Why is Taylor Series an exact match of the original function instead of an infinitely close approximation? If we must write the $f(x)$ in the form of a general power series, then I know all the following steps to calculate the coefficients.
However, how do we know that $f(x)$ can be written in such a form without losing any accuracy? There was once a linear approximation, then quadratic approximation. All of them are merely approximation. Why the series becomes the exact match when we raise the order of $n$ to $+\infty$?
Is there a convergence proof? By the way, I am also new to the "radius of convergence" concept. Probably they are related.
 A: If a function has a power series representation around $x_0:$
$$f(x)=\sum_{n=0}^{\infty} a_n(x-x_0)^n$$
for all $x\in (x_0-r,x_0+r)$ for some $r>0,$ then $f^{(n)}(x_0)$ exists for all $n$ and $$a_n=\frac{f^{(n)}(x_0)}{n!}.$$
If $f$ is a function defined on some interval $(a,b)$  such that $f(x)$ has a power series representation around every $x_0\in (a,b),$ then $f$ is called analytic on $(a,b).$
Note, Taylor series of analytic functions don’t necessarily converge for all $x.$ The function $f(x)=\frac1{1+x^2}$ is analytic on $(-\infty,\infty),$ but the maximum radius $r$ for any $x_0\in(-\infty,\infty)$ is $r=\sqrt{1+x_0^2}.$
Analytic functions are important in calculus and real analysis, but they are the heart of complex analysis. In complex analysis, analytic functions are the core of the topic.
Some properties of analytic functions:

*

*Polynomials are analytic.

*If $f,g$ are analytic on $(a,b),$ then $f+g, f\cdot g$ are analytic on $(a,b).$

*If $g(x)\neq 0$ for all $x\in(a,b)$ then $\frac fg$ is analytic.

*(Composition) If $f$ is analytic on $(a,b)$ and $g$ is analytic on $(c,d)$ with $(c,d)\subseteq f((a,b))$ then $h(x)=g(f(x))$ is analytic on $(a,b).$

*(Inverse functions) If $f(x)$ is analytic on $(a,b)$ and $x_0\in (a,b)$ with $f’(x_0)\neq 0,$ and $y=f(x_0)$ then there is an analytic function $g$ on some interval $(c,d)$ containing $y$ such that $g(f(x))=x.$

*If $f$ is analytic on $(a,b)$ then $f’$ exists and is analytic on $(a,b).$ Also, any anti-derivative of $f,$ $F’=f,$ on $(a,b)$ is analytic there.

The most important analytic function which doesn’t follow from the above is $f(x)=e^x.$ A lot of the most basic analytic functions, like $\log$ and the trigonometric functions, follow from $e^x$ and the above rules.
A: I don't know what you mean exactly by "infinitely close approximation", but an "infinitely close approxmation" sounds like an exact match. Example from a simpler domain: $0.9999...$ (infinitely many $9$s) is $1$, not an approximation to $1$.
There are cases when there is no convergence, ie step-functions. The error does not go to $0$. You can find more about it if you look into Gibbs phenomenon. As an illustration, here is an image from Wikipedia:

The error doesn't go away even if you send $N$ to infinity.
