IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Maclaurin Series of Composites & Products

Once you have the five standard Maclaurin series from the booklet, you can build series for almost anything else without re-deriving from first principles. For composites like sin(3x) or e^−x², substitute the “inside function” into the booklet series. For products like sin x · cos x or e^x · arctan x, multiply the two truncated series together. Both methods are much faster than working out derivatives manually.

📘 What you need to know

Composite f(g(x)): take the booklet series for f, then substitute g(x) wherever x appears. Expand brackets, simplify coefficients.
Inside function in IB problems is usually kx (a constant times x) or xⁿ (a power). Occasionally with a sign flip, like −x or −x².
Composite substitution: e.g., e^−x² means put −x² into the e^x series, giving 1 + (−x²) + (−x²)²/2! + …
Product f(x) · g(x): write both series as polynomials, multiply term-by-term, collect powers of x.
How many terms to keep: for a product up to xⁿ, keep enough terms in BOTH series so all products giving power ≤ n are captured. When unsure, keep MORE — extra terms can be discarded later, missing terms can’t be recovered.
Discard high powers as you multiply: any product giving x^m with m > n is irrelevant for an “xⁿ-truncation” — drop it as you go.
Watch for cancellations: in some products, certain coefficients vanish unexpectedly (e.g., the x⁴ term of e^x sin x is 0). Don’t assume — compute every term.
Verification trick: when possible, derive the answer another way and cross-check. e.g., sin x cos x = ½ sin(2x) — you can compute the product OR substitute 2x into sin and halve, and they must agree.

Composites — substitute the “inside” function

The substitution rule If f(x) = booklet series, then f(g(x)) = the SAME series with x replaced by g(x).

Example: ln(1 + x) = x − x²/2 + x³/3 − … ⟹ ln(1 + 3x) = 3x − (3x)²/2 + (3x)³/3 − … = 3x − 9x²/2 + 9x³ − …

Products — multiply two series term-by-term

Multiply the two booklet series the same way you’d multiply two polynomials. The visual organizer below uses a multiplication grid: every cell of the grid is a single product, and cells along the same anti-diagonal share the same power of x. Sum them to get the coefficient of that power.

Each grid cell is a single product. Cells with the same total power of x are coloured the same — sum these to get the coefficient of that power. For an x³ truncation, all cells of power ≥ 5 are discarded.

Two approaches — when to use which

Composite — substitute

f(g(x))

Examples: sin(2x), cos(3x), e^−x², ln(1 − x), arctan(x²). Take booklet series for f, replace x with g(x), simplify.

Product — multiply

f(x) · g(x)

Examples: e^x sin x, sin x cos x, (1+x) e^x, e^x arctan x. Write both series, multiply term-by-term, collect powers.

Mixed case: for something like e^x sin(2x), do substitution first to get the series of sin(2x), then multiply by the e^x series.

🧭 Recipe — composites and products

Identify the structure: is the function f(g(x)) (composite) or f(x) · g(x) (product)? Sometimes it’s both — do substitution first, then multiply.
For composites: write the booklet series for f, then substitute g(x) for every x. Expand brackets, simplify powers.
For products: write each booklet series, keeping enough terms (the safest rule: include all terms whose lowest power could combine to give xⁿ).
Multiply term-by-term (use a grid if it helps). Discard products that give powers higher than required.
Collect and simplify: gather all products with the same power of x into single coefficients. Watch for cancellations.

Worked examples

WE 1

Composite by substitution — cos(2x)

Find the Maclaurin series for cos(2x) up to and including the term in x⁴.

Step 1 — write the booklet series for cos cos x = 1 – x²/2! + x⁴/4! – … Step 2 — substitute 2x everywhere x appears cos(2x) = 1 – (2x)²/2! + (2x)⁴/4! Step 3 — expand and simplify (2x)² = 4x², so (2x)²/2 = 2x² (2x)⁴ = 16x⁴, so (2x)⁴/24 = 16x⁴/24 = 2x⁴/3 cos(2x) ≈ 1 − 2x² + 2x⁴3 notice the coefficients GREW after substitution: 1/2 became 2, and 1/24 became 2/3. This is typical when the inside function is kx with |k|>1 — the higher powers blow up faster.

WE 2

Composite with x² — e^−x²

Find the Maclaurin series for e^−x² up to and including the term in x⁶.

Step 1 — booklet series for e^x e^x = 1 + x + x²/2! + x³/3! + … Step 2 — substitute -x² for every x e^(-x²) = 1 + (-x²) + (-x²)²/2! + (-x²)³/3! Step 3 — simplify each power (-x²)¹ = -x² (-x²)² = x⁴, so /2! = x⁴/2 (-x²)³ = -x⁶, so /3! = -x⁶/6 e^−x² ≈ 1 − x² + x⁴2 − x⁶6 notice the series jumps in powers of 2 (no odd powers), because we substituted x² (even). The series alternates signs because we substituted -x² (negative sign flips on every odd-numbered power).

WE 3

Composite — sin(3x)

Find the Maclaurin series for sin(3x) up to and including the term in x⁵.

Step 1 — booklet series for sin x sin x = x – x³/3! + x⁵/5! – … Step 2 — substitute 3x sin(3x) = (3x) – (3x)³/6 + (3x)⁵/120 Step 3 — simplify each term (3x)³ = 27x³, /6 = 27x³/6 = 9x³/2 (3x)⁵ = 243x⁵, /120 = 243x⁵/120 = 81x⁵/40 sin(3x) ≈ 3x − 9x³2 + 81x⁵40 verify: at small x, sin(3x) ≈ 3x (the linear approximation, since sin θ ≈ θ for small θ). Our series starts with 3x ✓. The (3x)^k pattern means the k-th term gets a factor of 3^k, which blows up fast — only valid for small x.

