I' ve been asked to find the coefficient of $$(x-3)^6$$ Which is said to be in Taylor's expansion, in addition it is said that c=3 which is the point for Taylor's expansion.
Also, I' have the function $$f(x)=\frac{1}{x^2-6x+12}$$ for which the coefficient needs to be found.
I' m now confused. I've only seen tasks with functions which need to be expanded in series via Taylor's formula (where I take multiple derivatives and then form a series based on the derivatives pattern).
Could anyone clarify how to find the coefficient? I could neither find any reference on-line for such an example...
Hint:
$$f(x)=\frac{1}{x^2-6x+12}=\frac{1}{(x-3)^2+3}=\frac{1}{3}\frac{1}{(\frac{x-3}{\sqrt{3}})^2+1}$$
and use the geometric series $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty }(-1)^nx^{2n}$$
so that $$x\rightarrow \frac{x-3}{\sqrt{3}}$$ $$\frac{1}{x^2-6x+12}=\frac{1}{3}\sum_{n=0}^{\infty }(-1)^n\frac{(x-3)^{2n}}{3^n}$$ now let $n=3$ to get the coefficient