1+2x^{2}u_{n+2}}}}
__________________
* Schlömilch O . Ueber den Kettenbruch für tg x . Zs. Math. u. Phys. 2(1857), 137–165.[ keisti ] Skaičiuojant funkcijos
th
x
{\displaystyle \operatorname {th} x}
2
x
2
u
n
+
2
.
{\displaystyle 2x^{2}u_{n+2}.}
n paprastai laikomas lygiu 6 (n =6), o x reikšmių moduliai apribojami skaičiumi
π
4
≈
0.785398.
{\displaystyle {\frac {\pi }{4}}\approx 0.785398.}
Įvertinsime paklaidą su bet kokiu n .
Funkcijos
th
x
{\displaystyle \operatorname {th} x}
2
x
2
u
n
+
2
,
{\displaystyle 2x^{2}u_{n+2},}
th
¯
x
.
{\displaystyle {\overline {\operatorname {th} }}x.}
th
x
{\displaystyle \operatorname {th} x}
th
¯
x
{\displaystyle {\overline {\operatorname {th} }}x}
P
n
+
1
Q
n
+
1
{\displaystyle {\frac {P_{n+1}}{Q_{n+1}}}}
P
¯
n
+
1
Q
¯
n
+
1
.
{\displaystyle {\frac {{\overline {P}}_{n+1}}{{\overline {Q}}_{n+1}}}.}
Surašykime tų trupmenų dalinius skaitiklius
a
i
{\displaystyle a_{i}}
a
¯
i
{\displaystyle {\overline {a}}_{i}}
b
i
{\displaystyle b_{i}}
b
¯
i
{\displaystyle {\overline {b}}_{i}}
P
¯
n
+
1
Q
¯
n
+
1
{\displaystyle {\frac {{\overline {P}}_{n+1}}{{\overline {Q}}_{n+1}}}}
{
a
1
=
a
¯
1
=
x
,
a
2
=
a
¯
2
=
x
2
,
a
3
=
a
¯
3
=
x
2
,
.
.
.
,
a
n
+
1
=
a
¯
n
+
1
=
x
2
,
b
0
=
b
¯
0
=
0
,
b
1
=
b
¯
1
=
1
,
b
2
=
b
¯
2
=
3
,
b
3
=
b
¯
3
=
5
,
.
.
.
,
b
n
=
b
¯
n
=
2
n
−
1
,
b
n
+
1
=
2
n
+
1
+
2
x
2
u
n
+
2
,
b
¯
n
+
1
=
2
n
+
1.
(
8.106
)
{\displaystyle {\begin{cases}a_{1}={\overline {a}}_{1}=x,\;\;a_{2}={\overline {a}}_{2}=x^{2},\;\;a_{3}={\overline {a}}_{3}=x^{2},\;\;...,\;\;a_{n+1}={\overline {a}}_{n+1}=x^{2},&\\b_{0}={\overline {b}}_{0}=0,\;\;b_{1}={\overline {b}}_{1}=1,\;\;b_{2}={\overline {b}}_{2}=3,\;\;b_{3}={\overline {b}}_{3}=5,\;\;...,\;\;b_{n}={\overline {b}}_{n}=2n-1,&\\b_{n+1}=2n+1+2x^{2}u_{n+2},\;\;{\overline {b}}_{n+1}=2n+1.&\end{cases}}\quad \quad (8.106)}
Kadangi trupmenose
P
n
+
1
Q
n
+
1
{\displaystyle {\frac {P_{n+1}}{Q_{n+1}}}}
P
¯
n
+
1
Q
¯
n
+
1
{\displaystyle {\frac {{\overline {P}}_{n+1}}{{\overline {Q}}_{n+1}}}}
Q
−
1
=
Q
¯
−
1
=
0
,
Q
0
=
Q
¯
0
=
1
,
{\displaystyle Q_{-1}={\overline {Q}}_{-1}=0,\;\;Q_{0}={\overline {Q}}_{0}=1,}
{
Q
1
=
Q
¯
1
,
Q
2
=
Q
¯
2
,
Q
3
=
Q
¯
3
,
.
.
.
,
Q
n
=
Q
¯
n
,
Q
n
+
1
=
(
2
n
+
1
+
2
x
2
u
n
+
2
)
Q
¯
n
+
x
2
Q
¯
n
−
1
,
Q
¯
n
+
1
=
(
2
n
+
1
)
Q
¯
n
+
x
2
Q
¯
n
−
1
.
(
8.107
)
{\displaystyle {\begin{cases}Q_{1}={\overline {Q}}_{1},\;\;Q_{2}={\overline {Q}}_{2},\;\;Q_{3}={\overline {Q}}_{3},\;\;...,\;\;Q_{n}={\overline {Q}}_{n},&\\Q_{n+1}=(2n+1+2x^{2}u_{n+2}){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1},&\\{\overline {Q}}_{n+1}=(2n+1){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1}.&\end{cases}}\quad \quad (8.107)}
Dabar abi trupmenas
P
n
+
1
Q
n
+
1
{\displaystyle {\frac {P_{n+1}}{Q_{n+1}}}}
P
¯
n
+
1
Q
¯
n
+
1
{\displaystyle {\frac {{\overline {P}}_{n+1}}{{\overline {Q}}_{n+1}}}}
P
n
+
1
Q
n
+
1
−
P
¯
n
+
1
Q
¯
n
+
1
{\displaystyle {\frac {P_{n+1}}{Q_{n+1}}}-{\frac {{\overline {P}}_{n+1}}{{\overline {Q}}_{n+1}}}\;}
th
x
−
th
¯
x
,
{\displaystyle \operatorname {th} x-{\overline {\operatorname {th} }}x,}
th
x
−
th
¯
x
=
(
−
1
)
n
+
2
x
2
n
+
1
[
1
Q
n
Q
n
+
1
−
1
Q
¯
n
Q
¯
n
+
1
]
.
{\displaystyle \operatorname {th} x-{\overline {\operatorname {th} }}x=(-1)^{n+2}x^{2n+1}\left[{\frac {1}{Q_{n}Q_{n+1}}}-{\frac {1}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\right].}
Šią formulę, remiantis (8.107) lygybėmis, galima perrašyti šitaip:
th
x
−
th
¯
x
=
(
−
1
)
n
+
2
x
2
n
+
1
[
1
Q
¯
n
Q
n
+
1
−
1
Q
¯
n
Q
¯
n
+
1
]
=
(
−
1
)
n
+
2
x
2
n
+
1
[
Q
¯
n
Q
¯
n
+
1
−
Q
¯
n
Q
n
+
1
Q
¯
n
Q
n
+
1
Q
¯
n
Q
¯
n
+
1
]
=
(
−
1
)
n
+
2
x
2
n
+
1
[
Q
¯
n
+
1
−
Q
n
+
1
Q
¯
n
Q
n
+
1
Q
¯
n
+
1
]
=
{\displaystyle \operatorname {th} x-{\overline {\operatorname {th} }}x=(-1)^{n+2}x^{2n+1}\left[{\frac {1}{{\overline {Q}}_{n}Q_{n+1}}}-{\frac {1}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\right]=(-1)^{n+2}x^{2n+1}\left[{\frac {{\overline {Q}}_{n}{\overline {Q}}_{n+1}-{\overline {Q}}_{n}Q_{n+1}}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\right]=(-1)^{n+2}x^{2n+1}\left[{\frac {{\overline {Q}}_{n+1}-Q_{n+1}}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n+1}}}\right]=}
=
(
−
1
)
n
+
1
x
2
n
+
1
[
Q
n
+
1
−
Q
¯
n
+
1
Q
¯
n
Q
n
+
1
Q
¯
n
+
1
]
=
(
−
1
)
n
+
1
x
2
n
+
1
[
[
(
2
n
+
1
+
2
x
2
u
n
+
2
)
Q
¯
n
+
x
2
Q
¯
n
−
1
]
−
[
(
2
n
+
1
)
Q
¯
n
+
x
2
Q
¯
n
−
1
]
Q
¯
n
Q
n
+
1
Q
¯
n
+
1
]
=
{\displaystyle =(-1)^{n+1}x^{2n+1}\left[{\frac {Q_{n+1}-{\overline {Q}}_{n+1}}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n+1}}}\right]=(-1)^{n+1}x^{2n+1}\left[{\frac {[(2n+1+2x^{2}u_{n+2}){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1}]-[(2n+1){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1}]}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n+1}}}\right]=}
=
(
−
1
)
n
+
1
x
2
n
+
1
Q
¯
n
Q
¯
n
+
1
[
2
x
2
u
n
+
2
Q
¯
n
Q
n
+
1
]
=
{\displaystyle =(-1)^{n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{Q_{n+1}}}\right]=}
=
(
−
1
)
n
+
1
x
2
n
+
1
Q
¯
n
Q
¯
n
+
1
[
2
x
2
u
n
+
2
Q
¯
n
2
x
2
u
n
+
2
Q
¯
n
+
(
2
n
+
1
)
Q
¯
n
+
x
2
Q
¯
n
−
1
]
.
(
8.108
)
{\displaystyle =(-1)^{n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1}}}\right].\quad (8.108)}
Kad gautume reikalingą įvertį, pasinaudosime dviem nelygybėmis, kurias įrodysime vėliau.
Kai
x
≥
0
,
{\displaystyle x\geq 0,}
k
≥
1
{\displaystyle k\geq 1}
Q
k
≥
(
2
k
−
1
)
!
!
.
(
8.109
)
{\displaystyle Q_{k}\geq (2k-1)!!.\quad (8.109)}
Kai
x
>
0
,
{\displaystyle x>0,}
u
n
+
2
{\displaystyle u_{n+2}}
:
u
n
+
2
>
0.
(
8.110
)
{\displaystyle u_{n+2}>0.\quad (8.110)}
Dabar įvertinsime skirtumą
th
x
−
th
¯
x
,
{\displaystyle \operatorname {th} x-{\overline {\operatorname {th} }}x,}
x >0. Kadangi
u
n
+
2
{\displaystyle u_{n+2}}
Q
¯
k
{\displaystyle {\overline {Q}}_{k}}
x >0, tai reiškinys, parašytas (8.108) lygybės dešinės pusės laužtiniuose skliaustuose, nėra didesnis už vienetą. Be to, iš (8.109) gauname nelygybę
Q
¯
n
Q
¯
n
+
1
≥
[
(
2
n
−
1
)
!
!
]
2
(
2
n
+
1
)
.
{\displaystyle {\overline {Q}}_{n}{\overline {Q}}_{n+1}\geq [(2n-1)!!]^{2}(2n+1).}
Todėl, kai x >0, su bet kokiu numeriu n bus teisingas toks paklaidos įvertis:
|
th
x
−
th
¯
x
|
≤
x
2
n
+
1
[
(
2
n
−
1
)
!
!
]
2
(
2
n
+
1
)
.
(
8.111
)
{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}.\quad (8.111)}
Apskaičiuosime paklaidos įvertį, kai n =6, o x reikšmės tenkina nelygybes
0
<
x
<
π
4
.
{\displaystyle 0<x<{\frac {\pi }{4}}.}
n =6, skaičius
2
n
−
1
{\displaystyle 2n-1}
2
n
+
1
{\displaystyle 2n+1}
π
4
<
0.8
,
{\displaystyle {\frac {\pi }{4}}<0.8,}
x
13
<
(
0.8
)
13
<
5.6
⋅
10
−
2
.
{\displaystyle x^{13}<(0.8)^{13}<5.6\cdot 10^{-2}.}
11
!
!
=
10395.
{\displaystyle 11!!=10395.}
th
x
{\displaystyle \operatorname {th} x}
n =6, ne didesnė už
4
⋅
10
−
11
.
{\displaystyle 4\cdot 10^{-11}.}
[
0.8
13
=
0.0549755813888
<
5.5
⋅
10
−
2
.
