Maclaurin Series
A Maclaurin series expresses a function as a power series centred at x = 0, using successive derivatives at the origin:
Maclaurin's theorem
f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ··· = Σ f⁽ⁿ⁾(0)·xⁿ/n!
Valid only within the radius of convergence. The series may converge for all x, for |x| < R, or only at x = 0.
Standard Series — Must Memorise
These six series are the foundation. All others are derived from them by substitution, differentiation, or integration.
Deriving New Series — Three Methods
Method 1 — Substitution
Replace x with a function of x. E.g., sin(x²): replace x with x² in sin x series.
Method 2 — Differentiation
Differentiate a known series term by term. E.g., differentiate ln(1+x) to get 1/(1+x).
Method 3 — Direct Method
Compute f(0), f'(0), f''(0), … directly for functions not derivable by substitution.
E
Direct Method — Maclaurin series for f(x) = e^(sin x)
Find the Maclaurin series for e^(sin x) up to and including the term in x⁴.
f(0)
f(0) = e^(sin 0) = e⁰ = 1
f'(x)
f'(x) = cos x · e^(sin x) → f'(0) = 1·1 = 1
f''(x)
f''(x) = (−sin x)e^(sin x) + cos²x · e^(sin x) = e^(sin x)(cos²x − sin x)
f''(0) = 1·(1−0) = 1
f''(0) = 1·(1−0) = 1
f'''(x)
Differentiate f''(x): f'''(0) = 0 (compute at x=0 carefully — all sin x terms vanish)
f⁽⁴⁾(0)
f⁽⁴⁾(0) = −3 (verified by successive differentiation at x=0)
Series
e^(sin x) = 1 + x + x²/2! + 0·x³/3! + (−3)x⁴/4! + ···
= 1 + x + x²/2 − x⁴/8 + ···
E
Substitution Method — Series for sin(x²)
Write down the Maclaurin series for sin(x²) up to the term in x¹⁰.
Substitute
Replace x with x² in sin x = x − x³/3! + x⁵/5! − ···
sin(x²) = (x²) − (x²)³/6 + (x²)⁵/120 − ···
sin(x²) = (x²) − (x²)³/6 + (x²)⁵/120 − ···
= x² − x⁶/6 + x¹⁰/120 − ···
Integration use
∫₀¹ sin(x²) dx ≈ [x³/3 − x⁷/42 + x¹¹/1320]₀¹ = 1/3 − 1/42 + 1/1320 ≈ 0.3103
⚠️ Cambridge Exam Points — Maclaurin Series
- Always state the range of validity — Cambridge awards a mark for this.
- For product series: multiply term by term, collecting like powers. Work systematically up to the required degree.
- The direct method requires computing derivatives — use the chain rule carefully and substitute x=0 at each stage rather than carrying complicated expressions.
- Series can be used to evaluate limits: e.g. lim(x→0) sin x/x = lim(x→0) (x − x³/6 + ···)/x = 1.
Improper Integrals
An integral is improper when either the interval is unbounded (infinite limits) or the integrand is unbounded (singularity within the interval). Convergence must always be checked by taking a limit.
Type 1 — Infinite Limits
Definition
∫ₐ∞ f(x) dx = lim[t→∞] ∫ₐᵗ f(x) dx
The integral converges if this limit is finite. Otherwise it diverges. Always write the limit notation explicitly — do not substitute ∞ directly.
E
∫₁∞ 1/x² dx — Convergent
Limit form
∫₁∞ x⁻² dx = lim[t→∞] ∫₁ᵗ x⁻² dx = lim[t→∞] [−x⁻¹]₁ᵗ
Evaluate
= lim[t→∞] (−1/t + 1) = 0 + 1
= 1 (converges)
E
∫₁∞ 1/x dx — Divergent
Limit form
lim[t→∞] [ln x]₁ᵗ = lim[t→∞] ln t − 0 = ∞
Diverges — the harmonic series analogue.
Type 2 — Integrand Unbounded (Singularity)
Singularity at x = a
∫ₐᵇ f(x) dx = lim[ε→0⁺] ∫ₐ₊ε^b f(x) dx
The function blows up at x = a. Approach from the right. If the limit is finite, the integral converges.
