Question

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

Answer 1

Hint: 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 \[\left[ f,\left\{ x,0,n \right\} \right].\]