Calculus II: Power series expansion of a function

1. Definitions

Within its interval of convergence, a power series can represent the expansion of a function.

We recall Taylor and Maclaurin series are power series centred around a real c or 0 respectively.

Let f(x) a function having the following expansion in Taylor series (power series centred in c), within its interval of convergence ]c - R, c + R[, where R is the radius of convergence of the series:

f(x) = Σ_n=0^∞ a_n(x - c)ⁿ =

a₀ + a₁(x - c)ⁿ + ... + a_n(x - c)ⁿ + ...

Since a power series has the form of a polynomial function, we can take the derivative and an integral of this series using the usual methods.

The derivative of f(x) is:

f '(x) = [Σ_n=0^∞ a_n(x - c)ⁿ ]/dx =
Σ_n=0^∞ d[ a_n(x - c)ⁿ]/dx =
Σ_n=1^∞ n a_n(x - c)^{n - 1}


The integral of f(x) is:

∫ f (x) dx = ∫[Σ_n=0^∞ a_n(x - c)ⁿ ]dx =
Σ_n=0^∞ ∫[ a_n(x - c)ⁿ]dx =
Σ_n=0^∞ (a_n/(n + 1))(x - c)^{n + 1} + const.

2. Expansion of a function on a power series

2.1. Taylor and Maclaurin series

Some functions can be expanded in a power series Σ_n=0^∞ a_n(x - c)ⁿ, within their interval of convergence ]c - R, c + R[, where R is its radius of convergence.

Let f(x) a function infinitely differentiable. We want to find a power series that converges toward the function f(x) within an interval of convergence ]c - R, c + R[.

So, we can express f(x) as:

f(x) = Σ_n=0^∞ a_n(x - c)ⁿ in ]c - R, c + R[.

Therefore

f(c) = a₀
Then

a₀ = f(c)

f '(x) = Σ_n=1^∞ n a_n(x - c)^{n - 1}
f '(c) = a₁
Then

a₁ = f '(c)

f "(x) = Σ_n=2^∞ n (n - 1) a_n(x - c)^{n - 2}
f "(c) = 2 a₂
Then

a₂ = f "(c)/2

f⁽³⁾(x) = Σ_n=3^∞ n (n - 1) (n - 2)a_n(x - c)^{n - 3}
f⁽³⁾(c) = (3!) a₃
Then

a₃ = f "(c)/3!

and so on ...

f⁽ⁿ⁾(x) = Σ_n=2^∞ n! a_n(x - c)⁰
f⁽ⁿ⁾(c) = (n!) a_n
Then

a_n = f⁽ⁿ⁾(c)/n!

a_n = f⁽ⁿ⁾(c)/n!


f(x) = Σ_n=0^∞ (f⁽ⁿ⁾(c)/n!) (x - c)ⁿ
in ]c - R, c + R[.


The series
Σ_n=0^∞ (f⁽ⁿ⁾(c)/n!) (x - c)ⁿ =
f(c) + (f '(c)) (x - c)/1! + (f "(c)) (x - c)²/2! + (f⁽³⁾(c)) (x - c)³/3! + ... + (f⁽ⁿ⁾(c)/n!) (x - c)ⁿ + ...
is called Taylor series (Brook Taylor: 1685-1731).


The case where c = 0 , the series
Σ_n=0^∞ (f⁽ⁿ⁾(0)/n!) xⁿ =
f(0) + (f '(0)) (x)/1! + (f "(0)) (x)²/2! + (f⁽³⁾(0)) ()³/3! + ... + (f⁽ⁿ⁾(0)/n!) (x)ⁿ + ...
is called Maclaurin series (Colin Maclaurin: 1698-1746).

2.2. Examples

What is the expansion of the exponential function of base e f(x) = e^x in a Maclaurin series?

We have:

f(x) = f(0) + (f'(0))(x)/1! + (f"(0))(x)²/2! + (f⁽³⁾(0)) x³/3! + ... + (f⁽ⁿ⁾(0)/n!)(x)ⁿ + ...
= 1/0! + x/1! + x²/2! + x³/3! + ... + xⁿ/n! + ...
= Σ_n=0^∞ xⁿ/n!


In what interval of convergence?

The D'Alembert test gives:

lim |(xⁿ⁺¹/(n+1)!)/(xⁿ/n!)|
n → ∞
=
lim |x/(n+1)| = |x|/(n+1) = 0 < 1
n → ∞

Then the series converges for all x ∊ R.

Therefore

For all x ∊ R, e^x = Σ_n=0^∞ xⁿ/n!

For all x ∊ R
e^x = Σ_n=0^∞ xⁿ/n!

In particular, for x = 1, we have:

e = Σ_n=0^∞ 1/n!

3. Exercises

a) f(x) = 1/(1 - x) = 1 + x + x² + x³ + xⁿ + ... + ... =
= Σ_n=0^∞ xⁿ if |x| < 1

Using the derivative and the integral of f(x), find an expansion in Maclaurin series (power series centred at 0) for the two following functions:

g(x) = 1/(1 - x)²
and
h(x) = ln (1 - x)

Prove that ln 2 = Σ_n=1^∞ (- 1)^{n - 1}/n

b) Find the power series about x = 3 of the function f(x) = e^x.

Sol:
e^x = Σ_n=0^∞ e²(x - 3)ⁿ/n!
∀ x ∊ R.

c) Find the power series about x = 0 of the function f(x) = sin (x).

Sol:
sin (x) = Σ_n=0^∞ (- 1)ⁿ x²ⁿ⁺¹/(2n+1)!
∀ x ∊ R.

d) Using the above result, find the Maclaurin series of the function f(x) = cos(x).

Sol:
cos (x) = Σ_n=0^∞ (- 1)ⁿ x²ⁿ/(2n)!
∀ x ∊ R.

e) Change the variable in the latter result and find the Maclaurin series of the function f(x) = cos(3x)

f) Find the power series about x = 0 of the function f(x) = sin (x) + x cos(x).

g) Find the Taylor series on (x - 1) of the function f(x) = ln(x).

Sol:
ln(x) = Σ_n=1^∞ (- 1)^n-1 (x - 1)ⁿ/n ∀ x ∊ ]0,2[.