Matematika/Makloreno eilutės

Išvardinti kelių svarbių Makloreno eilučių išplėtimai.

Eksponentinė funkcija:

$\mathrm {e} ^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots {\text{ for all }}x.$

$\mathrm {e} ^{ix}=\sum _{n=0}^{\infty }{\frac {(ix)^{n}}{n!}}=1+ix+{\frac {(ix)^{2}}{2!}}+{\frac {(ix)^{3}}{3!}}+{\frac {(ix)^{4}}{4!}}+{\frac {(ix)^{5}}{5!}}+{\frac {(ix)^{6}}{6!}}+{\frac {(ix)^{7}}{7!}}+{\frac {(ix)^{7}}{7!}}\cdots =$

$=1+ix-{\frac {x^{2}}{2!}}-{\frac {ix^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {ix^{5}}{5!}}-{\frac {x^{6}}{6!}}-{\frac {ix^{7}}{7!}}\cdots =\cos x+i\sin x.$

Natūrinis logaritmas:

$\ln(1-x)=-(x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+{\frac {x^{6}}{6}}+\cdots )=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}{\text{ for }}-1\leq x<1.$

$\ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-{\frac {x^{6}}{6}}+...+(-1)^{n}{\frac {x^{n}}{n}}+...=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}{\text{ for }}-1<x\leq 1.$

• Pavyzdžiui, kai x=0.1:

$\ln(1+0.1)=\ln 1.1=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {0.1^{n}}{n}}=$

$=0.1-{\frac {0.1^{2}}{2}}+{\frac {0.1^{3}}{3}}-{\frac {0.1^{4}}{4}}+{\frac {0.1^{5}}{5}}-{\frac {0.1^{6}}{6}}+{\frac {0.1^{7}}{7}}-{\frac {0.1^{8}}{8}}+{\frac {0.1^{9}}{9}}=$

=0.09531017981349206349206349206349.

$\ln |1.1|=$

=0.09531017980432486004395212328077.

$\ln(x)=(x-1)-{\frac {(x-1)^{2}}{2}}+{\frac {(x-1)^{3}}{3}}-{\frac {(x-1)^{4}}{4}}\cdots ,\;{\text{kai}}\;0<x\leq 2.$

$\ln x=2\cdot (({\frac {x-1}{x+1}})+{\frac {1}{3}}\cdot ({\frac {x-1}{x+1}})^{3}+{\frac {1}{5}}\cdot ({\frac {x-1}{x+1}})^{5}+\cdots ),\;kai\;x>0.$

• Pavyzdžiui, kai x=50:

$\ln x=2\cdot (({\frac {50-1}{50+1}})+{\frac {1}{3}}\cdot ({\frac {50-1}{50+1}})^{3}+{\frac {1}{5}}\cdot ({\frac {50-1}{50+1}})^{5}+{\frac {1}{7}}\cdot ({\frac {50-1}{50+1}})^{7}+{\frac {1}{9}}\cdot ({\frac {50-1}{50+1}})^{9}+{\frac {1}{11}}\cdot ({\frac {50-1}{50+1}})^{11}+{\frac {1}{13}}\cdot ({\frac {50-1}{50+1}})^{13}+{\frac {1}{15}}\cdot ({\frac {50-1}{50+1}})^{15})=$

$=2\cdot (0.960784313+0.295635414+0.163741783+0.107965074+0.07751587+0.058545329+0.045729178+0.036584514)=2\cdot 1.746501478=3.493002957.$

$\ln x=2\cdot (({\frac {50-1}{50+1}})+{\frac {1}{3}}\cdot ({\frac {50-1}{50+1}})^{3}+{\frac {1}{5}}\cdot ({\frac {50-1}{50+1}})^{5}+{\frac {1}{7}}\cdot ({\frac {50-1}{50+1}})^{7}+{\frac {1}{9}}\cdot ({\frac {50-1}{50+1}})^{9}+{\frac {1}{11}}\cdot ({\frac {50-1}{50+1}})^{11}+{\frac {1}{13}}\cdot ({\frac {50-1}{50+1}})^{13}+{\frac {1}{15}}\cdot ({\frac {50-1}{50+1}})^{15}+$