{\displaystyle 0.8^{13}=0.0549755813888<5.5\cdot 10^{-2}.}
11!! = 11*9*7*5*3 = 10395.
x
2
n
+
1
[
(
2
n
−
1
)
!
!
]
2
(
2
n
+
1
)
=
0.0549755813888
10395
2
⋅
13
=
0.0549755813888
108056025
⋅
13
=
{\displaystyle {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {0.0549755813888}{10395^{2}\cdot 13}}={\frac {0.0549755813888}{108056025\cdot 13}}=}
= 0.0549755813888/(10395^2 * 13) = 3.9136095151210110325069440028555e-11
≈
3.9136
⋅
10
−
11
.
{\displaystyle \approx 3.9136\cdot 10^{-11}.}
Dabar įrodysime, kad (8.109) ir (8.110) nelygybės yra teisingos.
(8.109) nelygybės įrodymas.
Iš pradžių įrodysime, kad visi
Q
¯
k
{\displaystyle {\overline {Q}}_{k}}
a
¯
k
{\displaystyle {\overline {a}}_{k}}
b
¯
k
{\displaystyle {\overline {b}}_{k}}
x
≥
0
{\displaystyle x\geq 0}
k
≤
n
.
{\displaystyle k\leq n.}
Q
¯
−
1
=
0
,
Q
¯
0
=
1.
{\displaystyle {\overline {Q}}_{-1}=0,\;\;{\overline {Q}}_{0}=1.}
Q
¯
k
{\displaystyle {\overline {Q}}_{k}}
k
≤
n
.
{\displaystyle k\leq n.}
Iš antrosios (8.97) formulės ir iš to, kad
a
k
{\displaystyle a_{k}}
Q
¯
k
{\displaystyle {\overline {Q}}_{k}}
Q
¯
k
≥
b
k
Q
¯
k
−
1
.
(
8.112
)
{\displaystyle {\overline {Q}}_{k}\geq b_{k}{\overline {Q}}_{k-1}.\quad (8.112)}
Kadangi
Q
¯
0
=
1
,
{\displaystyle {\overline {Q}}_{0}=1,}
b
k
=
2
k
−
1
,
{\displaystyle b_{k}=2k-1,}
1
≤
k
≤
n
,
{\displaystyle 1\leq k\leq n,}
Q
¯
1
≥
1
,
Q
¯
2
≥
3
⋅
1
=
3
,
Q
¯
3
≥
5
⋅
3
=
15
,
Q
¯
4
≥
(
2
⋅
4
−
1
)
Q
¯
3
=
7
⋅
5
⋅
3
=
105
,
.
.
.
,
Q
¯
k
≥
(
2
k
−
1
)
!
!
.
{\displaystyle {\overline {Q}}_{1}\geq 1,\;\;{\overline {Q}}_{2}\geq 3\cdot 1=3,\;\;{\overline {Q}}_{3}\geq 5\cdot 3=15,\;\;{\overline {Q}}_{4}\geq (2\cdot 4-1){\overline {Q}}_{3}=7\cdot 5\cdot 3=105,\;\;...,\;\;{\overline {Q}}_{k}\geq (2k-1)!!.}
(8.110) nelygybės įrodymas.
Užtenka įrodyti, kad funkcijos
y
=
ch
x
{\displaystyle y=\operatorname {ch} {\sqrt {x}}}
x >0, yra teigiamos. Savaime aišku, kartu įrodysime ir (8.110) nelygybę, nes
u
n
+
2
=
y
(
n
+
2
)
y
(
n
+
1
)
.
{\displaystyle u_{n+2}={\frac {y^{(n+2)}}{y^{(n+1)}}}.}
Padauginę paskutinę (8.103) lygybę iš
1
4
x
n
−
1
2
,
{\displaystyle {\frac {1}{4}}x^{n-{\frac {1}{2}}},}
[
x
n
+
1
2
y
(
n
+
1
)
]
′
=
1
4
x
n
−
1
2
y
(
n
)
.
(
8.113
)
{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'={\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)}.\quad (8.113)}
[
4
x
y
(
n
+
2
)
+
(
4
n
+
2
)
y
(
n
+
1
)
−
y
(
n
)
=
0
|
⋅
1
4
x
n
−
1
2
,
{\displaystyle \;4xy^{(n+2)}+(4n+2)y^{(n+1)}-y^{(n)}=0\;\;|\cdot {\frac {1}{4}}x^{n-{\frac {1}{2}}},}
x
n
+
1
2
y
(
n
+
2
)
+
(
n
+
1
2
)
x
n
−
1
2
y
(
n
+
1
)
−
1
4
x
n
−
1
2
y
(
n
)
=
0.
{\displaystyle x^{n+{\frac {1}{2}}}y^{(n+2)}+(n+{\frac {1}{2}})x^{n-{\frac {1}{2}}}y^{(n+1)}-{\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)}=0.}
[
x
n
+
1
2
y
(
n
+
1
)
]
′
=
(
n
+
1
2
)
x
n
−
1
2
y
(
n
+
1
)
+
x
n
+
1
2
y
(
n
+
2
)
.
{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'=(n+{\frac {1}{2}})x^{n-{\frac {1}{2}}}y^{(n+1)}+x^{n+{\frac {1}{2}}}y^{(n+2)}.}
Pirmiausia įsitikinsime, kad
lim
x
→
0
+
0
[
x
n
+
1
2
y
(
n
+
1
)
(
x
)
]
=
0.
(
8.114
)
{\displaystyle \lim _{x\to 0+0}\left[x^{n+{\frac {1}{2}}}y^{(n+1)}(x)\right]=0.\quad (8.114)}
Tam reikalui užtenka įrodyti, kad funkcija
x
n
y
(
n
+
1
)
(
x
)
(
8.115
)
{\displaystyle x^{n}y^{(n+1)}(x)\quad (8.115)}
lieka aprėžta, kai
x
→
0
+
0
{\displaystyle x\to 0+0}
x artėja prie nulio iš dešinės). Iš formulių
y
=
ch
x
{\displaystyle y=\operatorname {ch} {\sqrt {x}}}
y
′
=
sh
x
2
x
{\displaystyle y'={\frac {\operatorname {sh} {\sqrt {x}}}{2{\sqrt {x}}}}}
y
(
x
)
{\displaystyle y(x)}
y
′
(
x
)
{\displaystyle y'(x)}
x
→
0
+
0.
{\displaystyle x\to 0+0.}
[Kad funkcija
y
′
(
x
)
{\displaystyle y'(x)}
y
′
(
x
)
{\displaystyle y'(x)}
Lopitalio taisyklę :
lim
x
→
0
+
0
sh
x
2
x
=
lim
x
→
0
+
0
(
sh
x
)
′
(
2
x
)
′
=
lim
x
→
0
+
0
1
2
x
ch
x
2
⋅
1
2
x
=
lim
x
→
0
+
0
1
2
ch
x
=
lim
x
→
0
+
0
1
2
⋅
e
x
+
e
−
x
2
=
1
4
(
e
0
+
e
−
0
)
=
1
4
(
1
+
1
e
0
)
=
1
2
.
{\displaystyle \lim _{x\to 0+0}{\frac {\operatorname {sh} {\sqrt {x}}}{2{\sqrt {x}}}}=\lim _{x\to 0+0}{\frac {(\operatorname {sh} {\sqrt {x}})'}{(2{\sqrt {x}})'}}=\lim _{x\to 0+0}{\frac {{\frac {1}{2{\sqrt {x}}}}\operatorname {ch} {\sqrt {x}}}{2\cdot {\frac {1}{2{\sqrt {x}}}}}}=\lim _{x\to 0+0}{\frac {1}{2}}\operatorname {ch} {\sqrt {x}}=\lim _{x\to 0+0}{\frac {1}{2}}\cdot {\frac {e^{\sqrt {x}}+e^{-{\sqrt {x}}}}{2}}={\frac {1}{4}}(e^{0}+e^{-0})={\frac {1}{4}}(1+{\frac {1}{e^{0}}})={\frac {1}{2}}.}
Tačiau tuomet iš (8.102) lygybės aišku, kad ir funkcija
x
y
″
(
x
)
{\displaystyle xy''(x)}
x
→
0
+
0.
{\displaystyle x\to 0+0.}
[
4
x
y
″
+
2
y
′
−
y
=
0.
(
8.102
)
{\displaystyle 4xy''+2y'-y=0.\quad (8.102)}
4
x
y
″
=
−
2
y
′
+
y
.
(
A
)
{\displaystyle 4xy''=-2y'+y.\quad (A)}
Lygybėje (A) dešinė pusė yra aprėžta, kai
x
→
0
+
0.
{\displaystyle x\to 0+0.}
lim
x
→
0
+
0
(
−
2
y
′
+
y
)
=
−
2
⋅
1
2
+
1
=
0.
{\displaystyle \lim _{x\to 0+0}(-2y'+y)=-2\cdot {\frac {1}{2}}+1=0.}
Toliau matematinės indukcijos metodu, remdamiesi paskutiniąją (8.103) lygybe, įrodome, kad funkcija
x
n
y
(
n
+
1
)
(
x
)
{\displaystyle x^{n}y^{(n+1)}(x)}
x
→
0
+
0
,
{\displaystyle x\to 0+0,}
n . Iš to aišku, kad (8.114) lygybė yra teisinga.
Dabar įsitikinsime, kad, imant bet kokį neneigiamą sveikąjį skaičių n , išvestinė
y
(
n
)
(
x
)
(
8.116
)
{\displaystyle y^{(n)}(x)\quad (8.116)}
yra teigiama pustiesėje x >0. Savaime aišku,
y
(
0
)
(
x
)
=
y
(
x
)
=
ch
x
{\displaystyle y^{(0)}(x)=y(x)=\operatorname {ch} {\sqrt {x}}\;}
x >0 yra teigiama. Tarkime, kad
y
(
n
)
(
x
)
{\displaystyle y^{(n)}(x)}
x >0 yra teigiama, kai n – koks nors fiksuotas numeris. Įsitikinsime, kad tuomet išvestinė
y
(
n
+
1
)
(
x
)
{\displaystyle y^{(n+1)}(x)}
x >0 irgi yra teigiama. Iš (8.113) lygybės matyti, kad jos kairėje pusėje parašyta išvestinė yra teigiama, kai x >0, o tai reiškia, kad funkcija
x
n
+
1
2
y
(
n
+
1
)
(
x
)
{\displaystyle x^{n+{1 \over 2}}y^{(n+1)}(x)}
x >0 didėja. Remdamiesi (8.114) lygybe, galime spręsti, kad ta funkcija pustiesėje x >0 yra teigiama. Vadinasi,
y
(
n
+
1
)
(
x
)
>
0
,
{\displaystyle y^{(n+1)}(x)>0,}
x >0, ir (8.110) nelygybė įrodyta. [ keisti ] Toliau simboliu
S
n
(
t
)
{\displaystyle S_{n}(t)}
S
n
(
t
)
=
1
+
t
3
+
t
5
+
t
7
+
(
8.117
)
{\displaystyle S_{n}(t)=1+{\cfrac {t}{3+{\cfrac {t}{5+{\cfrac {t}{7+\quad \quad }}}}}}\quad \quad (8.117)}
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
⋱
{\displaystyle \quad \ddots }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
+
t
2
n
+
1
{\displaystyle \quad \quad +{\frac {t}{2n+1}}}
Elektroninei skaičiavimo mašinai dažniausiai sudaroma tos trupmenos skaičiavimo programa. Ja naudojantis, galima lengvai sudaryti ir hiperbolinio tangento reikšmių skaičiavimo programą, nes, kaip išsiaiškinome praeitame skirsnyje, funkcijos
th
x
{\displaystyle \operatorname {th} x}
th
x
≈
x
S
n
(
x
2
)
.
