Taylor polynomial

Definition

About a general point

Suppose $n$ is a nonnegative integer. Suppose a function $f$ of one variable is defined and at least $n$ times differentiable at a point $x_0$ in its domain. The $n^{th}$ Taylor polynomial for a function $f$ at a point $x_0$ in the domain is the truncation of the Taylor series to powers up to the $n^{th}$ power. If we denote the polynomial by $P_n(f;x_0)$, it is given as:

$P_n(f;x_0)=x\mapsto \sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$

Note that this is a polynomial of degree at most $n$. The degree is exactly $n$ if and only if $f^{(n)}(x_0)\ne 0$.

About the point 0

Suppose a function $f$ of one variable is defined and at least $n$ times differentiable at a point $0$. The $n^{th}$ Taylor polynomial of $f$ at 0 is:

$P_n(f;0)=x\mapsto \sum _{k=0}^n\frac{f^{(k)}(0)}{k!}{x}^k$

Note that this is a polynomial of degree at most $n$. The degree is exactly $n$ if and only if $f^{(n)}(0)\ne 0$.

Intuition

The $n^{th}$ Taylor polynomial, intuitively, is an attempt to be the best local approximation of $f$ about $x_0$ among polynomials of degree $\le n$.

Particular cases

Value of $n$	$n^{th}$ Taylor polynomial about $x_0$	What it means	$n^{th}$ Taylor polynomial about $x_0$ case $x_0=0$
0	$f(x_0)$	This is a constant function whose value is the value of $f$ at $x_0$. Clearly, this is the best approximation for $f$ among approximations by constant functions.	$f(0)$
1	$f(x_0)+f^{\prime }(x_0)(x-x_0)$	The graph of the function is a straight line that equals the tangent line to the graph of $f$ at $(x_0,f(x_0))$. The tangent line is intuitively the best linear approximation of the graph of the function, so this makes sense.	$f(0)+f^{\prime }(0)x$.