b8-taylor and maclaurin series: problem 7\n(2 points)\nfind the maclaurin series for ( f(x)=ln left(1-8…

b8-taylor and maclaurin series: problem 7\n(2 points)\nfind the maclaurin series for ( f(x)=ln left(1-8 x^{3}\right) ).\n( ln left(1-8 x^{3}\right)=sum_{n = 1}^{infty} )\non what interval is the expansion valid? give your answer using interval notation. if you need to use ( infty ), 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.

b8-taylor and maclaurin series: problem 7\n(2 points)\nfind the maclaurin series for ( f(x)=ln left(1-8 x^{3}\right) ).\n( ln left(1-8 x^{3}\right)=sum_{n = 1}^{infty} )\non what interval is the expansion valid? give your answer using interval notation. if you need to use ( infty ), 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.

Answer

Answer:

$\sum_{n = 1}^{\infty}\frac{-(8x^{3})^{n}}{n}$; $\left[-\frac{1}{2},\frac{1}{2}\right)$

Explanation:

Step1: Recall the Maclaurin series for $\ln(1 + u)$

The Maclaurin series for $\ln(1 + u)=\sum_{n=1}^{\infty}\frac{(- 1)^{n + 1}u^{n}}{n}$, which converges for $-1<u\leq1$.

Step2: Substitute $u=-8x^{3}$

Substitute $u = - 8x^{3}$ into the series for $\ln(1 + u)$. Then $\ln(1-8x^{3})=\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}(-8x^{3})^{n}}{n}=\sum_{n = 1}^{\infty}\frac{-(8x^{3})^{n}}{n}$.

Step3: Find the interval of convergence

We know that for the series of $\ln(1 + u)$, the condition is $-1<u\leq1$. Substituting $u=-8x^{3}$, we have $-1<-8x^{3}\leq1$.

  • Solve $-1<-8x^{3}$: Divide both sides by $- 8$ (and reverse the inequality sign), we get $\frac{1}{8}>x^{3}$, i.e., $x<\frac{1}{2}$.
  • Solve $-8x^{3}\leq1$: Divide both sides by $-8$ (reverse the inequality sign), we get $x^{3}\geq-\frac{1}{8}$, i.e., $x\geq-\frac{1}{2}$.

So the interval of convergence is $\left[-\frac{1}{2},\frac{1}{2}\right)$.