E
∫₀¹ 1/√x dx — Convergent despite singularity at x=0
Limit form
lim[ε→0⁺] ∫ε¹ x⁻¹/² dx = lim[ε→0⁺] [2√x]ε¹
Evaluate
= lim[ε→0⁺] (2 − 2√ε) = 2 − 0
= 2 (converges)
Convergence Test — Comparison
| Integral | Behaviour as x→∞ | Converges? |
|---|---|---|
| ∫₁∞ xᵖ dx | p < −1 (i.e. f(x) → 0 fast enough) | ✓ iff p < −1 |
| ∫₁∞ 1/x dx | Decays too slowly (logarithm) | ✗ Diverges |
| ∫₁∞ e^(−x) dx | Exponential decay — very fast | ✓ Always |
| ∫₀¹ xᵖ dx | Singularity at 0 iff p < 0 | ✓ iff p > −1 |
⛔ Never Substitute ∞ Directly
Writing [−1/x]₁∞ = 0 − (−1) = 1 without the limit is technically incorrect and loses the method mark in Cambridge. Always write: lim[t→∞] [−1/x]₁ᵗ = lim[t→∞] (−1/t + 1) = 1.
E
∫₀∞ xe^(−x) dx — Integration by Parts + Limit
Evaluate ∫₀∞ xe^(−x) dx.
IBP
u=x, dv=e^(−x)dx → du=dx, v=−e^(−x)
∫xe^(−x)dx = −xe^(−x) + ∫e^(−x)dx = −xe^(−x) − e^(−x) + C = −e^(−x)(x+1) + C
∫xe^(−x)dx = −xe^(−x) + ∫e^(−x)dx = −xe^(−x) − e^(−x) + C = −e^(−x)(x+1) + C
Limit
lim[t→∞] [−e^(−x)(x+1)]₀ᵗ = lim[t→∞] [−e^(−t)(t+1)] − [−e⁰·1]
= 0 − (−1) [since te^(−t) → 0 by L'Hôpital]
= 0 − (−1) [since te^(−t) → 0 by L'Hôpital]
= 1
Reduction Formulae
A reduction formula expresses Iₙ (an integral with parameter n) in terms of Iₙ₋₁ or Iₙ₋₂ — reducing the problem step by step until a base case is reached. The method is always integration by parts.
📌 Standard Approach
- Define Iₙ = ∫ f(x)ⁿ g(x) dx (or a definite integral version).
- Apply IBP: choose u and dv to reduce the power by 1 or 2.
- Express the remaining integral as a multiple of Iₙ₋₁ (or Iₙ₋₂).
- Rearrange to get Iₙ = F(n)·Iₙ₋₁ + G(n).
- State the base cases I₀ or I₁ (computed directly).
- Apply the formula iteratively to reach the required value.
Full Derivation — Iₙ = ∫₀^(π/2) sinⁿx dx
Define Iₙ
Iₙ = ∫₀^(π/2) sinⁿx dx for n ≥ 2
Write sinⁿx
sinⁿx = sinⁿ⁻¹x · sin x
Choose: u = sinⁿ⁻¹x, dv = sin x dx
Then: du = (n−1)sinⁿ⁻²x cos x dx, v = −cos x
Choose: u = sinⁿ⁻¹x, dv = sin x dx
Then: du = (n−1)sinⁿ⁻²x cos x dx, v = −cos x
Apply IBP
Iₙ = [−sinⁿ⁻¹x cos x]₀^(π/2) + (n−1)∫₀^(π/2) sinⁿ⁻²x cos²x dx
Boundary term: at π/2: −sin^(n−1)(π/2)cos(π/2) = 0; at 0: 0. So boundary = 0.
Iₙ = (n−1)∫₀^(π/2) sinⁿ⁻²x (1−sin²x) dx = (n−1)[Iₙ₋₂ − Iₙ]
Boundary term: at π/2: −sin^(n−1)(π/2)cos(π/2) = 0; at 0: 0. So boundary = 0.
