The series $\ln ( x ) = ( x - 1 ) - \frac { ( x - 1 ) ^ { 2 } } { 2 } + \frac { ( x - 1 ) ^ { 3 } } { 3 } - \ldots + ( - 1 ) ^ { n - 1 } \frac { ( x - 1 ) ^ { n } } { n }$ can be used to estimate the value of ln(a) for any positive real number a such that $0

Question

The series $\ln ( x ) = ( x - 1 ) - \frac { ( x - 1 ) ^ { 2 } } { 2 } + \frac { ( x - 1 ) ^ { 3 } } { 3 } - \ldots + ( - 1 ) ^ { n - 1 } \frac { ( x - 1 ) ^ { n } } { n }$ can be used to estimate the value of ln(a) for any positive real number a such that $0 < \mathrm { a } \leq 2$. Use the first five terms of this series to estimate $\ln ( 1.17 )$. Round to the nearest thousandth.
A) $1.047$
B) $1.717$
C) $0.157$
D) $- 0.773$

Quizplus · Accepted Answer

The series given is the Taylor series expansion of $\ln(x)$ around $x=1$. To estimate $\ln(1.17)$, substitute $x=1.17$ into the series and use the first five terms: $\ln(1.17) \approx (1.17 - 1) - \frac{(1.17 - 1)^2}{2} + \frac{(1.17 - 1)^3}{3} - \frac{(1.17 - 1)^4}{4} + \frac{(1.17 - 1)^5}{5}$. Calculating this gives a value close to $0.157$, hence the correct answer is C.

The Series Can Be Used to Estimate the Value of Ln(a) for 1