Feb 24, 2018

${P}_{n} \left(x\right) = 2 x + \frac{2}{3} {x}^{3} + \frac{2}{5} {x}^{5} + \ldots = 2 {\sum}_{i = 0}^{\left[\frac{n}{2}\right] - 1} {x}^{2 i + 1} / \left\{2 i + 1\right\}$

$\ln \left(2\right) = f \left(\frac{1}{3}\right) \approx 0.6931$

#### Explanation:

$f \left(x\right) = \ln \left(1 + x\right) - \ln \left(1 - x\right)$

$f p r i m e \left(x\right) = \frac{1}{1 + x} + \frac{1}{1 - x}$

Let us consider one of the pieces : $\frac{1}{1 + x}$. Differentiating this repeatedly leads to
$\frac{d}{\mathrm{dx}} \left(\frac{1}{1 + x}\right) = \frac{- 1}{1 + x} ^ 2$
${d}^{2} / {\mathrm{dx}}^{2} \left(\frac{1}{1 + x}\right) = \frac{\left(- 1\right) \times \left(- 2\right)}{1 + x} ^ 3$
${d}^{3} / {\mathrm{dx}}^{3} \left(\frac{1}{1 + x}\right) = \frac{\left(- 1\right) \times \left(- 2\right) \times \left(- 3\right)}{1 + x} ^ 4$

The pattern is clear. We have

d^n/dx^n(1/{1+x}) = {(-1)^n n!}/(1+x)^{n+1}
d^n/dx^n(1/{1-x}) = {n!}/(1-x)^{n+1}
and similarly
This means that

${f}^{n + 1} \left(x\right) = {d}^{n} / {\mathrm{dx}}^{n} \left(\frac{1}{1 + x}\right) + {d}^{n} / {\mathrm{dx}}^{n} \left(\frac{1}{1 - x}\right)$
 = n![{(-1)^n }/(1+x)^{n+1}+1/(1+x)^{n+1}]

To find the Maclurin polynomial, we need to evaluate ${f}^{n} \left(0\right)$. This simply is
f^(n)(0) = (n-1)![1+(-1)^{n-1}]
So, ${f}^{n} \left(0\right)$ is 2(n-1)! for odd $n$ and it vanishes for even $n$, and the coefficients of the Maclurin polynomial are {f^(n)(0)} /{n!} = 2/n for odd $n$ and 0 for even $n$

Thus the Maclurin polynomial is
${P}_{n} \left(x\right) = 2 x + \frac{2}{3} {x}^{3} + \frac{2}{5} {x}^{5} + \ldots = 2 {\sum}_{i = 0}^{\left[\frac{n}{2}\right] - 1} {x}^{2 i + 1} / \left\{2 i + 1\right\}$

An alternative approach would be to start from the GP

$1 + x + {x}^{2} + \ldots + {x}^{n} = \frac{1 - {x}^{n + 1}}{1 - x}$
to get
$1 - x + {x}^{2} + \ldots + {\left(- 1\right)}^{n} {x}^{n} = \frac{1 - {\left(- 1\right)}^{n + 1} {x}^{n + 1}}{1 + x}$
so that
$2 + 2 {x}^{2} + \ldots + \left[1 + {\left(- 1\right)}^{n}\right] {x}^{n} = \frac{1}{1 + x} + \frac{1}{1 - x} - {x}^{n + 1} \left[\frac{1}{1 - x} + {\left(- 1\right)}^{n + 1} / \left\{1 + x\right\}\right]$

Integrating both sides gives the Maclurin series for $f \left(x\right)$ (the constant of integration being found from $f \left(0\right) = 0$. This approach also shows that the remainder term is

${R}_{n} \left(x\right) = {\int}_{0}^{x} {y}^{n + 1} \left[\frac{1}{1 - y} + {\left(- 1\right)}^{n + 1} / \left\{1 + y\right\}\right] \mathrm{dy}$

This form may be helpful in proving that the remainder vanishes as $n \to \infty$ for $| x | < 1$

To find $\ln \left(2\right)$ using this series, use $x = \frac{1}{3}$

$f \left(\frac{1}{3}\right) = \ln \left(\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}\right) = \ln \left(2\right) = 2 \left[\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot {3}^{3}} + \frac{1}{5 \cdot {3}^{5}} + \frac{1}{7 \cdot {3}^{7}} \ldots\right] \approx 0.6931$