# 1. Theory

This tutorial deals with the approximation of functions of x, ƒ(x), using power series expansions:

ƒ(x) = n=0 cixi = c0 + c1x + c2x2 + c3x3 + ...

- where ci are constant coefficients.

Power series open the door to the rapid calculation, manipulation and interpretation of analytical expressions that are, otherwise, difficult to deal with.

It will greatly simplify each calculation if, at an early stage, you manage to deduce how many terms are required from each standard series to give the final result to the specified accuracy (e.g. to x3).

Note also that writing the given function in terms of partial fractions, where appropriate, can result in a simpler derivation.

# 2. Exercises

Use the Standard Series, to expand the following functions in power series, as far as the terms shown. Also state the range of values of x for which the power series converges.

Click on questions to reveal their solutions

Exercise 1:

e−3x cos 2x, up to x3

Solution: Find the expansions for the two terms and then multiply them:

 1st term: eX = 1 + X + X2/2! + X3/3! + X4/4! + ... true for all X ∴ e−3x = 1 + (−3x) + (−3x)2/2! + (−3x)3/3! + (−3x)4/4! + ... i.e. e−3x = 1 − 3x + 9x2/2 − 9x3/2 + ... 2nd term: cos X = 1 − X2/2! + X4/4! − ... true for all X ∴ cos 2x = 1 − (2x)2/2! + ... (we only need terms up to x3) i.e. cos 2x = 1 − 2x2 + ... ∴ e−3x cos 2x = (1 − 3x + 9/2x2 − 9/2x3) (1 − 2x2) + ... = 1 − 3x + 9/2x2 − 9/2x3 − 2x2(1 − 3x) + ... (again, we only need terms up to x3) = 1 − 3x + 9/2 x2 − 9/2 x3 − 2x2 + 6x3 +... = 1 − 3x + (9 − 4)/2 x2 + (12 − 9)/2 x3 + ... ∴ e−3x cos 2x = 1 − 3x + 5/2 x2 + 3/2 x3 + ... (true for all x, ∴ converges for all x values).

Exercise 2:

(sin x) ln(1 − 2x), up to x4

Solution: Find the expansions for the two terms and then multiply them:

 1st term: sin x = x − x3/3! + ... = x − x3/6 + ... true for all x 2nd term: ln(1 + X) = X − X2/2 + X3/3 − X4/4 + ... true for −1 < X ≤ 1 ∴ ln(1 − 2x) = (−2x) − (−2x)2/2 + (−2x)3/3 − (−2x)4/4 + ... true for −1 < (−2x) ≤ 1 = − 2x − − 4x2/2 − 8x3/3 − 16x2/4 ... true for −1⁄2 ≤ x < 1⁄2 (note the switching of '<' and '≤' because of the minus sign of −2x) i.e. ln(1 − 2x) = − 2x − − 2x2 − 8/3x3 − 4x4 − ... ∴ (sin x) ln (1 − 2x) = (x − x3/6) (− 2x2 − 8/3x3 − 4x4) + ... = x(−2x − 2x2 − 8/3x3 − 4x4) −x3/6 (−2x − ...) + ... = −2x2 − 2x3 − 8/3x4 + x4/3 + ... ∴ (sin x) ln(1 − 2x) = −2x2 − 2x3 − 7/3x4 + ... (converges for −1⁄2 ≤ x < 1⁄2)

Note: We will assume convergence for the values of x that both the sin x and the ln(1 − 2x) power series are true.

Exercise 3:

1 + x ⋅ e−2x, up to x3

Solution:

 1st term: (1 + x)n = 1 + nx + n(n−1)/2! x2 + n(n−1)(n−2)/3! x3 + ... where n = 1⁄2 ∴ √1 + x = (1 + x)1⁄2 = 1 + 1/2x + 1/2 ⋅ (−1/2) x2/2 + 1/2 ⋅ (−1/2) ⋅ (−3/2) x3/6 + ... true for −1 < x < 1 ∴ √1 + x = 1 + 1/2x − 1/8x2 + 1/16x3 + ... 2nd term: eX = 1 + X + X2/2! + X3/3! + X4/4! + ... true for all X ∴ e−2x = 1 + (−2x) + (−2x)2/2! + (−2x)3/3! + (−2x)4/4! + ... i.e. e−2x = 1 − 2x + 2x2 − 4/3 x3 + ... ∴ √1 + x ⋅ e−2x = (1 + 1/2x − 1/8x2 + 1/16x3) ( 1 − 2x + 2x2 − 4/3 x3 ) + ... 1⋅(1 − 2x + 2x2 − 4/3 x3 ) + 1/2x ( 1 − 2x + 2x2 ) − 1/8x2 (1 − 2x ) + 1/16x3 + ... 1 − 2x + 2x2 − 4/3 x3 + 1/2x − x2 + x3 − 1/8x2 + 1/4x3 + 1/16x3 + ... ∴ √1 + x ⋅ e−2x = 1 − 3/2x + 7/8x2 − 1/48x3 + ... (converges for −1 < x < 1)

