Question

What makes the taylor polynomial of a function so accurate at x=0 when looking at a function which is differentiable and to the nth order?

Answer 1

the taylor polynomial shares the same point as the taylor polynomial at x= 0 , and shares the first derivative, second derivative.. up to nth derivative

Answer 2

It's pretty trippy imo that if you have two functions that have the same value for all their derivatives at a single point, then they are the same function EVERYWHERE.

Answer 3

Since derivatives can be used to represent velocity, acceleration, change in acceleration, etc. If you have all of these derivatives (infinite knowledge) of a particle at a point, then you have all knowledge of its future and past, at least classically speaking.

Answer 4

Of course in theory there's no difference between theory and practice, but in practice there is... so I suppose you should take this with a grain of salt haha. But Taylor series are still cool.

Answer 5

Are you saying, even though you do not know F(x) out right, as long as you have the values of the derivatives to the highest degree of F(x) then you can arrive at a polynomial that looks like F(x) using the taylor approximation?

Answer 6

Up to* the highest degree

Answer 7

@Empty Is it possible to have two functions, one analytic and one is not, such that their Taylor series coincide at one point but the two functions are not equal?

Answer 8

Something like this.
\[
\begin{align*}
f(x) &= \begin{cases}e^{-\frac{1}{1-x^2}} & \mbox{ if } |x| < 1, \\ 0 &\mbox{ otherwise }\end{cases}\\
g(x) &= 0
\end{align*}
\]

Answer 9

f(x) copied directly from "Smoothness" in Wikipedia.

Answer 10

Ok so I lied :X I'm a liar. But it was a white lie for good cause, plus I was kinda referencing physics. I guess if you hold a physicist at gunpoint he too will tell you that gauges aren't completely arbitrary, they must be analytic as well otherwise Gauss and Stokes theorems won't work out buuuuuuut...