数学外文翻译

1、精品文档精品文档power series expansion and its applications in 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 fu

2、nction. this section will discuss another issue, for an arbitrary function( )f x, can be expanded in a power series, and launched into. whether the power series ( )f xas and function? the following discussion will address this issue. 1 maclaurin (maclaurin) formula polynomial power series can be see

3、n as an extension of reality, so consider the function ( )f xcan expand into power series, you can from the function ( )f xand polynomials start to solve this problem. to this end, to give here without proof the following formula. taylor (taylor) formula, if the function( )f xat0 xxin a neighborhood

4、 that until the derivative of order 1n, then in the neighborhood of the following formula: 20000( )()()()()( )nnf xf xxxxxxxrx(9-5-1) among 10( )()nnrxxxthat ( )nrxfor the lagrangian remainder. that (9-5-1)-type formula for the taylor. if so00 x, get 2( )(0)( )nnf xfxxxrx, (9-5-2) at this point, (1)

5、(1)111( )()( )(1)!(1)!nnnnnffxrxxxnn(01). that (9-5-2) type formula for the maclaurin. formula shows that any function ( )f xas long as until the 1nderivative, ncan be equal to a polynomial and a remainder. we call the following power series ( )2(0)(0)( )(0)(0)2!nnfffxffxxxn(9-5-3) for the maclaurin

6、 series. so, is it to ( )f xfor the sum functions? if the order maclaurin series (9-5-3) the first 1nitems 精品文档精品文档and for1( )nsx, which ( )21(0)(0)( )(0)(0)2!nnnffsxffxxxnthen, the series (9-5-3) converges to the function ( )f xthe conditions 1lim( )( )nnsxf x. noting maclaurin formula (9-5-2) and

7、the maclaurin series (9-5-3) the relationship between the known 1( )( )( )nnfxsxrxthus, when ( )0nrxthere, 1( )( )nfxsxvice versa. that if 1lim( )( )nnsxf x, units must ( )0nrx. this indicates that the maclaurin series (9-5-3) to ( )fxand function as the maclaurin formula (9-5-2) of the remainder te

8、rm ( )0nrx(whenn). in this way, we get a function ( )f xthe power series expansion: ( )( )0(0)(0)( )(0)(0)!nnnnnfffxxffxxnn. (9-5-4) it is the function ( )f xthe power series expression, if, the function of the power series expansion is unique. in fact, assuming the function f(x) can be expressed as

9、 power series 20120( )nnnnnfxa xaa xa xa x, (9-5-5) well, according to the convergence of power series can be itemized within the nature of derivation, and then make 0 x(power series apparently converges in the 0 xpoint), it is easy to get ()2012(0)(0)(0),(0) ,2!nnnffafafx axaxn. 精品文档精品文档substitutin

10、g them into (9-5-5) type, income and ( )f xthe 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 star

11、t forming as (9-5-4) type of power series. power series ()20000000()()()( )()()()()1!2!nnfxfxfxfxf xxxxxxxn, known as the taylor series. second, primary function of power series expansion maclaurin formula using the function ( )f xexpanded in power series method, called the direct expansion method.

12、example 1 test the function( )xf xeexpanded in power series of x. solution because ( )( )nxfxe,(1,2,3,)ntherefore ( )(0)(0)(0)(0)1nffff, so we get the power series 21112!nxxxn, (9-5-6) obviously, (9-5-6)type convergence interval (,), as (9-5-6)whether type ( )xf xeis sum function, that is, whether i

13、t converges to ( )xf xe, but also examine remainder ( )nrx. because 1e( )(1)!xnnrxxn(01)，且xxx, therefore 11ee( )(1)!(1)!xxnnnrxxxnn, noting the value of any set x,xeis a fixed constant, while the series (9-5-6) is absolutely convergent, so the general when the item when n, 10(1)!nxn, so when n ,ther

14、e 精品文档精品文档10(1)!nxxen, from this lim( )0nnrxthis indicates that the series (9-5-6) does converge to( )xf xe, therefore 21112!xnexxxn(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

15、 convenient to use the following power series expansion method. prior to this, we have been a functionx11, xeand sin xpower 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

16、expansion method is called indirect expansion.example 2 find the function( )cosf xx,0 x,department in the power series expansion. solution because (sin)cosxx, and 3521111sin( 1)3!5!(21)!nnxxxxxn， （x）therefore, the power series can be itemized according to the rules of derivation can be 342111cos1( 1

17、)2!4!(2 )!nnxxxxn,（x）third, the function power series expansion of the application example the 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 estimatearctanxthe val

18、ue of. solution because arctan14because of 357arctan357xxxxx, (11x), so there 1114arctan14(1)357精品文档精品文档available right end of the first n items of the series and as an approximation of . however, the convergence is very slow progression to get enough items to get more accurate estimates of value. 此

19、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，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

20、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0 xx的某一邻域内，有直到1n阶的导数，则在这个邻域内有如下公式：( )20000000()()( )()()()()()( )2!nnnfxfxf xf xfxxxxxxxrxn,(9 5 1) 其中(1)10( )( )()(1)!nnnfrxxxn称( )nrx为拉格朗日型余项称(9 5 1)式为泰勒公式如果令00 x，就得到2( )(0)( )nnfxfxxxrx，(9 5 2) 此时，(1)(1)111( )()( )(1)!(1)!nnnnnffxrxxxnn,(01)精品文档精品文档称(9 5

21、2)式为马克劳林公式公式说明，任一函数( )f x只要有直到1n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( )2(0)(0)( )(0)(0)2!nnfffxffxxxn(9 5 3) 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(9 5 3)的前1n项和为1( )nsx，即( )21(0)(0)( )(0)(0)2!nnnffsxffxxxn，那么，级数 (9 5 3)收敛于函数( )fx的条件为1lim( )( )nnsxf x注意到马克劳林公式(9 5 2)与马克劳林级数(9 5 3)的关系，可知1( )( )( )nnfxsxrx于是，当(

22、)0nrx时，有1( )( )nfxsx反之亦然即若1lim( )( )nnsxf x则必有( )0nrx这表明，马克劳林级数(9 5 3)以( )f x为和函数马克劳林公式(9 5 2)中的余项( )0nrx(当n时)这样，我们就得到了函数( )f x的幂级数展开式：( )()20(0)(0)(0)( )(0)(0)!2!nnnnnffffxxffxxxnn(9 5 4) 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可以表示为幂级数精品文档精品文档20120( )nnnnnfxa xaa xa xa x，(9 5 5) 那么，根据幂级

23、数在收敛域内可逐项求导的性质，再令0 x(幂级数显然在0 x点收敛 )，就容易得到()2012(0)(0)(0),(0) ,2!nnnffafafx axaxn将它们代入 (9 5 5)式，所得与( )fx的马克劳林展开式(9 5 4)完全相同综上所述， 如果函数( )f x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( )00000()()( )()()()1!nnfxfxfxf xxxxxn, 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( )xf xe展开成x的幂级数解因为( )( )nxfxe,(1,2,3,)n所以( )(0)(0)(0)(0)1nffff，于是我们得到幂级数21112!nxxxn，(9 5 6) 显然， (9 5 6)式的收敛区间为(,)，至于 (9 5 6)