Notes PDF
Introductory Problems
1. Write 1 + 2x + 4x2 + 8x3 + 16x4 + … as a ratio of polynomials.
Answer
1. The series sums to 1/(1 – 2x).
Solution
1. Rewrite the terms of the series 1 + 2x + 4×2 + 8×3 + 16×4 + … = 1 + 2x + (2x)2 + (2x)3 + (2x)4 + …So we see this series converges to 1/(1 – 2x).
2. Write x + x2 + x3 + x4 + … as a ratio of polynomials.
Answer
1. The series sums to x/(1 – x).
Solution
1. The series is the geometric series for 1/(1 – x), missing the 1, so x + x2 + x3 + x4 + … = (1 + x + x2 + x3 + x4 + … ) – 1= 1/(1 – x) – 1 = x/(1 – x).
3. Write 2 + x2 + x3 + x4 + 2x6 + x8 + x9 + x10 + 2x12 + x14 + x15 + x16 + 2x18 + … as a ratio of polynomials.
Answer
1. The series sums to (2 – x2 – x3)/(1 – x2 – x3 + x5).
Solution
1. The terms with coefficient 2 are x0, x6, x12, x18, … , in general, x6n. This suggests rearranging the series into terms of the form x2n and terms of the form x3n, obtaining: 1 + x2 + x4 + x6 + x8 + x10 + x12 + x14 + x16 + … +1 + x3 + x6 + x9 + x12 + x15 + x18 + …= 1/(1 – x2) + 1/(1 – x3)(2 – x2 – x3)/(1 – x2 – x3 + x5)
4. Write 1 + 2x + 3x2 + 4x3 + 5x4 + … as a ratio of polynomials.
Answer
1. The series converges to 1/(1 – 2x + x2).
Solution
1. First observe the sum 1 + 2x + 3x2 + 4x3 + 5x4 +… can be reorganized:1 + x + x2 + x3 + x4 +…= 1/(1 – x) x + x2 + x3 + x4 +… = x/(1 – x) x2 + x3 + x4 +… = x2/(1 – x) x3 + x4 +… = x3/(1 – x) x4 +… = x4/(1 – x) Consequently, 1 + 2x + 3x2 + 4x3 + 5x4 + … = (1 + x + x2 + x3 + …)/(1 – x) = 1/(1 – x)2 = 1/(1 – 2x + x2)