Difference between Taylor's theorem and Taylor's series?

Question

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?

Answer 1

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).

Answer 2

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 $C^\omega$

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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 $|R_k(x)|\le M\frac{|x-a|^{k+1}}{(k+1)!}$