Question

Write the Taylor Series for f(x) = ln (x )centere

Answer 1

Answer:

Taylor series expansion for ln(x)

Step-by-step explanation:

Taylor series expansion:

$\sum_{n=0}^\infty f^n(a)\displaystyle\frac{(x-a)^n}{n!}$

Here, f(x) = ln (x)

$f(x) = \ln x, f(a) = \ln a\\\\f'(x) = \displaystyle\frac{1}{x}, f'(a) = \displaystyle\frac{1}{a} \\\\f''(x) = \displaystyle\frac{-1}{x^2}, f''(a) = \displaystyle\frac{-1}{a^2}\\\\f'''(x) = \displaystyle\frac{2}{x^3}, f'''(a) = \displaystyle\frac{2}{a^3}\\\\f^4(x) = \displaystyle\frac{-6}{x^4}, f^4(a) = \displaystyle\frac{-6}{a^4}\\\\f^5(x) = \displaystyle\frac{24}{x^5}, f^5(a) = \displaystyle\frac{24}{a^5}$

Thus, in general we get,

$f^n(x) = \displaystyle\frac{(-1)^n(n-1)!}{x^n}$

Putting all the values and expanding, we get,

$f(x) = f(a) + \displaystyle\frac{f'(a)(x-a)}{1!} + \displaystyle\frac{f''(a)(x-a)^2}{2!} + \displaystyle\frac{f'''(a)(x-a)^3}{3!} + \displaystyle\frac{f^4(a)(x-a)^4}{4!} + \displaystyle\frac{f^5(a)(x-a)^5}{5!} + ...\\\\= \ln a + \displaystyle\frac{x-a}{a} -\displaystyle\frac{(x-a)^2}{2a^2} + \displaystyle\frac{(x-a)^3}{3a^3} + ...$