数学专业外文翻译---幂级数的展开及其应用

1、Power 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. T

2、his section will discuss another issue, for an arbitraryfunction f (x) , can be expanded in a power series, and launched into.Whether the power seriesf ( x) as and function? The following discussion will address this issue.1 Maclaurin (Maclaurin) formulaPolynomial power series can be seen as an exte

3、nsion of reality, so consider the functionf ( x)canexpand into power series, you can from the functionf (x) and polynomials start to solve this problem. Tothis end, to give here without proof the following formula.Taylor (Taylor) formula, if the functionf ( x) at x x0in a neighborhood that until the

4、 derivativeof order n 1, then in the neighborhood of the following formula:f ( x) f ( x0 ) ( x x0 ) (x x0 ) 2 (x x0 ) nrn ( x) (9-5-1)Amongrn (x) ( x x0 )n 1Thatrn (x) for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.If so x00 , getf ( x)f (0)xx2xnrn ( x) ,(9-5-2)At this point,

5、rn 1( x)f ( n 1) ( ) xn 1f ( n 1) ( x) xn 1( 01).(n 1)!(n 1)!That (9-5-2) type formula for the Maclaurin.Formula shows that any functionf ( x) as long as until the n1derivative, n can be equal to apolynomial and a remainder.We call the following power seriesf ( x) f (0) f (0) xf (0) x2f ( n) (0) xn(

6、9-5-3)2!n!For the Maclaurin series.So, is it tof (x)for the Sum functions? If the order Maclaurin series (9-5-3) the firstn1 itemsand for Sn 1 ( x) , whichSn 1( x)f (0)f (0) xf (0) x2f ( n) (0) xn2!n!Then, the series (9-5-3) converges to the functionf ( x) the conditionslim sn 1 ( x)f (x) .nNoting M

7、aclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between the knownf ( x)Sn 1(x)rn (x)Thus, whenrn ( x)0There,f ( x)Sn 1 (x)Vice versa. That iflim sn 1 ( x)f (x) ,nUnits mustrn ( x) 0 .This indicates that the Maclaurinseries (9-5-3) to f ( x)and functionas the Maclaurin form

8、ula(9-5-2) of the remainder termrn (x)0(when n).In this way, we get a functionf (x) the power series expansion:f ( x)f ( n) (0) xnf (0)f (0) xf ( n) (0) xn . (9-5-4)n0n!n!It is the functionf ( x)the power series expression, if, the function of the power series expansion isunique. In fact, assuming t

9、he function f(x) can be expressed as power seriesf ( x)an xn a0a1x a2 x2 an xn ,(9-5-5)n0Well, according to the convergence of power series can be itemized within the nature of derivation,and then make x 0(power series apparently converges in thex0 point), it is easy to geta0f (0), a1f(0) x, a2f(0)

10、x2 , , anf ( n ) (0) xn , .2!n!Substituting them into (9-5-5) type, income andf ( x) the Maclaurin expansion of (9-5-4) identical.In summary, if the functionf(x) containszero in a range ofarbitrary order derivative,and in thisrange of Maclaurin formula in the remain der to zero as the limit (when n

11、,), then ,f(x)thecanfuctionstart forming as (9-5-4) type of power series.Power Seriesf ( x) f (x0 )f ( x0 ) ( xx0 )f ( x0 ) (xx0 ) 2 f (n) ( x0 ) ( xx0 ) n ,1!2!n!Known as the Taylor series.Second, primary function of power series expansionMaclaurin formulausingthe functionf ( x)expanded in power se

12、ries method, called the directexpansion method.Example 1Test the function f ( x)ex expanded in power series ofx .Solution becausef (n ) ( x)ex , (n1,2,3, )Thereforef (0)f (0)f(0) f (n ) (0)1,So we get the power series1 x1 x21 xn ,(9-5-6)2!n!Obviously, (9-5-6)type convergence interval(,) , As(9-5-6)w

13、hether type f (x) exis Sumfunction, that is, whether it converges tof (x)ex, but also examine remainder rn ( x) .Becausern ( x)e xxn 1( 01 )，且 xxx ,(n1)!Thereforern ( x)e xxn1exn1(n(nx,1)!1)!Noting the value of any setx , exis a fixed constant, while the series (9-5-6) is absolutely convergent, son1

14、the general when the item whennx0 , so when n, ,(n1)!theren 1exx0 ,(n 1)!From thislim rn ( x)0nThis indicates that the series (9-5-6) does converge to f ( x)ex , thereforeex1 x1 x21 xn(x).2!n!Such use of Maclaurin formula are expanded in power series method, although the procedure is clear, but oper

15、ators 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 function1, ex and sin xpower series expansion, the use of these1xknown expansion by power series of operations, we can achieve many functions of power s

