Unit 7, lesson 38

Analyticity and Convergence of Series Solutions

Why analytic coefficients of the normalized equation guarantee a radius of convergence for series solutions, how the nearest complex singularity sets it, and why an exact radius can be larger.

  • Video 38
  • Power-Series and Frobenius Solutions

Solve (1−x) y′′−2y′=0(1 - x)\,y'' - 2y' = 0 by a power series about 00, with xx as the independent variable and y′=dy/dxy' = dy/dx, and two very different series appear. The constant solution 11 is its own series and converges everywhere. The solution 1/(1−x)1/(1 - x) has the geometric series 1+x+x2+⋯1 + x + x^2 + \cdots, which diverges at x=−1x = -1 although 1/(1−x)1/(1 - x) equals 12\tfrac12 there. How far can a series solution be trusted? This lesson shows how the coefficients of the normalized equation, complex singular points included, guarantee a radius of convergence, and why that guarantee is only a lower bound for a particular solution.

Key terms

Normalized equation

Definition A second-order linear homogeneous equation P(x) y′′+Q(x) y′+R(x) y=0P(x)\,y'' + Q(x)\,y' + R(x)\,y = 0 is normalized by dividing by its leading coefficient wherever P(x)≠0P(x) \ne 0, which gives

y′′+p(x) y′+q(x) y=0,p=QP,q=RP.\begin{gathered} y'' + p(x)\,y' + q(x)\,y = 0, \\[0.4em] p = \frac{Q}{P}, \qquad q = \frac{R}{P}. \end{gathered}

The convergence theorem below is stated for the normalized coefficients pp and qq, so ordinary and singular points are read from them. When PP, QQ and RR are polynomials with no common factor, the singular points are exactly the zeros of PP, real and nonreal, because at each of them Q/PQ/P or R/PR/P has a pole.

In plain words Dividing by the leading coefficient puts the equation in the form the theorem talks about, and the places where that division fails are where trouble can start.

Example For (1−x) y′′−2y′=0(1 - x)\,y'' - 2y' = 0, division by 1−x1 - x is allowed for x≠1x \ne 1 and gives

y′′−21−x y′=0,y'' - \frac{2}{1 - x}\,y' = 0,

so p(x)=−2/(1−x)p(x) = -2/(1 - x) and q(x)=0q(x) = 0. The only singular point is x=1x = 1, one unit from the center 00. The original form is still worth keeping in view. At x=1x = 1 it reduces to −2y′(1)=0-2y'(1) = 0, so a solution defined across x=1x = 1 must have zero slope there, which is a restriction the normalized form cannot express.

Analytic coefficients

Definition A function ff is analytic at x0x_0 if f(x)=∑cn(x−x0)nf(x) = \sum c_n (x - x_0)^n for all xx in some open interval around x0x_0, with a series of positive radius. Polynomials, exe^x, sin⁡x\sin x and cos⁡x\cos x are analytic at every point, and so are sums, products and compositions of analytic functions, and quotients at points where the denominator is not zero. The normalized equation has analytic coefficients at x0x_0 when pp and qq are both analytic there, which is exactly what makes x0x_0 an ordinary point. How far a Taylor series reaches is decided in the complex plane. A power series that converges for real xx with ∣x−x0∣<R\lvert x - x_0 \rvert < R also converges for every complex xx in the disk ∣x−x0∣<R\lvert x - x_0 \rvert < R, and there its sum is analytic and finite. So the radius of the Taylor series of ff cannot exceed the distance from x0x_0 to a complex point where ff becomes infinite, such as a zero of the denominator of a rational function in lowest terms.

In plain words Analytic coefficients are coefficients that power series describe near the center. Being smooth on the real line is not enough to predict how far those series reach, because complex points count too.

Example The function 1/(1+x2)1/(1 + x^2) is analytic at every real xx, yet its Taylor series about 00,

11+x2=∑n=0∞(−1)nx2n,\frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n},

converges only for ∣x∣<1\lvert x \rvert < 1. At x=1.1x = 1.1 the terms (−1)n(1.21)n(-1)^n (1.21)^n grow without bound, although the function is finite there. The cause is the complex denominator, since 1+x2=01 + x^2 = 0 at x=ix = i and x=−ix = -i, both at distance 11 from 00.