$+{\frac {1}{17}}\cdot ({\frac {50-1}{50+1}})^{17}+{\frac {1}{19}}\cdot ({\frac {50-1}{50+1}})^{19}+{\frac {1}{21}}\cdot ({\frac {50-1}{50+1}})^{21}+{\frac {1}{23}}\cdot ({\frac {50-1}{50+1}})^{23}+{\frac {1}{25}}\cdot ({\frac {50-1}{50+1}})^{25})=3.600006127.$

$\ln 50=2\cdot (({\frac {50-1}{50+1}})+{\frac {1}{3}}\cdot ({\frac {50-1}{50+1}})^{3}+{\frac {1}{5}}\cdot ({\frac {50-1}{50+1}})^{5}+{\frac {1}{7}}\cdot ({\frac {50-1}{50+1}})^{7}+{\frac {1}{9}}\cdot ({\frac {50-1}{50+1}})^{9}+{\frac {1}{11}}\cdot ({\frac {50-1}{50+1}})^{11}+{\frac {1}{13}}\cdot ({\frac {50-1}{50+1}})^{13}+{\frac {1}{15}}\cdot ({\frac {50-1}{50+1}})^{15}+$

$+{\frac {1}{17}}\cdot ({\frac {50-1}{50+1}})^{17}+{\frac {1}{19}}\cdot ({\frac {50-1}{50+1}})^{19}+{\frac {1}{21}}\cdot ({\frac {50-1}{50+1}})^{21}+{\frac {1}{23}}\cdot ({\frac {50-1}{50+1}})^{23}+{\frac {1}{25}}\cdot ({\frac {50-1}{50+1}})^{25}+{\frac {1}{27}}\cdot ({\frac {50-1}{50+1}})^{27}+{\frac {1}{29}}\cdot ({\frac {50-1}{50+1}})^{29}+{\frac {1}{31}}\cdot ({\frac {50-1}{50+1}})^{31}+$

$+{\frac {1}{33}}\cdot ({\frac {50-1}{50+1}})^{33}+{\frac {1}{35}}\cdot ({\frac {50-1}{50+1}})^{35}+{\frac {1}{37}}\cdot ({\frac {50-1}{50+1}})^{37}+{\frac {1}{39}}\cdot ({\frac {50-1}{50+1}})^{39}+{\frac {1}{41}}\cdot ({\frac {50-1}{50+1}})^{41})=3.664129777.$

$\ln 50=3.664129777+2\cdot ({\frac {1}{43}}\cdot ({\frac {50-1}{50+1}})^{43}+{\frac {1}{45}}\cdot ({\frac {50-1}{50+1}})^{45}+{\frac {1}{47}}\cdot ({\frac {50-1}{50+1}})^{47}+{\frac {1}{49}}\cdot ({\frac {50-1}{50+1}})^{49}+{\frac {1}{51}}\cdot ({\frac {50-1}{50+1}})^{51}+$

$+{\frac {1}{53}}\cdot ({\frac {50-1}{50+1}})^{53}+{\frac {1}{55}}\cdot ({\frac {50-1}{50+1}})^{55}+{\frac {1}{57}}\cdot ({\frac {50-1}{50+1}})^{57}+{\frac {1}{59}}\cdot ({\frac {50-1}{50+1}})^{59}+{\frac {1}{61}}\cdot ({\frac {50-1}{50+1}})^{61})=3.664129777+0.051209316=3.715339094.$

$\ln 50=3.715339094+2\cdot ({\frac {1}{63}}\cdot ({\frac {50-1}{50+1}})^{63}+{\frac {1}{65}}\cdot ({\frac {50-1}{50+1}})^{65}+{\frac {1}{67}}\cdot ({\frac {50-1}{50+1}})^{67}+{\frac {1}{69}}\cdot ({\frac {50-1}{50+1}})^{69}+{\frac {1}{71}}\cdot ({\frac {50-1}{50+1}})^{71}+$

$+{\frac {1}{73}}\cdot ({\frac {50-1}{50+1}})^{73}+{\frac {1}{75}}\cdot ({\frac {50-1}{50+1}})^{75}+{\frac {1}{77}}\cdot ({\frac {50-1}{50+1}})^{77}+{\frac {1}{79}}\cdot ({\frac {50-1}{50+1}})^{79}+{\frac {1}{81}}\cdot ({\frac {50-1}{50+1}})^{81})=3.715339094+0.016400581=3.731739676.$

