Polynomials Hugging the Curve

This scene lays a polynomial over a curve. Raise the degree and it hugs the curve.

Match the value, the slope, and the bending at one point. The more you match, the better the imitation nearby. Far away, the imitation eventually breaks.

For example, knowing where a 🚗 car is, how fast it goes, and how it accelerates lets you predict its position a little later. More information predicts farther ahead.

Tn(x)=∑k=0nf(k)(a)k!(x−a)kT_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k

Examples & Exercises

Approximate ee with the degree-3 Taylor polynomial of exe^x, and bound the error with the Lagrange remainder.

T3(1)=83≈2.6667,∣R3(1)∣≤e24≈0.1133T_3(1) = \frac{8}{3} \approx 2.6667, \quad |R_3(1)| \le \frac{e}{24} \approx 0.1133
n=3n = 3: the actual error ≈0.0516\approx 0.0516 sits inside the bound 0.11330.1133.
Show proof

  1. Every derivative of exe^x is exe^x, so f(k)(0)=1f^{(k)}(0) = 1. Hence T3(x)=1+x+x22+x36T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} and T3(1)=83≈2.6667T_3(1) = \frac{8}{3} \approx 2.6667.

  2. By Taylor's theorem (Lagrange remainder), R3(1)=ec4!⋅14R_3(1) = \frac{e^c}{4!} \cdot 1^4 for some c∈(0,1)c \in (0, 1). Since ec<e<2.72e^c < e < 2.72, ∣R3(1)∣<2.7224≈0.1133|R_3(1)| < \frac{2.72}{24} \approx 0.1133.

  3. Indeed e−83≈0.0516e - \frac{8}{3} \approx 0.0516, inside the bound.

Prove from the derivatives at 0 why the Maclaurin series of sin⁡x\sin x has no even-degree terms.

sin⁡(2m)(0)=0  ⇒  T2m+1=T2m\sin^{(2m)}(0) = 0 \;\Rightarrow\; T_{2m+1} = T_{2m}
T3(x)=x−x3/6T_3(x) = x - x^3/6 — no x2x^2 term.
Show proof

  1. sin⁡′=cos⁡\sin' = \cos, sin⁡′′=−sin⁡\sin'' = -\sin, sin⁡′′′=−cos⁡\sin''' = -\cos, sin⁡(4)=sin⁡\sin^{(4)} = \sin, so the values at 0 repeat 0,1,0,−10, 1, 0, -1 with period 4.

  2. Hence every even-order derivative vanishes at 0, and the coefficients sin⁡(2m)(0)(2m)!=0\frac{\sin^{(2m)}(0)}{(2m)!} = 0.

  3. So T3(x)=x−x36T_3(x) = x - \frac{x^3}{6} has no x2x^2 term, and in general T2m+1=T2mT_{2m+1} = T_{2m} — because sin⁡\sin is odd.

Explain why Rn(x)→0R_n(x) \to 0 is the heart of Taylor convergence. (Idea level)

f(x)=Tn(x)+Rn(x)f(x) = T_n(x) + R_n(x)
n=9n = 9: the polynomial nearly overlaps the curve on [−π,π][-\pi, \pi].
Show hint

  1. Taylor's theorem splits f(x)=Tn(x)+Rn(x)f(x) = T_n(x) + R_n(x). The polynomial part is what we know; the remainder is everything we don't.

  2. Exactly when Rn(x)→0R_n(x) \to 0 do we get Tn(x)→f(x)T_n(x) \to f(x). The vanishing of the remainder IS the convergence of the series.

  3. Functions like sin⁡x\sin x, whose remainder vanishes, match their Taylor series globally — but smooth functions exist where it doesn't. Mentioned at the idea level only.

10. Differentiation