taylor series of sqrt(x) - Axtarish в Google
12 июл. 2016 г. · sqrt(x) = sum_(n=0)^infty (x-1)^n /(n!) For any function f(x) the taylor expansion of that function about a variable a will be: f(x) = f(a) ...
Продолжительность: 12:17
Опубликовано: 13 апр. 2021 г.
The Taylor series for √x at x=0 does not exist because the square root function is not differentiable at x=0. However, the Taylor series can be derived for √x ...
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