Q. 55 - Calculus

Question

 Find the Maclaurin series for the functions in the Given Exercises by substituting it into a known Maclaurin series. Also, give the interval of convergence for the series. 

      $\frac{1}{8+x^3}$

Step-by-Step Solution

Answer

Ans:  Hence, the solution to the inequality is $\left|x\right|<2$. Therefore, the series for $f\left(x\right)=\frac{1}{8+x^3}$ converges to the function on the interval $(-2,2)$.

Step 1. Given information.

given,    $\frac{1}{8+x^3}$

Step 2. Consider the given function,

We know that the Maclaurin series for the function is

So, to find the Maclaurin series for the function, we first rewrite this function in the form

Thus, if we substitute for x in the series for $g\left(x\right)=\frac{1}{8+x^3}$ and then multiply $\frac{1}{8}$, we get a series for

    $f\left(x\right)=\frac{1}{8+x^3}$

Therefore,

    $\frac{1}{8+x^3}=\frac{1}{8}\sum _0^{\mathrm{\infty }} {\left[{\left(-\frac{x}{2}\right)}^3\right]}^k$

Implies that,

      $\frac{1}{8+x^3}=\frac{1}{8}\sum _0^{\mathrm{\infty }} {\left(-\frac{1}{8}\right)}^k{x}^{3k}$

Step 3. 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 ${\left(\frac{-x}{2}\right)}^3$ in the inequality $\left|x\right|<1$
 So,

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

That is,

    $x^3<8$

Hence, the solution to the inequality is $\left|x\right|<2$. Therefore, the series for $f\left(x\right)=\frac{1}{(8+x^3)}$ converges to the function on the interval $(-2,2)$.