# Taylor Series, Maclaurin series

From Rogawski ET 2e section 10.7, exercise 4.

Find the Maclaurin series for $f(x) = \dfrac{x^2}{1-8x^8}$.

$$\dfrac{x^2}{1-8x^8} = \sum_{n=0}^{\infty} [\textrm{_______________}]$$

Hi! I am working on some online Calc2 homework problem and I am not quite sure how to go about solving this Taylor series. I know I should substitute $8x^8$ for $x$ in the Maclaurin series for $1 \over (1-x)$, but the $x^2$ in the numerator of the problem is throwing me off. If someone could help me find the Maclaurin series and on what interval the expansion is valid, I would greatly appreciate it!

Notice $\frac{1}{1-x} = \sum x^n$ . Hence

$$\frac{ x^2}{1 - 8x^8} = x^2 \sum (8x^8)^n = \sum 8^n x^{8n + 2}$$

and this is valid for $| 8x^8 | < 1$

• Thank you so much, I have more more question related to your answer…how would I write |x|<(1/8)^(1/8) in interval notation? – user124539 Apr 6 '14 at 21:59

You can just write $$\frac{x^2}{1-8x^8}=x^2\left(1+8x^8+(8x^8)^2+\cdots\right)=x^2+8x^{10}+64x^{18}+\cdots$$ The interval of convergence will be for all $x$ such that $$|8x^8|<1$$ which is equivalent to $$|x|<\left(\frac{1}{8}\right)^{{1\over 8}}$$

• Thank you so much, I have more more question related to your answer…how would I write |x|<(1/8)^(1/8) in interval notation? – user124539 Apr 6 '14 at 22:00
• It would be $$\left(-\left(\frac{1}{8}\right)^{{1\over 8}},\left(\frac{1}{8}\right)^{{1\over 8}}\right)$$ – user138335 Apr 6 '14 at 22:03

What you have got is almost the right answer. $$\frac{x^2}{1-8x^8} = x^2 \sum_{n=0}^{\infty}(8x^8)^n = \sum_{n=0}^{\infty}8^nx^{8n+2}$$, which is valid if $|8x^8|<1$,hence, $|x|<\frac{1}{8^{1/8}}$