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

Maclaurin Series of Standard Functions

A Maclaurin series writes a function as an infinite polynomial in x. If you include all terms, the series equals the function exactly. If you truncate at some power of x, you get a polynomial approximation — accurate near x = 0 and increasingly accurate as you add more terms. You can derive a Maclaurin series from first principles using the formula, or quote the standard series for e^x, ln(1+x), sin x, cos x, arctan x from the formula booklet.

📘 What you need to know

Maclaurin formula (formula booklet): f(x) = f(0) + x f′(0) + (x²/2!) f″(0) + (x³/3!) f‴(0) + …
Coefficient of xⁿ is f⁽ⁿ⁾(0)/n! — the nth derivative at 0 divided by n factorial.
Five standard series in the formula booklet: e^x, ln(1+x), sin x, cos x, arctan x. Use these instead of re-deriving when possible.
Truncation: stopping at xⁿ gives a polynomial approximation. Always exact at x = 0; accuracy degrades as |x| grows.
More terms → better approximation and accuracy over a wider range of x. A series up to x⁷ beats one up to x³.
Use “from first principles” when: the question says so, or the function isn’t in the booklet (e.g., tan x, sec x, (1+x)^1/2).
Binomial = Maclaurin for (1 + x)ⁿ: the binomial theorem expansion IS the Maclaurin series. Use whichever is faster.
GDC trick: many calculators compute f^(k)(0) directly via the kth-derivative function — saves you having to differentiate by hand.

The general Maclaurin formula

General Maclaurin series — formula booklet f(x) = f(0) + x f′(0) + x²2! f″(0) + x³3! f‴(0) + …

To use this: compute the derivatives of f at x = 0, plug them in, simplify coefficients. The accuracy improves as you keep more terms; near x = 0, just a handful of terms can give astonishing precision.

Sin x (navy) compared with truncated Maclaurin polynomials. T₁ = x matches only near 0; T₃ adds an x³ term and stays close until |x| ≈ 1.5; T₅ stays close further still. All four curves agree EXACTLY at x = 0.

The five standard series — formula booklet

Function	Maclaurin series (booklet)
e^x	1 + x + x²/2! + x³/3! + x⁴/4! + …
ln(1 + x)	x − x²/2 + x³/3 − x⁴/4 + …
sin x	x − x³/3! + x⁵/5! − x⁷/7! + …
cos x	1 − x²/2! + x⁴/4! − x⁶/6! + …
arctan x	x − x³/3 + x⁵/5 − x⁷/7 + …

Patterns to spot: sin and arctan have ODD powers only (because they’re odd functions); cos has EVEN powers only (because it’s even). The exponential has all powers; ln(1+x) misses the constant term (because ln 1 = 0). Recognizing these symmetries saves time.

Two approaches — when to use which

From first principles

use general formula

When the question says “use the Maclaurin formula” or the function isn’t in the booklet (e.g., tan x, sec x, (1+x)^1/2). Compute derivatives at 0, plug into formula.

From the booklet

quote standard series

When the function IS one of the five booklet series. Just write it out, then substitute or manipulate as needed. Much faster than re-deriving.

🧭 Recipe — finding a Maclaurin series

Check whether to use first principles or the booklet. If the function isn’t one of the five standard ones, you must derive from scratch.
For first principles: differentiate f as many times as you need (to find the term in xⁿ, you need up to f⁽ⁿ⁾). Evaluate each derivative at x = 0.
Substitute into the formula: f(x) = f(0) + xf′(0) + (x²/2!)f″(0) + … up to the required power.
Simplify each coefficient: combine the f^(k)(0) values with their 1/k! prefactors. Cancel where possible.
For numerical approximations: substitute the chosen value of x into the truncated polynomial. Compare with the actual value (calculator) to assess accuracy.

Worked examples

WE 1

First principles — function not in booklet

Use the Maclaurin formula to find the Maclaurin series of f(x) = (1 + x)^−1/2 up to and including the term in x³.

