# Math 201 Lecture 23: Power Series Method for Equations with Polynomial

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1 Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients Mar. 07, 2012 Many examples here are taken from the textbook. The first number in () refers to the problem number in the UA Custom edition, the second number in () refers to the problem number in the 8th edition. Radius of convergence. 0. Theoretical Issues Theorem 1. (Radius of convergence) For any power series a n (x x 0 ) n, there is a number ρ [0,] (meaning: ρ 0 and can be infinity) such that the power series converges for all x such that x x 0 <ρ; the power series diverges for all x such that x x 0 >ρ. This particular number ρ is called the radius of convergence. Remark 2. In other words, for any x x 0 <ρ, the infinite sum of numbers an (x x 0 ) n (1) indeed equals a number, while for any x x 0 >ρ the infinite sum is either infinity or does not have a limit at all. Consequently, inside x x 0 <ρ, the power series an (x x 0 ) n (2) indeed represent a well-defined function. How to calculate radius of convergence. We have two ways. The first is for an arbitrary power series, the 2nd deals with the situation when the power series is the result of solving a linear differential equation. Method 1: Ratio test. Theorem 3. (Ratio test) If, for n larger than a certain N, the coefficients a n satisfy lim a n+1 =L (0 L ) (3) n a n then the radius of convergence of the power series a n (x x 0 ) n is ρ=1/l, with ρ= if L=0 and ρ=0 if L=. That is, the power series converges 1 for x x 0 <ρ and diverges for x x 0 >ρ. Example 4. (8.2 2; 8.2 2) Determine the convergence set of a n 3 n n! xn. (4) Solution. We have a n = 3n. The ratio n! a n+1 = 3 n+1 /(n+1)! 3 n /n! = 3 n+1 0 as n. (5) So we have L=0 and the radius of convergence is. 1. In fact, absolutely converges. 1

2 +a +P 2 Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients Method 2: Checking singular points. Only applies when the power series is the solution to a linear differential equation. Analytic functions. Definition 5. A function is analytic at x 0 if, in an open interval about x 0 this function equals a power series about x 0. Theorem 6. If f(x) is analytic at x 0, then the power series it equals is exactly the Taylor expansion of f at x 0. Checking analyticity. Given f(x). Its analyticity at a point x 0 cannot be checked by computing the Taylor expansion of f and then study the radius of convergence of the latter. Use the following rules to check analyticity. 1. e x, sinx, cosx, polynomials are analytic everywhere. 2. If f(x) is analytic at x 0, so are f (x) and f(x). 3. If f(x),g(x) are analytic at x 0, so are f ±g,fg,f g. 4. If f,g are analytic at x 0, then f/g is analytic at x 0 if g(x 0 ) 0. Ordinary and singular points of a linear differential equation. For a linear homogeneous differential equation, a n (x)y (n) +a n 1 (x)y (n 1) + We can write it into standard form: y (n) +P n 1 (x)y (n 1) + Then a pointxis singular if at least one of P n 1, x is ordinary. Example 7. Find the singular points of the equation Solution. First write it into the standard form 1 (x)y +a 0 (x)y=0. (6) 0 (x)y=0. (7),P 0 is not analytic atx. Otherwise x 2 y +3y xy=0. (8) We have y + 3 x 2 y 1 y=0. (9) x p(x)= 3 x 2, q(x)= 1 x. (10) As 3,1,x 2 and x are all analytic everywhere, their ratios are analytic at all points except those making the denominator vanish. Therefore the only singular point is x=0. Criterion for radius of convergence. Theorem 8. (Existence of analytic solutions) Suppose x 0 is an ordinary point for equation y (x)+p(x)y (x)+q(x)y(x)=0, (11) then it has two linearly independent analytic solutions of the form y(x)= a n (x x 0 ) n. (12)

