Taylor polynomial of order $n$ for a polynomial of degree $n$ I noticed that the Taylor polynomial of order $n$ for a polynomial function of degree $n$ is identical to the function. I tried to understand the reason but couldn't really figure it out. Any input on why it should be so?
 A: The Taylorpolynomial of order $n$ is given by the value of the function and of its first $n$ derivatives at a given point. If your function is polynomial look at the difference of the taylorpolynomial and your function which will be a polynomial of degree at most $n$, and for some $x_0$ the difference takes the value $0$ and all it's derivative take 0 on this point. So which polynomial do we have ?
A: By definition the Taylor polynomial $P$ of order $n$ of a function $f$ at a point $a$ where it is $n$ times differentiable is (or should be) the following: $P$ is the unique polynomial of degree at most $n$ such that $P^{(i)}(a)=f^{(i)}(a)$ for $i=0,1,\ldots,n$ (where $P^{(i)}$ denotes the $i$-th derivative of the polynomial function assoicated to$~P$). [It is easy to see that writing $P=\sum_{i=0}^nc_i(X-a)^k$ one has $P^{(i)}(a)=i!c_i$, so the necessity of $c_i=\frac{f^{(i)}(a)}{i!}$ gives existence and uniqueness of$~P$, as well as the long-winded formula that usually serves as definition of a Taylor polynomial.]
Now if $f$ is itself a the polynomial function associated to some polynomial$~Q$ of degree at most $n$, then it is obvious that one has $P=Q$.
