b8-taylor and maclaurin series: problem 2 (2 points)\nfind the maclaurin series for ( f(x)=\frac{x^{4}}{1…

b8-taylor and maclaurin series: problem 2 (2 points)\nfind the maclaurin series for ( f(x)=\frac{x^{4}}{1 - 4x^{8}} ).\n( \frac{x^{4}}{1 - 4x^{8}}=sum_{n = 0}^{infty} )\non what interval is the expansion valid? give your answer using interval notation. if you need to use is the only point in the interval of convergence, you would answer with 0.\nthe expansion is valid on\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.\npage generated\nwebwork © 1996 - 2025 | th

b8-taylor and maclaurin series: problem 2 (2 points)\nfind the maclaurin series for ( f(x)=\frac{x^{4}}{1 - 4x^{8}} ).\n( \frac{x^{4}}{1 - 4x^{8}}=sum_{n = 0}^{infty} )\non what interval is the expansion valid? give your answer using interval notation. if you need to use is the only point in the interval of convergence, you would answer with 0.\nthe expansion is valid on\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.\npage generated\nwebwork © 1996 - 2025 | th

Answer

Answer:

$\sum_{n = 0}^{\infty}4^{n}x^{8n + 4}$; $\left(-\frac{1}{\sqrt[8]{4}},\frac{1}{\sqrt[8]{4}}\right)$

Explanation:

Step1: Recall the geometric series formula

The geometric series formula is $\frac{1}{1 - t}=\sum_{n = 0}^{\infty}t^{n}$, for $|t|\lt1$.

Step2: Substitute $t = 4x^{8}$ into the formula

We have $\frac{1}{1-4x^{8}}=\sum_{n = 0}^{\infty}(4x^{8})^{n}=\sum_{n = 0}^{\infty}4^{n}x^{8n}$.

Step3: Multiply by $x^{4}$

Since $f(x)=\frac{x^{4}}{1 - 4x^{8}}$, then $f(x)=x^{4}\sum_{n = 0}^{\infty}4^{n}x^{8n}=\sum_{n = 0}^{\infty}4^{n}x^{8n+4}$.

Step4: Find the interval of convergence

For the geometric - series $\sum_{n = 0}^{\infty}t^{n}$ with $t = 4x^{8}$, we need $|4x^{8}|\lt1$. Solve the inequality $|4x^{8}|\lt1$: First, $|x^{8}|\lt\frac{1}{4}$, then $|x|\lt\frac{1}{\sqrt[8]{4}}$. The interval of convergence is $x\in\left(-\frac{1}{\sqrt[8]{4}},\frac{1}{\sqrt[8]{4}}\right)$.