Unit 10 · Topic 10.15

Representing Functions as Power Series

New power series come from known ones by substitution, multiplication by powers of x , and term-by-term differentiation or integration.

Practice 1: Implementing Mathematical ProcessesPractice 2: Connecting Representations

Part 1 · Hook

Why this matters

Nobody memorizes the series for ln⁡(1+x)\ln(1 + x) or arctan⁡x\arctan x. They are built in a few lines from the geometric series, by the moves you already know: substitute, multiply, differentiate, integrate. A handful of known series becomes a whole library.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the Maclaurin series for 11−x\frac{1}{1 - x}, and where does it converge?

  1. 1+x+x2+⋯1 + x + x^2 + \cdots for |x|<1|x| < 1
  2. 1+x+x2+⋯1 + x + x^2 + \cdots for every xx
  3. 1−x+x2−⋯1 - x + x^2 - \cdots for |x|<1|x| < 1
  4. 1+x+x22!+⋯1 + x + \frac{x^2}{2!} + \cdots for every xx
Show the answer

It is the geometric series with ratio xx.

  • Correct: 1+x+x2+⋯1 + x + x^2 + \cdots for |x|<1|x| < 1:
  • 1+x+x2+⋯1 + x + x^2 + \cdots for every xx:
  • 1−x+x2−⋯1 - x + x^2 - \cdots for |x|<1|x| < 1:
  • 1+x+x22!+⋯1 + x + \frac{x^2}{2!} + \cdots for every xx:

2. Find ∫t4dt\int t^4\,dt.

  1. t55+C\frac{t^5}{5} + C
  2. 4t3+C4t^3 + C
  3. t5+Ct^5 + C
  4. t44+C\frac{t^4}{4} + C
Show the answer

Power rule for antiderivatives: raise the power by one and divide by the new power.

  • Correct: t55+C\frac{t^5}{5} + C:
  • 4t3+C4t^3 + C:
  • t5+Ct^5 + C:
  • t44+C\frac{t^4}{4} + C:

3. Find the radius of convergence of ∑(x−1)n4n\sum \frac{(x - 1)^n}{4^n}.

  1. 44
  2. 14\frac{1}{4}
  3. 11
  4. 55
Show the answer

It is geometric with ratio x−14\frac{x - 1}{4}: |x−1|<4|x - 1| < 4.

  • Correct: 44:
  • 14\frac{1}{4}:
  • 11:
  • 55:

Part 4 · See it

See it first

The graph of y = 1/(1 - x) from x = -1.6 to 1.6, with a vertical asymptote at x = 1, and three partial sums of 1 + x + x squared + ...: through x squared, through x to the fifth and through x to the tenth. A shaded strip marks -1 < x < 1, where the partial sums close in on the curve. To the left of -1 they swing up and down with growing size; to the right of 1 they climb, while 1/(1 - x) is negative there.
Partial sums of 1+x+x2+⋯1 + x + x^2 + \cdots close in on 11−x\frac{1}{1 - x} only for −1<x<1-1 < x < 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Start from a known series, such as 11−u=∑un\frac{1}{1 - u} = \sum u^n for |u|<1|u| < 1its sum and its interval are known
  2. Substitute (u=−x2u = -x^2), multiply by a power of xx, or add seriesa series for a new function, valid where the substituted value meets the old condition
  3. Differentiate or integrate term by terma series for the derivative or integral with the same radius; test the endpoints again

Part 6 · Key ideas

Key ideas

  • Substitution: replace xx by an expression in a known series, and the condition with it: 11+x2=∑(−x2)n\frac{1}{1 + x^2} = \sum (-x^2)^n for x2<1x^2 < 1.
  • Term by term: ddx∑cnxn\frac{d}{dx}\sum c_n x^n=∑ncnxn−1= \sum n c_n x^{n-1} and ∫0x∑cntndt\int_0^x \sum c_n t^n\,dt=∑cnxn+1n+1= \sum \frac{c_n x^{n+1}}{n + 1}.
  • The radius stays the same; endpoints may change, so test them again.
  • Differentiate with respect to xx, never nn; when integrating, fix the constant with a known value such as ln⁡1=0\ln 1 = 0.

Part 7 · Misconception

A common mistake

The wrong idea: ln⁡(1+x)\ln(1 + x)=x−x22+x33−⋯= x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots converges on (−1,1)(-1, 1), the same interval as 11+x\frac{1}{1 + x}.

What actually happens: The radius is the same, but at x=1x = 1 the integrated series is the alternating harmonic series, which converges, so the interval is (−1,1](-1, 1].

