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根据根值审敛法,则有柯西-阿达马公式。或者,复分析中的收敛半径将一个收敛半径是正数的幂级数的变量取为复数,就可以定义一个全纯函数。

最近点的取法是在整个复平面中,而不仅仅是在实轴上,即使中心和系数都是实数时也是如此。例如:函数没有复根。它在零处的泰勒展开为:运用达朗贝尔审敛法可以得到它的收敛半径为1。

扩展资料:

如果幂级数在a附近可展,并且收敛半径为r,那么所有满足 |za| =r的点的集合(收敛圆盘的边界)是一个圆,称为收敛圆。幂级数在收敛圆上可能收敛也可能发散。即使幂级数在收敛圆上收敛,也不一定绝对收敛。

幂级数的收敛半径是 1 并在整个收敛圆上收敛。设 h(z) 是这个级数对应的函数,那么 h(z) 是例2中的 g(z) 除以 z后的导数。

收敛区间英文

267 评论(15)

帅哥小蜜

本题中的等于号应该删去;

2、本题是典型的幂级数(Power series),解答收敛半径的方法有两种:

A、比值法;

B、根值法.

3、收敛半径是从英文Convergent Radius翻译而来,它本身是一个

牵强附会的概念,不涉及平面区域问题,无半径可言.它的准确

意思是:收敛区间长度的一半.

4、两种解法的具体过程如下:

84 评论(10)

哦、起床…

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270 评论(8)

鱿鱼女王

解:∵原式=∑(2/2^n)x^n+∑[(-1/2)^n]x^n,易得∑(2/2^n)x^n、∑[(-1/2)^n]x^n的收敛半径均为R=2,故原级数的收敛半径均为R=2。

1、本题中的等于号应该删去;

2、本题是典型的幂级数(Power series),解答收敛半径的方法有两种:

A、比值法;

B、根值法。

3、收敛半径是从英文Convergent Radius翻译而来,它本身是一个

牵强附会的概念,不涉及平面区域问题,无半径可言。它的准确

意思是:收敛区间长度的一半。

扩展资料:

收敛半径r是一个非负的实数或无穷大的数,使得在 | z -a| < r时幂级数收敛,在 | z -a| > r时幂级数发散。

具体来说,当 z和 a足够接近时,幂级数就会收敛,反之则可能发散。收敛半径就是收敛区域和发散区域的分界线。

收敛半径可以被如下定理刻画:

一个中心为 a的幂级数 f的收敛半径 R等于 a与离 a最近的使得函数不能用幂级数方式定义的点的距离。到 a的距离严格小于 R的所有点组成的集合称为收敛圆盘。

参考资料:百度百科-收敛半径

265 评论(12)

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