数学系英文论文翻译.doc

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，能否将其展开成一个幂级数，以及展开成的幂级数是否以为和函数?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，因此在考虑函数能否展开成幂级数时，可以从函数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 如果函数在的某一邻域内，有直到阶的导数，则在这个邻域内有如下公式：,(9-5-1)其中称为拉格朗日型余项称(9-5-1)式为泰勒公式如果令，就得到， (9-5-2)此时，, ()称(9-5-2)式为马克劳林公式公式说明，任一函数只要有直到阶导数，就可等于某个次多项式与一个余项的和我们称下列幂级数 (9-5-3)为马克劳林级数那么，它是否以为和函数呢?若令马克劳林级数(9-5-3)的前项和为，即，那么，级数(9-5-3)收敛于函数的条件为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，可知于是，当时，有反之亦然即若则必有这表明，马克劳林级数(9-5-3)以为和函数马克劳林公式(9-5-2)中的余项 (当时)这样，我们就得到了函数的幂级数展开式：(9-5-4)它就是函数的幂级数表达式，也就是说，函数的幂级数展开式是唯一的事实上，假设函数可以表示为幂级数， (9-5-5)那么，根据幂级数在收敛域内可逐项求导的性质，再令(幂级数显然在点收敛)，就容易得到将它们代入(9-5-5)式，所得与的马克劳林展开式(9-5-4)完全相同综上所述，如果函数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(当时)，那么，函数就可展开成形如(9-5-4)式的幂级数幂级数,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展开成幂级数的方法，称为直接展开法例1 试将函数展开成的幂级数解 因为, 所以，于是我们得到幂级数， (9-5-6)显然，(9-5-6)式的收敛区间为，至于(9-5-6)式是否以为和函数，即它是否收敛于，还要考察余项因为 ()，且，所以注意到对任一确定的值，是一个确定的常数，而级数(9-5-6)是绝对收敛的，因此其一般项当时，所以当时，有，由此可知这表明级数(9-5-6)确实收敛于，因此有 ()这种运用马克劳林公式将函数展开成幂级数的方法，虽然程序明确，但是运算往往过于繁琐，因此人们普遍采用下面的比较简便的幂级数展开法在此之前，我们已经得到了函数，及的幂级数展开式，运用这几个已知的展开式，通过幂级数的运算，可以求得许多函数的幂级数展开式这种求函数的幂级数展开式的方法称为间接展开法例2 试求函数在处的幂级数展开式解 因为，而,（），所以根据幂级数可逐项求导的法则，可得,（）三、 函数幂级数展开的应用举例幂级数展开式的应用很广泛，例如可利用它来对某些数值或定积分值等进行近似计算例3 利用的展开式估计的值解 由于，又因， ()，所以有可用右端级数的前n项之和作为的近似值但由于级数收敛的速度非常慢，要取足够多的项才能得到的较精确的估计值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. This section will discuss another issue, for an arbitrary function, can be expanded in a power series, and launched into.Whether the power series 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 can expand into power series, you can from the function and polynomials start to solve this problem. To this end, to give here without proof the following formula.Taylor (Taylor) formula, if the function at in a neighborhood that until the derivative of order , then in the neighborhood of the following formula: (9-5-1)Among That for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.If so, get , (9-5-2)At this point, ().That (9-5-2) type formula for the Maclaurin.Formula shows that any function as long as until the derivative, can be equal to a polynomial and a remainder.We call the following power series (9-5-3)For the Maclaurin series.So, is it to for the Sum functions? If the order Maclaurin series (9-5-3) the first items and for, whichThen, the series (9-5-3) converges to the function the conditions.Noting Maclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between the knownThus, whenThere,Vice versa. That if,Units must.This indicates that the Maclaurin series (9-5-3) to and function as the Maclaurin formula (9-5-2) of the remainder term (when).In this way, we get a function the power series expansion:. (9-5-4)It is the function 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, (9-5-5)Well, according to the convergence of power series can be itemized within the nature of derivation, and then make (power series apparently converges in the point), it is easy to get.Substituting them into (9-5-5) type, income and 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,Known as the Taylor series.Second, primary function of power series expansionMaclaurin formula using the function expanded in power series method, called the direct expansion method.Example 1 Test the functionexpanded in power series of .Solution because,Therefore,So we get the power series, (9-5-6)Obviously, (9-5-6)type convergence interval , As (9-5-6)whether type is Sum function, that is, whether it converges to , but also examine remainder . Because ()，且,Therefore,Noting the value of any set ,is a fixed constant, while the series (9-5-6) is absolutely convergent, so the general when the item when , , so when n , there,From thisThis indicates that the series (9-5-6) does converge to, therefore ().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 function, and 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 2 Find the function,Department in the power series expansion.Solution because,And，（）Therefore, the power series can be itemized according to the rules of derivation can be,（）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 estimatethe value of.Solution because Because of, (),So thereAvailable 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.