is the limit as k approaches infinity of a Taylor Polynomial of order k, that approximates a function f, the same as the function itself?

Question

Since Taylor polynomial approximation gets better as It's order gets bigger, so I was wondering, what happens when this order approaches infinity? Does the approximation equal the function itself and become perfect?

Thanks in advance. Loving calculus so far :)

Answer 1

In general, no. Take, for instance,$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}e^{-1/x^2}&\text{ if }x\ne0\\0&\text{ if }x=0.\end{cases}\end{array}$$You can check that $(\forall n\in\Bbb Z_+):f^{(n)}(0)=0$. So, $\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=0\ne f(x)$ (unless $x=0$).

Answer 2

There is a major difference between $$f(x)=\sum_{n=0}^p \frac {x^n}{n!}=\frac{ \Gamma (p+1,x)}{p!}e^x$$and $$g(x)=\sum_{n=0}^\infty \frac {x^n}{n!}=e^x$$ even if $p$ is very large.

Answer 3

The Taylor development will not converge to the function past singularities. For instance

$$\sum_{k=0}^\infty x^k$$ will converge to $$\frac1{1-x}$$ only when $|x|<1$.

And as shown by José, the development might not converge at all to the function.