Radius of convergence

Definition For every power series ∑an(x−x0)n\sum a_n (x - x_0)^n there is a number RR with 0≤R≤∞0 \le R \le \infty, its radius of convergence, such that the series converges absolutely for ∣x−x0∣<R\lvert x - x_0 \rvert < R and diverges for ∣x−x0∣>R\lvert x - x_0 \rvert > R. At the endpoints x0−Rx_0 - R and x0+Rx_0 + R anything can happen, so each endpoint is tested on its own. When ∣an+1/an∣\lvert a_{n+1}/a_n \rvert has a limit LL, possibly infinite, the ratio test gives R=1/LR = 1/L, read as R=∞R = \infty when L=0L = 0 and R=0R = 0 when L=∞L = \infty. If infinitely many coefficients are zero, as in a series of even powers only, apply the ratio test to the nonzero terms instead.

In plain words A power series converges on a symmetric interval around its center and diverges outside it. The radius is the half-width of that interval, and the two edges need their own check.

Example For (1−x) y′′−2y′=0(1 - x)\,y'' - 2y' = 0 and y=∑anxny = \sum a_n x^n, the coefficient of xnx^n in the equation is (n+2)(n+1)(an+2−an+1)(n + 2)(n + 1)(a_{n+2} - a_{n+1}), so an+2=an+1a_{n+2} = a_{n+1} for every n≥0n \ge 0. Every coefficient from a1a_1 on equals a1a_1, and

y=a0+a1∑n=1∞xn.y = a_0 + a_1 \sum_{n=1}^{\infty} x^n.

If a1≠0a_1 \ne 0, consecutive coefficients have ratio 11, so R=1R = 1. At x=1x = 1 every term is a1a_1, and at x=−1x = -1 the terms alternate between −a1-a_1 and a1a_1. In both cases the terms do not tend to 00, so the series diverges at both endpoints. If a1=0a_1 = 0, the series is the constant a0a_0 and R=∞R = \infty.

Nearest complex singularity

Definition Let x0x_0 be an ordinary point of y′′+p(x) y′+q(x) y=0y'' + p(x)\,y' + q(x)\,y = 0, and suppose the Taylor series of pp and qq about x0x_0 both converge for ∣x−x0∣<R\lvert x - x_0 \rvert < R, where 0<R≤∞0 < R \le \infty. Then for every choice of a0a_0 and a1a_1, the series solution with these initial coefficients converges at least for ∣x−x0∣<R\lvert x - x_0 \rvert < R and solves the equation there, and every solution on the interval x0−R<x<x0+Rx_0 - R < x < x_0 + R is one of these series. When pp and qq are rational functions in lowest terms, the Taylor series of each converges exactly out to its nearest pole in the complex plane. So RR can be taken as the distance from x0x_0 to the nearest complex singularity, the nearest real or nonreal singular point of the normalized equation. A point a+bia + bi lies at distance (a−x0)2+b2\sqrt{(a - x_0)^2 + b^2} from x0x_0. If pp and qq are polynomials, there is no singular point and R=∞R = \infty.

In plain words Draw the largest disk around the center in the complex plane that contains no singular point of the normalized equation. Every series solution converges at least inside that disk, which on the real line means at least on the interval the disk cuts out.

Example The equation (x−3)(x2+4) y′′+y=0(x - 3)(x^2 + 4)\,y'' + y = 0 has the singular points 33, 2i2i and −2i-2i. About x0=0x_0 = 0 their distances are 33, 22 and 22, so every series solution converges at least for ∣x∣<2\lvert x \rvert < 2, although the nearest real singular point is 33 units away. About x0=2x_0 = 2 the distances are 11, 222\sqrt{2} and 222\sqrt{2}, so the guarantee is ∣x−2∣<1\lvert x - 2 \rvert < 1, and this time the real point 33 is the nearest.

Guaranteed versus exact radius

Definition The radius RR from the theorem is a guaranteed radius, a lower bound that holds for every solution at once. The exact radius of a particular solution is the radius of convergence of its own Taylor series about x0x_0. It can be larger than RR, even infinite, because a solution need not be singular where a coefficient is, and in some equations every solution beats the guarantee. Neither number decides convergence at the endpoints. A solution can also exist far beyond its interval of convergence. The solution of an initial-value problem at x0x_0 exists on the whole largest open interval around x0x_0 that contains no real singular point, while its Taylor series about x0x_0 may converge on only part of that interval.

In plain words The singular points of the equation tell you how far every series solution is sure to converge. A particular solution may do better, and the only way to know is to look at that solution.

