Equations of motion

I don't know how many times I think about this idea, but each and every time I get some new perspective on its definition. And also it comes with some more questions.

I was just deriving the equations of motion from the basic definitions of mechanics that is from the definition of acceleration, $$ a = \frac{dv}{dt} \\ $$ On Integrating from initial velocity u to final velocity v in time t we get the first equation 
$$ \\ \int_0^t a \,dt = \int_u^v \,dv \\ v= u + at$$  
And velocity is rate of change of displacement,   $ \\ \frac{dx}{dt} = v = u + at \\ \int_{x_0}^x \,dx = \int_0^t u+at \,dt$  gives $$ x - x_0 = ut+ \frac{1}{2} at^2$$ and by squaring the first equation and applying the second result we get,
$$ v^2 = [u+at]^2 \\ v^2 = u^2 + a^2t^2+ 2uat \\ v^2 = u^2+ 2a[ut + \frac{1}{2}at^2] \\ v^2 = u^2 + 2a [x-x_0]$$
Thus we obtain the three fundamental equations of motion and this is the usual method of derivation. 
          But unfortunately my mind was struggling and thinking about the Taylor series and its derivation. 
  With surprise!! It just made me to realize that these equations are nothing but the simple Taylor expansion of respective variables. 
Taylor expansion of a function is,
$$ f(x) = f(x_0) + \frac{df}{dx}|_{x_0} x + \frac {d^2f}{dx^2}|_{x_0} \frac {x^2}{2!} + ...$$
Now, if we expand the velocity as function of time,
$$ v(t) = v(t_0)+ \frac{dv}{dt}|_{t_0} t + \frac {d^2v}{dt^2}|_{x_0} \frac {t^2}{2!} +...$$

Thus we get the firs equation in the general form,
$$v = u + at + \frac{a't^2}{2!} + \frac{a''t^3}{3!} + ...$$
where $a', a'',..$ are the higher derivatives of acceleration. When higher derivatives become zero this general form results into the above equation where acceleration is constant over time.

        Similarly the Taylor expansion for displacement follows as,
$$x = x_0 + \frac{dx}{dt}|_{x_0}t + \frac{d^2x}{dt^2}|_{x_0}\frac{t^2}{2} +\frac{d^3x}{dt^3}|_{x_0}\frac{t^3}{6} +...$$
As we derived in the first case, by squaring the first expansion and combining it with the second, the general derivation for the third equation of motion can be obtained.  

        Really it is astounding for me that why I didn't see this for these many days. But eventually there are questions and that goes with the understanding of the higher derivatives and the simplicity in Newton's laws. 

Question is simple, 
Why Newton's law is defined that 

$$F = ma$$ and not $$ F = ma^2$$ or $$ F = ma'+ma''+...$$}

The most probable answer is, "It is the way the law is defined."

Yes.. of course. It is the axiomatic definition. But from Taylor expansion, you will find that function at a point is determined by its neighbourhood points since derivatives depends on continuity between the points. 
             Only when the higher derivatives are negligibly small, you will arrive at simple solution. All these things itself appropriately tells us that there are intrinsic conditions in our definitions. 
             To understand those conditions, it should be tried out by changing the fundamental definitions and rules.