Find the Taylor series for the given function at the indicated point $x = a \text {. }$ $f ( x ) = \ln ( 5 x ) ; a = \frac { 1 } { 5 }$
A) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } 5 ^ { n } } { n } \left( x - \frac { 1 } { 5 } \right) ^ { n }$
B) $\sum _ { n = 1 } ^ { \infty } \frac { 5 ^ { n } } { n } \left( x - \frac { 1 } { 5 } \right) ^ { n }$
C) $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n 5 ^ { n } } \left( x - \frac { 1 } { 5 } \right) ^ { n }$
D) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } } { n 5 ^ { n } } \left( x - \frac { 1 } { 5 } \right) ^ { n }$

Question

Quizplus · Accepted Answer

The Taylor series for $f(x) = \ln(x)$ at $x = a$ is given by$$\ln(x) = \ln(a) + \frac{x-a}{a} - \frac{(x-a)^2}{2a^2} + \frac{(x-a)^3}{3a^3} - \cdots$$Substituting $a = \frac{1}{5}$ and $f(x) = \ln(5x)$, we get$$\ln(5x) = \ln\left(\frac{1}{5}\right) + \frac{5x-\frac{1}{5}}{\frac{1}{5}} - \frac{\left(5x-\frac{1}{5}\right)^2}{2\left(\frac{1}{5}\right)^2} + \frac{\left(5x-\frac{1}{5}\right)^3}{3\left(\frac{1}{5}\right)^3} - \cdots$$Simplifying, we get$$\ln(5x) = -\ln(5) + 5x - \frac{25}{2}\left(5x-\frac{1}{5}\right)^2 + \frac{125}{3}\left(5x-\frac{1}{5}\right)^3 - \cdots$$$$\ln(5x) = -\ln(5) + 5x - \frac{25}{2}\left(25x^2-\frac{10x}{5}+\frac{1}{25}\right) + \frac{125}{3}\left(125x^3-\frac{75x^2}{5}+\frac{15x}{25}-\frac{1}{125}\right) - \cdots$$$$\ln(5x) = -\ln(5) + 5x - \frac{625}{2}x^2 + 50x^3 - \frac{3125}{6}x^4 + \cdots$$$$\ln(5x) = -\ln(5) + \sum_{n=1}^\infty \frac{(-1)^{n+1}5^n}{n}\left(x-\frac{1}{5}\right)^n$$Therefore, the correct answer is A.

Find the Taylor Series for the Given Function at the Indicated