What is the significance behind taylor series? Why does taylor series have ample amount of importance in calculus? 
I like to know some insights behind taylor series. 
 A: Polynomials are about
the simplest type of functions there are.
They are closed under addition, multiplication,
integration, and differentiation.
Taylor series allow a function
to be approximated
in terms of polynomials
of a specified degree.
This gives a good indication
of how the function behaves locally.
Note that the Taylor series
of degree 1 is just 
the tangent.
The more terms of the Taylor series,
the closer the series fits the function
at the point where the series is computed.
This is a start.
There is lots more.
Also look up Chebychev approximation.
A: Taylor series use derivatives everywhere (that is one reason why they involve calculus).  It is interesting to know that values of a function $f(x)$ can be approximated using only derivative information at $x=0$. For a wide collection of functions, the approximation has increasing accuracy as more derivative information is used: 
$$f(x) \approx f(0) + xf’(0) + \frac{x^2}{2}f’’(0) + … + \frac{x^n}{n!} f^{(n)}(0)$$
Example (recall that $e^0=1$): 
\begin{align}
&e^{0.1} \approx e^0 + (0.1)e^0 = 1.1\\
&e^{0.1} \approx e^0 + (0.1)e^0 + \frac{(0.1)^2}{2}e^0 = 1.105\\
&e^{0.1} \approx e^0 + (0.1)e^0 + \frac{(0.1)^2}{2}e^0 +  \frac{(0.1)^3}{3!}e^0 = 1.105166666666667\\
&e^{0.1} \approx 1 + .1 + \frac{(.1)^2}{2} + \frac{(.1)^3}{3!} + \frac{(.1)^4}{4!} = 1.105170833333333
\end{align}
The approximations are converging to the true answer $e^{0.1} = 1.105170918075648…$. 
In the limit (in cases when the Taylor expansion applies) we get that $f(x)$ can be exactly written as an infinite sum of polynomial functions.  This is useful for understanding $f(x)$ in simpler ways.  For example, differentiating or integrating $f(x)$ can be done simply by differentiating or integrating each term, which involves differentiating/integrating polynomial functions, which is easy.  

Let $f(t)$ be the location of a car at time $t$. If you know where the car is at time $t=0$, and you know its velocity $v(0) = f'(0)$, then you can accurately predict where it will be at some small time later: 
$$ f(t) \approx f(0) + f'(0)t = f(0) + v(0)t $$
If you know both the velocity and the acceleration $a(0)=f''(0)$, then you can more accurately approximate: 
$$ f(t) \approx f(0) + f'(0)t + \frac{f''(0)}{2}t^2 = f(0)+v(0)t + \frac{a(0)t^2}{2} $$
You can see how this would be useful, for example, in a robotics control system that needs to guess where the robot will be in one second given its current location, velocity, and acceleration. 
