lnx的泰勒展开式可不可以用x-1代入ln(x+1)的展开式 ?

2025-03-16 16:58:07
推荐回答(1个)
回答1:

ln(1+x)泰勒展开式:

ln(1+x)=x-x^2/2+x^3/3-……+(-1)^(k-1)*(x^k)/k 式中(|x|<1)

这个泰勒展开式收缩太慢,如果|x|≤0.1,收缩较快;如果|x|≤0.01,收缩更快。

利用lnxy=lnx+lny可以把任意数化为易求的粗数和泰勒展开式细数。

尽管电脑附件计算器最多能算32位,但是计算很不方便。Excel表格最多能显示15位有效数,实际计算精度远远超过15位有效数。我的计算基本上在Excel表格中做,如果完全在Excel表格中做,往往看不到误差。

为了保留误差,只要记住:在做减法时,把两数复制到记事本,再复制回来,再做减法,很可能就没有15位有效数了,这就挖出了误差的主要来源。

例:求ln345

解:ln345=ln(3.45*10^2)=ln3.45+2ln10

3.45和10怎样分别分出易求的粗数和泰勒展开式细数?

可以用乘方,开方求出十几个基本的e^m,然后用3.45之类的数除以e^(m+n+...),直到商小于1.1,甚至小于1.01,再用泰勒展开式收缩很快。

e^x泰勒展开式

e^x = 1+x+x^2/2!+x^3/3!+……+x^n/n!+…

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