Power series and analyticity

Try

Inside |z|<1, the geometric series 1+z+z²+⋯ equals 1/(1−z). On |z|≤1/2, the tail after degree three has modulus at most (1/2)⁴/(1−1/2)=1/8. The error bound controls every point of that smaller disk, not just real inputs.

Record the smaller disk explicitly whenever quoting this error; using the same bound at the pole would have no meaning. Complex holomorphy has a remarkable converse to this example: every holomorphic function has such a convergent local power-series representation. The proof comes from expanding the kernel in Cauchy’s formula.

Predict

Predict whether knowing all derivatives at one point determines a holomorphic function nearby. Unlike a general infinitely differentiable real function, the answer is yes. The Taylor series actually converges to the function.

Predict separately whether convergence on an open disk guarantees convergence on its boundary circle. The geometric series at z=1 supplies a counterexample. Strict inequalities between radii must remain visible throughout the proof.

Experiment

For 1/(2−z), compare the first coefficients about zero.

Degree	Coefficient
0	1/2
1	1/4
2	1/8
3	1/16

Multiply a finite partial sum by 2−z to expose the remainder. This algebraic check confirms both the coefficient pattern and the radius-two obstruction at the pole.

Understand

Let f be holomorphic on a disk centered at a. Choose a circle of radius r strictly inside the disk and a point z with |z−a|<r. On that circle, 1/(ζ−z)=(1/(ζ−a))Σ((z−a)/(ζ−a))ⁿ. For z in a still smaller closed disk, the geometric series converges uniformly on the integration contour. Substituting it into Cauchy’s formula and integrating termwise gives f(z)=Σcₙ(z−a)ⁿ, where cₙ=(1/(2πi))∫f(ζ)/(ζ−a)^(n+1)dζ=f^(n)(a)/n!.

If |f|≤M on the radius-r circle, then |cₙ|≤M/rⁿ. For |z−a|≤qr with q<1, the tail after degree N is bounded by Mq^(N+1)/(1−q). This proves convergence to f with a uniform quantitative remainder.

Conversely, a power series is holomorphic strictly inside its radius of convergence. To justify differentiation, choose two radii r<s inside that radius. The derivative terms on the smaller disk are bounded by a constant times n r^(n−1)/sⁿ, whose sum converges. Uniform convergence of the derivative series then permits termwise differentiation. Thus holomorphic and locally analytic are equivalent, but their equivalence is a theorem, not a definition silently assumed in the contour proof.

The radius of a Taylor expansion is limited by obstructions to holomorphic extension. For rational functions, the nearest uncancelled pole gives the radius.

Apply

Expand 1/(1+z²) using the geometric series in −z². Explain why the radius is one even though the function has no real singularity. Its complex poles are i and −i, both one unit from the center.

Diagnose “The terms are differentiable, so differentiate an infinite series.” Finite differentiation rules do not automatically pass through an infinite limit. The nested-radius majorant supplies the missing uniform control.

Continue

Power series make isolated-zero and uniqueness arguments possible. Next, the first nonzero coefficient will identify a zero’s multiplicity. Optional depth: compare Taylor expansions of the same rational function about two different centers and compute how the distance to its nearest pole changes the available disk.