Smooth is not
analytic.
Two matched cartridges compare a profile with its Taylor reconstruction at−1. One converges with enough terms. One never does.
Left: Gaussian · right: bump. Accent: exact profile · grey: Taylor. Both windows x−1.5 to0.5, vertical−1.5 to1.5. Both dials 0–1.5. No autoscaling.
Reduced motion is enabled. Choose still frames with the controls.
What this comparator explains
B(x)=exp(1−1/(1−x²)) inside |x|<1 and0 outside. Every derivative at its left boundary−1 is zero, so every Taylor polynomial there is zero. Yet B(0)=1. Being infinitely smooth does not make a function analytic.
G(x)=exp(−(x+1)²). Its degree2N Taylor polynomial about−1 is Σⱼ₌₀ᴺ(−1)ʲ(x+1)²ʲ/j!. The degree is . As N grows, this analytic example is reconstructed across our fixed window. The first few errors need not decrease; the bump’s Taylor reconstruction stays zero.
The cartridges show only the left boundary and nearby interior; the full bump support is[−1,1]. A small bump value may underflow in floating-point arithmetic; its mathematical positivity is separate. Displayed Gaussian discrepancies include roundoff and are not certified error bounds.
This illustrates the smooth-versus-analytic distinction in episode76. It is not a Navier–Stokes solution, an implementation of the CIV obstruction, a general regularity result, or verification of any paper, Lean encoding or credit dispute.
MIT: smooth nonanalytic bumps ↗
CIV: conditional analytic-force obstruction ↗