Power Series Expansion and Its Applications幂级数的展开及其应用Maclaurin (Maclaurin) formulaPolynomial power series can be seen as an extension of reality, so consider the function ()f x can expand into power series, you can from the function ()f x and polynomials start to solve this problem. To this end, to give here without proof the following formula.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．为此，这里不加证明地给出如下的公式．Taylor (Taylor) formula, if the function ()f x at 0x x = in a neighborhood that until the derivative of order 1n +, 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)Among 10()()n n r x x x +=-That ()n r x for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.泰勒(Taylor)公式 如果函数()f x 在0x x =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!!n n n f x f x f x f x f x x x x x x x r x n '''=+-+-++-+…,(9-5-1) 其中 (1)10()()()(1)!n n n f r x x x n ξ++=-+． 称()n r x 为拉格朗日型余项．称(9-5-1)式为泰勒公式．If so 00x =, get 2()(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.如果令00x =，就得到 2()(0)()n n f x f x x x r x =+++++…， (9-5-2)此时， (1)(1)111()()()(1)!(1)!n n n n n f f x r x x x n n ξθ+++++==++, (01θ<<)．称(9-5-2)式为马克劳林公式．Formula shows that any function ()f x as long as until the 1n +derivative, n can be equal to a polynomial and a remainder.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．We call the following power series ()2(0)(0)()(0)(0)2!!n n f 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 +items and for 1()n S x +, which ()21(0)(0)()(0)(0)2!!n n n f f S x f f x x x n +'''=++++… 我们称下列幂级数 ()2(0)(0)()(0)(0)2!!n n f f f x f f x x x n '''=+++++…… (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()n S x +，即 ()21(0)(0)()(0)(0)2!!n n n f f S x f f x x x n +'''=++++…， Then, the series (9-5-3) converges to the function ()f x the conditions 1lim ()()n n s x f x +→∞=. 那么，级数(9-5-3)收敛于函数()f x 的条件为 1lim ()()n n s x f x +→∞=． Noting Maclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between the known 1()()()n n f x S x r x +=+ ， Thus, when ()0n r x = ， There, 1()()n f x S x+= ， Vice versa. That if 1l i m ()()n n s x f x +→∞=, Units must ()0n r 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()()()n n f x S x r x +=+． 于是，当 ()0n r x =时，有1()()n f x S x +=． 反之亦然．即若1lim ()()n n s x f x +→∞=则必有()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 n n 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)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()0n r x → (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!!n n n n n f f f f x x f f x x x n n ∞='''==+++++∑…… (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 20120()n n nn 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 n n 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.那么，根据幂级数在收敛域内可逐项求导的性质，再令0x =(幂级数显然在0x =点收敛)，就容易得到 ()2012(0)(0)(0),(0),,,,,2!!n n n f f a f a f x a x a x n '''====……。

将它们代入(9-5-5)式，所得与()f x 的马克劳林展开式(9-5-4)完全相同．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 remainde r 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.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!!n n f x f x f x f x x x x x n '=+-++-……, 称为泰勒级数．。