Step 1 — find derivatives f(x) = (1+x)^(-1/2) f(0) = 1 f'(x) = -1/2 (1+x)^(-3/2) f'(0) = -1/2 f”(x) = 3/4 (1+x)^(-5/2) f”(0) = 3/4 f”'(x)= -15/8 (1+x)^(-7/2) f”'(0)= -15/8 Step 2 — plug into Maclaurin formula f(x) = 1 + x(-1/2) + (x²/2!)(3/4) + (x³/3!)(-15/8) Step 3 — simplify coefficients (x²/2)(3/4) = 3x²/8 (x³/6)(-15/8) = -15x³/48 = -5x³/16 (1+x)^−1/2 ≈ 1 − x2 + 3x²8 − 5x³16 this is also the binomial expansion of (1+x)^(-1/2). Both methods give the same result — choose whichever is faster.

WE 2

Derive cos x from first principles (verify booklet)

Use the Maclaurin formula to find the series for cos x up to and including the term in x⁴, and verify it matches the formula booklet.

Step 1 — derivatives cycle through 4 values f(x) = cos x f(0) = 1 f'(x) = -sin x f'(0) = 0 f”(x) = -cos x f”(0) = -1 f”'(x) = sin x f”'(0) = 0 f””(x)= cos x f””(0)= 1 Step 2 — plug into Maclaurin formula cos x = 1 + x(0) + (x²/2!)(-1) + (x³/3!)(0) + (x⁴/4!)(1) Step 3 — simplify (odd-power coefficients vanish) cos x = 1 – x²/2! + x⁴/4! cos x ≈ 1 − x²2! + x⁴4! ✓ matches booklet since cos is an EVEN function, all odd-power coefficients are zero — no need to compute f'(0), f”'(0), etc. once you’ve spotted this. Same trick for sin (odd → all even-power coefficients zero).

WE 3

Binomial = Maclaurin for (1+x)^(1/2)

Find the Maclaurin series of (1 + x)^1/2 up to x³ using (a) the binomial theorem and (b) the Maclaurin formula. Show they agree.

(a) Binomial theorem with n = 1/2 (1+x)^(1/2) = 1 + (1/2)x + [(1/2)(-1/2)/2!]x² + [(1/2)(-1/2)(-3/2)/3!]x³ = 1 + x/2 + (-1/4)/2 x² + (3/8)/6 x³ = 1 + x/2 – x²/8 + x³/16 (b) Maclaurin formula — find derivatives f(x) = (1+x)^(1/2) f(0) = 1 f'(x) = 1/2 (1+x)^(-1/2) f'(0) = 1/2 f”(x) = -1/4 (1+x)^(-3/2) f”(0) = -1/4 f”'(x)= 3/8 (1+x)^(-5/2) f”'(0)= 3/8 Maclaurin: 1 + (1/2)x + (-1/4)/2 x² + (3/8)/6 x³ = 1 + x/2 – x²/8 + x³/16 Both give 1 + x2 − x²8 + x³16 ✓ same answer for (1+x)^n where n isn’t a positive integer, binomial and Maclaurin produce identical series — the binomial theorem IS the Maclaurin series in this case. Use binomial when faster.

WE 4

Numerical approximation — sin(0.3)

Use the formula-booklet Maclaurin series for sin x truncated at the x⁵ term to approximate sin(0.3). Compare with the calculator value 0.2955202067.

Step 1 — write the truncated series from the booklet sin x ≈ x – x³/3! + x⁵/5! = x – x³/6 + x⁵/120 Step 2 — substitute x = 0.3 sin(0.3) ≈ 0.3 – (0.3)³/6 + (0.3)⁵/120 ≈ 0.3 – 0.027/6 + 0.00243/120 ≈ 0.3 – 0.0045 + 0.00002025 ≈ 0.29552025 Step 3 — compare with actual value Actual: 0.2955202067 Error: ≈ 4.3 × 10⁻⁸ (accurate to 7 d.p.) sin(0.3) ≈ 0.2955 (accurate to 7+ decimal places) at x = 0.3 (small), even three terms give 7-d.p. accuracy. Adding x⁷/7! would push the accuracy beyond 10 d.p. The approximation works because |x| is small.

