数学与应用数学外文+翻译(有word版).pdf

2009200920092009no 8no 8no 8no 8 principleprincipleprincipleprinciple ofofofof mathematicalmathematicalmathematicalmathematical analysisanalysisanalysisanalysis 208208208208 powerpowerpowerpower seriesseriesseriesseries expansionexpansionexpansionexpansion andandandand itsitsitsitsapplicationsapplicationsapplicationsapplications lianghui 梁惠 major in pure andapplied mathematics053 hangzhou teacther s collage 310000 abstractabstractabstractabstractby learning some basic knowledge of power series and common elementary function of power series expansion and the function of the application of power series in elementary function is discussed keywordskeywordskeywordskeywordspower series maclaurin maclaurin formula taylor formula elementary function power series is a very important content of mathematical analysis and power series has a very wide range of applications can use the power series expansion it s easy to solve some of the more complex problems the purpose of this paper is to research the power series expansion and the application of elementary function frist maclaurin maclaurin formula polynomial power series can be seen 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 xat 0 xx in a neighborhood that until the derivative of order1n then in the neighborhood of the following formula 2 0000 n n f xf xxxxxxxr x 9 5 1 among 1 0 n n r xxx that n r xfor the lagrangian remainder that 9 5 1 type formula for the taylor if so 0 0 x get 2 0 n n f xfxxxr x 9 5 2 at this point 1 1 11 1 1 1 nn nn n ffx rxxx nn 01 that 9 5 2 type formula for the maclaurin formula shows that any function f xas long as until the1n derivative ncan be equal to a polynomial and a remainder we call the following power series 2009200920092009no 8no 8no 8no 8 principleprincipleprincipleprinciple ofofofof mathematicalmathematicalmathematicalmathematical analysisanalysisanalysisanalysis 209209209209 2 0 0 0 0 2 n n ff f xffxxx n 9 5 3 for the maclaurin series so is it to f xfor the sum functions if the order maclaurin series 9 5 3 the first1n items and for 1 n sx which 2 1 0 0 0 0 2 n n n ff sxffxxx n then the series 9 5 3 converges to the function f xthe conditions 1 lim n n sxf x noting maclaurin formula 9 5 2 and the maclaurin series 9 5 3 the relationship between the known 1 nn f xsxr x thus when 0 n r x there 1 n f xsx vice versa that if 1 lim n n sxf x units must 0 n r x this indicates that the maclaurin series 9 5 3 to f xand function as the maclaurin formula 9 5 2 of the remainder term 0 n r x whenn in this way we get a function f xthe power series expansion 0 0 0 0 0 nn nn n ff f xxffxx nn 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 functionf x can be expressed as power series 2009200920092009no 8no 8no 8no 8 principleprincipleprincipleprinciple ofofofof mathematicalmathematicalmathematicalmathematical analysisanalysisanalysisanalysis 210210210210 2 012 0 nn nn n f xa 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 make0 x power series apparently converges in the0 x point it is easy to get 2 012 0 0 0 0 2 n n n ff afafx axax n substituting them into 9 5 5 type income and f xthe maclaurin expansion of 9 5 4 identical in summary if the functionf 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 functionf x can start forming as 9 5 4 type of power series power series 2 000 0000 1 2 n n fxfxfx f xf xxxxxxx n 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 example 1 test the function x f xe expanded in power series ofx solution because nx fxe 1 2 3 n therefore 0 0 0 0 1 n ffff so we get the power series 2 11 