I'll admit I've always had a few confusions about the Taylor Series, as it itself looks completely unintuitive to me.

$$f(x) = f(x\_0)+f'(x\_0)(x-x\_0)+\frac{f''(x\_0)(x-x\_0)^2}{2!}+\frac{f'''(x\_0)(x-x\_0)^3}{3!}+ \ ...$$

This looks really weird to me for a couple reasons (just the general expansion of $f(x)$):

* Why is a point $x\_0$ required to expand the function?
* Why is the term $(x-x\_0)$ and not $(x+x\_0)$? I know shifting a graph to the right involves taking $f(x)$ to $f(x-x\_0)$ but I've never really understood why intuitively, and why this is needed to expand the series.

Taking the derivative of $f(x) = f'(x)$.

$$\therefore f'(x) = 
\frac{d}{dx}\left(f(x\_0)+f'(x\_0)(x-x\_0)+\frac{f''(x\_0)(x-x\_0)^2}{2!}+\frac{f'''(x\_0)(x-x\_0)^3}{3!}+ \ ...\right)$$
$$f'(x) = f'(x\_0)+ f''(x\_0)\frac{d}{dx}[(x-x\_0)]+ f'''(x\_0)/2!\frac{d}{dx}[(x-x\_0)]^2 \ + \ ...$$

$$f'(x) = f'(x\_0) - f''(x\_0)+f'''(x\_0)(x-x\_0)\ + \ ...$$

Now there are two constant terms, and the term to first have the $(x-x\_0)$ multiplied to it is now the third term. I don't see any intuitive pattern here or anything.

I suppose what I'm asking for is an explanation for the intuition going on here other than the expansion matches any function evaluated at a point for every derivative. I'm asking for the two bulleted points to be addressed, with an explanation. I've tried reading about it a bit now and it hasn't been coming nicely to me, but if someone instead thinks I should read about it somewhere they think is good with explaining it I welcome that as a comment too.