Let $f$ be an infinite times differentiable function. Is it true that:
the higher the degree $n$ of the Taylor polynomial $T_{n,f,x_0}$ of $f$ around $x_0$, the better the approximation?
Some thoughts. Given $n$, polynomial $T_{n,f,x_0}$ is the best approximation of $f$ near $x_0$ that fulfils the requirement of equal derivatives with $f$ at $x_0$. So regarding polynomials of degree at most $n$, $T_{n,f,x_0}$ is the winner.
On the other hand, although one would like $T_{n,f,x_0}$ to fit better function $f$, as $n$ grows larger, it seems to me that there is no reason for this to happen. Of course, one should define what "fit better" means. In our case, it would be something like:
$$\sup_{x\in I}|T_{n+1,f,x_0}(x)-f(x)|\leqslant \sup_{x\in I}|T_{n,f,x_0}(x)-f(x)|$$
in a neighborhood $I$ of $x_0$.
Of course, I must admit that the cases I see graphically most of the times, fulfill the last requirement, by fitting better and better the graph of $f.$
Thank in advance for the help.
Best Answer
WLOG $x_0=0.$ Claim: Given $n,$ there exists $r>0$ such that
$$|f(x)-T_{n+1}(x)| \le |f(x)-T_{n}(x)|,\,|x|<r.$$
To prove this, note first that if $f^{(n+1)}(0)=0,$ then $T_{n+1}=T_n,$ and there is nothing to prove. So assume $f^{(n+1)}(0)\ne 0.$ By the Langrange form of the remainder, we have
$$f(x)-T_{n}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1},$$
where $c$ is between $0$ and $x.$
Because $f^{(n+1)}(0)\ne 0,$ there is $s>0$ and a positive constant $A$ such that $|f(x)-T_{n}(x)| \ge A|x|^{n+1}$ for $|x|<s.$ But using Lagrange again, we have the standard estimate
$$|f(x)-T_{n+1}(x)|=O(x^{n+2})$$
as $x\to 0.$ Since $O(x^{n+2})$ is bounded above by $A|x|^{n+1}$ for small $x,$ we have the desired result.
We can say more: either $f^{(n+1)}(0)= 0,$ in which case $T_{n+1}=T_n,$ or $f^{(n+1)}(0)\ne 0,$ in which case there will exist $r>0$ such that
$$|f(x)-T_{n+1}(x)| < |f(x)-T_{n}(x)|,\,0<|x|<r.$$
I've taken a few liberties in getting the main points across. Ask if you have questions.