91Ó°ÊÓ

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

We know that the Maclaurin series for the function $g\left(x\right)=\frac{1}{1-x}$is $\frac{1}{1-x}=\underset{k=0}{\overset{∞}{∑}}{x}^k$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 localid="1650537976575" ${\left(\frac{x}{3}\right)}^2$for localid="1650537981240" $x$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 localid="1649867585498" $f\left(x\right)=\frac{x}{9-x^2}$

Therefore,$\frac{x}{9-x^2}=\frac{x}{9}\underset{k=0}{\overset{∞}{∑}}[{\left(\frac{x}{3}\right)}^2{]}^k$implies that,$\frac{x}{9-x^2}=\underset{k=0}{\overset{∞}{∑}}{\left(\frac{1}{9}\right)}^{k+1}{x}^{2k+1}$

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 $(-3,3)$.