Find a power-series approximation for ln((1+x)/(1-x)).

Step-by-step solution

1. 1

   Split the logarithm of the quotient first. Rather than expand the whole quotient, use $\ln(A/B)=\ln A-\ln B$: $\ln\!\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x).$ This is the move that makes the whole problem easy, because both pieces have standard Maclaurin series.

2. 2

   Write down the two standard series. For $|x|<1$, $\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{n}}{n},$ and replacing $x$ by $-x$, $\ln(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\cdots=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}.$

3. 3

   Subtract and watch the even powers cancel. Term by term, the $x^{2}$ contributions are $-\tfrac12 x^{2}$ and $-\tfrac12 x^{2}$, which cancel on subtraction; the $x^{3}$ contributions are $+\tfrac13 x^{3}$ and $-\tfrac13 x^{3}$, which reinforce. In general only odd powers survive and each is doubled: $\ln\!\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right)=2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}.$ Only odd powers is exactly what the function's symmetry demands: replacing $x$ by $-x$ inverts the argument and flips the sign of the logarithm, so the function is odd.

4. 4

   Record the useful truncations. Cutting the series off gives the practical approximations $\ln\!\left(\frac{1+x}{1-x}\right)\approx 2x,\qquad \approx 2x+\frac{2x^{3}}{3},\qquad \approx 2x+\frac{2x^{3}}{3}+\frac{2x^{5}}{5}.$ The first is the linear (tangent-line) approximation at $x=0$.

5. 5

   State the interval of validity and test it. Both parent series need $|x|<1$, and the quotient itself is only positive there, so the expansion holds for $|x|<1$. Numerical check at $x=0.2$: the exact value is $\ln(1.2/0.8)=\ln 1.5=0.405465108$, while $2x+\tfrac23 x^{3}+\tfrac25 x^{5}=0.4+0.005333333+0.000128=0.405461333$ - agreement to five decimal places, with the next term $\tfrac27 x^{7}=3.66\times10^{-6}$ accounting for the small gap.

6. 6

   Note why this form is preferred for computing logarithms. Any positive $r$ can be written as $(1+x)/(1-x)$ with $x=(r-1)/(r+1)$, and that $x$ is small even when $r$ is far from $1$. Because the series has only odd terms and converges twice as fast as the plain $\ln(1+x)$ series, this identity is the classical way to get logarithms to high precision by hand.