Taylor & Maclaurin Series

Zoom in on any smooth curve and it starts to look like a polynomial. Zoom a little more and the polynomial gets better. That’s the whole idea of a Taylor series: trade a messy function for an infinite polynomial that mimics it near one point. And here’s the payoff GATE leans on — once you’ve written the polynomial, every derivative at that point is just sitting there in a coefficient, waiting to be read.

The same “replace a hard function by its first couple of polynomial terms” move is everywhere in ML and numerics: it’s how gradient descent linearises a loss, how Newton’s method uses the quadratic term, and how libraries approximate exp/log fast — so this is a tool you’ll reuse well past the exam.

The Taylor and Maclaurin series

The Taylor series of f about the point a is

f(x) = Σ  f⁽ⁿ⁾(a)/n! · (x − a)ⁿ
       n=0
     = f(a) + f'(a)(x−a) + f''(a)/2! (x−a)² + f'''(a)/3! (x−a)³ + …

The Maclaurin series is the special, most-used case where a = 0:

f(x) = f(0) + f'(0)·x + f''(0)/2! · x² + f'''(0)/3! · x³ + …

You should memorise the five standard Maclaurin series — they appear constantly:

eˣ        = 1 + x + x²/2! + x³/3! + x⁴/4! + …
sin x     =     x        − x³/3! + x⁵/5! − …      (odd powers only)
cos x     = 1 − x²/2!     + x⁴/4! − …             (even powers only)
ln(1+x)   =     x − x²/2  + x³/3  − x⁴/4 + …
1/(1−x)   = 1 + x + x² + x³ + …                   (geometric)

Notice the structure: sin x and sinh x contain only odd powers of x; cos x and cosh x contain only even powers. That single fact answers a whole class of GATE questions, as we’ll see.

The picture below overlays cos x (solid) with its 2-term Taylor approximation 1 − x²/2 (dashed). Near x = 0 the parabola hugs the curve; it drifts off only as you move away.

The coefficient ↔ derivative relation

Look again at the Maclaurin form: the coefficient sitting in front of xⁿ is exactly f⁽ⁿ⁾(0)/n!. Turn that around and you get the relation GATE tests directly:

f⁽ⁿ⁾(0) = n! × (coefficient of xⁿ in the series)

So if you can write down (or recall) the series, you can read off any derivative at 0 without differentiating n times by hand. And if a particular power of x is missing from the series, its coefficient is 0 — so that derivative at 0 is 0. This is why “odd vs even powers” matters: sin x has no even powers, so all its even-order derivatives at 0 vanish; cos x has no odd powers, so all its odd-order derivatives at 0 vanish.

For example, the Maclaurin series of eˣ has 1/n! as the coefficient of xⁿ. Multiply by n! and you recover f⁽ⁿ⁾(0) = 1 for every n — consistent with the fact that every derivative of eˣ is eˣ, which equals 1 at x = 0. And the same series predicts that (eˣ − 1 − x)/x² → 1/2 as x → 0, because the leading surviving term of the numerator is x²/2.

How GATE asks this

The signature GATE DA question is a NAT that hands you a function, expects you to recall (or build) its Maclaurin series, and asks for f⁽ⁿ⁾(0) — an n-th derivative at the origin. You answer it without differentiating n times: find the coefficient of xⁿ, multiply by n!. The second flavour uses Taylor to evaluate a limit of the 0/0 form — expand the numerator a couple of terms and read off the leading behaviour. Both appeared in GATE DA 2024 and 2025.

Worked example — a real GATE DA 2025 question

Let f(x) = sinh x (hyperbolic sine, the odd-power cousin of eˣ: sinh x = (eˣ − e⁻ˣ)/2). Find f⁽¹⁰⁾(0).

First write the Maclaurin series of sinh x. It contains only odd powers:

sinh x = x + x³/3! + x⁵/5! + x⁷/7! + x⁹/9! + x¹¹/11! + …

Now apply the relation f⁽ⁿ⁾(0) = n! × (coefficient of xⁿ) with n = 10. The series has terms in x¹, x³, x⁵, x⁷, x⁹, x¹¹, … — there is no x¹⁰ term at all, so its coefficient is 0:

f⁽¹⁰⁾(0) = 10! × (coefficient of x¹⁰) = 10! × 0 = 0

So f⁽¹⁰⁾(0) = 0. This is a real GATE DA 2025 question, and the whole problem collapses the moment you notice that sinh (an odd function) has no even-power terms, so every even-order derivative at 0 — the 2nd, 4th, …, 10th — is zero.

A Taylor-for-a-limit companion. Evaluate lim_{x→0} (eˣ − 1 − x)/x². Expand the numerator using eˣ = 1 + x + x²/2! + …:

eˣ − 1 − x = (1 + x + x²/2 + x³/6 + …) − 1 − x = x²/2 + x³/6 + …

(eˣ − 1 − x)/x² = 1/2 + x/6 + …   →   1/2   as x → 0

The leading term of the numerator is x²/2, which cancels the x² below to leave 1/2 — matching the numeric run above.

Quick check