Properties of the remainder function for Taylor polynomials Considered $f$ differentiable at least $n$ times in $x_0$ and $P_{n,x_0}(x)$ the $n$ degree Taylor polynomial in $x_0$.
Defined the Remainder function $R(x)= f(x)-P_{n,x_0}(x)$
I can't understand the following properties of $R(x)$:


*

*If $f \in C^{n} $ then $R(x) \in C^{n}$

*$R(x)$ and all its derivatives are zero at $x_0$


Can anyone help me?
Thanks a lot in advice
 A: *

*Let's see how it works for a disc, $B\subset \mathbb{C}$, as this is the easiest domain for a taylor expansion.


If $f(x) \in C^n(B)$ then we may write for the truncated expansion around $x_0 \in B$ (and for $x\in B$)
$$P_{n,x_0}(x)=\sum_{i=1}^n \frac{f^{(i)}(x_0)}{i!}(x-x_0)^i. $$
This function is clearly in $C^n(B)$ (its just a polynomial). Therefore
$$R(x):=f(x)-P_{n,x_0} \in C^n(B)$$
as $C^n(B)$ is closed under addition. In english: add (or subtract) something differentiable to something differentiable and it is surely differentiable.
2.
If you evaluate $P_{n,x_0}$ at $x_0$ the only term that remains is $f(x_0)$ (the rest get killed by $(x-x_0)$), a similar result holds true for the derivatives of $P_{n,x_0}$ (try it yourself).
A: *

*The sum of two $C^n$ functions is $C^n$, hopefully for obvious reasons. A polynomial is smooth, so subtracting the polynomial $P_{n,x_0}(x)$ from the $C^n$ functions gives a $C^n$ function.

*This must be "first $n$ derivatives", since we are assuming that $R(x)$ could have only its first $n$ derivatives. You probably know how this works: $P_{n,x_0}$ is really defined so this is true:
$$ P_{n,x_0}(x) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k, $$
and the $r$th derivative of this is
$$ P_{n,x_0}^{(r)}(x) = f^{(r)}(x_0) + f^{(r+1)}(x_0)(x-x_0) + \frac{f^{(r+2)}(x_0)}{2!}(x-x_0)^2 + \dotsb, $$
if $0 \leqslant r \leqslant n$. Evaluating this at $x=x_0$, of course, gives $f^{(r)}(x_0)$. Hence $R^{(r)}(x_0) = f^{(r)}(x_0)-f^{(r)}(x_0) = 0$, for all $0 \leqslant r \leqslant n$.