(
8.118
)
{\displaystyle \operatorname {th} x\approx {\frac {x}{S_{n}(x^{2})}}.\quad (8.118)}
Be to, praeituose skirsniuose paaiškėjo, kad, didinant n , skaičiavimo tikslumas didėja ir paklaida artėja prie nulio.
Funkcijas
sh
2
x
,
ch
2
x
{\displaystyle \operatorname {sh} 2x,\;\;\operatorname {ch} 2x\;}
e
2
x
,
{\displaystyle e^{2x},}
sh
2
x
=
2
th
x
1
−
th
2
x
,
ch
2
x
=
1
+
th
2
x
1
−
th
2
x
,
e
2
x
=
1
+
th
x
1
−
th
x
,
{\displaystyle \operatorname {sh} 2x={\frac {2\operatorname {th} x}{1-\operatorname {th} ^{2}x}},\;\;\;\operatorname {ch} 2x={\frac {1+\operatorname {th} ^{2}x}{1-\operatorname {th} ^{2}x}},\;\;\;e^{2x}={\frac {1+\operatorname {th} x}{1-\operatorname {th} x}},}
galima išreikšti hiperbolinio tangento funkcijomis. Iš tų formulių ir (8.118) lygybės gauname šitokias formules išvardytųjų funkcijų reikšmių artiniams skaičiuoti:
sh
2
x
=
2
th
x
1
−
th
2
x
≈
(
2
x
S
n
(
x
2
)
)
:
(
1
−
x
2
S
n
2
(
x
2
)
)
=
2
x
S
n
(
x
2
)
:
S
n
2
(
x
2
)
−
x
2
S
n
2
(
x
2
)
=
2
x
S
n
2
(
x
2
)
−
x
2
S
n
(
x
2
)
=
2
S
n
(
x
2
)
⋅
x
S
n
2
(
x
2
)
−
x
2
,
{\displaystyle \operatorname {sh} 2x={\frac {2\operatorname {th} x}{1-\operatorname {th} ^{2}x}}\approx \left(2{\frac {x}{S_{n}(x^{2})}}\right):\left(1-{\frac {x^{2}}{S_{n}^{2}(x^{2})}}\right)={\frac {2x}{S_{n}(x^{2})}}:{\frac {S_{n}^{2}(x^{2})-x^{2}}{S_{n}^{2}(x^{2})}}={\frac {2x}{\frac {S_{n}^{2}(x^{2})-x^{2}}{S_{n}(x^{2})}}}={\frac {2S_{n}(x^{2})\cdot x}{S_{n}^{2}(x^{2})-x^{2}}},}
ch
2
x
=
1
+
th
2
x
1
−
th
2
x
≈
(
1
+
x
2
S
n
2
(
x
2
)
)
:
(
1
−
x
2
S
n
2
(
x
2
)
)
=
S
n
2
(
x
2
)
+
x
2
S
n
2
(
x
2
)
:
S
n
2
(
x
2
)
−
x
2
S
n
2
(
x
2
)
=
S
n
2
(
x
2
)
+
x
2
S
n
2
(
x
2
)
⋅
S
n
2
(
x
2
)
S
n
2
(
x
2
)
−
x
2
=
S
n
2
(
x
2
)
+
x
2
S
n
2
(
x
2
)
−
x
2
,
{\displaystyle \operatorname {ch} 2x={\frac {1+\operatorname {th} ^{2}x}{1-\operatorname {th} ^{2}x}}\approx \left(1+{\frac {x^{2}}{S_{n}^{2}(x^{2})}}\right):\left(1-{\frac {x^{2}}{S_{n}^{2}(x^{2})}}\right)={\frac {S_{n}^{2}(x^{2})+x^{2}}{S_{n}^{2}(x^{2})}}:{\frac {S_{n}^{2}(x^{2})-x^{2}}{S_{n}^{2}(x^{2})}}={\frac {S_{n}^{2}(x^{2})+x^{2}}{S_{n}^{2}(x^{2})}}\cdot {\frac {S_{n}^{2}(x^{2})}{S_{n}^{2}(x^{2})-x^{2}}}={\frac {S_{n}^{2}(x^{2})+x^{2}}{S_{n}^{2}(x^{2})-x^{2}}},}
e
2
x
=
1
+
th
x
1
−
th
x
≈
1
+
x
S
n
(
x
2
)
1
−
x
S
n
(
x
2
)
=
S
n
(
x
2
)
+
x
S
n
(
x
2
)
S
n
(
x
2
)
−
x
S
n
(
x
2
)
=
S
n
(
x
2
)
+
x
S
n
(
x
2
)
−
x
.
{\displaystyle e^{2x}={\frac {1+\operatorname {th} x}{1-\operatorname {th} x}}\approx {\frac {1+{\frac {x}{S_{n}(x^{2})}}}{1-{\frac {x}{S_{n}(x^{2})}}}}={\frac {\frac {S_{n}(x^{2})+x}{S_{n}(x^{2})}}{\frac {S_{n}(x^{2})-x}{S_{n}(x^{2})}}}={\frac {S_{n}(x^{2})+x}{S_{n}(x^{2})-x}}.}
Aišku, remiantis šiomis formulėmis ir
S
n
(
t
)
{\displaystyle S_{n}(t)}
sh
2
x
,
ch
2
x
{\displaystyle \operatorname {sh} 2x,\;\;\operatorname {ch} 2x\;}
e
2
x
{\displaystyle e^{2x}}
[ keisti ] Funkcijos
tg
x
{\displaystyle \operatorname {tg} x}
th
x
.
{\displaystyle \operatorname {th} x.}
Imkime funkciją
y
=
cos
x
,
{\displaystyle y=\cos {\sqrt {x}},}
x
>
0.
{\displaystyle x>0.}
2
x
y
′
=
−
sin
x
;
2
x
y
″
+
y
′
x
=
−
y
2
x
.
{\displaystyle 2{\sqrt {x}}y'=-\sin {\sqrt {x}};\quad 2{\sqrt {x}}y''+{\frac {y'}{\sqrt {x}}}=-{\frac {y}{2{\sqrt {x}}}}.}
[
y
′
=
(
cos
x
)
′
=
−
sin
x
⋅
1
2
x
.
{\displaystyle y'=(\cos {\sqrt {x}})'=-\sin {\sqrt {x}}\cdot {\frac {1}{2{\sqrt {x}}}}.}
Iš paskutinio sąryšio gauname tapatybę (padauginę tą paskutinį sąryšį iš
2
x
{\displaystyle 2{\sqrt {x}}}
4
x
y
″
+
2
y
′
+
y
=
0.
{\displaystyle 4xy''+2y'+y=0.}
Diferencijuodami tą tapatybę, gausime
{
4
x
y
‴
+
6
y
″
+
y
′
=
0
,
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4
x
y
(
n
+
2
)
+
(
4
n
+
2
)
y
(
n
+
1
)
+
y
(
n
)
=
0.
{\displaystyle {\begin{cases}4xy'''+6y''+y'=0,&\\.................................&\\4xy^{(n+2)}+(4n+2)y^{(n+1)}+y^{(n)}=0.&\end{cases}}}
Santykį
y
(
n
+
1
)
y
(
n
)
{\displaystyle {\frac {y^{(n+1)}}{y^{(n)}}}}
u
n
+
1
{\displaystyle u_{n+1}}
4
x
u
n
+
2
+
4
n
+
2
=
−
1
u
n
+
1
,
{\displaystyle 4xu_{n+2}+4n+2=-{\frac {1}{u_{n+1}}},}
u
n
+
1
{\displaystyle u_{n+1}}
u
n
+
1
=
−
1
4
x
u
n
+
2
+
4
n
+
2
=
−
1
2
2
n
+
1
+
2
x
u
n
+
2
.
{\displaystyle u_{n+1}=-{\frac {1}{4xu_{n+2}+4n+2}}={\frac {-{\frac {1}{2}}}{2n+1+2xu_{n+2}}}.}
Toliau samprotaujame visiškai panašiai, kaip ir dėstydami hiperbolinį tangentą, ir gauname šitokį funkcijos
tg
x
{\displaystyle \operatorname {tg} x}
tg
x
=
x
1
+
−
x
2
3
+
−
x
2
5
+
{\displaystyle \operatorname {tg} x={\cfrac {x}{1+{\cfrac {-x^{2}}{3+{\cfrac {-x^{2}}{5+\quad \quad }}}}}}}
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
⋱
{\displaystyle \quad \quad \ddots }
{\displaystyle \quad \quad \quad }
{\displaystyle \quad \quad \quad }
+
−
x
2
2
n
+
1
+
2
x
2
u
n
+
2
{\displaystyle \quad \quad \quad +{\frac {-x^{2}}{2n+1+2x^{2}u_{n+2}}}}
Funkcijos
tg
x
{\displaystyle \operatorname {tg} x}
2
x
2
u
n
+
2
.
{\displaystyle 2x^{2}u_{n+2}.}
tg
x
≈
x
S
n
(
−
x
2
)
.
(
8.119
)
{\displaystyle \operatorname {tg} x\approx {\frac {x}{S_{n}(-x^{2})}}.\quad (8.119)}
Kaip ir nagrinėjant hiperbolinį tangentą, galima įsitikinti, kad, didinant n , skaičiavimo pagal (8.119) formulę tikslumas didėja, o paklaida artėja prie nulio. Pasinaudoję iš elementariosios matematkos žinomomis formulėmis
sin
2
x
=
2
tg
x
1
+
tg
2
x
{\displaystyle \sin 2x={\frac {2\operatorname {tg} x}{1+\operatorname {tg} ^{2}x}}}
cos
2
x
=
1
−
tg
2
x
1
+
tg
2
x
{\displaystyle \cos 2x={\frac {1-\operatorname {tg} ^{2}x}{1+\operatorname {tg} ^{2}x}}\;}
sin
2
x
{\displaystyle \sin 2x}
cos
2
x
{\displaystyle \cos 2x}
sin
2
x
=
2
tg
x
1
+
tg
2
x
≈
2
x
S
n
(
−
x
2
)
1
+
x
2
S
n
2
(
−
x
2
)
=
2
x
S
n
(
−
x
2
)
S
n
2
(
−
x
2
)
+
x
2
S
n
2
(
−
x
2
)
=
2
x
S
n
2
(
−
x
2
)
+
x
2
S
n
(
−
x
2
)
=
2
S
n
(
−
x
2
)
⋅
x
S
n
2
(
−
x
2
)
+
x
2
;
{\displaystyle \sin 2x={\frac {2\operatorname {tg} x}{1+\operatorname {tg} ^{2}x}}\approx {\frac {2{\frac {x}{S_{n}(-x^{2})}}}{1+{\frac {x^{2}}{S_{n}^{2}(-x^{2})}}}}={\frac {\frac {2x}{S_{n}(-x^{2})}}{\frac {S_{n}^{2}(-x^{2})+x^{2}}{S_{n}^{2}(-x^{2})}}}={\frac {2x}{\frac {S_{n}^{2}(-x^{2})+x^{2}}{S_{n}(-x^{2})}}}={\frac {2S_{n}(-x^{2})\cdot x}{S_{n}^{2}(-x^{2})+x^{2}}};}
cos
2
x
=
1
−
tg
2
x
1
+
tg
2
x
≈
1
−
x
2
S
n
2
(
−
x
2
)
1
+
x
2
S
n
2
(
−
x
2
)
=
S
n
2
(
−
x
2
)
−
x
2
S
n
2
(
−
x
2
)
S
n
2
(
−
x
2
)
+
x
2
S
n
2
(
−
x
2
)
=
S
n
2
(
−
x
2
)
−
x
2
S
n
2
(
−
x
2
)
+
x
2
.
