Having trouble understanding Taylor Series I'm having trouble interpreting the Taylor series formula. The nth term of the Taylor series looks like the nth integral of f(x). Is this correct? If so, I don't quite understand the meaning of the nth integral, and how it is able approximate f(x) at higher values of n. 
Edit: I was looking at an example where f(x) = e^x, which looked like the nth integral as n increased. Specifically, I'm trying to make sense of the division by n! and how this helps approximating the original function. 
 A: From a purely symbol manipulation point of view, you can easily obtain the Taylor series formula in the following way.
Start by assuming
$$f(x) \; = \; a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5 + a_6x^6 + \; \cdots $$
Plugging $x=0$ into this tells us that $f(0) = a_0.$
Now differentiate both sides, assuming we can differentiate a sum of infinitely many terms like we can differentiate a sum of finitely many terms:
$$f'(x) \; = \; a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + 5a_5x^4 + 6a_6x^5 + \; \cdots $$
Plugging $x=0$ into this tells us that $f'(0) = a_1.$
Differentiate again:
$$f''(x) \; = \; 2a_2 + (3)(2)a_3x + (4)(3)a_4x^2 + (5)(4)a_5x^3 + (6)(5)a_6x^4 + \; \cdots $$
Plugging $x=0$ into this tells us that $f''(0) = 2a_2,\;$ or $\;a_2 = \frac{1}{2}f''(0).$
Differentiate again:
$$f^{(3)}(x) \; = \; (3)(2)a_3 + (4)(3)(2)a_4x + (5)(4)(3)a_5x^2 + (6)(5)(4)a_6x^3 + \; \cdots $$
Plugging $x=0$ into this tells us that $f^{(3)}(0) = (3)(2)a_3,\;$ or $\;a_3 = \frac{1}{3!}f^{(3)}(0).$
Differentiate again:
$$f^{(4)}(x) \; = \; (4)(3)(2)a_4 + (5)(4)(3)(2)a_5x + (6)(5)(4)(3)a_6x^2 + \; \cdots $$
Plugging $x=0$ into this tells us that $f^{(4)}(0) = (4)(3)(2)a_4,\;$ or $\;a_4 = \frac{1}{4!}f^{(4)}(0).$
Keep going in this manner.
