Calculus II: Power series

1. Definitions

A power series is an infinite series where the terms are variable. Its geral form is:

f(x) = Σ_n=0^∞ a_n (x - c)ⁿ
where:
a_n is the coefficient of the n^th term, x is a variable and c is a real constant.

This series is called the Taylor series.

When c is equal to zero, the series becomes:

f(x) = Σ^∞_n=0 a_n xⁿ,
and called Mac Laurin series.

When c is equal to zero, and all the coefficients are equal to 1, the series becomes:

f(x) = Σ^∞_n=0 xⁿ,
and called geometric series.

A power series is an a polynomial with an infinite degree .

The set of values of the variable x for which the power series congerges is called the interval of convergence of the series

The number c is the center of the disk of convergence of the series.

The radius of convergence R of the series is the positive real number (including + ∞) such that the series converges if
|x - c| < R
and diverges if
|x - c| > R

The radius of convergence of a power series is equal to the half of the length of its interval of convergence.

If a power series is geometric, that is its ratio a_n+1/a_n is independent of n, then this series can be written as a function f(x) in its interval of convergence.

2. Examples

2.1. Example 1

The geometric series:

f(x) = Σ^∞_n=0 xⁿ,
Σ^∞_n=0 xⁿ = 1 + x + x² + x³ + x⁴ + ... + xⁿ + ...

has the first term equal to 1, and the ratio equal to x.

The sum of the n first terms of this series is:

s_n = Σ₀ⁿ x^k = (1 - xⁿ)/(1 - x)

Therefore

If |x| >= 1
lim s_n = + ∞. The series is divergent.
n → + ∞

If |x| < 1
lim s_n = 1/(1 - x)
n → + ∞
The series is convergent.

Hence

For - 1 < x < 1, the function f(x) = 1/(1 - x) is expanded in the power series Σ^∞_n=0 xⁿ

The interval of convergence of the geometric series is ]- 1, + 1[

2.2. Example 2

We want to calculate the interval of convergence, and the radius of convergence R of the following power series:

Σ_n=0^∞ xⁿ/(n + 1)

The d'Alembert test is:

L = |a_n+1/a_n| = |xⁿ⁺¹/(n + 2) / xⁿ/(n + 1)| =
|x (n + 1)/(n + 2)| = |x| (n + 1)/(n + 2)

L = |x| (n + 1)/(n + 2)

lim L = |x| . 1 = |x|
n → + ∞

Therefore
if |x| < 1, the series converges
if |x| > 1, the series diverges
if |x| = 1, we cannot conclude

For |x| = ±1, we have:

If x = + 1 → , the series becomes an harmonic series :
Σ_n=0^∞ 1/(n + 1) that diverges.

If x = - 1 → , the series becomes an alternating harmonic series:
Σ_n=0^∞ (- 1)ⁿ/(n + 1) that converges (ln 2).

Finally, for the power series
Σ_n=0^∞ xⁿ/(n + 1)
The interval of the convergence is: [-1, +1[
And since the series converges if |x| < 1, and diverges if |x| > 1, the radius of convergence is R = 1

2.3. Example 3

Σ_n=0^∞ ((x - 1)/2)ⁿ

The d'Alembert test is:

L = |a_n+1/a_n| = |((x - 1)/2)ⁿ⁺¹/((x - 1)/2)ⁿ| = |(x - 1)/2|

lim L = |(x - 1)/2|
n → + ∞

Therefore
if |x - 1| < 2, that is
- 2 < x - 1 < 2, → - 1 < x < 3,
the series converges.

if |x - 1| > 2, that is
- 2 < x - 1 or x - 1 > 2, → - 1 < x or x > 3,
the series diverges.


if |x - 1| = 2, that is
x - 1 = 2 or x - 1 = - 2
we cannot conclude.

x - 1 = 2
Σ_n=0^∞ (1)ⁿ
The series diverges.

x - 1 = - 2
Σ_n=0^∞ (- 1)ⁿ
The series diverges.

Therefore, the interval of convergence of the series is: ]- 1, + 3[. The radius of convergence is 4/2 = 2.

Since the series is geometric of first term a equal to 1, and of ratio r = (x - 1)/2, it converges to
a/(1 - r) = 1/(1 - (x - 1)/2) = 2/(2 - (x - 1)) = 2/(3 - x))

Hence, in the the interval ]- 1, + 3[, the series Σ_n=0^∞ ((x - 1)/2)ⁿ can be written as the fuction 2/(3 - x)).

The function f(x) = 2/(3 - x)) has for expansion the series Σ_n=0^∞ ((x - 1)/2)ⁿ.

2.4. Example 4

The function f(x) = 7/(2 - x) = (7/2)/(1 - x/2) can be expanded in a geometric series of first term a = 7/2 and ratio r = x/2 in the interval |x/2| < 1, that is ]-2, + 2[:
Σ_n=1^∞ ar^{n - 1} = (7/2)Σ_n=1^∞ (x/2)ⁿ .

3. Exercises