How do you find the maclaurin series expansion of #ln((1+x)/(1-x))#?

Question

Isaiah Allen · Accepted Answer

#ln((1+x)/(1-x)) = 2(x + 1/(3) x^3 + 1/(5) x^5 + ...) = 2sum_(n=0)^(∞) x^(2n+1) / (2n+1)#

To find a Maclaurin series for #ln((1+x)/(1-x))# from scratch, we first need to take note of expressing a function as an infinite sum centered at #x= 0#.

To accomplish this, we write

#f(x) = #

#f(0) + f^(1)(0)/(1!)x + f^(2)(0)/(2!)x^2 + f^(3)(0)/(3!)x^3 + ... = sum_(n=0)^(∞) f^(n)(0) x^(n) / (n!)#

We would need to compute some derivatives, as this infinite sum implies. I will do this for three terms, hopefully enough to identify a pattern.

Due to logarithm properties, we have

#f(x) = ln((1+x)/(1-x)) = ln(1+x) - ln(1-x)#

Calculating derivatives:

#f^(1)(x) = 1/(1+x) - (-1)/(1-x) = 1/(1+x) + 1/(1-x)#

# = (1cancel(-x)+1cancel(+x))/(1cancel(-x+x)-x^2) = 2/(1-x^2) = 2(1-x^2)^(-1)#

#f^(2)(x) =-2(1-x^2)^(-2)(-2x) = 4x(1-x^2)^(-2) = (4x)/(1-x^2)^(2)#

#f^(3)(x) =4[1(1-x^2)^(-2)+x*-2(1-x^2)^(-3)(-2x)]#

#= 4/(1-x^2)^(2) + (16x^2)/(1-x^2)^(3)#

Substituting #0# into our derivatives we get

#f(0) = 0# #f^(1)(0) = 2# #f^(2)(0) = 0# #f^(3)(0) = 4#

#...#

#f^(4)(0) = 0# #f^(5)(0) = 48#

Actually, it turns out that only odd derivatives at #x=0# give an actual value, and so to make up for a missing term, I've used a computing device to calculate the fourth and fifth derivative, as shown above.

Using our #f^(n)(0)#-values to construct a Maclaurin series, we write

#ln((1+x)/(1-x)) = 0 + 2/(1!) x + 0/(2!) x^2 + 4/(3!) x^3 + 0/(4!) x^4 + 48/(5!) x^5 + ...#

Observing that some terms simply cancel each other out, we are left with

#ln((1+x)/(1-x)) = 2/(1!) x + 4/(3!) x^3 + 48/(5!) x^5 + ...#

When we reduce the denominators of each term (the factorials), we obtain

#ln((1+x)/(1-x)) = 2x + 4/(6) x^3 + 48/(120) x^5 + ...#

We can actually simplify a few fractions and factor out a #2# out of each term, leaving us with

#ln((1+x)/(1-x)) = 2(x + 1/(3) x^3 + 1/(5) x^5 + ...)#

As we may notice, we can see that the power and the denominator on each term is increasing by #2#, so we end up with

#ln((1+x)/(1-x)) = 2(x + 1/(3) x^3 + 1/(5) x^5 + ...) = 2sum_(n=0)^(∞) x^(2n+1) / (2n+1)#

Luke Alvarez · Answer

undefined To find the Maclaurin series expansion of $\ln\left(\frac{1+x}{1-x}\right)$, we can start by using the properties of logarithms to rewrite the expression as $\ln(1+x) - \ln(1-x)$. Then, we can use the Maclaurin series expansions of $\ln(1+x)$ and $\ln(1-x)$, which are:

$$
\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots
$$
$$
\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots
$$

Substitute these series expansions into $\ln\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x)$ and simplify to get the Maclaurin series expansion.