Difference between Taylor's theorem and Taylor's series? What does Taylor's theorem say? How do we use Taylor's theorem to get to Taylor's series? I need the basic idea behind these two. What's going on between them?
 A: Rougly, the Taylor theorem says: if $f$ has enough derivatives in a nhood of the point $a$, then $f$ is the sum of a polynomial and a remainder:
$$f(x) = \sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x - a)^k + R_k(x).$$
If you can calculate the derivatives (see What do Taylor series accomplish?) and the remainder is "small",
$$f(x) \approx\hbox{ a polynomial.}$$
Why this is important? Because polynomials are calculable.
If $f$ is $C^\infty$, you can write the Taylor series
$$
\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x - a)^k,
$$
but
$$
f(x) = \sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x - a)^k
$$
can be false. When is true (this depends obviously of the behavior of the remainder) we say that the function is analytic. But being $C^\infty$ isn't enough. The function $f(x) = e^{-1/x^2}$ for $x\ne 0$, $f(0) = 0$ is $C^\infty$ but its Taylor series at $x = 0$ is zero. Even worse: there are examples of functions with divergent Taylor series (Real Mathematical Analysis by C. Pugh).
A: Although the exact syntax of Taylor theorem is not agreed upon generally the basic of the theorem is:
For every $k\in\Bbb N$ differentiable function at some point $a$, $f$, exists $h_k$ such that $$\lim_{x\to a}h_k(x)=0,\;f(x)=\sum_{n=0}^k\frac{f^{(n)}(a)}{n!}(x-a)^n+h_k(x)(x-a)^k$$
Taylor series is:

  
*
  
*if $f$ is of class $C^\omega$, then $h_k(x)=\sum_{n=k+1}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^{n-k-1}$
  

The difference is that not all functions are class $C^\omega$

Few more things things that the theorem says is that if there exists polynomial, $p(x)$, such that $f(x)=p(x)+h_k(x)(x-a)^k$, then $p(x)=\sum_{n=0}^k\frac{f^{(n)}(a)}{n!}(x-a)^{n}$(uniqueness)
If so we can calculate the approximation error: $R_k(x)=f(x)-p(x)$ and $\lim_{x\to a}R_k(x)=o(|x-a|^k)$.
Using this we can get $|R_k(x)|\le M\frac{|x-a|^{k+1}}{(k+1)!}$
