2
$\begingroup$

If I want to find the power series representation of the following function:

$$ \ln \frac{1+x}{1-x} $$

I understand that it can be written as

$$ \ln (1+x) - \ln(1-x) $$

And I understand that if I now write in the power series representations for $ln(1+x)$ and $ln(1-x)$:

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n} $$

My textbook solution does an odd thing where it writes it out as

$$\sum_{n=1}^\infty \frac{x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n} $$

$$2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1} $$

I have no idea how it got from the line where I have the power series representation for $ln(1+x)$ and $ln(1-x)$ to the last two lines. If anyone could help me link my part to the textbook solution I would really appreciate it! Thank you!

  • 0
    Well, $(-1)^{n-1}(-x)^n=(-1)^{n-1}(-1)^nx^n=(-1)^{2n-1}x^n=-x^n$ – 2017-02-05

2 Answers 2

0

$$\begin{align}\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n}&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-1)^nx^{n}}{n}\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{2n-1}x^{n}}{n}\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{-x^{n}}{n}\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} + \sum_{n=1}^\infty \frac{x^{n}}{n}\\ &=\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\right]+ \left[x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots\right]\\ &=2\left[\frac{x^1}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\dots\right]\\ &=2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1} \end{align}$$

  • 0
    Did you get the $\frac{x^{2n-1}}{2n-1}$ By looking at the series itself and finding a pattern? Is that the only way I could figure out the $\frac{x^{2n-1}}{2n-1}$ part? – 2017-02-05
  • 0
    Yeah, based on the pattern. Observed that $1,3,5,7,\dots$ are odd numbers that can be represented by $2n-1$ for $n=1,2,3,4,\dots$ – 2017-02-05
  • 0
    Lovely! Thank you very much I understand now – 2017-02-05
  • 0
    Always happy to help. – 2017-02-05
0

Assuming $|x|<1$, by absolute convergence, one is allowed to rearrange terms, obtaining $$ \sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n}=\sum_{n=1}^\infty \frac{(-1)^{n-1}+1}{n} \cdot x^{n}=2\sum_{k=1}^\infty \frac{x^{2k-1}}{2k-1} $$ since term corresponding to $n=2,4,\cdots,$ cancel and since $(-1)^{n-1}+1=2$ for $n=1,3,5\cdots$.