幂级数的展开及其应用大学毕业论文外文文献翻译及原文

毕业设计（论文）外文文献翻译文献、资料中文题目：幂级数的展开及其应用文献、资料英文题目：文献、资料来源：文献、资料发表（出版）日期：院（部）：专业：班级：姓名：学号：指导教师：翻译日期： 2017.02.14Power Series Expansion and Its ApplicationsIn the previous section, we discuss the convergence of power series, in its convergence region, the power series always converges to a function. For the simple power series, but also with itemized derivative, or quadrature methods, find this and function. This section will discuss another issue, for an arbitrary function ()x f , can be expanded in a power series, and launched into.Whether the power series ()x f as and function? The following discussion will address this issue. 1 Maclaurin (Maclaurin) formulaPolynomial power series can be seen as an extension of reality, so consider the function ()x f can expand into power series, you can from the function ()x f and polynomials start to solve this problem. To this end, to give here without proof the following formula.Taylor (Taylor) formula, if the function ()x f at 0x x = in a neighborhood that until the derivative of order 1+n , then in the neighborhood of the following formula :20000()()()()()()n n f x f x x x x x x x r x =+-+-++-+… (9-5-1)Among10()()n n r x x x +=-That ()n r x for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.If so 00x =, get2()(0)()n n f x f x x x r x=+++++…, (9-5-2) At this point,(1)(1)111()()()(1)!(1)!n n n n n f f x r x x x n n ξθ+++++==++ (01θ<<).That (9-5-2) type formula for the Maclaurin.Formula shows that any function ()f x as long as until the 1n +derivative, n can be equal to a polynomial and a remainder.We call the following power series()2(0)(0)()(0)(0)2!!n nf f f x f f x x x n '''=+++++…… (9-5-3) For the Maclaurin series.So, is it to ()f x for the Sum functions? If the order Maclaurin series (9-5-3) the first 1n + itemsand for 1()n S x +, which()21(0)(0)()(0)(0)2!!n nn f f S x f f x x x n +'''=++++…Then, the series (9-5-3) converges to the function ()f x the conditions1lim ()()n n s x f x +→∞=.Noting Maclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between theknown1()()()n n f x S x r x +=+Thus, when()0n r x =There,1()()n f x S x +=Vice versa. That if1lim ()()n n s x f x +→∞=,Units must()0n r x =.This indicates that the Maclaurin series (9-5-3) to ()f x and function as the Maclaurin formula (9-5-2) of the remainder term ()0n r x → (when n →∞).In this way, we get a function ()f x the power series expansion:()()0(0)(0)()(0)(0)!!n n n nn f f f x x f f x x n n ∞='==++++∑……. (9-5-4) It is the function ()f x 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 series20120()n n n n n f x a x a a x a x a x ∞===+++++∑……, (9-5-5)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()2012(0)(0)(0),(0),,,,,2!!n nn f f a f a f x a x a x n '''====…….Substituting them into (9-5-5) type, income and ()f x the Maclaurin expansion of (9-5-4) 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 (9-5-4) type of power series.Power Series()20000000()()()()()()()()1!2!!n n f x f x f x f x f x x x x x x x n '''=+-+-++-……,Known as the Taylor series.Second, primary function of power series expansionMaclaurin formula using the function ()f x expanded in power series method, called the direct expansion method.Example 1Test the function ()x f x e =expanded in power series of x . Solution because()()n x f x e =,(1,2,3,)n =…Therefore()(0)(0)(0)(0)1n f f f f '''====…,So we get the power series21112!!n x x x n +++++……, (9-5-6) Obviously, (9-5-6)type convergence interval (,)-∞+∞, As (9-5-6)whether type ()xf x e = is Sumfunction, that is, whether it converges to ()xf x e = , but also examine remainder ()n r x . Because1e ()(1)!xn n r x x n θ+=+ (01θ<<)，且x x x θθ≤≤,Therefore11e e ()(1)!(1)!xx n n n r x x x n n θ++=<++,Noting the value of any set x ,xe is a fixed constant, while the series (9-5-6) is absolutely convergent, so the general when the item when n →∞, 10(1)!n xn +→+ , so when n → ∞,there10(1)!n xxen +→+,From thislim ()0n n r x →∞=This indicates that the series (9-5-6) does converge to ()x f x e =, therefore21112!!x n e x x x n =+++++…… (x -∞<<+∞). Such use of Maclaurin formula are expanded in power series method, although the procedure is clear,but operators are often too Cumbersome, so it is generally more convenient to use the following power series expansion method.Prior to this, we have been a functionx-11, xe and sin x power series expansion, the use of these known expansion by power series of operations, we can achieve many functions of power series expansion. This demand function of power series expansion method is called indirect expansion .Example 2Find the function ()cos f x x =,0x =,Department in the power series expansion. Solution because(sin )cos x x '=,And3521111sin (1)3!5!(21)!n n x x x x x n +=-+-+-++……，（x -∞<<+∞）Therefore, the power series can be itemized according to the rules of derivation can be342111cos 1(1)2!4!(2)!n nx x x x n =-+-+-+……,（x -∞<<+∞） Third, the function power series expansion of the application exampleThe application of power series expansion is extensive, for example, can use it to set some numerical or other approximate calculation of integral value.Example 3 Using the expansion to estimate arctan x the value of π.Solution because πarctan14= Because of357arctan 357x x x x x =-+-+…, (11x -≤≤),So there1114arctan14(1)357π==-+-+…Available right end of the first n items of the series and as an approximation of π. However, the convergence is very slow progression to get enough items to get more accurate estimates of πvalue.此外文文献选自于：Walter.Rudin.数学分析原理(英文版)[M].北京：机械工业出版社.。