Taylor polynomial of ln x at x₀=1

Problem

Third-order Taylor polynomial of $\ln x$ at $x_0 = 1$ (note: $x_0\ne 0$, expand in powers of $(x-1)$).

Solution

I compute the value and the derivatives at $x_0=1$: $f(x)=\ln x,\ f(1)=0; \quad f'(x)=\tfrac1x,\ f'(1)=1; \quad f''(x)=-\tfrac1{x^2},\ f''(1)=-1; \quad f'''(x)=\tfrac2{x^3},\ f'''(1)=2.$ Substituting into the formula $T_3(x)=\sum_{k=0}^3 \frac{f^{(k)}(1)}{k!}(x-1)^k$: $T_3(x) = 0 + 1\cdot(x-1) + \frac{-1}{2}(x-1)^2 + \frac{2}{6}(x-1)^3 = \boxed{(x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}}.$

Links

Topics: Derivatives
Concepts: Higher-order derivative · Taylor polynomial
Functions: Logarithmic function
Methods: Taylor polynomial
Skills: Use formulae
Exercise type: Derivative computation