Q. 56 - Calculus

Question

Find the Maclaurin series for the functions in Exercises 51–60

by substituting into a known Maclaurin series. Also, give the

interval of convergence for the series.

$\frac{x}{9-x^2}$

Step-by-Step Solution

Answer

The solution to the inequality is $(-3, 3)$

Step 1. Given Information

Consider the function $f\left(x\right)=\frac{x}{9-x^2}$

Step 2 : Simplification

We know that the Maclaurin series for the function $g\left(x\right)=\frac{1}{1-x}$ is $\frac{1}{1-x}=\sum _{k=0}^{\infty }{x}^k$ So to find the Maclaurin series for the function $f\left(x\right)=\frac{x}{9-x^2}$, we first rewrite the function in the form, $\frac{x}{9-x^2}=\frac{x}{9}\left[\frac{1}{1-{\left({\displaystyle \frac{x}{3}}\right)}^2}\right]$

Thus if we substitute ${\left(\frac{x}{3}\right)}^2$ for $x$ in the series of $g\left(x\right)=\frac{1}{1-x}$ and then multiply it by $\frac{x}{9}$ we get the series for $f\left(x\right)=\frac{x}{9-x^2}$

Therefore, $\frac{x}{9-x^2}=\frac{x}{9}\sum _{k=0}^{\infty }[{\left(\frac{x}{3}\right)}^2{]}^k$ implies that, $\frac{x}{9-x^2}=\sum _{k=0}^{\infty }{\left(\frac{1}{9}\right)}^{k+1} x^{2k+1}$

Step 3: Find the interval of convergence for the series.

Also, to find the interval of convergence of the new series we make the substitution in the inequality that defines the interval of convergence of the original series .

Therefore, we replace $x$ by ${\left(\frac{x}{3}\right)}^2$ in the inequality $\left|x\right|<1$

So, $\left|{\left(\frac{x}{3}\right)}^2\right|<1$

that is, $x^2<9$

Hence, the solution of the inequality is $\left|x\right|<3$

Therefore, the series for $f\left(x\right)=\frac{x}{9-x^2}$ converges to the function on the interval $(-3,3)$.