MathLab with Mr. Ahmed

MathLab with Mr. Ahmed is a math learning website by Ahmed Amasha, a mathematics teacher teaching the American curriculum. Explore Algebra 2, Precalculus, Calculus, and SAT and PSAT Math Prep through lessons, focused practice and printable resources.

MathLabwith Mr. Ahmed
Infinite Series

9.9 Representation of Functions by Power Series

Focus on represent rational functions, term-by-term integration, term-by-term differentiation.

Free preview Explore the lesson before creating an account.

Play games for this section

Before you solve

Ideas in this lesson

01Represent rational functionsWorked steps · five questions02Term-by-term integrationWorked steps · five questions03Term-by-term differentiationWorked steps · five questions
Idea 1 of 3

Represent rational functions

Transform a rational function into a scaled geometric-series form and retain its convergence restriction.

Worked example 1

In the Maclaurin expansion of 11−2x\dfrac{1}{1 - 2x}, find the coefficient of x2x^{2}.
  1. 11−2x=∑(2x)n,∣x∣<12\dfrac{1}{1 - 2x} = \sum (2x)^{n}, |x| < \dfrac{1}{2}
  2. (2x)n=2nxn(2x)^{n} = 2^{n}x^{n}
  3. Coefficient of x2=22=4x^{2} = 2^{2} = 4

Answer: 44

Worked example 2

In the Maclaurin expansion of 11−3x\dfrac{1}{1 - 3x}, find the coefficient of x3x^{3}.
  1. 11−3x=∑(3x)n,∣x∣<13\dfrac{1}{1 - 3x} = \sum (3x)^{n}, |x| < \dfrac{1}{3}
  2. (3x)n=3nxn(3x)^{n} = 3^{n}x^{n}
  3. Coefficient of x3=33=27x^{3} = 3^{3} = 27

Answer: 2727

Idea 2 of 3

Term-by-term integration

A power series may be integrated inside its radius. Determine the integration constant separately and retest endpoints.

Worked example 1

For −2ln⁡(1−x)=∑(n=1-2 \ln (1 - x) = \sum (n = 1 to ∞)2xnn\infty ) \dfrac{2x^{n}}{n}, find the coefficient of x2x^{2}.
  1. ∫0x2∑tndt=∑2xn+1n+1=−2ln⁡(1−x),∣x∣<1\int _{0}^{x} 2\sum t^{n} dt = \sum \dfrac{2x^{n+1}}{n + 1} = -2 \ln (1 - x), |x| < 1
  2. x=0  ⟹  C=0x = 0 \implies C = 0
  3. Coefficient of x2=22=1x^{2} = \dfrac{2}{2} = 1

Answer: 11

Worked example 2

For −3ln⁡(1−x)=∑(n=1-3 \ln (1 - x) = \sum (n = 1 to ∞)3xnn\infty ) \dfrac{3x^{n}}{n}, find the coefficient of x3x^{3}.
  1. ∫0x3∑tndt=∑3xn+1n+1=−3ln⁡(1−x),∣x∣<1\int _{0}^{x} 3\sum t^{n} dt = \sum \dfrac{3x^{n+1}}{n + 1} = -3 \ln (1 - x), |x| < 1
  2. x=0  ⟹  C=0x = 0 \implies C = 0
  3. Coefficient of x3=33=1x^{3} = \dfrac{3}{3} = 1

Answer: 11

Idea 3 of 3

Term-by-term differentiation

Differentiate each monomial within the radius. The radius stays the same, but endpoint behavior may change.

Worked example 1

For ∣x∣<2,f(x)=12−x=∑|x| < 2, f(x) = \dfrac{1}{2 - x} = \sum from n=0n = 0 to ∞\infty of xn2n+1\dfrac{x^{n}}{2^{n + 1}}. Find the coefficient of x2x^{2} in f′(x)f'(x).
  1. f′(x)=∑f'(x) = \sum from n=1n = 1 to ∞\infty of nxn−12n+1\dfrac{nx^{n - 1}}{2^{n + 1}}
  2. n−1=2  ⟹  n=3n - 1 = 2 \implies n = 3
  3. Coefficient =324= \dfrac{3}{2^{4}}

Answer: 316≈0.19\dfrac{3}{16} \approx 0.19

Worked example 2

For ∣x∣<3,f(x)=13−x=∑|x| < 3, f(x) = \dfrac{1}{3 - x} = \sum from n=0n = 0 to ∞\infty of xn3n+1\dfrac{x^{n}}{3^{n + 1}}. Find the coefficient of x3x^{3} in f′(x)f'(x).
  1. f′(x)=∑f'(x) = \sum from n=1n = 1 to ∞\infty of nxn−13n+1\dfrac{nx^{n - 1}}{3^{n + 1}}
  2. n−1=3  ⟹  n=4n - 1 = 3 \implies n = 4
  3. Coefficient =435= \dfrac{4}{3^{5}}

Answer: 4243≈0.02\dfrac{4}{243} \approx 0.02

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §9.9,p9.9, p. 657657. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.