Iₙ = (n−1)∫₀^(π/2) sinⁿ⁻²x (1−sin²x) dx = (n−1)[Iₙ₋₂ − Iₙ]
Rearrange
Iₙ + (n−1)Iₙ = (n−1)Iₙ₋₂ → nIₙ = (n−1)Iₙ₋₂
Iₙ = ((n−1)/n) · Iₙ₋₂
Iₙ = ((n−1)/n) · Iₙ₋₂
Base cases
I₀ = ∫₀^(π/2) 1 dx = π/2 I₁ = ∫₀^(π/2) sin x dx = [−cos x]₀^(π/2) = 1
Example: I₄
I₄ = (3/4)·I₂ = (3/4)·(1/2)·I₀ = (3/4)·(1/2)·(π/2)
I₄ = 3π/16
I₄ = 3π/16
Wallis's Formula — A Beautiful Result
From the sinⁿ reduction formula
∫₀^(π/2) sinⁿx dx = { (n−1)!!/(n!!) · π/2 if n even; (n−1)!!/(n!!) if n odd }
Where n!! = n·(n−2)·(n−4)···2 (or ···1). Example: 6!! = 6·4·2 = 48, 5!! = 5·3·1 = 15.
Reduction Formula — Iₙ = ∫ xⁿeˣ dx
IBP setup
u = xⁿ, dv = eˣ dx → du = nxⁿ⁻¹ dx, v = eˣ
Result
Iₙ = xⁿeˣ − n·Iₙ₋₁
Base case
I₀ = ∫eˣ dx = eˣ + C
Example: I₂
I₂ = x²eˣ − 2I₁ = x²eˣ − 2(xeˣ − I₀) = x²eˣ − 2xeˣ + 2eˣ + C
= eˣ(x²−2x+2) + C
= eˣ(x²−2x+2) + C
Other Common Reduction Formulae
| Iₙ | Reduction formula | Base case |
|---|---|---|
| ∫cosⁿx dx | Iₙ = ((n−1)/n)·Iₙ₋₂ [same as sin] | I₀=π/2, I₁=1 (over [0,π/2]) |
| ∫tanⁿx dx | Iₙ = tanⁿ⁻¹x/(n−1) − Iₙ₋₂ | I₀=x+C, I₁=ln|sec x|+C |
| ∫₀¹ xⁿe^(−x) dx | Iₙ = −e⁻¹ + n·Iₙ₋₁ | I₀ = 1−e⁻¹ |
| ∫secⁿx dx | Iₙ = [secⁿ⁻²x tanx]/(n−1) + ((n−2)/(n−1))·Iₙ₋₂ | I₁=ln|sec x+tan x|+C |
Arc Length & Surface Area
Arc Length — Cartesian Form
Length of curve y = f(x) from x = a to x = b
L = ∫ₐᵇ √(1 + (dy/dx)²) dx
Derived from the approximation ds ≈ √(dx² + dy²) = √(1+(dy/dx)²) dx for an infinitesimal arc element.
Arc Length — Parametric Form
Parametric curve x=x(t), y=y(t) from t=α to t=β
L = ∫α^β √((dx/dt)² + (dy/dt)²) dt
Arc Length — Polar Form
Polar curve r = f(θ) from θ = α to θ = β
L = ∫α^β √(r² + (dr/dθ)²) dθ
Surface of Revolution
Rotation about x-axis
S = 2π ∫ₐᵇ y √(1+(dy/dx)²) dx
Area of surface generated by revolving y=f(x) about the x-axis.
Rotation about y-axis
S = 2π ∫ₐᵇ x √(1+(dy/dx)²) dx
E
Arc Length — y = ln(cos x) from x=0 to x=π/4
Find the length of the curve y = ln(cos x) for 0 ≤ x ≤ π/4.
dy/dx
dy/dx = −sin x / cos x = −tan x
1 + (dy/dx)²
1 + tan²x = sec²x
Integrand
√(sec²x) = |sec x| = sec x (since 0 ≤ x < π/2)
Integrate
L = ∫₀^(π/4) sec x dx = [ln|sec x + tan x|]₀^(π/4)
= ln(√2 + 1) − ln(1 + 0)
= ln(√2 + 1) − ln(1 + 0)
= ln(1 + √2)
E
Surface of Revolution — y = √x about x-axis, 0 ≤ x ≤ 4
Find the surface area when y = √x is rotated about the x-axis from x=0 to x=4.
dy/dx
dy/dx = 1/(2√x) → 1+(dy/dx)² = 1 + 1/(4x) = (4x+1)/(4x)
Integral
S = 2π∫₀⁴ √x · √((4x+1)/(4x)) dx = 2π∫₀⁴ √((4x+1)/4) dx = π∫₀⁴ √(4x+1) dx
Evaluate
= π · [√(4x+1)³/(3·4/2)]₀⁴ = π · [(4x+1)^(3/2)/6]₀⁴
= π/6 · (17^(3/2) − 1)
= π/6 · (17^(3/2) − 1)
= π(17√17 − 1)/6
Worked Examples
1
Maclaurin Series — Product and Limit
(i) Write down the Maclaurin series for eˣ and cos x up to x⁴.
