Taylor-Maclaurin Series
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Direct answer
The Maclaurin series writes a function as its own polynomial around zero — f(x) = f(0) + f′(0)x + f″(0)x²/2! + ... — and the five worth carrying into the hall are e^x = 1 + x + x²/2! + ...; sin x = x − x³/3! + ...; cos x = 1 − x²/2! + ...; ln(1 + x) = x − x²/2 + ...; and (1 + x)^n = 1 + nx + n(n − 1)x²/2! + .... Taylor's version shifts the centre to any point a, but JEE lives almost entirely at a = 0. The currency is threefold: limits without repeated L'Hôpital, numerical approximations by hand, and coefficient comparison when a function satisfies a differential equation.
What you must remember
- The five classics: e^x, sin x, cos x, ln(1 + x) (−1 < x ≤ 1) and the general binomial (1 + x)^n — the exponential and binomial series appear in NCERT; the trigonometric expansions are coaching-module staples that solve Main limits faster than any rule.
- Less famous but scoring: tan x = x + x³/3 + 2x⁵/15; arctan x = x − x³/3 + x⁵/5; e^(−x²) integrable only through its series.
- Order discipline: for a limit with denominator x³, carry the numerator to x³ — stopping one term early is the standard source of wrong options.
- Alternating signs: sin and cos alternate; ln(1 + x) alternates; e^x never does — sign slips are the commonest arithmetic failure.
- Validity matters: ln(1 + x) and the binomial series live on −1 < x ≤ 1; e^x, sin, cos are valid for all x.
- Licence to integrate: ∫₀^x e^(−t²) dt has no elementary closed form, yet its series integrates term by term to x − x³/3 + x⁵/(5 × 2!) − ...
- Recognising the answer: lim (sin x/x)^(1/x²) = e^(−1/6) appears as a final value often enough to be worth knowing cold.
A series doing a limit's job
Evaluate lim x→0 (sin x/x)^(1/x²). Write sin x/x = 1 − x²/6 + x⁴/120 − ..., so the expression is (1 + u)^(1/x²) with u = −x²/6 + ... tending to 0. Take logarithms: (1/x²) ln(1 + u), and since ln(1 + u) ≈ u for small u, the exponent tends (−x²/6)/x² = −1/6. The limit is e^(−1/6) ≈ 0.846. Notice what the series bought: one line, no L'Hôpital rounds, no 1^∞ ceremony beyond the log step. Contrast a second use: approximate (1.05)^10 by the binomial series — 1 + 10(0.05) + 45(0.05)² + 120(0.05)³ = 1 + 0.5 + 0.1125 + 0.015 = 1.6275 against the true value 1.6289, with the next term 210(0.05)⁴ ≈ 0.0013 closing most of the gap. The series is not just a limit tool; it is a computational instrument whose accuracy you control by choosing how many terms to carry.
Where marks leak
JEE Main asks series-based limits where the distractors are one-term-short answers — e^(−1/6) versus e^(1/6) versus e^(−1/12), the sign and the coefficient each catching a different cohort. JEE Advanced asks for coefficients rather than limits: given that e^x sin x equals a power series, the x³ coefficient comes from matching (1 + x + x²/2 + ...)(x − x³/6 + ...) — collecting the x³ terms gives 1 − 1/6 = 5/6. The recurring failure is validity amnesia: substituting x = 2 into the ln(1 + x) series produces nonsense, and one option per such question is built from exactly that nonsense. Term-wise differentiation is legal inside the interval but candidates differentiate outside it for divergent series, and truncation discipline fails in the (tan x − sin x)/x³ tier where the x³ coefficient must survive intact. Finally, memorise with structure not digits: sin carries odd powers over odd factorials, cos even over even — recalling the pattern rescues the formula even when memory of the exact series wobbles.
Frequently asked questions
What is the Maclaurin series of a function?
f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x³/3! + ... — the Taylor series centred at zero, one derivative per term.
Which expansions must be carried into the exam?
e^x, sin x, cos x, ln(1 + x) and (1 + x)^n, plus tan x and arctan x for the harder coefficient questions.
Where is the ln(1 + x) series valid?
Only on −1 < x ≤ 1; at x = 1 it converges to ln 2, and beyond it the series diverges outright.
How does a series evaluate (sin x/x)^(1/x²)?
sin x/x = 1 − x²/6 + ..., so the logarithm's limit is −1/6 and the answer is e^(−1/6).
Can a power series be differentiated term by term?
Inside its interval of validity, yes — the derivative series has the same radius, which is how differential-equation coefficient matching works.