I was playing around trying to check if the $n$th order Taylor approximation about $0$ can be explained and obtained by using a physical reasoning argument like below for the case $n = 2$ (that is given $f(0),f’(0)$ and $f’’(0)$ and thinking of $f$ as the position of a car after $t$ seconds).To estimate $f(t),$ with $f’(0),f’’(0)$, divide $t$ by $2$ (ideally $n$ in the general case) and estimate $f$ at those time steps.So after $t/2$ seconds, the position can be estimated as $f(0)+f’(0)t/2$.Also, its velocity is approximately $f’(0)+f’’(0)t/2$.Now, having lost info about the acceleration, the best we can do is use the estimated position and velocity after $t/2$ seconds to estimate $f(t)$ as\begin{align*}f(t/2)+f’(t/2)t/2 &= f(0)+f’(0)t/2 [f’(0)+f’’(0)t/2]t/2 \\&= f(0)+f’(0)+f’’(0)/2\end{align*}which checks out with the $n =2$ Taylor expansion. Using a similar logic for $n = 3$ (take steps of size $t/3$) I think that the pattern breaks and it doesn’t match what it should be for Taylor. I concretely got as an estimate $f(0)+f’(0)+f’’(0)(t^2)/3+f’’’(0)(t)^3/9.$ Could someone else do it and let me known if it breaks in fact?
↧