16、eries expansion.This demand function of power series expansion method is called indirect expansion.Example 2Find the function f (x)cos x, x0 ,Department in the power series expansion.Solution because(sin x)cos x ,Andsin x x1x31x5 ( 1)n1x2 n 1 ，（x）3!5!(2n1)!Therefore, the power series can be itemized

17、 according to the rules of derivation can becos x 11 x31 x4 ( 1)n1x2n ,（x）2!4!(2n)!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 v

18、alue.Example 3Using the expansion to estimate arctanx the value of.Solution becausearctan14Because ofx3x5x7arctan x x5 , ( 1 x 1 ),37So there1114arctan1 4(15)37Available right end of the first n items of the series and as an approximation of convergence is very slow progression to get enough items t

19、o get more accurate estimates of. However,value.the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.幂级数的展开及其应用在上一节中，我们讨论了幂级数的收敛性，在其收敛域内，幂级数总是收敛于一个和函数对于一些简单的幂级数，还可以借助逐项求导或求积分的方法，求出这个和函数本节将要讨论另外一个问题， 对于任意一个函数f ( x) ，能否将其展开成一个幂级数，以及展开成的幂级数是否以f ( x)为和函数 ?下面的讨论将解决这一问题一、马克劳林 (Maclaurin) 公式幂级数实际上可以视为多项式的延伸

20、，因此在考虑函数f ( x) 能否展开成幂级数时，可以从函数f ( x) 与多项式的关系入手来解决这个问题为此，这里不加证明地给出如下的公式泰勒 (Taylor) 公式 如果函数f ( x) 在 xx0 的某一邻域内，有直到n1 阶的导数，则在这个邻域内有如下公式：f (x) f (x0 )f( x0 )( xx0 )f (x)( xx0 )2f (n ) ( x )x0 )nrn ( x) ,(9 5 1)00 ( x2!n!其中rn ( x)f (n 1) ( ) ( x x0 )n 1 (n 1)!称 rn ( x) 为拉格朗日型余项称(9 51)式为泰勒公式如果令 x00，就得到f (

21、 x)f (0)x x2xnrn ( x) ，(9 52)此时，rn 1 (x)f (n 1) ( ) xn 1f (n 1) ( x) xn 1 ,( 01)(n1)!(n1)!称 (95 2)式为马克劳林公式公式说明，任一函数f ( x) 只要有直到 n1阶导数，就可等于某个n 次多项式与一个余项的和我们称下列幂级数f ( x)f (0)f (0) xf(0)x2f ( n) (0)xn(9 53)2!n!为马克劳林级数那么，它是否以f (x) 为和函数呢 ?若令马克劳林级数(95 3)的前 n 1项和为Sn 1 ( x) ，即Sn 1( x)f (0)f (0) xf(0)2f ( n)

22、 (0)xn2!x，n!那么，级数 (9 5 3)收敛于函数f ( x) 的条件为lim sn 1 ( x)f (x) n注意到马克劳林公式 (9 52)与马克劳林级数 (953)的关系，可知f ( x)Sn 1(x) rn (x) 于是，当rn ( x) 0时，有f ( x)Sn 1 (x) 反之亦然即若lim sn 1 ( x)f (x)n则必有rn ( x)0这表明，马克劳林级数(953) 以 f ( x) 为和函数马克劳林公式 (952)中的余项 rn (x)0(当 n时 )这样，我们就得到了函数f (x) 的幂级数展开式：f ( x)f ( n) (0)nf (0)f (0) xf

23、(0)2f (n ) (0)n (954)xxxn0n!2!n!它就是函数f (x) 的幂级数表达式，也就是说，函数的幂级数展开式是唯一的事实上，假设函数 f ( x) 可以表示为幂级数f ( x)an xn a0 a1x a2 x2 an xn ，(955)n 0那么，根据幂级数在收敛域内可逐项求导的性质，再令x0 (幂级数显然在 x0 点收敛 )，就容易得到a0f (0), a1f (0) x, a2f (0) x2 , , anf ( n ) (0) xn , 2!n!将它们代入 (9 55)式，所得与 f ( x) 的马克劳林展开式 (95 4)完全相同综上所述， 如果函数 f ( x

24、) 在包含零的某区间内有任意阶导数，且在此区间内的马克劳林公式中的余项以零为极限(当 n时 )，那么，函数f (x) 就可展开成形如 (9 5 4)式的幂级数幂级数f ( x)f ( x0 )f ( x0 ) ( x x0 ) f (n ) ( x0 ) ( x x0 ) n ,1!n!称为泰勒级数二、初等函数的幂级数展开式利用马克劳林公式将函数f (x) 展开成幂级数的方法，称为直接展开法例 1试将函数 f ( x)ex 展开成 x 的幂级数解因为f (n ) ( x)ex ,( n1,2,3, )所以f (0)f (0)f(0) f (n ) (0)1，于是我们得到幂级数1 x1 x21 xn ，(956)2!n!显然， (95 6)式的收敛区间为(,) ，至于 (956)式是否以 f ( x)ex 为和函数， 即它是否