Example For (1−x) y′′−2y′=0(1 - x)\,y'' - 2y' = 0 about 00 the guaranteed radius is 11. On each interval that avoids x=1x = 1 the solutions are

y=C0+C11−x.y = C_0 + \frac{C_1}{1 - x}.

The constant solution 11 has exact radius ∞\infty. The solution 1/(1−x)1/(1 - x) has the geometric series as its Taylor series, with exact radius 11, and it really is singular at x=1x = 1. It is defined on all of (−∞,1)(-\infty, 1), yet its series about 00 diverges at x=−1x = -1, where the function equals 12\tfrac12. A new center reaches farther. About x0=−2x_0 = -2, which is 33 units from the pole,

11−x=∑n=0∞(x+2)n3n+1,\frac{1}{1 - x} = \sum_{n=0}^{\infty} \frac{(x + 2)^n}{3^{n+1}},

valid for ∣x+2∣<3\lvert x + 2 \rvert < 3. At x=−1x = -1 this series is ∑1/3n+1=12\sum 1/3^{n+1} = \tfrac12, the correct value.

Check yourself

  1. Without solving, find the radius that the theorem guarantees for series solutions of (x2−2x+5) y′′+xy′−y=0(x^2 - 2x + 5)\,y'' + xy' - y = 0 about x0=0x_0 = 0 and about x0=1x_0 = 1.
Show the answer

The leading coefficient has the roots x=1±2ix = 1 \pm 2i, which are the singular points of the normalized equation, since the polynomials xx and −1-1 do not vanish there. Their distance from 00 is 12+22=5\sqrt{1^2 + 2^2} = \sqrt{5}, so the series about 00 converge at least for ∣x∣<5\lvert x \rvert < \sqrt{5}. Their distance from 11 is 22, so the series about 11 converge at least for ∣x−1∣<2\lvert x - 1 \rvert < 2. The equation has no real singular point at all, yet both guarantees are finite.

  1. The equation (1+x2) y′′+2xy′=0(1 + x^2)\,y'' + 2xy' = 0 has no real singular points. Find the solution with y(0)=0y(0) = 0 and y′(0)=1y'(0) = 1, the exact radius of its series about 00, and decide whether that series converges at x=1x = 1 and at x=2x = 2.
Show the answer

The left side is the derivative of (1+x2) y′(1 + x^2)\,y', so (1+x2) y′=C(1 + x^2)\,y' = C with C=1C = 1 from the data. Then y′=1/(1+x2)y' = 1/(1 + x^2) and y=arctan⁡xy = \arctan x, whose series about 00 is

arctan⁡x=∑k=0∞(−1)kx2k+12k+1\arctan x = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k + 1}

The same coefficients follow from the recurrence an+2=−nn+2 ana_{n+2} = -\tfrac{n}{n+2}\,a_n. The absolute ratio of consecutive nonzero terms tends to x2x^2, so the exact radius is 11, equal to the guarantee from the singular points ±i\pm i. At x=1x = 1 the series 1−13+15−⋯1 - \tfrac13 + \tfrac15 - \cdots converges by the alternating series test, to π4\tfrac{\pi}{4}. At x=2x = 2 the terms grow without bound, so it diverges, even though arctan⁡2\arctan 2 is defined. Ignoring the complex singular points would wrongly suggest convergence for every xx.

  1. The functions xx and x2x^2 solve x2y′′−2xy′+2y=0x^2y'' - 2xy' + 2y = 0. What radius does the theorem guarantee for series solutions about x0=1x_0 = 1? Does the theorem promise that these series converge at x=3x = 3, and do they?
Show the answer

The normalized coefficients −2/x-2/x and 2/x22/x^2 are singular only at 00, which is 11 unit from the center, so the guaranteed radius is 11 and the theorem promises nothing at x=3x = 3. Both functions do solve the equation, since 0−2x+2x=00 - 2x + 2x = 0 and 2x2−4x2+2x2=02x^2 - 4x^2 + 2x^2 = 0, and they are independent, so every solution on an interval around 11 inside (0,∞)(0, \infty) is C1x+C2x2C_1x + C_2x^2. Its Taylor series about 11 is a polynomial of degree at most two, because x=1+(x−1)x = 1 + (x - 1) and x2=1+2(x−1)+(x−1)2x^2 = 1 + 2(x - 1) + (x - 1)^2. So every solution has exact radius ∞\infty and the series converge at x=3x = 3. The guarantee is only a lower bound, and here every solution beats it.

How this lesson connects

It also uses the ratio test.