(ii) Find the series for eˣ cos x up to x⁴.
(iii) Hence evaluate lim(x→0) (eˣ cos x − 1)/x.
(ii) Find the series for eˣ cos x up to x⁴.
(iii) Hence evaluate lim(x→0) (eˣ cos x − 1)/x.
Part (i)
eˣ = 1 + x + x²/2 + x³/6 + x⁴/24 + ···
cos x = 1 − x²/2 + x⁴/24 + ···
cos x = 1 − x²/2 + x⁴/24 + ···
Part (ii)
Multiply term by term, collecting up to x⁴:
constant: 1·1 = 1
x¹: 1·x + x·1 = x → wait, cos has no x term → x coefficient: x·1 = x
x²: (x²/2)·1 + 1·(−x²/2) = 0
x³: (x³/6)·1 + x·(−x²/2) = x³/6 − x³/2 = −x³/3
x⁴: (x⁴/24)·1 + (x²/2)·(−x²/2) + 1·(x⁴/24) = x⁴/24 − x⁴/4 + x⁴/24 = −x⁴/6
constant: 1·1 = 1
x¹: 1·x + x·1 = x → wait, cos has no x term → x coefficient: x·1 = x
x²: (x²/2)·1 + 1·(−x²/2) = 0
x³: (x³/6)·1 + x·(−x²/2) = x³/6 − x³/2 = −x³/3
x⁴: (x⁴/24)·1 + (x²/2)·(−x²/2) + 1·(x⁴/24) = x⁴/24 − x⁴/4 + x⁴/24 = −x⁴/6
eˣ cos x = 1 + x − x³/3 − x⁴/6 + ···
Part (iii)
(eˣ cos x − 1)/x = (x − x³/3 − x⁴/6 + ···)/x = 1 − x²/3 − x³/6 + ···
As x→0:
As x→0:
lim = 1
2
Reduction Formula — Cambridge Style
Let Iₙ = ∫₀^1 xⁿ e^x dx. Show that Iₙ = e − n·Iₙ₋₁. Hence find I₃.
IBP
u = xⁿ, dv = eˣ dx → du = nxⁿ⁻¹dx, v = eˣ
Iₙ = [xⁿeˣ]₀¹ − n∫₀¹ xⁿ⁻¹eˣ dx = e − n·Iₙ₋₁ ✓
Iₙ = [xⁿeˣ]₀¹ − n∫₀¹ xⁿ⁻¹eˣ dx = e − n·Iₙ₋₁ ✓
Base: I₀
I₀ = ∫₀¹ eˣ dx = [eˣ]₀¹ = e − 1
I₁
I₁ = e − 1·I₀ = e − (e−1) = 1
I₂
I₂ = e − 2·I₁ = e − 2
I₃
I₃ = e − 3·I₂ = e − 3(e−2) = e − 3e + 6 =
6 − 2e
3
Improper Integral with Parameter
Find the values of p for which ∫₁∞ x^p dx converges, and evaluate it for those values.
Case p ≠ −1
lim[t→∞] [x^(p+1)/(p+1)]₁ᵗ = lim[t→∞] t^(p+1)/(p+1) − 1/(p+1)
This limit is finite iff p+1 < 0, i.e. p < −1.
Value = 0 − 1/(p+1) = −1/(p+1) = 1/(−p−1)
This limit is finite iff p+1 < 0, i.e. p < −1.
Value = 0 − 1/(p+1) = −1/(p+1) = 1/(−p−1)
Case p = −1
lim[t→∞] ln t = ∞ → diverges.
Conclusion
Converges iff p < −1. Value = −1/(p+1).
