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 $f(x)$):
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Why is a point $x_0$ required to expand the function?
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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.
Best Answer
One way to think of Taylor series is as a generalization of the tangent line to a curve. At the point $x_0$ the tangent line to the graph of $y = f(x)$ can be written as
$$y = f(x_0) + f'(x_0)(x - x_0)$$
which is also called the linearization of $f$ at $x_0$: approximating the function by a linear/degree $1$ polynomial. If you want to approximate the function with a tangent "parabola" add higher-order derivatives and powers:
$$y = f(x_0) + f'(x_0)(x - x_0) + \dfrac{f''(x_0)}{2!}(x - x_0)^2$$
approximates the function with a degree $2$ polynomial at $x_0$.
If you want to approximate the function with a tangent "cubic" then continue the pattern:
$$y = f(x_0) + f'(x_0)(x - x_0) + \dfrac{f''(x_0)}{2!}(x - x_0)^2 + \dfrac{f'''(x_0)}{3!}(x - x_0)^3$$
Taylor series allow you to continue this to a degree $n$ polynomial:
$$y = f(x_0) + f'(x_0)(x - x_0) + \dfrac{f''(x_0)}{2!}(x - x_0)^2 + \dots +\dfrac{f^{(n)}(x_0)}{n!}(x - x_0)^n$$