{\displaystyle \cos 2x={\frac {1-\operatorname {tg} ^{2}x}{1+\operatorname {tg} ^{2}x}}\approx {\frac {1-{\frac {x^{2}}{S_{n}^{2}(-x^{2})}}}{1+{\frac {x^{2}}{S_{n}^{2}(-x^{2})}}}}={\frac {\frac {S_{n}^{2}(-x^{2})-x^{2}}{S_{n}^{2}(-x^{2})}}{\frac {S_{n}^{2}(-x^{2})+x^{2}}{S_{n}^{2}(-x^{2})}}}={\frac {S_{n}^{2}(-x^{2})-x^{2}}{S_{n}^{2}(-x^{2})+x^{2}}}.}
Baigdami pastebėsime, kad visų paskutiniuose dviejuose skirsniuose (4. ir 5.) aptartų funkcijų reikšmių skaičiavimo tikslumas, atliekant šešias iteracijas (n =6), bus ne mažesnis kaip
10
−
11
,
{\displaystyle 10^{-11},}
x modulis ne didesnis už
π
4
.
{\displaystyle {\frac {\pi }{4}}.}
Apskaičiuosime
th
x
,
{\displaystyle \operatorname {th} x,}
x =0.5, o n =4. Tada, pagal (8.105) formulę,
th
x
≈
x
1
+
x
2
3
+
x
2
5
;
{\displaystyle \operatorname {th} x\approx {\cfrac {x}{1+{\cfrac {x^{2}}{3+{\cfrac {x^{2}}{5}}}}}};}
th
0.5
≈
0.5
1
+
0.5
2
3
+
0.5
2
5
=
0.5
1
+
0.25
3
+
0.25
5
=
0.5
1
+
0.25
15
+
0.25
5
=
0.5
1
+
1.25
15.25
=
{\displaystyle \operatorname {th} 0.5\approx {\cfrac {0.5}{1+{\cfrac {0.5^{2}}{3+{\cfrac {0.5^{2}}{5}}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{3+{\cfrac {0.25}{5}}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{\cfrac {15+0.25}{5}}}}}={\cfrac {0.5}{1+{\cfrac {1.25}{15.25}}}}=}
=
0.5
⋅
15.25
15.25
+
1.25
=
7.625
16.5
=
0.46
(
21
)
=
{\displaystyle ={\cfrac {0.5\cdot 15.25}{15.25+1.25}}={\frac {7.625}{16.5}}=0.46(21)=}
= 0.46212121212121212121212121212121.
Įvertinsime paklaidą, pagal (8.111) formulę.
|
th
x
−
th
¯
x
|
≤
x
2
n
+
1
[
(
2
n
−
1
)
!
!
]
2
(
2
n
+
1
)
=
0.5
2
⋅
4
+
1
[
(
2
⋅
4
−
1
)
!
!
]
2
(
2
⋅
4
+
1
)
=
0.5
9
[
7
!
!
]
2
⋅
9
=
{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {0.5^{2\cdot 4+1}}{[(2\cdot 4-1)!!]^{2}(2\cdot 4+1)}}={\frac {0.5^{9}}{[7!!]^{2}\cdot 9}}=}
=
0.5
9
[
7
⋅
5
⋅
3
]
2
⋅
9
=
0.001953125
105
2
⋅
9
=
{\displaystyle ={\frac {0.5^{9}}{[7\cdot 5\cdot 3]^{2}\cdot 9}}={\frac {0.001953125}{105^{2}\cdot 9}}=}
= 0.001953125/(105^2 *9) = 1.9683799445704207608969513731418e-8
≈
1.96837994457
⋅
10
−
8
.
{\displaystyle \approx 1.96837994457\cdot 10^{-8}.}
Iš kalkuliatoriaus:
th
0.5
=
{\displaystyle \operatorname {th} 0.5=}
0.46211715726000975850231848364367 - 0.46212121212121212121212121212121 =
= -0.00000405486120236270980272847754
≈
−
4.0548612
⋅
10
−
6
.
{\displaystyle \approx -4.0548612\cdot 10^{-6}.}
Čia pas mus buvo n =2. Todėl paklaidos įvertis yra skaičiui n =2 toks:
|
th
x
−
th
¯
x
|
≤
x
2
n
+
1
[
(
2
n
−
1
)
!
!
]
2
(
2
n
+
1
)
=
0.5
2
⋅
2
+
1
[
(
2
⋅
2
−
1
)
!
!
]
2
(
2
⋅
2
+
1
)
=
0.5
5
[
3
!
!
]
2
⋅
5
=
0.5
5
3
2
⋅
5
=
0.03125
9
⋅
5
=
{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {0.5^{2\cdot 2+1}}{[(2\cdot 2-1)!!]^{2}(2\cdot 2+1)}}={\frac {0.5^{5}}{[3!!]^{2}\cdot 5}}={\frac {0.5^{5}}{3^{2}\cdot 5}}={\frac {0.03125}{9\cdot 5}}=}
= 0.03125/45 = 6.9444444444444444444444444444444e-4
≈
6.9444444
⋅
10
−
4
.
{\displaystyle \approx 6.9444444\cdot 10^{-4}.}
Taigi,
|
−
4.0548612
⋅
10
−
6
|
<
6.9444444
⋅
10
−
4
{\displaystyle |-4.0548612\cdot 10^{-6}|<6.9444444\cdot 10^{-4}}
n =2, gautas teisingai. Toliau jau tikrai apskaičiuosime artinį
th
0.5
,
{\displaystyle \operatorname {th} 0.5,}
n =4.
th
x
≈
x
1
+
x
2
3
+
x
2
5
+
x
2
7
+
x
2
9
;
{\displaystyle \operatorname {th} x\approx {\cfrac {x}{1+{\cfrac {x^{2}}{3+{\cfrac {x^{2}}{5+{\cfrac {x^{2}}{7+{\cfrac {x^{2}}{9}}}}}}}}}};}
th
0.5
≈
0.5
1
+
0.5
2
3
+
0.5
2
5
+
0.5
2
7
+
0.5
2
9
=
0.5
1
+
0.25
3
+
0.25
5
+
0.25
7
+
0.25
9
=
0.5
1
+
0.25
3
+
0.25
5
+
0.25
63
+
0.25
9
=
{\displaystyle \operatorname {th} 0.5\approx {\cfrac {0.5}{1+{\cfrac {0.5^{2}}{3+{\cfrac {0.5^{2}}{5+{\cfrac {0.5^{2}}{7+{\cfrac {0.5^{2}}{9}}}}}}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{3+{\cfrac {0.25}{5+{\cfrac {0.25}{7+{\cfrac {0.25}{9}}}}}}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{3+{\cfrac {0.25}{5+{\cfrac {0.25}{\cfrac {63+0.25}{9}}}}}}}}}=}
=
0.5
1
+
0.25
3
+
0.25
5
+
0.25
⋅
9
63.25
=
0.5
1
+
0.25
3
+
0.25
5
⋅
63.25
+
0.25
⋅
9
63.25
=
0.5
1
+
0.25
3
+
0.25
318.5
63.25
=
{\displaystyle ={\cfrac {0.5}{1+{\cfrac {0.25}{3+{\cfrac {0.25}{5+{\cfrac {0.25\cdot 9}{63.25}}}}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{3+{\cfrac {0.25}{\cfrac {5\cdot 63.25+0.25\cdot 9}{63.25}}}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{3+{\cfrac {0.25}{\cfrac {318.5}{63.25}}}}}}}=}
=
0.5
1
+
0.25
3
+
0.25
⋅
63.25
318.5
=
0.5
1
+
0.25
3
⋅
318.5
+
0.25
⋅
63.25
318.5
=
0.5
1
+
0.25
971.3125
318.5
=
{\displaystyle ={\cfrac {0.5}{1+{\cfrac {0.25}{3+{\cfrac {0.25\cdot 63.25}{318.5}}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{\cfrac {3\cdot 318.5+0.25\cdot 63.25}{318.5}}}}}={\cfrac {0.5}{1+{\cfrac {0.25}{\cfrac {971.3125}{318.5}}}}}=}
=
0.5
1
+
0.25
⋅
318.5
971.3125
=
0.5
971.3125
+
79.625
971.3125
=
0.5
⋅
971.3125
1050.9375
=
485.65625
1050.9375
=
{\displaystyle ={\cfrac {0.5}{1+{\cfrac {0.25\cdot 318.5}{971.3125}}}}={\cfrac {0.5}{\cfrac {971.3125+79.625}{971.3125}}}={\cfrac {0.5\cdot 971.3125}{1050.9375}}={\frac {485.65625}{1050.9375}}=}
= 485.65625/1050.9375 = 0.46211715730002973535533749628308.
Atėmę šią gautą
th
0.5
{\displaystyle \operatorname {th} 0.5}
0.46211715726000975850231848364367 - 0.46211715730002973535533749628308 =
= -0.00000000004001997685301901263941
≈
−
4.0019976853
⋅
10
−
11
.
{\displaystyle \approx -4.0019976853\cdot 10^{-11}.}
[Kad "atrast", kad 10 pakelta -11, reikia paskaičiuoti kelintu numeriu stovi pirmas skaičius po kablelio nelygus nuliui; čia prieš skaičių "4" yra 10 nuliu po kablelio (taško), o pats ketvertas stovi 11-tas po kablelio (pirmas nulis prieš kablelį neskaičiuojamas).]
Taigi,
|
−
4.0019976853
⋅
10
−
11
|
≈
|
th
x
−
th
¯
x
|
<
1.96837994457
⋅
10
−
8
.
{\displaystyle |-4.0019976853\cdot 10^{-11}|\approx |\operatorname {th} x-{\overline {\operatorname {th} }}x|<1.96837994457\cdot 10^{-8}.}
Vadinasi, paklaidos įvertinimas gautas teisingai. Beveik trim eilėm reali paklaida mažesnė nei nustatytas paklaidos įvertis.
Apskaičiuosime
tg
x
,
{\displaystyle \operatorname {tg} x,}
x =0.5 radiano, o n =4.
tg
x
≈
x
1
+
−
x
2
3
+
−
x
2
5
+
−
x
2
7
+
−
x
2
9
;
{\displaystyle \operatorname {tg} x\approx {\cfrac {x}{1+{\cfrac {-x^{2}}{3+{\cfrac {-x^{2}}{5+{\cfrac {-x^{2}}{7+{\cfrac {-x^{2}}{9}}}}}}}}}};}
tg
0.5
≈
0.5
1
+
−
0.5
2
3
+
−
0.5
2
5
+
−
0.5
2
7
+
−
0.5
2
9
=
0.5
1
+
−
0.25
3
+
−
0.25
5
+
−
0.25
7
+
−
0.25
9
=
0.5
1
+
−
0.25
3
+
−
0.25
5
+
−
0.25
63
−
0.25
9
=
{\displaystyle \operatorname {tg} 0.5\approx {\cfrac {0.5}{1+{\cfrac {-0.5^{2}}{3+{\cfrac {-0.5^{2}}{5+{\cfrac {-0.5^{2}}{7+{\cfrac {-0.5^{2}}{9}}}}}}}}}}={\cfrac {0.5}{1+{\cfrac {-0.25}{3+{\cfrac {-0.25}{5+{\cfrac {-0.25}{7+{\cfrac {-0.25}{9}}}}}}}}}}={\cfrac {0.5}{1+{\cfrac {-0.25}{3+{\cfrac {-0.25}{5+{\cfrac {-0.25}{\cfrac {63-0.25}{9}}}}}}}}}=}
=
0.5
1
+
−
0.25
3
+
−
0.25
5
+
−
0.25
⋅
9
62.75
=
0.5
1
+
−
0.25
3
+
−
0.25
⋅
62.75
5
⋅
62.75
−
0.25
⋅
9
=
0.5
1
+
−
0.25
3
+
−
15.6875
311.5
=
{\displaystyle ={\cfrac {0.5}{1+{\cfrac {-0.25}{3+{\cfrac {-0.25}{5+{\cfrac {-0.25\cdot 9}{62.75}}}}}}}}={\cfrac {0.5}{1+{\cfrac {-0.25}{3+{\cfrac {-0.25\cdot 62.75}{5\cdot 62.75-0.25\cdot 9}}}}}}={\cfrac {0.5}{1+{\cfrac {-0.25}{3+{\cfrac {-15.6875}{311.5}}}}}}=}
=
0.5
1
+
−
0.25
⋅
311.5
3
⋅
311.5
−
15.6875
=
0.5
1
+
−
77.875
918.8125
=
0.5
⋅
918.8125
918.8125
−
77.875
=
459.40625
840.9375
=
{\displaystyle ={\cfrac {0.5}{1+{\cfrac {-0.25\cdot 311.5}{3\cdot 311.5-15.6875}}}}={\cfrac {0.5}{1+{\cfrac {-77.875}{918.8125}}}}={\cfrac {0.5\cdot 918.8125}{918.8125-77.875}}={\cfrac {459.40625}{840.9375}}=}
= 459.40625/840.9375 = 0.54630248978075065031586770717205.
