By Piyush Wairale · GATE DA Educator & Course Instructor, IIT Madras BS Programme · Updated August 2026

Taylor and Maclaurin Series: the Definition That Generates Everything

If f has derivatives of all orders at a, its Taylor series there is f(x) = Σ_n=0^∞ f⁽ⁿ⁾(a)(x − a)ⁿ/n! — a polynomial rebuilt entirely from local derivative information. Truncating after n terms leaves the remainder Rₙ = f⁽ⁿ⁺¹⁾(ξ)(x−a)ⁿ⁺¹/(n+1)! for some ξ between a and x, which is how error bounds get computed in the worked examples below. Set a = 0 and you have the Maclaurin series, the form GATE almost always uses. Both this topic and optimisation are named in the official GATE DA syllabus; the calculus pillar notes cover the limits and derivatives this builds on.

The Five Standard Expansions

Substitution generates everything else: replace x by 2x, −x², etc. — the coefficient questions in the solved section run on exactly this trick.

Two Approximations, Fully Worked

e⁰⋅¹ to four terms: 1 + 0.1 + 0.01/2 + 0.001/6 = 1 + 0.1 + 0.005 + 0.000167 = 1.105167. Remainder bound: next term × safety, R₃ ≤ e⁰⋅¹(0.1)⁴/4! < 1.2 × 0.0001/24 ≈ 5×10⁻⁶ — so all four printed decimals are trustworthy (true value 1.10517…).

cos(0.2) to two terms: 1 − (0.2)²/2 = 1 − 0.02 = 0.98. The series alternates, so the error is at most the first omitted term: (0.2)⁴/24 = 0.0000667 — giving 0.98007 to five decimals (true value 0.980067). That “error ≤ first omitted term” rule for alternating series is itself a GATE MCQ.

Single-Variable Optimisation, Worked Fully

Take f(x) = x³ − 3x² + 4.

Step 1 — critical points: f′(x) = 3x² − 6x = 3x(x − 2) = 0 ⇒ x = 0 or x = 2.

Step 2 — classify with f″: f″(x) = 6x − 6. f″(0) = −6 < 0 ⇒ local maximum at x = 0, value f(0) = 4. f″(2) = 6 > 0 ⇒ local minimum at x = 2, value f(2) = 8 − 12 + 4 = 0.

If f″ = 0 at a critical point the test is inconclusive — fall back to the first-derivative sign chart (e.g. x³ at 0 is neither max nor min).

Closed-Interval Max/Min: Don’t Forget the Endpoints

Same f on [−1, 4]. Candidates = critical points inside + endpoints: f(−1) = −1 − 3 + 4 = 0; f(0) = 4; f(2) = 0; f(4) = 64 − 48 + 4 = 20.

Global maximum 20 at x = 4 (an endpoint — the interior local max of 4 loses!), global minimum 0 attained twice, at x = −1 and x = 2. Endpoint-forgetting is the single most common error in this question type.

Two Variables: Gradient, Hessian and the Second-Partials Test

Critical points now solve ∇f = (fₓ, fᵢ) = (0, 0), and classification uses the discriminant of the Hessian matrix:

Worked (minimum): f(x, y) = x² + y² − 2x − 4y. fₓ = 2x − 2 = 0 ⇒ x = 1; fᵢ = 2y − 4 = 0 ⇒ y = 2. Second partials: fₓₓ = 2, fᵢᵢ = 2, fₓᵢ = 0 ⇒ D = 4 > 0 with fₓₓ > 0 ⇒ local (indeed global) minimum at (1, 2), value f(1,2) = 1 + 4 − 2 − 8 = −5.

Worked (saddle): f(x, y) = x² − y² at (0, 0): D = (2)(−2) − 0 = −4 < 0 → saddle point — a min along the x-axis, a max along the y-axis, an optimum along neither.

Gradient Descent and Convexity: the ML Connection

When ∇f = 0 can’t be solved in closed form — a neural network’s loss over millions of weights — you walk downhill numerically: w ← w − η∇f(w). That’s the update rule trained step by step in the neural networks deep dive and behind every model in the ML pillar. The guarantee question GATE likes: for a convex function (f″ ≥ 0 everywhere, or Hessian positive semi-definite), every local minimum is global — so gradient descent with a suitable step size converges to the optimum. Non-convex losses (deep nets) offer no such promise, only good practice.

Three GATE-Style Problems, Solved

Problem 1 (NAT). The coefficient of x³ in the Maclaurin series of e²ₓ?

Solution. Substitute 2x into eₓ: term = (2x)³/3! = 8x³/6 ⇒ coefficient = 4/3 ≈ 1.33. (Equivalently f‴′(0)/3! = 8/6.)

Problem 2 (MCQ). For f(x) = x⁴ − 4x³, classify the critical points.

Solution. f′ = 4x³ − 12x² = 4x²(x − 3) ⇒ x = 0, 3. f″ = 12x² − 24x: f″(3) = 36 > 0 → local min at x = 3 (value −27). f″(0) = 0 → inconclusive; sign chart: f′ < 0 on both sides of 0 → x = 0 is neither max nor min (an inflection with horizontal tangent). Options claiming “max at 0” are the planted trap.

Problem 3 (NAT). f(x, y) = x³ − 3x + y². Classify the critical points and give D at each.

Solution. fₓ = 3x² − 3 = 0 ⇒ x = ±1; fᵢ = 2y = 0 ⇒ y = 0. Partials: fₓₓ = 6x, fᵢᵢ = 2, fₓᵢ = 0. At (1, 0): D = 12 > 0, fₓₓ = 6 > 0 → local min (f = −2). At (−1, 0): D = −12 < 0 → saddle. Both the sign of D and the missed x = −1 root are standard traps.

Common Mistakes to Avoid

Forgetting endpoints on closed intervals. The global max of our worked example lived at x = 4, not at any critical point.

Treating D = 0 or f″ = 0 as “saddle” or “neither”. Both are inconclusive — resolve with a sign chart or higher derivatives.

Missing negative roots of f′ = 0. 3x² − 3 = 0 has two solutions; NAT answers change completely if you drop x = −1.

Using ln(1+x) outside its radius. The series only converges for −1 < x ≤ 1 — substituting x = 2 produces confident nonsense.

Adding factorials to ln(1+x) or dropping them from eₓ. The log series has plain denominators n; the exponential has n!. Mixing them is the top coefficient-question error.

How GATE DA Asks This Topic

Recurring shapes: (1) NAT — a Maclaurin coefficient after substitution, or a two-to-four-term approximation of eₓ/sin/cos at a small value; (2) NAT/MCQ — locate and classify critical points in one or two variables (the D-rule table is the whole answer); (3) NAT — global max/min on a closed interval; (4) MCQ — convexity, remainder/error bounds, or why gradient descent needs the gradient at all. Arithmetic stays small; the marks reward the checklist: candidates → classify → compare values.

FAQs: Taylor Series & Optimisation for GATE DA

Round out the maths: revise limits and derivatives in the calculus pillar, see the optimisation machinery power training in the ML pillar and the neural networks deep dive, and track coverage against the GATE DA 2027 syllabus. New calculus problem-solving sessions drop regularly on my YouTube channel — subscribe so you don’t miss them.