Maclaurin futhi yokubola abanye imisebenzi

Ukutadisha izibalo zezinga eliphezulu kufanele uqaphele ukuthi isamba sochungechunge lwamandla ku Isikhawu convergence eziningi kithi, iyinombolo okuqhubekayo kanye enganqunyelwe yezikhathi umsebenzi umahluko. Lo mbuzo ubuzwa: Kungenzeka ukuphikisana ngokuthi inikezwe ongumashiqela umsebenzi f (x) - yinani sochungechunge lwamandla? Okungukuthi, ngaphansi kwaziphi izimo i-f-ii- f (x) kungenziwa emelelwa uchungechunge amandla? Ukubaluleka kwalokhu inkinga ukuthi kungenzeka ukuba esikhundleni cishe £ Theological f (x) yinani imigomo embalwa yokuqala sochungechunge lwamandla, ukuthi kuyinto polynomial. umsebenzi okunjalo esikhundleni ulula Inkulumo - polynomial - kuyinto elula futhi ekuxazululeni izinkinga ezithile e izingabunjalo, okungukuthi ekuxazululeni integrals lapho kubalwa umehluko zibalo , njll ...

It is wafakazela, ukuthi kwabanye f f-ii (x), lapho amagama ahlobene we (n + 1) oda -th ingabalwa, kuhlanganise zakamuva eduze (α - R; x ₀ + R) ngephuzu x = α ifomula ngobulungisa:

Lokhu ifomula yaqanjwa usosayensi odumile Brooke Taylor. Kunamaphuzu amaningana elisuselwa sangaphambilini, ibizwa ngokuthi uchungechunge Maclaurin:

A Ukubusa eyenza ukuba kukhiqizwe ukunwetshwa ochungechungeni Maclaurin:

1. Nquma amagama ahlobene of kuqala, kwesibili, kwesithathu, ... oda.
2. Bala iziphi nemikhiqizo at x = 0.
3. Record Maclaurin uchungechunge ngoba lo msebenzi, bese ukucacisa isikhawu convergence.
4. Nquma isikhawu (-R; R), lapho ingxenye bokungasetshenziswa ifomula Maclaurin

R _n (x) -> 0 I-n -> okufanekisa ingunaphakade. Uma ikhona, ke umsebenzi f (x) kufanele alingane isamba Maclaurin chungechunge.

Manje cabangela Maclaurin uchungechunge imisebenzi ngabanye.

1. Ngakho, umuntu wokuqala ukuba f (x) = e ^x. Yiqiniso, ukuthi izici zabo ngakho f-IA uye etholakala ezihlukahlukene oda, futhi f ^{(k) (x)} = e ^x, lapho k ilingana bonke izinombolo zemvelo. Obambele x = 0. We ukuthola f ^{(k) (0)} = e ⁰ = 1, k = 1,2 ... Ngokusekelwe osekushiwo, eziningi e ^x Kuyoba ngendlela elandelayo:

2. Maclaurin uchungechunge ngenxa f umsebenzi (x) = isono x. Ngokushesha ucacise ukuthi f-ii- kuwo wonke amagama ahlobene engaziwa kuzodingeka, ngaphandle f ^'(x) = cos x = isono (x + n / 2), ^{f' '(x)} = -sin x = isono (x + 2 * n / ²⁾ ..., f ^{(k) (x)} = isono (x + n * k / 2), lapho k Unamandla okuxazulula noma isiphi nenombolo evumayo ephelele. Lokho, okwenza izibalo ezilula, singaphetha ngokuthi uchungechunge for f (x) = isono x kuyoba kanje:

3. Manje ake sicabangele iju f-f (x) = cos x. Kuyinto ongaziwa zonke nemikhiqizo oda kabi amandla akhe, futhi ^{| f (k) (x)} | = | Cos (x + k * n / 2) | <= 1, k = 1,2 ... Again, Esenze ezinye izibalo, sithola ukuthi uchungechunge for f (x) = cos x izobukeka kanje:

Ngakho, siye ohlwini izici ezibaluleke kakhulu ukuthi kungenziwa landiswa ochungechungeni Maclaurin, kodwa umphelelisi Taylor uchungechunge kwabanye imisebenzi. Manje sizokwenza zibhale futhi. Kufanele kuphawulwe ukuthi Taylor uchungechunge futhi Maclaurin uchungechunge ziyingxenye ebalulekile yochungechunge workshop sezinqumo mathematics ephakeme. Ngakho, Taylor chungechunge.

1. Esokuqala uchungechunge f-ii f (x) = ln (1 + x). Njengoba sibonile ezibonelweni langaphambilini, kulolu f thina (x) = ln (1 + x) kungenziwa eligoqiwe inombolo, ngokusebenzisa ifomu jikelele Maclaurin chungechunge. kodwa lesi sici Maclaurin ingatholakala lula kakhulu. Ukuhlanganisa uchungechunge weJiyomethri, sithola nombolo f (x) = ln (1 + x) isampula:

2. Futhi yesibili, okuzoba sokugcina kulesi sihloko, kuyoba nochungechunge ngoba f (x) = arctg x. Ukuze x okuqondene isikhawu [-1; 1] kuyinto noqhekeko evumelekile:

Yilokho. Kulesi sihloko I baye ehlola esetshenziswa kakhulu Taylor uchungechunge futhi Maclaurin uchungechunge mathematics ephakeme, ikakhulukazi colleges kwezomnotho kanye lobuchwepheshe.