Atėmę šią gautą
tg
x
{\displaystyle \operatorname {tg} x}
tg
x
{\displaystyle \operatorname {tg} x}
0.54630248984379051325517946578029 - 459.40625/840.9375 = 6.3039862939311758608230385155597e-11
≈
6.30398629393
⋅
10
−
11
.
{\displaystyle \approx 6.30398629393\cdot 10^{-11}.}
Paklaidos įvertinimas tangentui yra panašus kaip hiperboliniam tangentui.
[Tangentui, (8.107) lygybės pavirsta į tokias:
{
Q
1
=
Q
¯
1
,
Q
2
=
Q
¯
2
,
Q
3
=
Q
¯
3
,
.
.
.
,
Q
n
=
Q
¯
n
,
Q
n
+
1
=
(
2
n
+
1
+
2
x
2
u
n
+
2
)
Q
¯
n
−
x
2
Q
¯
n
−
1
,
Q
¯
n
+
1
=
(
2
n
+
1
)
Q
¯
n
−
x
2
Q
¯
n
−
1
.
(
8.107
for
tg
x
)
{\displaystyle {\begin{cases}Q_{1}={\overline {Q}}_{1},\;\;Q_{2}={\overline {Q}}_{2},\;\;Q_{3}={\overline {Q}}_{3},\;\;...,\;\;Q_{n}={\overline {Q}}_{n},&\\Q_{n+1}=(2n+1+2x^{2}u_{n+2}){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1},&\\{\overline {Q}}_{n+1}=(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}.&\end{cases}}\quad \quad (8.107\;\;{\text{for}}\;\;{\text{tg}}\;x)}
Tada (8.108) formulė pavirsta į tokią (padauginome viską dar iš
(
−
1
)
n
{\displaystyle (-1)^{n}}
n koeficientų
−
x
2
{\displaystyle -x^{2}}
a
1
=
x
,
a
2
=
−
x
2
,
a
3
=
−
x
2
,
.
.
.
,
a
n
+
1
=
−
x
2
{\displaystyle a_{1}=x,\;a_{2}=-x^{2},\;a_{3}=-x^{2},\;...,\;a_{n+1}=-x^{2}}
tg
x
−
tg
¯
x
=
(
−
1
)
n
(
−
1
)
n
+
1
x
2
n
+
1
Q
¯
n
Q
¯
n
+
1
[
2
x
2
u
n
+
2
Q
¯
n
Q
n
+
1
]
=
(
−
1
)
2
n
+
1
x
2
n
+
1
Q
¯
n
Q
¯
n
+
1
[
2
x
2
u
n
+
2
Q
¯
n
2
x
2
u
n
+
2
Q
¯
n
+
(
2
n
+
1
)
Q
¯
n
−
x
2
Q
¯
n
−
1
]
.
(
8.108
for
tg
x
)
{\displaystyle \operatorname {tg} x-{\overline {\operatorname {tg} }}x=(-1)^{n}(-1)^{n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{Q_{n+1}}}\right]=(-1)^{2n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right].\;\;(8.108\;\;{\text{for}}\;\;{\text{tg}}\;x)}
Taigi, nėra funkcijai tg(x) griežto paklaidos nustatymo būdo, nes vardiklis
2
x
2
u
n
+
2
Q
¯
n
+
(
2
n
+
1
)
Q
¯
n
−
x
2
Q
¯
n
−
1
{\displaystyle 2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}
th
x
{\displaystyle \operatorname {th} x}
th
x
{\displaystyle \operatorname {th} x}
tg
x
{\displaystyle \operatorname {tg} x}
Todėl paklaida vertinama pagal (8.111) formulę. Pagal (8.111) formulę, hiperboliniam tangentui gavome, kad
|
th
x
−
th
¯
x
|
≤
1.96837994457
⋅
10
−
8
.
{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq 1.96837994457\cdot 10^{-8}.}
Todėl
6.30398629393
⋅
10
−
11
≈
|
tg
x
−
tg
¯
x
|
≤
1.96837994457
⋅
10
−
8
.
{\displaystyle 6.30398629393\cdot 10^{-11}\approx |\operatorname {tg} x-{\overline {\operatorname {tg} }}x|\leq 1.96837994457\cdot 10^{-8}.}
Paklaidos įvertinimas gautas teisingai.
Apskaičiuosime
th
x
,
{\displaystyle \operatorname {th} x,}
x =0.8, o n =6.
th
x
≈
x
1
+
x
2
3
+
x
2
5
+
x
2
7
+
x
2
9
+
x
2
11
+
x
2
13
;
{\displaystyle \operatorname {th} x\approx {\cfrac {x}{1+{\cfrac {x^{2}}{3+{\cfrac {x^{2}}{5+{\cfrac {x^{2}}{7+{\cfrac {x^{2}}{9+{\cfrac {x^{2}}{11+{\cfrac {x^{2}}{13}}}}}}}}}}}}}};}
th
0.8
≈
0.8
1
+
0.8
2
3
+
0.8
2
5
+
0.8
2
7
+
0.8
2
9
+
0.8
2
11
+
0.8
2
13
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
7
+
0.64
9
+
0.64
11
+
0.64
13
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
7
+
0.64
9
+
0.64
11
⋅
13
+
0.64
13
=
{\displaystyle \operatorname {th} 0.8\approx {\cfrac {0.8}{1+{\cfrac {0.8^{2}}{3+{\cfrac {0.8^{2}}{5+{\cfrac {0.8^{2}}{7+{\cfrac {0.8^{2}}{9+{\cfrac {0.8^{2}}{11+{\cfrac {0.8^{2}}{13}}}}}}}}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{7+{\cfrac {0.64}{9+{\cfrac {0.64}{11+{\cfrac {0.64}{13}}}}}}}}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{7+{\cfrac {0.64}{9+{\cfrac {0.64}{\cfrac {11\cdot 13+0.64}{13}}}}}}}}}}}}}=}
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
7
+
0.64
9
+
0.64
⋅
13
143.64
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
7
+
0.64
9
⋅
143.64
+
0.64
⋅
13
143.64
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
7
+
0.64
1301.08
143.64
=
{\displaystyle ={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{7+{\cfrac {0.64}{9+{\cfrac {0.64\cdot 13}{143.64}}}}}}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{7+{\cfrac {0.64}{\cfrac {9\cdot 143.64+0.64\cdot 13}{143.64}}}}}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{7+{\cfrac {0.64}{\cfrac {1301.08}{143.64}}}}}}}}}}}=}
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
7
+
0.64
⋅
143.64
1301.08
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
7
⋅
1301.08
+
0.64
⋅
143.64
1301.08
=
0.8
1
+
0.64
3
+
0.64
5
+
0.64
9199.4896
1301.08
=
{\displaystyle ={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{7+{\cfrac {0.64\cdot 143.64}{1301.08}}}}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{\cfrac {7\cdot 1301.08+0.64\cdot 143.64}{1301.08}}}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{5+{\cfrac {0.64}{\cfrac {9199.4896}{1301.08}}}}}}}}}=}
=
0.8
1
+
0.64
3
+
0.64
5
⋅
9199.4896
+
0.64
⋅
1301.08
9199.4896
=
0.8
1
+
0.64
3
+
0.64
46830.1392
9199.4896
=
0.8
1
+
0.64
3
⋅
46830.1392
+
0.64
⋅
9199.4896
46830.1392
=
0.8
1
+
0.64
146378.090944
46830.1392
=
{\displaystyle ={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{\cfrac {5\cdot 9199.4896+0.64\cdot 1301.08}{9199.4896}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{3+{\cfrac {0.64}{\cfrac {46830.1392}{9199.4896}}}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{\cfrac {3\cdot 46830.1392+0.64\cdot 9199.4896}{46830.1392}}}}}={\cfrac {0.8}{1+{\cfrac {0.64}{\cfrac {146378.090944}{46830.1392}}}}}=}
=
0.8
146378.090944
+
0.64
⋅
46830.1392
146378.090944
=
0.8
176349.380032
146378.090944
=
0.8
⋅
146378.090944
176349.380032
=
117102.4727552
176349.380032
=
{\displaystyle ={\cfrac {0.8}{\cfrac {146378.090944+0.64\cdot 46830.1392}{146378.090944}}}={\cfrac {0.8}{\cfrac {176349.380032}{146378.090944}}}={\cfrac {0.8\cdot 146378.090944}{176349.380032}}={\cfrac {117102.4727552}{176349.380032}}=}
= 117102.4727552/176349.380032 = 0.66403677026792395500016180062552.
Atėmę gautą apytikslią
th
0.8
{\displaystyle \operatorname {th} 0.8}
th
0.8
{\displaystyle \operatorname {th} 0.8}
0.66403677026784896368484465640024 - 0.66403677026792395500016180062552 = -0.00000000000007499131531714422528 arba
tanh(0.8) - 117102.4727552/176349.380032 = -7.4991315317144225281449358567679e-14.
Taigi, kaip skaičiavome pagal (8.111) 3. skirsnyje su x =0.8 ir n =6, rezultatas buvo
|
th
x
−
th
¯
x
|
≤
3.9136
⋅
10
−
11
.
{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq 3.9136\cdot 10^{-11}.}
Matome, kad reali paklaida
≈
|
−
7.49913153
⋅
10
−
14
|
{\displaystyle \approx |-7.49913153\cdot 10^{-14}|}
3.9136
⋅
10
−
11
,
{\displaystyle 3.9136\cdot 10^{-11},}
Tangento paklaidos įvertinimas. Kai x =0.5, o n =4, gavome, kad th(x) paklaida yra
≈
−
4.0019976853
⋅
10
−
11
,
{\displaystyle \approx -4.0019976853\cdot 10^{-11},}
≈
6.30398629393
⋅
10
−
11
.
{\displaystyle \approx 6.30398629393\cdot 10^{-11}.}
Nystatysime funkcijos tg(x) paklaidą, kai x =1, o n =6. Truputi aukščiau gavome tokia (8.108 for tg(x)) paklaidos formulę:
tg
x
−
tg
¯
x
=
(
−
1
)
2
n
+
1
x
2
n
+
1
Q
¯
n
Q
¯
n
+
1
[
2
x
2
u
n
+
2
Q
¯
n
2
x
2
u
n
+
2
Q
¯
n
+
(
2
n
+
1
)
Q
¯
n
−
x
2
Q
¯
n
−
1
]
=
{\displaystyle \operatorname {tg} x-{\overline {\operatorname {tg} }}x=(-1)^{2n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right]=}
=
−
x
2
n
+
1
Q
¯
n
Q
¯
n
+
1
[
2
x
2
u
n
+
2
Q
¯
n
2
x
2
u
n
+
2
Q
¯
n
+
(
2
n
+
1
)
Q
¯
n
−
x
2
Q
¯
n
−
1
]
.
(
8.108
for
tg
x
)
{\displaystyle =-{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right].\;\;(8.108\;\;{\text{for}}\;\;{\text{tg}}\;x)}
Iš aukščiau žinome, kad
P
−
1
=
1
,
Q
−
1
=
0.
