Taylor Polynomials and the Remainder
A Taylor polynomial is the best polynomial impostor of a function near a chosen point. The degree-$n$ one, $P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$, is built to match $f$ in value and in its first $n$ derivatives at the centre $a$, so right there the two are indistinguishable. Away from $a$ they part company, and the gap is the remainder $R_n = f - P_n$, shaded here. Watch a term fade in: each higher power bends the polynomial to hug the curve over a wider stretch, and the shaded gap is squeezed outward. Taylor's theorem pins the remainder down with the Lagrange bound, $|R_n(x)| \le \max|f^{(n+1)}|\,|x-a|^{n+1}/(n+1)!$, and the factorial in the denominator is why the approximation can improve so fast. But only inside the radius of convergence: for $1/(1-x)$ or $\ln(1+x)$ there is a wall (the green band) past which no amount of degree helps, the polynomial sailing off as the true function blows up or simply refuses to be matched. Drag the centre $a$ along the curve, drag the test point $x$, and read the error against degree on the log plot below: a clean downhill line inside the radius, an uphill one outside.
WHAT TO TRY
- Watch the build-up: each term fades in and the polynomial wraps further around the curve, the shaded remainder squeezed outward from the centre.
- Drag the centre $a$ along the curve: the polynomial re-anchors there, hugging the function around the new point.
- Switch to $1/(1-x)$ or $\ln(1+x)$: a green wall (the radius of convergence) appears, and past it the polynomial diverges no matter how high the degree.
- Drag the test point $x$ across that wall and watch the error plot flip from a downhill line to an uphill one.