数学外文翻译

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 function This section will discuss another issue for an arbitrary function can be expanded in a power series and launched into f x Whether the power series as and function The following discussion will address this issue f x 1 Maclaurin Maclaurin formula Polynomial power series can be seen as an extension of reality so consider the function can f x expand into power series you can from the function and polynomials start to solve this problem To f x 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 f x 0 xx order then in the neighborhood of the following formula 1n 9 5 1 2 0000 n n f xf xxxxxxxr x Among 1 0 n n r xxx That for the Lagrangian remainder That 9 5 1 type formula for the Taylor n r x If so get 0 0 x 9 5 2 2 0 n n f xfxxxr x 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 as long as until the derivative can be equal to a f x1n n polynomial and a remainder We call the following power series 9 5 3 2 0 0 0 0 2 n n ff f xffxxx n For the Maclaurin series So is it to for the Sum functions If the order Maclaurin series 9 5 3 the first items and f x1n for which 1 n Sx 2 1 0 0 0 0 2 n n n ff Sxffxxx n Then the series 9 5 3 converges to the function the conditions f x 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 and function as the Maclaurin formula 9 5 2 f x of the remainder term when 0 n r x n In this way we get a function the power series expansion f x 9 5 4 0 0 0 0 0 nn nn n ff f xxffxx nn It is the function the power series expression if the function of the power series expansion is f x unique In fact assuming the function f x can be expressed as power series 9 5 5 2 012 0 nn nn n f xa xaa xa xa x 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 get0 x 0 x 2 012 0 0 0 0 2 n n n ff afafx axax n Substituting them into 9 5 5 type income and the Maclaurin expansion of 9 5 4 identical f x 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 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 expanded in power series method called the direct f x expansion method Example 1 Test the functionexpanded in power series of x f xe x Solution because nx fxe 1 2 3 n Therefore 0 0 0 0 1 n ffff So we get the power series 9 5 6 2 11 1 2 n xxx n Obviously 9 5 6 type convergence interval As 9 5 6 whether type is Sum x f xe function that is whether it converges to but also examine remainder x f xe n r x Because 且 1 e 1 x n n r xx n 01 xx x Therefore 11ee 1 1 xx nn n r xxx nn Noting the value of any set is a fixed constant while the series 9 5 6 is absolutely convergent so x x e the general when the item when so when n n 1 0 1 n x 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 therefore x f xe 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 and power series expansion the use of these x 1 1 x esin x 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 cosf xx 0 x 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 3 Using the expansion to estimatethe value of arctan x 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 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 f x 为和函数 下面的讨论将解决这一问题 f x 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 f x 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 f x 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 f x 0 xx 1n 域内有如下公式 9 5 1 2 00 00000 2 n n n fxfx f xf xfxxxxxxxr x n 其中 1 1 0 1 n n n f r xxx n 称为拉格朗日型余项 称 9 5 1 式为泰勒公式 n r x 如果令 就得到 0 0 x 9 5 2 2 0 n n f xfxxxr x 此时 1 1 11 1 1 1 nn nn n ffx rxxx nn 01 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 f x1n n 和 我们称下列幂级数 9 5 3 2 0 0 0 0 2 n n ff f xffxxx n 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 f x1n 即 1 n Sx 2 1 0 0 0 0 2 n n n ff Sxffxxx n 那么 级数 9 5 3 收敛于函数的条件为 f x 1 lim n n sxf 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 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 9 5 3 以为和函数马克劳林公式 9 5 2 中的余项 f x 0 n r x 当时 n 这样 我们就得到了函数的幂级数展开式 f x 9 5 4 2 0 0 0 0 0 0 2 nn nn n fff f xxffxxx nn 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 f x 函数可以表示为幂级数 f x 9 5 5 2 012 0 nn nn n f xa xaa xa xa x 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0 x 0 x 容易得到 2 012 0 0 0 0 2 n n n ff afafx axax n 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 f x 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 f x 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 n f x 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展开成幂级数的方法 称为直接展开法 f x 例 1 试将函数展开成的幂级数 x f xe x 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 9 5 6 2 11 1 2 n xxx n 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 x f xe 否收敛于 还要考察余项 x f xe n r x 因为 且 1 e 1 x n