Reuse, don't re-derive: Once you know eˣ, sin x, cos x and ln(1 + x), most new series are just bookwork on the known ones.
Substitution: to get ex², write the eˣ series and replace every x by x². Multiplication: to get x·sin x or eˣ·cos x, multiply the two series and collect like powers up to the term you need.
IB-style question — substitution
Find the Maclaurin series of f(x) = ex^{2} up to and including the term in x⁴.
Step by step
- Start from the known eˣ series.
- Replace u by x² (so u² = x⁴).
- Simplify (x²)² = x⁴ and 2! = 2.
Final answer
ex² = 1 + x² + x⁴/2 + … (substitute x² into eˣ — no differentiation needed).
IB-style question — multiplication
Find the Maclaurin series of f(x) = x·sin x up to and including the term in x⁴.
Step by step
- Write the sin x series (only the terms that will reach x⁴ after multiplying by x).
- Multiply every term by x.
- Distribute.
Final answer
x·sin x = x² − x⁴/6 + … (multiply the sin x series by x).
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Series turn 0/0 into easy algebra: A limit like (sin x − x)/x³ gives 0/0 at x = 0 — undefined as it stands. Replace the top with its Maclaurin series: the leading terms cancel, and the lowest surviving power matches the bottom.
Divide every term by that power, then let x → 0 — only the constant term is left, and that's the limit.
IB-style question — a 0/0 limit
A student needs the value of a limit that the GDC reports as 0/0.
Use a Maclaurin series to evaluate .
Step by step
- Replace cos x by its series.
- The 1's cancel; the lowest surviving power is x².
- Divide every term by x² (the bottom).
- Let x → 0: every term with an x vanishes.
Final answer
The limit is 1/2 — the constant term left after dividing by x².
IB-style question — a cubic-denominator limit
Evaluate using a Maclaurin series.
Step by step
- Replace sin x by its series, then subtract x.
- The x's cancel; the lowest surviving power is x³.
- Divide every term by x³.
- Let x → 0.
Final answer
The limit is −1/6.