Find the Taylor series generated by f at x=a.

Question

$f(x) = \frac 1 {9 - x}, a = 3$.

The answer in the book is $$\sum_{n = 0}^{\infty} \frac{(x - 3)^n}{6^{n + 1}}$$but I'm not sure how to get the above.

Answer 1

We know that:

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

I assume that $a$ is the center that you want to write the Taylor expansion around it.

Therefore:

$$\frac{1}{9-x}=\frac{1}{6}\frac{1}{1-\frac{(x-3)}{6}}=\frac{1}{6} \cdot (1+\frac{(x-3)}{6}+\cdots+\frac{(x-3)^n}{6^n})$$

Answer 2

Hint You can write $$ \frac{1}{9-x} = \frac{1}{6 - (x - 3)} = \frac16 \frac{1}{1 - \left( \frac{x-3}{6} \right)} $$ Then use the fact that $$ \frac{1}{1 - y} = \sum_{k \ge 0} y^k. $$