Solve by a power series about , with as the independent variable and , and two very different series appear. The constant solution is its own series and converges everywhere. The solution has the geometric series , which diverges at although equals 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 is normalized by dividing by its leading coefficient wherever , which gives
The convergence theorem below is stated for the normalized coefficients and , so ordinary and singular points are read from them. When , and are polynomials with no common factor, the singular points are exactly the zeros of , real and nonreal, because at each of them or 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 , division by is allowed for and gives
so and . The only singular point is , one unit from the center . The original form is still worth keeping in view. At it reduces to , so a solution defined across must have zero slope there, which is a restriction the normalized form cannot express.
Analytic coefficients
Definition A function is analytic at if for all in some open interval around , with a series of positive radius. Polynomials, , and 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 when and are both analytic there, which is exactly what makes an ordinary point. How far a Taylor series reaches is decided in the complex plane. A power series that converges for real with also converges for every complex in the disk , and there its sum is analytic and finite. So the radius of the Taylor series of cannot exceed the distance from to a complex point where 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 is analytic at every real , yet its Taylor series about ,
converges only for . At the terms grow without bound, although the function is finite there. The cause is the complex denominator, since at and , both at distance from .
Radius of convergence
Definition For every power series there is a number with , its radius of convergence, such that the series converges absolutely for and diverges for . At the endpoints and anything can happen, so each endpoint is tested on its own. When has a limit , possibly infinite, the ratio test gives , read as when and when . 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 and , the coefficient of in the equation is , so for every . Every coefficient from on equals , and
If , consecutive coefficients have ratio , so . At every term is , and at the terms alternate between and . In both cases the terms do not tend to , so the series diverges at both endpoints. If , the series is the constant and .
Nearest complex singularity
Definition Let be an ordinary point of , and suppose the Taylor series of and about both converge for , where . Then for every choice of and , the series solution with these initial coefficients converges at least for and solves the equation there, and every solution on the interval is one of these series. When and are rational functions in lowest terms, the Taylor series of each converges exactly out to its nearest pole in the complex plane. So can be taken as the distance from to the nearest complex singularity, the nearest real or nonreal singular point of the normalized equation. A point lies at distance from . If and are polynomials, there is no singular point and .
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 has the singular points , and . About their distances are , and , so every series solution converges at least for , although the nearest real singular point is units away. About the distances are , and , so the guarantee is , and this time the real point is the nearest.
Guaranteed versus exact radius
Definition The radius 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 . It can be larger than , 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 exists on the whole largest open interval around that contains no real singular point, while its Taylor series about 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 about the guaranteed radius is . On each interval that avoids the solutions are
The constant solution has exact radius . The solution has the geometric series as its Taylor series, with exact radius , and it really is singular at . It is defined on all of , yet its series about diverges at , where the function equals . A new center reaches farther. About , which is units from the pole,
valid for . At this series is , the correct value.
Check yourself
- Without solving, find the radius that the theorem guarantees for series solutions of about and about .
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The leading coefficient has the roots , which are the singular points of the normalized equation, since the polynomials and do not vanish there. Their distance from is , so the series about converge at least for . Their distance from is , so the series about converge at least for . The equation has no real singular point at all, yet both guarantees are finite.
- The equation has no real singular points. Find the solution with and , the exact radius of its series about , and decide whether that series converges at and at .
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The left side is the derivative of , so with from the data. Then and , whose series about is
The same coefficients follow from the recurrence . The absolute ratio of consecutive nonzero terms tends to , so the exact radius is , equal to the guarantee from the singular points . At the series converges by the alternating series test, to . At the terms grow without bound, so it diverges, even though is defined. Ignoring the complex singular points would wrongly suggest convergence for every .
- The functions and solve . What radius does the theorem guarantee for series solutions about ? Does the theorem promise that these series converge at , and do they?
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The normalized coefficients and are singular only at , which is unit from the center, so the guaranteed radius is and the theorem promises nothing at . Both functions do solve the equation, since and , and they are independent, so every solution on an interval around inside is . Its Taylor series about is a polynomial of degree at most two, because and . So every solution has exact radius and the series converge at . The guarantee is only a lower bound, and here every solution beats it.