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 (so we use the scalars here instead of vectors):
$$ \begin{aligned} \frac{\mathrm{d} x}{\mathrm{d} t} &=v_t \\ \\ \mathrm{d} x&=v_t \times \mathrm{d} t \\ \int_{t_0}^{t} \mathrm{d} x&= \int_{t_0}^{t}\left(v_t \right) \ \mathrm{d} t \\ x_t&=\int_{t_0}^{t}\left(v_t \right) \ \mathrm{d} t + x_0 \end{aligned} $$
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:
$$ x_t=\int_{t_0}^{t}\left(\int_{t_0}^{t}\left(\int_{t_0}^{t}\left( j_t\right) \ \mathrm{d} t \right) \ \mathrm{d} t \right) \ \mathrm{d} t + \dfrac{\ddot{x}_t|_{t=t_0}(t-t_0)^2}{2!}+\dfrac{\dot{x}_t|_{t=t_0}(t-t_0)}{1!}+ \dfrac{x_{t=t_0}}{0!} $$
$$ \text{Where } \dot{x}_{t} = \dfrac{\mathrm{d} x}{\mathrm{d} t}, \ddot{x}_{t} =\dfrac{\mathrm{d}^2 x}{(\mathrm{d} t)^2} \text{ and so on.} $$
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:
$$ f(t)=\sum_{n=0}^{\infty }\left( \frac{f^{(n)}(t)|_{t=t_0}}{n!} \left(t-t_0\right)^{n}\right) $$
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 expression of a function.
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