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.
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.