To find the Maclaurin series for sin(x²), a student should...
ASubstitute x² for x in the known Maclaurin series for sin(x)
BDifferentiate the known Maclaurin series for cos(x) term by term
CIntegrate the known Maclaurin series for sin(x) term by term
DCompute all derivatives of sin(x²) at x = 0 from scratch using the chain rule
The most efficient approach is substitution: since sin(x) = x − x³/3! + x⁵/5! − ⋯, replacing every x with x² gives sin(x²) = x² − x⁶/3! + x¹⁰/5! − ⋯. This takes seconds and avoids the escalating complexity of computing higher derivatives of sin(x²) via repeated chain and product rules. Option D represents the common but inefficient approach — the whole point of memorizing the standard series is to avoid re-deriving from scratch every time.
Question 2 Multiple Choice
A classmate claims: 'The Maclaurin series is a generalization of the Taylor series — it works at more points.' What is wrong with this statement?
AIt reverses the relationship: a Maclaurin series is a special case of a Taylor series, specifically the one centered at a = 0
BNothing — Maclaurin series do converge for a wider range of x than Taylor series in general
CThe Maclaurin series only works for trigonometric functions, not for general functions
DTaylor series centered at a ≠ 0 are undefined; all Taylor series are centered at 0
A Maclaurin series is simply a Taylor series with the center set to a = 0. It is more restricted, not more general — you sacrifice the ability to center the expansion near other points. The Taylor series formula Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ becomes Σ f⁽ⁿ⁾(0)/n! · xⁿ when a = 0. The Maclaurin series is used frequently because centering at 0 simplifies the algebra for many common functions, not because it has broader applicability.
Question 3 True / False
Differentiating the Maclaurin series for sin(x) term by term yields the Maclaurin series for cos(x).
TTrue
FFalse
Answer: True
Yes. sin(x) = x − x³/3! + x⁵/5! − x⁷/7! + ⋯. Differentiating term by term: 1 − 3x²/3! + 5x⁴/5! − 7x⁶/7! + ⋯ = 1 − x²/2! + x⁴/4! − x⁶/6! + ⋯, which is exactly the Maclaurin series for cos(x). This works because within the radius of convergence, power series can be differentiated term by term — and this is actually how the cos(x) series can be derived from sin(x) rather than by computing cos's derivatives independently.
Question 4 True / False
Most Maclaurin series converges for most real numbers x.
TTrue
FFalse
Answer: False
This is false. Some Maclaurin series have limited radii of convergence. The series for e^x, sin(x), and cos(x) do converge for all x, but 1/(1−x) = 1 + x + x² + ⋯ only converges for |x| < 1, and ln(1+x) only converges for −1 < x ≤ 1. When you manipulate a known series (by substitution, differentiation, or integration), you must track what happens to the radius of convergence — it can shrink but never grow.
Question 5 Short Answer
Why is it generally more efficient to find the Maclaurin series for e^(−x²) by substituting into the known series for e^x rather than computing derivatives of e^(−x²) directly?
Think about your answer, then reveal below.
Model answer: Substituting −x² for x in the known series e^x = 1 + x + x²/2! + x³/3! + ⋯ gives 1 − x² + x⁴/2! − x⁶/3! + ⋯ in one step. Computing derivatives of e^(−x²) directly requires repeatedly applying the chain and product rules — the n-th derivative at x = 0 grows rapidly in complexity. Both methods yield the same result, but substitution leverages already-known information and avoids computational errors.
The point of memorizing the standard Maclaurin series is precisely to make this substitution strategy available. The key insight is that the Maclaurin series for a function is uniquely determined by its derivatives at 0 — so if you can algebraically transform a known series into the target function, you have the correct series without any differentiation. Substitution, differentiation, and integration of series are the core manipulation toolkit.