Practice Questions
Question 1 — Maclaurin Series by Substitution
[4 marks]
Write down the Maclaurin series for ln(1+2x) up to and including the term in x⁴. State the range of validity.
Replace x with 2x in the series for ln(1+x). Remember to raise 2x to each power — e.g. the x³ term becomes (2x)³/3 = 8x³/3.
✓ Solution
ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + ···Replace x with 2x:
ln(1+2x) = (2x) − (2x)²/2 + (2x)³/3 − (2x)⁴/4 + ···
= 2x − 2x² + 8x³/3 − 4x⁴ + ···
Valid for −1 < 2x ≤ 1, i.e. −½ < x ≤ ½
Question 2 — Improper Integral
[4 marks]
Determine whether ∫₀¹ x⁻¹/³ dx converges. If it does, evaluate it.
The integrand has a singularity at x=0. Write as lim[ε→0⁺] ∫ε¹ x^(−1/3) dx. Since p = −1/3 > −1, it converges.
✓ Solution
Singularity at x=0 since x^(−1/3)→∞.lim[ε→0⁺] ∫ε¹ x^(−1/3) dx = lim[ε→0⁺] [x^(2/3)/(2/3)]ε¹ = lim[ε→0⁺] (3/2)[1 − ε^(2/3)]
= (3/2)(1 − 0)
= 3/2 (converges)
Question 3 — Reduction Formula
[7 marks]
Let Iₙ = ∫₀^(π/2) cos^n x dx.
(i) Show that Iₙ = ((n−1)/n) Iₙ₋₂ for n ≥ 2.
(ii) Hence find I₅.
(i) Show that Iₙ = ((n−1)/n) Iₙ₋₂ for n ≥ 2.
(ii) Hence find I₅.
Write cosⁿx = cosⁿ⁻¹x · cos x. Set u = cosⁿ⁻¹x, dv = cos x dx → du = −(n−1)cosⁿ⁻²x sin x dx, v = sin x. The boundary term vanishes at both limits.
✓ Solution
(i) IBP: u=cosⁿ⁻¹x, dv=cos x dxIₙ = [cosⁿ⁻¹x sin x]₀^(π/2) + (n−1)∫₀^(π/2) cosⁿ⁻²x sin²x dx
Boundary = 0 (sin 0=0, cos(π/2)=0).
Iₙ = (n−1)∫cosⁿ⁻²x(1−cos²x)dx = (n−1)(Iₙ₋₂ − Iₙ)
nIₙ = (n−1)Iₙ₋₂ → Iₙ = ((n−1)/n)Iₙ₋₂ ✓
(ii) I₁ = ∫₀^(π/2) cos x dx = 1
I₃ = (2/3)I₁ = 2/3
I₅ = (4/5)I₃ = (4/5)(2/3) =
8/15
Question 4 — Arc Length
[5 marks]
Find the length of the curve y = cosh x for 0 ≤ x ≤ ln 2.
dy/dx = sinh x. 1 + sinh²x = cosh²x (hyperbolic identity). So √(1+(dy/dx)²) = cosh x.
✓ Solution
dy/dx = sinh x → 1+(dy/dx)² = 1+sinh²x = cosh²x√(cosh²x) = cosh x (since cosh x > 0 always)
L = ∫₀^(ln2) cosh x dx = [sinh x]₀^(ln2)
sinh(ln 2) = (e^(ln2) − e^(−ln2))/2 = (2 − 1/2)/2 = 3/4
sinh(0) = 0
L = 3/4
Question 5 — Maclaurin Series and Approximation
[6 marks]
(i) Use the series for eˣ and sin x to find the Maclaurin series for e^(sin x) up to x³.
(ii) Hence find an approximation for ∫₀^(0.1) e^(sin x) dx, giving your answer to 6 decimal places.
(ii) Hence find an approximation for ∫₀^(0.1) e^(sin x) dx, giving your answer to 6 decimal places.
Write sin x ≈ x − x³/6, then expand e^(sin x) ≈ e^(x − x³/6). Use eᵘ = 1+u+u²/2+u³/6, substituting u = x − x³/6 and collecting powers.
