How to represent $\ln(5-x)$ as a power series?

Question

I know that $$ \ln(1+x)=\sum _{n=1}^{\infty }\:\left(-1\right)^{n-1}\frac{x^{n}}{n} $$

score 7 · Accepted Answer · edited Jun 11 '14 at 15:36

7

Hint: $$\ln(5-x)=\ln\left[5\left(1-\frac{x}{5}\right)\right]=\ln 5+\ln\left(1-\frac{x}{5}\right). $$

edited Jun 11 '14 at 15:36

answered Jun 11 '14 at 14:38

user1337

But then we use $\sum _{n=0}^{\infty }:\frac{\left(\frac{x}{5}\right)^{n-1}}{n-1}\left(-1\right)^n$ instead of $\sum _{n=0}^{\infty }:\frac{\left(x\right)^{n-1}}{n-1}\left(-1\right)^n$ ? – DDDD Jun 11 '14 at 14:43

score 2 · Answer 2 · answered Jun 11 '14 at 14:45

2

In your case it's more convenient to use $$ \log \bigg( \frac{1}{1-w} \bigg) = \sum_{k=1}^{n} \frac{w^k}{k} $$ just multiply by $-1$ and set $w=5x$.

answered Jun 11 '14 at 14:45

Alex

2 Answers2