Taylor Polynomials and Remainders

Match derivatives at a center

The degree- $n$ Taylor polynomial of $f$ about $a$ is

P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k.

It matches the function and its first $n$ derivatives at the center. Centering at zero gives a Maclaurin polynomial. Matching derivatives is a local construction; to control the error away from the center, a remainder theorem is needed.

If $f$ has the required continuous derivatives between $a$ and $x$ , and $|f^{(n+1)}(t)|\le M$ there, Taylor's theorem gives

|f(x)-P_n(x)|\le\frac{M|x-a|^{n+1}}{(n+1)!}.

Worked example: exponential approximation

Every derivative of $e^x$ at zero is $1$ , so $P_3(x)=1+x+x^2/2+x^3/6$ . At $x=0.2$ , this gives $1.221333\ldots$ . On $[0,0.2]$ , the fourth derivative is at most $e^{0.2}$ ; the error is at most $e^{0.2}(0.2)^4/24<0.000082$ .

The exact error is smaller than this bound. A bound is a guarantee, not an estimate that must equal the observed discrepancy. Moving farther from the center usually needs more terms for comparable accuracy.

Explore

Taylor approximation of the exponential

Try this. Start at degree 1 and x = 2, then increase the degree. Move the probe to 0: every polynomial agrees there. Compare the error at −2 and 2.

Polynomial degree3Probe x1

Centered at 0; degree 3. At x = 1: eˣ = 2.71828, polynomial = 2.66667, signed error = -0.05162. The orange segment shows the difference at the probe. Fixed plot: −2 ≤ x ≤ 2. Values rounded to five decimals.

Compare the polynomial with $e^x$ on $[-2,2]$ . Change both the degree and the probe point. The readout measures actual error at the probe; it is not the Taylor-theorem upper bound.

Worked example: sine and a warning

For sine, $P_3(x)=x-x^3/6$ . Since the fourth derivative has magnitude at most $1$ , the degree-three remainder is bounded by $|x|^4/24$ . Using the degree-four Taylor polynomial, whose fourth-degree coefficient is zero, improves the bound to $|x|^5/120$ .

An infinitely differentiable function need not equal its Taylor series. The function $f(x)=e^{-1/x^2}$ for $x\ne0$ , with $f(0)=0$ , has all derivatives zero at zero but is positive elsewhere. Equality with an infinite Taylor series requires the remainder to tend to zero.

Practice

Find $P_2$ for $e^x$ centered at zero.
Find the linear Taylor polynomial of $\ln x$ centered at $1$ .
Bound the error of $1-x^2/2$ for $\cos x$ at $x=0.1$ using the degree-three polynomial.

Show worked solutions

$P_2=1+x+x^2/2$ .
$P_1=x-1$ , since $\ln1=0$ and $(\ln x)'|_{x=1}=1$ .
The cubic coefficient is zero and the fourth derivative is bounded by $1$ , so error is at most $0.1^4/24\approx0.00000417$ .