Taylor polynomials

Differentiation, derivatives and Taylor approximations: Taylor approximations

Taylor polynomials

We have already seen that we can approximate the graph of a neat function $f$ near a point $x=a$ with the tangent line given by the equation

$y=f(a)+f'(a)\cdot (x-a)\tiny.$ This is precisely the straight line near $\bigl(a, f(a)\bigr)$ which is the closest to the graph of $f$ .
The straight line is the graph of the following polynomial function

$P(x) = f(a)+f'(a)\cdot (x-a)\tiny.$ Note that $P$ is the only first-degree polynomial with the property that

$P(a)=f(a),\quad P'(a)=f'(a)\tiny.$

Once the dam breaks, the flood comes: we can also do the above for second-degree polynomials.

We assume that the second derivative of $f$ exists (this is part of being a neat function).
Define the polynomial

$P(x) = f(a)+f'(a)\cdot (x-a)+ \frac{1}{2}f''(a)\cdot (x-a)^2\tiny.$ Then:

$P(a)=f(a),\quad P'(a)=f'(a),\quad P''(a)=f''(a)\tiny.$ The graph of $P$ is precisely the parabola that is near $\bigl(a, f(a)\bigr)$ closest to the graph of $f$ . This parabola is near $\bigl(a, f(a)\bigr)$ closer to the graph of $f$ than the tangent line at this point.

Approximating a mathematical function by a cubic function You can continue in this way with a third-degree polynomial and define

$P(x) = f(a)+f'(a)\cdot (x-a)+ \frac{1}{2}f''(a)\cdot (x-a)^2+ \frac{1}{6}f'''(a)\cdot (x-a)^3\tiny.$ The graph of the cubic polynomial is near $\bigl(a, f(a)\bigr)$ closer to the graph of $f$ than the two previous approximations.

The following theorem makes the above statements mathematically more precise; moreover, it works for polynomials of any degree. We say that a function is $k$ times differentiable if the derivatives $f', f'', f''', \ldots f^{(k)}$ exist.

Suppose that $k$ is a natural number and the function $f$ is at least $k+1$ times differentiable. Then

$f(x)=P(x)+R(x)$ with

$P(x) = f(a)+f'(a)\cdot (x-a)+ \frac{1}{2}f''(a)\cdot (x-a)^2+ \cdots + \frac{1}{k!}f^{(k)}(a)\cdot (x-a)^k$ is a polynomial function of degree $k$ and

$R(x)=\frac{1}{(k+1)!}f^{(k+1)}(\xi)\cdot (x-a)^{k+1}$ for some $\xi$ that is situated between $a$ and $x$ and that depends on $x$ .
The polynomial $P$ is called the Taylor polynomial for $f$ about $a$ of degree $k$ ; $R$ is called the (Lagrange) remainder term of order $k$ .

Actually we should write $P_k$ and $R_k$ to express the dependence of $k$ , but that is usually clear from the context. For numerical approximation methods, the following two points are important:

$\frac{R_k(x)}{(x-a)^k}\rightarrow 0\;\;\text{when}\;\;x\rightarrow a$ and

$|R_k(x)|\le M\cdot |x-a|^{k+1}\;\;\text{if } |x-a|\le r\text{ for some }M\text{ and }r\tiny.$ To save ink, but yet to give an indication of the order of a truncation, one often uses the (big- $O$ ) $O$ symbol of Landau and one writes

$R_k(x)=O\bigl((x-a)^{k+1}\bigr)\tiny.$

The table below shows the quadratic approximations of some functions about the origin.

$\begin{array}{ccccccccc} e^x &= &1 &+ &x &+ &\frac{1}{2}x^2 &+ &O(x^3) \\ \sin x &= & & &x & & &+ &O(x^3) \\ \cos x &= &1 & & &- &\frac{1}{2}x^2 &+ &O(x^3) \\ \dfrac{1}{1-x} &= &1 &+ &x &+ &x^2 &+ &O(x^3)\\ \sqrt{1+x} &= &1 &+ &\frac{1}{2} x &- &\frac{1}{8}x^2 &+ &O(x^3)\\ \ln(1+x) &= & & &x &- &\frac{1}{2}x^2 &+ &O(x^3)\end{array}$

Taylor approximation of a function In the interactive diagram below you can visually inspects how well a Taylor polynomial ff certain degree about some expansion point approxiamtes the given function.

GeoGebra

A direct calculation shows that the 6th-order Taylor approximation of $f(x) = \sin(x)$ around the point $0$ is given by $P_6(x) = x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5$ . The absolute value of the seventh derivative $\left(\frac{d}{dx}\right)^7\sin (x) = -\cos(x)$ is at most 1. Therefore:

$|\sin(x) - x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5| \le \frac{1}{7!} |x|^{7}\tiny.$ We conclude that $\frac{1}{7!} |x|^{7} < \frac{1}{10000}$ for $|x| < \sqrt[7]{\frac{7!}{10000}} \approx 0.821$ . Thus, on the interval $[-0.8,0.8]$ , the sine function is approximated by the polynomial $P_6(x)$ with an error less than $0.0001$ .