✓ Solution
(i) Let u = sin x = x − x³/6 + ···e^u = 1 + u + u²/2 + u³/6 + ···
= 1 + (x−x³/6) + (x−x³/6)²/2 + (x)³/6 + ···
= 1 + x − x³/6 + x²/2 + x³/6 + ··· (collecting up to x³, cross terms cancel)
e^(sin x) = 1 + x + x²/2 + 0·x³ + ··· = 1 + x + x²/2 + ···
(ii) ∫₀^(0.1)(1 + x + x²/2)dx = [x + x²/2 + x³/6]₀^(0.1)
= 0.1 + 0.01/2 + 0.001/6
= 0.1 + 0.005 + 0.000167
≈ 0.105167
Interactive Maclaurin Series Visualiser
Select a function and adjust the number of terms. The tool shows the polynomial approximation converging to the true function.
Maclaurin Series Convergence
3 terms
±3
Select a function above.
Formula Sheet — Further Calculus
Maclaurin's Theorem
f(x) = Σ f⁽ⁿ⁾(0)xⁿ/n!n=0 to ∞
eˣ1+x+x²/2!+x³/3!+… all x
sin xx−x³/3!+x⁵/5!−… all x
cos x1−x²/2!+x⁴/4!−… all x
ln(1+x)x−x²/2+x³/3−… |x|≤1, x≠−1
arctan xx−x³/3+x⁵/5−… |x|≤1
(1+x)ⁿ1+nx+n(n−1)x²/2!+… |x|<1
Improper Integrals
∫ₐ∞ f dxlim[t→∞] ∫ₐᵗ f dx
∫ₐᵇ f dx (sing. at a)lim[ε→0⁺] ∫ₐ₊ε^b f dx
∫₁∞ xᵖ dxConverges iff p < −1
∫₀¹ xᵖ dxConverges iff p > −1
Always write limitNever substitute ∞
Key Reduction Formulae
∫₀^(π/2) sinⁿx dx= ((n−1)/n)·Iₙ₋₂
∫₀^(π/2) cosⁿx dx= ((n−1)/n)·Iₙ₋₂
∫xⁿeˣ dx= xⁿeˣ − n·Iₙ₋₁
∫₀¹ xⁿeˣ dx= e − n·Iₙ₋₁
Arc Length
Cartesian y=f(x)∫√(1+(dy/dx)²) dx
Parametric∫√((ẋ)²+(ẏ)²) dt
Polar r=f(θ)∫√(r²+(dr/dθ)²) dθ
Surface of Revolution
About x-axis2π∫y√(1+(y')²) dx
About y-axis2π∫x√(1+(y')²) dx
Useful Limits (from series)
lim(x→0) sin x/x1
lim(x→0) (1−cos x)/x²1/2
lim(x→0) (eˣ−1)/x1
lim(x→0) ln(1+x)/x1
lim(x→∞) xⁿe^(−x)0 (all n)
📋 Cambridge Exam Strategy — Further Calculus
- Maclaurin series: Always state the range of validity — this earns a separate mark. For product series, multiply term by term and collect like powers carefully up to the required degree.
- Limits using series: Substitute the series, cancel, and take x→0. Far faster than L'Hôpital for higher-order limits.
- Improper integrals: Always introduce lim[t→∞] or lim[ε→0⁺] explicitly. State whether the integral converges or diverges — Cambridge awards this conclusion as a separate mark.
- Reduction formulae: Show the IBP working fully. The boundary evaluation and the use of a trig identity (sin²=1−cos², etc.) are where marks are allocated.
- Arc length: The key step is simplifying 1+(dy/dx)² to a perfect square — look for sec²x = 1+tan²x, cosh²x = 1+sinh²x, or similar. This makes the integral tractable.
- Convergence of series: State whether each standard series converges for all x or only for |x| < R. Substituted series inherit the convergence condition (scaled).
🎓 P2 COMPLETE
Further Pure Mathematics P2 — All 6 Lessons Done
You have now covered the full 9231 P2 syllabus: Summation of Series, Mathematical Induction, Matrices, Eigenvalues & Eigenvectors, Complex Numbers, and Further Calculus. Upload all six files to gpmeducation.org and link them from your Further Maths course index page.
2.1 Summation of Series ✓
2.2 Mathematical Induction ✓
2.3 Matrices ✓
2.4 Eigenvalues & Eigenvectors ✓
2.5 Complex Numbers ✓
2.6 Further Calculus ✓