Exercise 4:

sin x/ 1 + 2x  , up to x4

Solution:

 1st term: sin x = x − x3/3! + ... true for all x 2nd term: (1 + X)n = 1 + nX + n(n−1)/2! X2 + n(n−1)(n−2)/3! X3 + ... true for −1 < X < 1 ∴ (1 + 2x)1⁄2 = 1 − 1/2 (2x) + 3/22 (2x)2/2 − 15/23 (2x)3/6 + ... true for −1 < 2x < 1 = 1 − x + 3/2x2 −15/6x3 + ... true for −1⁄2 < x < 1⁄2 Note: in the above we only need the denominator term expressed up to x3 since sin x ≈ x − x3⁄6 will multiply this by at least x. ∴ sin x/ √1 + 2x = (x − x3/6) ( 1 − x + 3/2x2 − 15/6x3 ) + ... = x(1 − x + 3/2x2 − 5/2x3) − x3/6(1 − x) + ... = x − x2 + 3/2x3 − 5/2x4 − 1/6x3 + 1/6x4 + ... = x − x2 + (9 − 1)/6x3 + (1 − 15)/6x4 + ... ∴ sin x/ √1 + 2x = x − x2 + 4/3x3 − 7/3x4 + ... (converges for − 1⁄2 < x < 1⁄2)

Exercise 5:

(cos 4x) ln(1 + 2x), up to x4

Solution:

 1st term: cos X = 1 − X2/2! + X4/4! − ... true for all X = 1 − X2/2 + X4/24 − ... ∴ cos 4x = 1 − (4x)2/2 + (4x)4/24 − ... = 1 − 8x2 + 64/6x4 − . i.e. cos 4x = 1 − 8x2 + 32/3x4 − ... 2nd term: ln(1 + X) = X − X2/2 + X3/3 − X4/4 + ... true for −1 < X ≤ 1 ∴ ln(1 + 2x) = (2x) − (2x)2/2 + (2x)3/3 − (2x)4/4 + ... true for −1 < 2x ≤ 1 i.e. ln(1 + 2x) = 2x − 2x2 + 8/3x3 − 4x4 + ... true for −1⁄2 < x ≤ 1⁄2 ∴ (cos 4x) ln(1 + 2x) = (1 − 8x2) (2x − 2x2 + 8/3x3 − 4x4) + ... = 2x − 2x2 + 8/3x3 − 4x4 − 8x2(2x − 2x2) + ... i.e. (cos 4x) ln(1 + 2x) = 2x − 2x2 − 40/3x3 + 12x4 + ... (converges for −1⁄2 < x ≤ 1⁄2)

Exercise 6:

5 + 4x / (1 + 2x)(1 − x) , up to x3

Solution:

First, express the equation in terms of partial fractions:

5 + 4x / (1 + 2x)(1 − x) = A / 1 + 2x + B / 1 − x = A(1 − x) + B(1 + 2x) / (1 + 2x)(1 − x)
5 + 4x = A(1 − x) + B(1 + 2x)   [Identity, true for all x]
When x = 1:  5 + 4 = 0 + B(3)   i.e. B = 3
When x = −12:  5 − 2 = A( 32) + 0     i.e. A = 2

So, 5 + 4x / (1 + 2x)(1 − x) = 2 / 1 + 2x + 3 / 1 − x

[Note: this form allows addition of series rather than multiplication, i.e. it makes the calculation easier]