WE 5

Numerical approximation — ln(1.2)

Use the booklet series for ln(1+x) truncated at x⁴ to approximate ln(1.2). Compare with the calculator value 0.1823215568.

Step 1 — booklet series ln(1+x) ≈ x – x²/2 + x³/3 – x⁴/4 Step 2 — substitute x = 0.2 (so that 1+x = 1.2) ln(1.2) ≈ 0.2 – (0.2)²/2 + (0.2)³/3 – (0.2)⁴/4 = 0.2 – 0.04/2 + 0.008/3 – 0.0016/4 = 0.2 – 0.02 + 0.002667 – 0.0004 = 0.182267 Step 3 — compare with actual Actual: 0.1823215568 Error: ≈ 5.5 × 10⁻⁵ ln(1.2) ≈ 0.182 (accurate to 3 d.p.) the ln series converges slower than e^x or sin x because its coefficients don’t have factorials in the denominator. To get more accuracy you’d need many more terms — or pick smaller x.

WE 6

First principles for tan x (not in booklet)

tan x is not in the formula booklet. Use the Maclaurin formula to find its series up to the x³ term.

Step 1 — find derivatives f(x) = tan x f(0) = 0 f'(x) = sec²x f'(0) = 1 f”(x) = 2 sec²x tan x f”(0) = 0 f”'(x) = 2[2 sec²x tan²x + sec⁴x] At x = 0: 2[0 + 1] = 2 f”'(0) = 2 Step 2 — plug into Maclaurin formula tan x = 0 + x(1) + (x²/2!)(0) + (x³/3!)(2) Step 3 — simplify tan x = x + 2x³/6 = x + x³/3 tan x ≈ x + x³3 (for small x) verify: tan(0.1) ≈ 0.1 + 0.001/3 ≈ 0.10033. Calculator: tan(0.1) ≈ 0.10033 ✓. Since tan is odd, all even-power coefficients are 0 — could have skipped the f”(0) computation.

💡 Top tips

Check the booklet first: if your function is e^x, ln(1+x), sin x, cos x, or arctan x, just quote the series. Don’t re-derive when you don’t have to.
Use parity: if f is even, only even powers of x appear; if odd, only odd powers. Cuts the work in half.
Coefficient of xⁿ is f⁽ⁿ⁾(0)/n! — keep this clean: compute the derivative value, then divide by n! at the end.
“Up to and including xⁿ“ means find ALL terms up to that power. Don’t stop early just because some are zero.
Approximation accuracy depends on (i) how many terms you keep, and (ii) how small |x| is. Both matter. For better accuracy: either add terms or use a smaller x.

⚠ Common mistakes

Forgetting the factorial denominators: the coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, NOT just f⁽ⁿ⁾(0). Easy slip that kills the whole series.
Sign errors when differentiating sin/cos repeatedly: the cycle is sin → cos → −sin → −cos → sin… Track signs carefully.
Using the booklet series for a composite: ln(1 + 2x) is NOT the same as the booklet ln(1+x) series — you have to substitute 2x wherever x appears (next note).
Stopping too early: “up to x⁴” means include the x⁴ term. For sin x, the x⁴ coefficient is 0, but you still need to compute f⁽⁴⁾(0) to know that.
Mixing degrees and powers: a “third-order” approximation usually means up to x³, not up to f‴. Read the question carefully.

Up next: Maclaurin Series of Composites & Products. Once you have the booklet’s standard series, you can build series for composite functions like sin(2x), e^x², ln(1 + 3x) by substitution; for products like e^x · sin x by multiplying truncated series together. No need to redo first-principles for every new function — just remix the standard ones.

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