Maclaurin and Taylor Series

Maclaurin and Taylor Series

Series expansions allow us to approximate functions as polynomials, making them easier to differentiate, integrate, and evaluate. The Maclaurin series is the special case centred at x = 0; the Taylor series generalises to any centre point.

Maclaurin Series

The Maclaurin series of f(x) is:

f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ... = Σ f⁽ⁿ⁾(0)xⁿ/n!

To find it:

1. Compute f(0), f'(0), f''(0), f'''(0), ... up to the required number of terms

2. Substitute into the formula

Standard Maclaurin Series (Must Know)

Function	Series	Validity
eˣ	1 + x + x²/2! + x³/3! + ...	All x
sin x	x − x³/3! + x⁵/5! − x⁷/7! + ...	All x
cos x	1 − x²/2! + x⁴/4! − x⁶/6! + ...	All x
ln(1 + x)	x − x²/2 + x³/3 − x⁴/4 + ...	−1 < x ≤ 1
(1 + x)ⁿ	1 + nx + n(n−1)x²/2! + ...	\	x\	< 1 (if n not a +ve integer)
1/(1 − x)	1 + x + x² + x³ + ...	\	x\	< 1
tan⁻¹ x	x − x³/3 + x⁵/5 − x⁷/7 + ...	\	x\	≤ 1
sinh x	x + x³/3! + x⁵/5! + ...	All x
cosh x	1 + x²/2! + x⁴/4! + ...	All x

Taylor Series

The Taylor series of f(x) centred at x = a is:

f(x) = f(a) + f'(a)(x−a) + f''(a)(x−a)²/2! + f'''(a)(x−a)³/3! + ...

Or equivalently, with h = x − a:

f(a + h) = f(a) + f'(a)h + f''(a)h²/2! + ...

Example: Taylor series of ln x about x = 1 (i.e., a = 1):

• f(x) = ln x, f(1) = 0
• f'(x) = 1/x, f'(1) = 1
• f''(x) = −1/x², f''(1) = −1
• f'''(x) = 2/x³, f'''(1) = 2

ln x = (x−1) − (x−1)²/2 + (x−1)³/3 − ...

This is valid for 0 < x ≤ 2.

Deriving Series by Substitution

Rather than differentiating from scratch, use known series with substitutions:

Example: Find the Maclaurin series for e^(−x²) up to x⁶.

• Start with eᵘ = 1 + u + u²/2! + u³/3! + ...
• Substitute u = −x²:
• e^(−x²) = 1 + (−x²) + (−x²)²/2! + (−x²)³/3!
• = 1 − x² + x⁴/2 − x⁶/6 + ...

Deriving Series by Multiplication and Division

Example: Find the first three non-zero terms of tan x = sin x / cos x.

• sin x = x − x³/6 + x⁵/120 − ...
• cos x = 1 − x²/2 + x⁴/24 − ...
• Use long division or multiply tan x · cos x = sin x and equate coefficients
• Result: tan x = x + x³/3 + 2x⁵/15 + ...

Deriving Series by Integration

Example: Find the Maclaurin series for sin⁻¹ x.

• We know d/dx(sin⁻¹ x) = 1/√(1 − x²) = (1 − x²)^(−1/2)
• Expand: (1 − x²)^(−1/2) = 1 + x²/2 + 3x⁴/8 + 5x⁶/16 + ...
• Integrate term by term:
• sin⁻¹ x = x + x³/6 + 3x⁵/40 + 5x⁷/112 + ... (with C = 0 since sin⁻¹(0) = 0)

Using Series for Limits

Series expansions make L'Hôpital-resistant limits straightforward.

Example: Find lim(x→0) (sin x − x)/(x³)

• sin x = x − x³/6 + x⁵/120 − ...
• sin x − x = −x³/6 + x⁵/120 − ...
• (sin x − x)/x³ = −1/6 + x²/120 − ...
• As x → 0: limit = −1/6

Using Series for Approximation

Example: Estimate ∫₀^(0.5) e^(−x²) dx.

• e^(−x²) ≈ 1 − x² + x⁴/2 − x⁶/6
• ∫₀^(0.5) (1 − x² + x⁴/2 − x⁶/6) dx = [x − x³/3 + x⁵/10 − x⁷/42]₀^(0.5)
• = 0.5 − 0.04167 + 0.003125 − 0.000186 ≈ 0.4613

Convergence and Validity

Every power series has a radius of convergence R:

• The series converges for |x| < R (or |x − a| < R for Taylor series)
• It diverges for |x| > R
• Behaviour at |x| = R must be checked separately

The ratio test can find R: if the nth term is aₙxⁿ, then R = lim|aₙ/aₙ₊₁| as n → ∞.

Exam Tips

• Memorise the standard series — many questions are just substitution into these
• When asked for "the first three non-zero terms", discard zero terms and count only actual terms
• Always state the range of validity if asked or if using a series with restricted convergence
• For Taylor series, clearly state your centre point a and expand in powers of (x − a)
• Use the Maclaurin series approach for limits — it is often faster and less error-prone than L'Hôpital's rule
• When multiplying series, you only need terms up to the required power — discard higher-order terms early to save time