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