Maclaurin nokucazululwa kuka eminye imisebenzi

Izifundo zemathematika oluphambili kufuneka bazi ukuba inani lothotho uyilawulayo lokuphumla kukuhlangana inani kuthi, yinto eqhubekayo kwaye unlimited inani wambi umsebenzi enomahluko. Umbuzo ngulo: ngaba kunokwenzeka ukuphikisana ukuba anikwe engenasizathu umsebenzi f (x) - i sine ye x, apho ux enikwe kwisangqa? Oko kukuthi, phantsi kwaziphi iimeko apho eli f f-sweni (x) angamelwa apho ux enikwe kwisangqa? Ukubaluleka kwalo mba kukuba kunokwenzeka ukubuyisela malunga £ Theological f (x) yeyona udibaniso kwiikota zokuqala ezimbalwa kuthotho lwamandla, oko polynomial. Loo umsebenzi nokutshintsha ibinzana kakhulu ilula - polynomial - kulula kwaye ekucombululeni iingxaki ezithile kuhlalutyo zemathematika, ezizezi ekusombululeni integrals xa kubalwa umehluko zibalo , etc ...

Oku kubonakaliswa, ukuba ezinye f-ii f (x), apho yezabelo of the (n + 1) ukuze -th kungabalelwa, kuquka nolwazi endhawini (α - R; x ₀ + R) yencopho x = α fomula ngobulungisa:

Le fomyula ithiywe emva kokuba nzulu esidumileyo Brooke Taylor. A Inani ethatyathwe kowangaphambili, ebizwa Maclaurin series:

Umthetho eyenza kube lula ukuvelisa ukwanda kuthotho Maclaurin:

1. Misela ahlobene of yokuqala, eyesibini, eyesithathu, ... umyalelo.
2. Bala into avela kwi x = 0.
3. Irekhodi Maclaurin series lo msebenzi, yaye emva koko ukujonga lokuphumla convergence.
4. Fumana kwithuba lokuphumla (-R; R), apho inxalenye intsalela ubisi Maclaurin

R _n (x) -> 0 ngenxa n -> obungapheliyo. Ukuba ikhona, oko umsebenzi f (x) kufuneka ilingane umdibaniso Maclaurin series.

Ngoku makhe siqwalasele series Maclaurin ukuba imisebenzi ngamnye.

1. Ngoko ke, ngowokuqala ukuba f (x) = e ^x. Kakade ke, ukuba iimpawu zabo ukuze f-Ia uye wafumana iintlobo kwee-odolo, kunye f ^{(k) (x)} = e ^x, apho k ilingana onke amanani zendalo. Tshintsha x = 0. We ukufumana f ^{(k) (0)} = e ⁰ = 1, k = 1,2 ... Ngokusekelwe kule nkcazelo ingasentla, iqela e ^x Kuya kuba ngolu hlobo lulandelayo:

2. Maclaurin series ukuba umsebenzi f (x) = isono x. Ngoko nangoko ucacise ukuba f-zisa zonke yezabelo ezingaziwayo kuya kufuneka, ngaphandle f ^'(x) = cos x = nesono (x + n / 2), ^{f' '(x)} = -sin x = isono (x + 2 * n / ²⁾ ..., f ^{(k) (x)} = nesono (x + n * k / 2), apho k ilingana nayiphi kweqanda. Oko kukuthi, ukwenza izibalo ezilula, sinokugqiba ukuba uthotho ukuze f (x) = isono x kuya kuba ngolu hlobo:

3. Ngoku makhe sihlolisise iju f-f (x) = cos x. ^{f (k) (x) |} ayaziwa ukuba zonke yezabelo ucwangco ngendlela engaqondakaliyo, kwaye | = | EKosi (x + k * n / 2) | <= 1, k = 1,2 ... Kwakhona, oko bemthomalalisile ezinye izibalo, sifumanisa ukuba uthotho ukuze f (x) = cos x iza kukhangeleka ngolu hlobo:

Ngoko ke, siye kuluhlu zinto zibalulekileyo ekwayiyo kuthotho Maclaurin, kodwa ukuncedisana series Taylor kwimisebenzi ethile. Ngoku siza zibhale ngokunjalo. Kufuneka kuqatshelwe ukuba Taylor series Maclaurin series ziyinxalenye ebalulekileyo workshop uthotho lwezigqibo kwimathematika eziphezulu. Ngoko ke, Taylor series.

1. Okokuqala uthotho f-ii f (x) = ln (1 + x). Njengokuba le mizekelo zangaphambili, kuba oku f (x) = ln (1 + x) ngokuthi isongwe kusetyenziswa inombolo, usebenzisa ifomu jikelele Maclaurin series. kodwa ukuba lo msebenzi osemqoka unokuba Maclaurin zingafumaneka lula kakhulu. Ekuhlanganiseni indlela uthotho zejometri, siya ufumane inombolo ye f (x) = ln (1 + x) isampula:

2. Kwaye ke wesibini, apho kuya kubakho lokugqibela kweli nqaku, uya kuba kolandelayo kube f (x) = arctg x. Ukuze x lilungu lokuphumla [-1; 1] yi ibole esebenzayo:

Kuko konke. Kweli nqaku mna, wahlola Taylor series lusetyenziswa kunye Maclaurin series kwimathematika eziphezulu, ingakumbi iikholeji kwezoqoqosho kunye nobugcisa.