I still remember the first time that I meet the concept of Taylor series was in one AP calculus textbook. It’s kind of a pity that the AP calculus textbook I was reading at that time didn’t give any further explanation about why and how this formula came up, etc. So I just questioned myself about its formation and finally I did figured it out at that time by using the knowledge I’ve learned in Physics. I guess you may wonder why wouldn’t I just search on the Internet and see the proof of Taylor series, in fact, I did do that, but didn’t work for me at that point, since I lack too much Maths knowledge to understand that proof. Therefore I found a new way to help myself understand the Taylor series at that time.

Firstly, let us derive the displacement formula with acceleration in one dimension:

$$ \begin{aligned} \frac{\mathrm{d} x}{\mathrm{d} t} &=v_t \\ \\ \mathrm{d} x&=v_t \times \mathrm{d} t \\ \int_{t_0}^{t} \mathrm{d} x&= \int_{t_0}^{t}\left(v_t \right) \ \mathrm{d} t \\ x_t&=\int_{t_0}^{t}\left(v_t \right) \ \mathrm{d} t + x_0 \end{aligned} $$

Similarly, we can use the same deriving method of:

$$ \frac{\mathrm{d} x}{\mathrm{d} t} = v_t \quad \Rightarrow \quad x_t=\int_{t_0}^{t}\left(v_t \right) \ \mathrm{d} t + x_0 $$

To get:

$$ \frac{\mathrm{d} v}{\mathrm{d} t} = a_t \quad \Rightarrow \quad v_t=\int_{t_0}^{t}\left(a_t \right) \ \mathrm{d} t + v_0 $$

Then we substitute $v_t$ into $x_t$ , we would have:

$$ x_t=\int_{t_0}^{t}\left(\int_{t_0}^{t}\left(a_t \right) \ \mathrm{d} t \right)\ \mathrm{d} t + v_0(t-t_0) + x_0 \tag{1} $$

Note that since $a_t$ is a constant (In fact, this is related to the Principle of least action in Physics: We assume that the accleration of any system must be a constant, in other words: $\dfrac{\mathrm{d}^3 (x)}{(\mathrm{d} t)^3} = 0$), therefore we could get our displacement formula with acceleration:

$$ x_t=\dfrac{a_t(t-t_0)^2}{2}+v_0(t-t_0)+x_0 $$

But the problem is: What if $a_t$ is not a constant anymore (And that situations do happen a lot in real life), for example, have you heard about presence of jerk ( $j_t =\dfrac{\mathrm{d} a}{\mathrm{d} t}$ ) before?

Then we need to deal with the equation $(1)$ again, through using the same deriving method stated above, we could get:

$$ \frac{\mathrm{d} a}{\mathrm{d} t} = j_t \quad \Rightarrow \quad a_t=\int_{t_0}^{t}\left(j_t \right) \ \mathrm{d} t + a_0 $$

Next we substitute $a_t$ into $x_t$ , we can have:

$$ x_t=\int_{t_0}^{t}\left(\int_{t_0}^{t}\left(\int_{t_0}^{t}\left( j_t\right) \ \mathrm{d} t \right) \ \mathrm{d} t \right) \ \mathrm{d} t + \dfrac{a_0(t-t_0)^2}{2}+ v_0(t-t_0)+ x_0 $$

And using some derivatives form to substitute in that equation:

$$ x_t=\int_{t_0}^{t}\left(\int_{t_0}^{t}\left(\int_{t_0}^{t}\left( j_t\right) \ \mathrm{d} t \right) \ \mathrm{d} t \right) \ \mathrm{d} t + \dfrac{\ddot{x}_{t=t_0}(t-t_0)^2}{2!}+\dfrac{\dot{x}_{t=t_0}(t-t_0)}{1!}+ \dfrac{x_{t=t_0}}{0!} $$
$$ \text{Where } \dot{x}_{t} = \dfrac{\mathrm{d} x}{\mathrm{d} t}, \ddot{x}_{t} =\dfrac{\mathrm{d}^2 x}{(\mathrm{d} t)^2} \text{ and so on.} $$

Now I believe that you could clearly see the presence of the Taylor series in this equation right?

Therefore if we keep iterating using the same method and let $x_{t}=f(t)$ , we would be able to get the real Taylor series formula:

$$ f(t)=\sum_{n=0}^{\infty }\left( \frac{f^{(n)}{'}({t=t_0})}{n!} \left(t-t_0\right)^{n}\right) $$

This also indicate us the essence of Taylor series: Knowing all of its $n$th ($n$ is natural numbers) derivatives value at one point is equivalent to Knowing the primitive formula of a function.