{\displaystyle P_{-1}=1,\;\;Q_{-1}=0.}
P
0
=
b
0
,
Q
0
=
1.
{\displaystyle P_{0}=b_{0},\;\;Q_{0}=1.}
{
P
k
=
b
k
P
k
−
1
+
a
k
P
k
−
2
,
Q
k
=
b
k
Q
k
−
1
+
a
k
Q
k
−
2
.
(
8.97
)
{\displaystyle {\begin{cases}P_{k}=b_{k}P_{k-1}+a_{k}P_{k-2},&\\Q_{k}=b_{k}Q_{k-1}+a_{k}Q_{k-2}.&\end{cases}}\quad \quad (8.97)}
reikia išreikšti visus
Q
k
,
{\displaystyle Q_{k},}
k yra nuo 1 iki n +1. Mes žinome, kad
a
1
=
x
,
a
2
=
−
x
2
,
a
3
=
−
x
2
,
.
.
.
,
a
n
+
1
=
−
x
2
.
{\displaystyle a_{1}=x,\;a_{2}=-x^{2},\;a_{3}=-x^{2},\;...,\;a_{n+1}=-x^{2}.}
b
1
=
1
,
b
2
=
3
,
b
3
=
5
,
.
.
.
,
b
n
=
2
n
−
1
,
b
¯
n
+
1
=
2
n
+
1.
{\displaystyle b_{1}=1,\;b_{2}=3,\;b_{3}=5,\;...,\;b_{n}=2n-1,\;{\overline {b}}_{n+1}=2n+1.}
Q
1
=
b
1
Q
1
−
1
+
a
1
Q
1
−
2
=
1
⋅
Q
0
+
x
Q
−
1
=
1
⋅
1
+
1
⋅
0
=
1
;
{\displaystyle Q_{1}=b_{1}Q_{1-1}+a_{1}Q_{1-2}=1\cdot Q_{0}+xQ_{-1}=1\cdot 1+1\cdot 0=1;}
Q
2
=
b
2
Q
2
−
1
+
a
2
Q
2
−
2
=
3
⋅
Q
1
+
(
−
x
2
)
Q
0
=
3
⋅
1
+
(
−
1
2
)
⋅
1
=
3
−
1
=
2
;
{\displaystyle Q_{2}=b_{2}Q_{2-1}+a_{2}Q_{2-2}=3\cdot Q_{1}+(-x^{2})Q_{0}=3\cdot 1+(-1^{2})\cdot 1=3-1=2;}
Q
3
=
b
3
Q
3
−
1
+
a
3
Q
3
−
2
=
5
⋅
Q
2
+
(
−
x
2
)
Q
1
=
5
⋅
2
+
(
−
1
2
)
⋅
1
=
10
−
1
=
9
;
{\displaystyle Q_{3}=b_{3}Q_{3-1}+a_{3}Q_{3-2}=5\cdot Q_{2}+(-x^{2})Q_{1}=5\cdot 2+(-1^{2})\cdot 1=10-1=9;}
Q
4
=
b
4
Q
4
−
1
+
a
4
Q
4
−
2
=
7
⋅
Q
3
+
(
−
x
2
)
Q
2
=
7
⋅
9
+
(
−
1
2
)
⋅
2
=
63
−
2
=
61
;
{\displaystyle Q_{4}=b_{4}Q_{4-1}+a_{4}Q_{4-2}=7\cdot Q_{3}+(-x^{2})Q_{2}=7\cdot 9+(-1^{2})\cdot 2=63-2=61;}
Q
5
=
b
5
Q
5
−
1
+
a
5
Q
5
−
2
=
9
⋅
Q
4
+
(
−
x
2
)
Q
3
=
9
⋅
61
+
(
−
1
2
)
⋅
9
=
549
−
9
=
540
;
{\displaystyle Q_{5}=b_{5}Q_{5-1}+a_{5}Q_{5-2}=9\cdot Q_{4}+(-x^{2})Q_{3}=9\cdot 61+(-1^{2})\cdot 9=549-9=540;}
Q
n
=
Q
6
=
b
6
Q
6
−
1
+
a
6
Q
6
−
2
=
11
⋅
Q
5
+
(
−
x
2
)
Q
4
=
11
⋅
540
+
(
−
1
2
)
⋅
61
=
5940
−
61
=
5879
;
{\displaystyle Q_{n}=Q_{6}=b_{6}Q_{6-1}+a_{6}Q_{6-2}=11\cdot Q_{5}+(-x^{2})Q_{4}=11\cdot 540+(-1^{2})\cdot 61=5940-61=5879;}
Q
¯
n
+
1
=
Q
¯
7
=
b
¯
7
Q
7
−
1
+
a
7
Q
7
−
2
=
13
⋅
Q
6
+
(
−
x
2
)
Q
5
=
13
⋅
5879
+
(
−
1
2
)
⋅
540
=
76427
−
540
=
75887.
{\displaystyle {\overline {Q}}_{n+1}={\overline {Q}}_{7}={\overline {b}}_{7}Q_{7-1}+a_{7}Q_{7-2}=13\cdot Q_{6}+(-x^{2})Q_{5}=13\cdot 5879+(-1^{2})\cdot 540=76427-540=75887.}
Dabar galime pagal (8.108 for tg(x)) formulę nustatyti paklaidą, kai x =1, o n =6. Taigi,
tg
x
−
tg
¯
x
=
−
x
2
n
+
1
Q
¯
n
Q
¯
n
+
1
[
2
x
2
u
n
+
2
Q
¯
n
2
x
2
u
n
+
2
Q
¯
n
+
(
2
n
+
1
)
Q
¯
n
−
x
2
Q
¯
n
−
1
]
=
{\displaystyle \operatorname {tg} x-{\overline {\operatorname {tg} }}x=-{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right]=}
=
−
1
2
⋅
6
+
1
Q
¯
6
Q
¯
6
+
1
[
2
⋅
1
2
u
n
+
2
Q
¯
6
2
⋅
1
2
u
n
+
2
Q
¯
6
+
(
2
⋅
6
+
1
)
Q
¯
6
−
1
2
Q
¯
6
−
1
]
=
−
1
13
Q
¯
6
Q
¯
7
[
2
u
n
+
2
Q
¯
6
2
u
n
+
2
Q
¯
6
+
13
Q
¯
6
−
Q
¯
5
]
=
{\displaystyle =-{\frac {1^{2\cdot 6+1}}{{\overline {Q}}_{6}{\overline {Q}}_{6+1}}}\left[{\frac {2\cdot 1^{2}u_{n+2}{\overline {Q}}_{6}}{2\cdot 1^{2}u_{n+2}{\overline {Q}}_{6}+(2\cdot 6+1){\overline {Q}}_{6}-1^{2}{\overline {Q}}_{6-1}}}\right]=-{\frac {1^{13}}{{\overline {Q}}_{6}{\overline {Q}}_{7}}}\left[{\frac {2u_{n+2}{\overline {Q}}_{6}}{2u_{n+2}{\overline {Q}}_{6}+13{\overline {Q}}_{6}-{\overline {Q}}_{5}}}\right]=}
=
−
1
13
5879
⋅
75887
[
2
u
n
+
2
⋅
5879
2
u
n
+
2
⋅
5879
+
13
⋅
5879
−
540
]
=
−
1
446139673
[
2
u
n
+
2
⋅
5879
2
u
n
+
2
⋅
5879
+
75887
]
.
{\displaystyle =-{\frac {1^{13}}{5879\cdot 75887}}\left[{\frac {2u_{n+2}\cdot 5879}{2u_{n+2}\cdot 5879+13\cdot 5879-540}}\right]=-{\frac {1}{446139673}}\left[{\frac {2u_{n+2}\cdot 5879}{2u_{n+2}\cdot 5879+75887}}\right].}
Matome, kad reiškinys laužtiniuose skliaustuose tikrai mažesnis už 1, kai
u
n
+
2
>
0.
{\displaystyle u_{n+2}>0.}
u
n
+
2
>
0
,
{\displaystyle u_{n+2}>0,}
x =1, o n =6, yra mažesnė už
|
−
1
446139673
|
=
1
446139673
=
{\displaystyle \left|-{\frac {1}{446139673}}\right|={\frac {1}{446139673}}=}
= 1/446139673 = 2.2414505154308480429625454986156e-9
≈
2.24145051543
⋅
10
−
9
.
{\displaystyle \approx 2.24145051543\cdot 10^{-9}.}
Jei x būtų lygus 0.8, tai paklaida (tiksliau, jos įvertinimas) gauta šiuo būdu būtų kažkiek mažesnė (gal panaši į th(x) paklaidą [su x =0.8]; tiksliau, panaši į jos įvertinimą). Įvertinsime th(x) paklaidą, kai x =1, o n =6. Pagal (8.111) formulę
|
th
x
−
th
¯
x
|
≤
x
2
n
+
1
[
(
2
n
−
1
)
!
!
]
2
(
2
n
+
1
)
(
8.111
)
{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}\quad (8.111)}
gauname
x
2
n
+
1
[
(
2
n
−
1
)
!
!
]
2
(
2
n
+
1
)
=
x
13
[
11
!
!
]
2
(
2
n
+
1
)
=
1
13
10395
2
⋅
13
=
1
108056025
⋅
13
=
1
1404728325
=
{\displaystyle {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {x^{13}}{[11!!]^{2}(2n+1)}}={\frac {1^{13}}{10395^{2}\cdot 13}}={\frac {1}{108056025\cdot 13}}={\frac {1}{1404728325}}=}
= 1/1404728325 = 7.1188142376213564284752355940427e-10
≈
7.11881423762
⋅
10
−
10
.
{\displaystyle \approx 7.11881423762\cdot 10^{-10}.}
Matome, kad funkcijos th(x), kai x =1, o n =6 paklaidos nustatymas gaunamas su didesniu tikslumu negu funckijos tg(x) (kai x =1, n =6). Įrodysime arba paneigsime, kad funkcijai tg(x)
u
n
+
2
>
0.
(
8.110
)
{\displaystyle u_{n+2}>0.\quad (8.110)}
Užtenka įrodyti, kad funkcijos
y
=
cos
x
{\displaystyle y=\cos {\sqrt {x}}}
π
2
>
x
>
0
{\displaystyle {\frac {\pi }{2}}>x>0}
u
n
+
2
=
y
(
n
+
2
)
y
(
n
+
1
)
.
{\displaystyle u_{n+2}={\frac {y^{(n+2)}}{y^{(n+1)}}}.}
Padauginę lygybę
4
x
y
(
n
+
2
)
+
(
4
n
+
2
)
y
(
n
+
1
)
+
y
(
n
)
=
0
{\displaystyle 4xy^{(n+2)}+(4n+2)y^{(n+1)}+y^{(n)}=0}
iš 5. skirsnio iš
1
4
x
n
−
1
2
,
{\displaystyle {\frac {1}{4}}x^{n-{\frac {1}{2}}},}
[
x
n
+
1
2
y
(
n
+
1
)
]
′
=
−
1
4
x
n
−
1
2
y
(
n
)
.
(
8.113
for
tg
x
)
{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'=-{\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)}.\quad (8.113\;\;{\text{for}}\;\;\operatorname {tg} x)}
[
4
x
y
(
n
+
2
)
+
(
4
n
+
2
)
y
(
n
+
1
)
+
y
(
n
)
=
0
|
⋅
1
4
x
n
−
1
2
,
{\displaystyle \;4xy^{(n+2)}+(4n+2)y^{(n+1)}+y^{(n)}=0\;\;|\cdot {\frac {1}{4}}x^{n-{\frac {1}{2}}},}
x
n
+
1
2
y
(
n
+
2
)
+
(
n
+
1
2
)
x
n
−
1
2
y
(
n
+
1
)
+
1
4
x
n
−
1
2
y
(
n
)
=
0.
