Power series and local uniform convergence

Try

The geometric series Σ from k=0 of x^k equals 1/(1−x) for |x|<1. On any fixed interval |x|≤r<1, its tail is uniformly bounded by r^(N+1)/(1−r). On the entire open interval (−1,1), that bound cannot be made uniform as r approaches one. “Inside the radius” means local uniform control on smaller closed sets, not automatically one bound reaching the boundary.

Predict

Predict what differentiation does to the terms. The derivative series has terms kx^(k−1), introducing a growing factor k. A geometric factor r^k still defeats this polynomial growth when r<1. At the endpoint x=1, however, there is no geometric decay and the derivative terms grow. Interior calculus and endpoint convergence are separate questions.

Experiment

Compare behavior of Σ from k=1 of x^k/k.

Location	Original series	Derivative series
|x|≤r<1	Uniformly convergent	Uniformly convergent
x=1	Harmonic divergence	Terms equal one
x=−1	Alternating convergence	Terms have magnitude one

The last row disproves automatic inheritance of endpoint convergence under differentiation. An endpoint may belong to the original convergence set while the differentiated series fails even the necessary term test.

Understand

Here is the majorant argument. If a power series Σa_k s^k converges at a positive radius s, its terms a_k s^k are bounded in magnitude by M. For |x|≤r<s, |a_kx^k|≤M(r/s)^k. The geometric majorant proves absolute uniform convergence on the smaller interval. This argument underlies the radius-of-convergence structure; outside the radius, scalar convergence fails, while individual boundary points need separate testing.

To control derivatives choose 0<r<s<R, where R is the radius. Then |ka_kx^(k−1)|≤(M/s)k(r/s)^(k−1). The numerical series converges because its ratio tends to r/s<1. Thus derivative partial sums converge uniformly on |x|≤r, and the original partial sums converge at zero. The derivative-interchange theorem justifies termwise differentiation. Repeating the argument handles higher derivatives on still smaller compact intervals, whose radii can be chosen around any interior point.

Apply

Integrating a power series term by term over a compact interval inside its radius follows from uniform convergence and the integral-interchange theorem. For the geometric series, differentiating gives Σkx^(k−1)=1/(1−x)². Integrating from zero gives Σx^(k+1)/(k+1)=−log(1−x) for −1<x<1. The logarithm identity is justified here by a valid interior integral, not by evaluating a formal expression at a singular endpoint.

Repair “a power series has the same convergence interval after differentiation.” The radius stays the same, but endpoints can change. Distinguish the open radius interval from the full convergence set, which may contain neither, one or both boundary points.

Construct an ODE solution from a series

Define φ(x)=Σ from n=0 to infinity of aⁿx²ⁿ/n!, where a is either one or minus one. We will prove φ′=2axφ and φ(0)=1 without first naming the sum. This assembles the series-defined ODE action in MIT 18.100B problem set 10, question 1, using the existing uniform derivative theorem.

Does a small residual prove the whole series is accurate?

Coefficient a +1−1 Compact interval radius R 0.512 Last term index N 24816 Inspected x/R -1-0.500.51

Partial sum Pₙ(x)=Σ from k=0 to N of aᵏx²ᵏ/k!. Predict what happens to the certificates when the interval widens. All variables are dimensionless; no solver uses the exponential reference to construct this polynomial.

At x=1: P=2.70833333, P′=5.33333333, ODE residual P′−2axP=-0.08333333. Proved interval residual supremum: 0.0833333.

Uniform value-tail bound: 0.0100000 (ratio 0.166667). Uniform derivative-tail bound: 0.104167 (ratio 0.200000).

For primitives anchored at zero, the uniform primitive-error bound is 0.0100000. A zero residual at x=0 is automatic for every truncation and says nothing about the maximum error elsewhere.

Independent closed-form check, after the series proof

The proved solution can subsequently be identified as exp(ax²). At the inspected point, observed value error is 0.00994850 and derivative error is 0.103230. These floating-point comparisons are separate from the analytic majorants and are not their proof. Roundoff may dominate very small reported errors.