Tuo tarpu skaičiuotuvo reikšmė yra $\ln(50)=3.912023005.$

Į šią formulę

$\ln x=2\cdot ({\frac {x-1}{x+1}}+{\frac {1}{3}}\cdot ({\frac {x-1}{x+1}})^{3}+{\frac {1}{5}}\cdot ({\frac {x-1}{x+1}})^{5}+\cdots ),\;kai\;x>0.$

įrašę ${\frac {1+x}{1-x}}$ vietoje x, gauname, kad

${\frac {x-1}{x+1}}={\frac {{\frac {1+x}{1-x}}-1}{{\frac {1+x}{1-x}}+1}}={\frac {\frac {1+x-(1-x)}{1-x}}{\frac {1+x+(1-x)}{1-x}}}={\frac {\frac {2x}{1-x}}{\frac {2}{1-x}}}={\frac {2x}{2}}=x.$

Tada

$\ln {\frac {1+x}{1-x}}=2\cdot (x+{\frac {1}{3}}\cdot x^{3}+{\frac {1}{5}}\cdot x^{5}+...+{\frac {x^{2n+1}}{2n+1}}+...),\;{\text{kai}}\;\;x>0.$

Ne begalinės geometrinės eilutės:

${\frac {1-x^{m+1}}{1-x}}=\sum _{n=0}^{m}x^{n}\quad {\mbox{ for }}x\not =1{\text{ and }}m\in \mathbb {N} _{0}\!$

Begalinės geometrinės eilutės:

${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}{\text{ for }}|x|<1\!$

Variantai begalinių geometrinių eilučių:

${\frac {x^{m}}{1-x}}=\sum _{n=m}^{\infty }x^{n}\quad {\mbox{ for }}|x|<1{\text{ and }}m\in \mathbb {N} _{0}\!$

${\frac {x}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n}\quad {\text{ for }}|x|<1\!$

Šaknis:

${\sqrt {1+x}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)(n!)^{2}(4^{n})}}x^{n}=1+\textstyle {\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+\dots {\text{ for }}-1<x\leq 1$

Binomo eilutė (įskaitant šaknį alfai α = 1/2 ir begalinės geometrinės eilutės alfai α = −1):

$(1+x)^{\alpha }=\sum _{n=0}^{\infty }{\alpha \choose n}x^{n}\quad {\mbox{ for all }}|x|<1{\text{ and all complex }}\alpha \!$

su apibendrintais binominiais koeficientais

${\alpha \choose n}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}$

Trigonometrinės funkcijos:

$\sin x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots {\text{ for all }}x\!$

$\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots {\text{ for all }}x\!$

$\tan x=\sum _{n=1}^{\infty }{\frac {B_{2n}(-4)^{n}(1-4^{n})}{(2n)!}}x^{2n-1}=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+\cdots {\text{ for }}|x|<{\frac {\pi }{2}}\!$

where the B_s are Bernoulli numbers.

$\sec x={1 \over \cos x}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}{\text{ for }}|x|<{\frac {\pi }{2}}\!$

$\arcsin x=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}{\text{ for }}|x|\leq 1\!$

$\arccos x={\pi \over 2}-\arcsin x={\pi \over 2}-\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}{\text{ for }}|x|\leq 1\!$

$\arctan x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}{\text{ for }}|x|\leq 1\!$

Hiperbolinės funkcijos:

$\sinh x=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{(2n+1)!}}=x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+\cdots {\text{ for all }}x\!$

$\cosh x=\sum _{n=0}^{\infty }{\frac {x^{2n}}{(2n)!}}=1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+\cdots {\text{ for all }}x\!$

$\tanh x=\sum _{n=1}^{\infty }{\frac {B_{2n}4^{n}(4^{n}-1)}{(2n)!}}x^{2n-1}=x-{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}-{\frac {17}{315}}x^{7}+\cdots {\text{ for }}|x|<{\frac {\pi }{2}}\!$

$\mathrm {arsinh} (x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}{\text{ for }}|x|\leq 1\!$

$\mathrm {artanh} (x)=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{2n+1}}{\text{ for }}|x|<1\!$

Lamberto W funkcija:

$W_{0}(x)=\sum _{n=1}^{\infty }{\frac {(-n)^{n-1}}{n!}}x^{n}{\text{ for }}|x|<{\frac {1}{\mathrm {e} }}\!$

Skaičiai B_k esantis sudeties išpletime tan(x) ir tanh(x) yra Bernulio skaičiai. E_k išpletime sec(x) yra Eulerio skaičiai.

