Question

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

Answer 1

Taylor series expansion for ln(x)

Explanation:

Taylor series expansion:

`\sum_(n=0)^\infty f^n(a)\displaystyle((x-a)^n)/(n!)`

Here, f(x) = ln (x)

`f(x) = \ln x, f(a) = \ln a\\\\f'(x) = \displaystyle(1)/(x), f'(a) = \displaystyle(1)/(a) \\\\f''(x) = \displaystyle(-1)/(x^2), f''(a) = \displaystyle(-1)/(a^2)\\\\f'''(x) = \displaystyle(2)/(x^3), f'''(a) = \displaystyle(2)/(a^3)\\\\f^4(x) = \displaystyle(-6)/(x^4), f^4(a) = \displaystyle(-6)/(a^4)\\\\f^5(x) = \displaystyle(24)/(x^5), f^5(a) = \displaystyle(24)/(a^5)`

Thus, in general we get,

`f^n(x) = \displaystyle((-1)^n(n-1)!)/(x^n)`

Putting all the values and expanding, we get,

`f(x) = f(a) + \displaystyle(f'(a)(x-a))/(1!) + \displaystyle(f''(a)(x-a)^2)/(2!) + \displaystyle(f'''(a)(x-a)^3)/(3!) + \displaystyle(f^4(a)(x-a)^4)/(4!) + \displaystyle(f^5(a)(x-a)^5)/(5!) + ...\\\\= \ln a + \displaystyle(x-a)/(a) -\displaystyle((x-a)^2)/(2a^2) + \displaystyle((x-a)^3)/(3a^3) + ...`