{\displaystyle x^{n+{\frac {1}{2}}}y^{(n+2)}+(n+{\frac {1}{2}})x^{n-{\frac {1}{2}}}y^{(n+1)}+{\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)}=0.}
[
x
n
+
1
2
y
(
n
+
1
)
]
′
=
(
n
+
1
2
)
x
n
−
1
2
y
(
n
+
1
)
+
x
n
+
1
2
y
(
n
+
2
)
.
{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'=(n+{\frac {1}{2}})x^{n-{\frac {1}{2}}}y^{(n+1)}+x^{n+{\frac {1}{2}}}y^{(n+2)}.}
Pirmiausia įsitikinsime, kad
lim
x
→
0
+
0
[
x
n
+
1
2
y
(
n
+
1
)
(
x
)
]
=
0.
(
8.114
)
{\displaystyle \lim _{x\to 0+0}\left[x^{n+{\frac {1}{2}}}y^{(n+1)}(x)\right]=0.\quad (8.114)}
Tam reikalui užtenka įrodyti, kad funkcija
x
n
y
(
n
+
1
)
(
x
)
(
8.115
)
{\displaystyle x^{n}y^{(n+1)}(x)\quad (8.115)}
lieka aprėžta, kai
x
→
0
+
0
{\displaystyle x\to 0+0}
x artėja prie nulio iš dešinės). Iš formulių
y
=
cos
x
{\displaystyle y=\cos {\sqrt {x}}}
y
′
=
−
sin
x
2
x
{\displaystyle y'=-{\frac {\sin {\sqrt {x}}}{2{\sqrt {x}}}}}
y
(
x
)
{\displaystyle y(x)}
y
′
(
x
)
{\displaystyle y'(x)}
x
→
0
+
0.
{\displaystyle x\to 0+0.}
[Kad funkcija
y
′
(
x
)
{\displaystyle y'(x)}
y
′
(
x
)
{\displaystyle y'(x)}
Lopitalio taisyklę :
lim
x
→
0
+
0
−
sin
x
2
x
=
lim
x
→
0
+
0
−
(
sin
x
)
′
(
2
x
)
′
=
lim
x
→
0
+
0
−
1
2
x
cos
x
2
⋅
1
2
x
=
lim
x
→
0
+
0
−
1
2
cos
x
=
−
1
2
⋅
cos
0
=
−
1
2
.
{\displaystyle \lim _{x\to 0+0}{\frac {-\sin {\sqrt {x}}}{2{\sqrt {x}}}}=\lim _{x\to 0+0}{\frac {-(\sin {\sqrt {x}})'}{(2{\sqrt {x}})'}}=\lim _{x\to 0+0}{\frac {-{\frac {1}{2{\sqrt {x}}}}\cos {\sqrt {x}}}{2\cdot {\frac {1}{2{\sqrt {x}}}}}}=\lim _{x\to 0+0}{\frac {-1}{2}}\cos {\sqrt {x}}=-{\frac {1}{2}}\cdot \cos 0=-{\frac {1}{2}}.}
Tačiau tuomet iš 5. skirsnio lygybės
4
x
y
″
+
2
y
′
+
y
=
0
{\displaystyle 4xy''+2y'+y=0}
x
y
″
(
x
)
{\displaystyle xy''(x)}
x
→
0
+
0.
{\displaystyle x\to 0+0.}
[
4
x
y
″
+
2
y
′
+
y
=
0.
(
8.102
for
tg
x
)
{\displaystyle 4xy''+2y'+y=0.\quad (8.102\;\;{\text{for}}\;\;\operatorname {tg} x)}
4
x
y
″
=
−
2
y
′
−
y
.
(
B
)
{\displaystyle 4xy''=-2y'-y.\quad (B)}
Lygybėje (B) dešinė pusė yra aprėžta, kai
x
→
0
+
0.
{\displaystyle x\to 0+0.}
lim
x
→
0
+
0
(
−
2
y
′
−
y
)
=
−
2
⋅
−
1
2
−
1
=
0.
{\displaystyle \lim _{x\to 0+0}(-2y'-y)=-2\cdot {\frac {-1}{2}}-1=0.}
Toliau matematinės indukcijos metodu, remdamiesi
4
x
y
(
n
+
2
)
+
(
4
n
+
2
)
y
(
n
+
1
)
+
y
(
n
)
=
0
{\displaystyle 4xy^{(n+2)}+(4n+2)y^{(n+1)}+y^{(n)}=0}
x
n
y
(
n
+
1
)
(
x
)
{\displaystyle x^{n}y^{(n+1)}(x)}
x
→
0
+
0
,
{\displaystyle x\to 0+0,}
n . Iš to aišku, kad (8.114) lygybė yra teisinga.
Dabar įsitikinsime, kad, imant bet kokį neneigiamą sveikąjį skaičių n , išvestinė
y
(
n
)
(
x
)
(
8.116
)
{\displaystyle y^{(n)}(x)\quad (8.116)}
gali būti neigiama ir gali būti teigiama (išvestinių ženklai kaitoliojaisi) su
π
2
>
x
>
0.
{\displaystyle {\frac {\pi }{2}}>x>0.}
y
(
0
)
(
x
)
=
y
(
x
)
=
cos
x
{\displaystyle y^{(0)}(x)=y(x)=\cos {\sqrt {x}}\;}
π
2
>
x
>
0
{\displaystyle {\frac {\pi }{2}}>x>0}
π
2
>
x
>
0
{\displaystyle {\frac {\pi }{2}}>x>0}
[
x
n
+
1
2
y
(
n
+
1
)
]
′
=
−
1
4
x
n
−
1
2
y
(
n
)
;
(
8.113
for
tg
x
)
{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'=-{\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)};\quad (8.113\;\;{\text{for}}\;\;\operatorname {tg} x)}
[
x
0
+
1
2
y
(
0
+
1
)
]
′
=
−
1
4
x
0
−
1
2
y
(
0
)
,
{\displaystyle \left[x^{0+{\frac {1}{2}}}y^{(0+1)}\right]'=-{\frac {1}{4}}x^{0-{\frac {1}{2}}}y^{(0)},}
[
x
1
2
y
(
1
)
]
′
=
−
1
4
x
−
1
2
cos
x
.
(
C
)
{\displaystyle \left[x^{\frac {1}{2}}y^{(1)}\right]'=-{\frac {1}{4}}x^{-{\frac {1}{2}}}\cos {\sqrt {x}}.\quad (C)}
Pagal lygybę (C) funkcijos
x
1
2
y
(
1
)
{\displaystyle x^{\frac {1}{2}}y^{(1)}}
x
1
2
y
(
1
)
{\displaystyle x^{\frac {1}{2}}y^{(1)}}
(
0
;
π
2
)
{\displaystyle (0;\;{\frac {\pi }{2}})}
(
0
;
π
2
)
{\displaystyle (0;\;{\frac {\pi }{2}})}
y
(
1
)
<
0
{\displaystyle y^{(1)}<0}
Toliau pagal (8.113 for tg(x)) formulę, kai n=1, gauname (kai
π
2
>
x
>
0
{\displaystyle {\frac {\pi }{2}}>x>0}
[
x
1
+
1
2
y
(
1
+
1
)
]
′
=
−
1
4
x
1
−
1
2
y
(
1
)
,
{\displaystyle \left[x^{1+{\frac {1}{2}}}y^{(1+1)}\right]'=-{\frac {1}{4}}x^{1-{\frac {1}{2}}}y^{(1)},}
[
x
1
+
1
2
y
(
2
)
]
′
=
−
1
4
x
1
−
1
2
y
(
1
)
.
(
D
)
{\displaystyle \left[x^{1+{\frac {1}{2}}}y^{(2)}\right]'=-{\frac {1}{4}}x^{1-{\frac {1}{2}}}y^{(1)}.\quad (D)}
Lygybėje (D) dešinė pusė yra teigiama. Todėl funkcijos
x
1
+
1
2
y
(
2
)
{\displaystyle x^{1+{\frac {1}{2}}}y^{(2)}}
x
1
+
1
2
y
(
2
)
{\displaystyle x^{1+{\frac {1}{2}}}y^{(2)}}
(
0
;
π
2
)
{\displaystyle (0;\;{\frac {\pi }{2}})}
y
(
2
)
>
0.
{\displaystyle y^{(2)}>0.}
Toliau pagal (8.113 for tg(x)) formulę, kai n=2, gauname (kai
π
2
>
x
>
0
{\displaystyle {\frac {\pi }{2}}>x>0}
[
x
2
+
1
2
y
(
2
+
1
)
]
′
=
−
1
4
x
2
−
1
2
y
(
2
)
,
{\displaystyle \left[x^{2+{\frac {1}{2}}}y^{(2+1)}\right]'=-{\frac {1}{4}}x^{2-{\frac {1}{2}}}y^{(2)},}
[
x
2
+
1
2
y
(
3
)
]
′
=
−
1
4
x
2
−
1
2
y
(
2
)
.
(
E
)
{\displaystyle \left[x^{2+{\frac {1}{2}}}y^{(3)}\right]'=-{\frac {1}{4}}x^{2-{\frac {1}{2}}}y^{(2)}.\quad (E)}
Lygybėje (E) dešinė pusė yra neigiama. Todėl funkcijos
x
2
+
1
2
y
(
3
)
{\displaystyle x^{2+{\frac {1}{2}}}y^{(3)}}
x
2
+
1
2
y
(
3
)
{\displaystyle x^{2+{\frac {1}{2}}}y^{(3)}}
(
0
;
π
2
)
{\displaystyle (0;\;{\frac {\pi }{2}})}
y
(
3
)
<
0.
{\displaystyle y^{(3)}<0.}
Toliau pagal (8.113 for tg(x)) formulę, kai n=3, gauname (kai
π
2
>
x
>
0
{\displaystyle {\frac {\pi }{2}}>x>0}
[
x
3
+
1
2
y
(
3
+
1
)
]
′
=
−
1
4
x
3
−
1
2
y
(
3
)
,
{\displaystyle \left[x^{3+{\frac {1}{2}}}y^{(3+1)}\right]'=-{\frac {1}{4}}x^{3-{\frac {1}{2}}}y^{(3)},}
[
x
3
+
1
2
y
(
4
)
]
′
=
−
1
4
x
3
−
1
2
y
(
3
)
.
(
F
)
{\displaystyle \left[x^{3+{\frac {1}{2}}}y^{(4)}\right]'=-{\frac {1}{4}}x^{3-{\frac {1}{2}}}y^{(3)}.\quad (F)}
Lygybėje (F) dešinė pusė yra teigiama. Todėl funkcijos
x
3
+
1
2
y
(
4
)
{\displaystyle x^{3+{\frac {1}{2}}}y^{(4)}}
x
3
+
1
2
y
(
4
)
{\displaystyle x^{3+{\frac {1}{2}}}y^{(4)}}
(
0
;
π
2
)
{\displaystyle (0;\;{\frac {\pi }{2}})}
y
(
4
)
>
0.
{\displaystyle y^{(4)}>0.}
Taigi, indukcijos metodu gavome, kad išvestinių ženklai kaitoliojasi (
y
(
0
)
>
0
,
{\displaystyle y^{(0)}>0,\;}
y
(
1
)
<
0
,
{\displaystyle y^{(1)}<0,\;}
y
(
2
)
>
0
,
{\displaystyle y^{(2)}>0,\;}
y
(
3
)
<
0
,
{\displaystyle y^{(3)}<0,\;}
y
(
4
)
>
0
{\displaystyle y^{(4)}>0}
0
<
x
<
π
2
.