Įrodymas per Pitagoro teoremą

Bus per Pitagoro teoremą įrodyta, kad sinuso, kosinuso eilutės ir skaičiuotuvo reikšmės yra teisingos.

Iš pradžiu, paliginsime kalkuliatoriaus reikšme, tam tikram kampui k su $\sin k$ Tailoro eilutės rezultatu. Tarkim, kampas k=60 laipsnių arba $k=1,047197551$ radianų. Tuomet kalkuliatoriaus reikšmė:

$\sin k=\sin 1.047197551={{\sqrt {3}} \over 2}=0.866025403.$

$\sin k=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}k^{2n+1}=k-{\frac {k^{3}}{3!}}+{\frac {k^{5}}{5!}}-{\frac {k^{7}}{7!}}+{\frac {k^{9}}{9!}}-...=$

$=1.047197551-{\frac {1.047197551^{3}}{3!}}+{\frac {1.047197551^{5}}{5!}}-{\frac {1.047197551^{7}}{7!}}+{\frac {1.047197551^{9}}{9!}}-{\frac {1.047197551^{11}}{11!}}=$

$=0.866025403.$

Net po salyginai trumpos eilutės atsakymo tikslumas gavosi =>9 skaičiai po kablelio.

Kai kampas k=1 radianas, tada $\sin k=\sin 1=0,841470984$.

$\sin k=\sin 1\approx 1-{\frac {1^{3}}{3!}}+{\frac {1^{5}}{5!}}-{\frac {1^{7}}{7!}}=1-{1 \over 6}+{1 \over 120}-{1 \over 5040}=0.841468254.$

$\sin k=\sin 1=1-{\frac {1^{3}}{3!}}+{\frac {1^{5}}{5!}}-{\frac {1^{7}}{7!}}+{\frac {1^{9}}{9!}}=1-{1 \over 6}+{1 \over 120}-{1 \over 5040}+{\frac {1}{362880}}=0.841471009.$

$\sin 1=1-{\frac {1^{3}}{3!}}+{\frac {1^{5}}{5!}}-{\frac {1^{7}}{7!}}+{\frac {1^{9}}{9!}}-{\frac {1^{11}}{11!}}=1-{1 \over 6}+{1 \over 120}-{1 \over 5040}+{\frac {1}{362880}}-{1 \over 39916800}=0.841470984.$

$\cos k=\cos 1=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}k^{2n}=1-{\frac {k^{2}}{2!}}+{\frac {k^{4}}{4!}}-\cdots =1-{\frac {1^{2}}{2!}}+{\frac {1^{4}}{4!}}-{\frac {1^{6}}{6!}}+{\frac {1^{8}}{8!}}=$

$=1-{1 \over 2}+{1 \over 24}-{1 \over 720}+{1 \over 40320}=0.540302579.$ Kalkuliatoriaus reikšmė: $\cos 1=0.540302305$.

Tariam, kad $\sin k=b$, o $\cos k=1-a$. Kampas k šioje užduotyje yra lygus 1 radianui (k=1). Tiesės a reikšmės gali būti nuo 0 iki 1. Tiesės b reikšmės gali būti nuo 0 iki 1.

