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Radius of Convergence

The Taylor series of 1/(1 + x²) about a converges only within a distance R of a. Move the degree, the point x₀ and the centre, and watch the real graph and the complex plane together: the disc on the plane decides where the curve can be trusted.

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Lesson

The theory — Radius of Convergence

The radius of convergence of a power series Σ cn(x − a)n is the number R for which the series converges when |x − a| < R and diverges when |x − a| > R. For 1/(1 + x²), R is the distance from a to the nearest point of the complex plane where the function blows up, i or −i, so R = √(1 + a²).

What each symbol means

R
the radius of convergence about a, |a − i| = √(1 + a²)
q
(a − x)/(a − i); its size |x − a|/R says how many radii x lies from the centre

Where the formula comes from

  1. Where the coefficients come from. Split 1/(1 + z²) into (1/2i)[1/(z − i) − 1/(z + i)]. Each piece is a geometric series about a: 1/(z − i) = 1/((a − i) + (z − a)) = Σ (−1)n(z − a)n/(a − i)n+1, which converges exactly when |z − a| < |a − i|. The other piece is the same with a + i, at the same distance, so both stop at R = √(1 + a²).
  2. Why the error is a power of q. A geometric series stopped after N + 1 terms leaves qN+1 times the whole as its remainder, so f(x) − TN(x) = Im[qN+1/(x − i)] exactly. Its size is at most |q|N+1/√(1 + x²): it dies like a geometric sequence when |q| < 1, grows like one when |q| > 1, and on |q| = 1 it neither dies nor grows.

No trouble anywhere on the real line, and the series still dies at 1

Press Just past 1. 1/(1 + x²) is as tame as a function gets: positive, never above 1, with a derivative of every order at every real number. Yet degree 80 about 0 gives 1,121.975 at x = 1.1, where the curve is 0.4524887, and adding terms makes it worse: every two degrees multiply the error by 1.21. Press Just inside 1: 0.9 is the same distance from 1 on the other side, and the same degree is off by 9.777×10⁻⁵. Nothing at x = 1 explains the difference; the curve passes through it at ½ without a kink. The lower panel does. Let x be a complex number z, and 1/(1 + z²) is infinite where z² = −1, at i and −i, both exactly 1 from the centre.

The radius is the distance to i, wherever you centre the series

Drag the centre a along the real axis in the lower panel. The disc grows and shrinks so that its edge always passes through i and −i, R = √(1 + a²), and the shaded band on the real graph above is the disc’s shadow on the real axis. Press Centred at 1: R = 1.414214, so the series converges from −0.414214 to 2.414214, and at x = 2, where the series about 0 gave 9.671×10²³, degree 80 is off by 2.728×10⁻¹³. Neither end of that interval is special on the real line. Each is where a circle round 1 through the two poles happens to cross the real axis.

On the circle itself the series neither settles nor runs away

Press On the edge. At x = 1 the series about 0 becomes 1 − 1 + 1 − 1 + …, Grandi’s series: degree by degree its partial sums are 1, 1, 0, 0, 1, 1, …, and they never settle and never grow. The function there is ½. The last row of the results averages the partial sums, and the average, 0.5061728 at degree 80, closes on exactly that ½.

Problem solved in full

  1. The series fails at x = 1. Does its integral? 4 steps

    Integrate the degree-80 polynomial from 0 to 1, term by term, and compare it with the true area under 1/(1 + x²). At x = 1 the series itself never settles. What happens to the area?

    Open this problem in the tool

    1. The true area is an arctangent:

      ∫01dx1+x2=arctan⁡1=π4=0.7853982\int_0^1\frac{dx}{1+x^2}=\arctan 1=\frac{\pi}{4}=0{.}7853982

    2. Term by term, each x2k becomes 1/(2k + 1):

      ∫01T80 dx=∫01(1−x2+x4−⋯+x80) dx=1−13+15−⋯+181=0.7914948\int_0^1 T_{80}\,dx=\int_0^1\big(1-x^2+x^4-\dots+x^{80}\big)\,dx=1-\tfrac13+\tfrac15-\dots+\tfrac{1}{81}=0{.}7914948

    3. The difference is the integral of the error, which the geometric series gives exactly, and 1/(1 + x²) lies between ½ and 1 on [0, 1]:

      ∫01(T80−f) dx=∫01x821+x2 dx∈(1166, 183)=(0.0060241, 0.0120482);0.0060967\int_0^1\big(T_{80}-f\big)\,dx=\int_0^1\frac{x^{82}}{1+x^2}\,dx\in\Big(\tfrac{1}{166},\,\tfrac{1}{83}\Big)=(0{.}0060241,\ 0{.}0120482);\quad 0{.}0060967

    4. That bound goes to 0 as the degree grows, so the alternating series of odd reciprocals converges, and to π/4:

      0<∫01x2k+21+x2 dx<12k+3→0 ⇒ 1−13+15−17+⋯=π40<\int_0^1\frac{x^{2k+2}}{1+x^2}\,dx<\frac{1}{2k+3}\to0\ \Rightarrow\ 1-\tfrac13+\tfrac15-\tfrac17+\dots=\frac{\pi}{4}

    Answer

    The area converges where the series did not: 0.7914948 against π/4 = 0.7853982, and every extra pair of degrees closes the gap. It converges slowly, the error after n terms being close to 1/(4n), so an error of a millionth takes about 250,000 terms. It also converges only conditionally, since 1 + ⅓ + ⅕ + … diverges, which is exactly the kind of series Riemann showed can be re-ordered to add up to anything. In the same 1854 essay he noted that series in powers of a variable converge absolutely, single values of the variable excepted; x = 1 is one of those values.

    0.7914948−0.7853982=0.0060967≈14×41;10−6 needs≈250,000 terms0{.}7914948-0{.}7853982=0{.}0060967\approx\tfrac{1}{4\times41};\qquad 10^{-6}\ \text{needs}\approx250{,}000\ \text{terms}

Learning path

Infinite series and convergence

References (2)

· Reviewed against published sources. Corrections welcome.