Review of Power Series
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1 Review of Power Series MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018
2 Introduction In addition to the techniques we have studied so far, we may use power series to solve ODEs/IVPs. Today we will review the properties of power series and techniques for manipulating power series.
3 Power Series Definition A power series in x x 0 where x 0 is a constant is an infinite series of the form a n (x x 0 ) n where the a n are constant coefficients. Note: if f (x) = a n (x x 0 ) n then f (x 0 ) = a 0.
4 Convergence Definition A power series f (x) = x x 0 if exists. a n (x x 0 ) n is said to converge at lim N N a n (x x 0 ) n
5 Convergence Definition A power series f (x) = x x 0 if exists. a n (x x 0 ) n is said to converge at lim N The power series may converge: only for x = x 0, N a n (x x 0 ) n for all x R, only for x (x 0 ρ, x 0 + ρ) where ρ > 0 (or x (x 0 ρ, x 0 + ρ], or x [x 0 ρ, x 0 + ρ), or x [x 0 ρ, x 0 + ρ]).
6 Absolute Convergence Definition A power series f (x) = a n (x x 0 ) n is said to converge absolutely at x x 0 if converges. a n (x x 0 ) n = a n x x 0 n Note: absolute convergence implies convergence, but the converse is not true.
7 Ratio Test Theorem (Ratio Test) Given the power series f (x) = that n a n (x x 0 ) n and supposing lim a n+1 (x x 0 ) a n = L x x 0 1. if L x x 0 < 1 then f (x) = absolutely for x x 0 < 1/L, then a n (x x 0 ) n converges 2. if L x x 0 > 1 then the series diverges, 3. if L x x 0 = 1 the test is inconclusive. Note: ρ = 1/L is the radius of convergence of the power series.
8 Examples Find the radius of convergence of each of the following power series. x n n!x n x n n!
9 Examples Find the radius of convergence of each of the following power series. x n (ρ = 1) n!x n x n n!
10 Examples Find the radius of convergence of each of the following power series. x n (ρ = 1) n!x n (ρ = 0) x n n!
11 Examples Find the radius of convergence of each of the following power series. x n (ρ = 1) n!x n (ρ = 0) x n n! (ρ = )
12 Equating Power Series If a n (x x 0 ) n = b n (x x 0 ) n for all x (x 0 ρ, x 0 + ρ) with ρ > 0 then a n = b n for n = 0, 1, 2,....
13 Arithmetic of Power Series Suppose a n (x x 0 ) n and b n (x x 0 ) n converge to f (x) and g(x) respectively for x x 0 < ρ with ρ > 0. f (x) ± g(x) = (a n ± b n )(x x 0 ) n converges at least for x x 0 < ρ. f (x)g(x) = c n (x x 0 ) n where c n = a 0 b n + a 1 b n a n 1 b 1 + a n b 0 and the series converges at least for x x 0 < ρ. f (x) g(x) = d n (x x 0 ) n provided g(x 0 ) 0. The radius of convergence of the series may be less than ρ.
14 Differentiating Power Series If f (x) = a n (x x 0 ) n has radius of convergence ρ > 0 then f is continuous for x x 0 < ρ, f has derivatives of all orders for x x 0 < ρ. f (x) = f (x) = na n (x x 0 ) n 1 n=1 n(n 1)a n (x x 0 ) n 2 n=2. The derivatives converge absolutely for x x 0 < ρ.
15 Integrating Power Series If f (x) = a n (x x 0 ) n has radius of convergence ρ > 0 then the power series has radius of convergence ρ > 0. a x n n + 1 (x x 0) n+1 = f (t) dt x 0
16 Taylor Series If f (x) = and a n (x x 0 ) n has radius of convergence ρ > 0 then f (x) = a n = f (n) (x 0 ) n! f (n) (x 0 ) (x x 0 ) n. n! This is called the Taylor Series for f (x) about x = x 0. The function f is said to be analytic at x = x 0.
17 Taylor Polynomials Definition The N th partial sum of a Taylor series is polynomial of degree N P N (x) = N f (n) (x 0 ) (x x 0 ) n n! = f (x 0 ) + f (x 0 )(x x 0 ) + + f (N) (x 0 ) (x x 0 ) N N! called the Taylor polynomial of degree N for f expanded about x = x 0.
18 Taylor s Theorem Theorem (Taylor s Theorem) Suppose that f has N + 1 derivatives on the interval (x 0 r, x 0 + r) for some r > 0. Then for x (x 0 r, x 0 + r), f (x) P N (x) and the error in using P N (x) to approximate f (x) is R N (x) = f (x) P N (x) = f (N+1) (z) (N + 1)! (x x 0) N+1 for some z between x and x 0.