Dabar toliau labai svarbu apčiupti bet kuriuos taškus ant apskritimo, kurio spindulys r=1, bet, kad tie taškai sujungtų daug trumpų tiesių taip, kad tos sujungtos tiesės labai primintų apskritimo lanko formą ir kad kiekviena tiesė nebūtų ilgesnė 3 kartus už betkurią kitą tiesę, kuri jungia bet kuriuos 2 taškus. Taigi, pradedame rinkti taškus ant Ox ašies: $x_{1}=1$; $x_{2}=0.9$; $x_{3}=0.8$; $x_{4}=0.7$; $x_{5}=0.6$; $x_{6}=0.5$; $x_{7}=0.4$; $x_{8}=0.3$; koordinatės $x_{9}$ gali ir neprireikti, nes $\cos 1$ negali būti labai maža reikšmė. Kiekvieno taško koordinatės ant apskritimo lanko bus užrašytos šitaip: $(x_{1};y_{1})$, $(x_{2};y_{2})$, $(x_{3};y_{3}),$ $(x_{4};y_{4}),$ $(x_{5};y_{5}),$ $(x_{6};y_{6}),$ $(x_{7};y_{7}).$ Dalį koordinačių jau galima užrašyti dabar: $(1;y_{1}),$ $(0,9;y_{2}),$ $(0,8;y_{3}),$ $(0,7;y_{4}),$ $(0,6;y_{5}),$ $(0,5;y_{6})$, $(0,4;y_{7})$. Likusią dalį koordinačių gausime pasinaudoję Pitagoro teorema:

$y_{1}={\sqrt {r^{2}-x_{1}^{2}}}={\sqrt {1^{2}-1^{2}}}=0;$

$y_{2}={\sqrt {r^{2}-x_{2}^{2}}}={\sqrt {1^{2}-0.9^{2}}}={\sqrt {1-0.81}}={\sqrt {0.19}}=0.435889894;$

$y_{3}={\sqrt {r^{2}-x_{3}^{2}}}={\sqrt {1^{2}-0.8^{2}}}={\sqrt {0.36}}=0.6;$

$y_{4}={\sqrt {r^{2}-x_{4}^{2}}}={\sqrt {1^{2}-0.7^{2}}}={\sqrt {0.51}}=0.714142842;$

$y_{5}={\sqrt {r^{2}-x_{5}^{2}}}={\sqrt {1^{2}-0.6^{2}}}={\sqrt {0.64}}=0.8;$

$y_{6}={\sqrt {r^{2}-x_{6}^{2}}}={\sqrt {1^{2}-0.5^{2}}}={\sqrt {0.75}}=0.866025403;$

$y_{7}={\sqrt {r^{2}-x_{7}^{2}}}={\sqrt {1^{2}-0.4^{2}}}={\sqrt {0.84}}=0.916515139.$

Dabar žinome visų reikiamų taškų, esančių ant apskritimo lanko, koordinates: $(1;0),$ $(0,9;0,435889894),$ $(0,8;0.6),$ $(0,7;0.714142842),$ $(0,6;0.8),$ $(0,5;0.866025403)$, $(0,4;0.916515139)$.

Jeigu kalkuliatorius suranda teisingai $\cos k$ ir $\sin k$ reikšmes, tai sudėję visus tiesių ilgius (šių tiesių ilgiai bus surasti), kuriuos sudaro taškai turėtume gauti kampą k .

Yra žinoma, kad atstumas h nuo vieno taško $(x_{1};y_{1})$ iki kito taško $(x_{2};y_{2})$ yra randamas pagal formulę:

$h={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}.$

Todėl:

$k_{1}={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}={\sqrt {(1-0.9)^{2}+(0-{\sqrt {0.19}})^{2}}}={\sqrt {0.01+0.19}}={\sqrt {0.2}}=0.447213595;$

$k_{2}={\sqrt {(x_{2}-x_{3})^{2}+(y_{2}-y_{3})^{2}}}={\sqrt {(0.9-0.8)^{2}+({\sqrt {0.19}}-0.6)^{2}}}={\sqrt {0.1^{2}+(-0.164110105)^{2}}}=$

$={\sqrt {0.01+0.026932126}}={\sqrt {0.036932126}}=0.192177331;$

$k_{3}={\sqrt {(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}}={\sqrt {(0.8-0.7)^{2}+(0.6-{\sqrt {0.51}})^{2}}}={\sqrt {0.1^{2}+(-0.114142842)^{2}}}=$

$={\sqrt {0.023028588}}=0.151751733;$

$k_{4}={\sqrt {(x_{4}-x_{5})^{2}+(y_{4}-y_{5})^{2}}}={\sqrt {(0.7-0.6)^{2}+({\sqrt {0.51}}-0.8)^{2}}}={\sqrt {0.1^{2}+(-0.085857157)^{2}}}=$

