Question

How do you find the maclaurin series expansion of $\int{\left( \dfrac{1+x}{1-x} \right)?}$

Answer 1

To find maclaurin series for $\int{\left( \dfrac{1+x}{1-x} \right)}$ from scratch, we first need to take note expressing a function as an infinite sum centred at $x=0.$

Complete step by step solution:
As we know that, you have to find the maclaurin series expansion of $\int{\left( \dfrac{1+x}{1-x} \right).}$
Maclaurin series is $\int{\left( \dfrac{1+x}{1-x} \right),}$ we have to express a given function as an infinite sum centred at $x=0.$
In order to do this, we write
$f\left( x \right)=f\left( 0 \right)+\dfrac{{{f}^{1}}\left( 0 \right)}{1!}x+\dfrac{{{f}^{2}}\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{{f}^{3}}\left( 0 \right)}{3!}{{x}^{3}}+......=\sum\limits_{n=0}^{\infty }{{{f}^{n}}\left( 0 \right)\dfrac{{{x}^{n}}}{n!}}$
This infinite sum suggests that we would have to calculate some derivatives, which I will do for three terms.
By the properties of logarithm, we have
$\Rightarrow f\left( x \right)=\int{\left( \dfrac{1+x}{1-x} \right)=\int{\left( 1+x \right)-\int{\left( 1-x \right)}}}$
Derivative calculations:
$\Rightarrow {{f}^{1}}\left( x \right)=\dfrac{1}{1+x}-\dfrac{-1}{1-x}$
$\Rightarrow {{f}^{1}}\left( x \right)=\dfrac{1}{1+x}+\dfrac{1}{1-x}$
$\Rightarrow {{f}^{1}}\left( x \right)=\dfrac{1-x+1+x}{1-{{x}^{2}}}$
$\Rightarrow {{f}^{1}}\left( x \right)=\dfrac{2}{1-{{x}^{2}}}$
You can also write it as,
$\Rightarrow {{f}^{1}}\left( x \right)=2{{\left( 1-{{x}^{2}} \right)}^{-1}}$
$\Rightarrow {{f}^{2}}\left( x \right)=-2{{\left( 1-{{x}^{2}} \right)}^{-2}}\left( -2x \right)$
$\Rightarrow {{f}^{2}}\left( x \right)=4x{{\left( 1-{{x}^{2}} \right)}^{-2}}$
$\Rightarrow {{f}^{2}}\left( x \right)=\dfrac{4x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow {{f}^{3}}\left( x \right)=4\left[ 1\left( 1-{{x}^{2}} \right)+x....-2{{\left( 1-{{x}^{2}} \right)}^{-3}}\left( -2x \right) \right]$
$\Rightarrow {{f}^{3}}\left( x \right)=\dfrac{4}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{16{{x}^{2}}}{{{\left( 1-{{x}^{2}} \right)}^{3}}}$
Substituting $'0'$ into our derivative we get,
$\Rightarrow f\left( 0 \right)=0$
$\Rightarrow {{f}^{1}}\left( 0 \right)=2$
$\Rightarrow {{f}^{2}}\left( 0 \right)=0$
$\Rightarrow {{f}^{3}}\left( 0 \right)=4$
$\Rightarrow {{f}^{4}}\left( 0 \right)=0$
$\Rightarrow {{f}^{5}}\left( 0 \right)=48$
Actually it twins out that only odd derivatives at $x=0$ given an actual value, and 0 to make up for a missing term, you have 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.
$\Rightarrow \int{\left( \dfrac{1+x}{1-x} \right)=0+\dfrac{2}{1!}x+\dfrac{0}{2!}{{x}^{2}}+\dfrac{4}{3!}{{x}^{3}}+\dfrac{0}{4!}{{x}^{4}}+\dfrac{48}{5!}{{x}^{5}}}$
As we can see, a few terms just cancel out, leaving us with
$\Rightarrow \int{\left( \dfrac{1+x}{1-x} \right)=\dfrac{2}{1!}x+\dfrac{4}{3!}{{x}^{3}}+\dfrac{48}{5!}{{x}^{5}}+..........}$
Simplifying the denominator on each term the factorials, that is, we end up with
$\Rightarrow \int{\left( \dfrac{1+x}{1-x} \right)=2x+\dfrac{4}{6}{{x}^{3}}+\dfrac{48}{120}{{x}^{5}}+.........}$
We can actually simplify a few fraction and factor out a $2$ out of each term, leaving us with
$\Rightarrow \int{\left( \dfrac{1+x}{1-x} \right)=2\left( x+\dfrac{1}{3}{{x}^{3}}+\dfrac{1}{5}{{x}^{5}}+.......... \right)}$
As we may notice, we can see that the power and the denominator on each term is increasing by $2,$ so we end we with
$\Rightarrow \int{\left( \dfrac{1+x}{1-x} \right)=2\left( x+\dfrac{1}{3}{{x}^{3}}+\dfrac{1}{5}{{x}^{5}}+.......... \right)}$
$\Rightarrow \int{\left( \dfrac{1+x}{1-x} \right)=2\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{2n+1}}}{2n+1}}}$

Note: Remember that a maclaurin series is a function that has an expansion series that gives the sum of derivatives of that function. The maclaurin series of a function $f\left( x \right)$ up to order $'n'$ may be found using series \left[ f,\left\\{ x,0,n \right\\} \right].