Index k	Term aᵏx²ᵏ/k!	Partial sum through k
0	1.000000	1.00000000
1	1.000000	2.00000000
2	0.5000000	2.50000000
3	0.1666667	2.66666667
4	0.04166667	2.70833333

Reset

At most seventeen terms per evaluation and sixty-five curve samples. The dashed reference is a verification aid, not an assumed series identity. Temporary controls award no mastery.

Prove both uniform convergences

Fix any finite R>0. On |x|≤R, absolute term values are bounded by tₙ=R²ⁿ/n!. Their ratio tₙ₊₁/tₙ=R²/(n+1) is eventually below, for example, one half. A finite initial segment plus a geometric tail proves summability, so the M-test gives uniform convergence on this compact interval.

For n≥1, the derivative term is 2a x times aⁿ⁻¹x²⁽ⁿ⁻¹⁾/(n−1)!. Its magnitude is at most 2R tₙ₋₁, also summable. Thus derivative partial sums converge uniformly, and the polynomial partial sums all equal one at zero. The uniform derivative theorem now permits termwise differentiation. Reindexing the convergent derivative series gives φ′(x)=2axφ(x). Every real x lies inside some such compact interval, so the equation holds on all of R. This proves an everywhere-defined differentiable solution without assuming its closed form.

Only afterward may we identify it: the derivative of exp(−ax²)φ(x) is zero, so MVT makes that product constant on R. Its value at zero is one. Hence φ(x)=exp(ax²). The independent exponential curve in the inspector checks a finite truncation; it is not used to construct or justify the infinite series.

Turn majorants into finite certificates

Let Pₙ include terms indexed zero through N. For the value tail, the first omitted majorant is tₙ₊₁ and every subsequent ratio is at most qᵥ=R²/(N+2). If qᵥ<1, the uniform value error is at most tₙ₊₁/(1−qᵥ). For the derivative tail the first omitted majorant is 2R tₙ, with subsequent ratio at most q𝒹=R²/(N+1). If q𝒹<1, its uniform error is at most 2R tₙ/(1−q𝒹).

When one of these ratios is at least one, that particular geometric bound is unavailable; the series has not been shown divergent. Increase N to move beyond the large initial ratios. At R=2 and N=2 both certificates are withheld. At R=1 and N=4 they give value bound 1/100 and derivative bound 5/48.

Exact polynomial cancellation gives Pₙ′−2axPₙ=−2a⁽ᴺ⁺¹⁾x²ᴺ⁺¹/N!. Its interval supremum is 2R tₙ. At x=0 the residual is always zero, even for a poor approximation elsewhere. A pointwise residual is not an interval error certificate; the derivative and value tails are separately proved quantities.

Primitive convergence and the domain boundary

The existing derivative-interchange proof already contains the variable-upper-limit estimate requested in PS10 question 2. For functions gₙ→g uniformly on [a,b], set Fₙ(x)=∫ₐˣgₙ and F(x)=∫ₐˣg. Then sup |Fₙ−F| ≤ (b−a) sup |gₙ−g|, by the integral absolute bound for every x. For our interval [−R,R] with anchor zero, the sharper length R applies, giving primitive error at most R times the value-tail bound. This is a supremum estimate for all upper limits, not only convergence of one endpoint integral.

Compact-uniform convergence does not imply uniform convergence on the whole real line. Uniform convergence of a series would force its individual terms to tend uniformly to zero, by subtracting successive partial sums. Here every positive-index term aⁿx²ⁿ/n! is unbounded on R, so that necessary condition fails. Each fixed x still has a convergent series. Keep the domain attached to every uniformity claim.

Continue

Optional depth notes that an interior power-series sum is infinitely differentiable and its coefficients equal derivatives at the center divided by factorials. The converse from smoothness to a power-series representation remains false, as the flat exponential example in Analysis I shows. Analyticity is an additional property.

Save a proof with an intermediate radius s between the target compact interval and R. That extra room is what absorbs derivative factors. Next, the space carrying these uniform estimates will be generalized from real inputs to arbitrary metric spaces, where neighborhoods, convergence and continuity still make sense without coordinates.