Q 30 - Calculus

Question

Use the Maclaurin series for $\frac{1}{1-x}$to find a power series representation for $\frac{x^2}{{(1-x^3)}^2}$

Step-by-Step Solution

Answer

The power series for the function is $f\left(x\right)={\left(\sum _{k=0}^{\infty }\frac{x^{k+1}}{1+x^2+x}\right)}^2$

Step 1: Given information

$f\left(x\right)=\frac{x^2}{{\left(1-x^3\right)}^2}$

Step 2: Find the Maclaurin series for the function

$g\left(x\right)=\frac{1}{1-x}$at $x=0$is

$1+x+x^2+x^3+\cdots +x^k+\cdots$

Or $f\left(x\right)=\sum _{k=0}^{\infty }{x}^k$

Step 3: The function f ( x ) = x 2 1 - x 3 2   is written as:

$$\begin{gathered} \frac{x^2}{{(1-x^3)}^2}=\frac{x^2}{(1-x^2){(1+x^2+x)}^2} \\ ={\left(\frac{x}{1+x^2+x}\right)}^2\frac{1}{(1-x)2} \end{gathered}$$

Step 4: Find the power series for the function

The power series for the function $f\left(x\right)=\frac{x^2}{{\left(1-x^3\right)}^2}$is

$\frac{x^2}{{\left(1-x^3\right)}^2}={\left(\frac{x}{1+x^2+x}\right)}^2{\left(\sum _{k=0}^{\infty }{x}^k\right)}^2$

That is

 $f\left(x\right)={\left(\sum _{k=0}^{\infty }\frac{x^{k+1}}{1+x^2+x}\right)}^2$