Taylor & Maclaurin Series

A few lessons ago you zoomed in on a smooth curve until it looked like a straight line — the tangent, the first derivative. The last lesson then gave you fast ways to compute that derivative, and the second, and the third. Now put those together and zoom more carefully.

Zoom in on any smooth curve and, yes, it first looks like a line; keep one more term and it looks like a parabola; keep more terms still and it looks like a polynomial that hugs the curve ever more tightly.

That is the whole idea of a Taylor series: trade a messy function for an infinite polynomial that mimics it near one point. And here is the payoff GATE leans on: once the polynomial is written, every derivative at that point is already sitting there, folded into a coefficient, waiting to be read off.

The same “replace a hard function by its first couple of polynomial terms” move is everywhere in machine learning and numerics:

• It is how gradient descent linearises a loss.
• It is how Newton’s method uses the quadratic term.
• It is how libraries approximate exp and log fast.

So this is a tool you will reuse long 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 + …

Read it term by term and the earlier lessons line up:

• f(a) is the height.
• f'(a)(x−a) is the tangent line.
• f''(a)/2!(x−a)² adds the bend.
• The pattern continues, and so on.

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 carry only odd powers of x; cos x and cosh x carry only even powers.

Those two words have precise meanings worth having:

• An odd function is one with f(−x) = −f(x), its graph unchanged by a half-turn about the origin.
• An even function has f(−x) = f(x), mirror-symmetric across the y-axis.

Only odd powers of x can assemble the first kind, only even powers the second — which is why the series come out the way they do. That single fact answers a whole class of GATE questions, as we are about to 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 visual proof that more terms would be needed further out.

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 once you can write (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. That derivative at 0 is 0.

This is exactly 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.

Take eˣ as a check. Its series has 1/n! as the coefficient of xⁿ. Multiply by n! and you recover f⁽ⁿ⁾(0) = 1 for every n. This is consistent with the fact that every derivative of eˣ is eˣ, which equals 1 at x = 0.

The same series even 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. It 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 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 carries 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. The whole problem collapses the moment you notice that sinh (an odd function) has no even-power terms.

That means every even-order derivative at 0 — the 2nd, 4th, …, 10th — is zero. The predict-prompt was the shortcut: parity, not computation.

Run the other case once too, so “parity” does not quietly become “the answer is always zero” in your memory. Ask the same question for f⁽⁹⁾(0).

Here the series does have an x⁹ term, and its coefficient is 1/9!:

f⁽⁹⁾(0) = 9! × (coefficient of x⁹) = 9! × 1/9! = 1

So every odd-order derivative of sinh at 0 is exactly 1, and every even-order one is 0. Parity tells you which of those two worlds you are in; the n! factor then supplies the actual number.

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 — the same answer L’Hôpital reached two lessons back, now read straight off the series.

A question to carry forward

Look once more at the first two Taylor terms, f(a) + f'(a)(x − a). The slope f'(a) is what tilts the approximation.

Now picture standing at the very top of a hill or the bottom of a valley on the graph: for an instant the curve is flat, neither rising nor falling, so that slope is momentarily zero and the tangent term drops out entirely.

Those flat spots are exactly the peaks and troughs that optimisation — the heart of training any model — cares about most. Here is the thread onward: if “slope = 0” marks where a function levels off, can we use it to find a function’s maxima and minima? Once found, how do we tell a peak from a trough?

In one breath

• Taylor series: f(x) = Σ f⁽ⁿ⁾(a)/n! (x−a)ⁿ — a polynomial that mimics f near a; Maclaurin is the a = 0 case.
• Memorise five: eˣ, sin x (odd powers), cos x (even powers), ln(1+x), 1/(1−x) (geometric).
• Coefficient ↔ derivative: f⁽ⁿ⁾(0) = n! × (coeff of xⁿ) — read any derivative off the series; a missing power ⇒ that derivative is 0.
• Parity: odd functions ⇒ even-order derivatives at 0 vanish; even functions ⇒ odd-order vanish. sinh⁽¹⁰⁾(0) = 0 (GATE DA 2025) is parity alone.
• Taylor-for-a-limit: expand a couple of terms and read the leading behaviour — (eˣ − 1 − x)/x² → 1/2.

Practice