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

From a geometric series to arctan⁡x\arctan x

Building a power series for arctan⁡x\arctan x from the series for 11−u\frac{1}{1 - u}
StepSeries
Start11−u\frac{1}{1 - u}=1+u+u2+u3+⋯= 1 + u + u^2 + u^3 + \cdots for |u|<1|u| < 1
Substitute u=−x2u = -x^211+x2\frac{1}{1 + x^2}=1= 1+(−x2)+ (-x^2)+(−x2)2+ (-x^2)^2+⋯+ \cdots
Integrate from 00 to xxarctan⁡x\arctan x=∫0x11+t2dt= \int_0^x \frac{1}{1 + t^2}\,dt

1. Write the first four nonzero terms of the Maclaurin series for 11+x2\frac{1}{1 + x^2}.

Type an expression for 11+x2≈\frac{1}{1 + x^2} \approx.

Show the answer

With u=−x2u = -x^2: 11+(−x2)+ (-x^2)+(−x2)2+ (-x^2)^2+(−x2)3+ (-x^2)^3+⋯+ \cdots=1−x2= 1 - x^2+x4−x6+ x^4 - x^6+⋯+ \cdots

  • Answer: 11+x2≈\frac{1}{1 + x^2} \approx 1−x2+x4−x61 - x^{2} + x^{4} - x^{6}

2. What is the radius of convergence of the series for 11+x2\frac{1}{1 + x^2}?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The geometric series needs |u|<1|u| < 1, that is x2<1x^2 < 1, so |x|<1|x| < 1 and the radius is 1.

  • Answer: 1

3. Integrate term by term from 00 to xx: write the first four nonzero terms of the Maclaurin series for arctan⁡x\arctan x.

Type an expression for arctan⁡x≈\arctan x \approx.

Show the answer

∫0x(1−t2+t4−t6)dt\int_0^x (1 - t^2 + t^4 - t^6)\,dt=x−x33= x - \frac{x^3}{3}+x55−x77+ \frac{x^5}{5} - \frac{x^7}{7}, and the constant is 0 because arctan⁡0=0\arctan 0 = 0.

  • Answer: arctan⁡x≈\arctan x \approx x−x33+x55−x77x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7}

4. Integrate the series for 11+t\frac{1}{1 + t} from 00 to xx to write the first four nonzero terms of the Maclaurin series for ln⁡(1+x)\ln(1 + x).

Type an expression for ln⁡(1+x)≈\ln(1 + x) \approx.

Show the answer

11+t\frac{1}{1 + t}=1−t+t2−t3+⋯= 1 - t + t^2 - t^3 + \cdots for |t|<1|t| < 1. Integrating from 0 to xx: ln⁡(1+x)\ln(1 + x)=x−x22= x - \frac{x^2}{2}+x33−x44+ \frac{x^3}{3} - \frac{x^4}{4}+⋯+ \cdots

  • Answer: ln⁡(1+x)≈\ln(1 + x) \approx x−x22+x33−x44x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}

5. Give the interval of convergence of ln⁡(1+x)\ln(1 + x)=∑n=1∞(−1)n+1xnn= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}.

Type the interval, in interval notation such as [1, 5) or as an inequality such as 1 ≤ x < 5.

Show the answer

The radius is 1, as for 11+x\frac{1}{1 + x}. At x=1x = 1 the series is the alternating harmonic series, which converges; at x=−1x = -1 it is −∑1n-\sum \frac{1}{n}, which diverges. The interval is (−1,1](-1, 1].

  • Answer: (−1,1](-1, 1]

6. What happens to the radius of convergence when a power series is differentiated or integrated term by term?

  1. It doubles when the series is integrated
  2. It shrinks by 1 when the series is differentiated
  3. It stays the same, and so does each endpoint
  4. It stays the same, though an endpoint may change
Show the answer

Term-by-term differentiation and integration keep the radius of convergence. Behavior at the endpoints can change, so test them again.

  • It doubles when the series is integrated: The radius does not change.
  • It shrinks by 1 when the series is differentiated: The radius does not change.
  • It stays the same, and so does each endpoint: Endpoints can change: ∑(−1)n+1xnn\sum \frac{(-1)^{n+1}x^n}{n} converges at x=1x = 1, but its derivative series does not.
  • Correct: It stays the same, though an endpoint may change: Right: the radius is kept; test the endpoints again.

Part 9 · Summary

Summary

New power series come from known ones by substitution, multiplication by powers of xx, and term-by-term differentiation or integration. These operations keep the radius of convergence, but the endpoints must be tested again. This is how the series for ln⁡(1+x)\ln(1 + x) and arctan⁡x\arctan x are derived.

Part 10 · Up next

What comes next

This is the last topic in the course.

Part 11 · Connections

Connections

Question set for this topic