 1 / 1 − X = 1 + X + X2 + X3 + ... true for −1 < X < 1; ∴ 1 / 1 + 2x = 1 + (−2x) + (−2x)2 + (−2x)3 + ... (true for −1 < (−2x) < 1) i.e. 1 / 1 + 2x = 1 − 2x + 4x2 − 8x3 + ... true for −1⁄2 < x < 1⁄2 ∴ 1 / 1 + 2x + 3 / 1 − x = 2(1 − 2x + 4x2 − 8x3 + ...) + 3(1 + x + x2 + x3 + ...) = 5 − x + 11x2 − 13x3 + ... (converges for −1⁄2 < x < 1⁄2)

Exercise 7:

3 + x / (2 + x)(1 + x) , up to x3

Solution:

Express the equation in terms of partial fractions:

3 + x / (2 + x)(1 + x) = A / 2 + x + B / 1 + x = A(1 + x) + B(2 + x) / (2 + x)(1 + x)
3 + x = A(1 + x) + B(2 + x)   [Identity, true for all x]
When x = −1:  3 − 1 = 0 + B(2 − 1)   i.e. B = 2
When x = −2:  3 − 2 = A(1 − 2) + 0    i.e. A = −1

So:    3 + x / (2 + x)(1 + x) = −1 / 2 + x + 2 / 1 + x

 1 / 1 − X = 1 + X + X2 + X3 + ... true for −1 < X < 1; ∴ 1 / 1 + x = 1 + (−x) + (−x)2 + (−x)3 + ... (true for −1 < x < 1) = 1 − x + x2 − x3 + ... true for −1 < x < 1; and: 1 / 2 + x = (1/2) 1 / 1 + ( x/ 2) = (1/2) [ 1 − ( x/ 2) + ( x/ 2)2 − ( x/ 2)3 + ... ] (true for −1 < x/ 2 < 1) i.e. − 1 / 2 + x = − 1 / 2 + x  / 4 - x2 / 8 + x3 / 16 + ... true for −1 < x < 2 ∴ 3 + x / (2 + x)(1 + x) = −1 / 2 + x + 2 / 1 + x = − 1 / 2 + x  / 4 - x2 / 8 + x3 / 16 + ... + 2( 1 − x + x2 − x3 + ...) = 3/ 2 − 7/ 4x + 15/ 8x2 − 31/ 16x3 + ... (converges for −1 < x < 1)

Exercise 8:

4 + 5x / (2 + x)(1 − x) , up to x3

Solution:

Express the equation in terms of partial fractions:

4 + 5x/ (2 + x)(1 − x) = A/ 2 + x + B/ 1 − x = A(1 − x) + B(2 + x) / (2 + x)(1 − x)
4 + 5x = A(1 − x) + B(2 + x)   [Identity, true for all x]
True for x = +1:  4 + 5 = 0 + B⋅(3)  i.e. B = 3
True for x = −2:  4 − 10 = A⋅(3) + 0   i.e. A = −2

So:    4 + 5x/ (2 + x)(1 − x) = − 2 / 2 + x + 3 / 1 − x

 Now: 1 / 1 − x = 1 + x + x2 + x3 + ... true for −1 < x < 1; and: 1 / 2 + x = 1 / 2 1 / 1 + x / 2 = 1 / 2 [ 1 − ( x/ 2) + ( x/ 2)2 − ( x/ 2)3 + ... ]  (true for −1 < x⁄2 < 1) = 1 / 2 − x/ 4 + x2/ 8 − x3/ 16 + ... true for −2 < x < 2 ∴ 4 + 5x / (2 + x)(1 − x) = −2 ( 1 / 2 − x/ 4 + x2/ 8 − x3/ 16) + 3 ( 1 + x + x2 + x3) + ... = 2 + 7 / 2x + 11 / 4x2 + 25 / 8x3 + ...  (converges for −1 < x < 1)

Exercise 9:

1 + 2x / 1 − 3x , up to x3

Solution:

 Numerator: (1 + X)n = 1 + nX + n(n−1)/2! X2 + n(n−1)(n−2)/3! X3 + ...   (true for −1 < X < 1) so: √1 + X) = 1 + 1/2X + (1/2)⋅ (−1/2) X2/2! + (1/2)⋅ (−1/2)⋅ (−3/2) X3/3! + ... ∴ √1 + 2x) = 1 + 1/2(2x) + (1/2)⋅ (−1/2) (2x)2/2! + (1/2)⋅ (−1/2)⋅ (−3/2) (2x)3/3! + ...  (true for −1 < 2x < 1) = 1 + x − x2/2 + x3/2 − ... true for −1⁄2 < x < 1⁄2 Denominator: 1/1 − X = 1 + X + X2 + X3 + ...  (true for −1 < X < 1) ∴ 1/1 − 3x = 1 + (3x) + (3x)2 + (3x)3 + ...  (true for −1 < 3x < 1) = 1 + 3x + 9x2 + 27x3 + ...  (true for −1⁄3 < x < 1⁄3) ∴ √1 + 2x / 1 − 3x = (1 + x − x2/2 + x3/2) (1 + 3x + 9x2 + 27x3) + ... = 1 + 3x + 9x2 + 27x3 +x(1 + 3x + 9x2) −x2/2 (1 + 3x) +x3/2 (1) = 1 + 3x + 9x2 + 27x3 +x + 3x2 + 9x3 −1/2x2 −3/2x3 +1/2x3 = 1 + 4x + 23/2x2 + 35x3 + ...  (converges for −1⁄3 < x <1⁄3)

Exercise 10:

1 + x / 1 − 2x  , up to x3

Solution:

 Numerator: √1 + x) = (1 + x)1⁄2 = 1 + (1/2)x + (1/2)⋅ (−1/2) x2/2! + (1/2)⋅ (−1/2)⋅ (−3/2) x3/3! + ... = 1 + 1/2x − 1/8x2 + 1/16x3 + ...  true for −1 < x < 1 Denominator: 1/√1 + X) = (1 + X)1⁄2 = 1 − (1/2)X + (−1/2)⋅ (−3/2) X2/2! + (−1/2)⋅ (−3/2)⋅ (−5/2)⋅ X3/3! + ... (true for −1 < X < 1) ∴ (1 − 2x)1⁄2 = 1 − 1/2(−2x) + (1/2)⋅ (3/2) (−2x)2/2! − (1/2)⋅ (3/2)⋅ (5/2) (−2x)3/3! + ... (true for −1 < (−2x) < 1) = 1 + x + 3/2x2 + 5/2x3 + ... true for −1⁄2 < x < 1⁄2 ∴ (1 + x)1⁄2(1 − 2x)−1⁄2 = (1 + 1/2x − 1/8x2 + 1/16x3) (1 + x + 3/2x2 + 5/2x3) + ... = 1 + x + 3/2x2 + 5/2x3 + 1/2 x(1 + x + 3/2x2) −1/8x2(1 + x) +1/16x3 (1) + ... = 1 + 3/2x + 15/8x2 + 51/16x3 + ...  (converges for −1⁄2 < x < 1⁄2)

# 3. Standard Series

ƒ(x) Power Series Convergence
1/1 − x 1 + x + x2 + x3 + x4 + ... (−1 < x < 1)
(1 + x)n 1 + nx + n(n−1)/2! x2 + n(n−1)(n−2)/3! x3 + ... (−1 < x < 1 when n ≠ 1, 2, 3, ...)
(for all x when n = 1, 2, 3, ...)
ex 1 + x + x2/2! + x3/3! + x4/4! + ... (for all x)
ln(1 + x) xx2/2 + x3/3x4/4 + ... (−1 < x ≤ 1)
sin x xx3/3! + x5/5!x7/7! + ... (for all x)
cos x 1 − x2/2! + x4/4!x6/6! + ... (for all x)
tan x x + x3/3! + 2x5/15 + 17x7/315 + ... (−π2 < x < π2)
sinh x x + x3/3! + x5/5! + x7/7! + ... (for all x)
cosh x 1 + x2/2! + x4/4! + x6/6! + ... (for all x)

Note: 2! = 2 × 1 = 2; 3! = 3 × 2 × 1 = 6; 4! = 4 × 3 × 2 × 1 = 24; etc.

# 4. Alternative Notation

The power series for (1 + x)n is an example of a binomial series.

When n is not a whole number (i.e. n ≠ 0, 1, 2, 3, ...) then the series is an infinite series and it is only true for −1 < x < 1.

On the other hand, when n is a whole number (i.e. n = 0, 1, 2, 3, ...) then the series no longer has an infinite number of terms and it is valid for all values of x.

Some books use notation to distinguish the above two cases, where:

(1 + x)n is written when the power is a whole number, and

(1 + x)r is written otherwise.

In this Tutorial we use the symbol n to denote the power in either case.