19 Taylor s Theorem Theorem (Taylor s Theorem) Suppose that f has N + 1 derivatives on the interval (x 0 r, x 0 + r) for some r > 0. Then for x (x 0 r, x 0 + r), f (x) P N (x) and the error in using P N (x) to approximate f (x) is R N (x) = f (x) P N (x) = f (N+1) (z) (N + 1)! (x x 0) N+1 for some z between x and x 0. Remark: R N (x) is called the Taylor remainder.
20 Taylor s Theorem and Taylor Series Theorem Suppose that f has derivatives of all orders on the interval (x 0 r, x 0 + r) for some r > 0 and that for all x in (x 0 r, x 0 + r). Then for all x in (x 0 r, x 0 + r). lim R N(x) = 0 N f (n) (x 0 ) (x x 0 ) n = f (x) n!
21 Example Find the Taylor series for f (x) = x about x 0 = 1.
22 Example Find the Taylor series for f (x) = x about x 0 = 1. f (x 0 ) = f (1) = 1 f (x 0 ) = f (1) = 1 f (x 0 ) = f (1) = 0 f (x 0 ) = f (1) = 0.
23 Example Find the Taylor series for f (x) = x about x 0 = 1. f (x 0 ) = f (1) = 1 f (x 0 ) = f (1) = 1 f (x 0 ) = f (1) = 0 f (x 0 ) = f (1) = 0. The Taylor remainder R N (x) = 0 for N > 1,
24 Example Find the Taylor series for f (x) = x about x 0 = 1. f (x 0 ) = f (1) = 1 f (x 0 ) = f (1) = 1 f (x 0 ) = f (1) = 0 f (x 0 ) = f (1) = 0 The Taylor remainder R N (x) = 0 for N > 1, thus f (x) =. f (n) (x 0 ) (x x 0 ) n = 1 + (x 1). n!
25 Example Find the Taylor series for f (x) = e x about x 0 = 0.
26 Example Find the Taylor series for f (x) = e x about x 0 = 0. f (x 0 ) = f (0) = 1 f (x 0 ) = f (0) = 1 f (x 0 ) = f (0) = 1.
27 Example Find the Taylor series for f (x) = e x about x 0 = 0. The Taylor remainder R N (x) = N, f (x 0 ) = f (0) = 1 f (x 0 ) = f (0) = 1 f (x 0 ) = f (0) = 1. e z (N + 1)! x N+1 0 for all x as
28 Example Find the Taylor series for f (x) = e x about x 0 = 0. The Taylor remainder R N (x) = N, thus e x = f (x 0 ) = f (0) = 1 f (x 0 ) = f (0) = 1 f (x 0 ) = f (0) = 1. e z (N + 1)! x N+1 0 for all x as f (n) (x 0 ) (x x 0 ) n = n! x n n!
29 Geometric Series Definition A geometric series with ratio r has the form a + a r + a r 2 + = If a 0 the series converges to otherwise. a 1 r a r n. for r < 1 and diverges
30 Power Series for Common Functions 1 1 x = e x = sin x = cos x = ln x = x n for 1 < x < 1 x n n! for < x < ( 1) n (2n + 1)! x 2n+1 for < x < ( 1) n (2n)! x 2n for < x < ( 1) n+1 (x 1) n for 0 < x 2 n n=1
31 Re-indexing a Power Series (1 of 2) Re-index the series starts at 0. n=3 n n x n so that the summation index
32 Re-indexing a Power Series (1 of 2) Re-index the series starts at 0. n=3 Replace every n with n + 3. n=3 n n x n = n+3=3 n n x n so that the summation index n + 3 (n + 3) x n+3 = n + 3 (n + 3) x n+3
33 Re-indexing a Power Series (2 of 2) Re-index the series below so that the power series may be subtracted. 2a n x n+1 b n x n n=1
34 Re-indexing a Power Series (2 of 2) Re-index the series below so that the power series may be subtracted. 2a n x n+1 b n x n n=1 2a n x n+1 b n x n = n=1 = = 2a n x n+1 b n+1 x n+1 n+1=1 2a n x n+1 b n+1 x n+1 (2a n b n+1 )x n+1
35 Example Find the Taylor series about x 0 = 0 for (1 x + x 2 )e x.
36 Solution (1 x + x 2 )e x = (1 x + x 2 ) = = = = = k=0 k=0 k=0 k=0 x k k! k=0 x k k! k=1 x k k! k=1 x k k! k=0 k=0 (k 1) 2 x k k=0 k! x k k! x k+1 k! + k=0 x k+2 k! x k (k 1)! + kx k k! kx k k! + + k=2 x k (k 2)! k(k 1)x k k=2 k! k(k 1)x k k=0 k!
37 Homework Read Section 5.1 Exercises: 1 27 odd
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