which series is the function $\frac{1}{1-6x^2}$ generating?

Question

which series is the function $\frac{1}{1-6x^2}$ generating? I think it should be $f(n)=6^n$

can you define the notion of a sequence generated by a function? Usually it just means to evaluate at $x=n$ in which case you get $\frac{1}{1-6n^2}$. — Yanko, Jul 06 '19 at 15:04
Suggest getting first several terms of Taylor series and seeing what pattern. — coffeemath, Jul 06 '19 at 15:04
it's not a series it's generating, but a sequence. That difference matters. — Henno Brandsma, Jul 06 '19 at 15:24

score 2 · Accepted Answer · answered Jul 06 '19 at 15:15

$$\frac{1}{1-\alpha}=\sum_{n=0}^\infty \alpha^n$$

is the foundation. Apply with $\alpha=6x^2$ and we get

$$\frac{1}{1-6x^2}=\sum_{n=0}^\infty (6x^2)^n = \sum_{n=0}^\infty 6^n x^{2n}$$

so this corresponds to the sequence $1, 0, 6, 0, 36, \ldots$, or in a formula

$$a_n = \begin{cases} 0 & n \text{ odd }\\ 6^{\frac{n}{2}} & n \text{ even } \end{cases}$$

if you prefer.

1 Answers1