WE 4

Composite with sign flip — ln(1 − x)

Find the Maclaurin series for ln(1 − x) up to and including the term in x⁴.

Step 1 — booklet series for ln(1+x) ln(1+x) = x – x²/2 + x³/3 – x⁴/4 + … Step 2 — substitute (-x) for x ln(1 + (-x)) = (-x) – (-x)²/2 + (-x)³/3 – (-x)⁴/4 Step 3 — simplify powers of (-x) (-x)¹ = -x (-x)² = x², with minus sign in front: -x²/2 (-x)³ = -x³, with plus sign in front: -x³/3 (-x)⁴ = x⁴, with minus sign in front: -x⁴/4 ln(1 − x) ≈ −x − x²2 − x³3 − x⁴4 all signs are negative — the alternating signs of ln(1+x) ALL flip to give a series of negatives. This makes sense: ln(1-x) approaches -∞ as x→1, so the series should be dominated by negative growth.

WE 5

Product — sin x · cos x (cross-check via composite)

Find the Maclaurin series for sin x · cos x up to and including the term in x⁵, and cross-check using the identity sin x cos x = ½ sin(2x).

Step 1 — booklet series, enough terms in each sin x = x – x³/6 + x⁵/120 (need up to x⁵) cos x = 1 – x²/2 + x⁴/24 (need up to x⁴) Step 2 — multiply, keeping only terms up to x⁵ x · 1 = x x · (-x²/2) = -x³/2 x · (x⁴/24) = x⁵/24 (-x³/6) · 1 = -x³/6 (-x³/6) · (-x²/2) = x⁵/12 (x⁵/120) · 1 = x⁵/120 (other products give x⁶ or higher — discard) Step 3 — collect by power x¹: x x³: -x³/2 – x³/6 = -3x³/6 – x³/6 = -4x³/6 = -2x³/3 x⁵: x⁵/24 + x⁵/12 + x⁵/120 = 5x⁵/120 + 10x⁵/120 + x⁵/120 = 16x⁵/120 = 2x⁵/15 sin x · cos x ≈ x − 2x³3 + 2x⁵15 Cross-check: ½ sin(2x) using substitution sin(2x) = 2x – (2x)³/6 + (2x)⁵/120 = 2x – 4x³/3 + 4x⁵/15 ½ · sin(2x) = x – 2x³/3 + 2x⁵/15 ✓ matches! the cross-check is powerful: same answer from two different methods. If they disagree, you’ve made an arithmetic error somewhere.

WE 6

Product — e^x · arctan x

Find the Maclaurin series for e^x · arctan x up to and including the term in x⁴.

Step 1 — booklet series with enough terms e^x = 1 + x + x²/2 + x³/6 + x⁴/24 (need up to x³, since arctan starts at x) arctan x = x – x³/3 + x⁵/5 – … (need up to x³) Step 2 — multiply, keeping only terms up to x⁴ 1 · x = x 1 · (-x³/3) = -x³/3 x · x = x² x · (-x³/3) = -x⁴/3 (x²/2) · x = x³/2 (x²/2) · (-x³/3) = …x⁵, discard (x³/6) · x = x⁴/6 (x⁴/24) · x = …x⁵, discard Step 3 — collect by power x¹: x x²: x² x³: -x³/3 + x³/2 = -2x³/6 + 3x³/6 = x³/6 x⁴: -x⁴/3 + x⁴/6 = -2x⁴/6 + x⁴/6 = -x⁴/6 e^x · arctan x ≈ x + x² + x³6 − x⁴6 notice the x³ and x⁴ coefficients each come from TWO products. Adding fractions with common denominators (here 6) makes the arithmetic clean. Don’t try to add 1/3 + 1/2 in your head — convert to common denominator first.

💡 Top tips

Always check whether the function is a composite, product, or both before deciding which technique. Sometimes you need both in sequence.
For composites, just substitute: write the booklet series, replace x with the inside function, then expand. Don’t re-derive from f⁽ⁿ⁾(0).
For products, keep more terms than you need in each factor while multiplying. Extra terms are easily discarded; missing ones cost you marks.
Use common denominators when adding fractions like 1/3 + 1/2 — convert both to /6 first (= 2/6 + 3/6). Reduces arithmetic errors.
Cross-check when possible: identities like sin x cos x = ½ sin(2x) give an independent verification path.

⚠ Common mistakes

Forgetting to substitute everywhere: ln(1+3x) must have 3x in EVERY term — including the (3x)² and (3x)³ parts, not just the first one.
Sign errors in (−x)ⁿ: (−x)² = +x², (−x)³ = −x³. Even powers stay positive, odd powers stay negative.
Stopping multiplication too early in products: when the question asks “up to x⁴”, you must include EVERY pair of terms that could combine to give x⁴ — not just the first few.
Missing cancellations: in products, sometimes coefficients vanish (e.g., the x⁴ term of e^x sin x is exactly 0). Compute all contributing terms — don’t assume non-zero.
Forgetting to simplify (kx)ⁿ: it equals kⁿ · xⁿ, not just k · xⁿ. (2x)³ = 8x³, not 2x³.

Up next: Differentiating & Integrating Maclaurin Series. If you differentiate a Maclaurin series term-by-term, you get the series for the function’s derivative; if you integrate term-by-term, you get the series for the antiderivative (plus a constant of integration). This lets you build new Maclaurin series from existing ones — e.g., differentiate the arctan x series to get 1/(1+x²); integrate −sin x to get cos x.

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