Polynomials Hugging the Curve

Lay the Maclaurin polynomial Tn(x)T_n(x) over sin⁡x\sin x. Raise the degree and watch it hug the curve.

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 proof
  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.

9. Differentiation