{\displaystyle 0<x<{\frac {\pi }{2}}.}
u
n
+
2
=
y
(
n
+
2
)
y
(
n
+
1
)
,
{\displaystyle u_{n+2}={\frac {y^{(n+2)}}{y^{(n+1)}}},}
u
n
+
2
<
0
{\displaystyle u_{n+2}<0\;}
0
<
x
<
π
2
{\displaystyle 0<x<{\frac {\pi }{2}}}
![{\displaystyle P_{k}Q_{k-1}-Q_{k}P_{k-1}=(b_{k}P_{k-1}+a_{k}P_{k-2})Q_{k-1}-(b_{k}Q_{k-1}+a_{k}Q_{k-2})P_{k-1}=-a_{k}[P_{k-1}Q_{k-2}-Q_{k-1}P_{k-2}].\quad (8.99)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac3219a89491f05de06116e6f4174abd923425ae)
![{\displaystyle {\frac {P_{k}}{Q_{k}}}-{\frac {P_{k-1}}{Q_{k-1}}}={\frac {-a_{k}[P_{k-1}Q_{k-2}-Q_{k-1}P_{k-2}]}{Q_{k-1}Q_{k}}}={\frac {-a_{k}(-a_{k-1})[P_{k-2}Q_{k-3}-Q_{k-2}P_{k-3}]}{Q_{k-1}Q_{k}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f815ab080c258b8e9e05393e8b6e080052b1cab)
![{\displaystyle ={\frac {-a_{k}(-a_{k-1})(-a_{k-2})[P_{k-3}Q_{k-4}-Q_{k-3}P_{k-4}]}{Q_{k-1}Q_{k}}}=(-1)^{k+1}a_{k}a_{k-1}a_{k-2}\;...\;a_{1}{\frac {1}{Q_{k-1}Q_{k}}}.\quad (8.100)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0104a39dc8d3af64ea7ffa84160cb8adbda80dee)
![{\displaystyle P_{k}Q_{k-1}-Q_{k}P_{k-1}=-a_{k}[P_{k-1}Q_{k-2}-Q_{k-1}P_{k-2}].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/467e1a1bba66dac2ba4812231b608748c675e967)
![{\displaystyle P_{1}Q_{0}-Q_{1}P_{0}=-a_{1}[P_{0}Q_{-1}-Q_{0}P_{-1}]=-a_{1}[b_{0}\cdot 0-1\cdot 1]=a_{1}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68086114f944f5facb373f429295319b7bdad080)
![{\displaystyle \operatorname {th} x-{\overline {\operatorname {th} }}x=(-1)^{n+2}x^{2n+1}\left[{\frac {1}{Q_{n}Q_{n+1}}}-{\frac {1}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33f0ae703f33550bb93dd5c1f6ac2898fbf5f278)
![{\displaystyle \operatorname {th} x-{\overline {\operatorname {th} }}x=(-1)^{n+2}x^{2n+1}\left[{\frac {1}{{\overline {Q}}_{n}Q_{n+1}}}-{\frac {1}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\right]=(-1)^{n+2}x^{2n+1}\left[{\frac {{\overline {Q}}_{n}{\overline {Q}}_{n+1}-{\overline {Q}}_{n}Q_{n+1}}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\right]=(-1)^{n+2}x^{2n+1}\left[{\frac {{\overline {Q}}_{n+1}-Q_{n+1}}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n+1}}}\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9616b9473dec206b8030381e77ba8cbdeec1b77)
![{\displaystyle =(-1)^{n+1}x^{2n+1}\left[{\frac {Q_{n+1}-{\overline {Q}}_{n+1}}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n+1}}}\right]=(-1)^{n+1}x^{2n+1}\left[{\frac {[(2n+1+2x^{2}u_{n+2}){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1}]-[(2n+1){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1}]}{{\overline {Q}}_{n}Q_{n+1}{\overline {Q}}_{n+1}}}\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/598e04a1a401dd7b3acb48b8b6dad019b4463be6)
![{\displaystyle =(-1)^{n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{Q_{n+1}}}\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/898bb7e41122439384b4188f4ea4722e6d0510d8)
![{\displaystyle =(-1)^{n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}+x^{2}{\overline {Q}}_{n-1}}}\right].\quad (8.108)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/156c2d60fe0f96073ce3c79c50706b0d8a3ec51a)
![{\displaystyle {\overline {Q}}_{n}{\overline {Q}}_{n+1}\geq [(2n-1)!!]^{2}(2n+1).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc523f78634f66eccea987970c1a063b5ad8b88f)
![{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}.\quad (8.111)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c74d25f8ee5e8c8ebc2cabdcc9dea420be60c02)
![{\displaystyle {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {0.0549755813888}{10395^{2}\cdot 13}}={\frac {0.0549755813888}{108056025\cdot 13}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33f51e56d4a02374fa71168e3211df1025eba318)
![{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'={\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)}.\quad (8.113)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b95fbe2c735e0ad5263252342a85f4ffc44138aa)
![{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'=(n+{\frac {1}{2}})x^{n-{\frac {1}{2}}}y^{(n+1)}+x^{n+{\frac {1}{2}}}y^{(n+2)}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17ef84383f0c151882855c5a299b8e86f13ca84c)
![{\displaystyle \lim _{x\to 0+0}\left[x^{n+{\frac {1}{2}}}y^{(n+1)}(x)\right]=0.\quad (8.114)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bc2346b0b846f25b59ca66a8c73e8d2b47085cb)
![{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {0.5^{2\cdot 4+1}}{[(2\cdot 4-1)!!]^{2}(2\cdot 4+1)}}={\frac {0.5^{9}}{[7!!]^{2}\cdot 9}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2099c7b8daefee06558e8c311a72aa6c8deca003)
![{\displaystyle ={\frac {0.5^{9}}{[7\cdot 5\cdot 3]^{2}\cdot 9}}={\frac {0.001953125}{105^{2}\cdot 9}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/783938e62635195e4c3b257dcd4700fb270e4ff2)
![{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {0.5^{2\cdot 2+1}}{[(2\cdot 2-1)!!]^{2}(2\cdot 2+1)}}={\frac {0.5^{5}}{[3!!]^{2}\cdot 5}}={\frac {0.5^{5}}{3^{2}\cdot 5}}={\frac {0.03125}{9\cdot 5}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/397ae2183b1ecfc9ff645ceb2fc7621485a821a0)
![{\displaystyle \operatorname {tg} x-{\overline {\operatorname {tg} }}x=(-1)^{n}(-1)^{n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{Q_{n+1}}}\right]=(-1)^{2n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right].\;\;(8.108\;\;{\text{for}}\;\;{\text{tg}}\;x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33f4a5db1680f9a9065be8b72a68a4fd7013b7b2)
![{\displaystyle \operatorname {tg} x-{\overline {\operatorname {tg} }}x=(-1)^{2n+1}{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4814e53f66725b1a2b41ae5b30175ccfc23f32cb)
![{\displaystyle =-{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right].\;\;(8.108\;\;{\text{for}}\;\;{\text{tg}}\;x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4aad1db54442e5e95fa5658b8f0534750da1af44)
![{\displaystyle \operatorname {tg} x-{\overline {\operatorname {tg} }}x=-{\frac {x^{2n+1}}{{\overline {Q}}_{n}{\overline {Q}}_{n+1}}}\left[{\frac {2x^{2}u_{n+2}{\overline {Q}}_{n}}{2x^{2}u_{n+2}{\overline {Q}}_{n}+(2n+1){\overline {Q}}_{n}-x^{2}{\overline {Q}}_{n-1}}}\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1409cfbcae46c1aac4435c2105bee35384eb807)
![{\displaystyle =-{\frac {1^{2\cdot 6+1}}{{\overline {Q}}_{6}{\overline {Q}}_{6+1}}}\left[{\frac {2\cdot 1^{2}u_{n+2}{\overline {Q}}_{6}}{2\cdot 1^{2}u_{n+2}{\overline {Q}}_{6}+(2\cdot 6+1){\overline {Q}}_{6}-1^{2}{\overline {Q}}_{6-1}}}\right]=-{\frac {1^{13}}{{\overline {Q}}_{6}{\overline {Q}}_{7}}}\left[{\frac {2u_{n+2}{\overline {Q}}_{6}}{2u_{n+2}{\overline {Q}}_{6}+13{\overline {Q}}_{6}-{\overline {Q}}_{5}}}\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ee8f6675582f0cf2fbb5fe05073903036d02617)
![{\displaystyle =-{\frac {1^{13}}{5879\cdot 75887}}\left[{\frac {2u_{n+2}\cdot 5879}{2u_{n+2}\cdot 5879+13\cdot 5879-540}}\right]=-{\frac {1}{446139673}}\left[{\frac {2u_{n+2}\cdot 5879}{2u_{n+2}\cdot 5879+75887}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbd6b0c6bc7bee5fce92b96dd178eb64ff3035e6)
![{\displaystyle |\operatorname {th} x-{\overline {\operatorname {th} }}x|\leq {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}\quad (8.111)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/032eabc7045befbd41082d78d565128ce831f288)
![{\displaystyle {\frac {x^{2n+1}}{[(2n-1)!!]^{2}(2n+1)}}={\frac {x^{13}}{[11!!]^{2}(2n+1)}}={\frac {1^{13}}{10395^{2}\cdot 13}}={\frac {1}{108056025\cdot 13}}={\frac {1}{1404728325}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd4d4c78aaa2d36db97e98005ad813c2d129ee04)
![{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'=-{\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)}.\quad (8.113\;\;{\text{for}}\;\;\operatorname {tg} x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cf511a09bf134c652701347207f5a07af2318f8)
![{\displaystyle \left[x^{n+{\frac {1}{2}}}y^{(n+1)}\right]'=-{\frac {1}{4}}x^{n-{\frac {1}{2}}}y^{(n)};\quad (8.113\;\;{\text{for}}\;\;\operatorname {tg} x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb36428474daeef6b3b762c58e1ef03910e4c43d)
![{\displaystyle \left[x^{0+{\frac {1}{2}}}y^{(0+1)}\right]'=-{\frac {1}{4}}x^{0-{\frac {1}{2}}}y^{(0)},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f282dae6ffabe9b43f074c34bd8121768412e9f5)
![{\displaystyle \left[x^{\frac {1}{2}}y^{(1)}\right]'=-{\frac {1}{4}}x^{-{\frac {1}{2}}}\cos {\sqrt {x}}.\quad (C)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69af3a6439c8e17e32da0806eaebfd751d872389)
![{\displaystyle \left[x^{1+{\frac {1}{2}}}y^{(1+1)}\right]'=-{\frac {1}{4}}x^{1-{\frac {1}{2}}}y^{(1)},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2b6402d6102ef047c15c7a6fff6bf32501b9538)
![{\displaystyle \left[x^{1+{\frac {1}{2}}}y^{(2)}\right]'=-{\frac {1}{4}}x^{1-{\frac {1}{2}}}y^{(1)}.\quad (D)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ffe96feba04724fb3d91e9cd46490bfae617a05)
![{\displaystyle \left[x^{2+{\frac {1}{2}}}y^{(2+1)}\right]'=-{\frac {1}{4}}x^{2-{\frac {1}{2}}}y^{(2)},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec3e4791ac2a1656719806e568f1c940d4e196f2)
![{\displaystyle \left[x^{2+{\frac {1}{2}}}y^{(3)}\right]'=-{\frac {1}{4}}x^{2-{\frac {1}{2}}}y^{(2)}.\quad (E)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc4bf53ca858125e4d1a4bedc27e2fc724a85457)
![{\displaystyle \left[x^{3+{\frac {1}{2}}}y^{(3+1)}\right]'=-{\frac {1}{4}}x^{3-{\frac {1}{2}}}y^{(3)},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4da0ffa74257b8b1115b8f2b4e7a96325a28e2ff)
![{\displaystyle \left[x^{3+{\frac {1}{2}}}y^{(4)}\right]'=-{\frac {1}{4}}x^{3-{\frac {1}{2}}}y^{(3)}.\quad (F)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b80a8d7f40ba56558e2dff12f214c5674dd510dc)
