f(x) = ln(1+x)-ln(1-x)
Differentiation leads to
fprime (x) = 1/{1+x}+1/{1-x}
Let us consider one of the pieces : 1/{1+x}. Differentiating this repeatedly leads to
d/dx(1/{1+x}) = {-1}/(1+x)^2
d^2/dx^2(1/{1+x}) = {(-1) times (-2)}/(1+x)^3
d^3/dx^3(1/{1+x}) = {(-1) times (-2) times (-3)}/(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)(x) =d^n/dx^n(1/{1+x})+d^n/dx^n(1/{1-x})
= n![{(-1)^n }/(1+x)^{n+1}+1/(1+x)^{n+1}]
To find the Maclurin polynomial, we need to evaluate f^(n)(0). This simply is
f^(n)(0) = (n-1)![1+(-1)^{n-1}]
So, f^(n)(0) 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(x) = 2x+2/3x^3+2/5x^5+... = 2 sum_{i=0}^{[n/2]-1}x^{2i+1}/{2i+1}
An alternative approach would be to start from the GP
1+x+x^2+...+x^n = {1-x^{n+1}}/{1-x}
to get
1-x+x^2+...+(-1)^nx^n = {1-(-1)^{n+1}x^{n+1}}/{1+x}
so that
2 + 2x^2 + ...+[1+(-1)^n]x^n = 1/{1+x}+1/{1-x}-x^{n+1}[1/{1-x}+(-1)^{n+1}/{1+x}]
Integrating both sides gives the Maclurin series for f(x) (the constant of integration being found from f(0)=0. This approach also shows that the remainder term is
R_n(x) = int_0^x y^{n+1}[1/{1-y}+(-1)^{n+1}/{1+y}]dy
This form may be helpful in proving that the remainder vanishes as n to infty for |x|<1
To find ln(2) using this series, use x= 1/3
f(1/3) = ln({1+1/3}/{1-1/3})= ln(2) = 2[1/{1cdot3}+1/{3cdot 3^3}+1/{5cdot 3^5}+1/{7cdot3^7}...] ~~ 0.6931