1 2 n xxx n 9 5 6 obviously 9 5 6 type convergence interval as 9 5 6 whether type x f xe is sum function that is whether it converges to x f xe but also examine remainder n r x because 1 e 1 x n n r xx n 01 且 xx x therefore 2009200920092009no 8no 8no 8no 8 principleprincipleprincipleprinciple ofofofof mathematicalmathematicalmathematicalmathematical analysisanalysisanalysisanalysis 211211211211 11ee 1 1 x x nn n r xxx nn noting the value of any setx x eis a fixed constant while the series 9 5 6 is absolutely convergent so the general when the item whenn 1 0 1 n x n so when n there 1 0 1 n x x e n from this lim 0 n n r x this indicates that the series 9 5 6 does converge to x f xe therefore 2 11 1 2 xn exxx 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 function x 1 1 x eandsinxpower 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 cosf xx 0 x department in the power series expansion solution because sin cosxx and 3521 111 sin 1 3 5 21 nn xxxxx n x therefore the power series can be itemized according to the rules of derivation can be 342 111 cos1 1 2 4 2 nn xxxx n 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 3using the expansion to estimatearctanxthe value of 2009200920092009no 8no 8no 8no 8 principleprincipleprincipleprinciple ofofofof mathematicalmathematicalmathematicalmathematical analysisanalysisanalysisanalysis 212212212212 solution because arctan1 4 because of 357 arctan 357 xxx xx 11x so there 111 4arctan14 1 357 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 referencesreferencesreferencesreferences 1 cheng chuanzhang ouyang guangzhong mathematical analysis part ii m higher education press beijing chinese 1983 180 192 2 shi jihang generating function m shanghai education publishing house shanghai chinese 1980 3 zhong yuquan theory of complex function m higher education press beijing chinese 1988 14l 一 162 2009 no 8 数学分析原理数学分析原理 数学分析原理数学分析原理 208 幂级数的展开及其应用幂级数的展开及其应用 梁慧梁慧 杭州师范大学理学院数学与应用数学数学杭州师范大学理学院数学与应用数学数学 053053 浙江杭州 浙江杭州 310000310000 摘要摘要 通过学习幂级数的一些基本知识通过学习幂级数的一些基本知识 得出常用初等函数幂级数的展开式 并且探讨函数幂级数得出常用初等函数幂级数的展开式 并且探讨函数幂级数 在在初等函数初等函数的的应用 应用 关键词关键词 幂级数 马克劳林公式 泰勒公式 初等函数幂级数 马克劳林公式 泰勒公式 初等函数 幂级数是数学分析中的 个非常重要的内容 而且幂级数的应用也非常广泛 可以借助 幂级数的展开形式 很容易的解决一些较为复杂的问题 本文旨在研究幂级数的展开形式及 其在初等函数的应用 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 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 2 00 00000 2 n n n fxfx f xf xfxxxxxxxr x n 9 51 其中 1 1 0 1 n n n f r xxx n 称 n r x为拉格朗日型余项 称 951 式为泰勒公式 如果令 0 0 x 就得到 2 0 n n f xfxxxr x 952 此时 1 1 11 1 1 1 nn nn n ffx rxxx nn 01 称 952 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 2009 no 8 数学分析原理数学分析原理 数学分析原理数学分析原理 209 2 0 0 0 0 2 n n ff f xffxxx n 953 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 953 的前1n 项和为 1 n sx 即 2 1 0 0 0 0 2 n n n ff sxffxxx n 那么 级数 953 收敛于函数 f x的条件为 1 lim n n sxf x 注意到马克劳林公式 952 与马克劳林级数 953 的关系 可知 1 nn f xsxr x 于是 当 0 n r x 时 有 1 n f xsx 反之亦然 即若 1 lim n n sxf x 则必有 0 n r x 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 0 n r x 当n 时 这样 我们就得到了函数 f x的幂级数展开式 2 0 0 0 0 0 0 2 nn nn n fff f xxffxxx nn 954 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 2009 no 8 数学分析原理数学分析原理 数学分析原理数学分析原理 210 2 012 0 nn nn n f xa xaa xa xa x 9 55 那么 根据幂级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 2 012 0 0 0 0 2 n n n ff afafx axax n 将它们代入 955 式 所得与 f x的马克劳林展开式 954 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 954 式的幂级数 幂级数 00 000 1 n n fxfx f xf xxxxx 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 x f xe 展开成x的幂级数 解 因为 nx fxe 1 2 3 n 所以 0 0 0 0 1 n ffff 于是我们得到幂级数 2 11 1 2 n xxx n 956 显然 956 式的收敛区间为 至于 956 式是否以 x f xe