數(shù)學(xué)系英文論文翻譯(已改無錯字)
2022-10-21 13:57:52 本頁面
【正文】 (952) of the remainder term ( ) 0nrx? (whenn?? ). In this way, we get a function ()fx the power series expansion: ( ) ( )0( 0 ) ( 0 )( ) ( 0 ) ( 0 )!!nnnnnfff x x f f x x?? ?? ? ? ? ? ?? … …. (954) It is the function ()fx the power series expression, if, the function of the power series expansion is unique. In fact, assuming the function f(x) can be expressed as power series 20 1 20()nnnf x a x a a x a x a x??? ? ? ? ? ? ?? … …, (955) Well, according to the convergence of power series can be itemized within the nature of derivation, and then make 0x? (power series apparently converges in the 0x? point), it is easy to get ()20 1 2 ( 0 ) ( 0 )( 0 ) , ( 0 ) , , , , ,2 ! !n nnffa f a f x a x a xn???? ? ? ?… …. Substituting them into (955) type, ine and ()fx the Maclaurin expansion of (954) identical. In summary, if the function f(x) contains zero in a range of arbitrary order derivative, and in this range of Maclaurin formula in the remainder to zero as the limit (when n → ∞,), then , the function f(x) can start forming as (954) type of power series. Power Series ()20 0 00 0 0 0( ) ( ) ( )( ) ( ) ( ) ( ) ( )1 ! 2 ! !n nf x f x f xf x f x x x x x x xn? ??? ? ? ? ? ? ? ?… …, Known as the Taylor series. Second, primary function of power series expansion Maclaurin formula using the function ()fx expanded in power series method, called the direct expansion method. Example 1 Test the function ()xf x e? expanded in power series of x . Solution because ()()nxf x e? , ( 1,2,3, )n ? … Therefore ()( 0) ( 0) ( 0) ( 0) 1nf f f f? ??? ? ? ?…, So we get the power series 2111 2 ! ! nx x xn? ? ? ? ?… …, (956) Obviously, (956)type convergence interval ( , )???? , As (956)whether type ()xf x e? is Sum function, that is, whether it converges to ()xf x e? , but also examine remainder ()nrx. Because 1e() ( 1)!x nnr x xn ? ?? ? (01???)，且 xxx????, Therefore 11ee() ( 1
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