$={\sqrt {0.017371451}}=0.131800802;$

$k_{5}={\sqrt {(x_{5}-x_{6})^{2}+(y_{5}-y_{6})^{2}}}={\sqrt {(0.6-0.5)^{2}+(0.8-{\sqrt {0.75}})^{2}}}={\sqrt {0.1^{2}+(-0.066025403)^{2}}}=$

$={\sqrt {0.014359353}}=0.119830521;$

$k_{6}={\sqrt {(x_{6}-x_{7})^{2}+(y_{6}-y_{7})^{2}}}={\sqrt {(0.5-0.4)^{2}+({\sqrt {0.75}}-{\sqrt {0.84}})^{2}}}={\sqrt {0.1^{2}+(-0.050489735)^{2}}}=$

$={\sqrt {0.012549213}}=0.112023271.$

$a_{1}=(x_{1}-x_{2})=1-0.9=0.1;$

$a_{2}=(x_{2}-x_{3})=0.9-0.8=0.1;$

$a_{3}=(x_{3}-x_{4})=0.8-0.7=0.1;$

$a_{4}=(x_{4}-x_{5})=0.7-0.6=0.1;$

$a_{5}=(x_{5}-x_{6})=0.6-0.5=0.1;$

$a_{6}=(x_{6}-x_{7})=0.5-0.4=0.1.$

Matome, kad $\cos 1=0.540302305$, bet sudėjus 5 dalis gaunama $a=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0.1+0.1+0.1+0.1+0.1=0.5$ arba sudėjus 6 dalis gaunama $a=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=0.1+0.1+0.1+0.1+0.1+0.1=0.6$ (todėl tikslingiau būtų skirstyti po 0,05 tiesę a, kad gautusi 0,55). Bet $\cos k=1-a$, todėl arba 1-0,5=0.5 arba 1-0.6=0.4. Su sinusu reikalai yra tokie $\sin 1=0.841470984$.

$b_{1}=(y_{2}-y_{1})={\sqrt {0.19}}-0=0.435889894;$

$b_{2}=(y_{3}-y_{2})=0.6-0.435889894=0.164110105;$

$b_{3}=(y_{4}-y_{3})=0.714142842-0.6=0.114142842;$

$b_{4}=(y_{5}-y_{4})=0,8-0.714142842=0.085857157;$

$b_{5}=(y_{6}-y_{5})=0.866025403-0.8=0.066025403;$

$b_{6}=(y_{7}-y_{6})=0.916515139-0.866025403=0,050489735.$

$b=b_{1}+b_{2}+b_{3}+b_{4}+b_{5}=0,866025401.$

$b_{1}+b_{2}+b_{3}+b_{4}+b_{5}+b_{6}=0,916515136.$

Matome, kad sudejus tik penkias dalis, gaunami atsakymai artimesni skaičiuotuvu gautai reikšmei nei sudėjus 6 dalis.

$k=k_{1}+k_{2}+k_{3}+k_{4}+k_{5}=0.447213595+0.192177331+0.151751733+0.131800802+0.119830521=1.042773983.$

$k_{1}+k_{2}+k_{3}+k_{4}+k_{5}+k_{6}=0.447213595+0.192177331+0.151751733+0.131800802+0.119830521+0.112023271=1.154797254.$

Štai dar kalkuliatoriumi gautos reikšmės palygintos su per Pitagoro teoremą gautomis reikšmėmis:

$a=\cos 1.042773983=0.503826018$ prieš 0,5.

$b=\sin 1.042773983=0.863805153$ prieš 0,866025401.

$a=\cos 1.154797254=0.404103996$ prieš 0,4.

$b=\sin 1.154797254=0.914713047$ prieš 0,916515136.

Nuorodos

• http://www.efunda.com/math/taylor_series/taylor_series.cfm
• http://www.sosmath.com/calculus/powser/powser01.html
• Natūrinio logoritmo išvedimas, pasinaudojant Teiloro teorema, kai $0<x\leq 2;$ puslapis 5
• http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/series/taylor.html