]> Exercise 1

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 1: Indefinite Integrals; page 1

Definitions and Basics, Exercise 1

Question

Use (2.1.1.10) and D'Alambert's test to prove that:

A power series can be integrated term by term within the interval of absolute convergence, without affecting the convergence in the following cases of power series:
  1. ∑ n=0 ∞ b n x n
     
  2. ∑ n=0 ∞ b n x 2n+1

Reminder

The D'alambert's test of the series with terms an states:

If { lim⁡ n→∞  | a n+1 a n | = ρ < 1 the series converges absolutely lim⁡ n→∞  | a n+1 a n | = ρ > 1 the series diverges lim⁡ n→∞  | a n+1 a n | = ρ = 1 the test is indecisive } (1.3.5.22)

... rules for integration:

∫ c f( x )dx=c ∫  f( x )dx ( where c=constant ) (2.1.1.8)
∫ [ f 1 ( x )+ f 2 ( x ) ] dx=∫ f 1 ( x ) dx+∫ f 2 ( x ) dx (2.1.1.9)
∫ x p dx= x p+1 p+1 +C (2.1.1.10)

Parts 1-4

Solution of question 1

  1. For a power series written as ∑ n=0 ∞ b n x n the limit for the D'Alambert test is according to (1.3.5.22): lim⁡ n→∞ | a n+1 a n |= lim⁡ n→∞ | b n+1 x b n |=ρ
  2. The integration of the series is obtained by use of the rules (2.1.1.8-10): ∫ dx ∑ n=0 ∞ b n x n = ∑ n=0 ∞ b n ∫ x n dx =C+ ∑ n=0 ∞ b n x n+1 n+1
  3. The additive constant C from part 2 does not affect the convergence of the series.
  4. The D'Alambert's test for this new series is: lim⁡ n→∞ | b n+1 x b n n n+1 |= lim⁡ n→∞ n n+1   lim⁡ n→∞ | b n+1 x b n |= lim⁡ n→∞ | b n+1 x b n | exactly as the result of part 1, which proves the statement as required.

Parts 5-7

Solution of question 2

  1. For a power series written as ∑ n=0 ∞ b n x 2n+1 the limit of the D'Alambert's test is lim⁡ n→∞ | a n+1 a n |= lim⁡ n→∞ | b n+1 x 2 b n |
  2. The integration of the series yields ∫ dx ∑ n=0 ∞ b n x 2n+1 = ∑ n=0 ∞ b n ∫ x 2n+1 dx =C+ ∑ n=0 ∞ b n x 2n+2 2n+2 where C as previously does not affect the convergence of the series.
  3. The D'Alambert's test for this new series is: lim⁡ n→∞ | b n+1 x 2 b n 2n+2 2n+4 |= lim⁡ n→∞ | b n+1 x 2 b n | lim⁡ n→∞ | n+1 n+2 |= lim⁡ n→∞ | b n+1 x 2 b n | in full agreement with part 5.

Score

Questions 1 gives credit of 6 points.

Question 2 gives credit of 4 points.

By parts:

Parts 1,3,5,6 are worth 1 point each.
Parts 2,4,7 are worth 2 points each.