section_path: Calculation/Power series/Taylor series
section_name: Taylor series
paragraph_text: Taylor series
  
  For any real number z that satisfies 0 < z ≤ 2, the following formula holds:
  
  ln (z) = (z − 1) 1 1 − (z − 1) 2 2 + (z − 1) 3 3 − (z − 1) 4 4 + ⋯ = ∑ k = 1 ∞ (− 1) k + 1 (z − 1) k k. {\displaystyle {\begin{aligned}\ln(z)&={\frac {(z-1)^{1}}{1}}-{\frac {(z-1)^{2}}{2}}+{\frac {(z-1)^{3}}{3}}-{\frac {(z-1)^{4}}{4}}+\cdots \\&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {(z-1)^{k}}{k}}.\end{aligned}}}
  
  Equating the function ln(z) to this infinite sum (series) is shorthand for saying that the function can be approximated to a more and more accurate value by the following expressions (known as partial sums):
  
  (z − 1), (z − 1) − (z − 1) 2 2, (z − 1) − (z − 1) 2 2 + (z − 1) 3 3, … {\displaystyle (z-1),\ \ (z-1)-{\frac {(z-1)^{2}}{2}},\ \ (z-1)-{\frac {(z-1)^{2}}{2}}+{\frac {(z-1)^{3}}{3}},\ \ldots }
  
  For example, with z = 1.5 the third approximation yields 0.4167, which is about 0.011 greater than ln(1.5) = 0.405465, and the ninth approximation yields 0.40553, which is only about 0.0001 greater. The n th partial sum can approximate ln(z) with arbitrary precision, provided the number of summands n is large enough.
  
  In elementary calculus, the series is said to converge to the function ln(z), and the function is the limit of the series. It is the Taylor series of the natural logarithm at z = 1. The Taylor series of ln(z) provides a particularly useful approximation to ln(1 + z) when z is small, | z | < 1, since then ln (1 + z) = z − z 2 2 + z 3 3 − ⋯ ≈ z. {\displaystyle \ln(1+z)=z-{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}-\cdots \approx z.}
  
  For example, with z = 0.1 the first-order approximation gives ln(1.1) ≈ 0.1, which is less than 5% off the correct value 0.0953.