3 +a Mar. 07, Moreover, the radius of convergence of any power series solution of the form given above is at least as large as the distance from x 0 to the nearest singular point (real or complexvalued). Remark 9. It is important, when determining the radius of convergence, to remember count in complex singular points. Remark 10. The distance between any two complex numbers a + b i, c + d i can be computed as (a c) 2 +(b d) 2. (13) Example 11. Find a minimum value for the radius of convergence of a power series solution about x 0. (x+1)y 3xy +2y=0, x 0 =1. (14) Solution. Write the equation to standard form y 3x x+1 y + 2 y=0. (15) x+1 The only singular point is x = 1. Thus the minimum radius of convergence is the distance between x 0 =1 and 1, which is 2. Example 12. Find a minimum value for the radius of convergence of a power series solution about x 0. (1+x+x 2 )y 3y=0; x 0 =1. (16) Solution. Write the equation to standard form y 3 y=0. (17) 1+x+x2 The singular points are roots of 1+x+x 2, which are x 1,2 = 1± 3 i. (18) 2 To find out the closest singular point to x 0, we compute x 1 x 0 = i = 3 ; x2 x 0 = 3. (19) So both are 3 away from x0. As a consequence, the minimum radius of convergence is Basic Information The Equation We consider the following type of equations: with a n, a n (x)y (n) +,a 0 and b all polynomials. How to get general solution 1. Set ansatz Substitute into the equation. 2. Figure out the recurrence relation. 1 (x)y +a 0 (x)y=b(x) (20) y= a n (x x 0 ) n. (21)

4 (2k 2) 4 Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients 3. Solve the recurrence relation. 4. Simplify. How to solve initial value problem (IVP) Keep in mind that if the initial conditions are y(x 0 )=A 0, y (x 0 )=A 1, (22) then necessarily a 0 =A 0,a 1 =A 1, (23) How to check solutions General formula required. 2. More Examples Example 13. (8.3 21; ) Find a power series expansion about x=0 for a general solution to the given differential equation. Your answer should include a general formula for the coefficients. Solution. We write Substituting into the equation, we have This gives Therefore 0 = y xy +4y = n=2 = y xy +4y=0. (24) y(x)= a n n(n 1)x n 2 x n=1 a n x n. (25) a n+2 (n+2)(n+1)x n n=1 = (2a 2 +4a 0 )+ n=1 na n x n 1 + na n x n + 4a n x n 4a n x n [(n+2)(n+1)a n+2 (n 4)a n ]x n. (26) 2a 2 +4a 0 = 0 (27) (n+2)(n+1)a n+2 (n 4)a n = 0. (28) a 2 = 2a 0, (29) n 4 a n+2 = (n+2)(n+1) a n. (30) It is clear that we should discuss n=2k and n=2k 1 separately. For even n, we have For odd n, we have a 2 = 2a 0, a 4 = 1 6 a 2= 1 3 a 0, a 6 =0, a 8 =0, a 10 =0, (31) 2k 5 a 2k+1 = (2k+1)(2k) a 2k 1= (2k 5)(2k 7) (2k+1) a = (2k 5) ( 3) a 2k 3= 1. (32) (2k+1)! Summarizing, we have [ y(x)=a 0 1 2x ] [ 3 x4 +a 1 x+ (2k 5) ( 3) (2k+1)! k=1 x 2k+1 ]. (33)

5 Mar. 07, Remark 14. For this problem it is clear that the radius of convergence is. 3. Notes and Comments Whether the power series converges at x = x 0 ± ρ is tricky to determine. Different approaches are needed for different power series. In the textbook analytic at x 0 is defined as if, in an open interval about x 0, this function is the sum of a power series a n (x x 0 ) n that has a positive radius of convergence. The last part that has a positive radius of convergence is redundant. If a power series sums up to a well-defined function in an interval containing x 0, then the radius of convergence is positive by definition. Note that, f(x) is analytic at x 0 is not the same as the Taylor expansion of f at x 0 has positive radius of convergence. A classical example is the following: { f(x)= e 1/x x>0 0 x 0. (34) Turns out we have for all n. Consequently its Taylor expansion is f (n) (0)=0 (35) 0+0x+0x 2 +0x 3 + (36) which has radius of convergence. But obviously this power series sum up to 0 which is different from f. Note that it is easy to modify the example so that the power series looks less artificial. For example if we let then the Taylor expansion is f(x)= which of course sum up to e x, not f(x). { e x +e 1/x x>0 e x x 0. (37) x n n! (38)

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