Find the Taylor series for ( ) lnf x x= at 1a = by usingthe above result.(第二頁)
Question
Find the Taylor series for ( ) lnf x x= at 1a = by usingthe above result.(第二頁)
Solution
The Taylor series for a function f(x) about a point a is given by:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
Given f(x) = ln(x), we want to find the Taylor series at a = 1.
First, we need to find the derivatives of f(x) at a = 1.
f'(x) = 1/x, so f'(1) = 1. f''(x) = -1/x^2, so f''(1) = -1. f'''(x) = 2/x^3, so f'''(1) = 2. f''''(x) = -6/x^4, so f''''(1) = -6.
We can see a pattern here. The nth derivative at x = 1 is (-1)^(n+1) * (n-1)!, for n >= 2.
So, the Taylor series for ln(x) at a = 1 is:
ln(x) = (x-1) - (x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4 + ...
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