# Taylor approximation

Suppose we're given the position $x$ , velocity $\dot{x}$ , and acceleration $\ddot{x}$ at $t_1$.

We need to approximate the position $x$ at $t_2$.

In below picture, I understand how $x_1$ represents the vertical black length, and how $\color{blue}{\dot{x_1}(t_1)}h$ represents the vertical blue length. However I don't get how $\color{green}{\ddot{x}(t_1)}\dfrac{h^2}{2}$ represents the green length. I'm wondering if there a simple derivation for proving that the green vertical length equals $\color{green}{\ddot{x}(t_1)}\dfrac{h^2}{2}$ . Any ideas ? Thanks !

• This question is from signals and systems, but I completely understand if this question belongs in math and I'll gladly post it in math if you think so.. (I got so used to electronics page that I didn't realize this till after finishing composing the q..) Commented Jun 15, 2018 at 7:40
• I think this is indeed better suited to Math SE. I'm usually not too strict on this type of stuff, but I believe the connection to electronics is too weak for this question ;-) Commented Jun 15, 2018 at 8:14
• Integral of h is.......? Commented Jun 15, 2018 at 8:44
• @Andyaka Integral of h is h^2/2. That quadratic is what we get by integrating x''(t) = c. I now see clearly how the taylor quadratic looks geometrically. Thank you so much :) Commented Jun 15, 2018 at 9:09
• Yes @LongPham if I recall correctly, physicists use $\dot{x}$ to represent derivative of $x$ with respect to time, and mathematicians use $x'$... Not sure about engineers hmm Commented Jun 15, 2018 at 9:59

$$ah^2/2$$
• Ah so $ah^2/2$ is $a(t_2-t_1)^2/2$, and this term is part of the whole quadratic $x(t_1) + x'(t_1)(t_2-t_1) + x''(t_1)(t_2-t_1)^2/2$ . I see... that average acceleration interpretation is clever! Can we interpret something similar for the fourth term too ? $ah^3/6$ Commented Jun 15, 2018 at 9:10