GATE DA

Taylor series

A Taylor series writes a smooth function of one real variable as an infinite polynomial in powers of (x - a), built from its derivatives at the point a. For GATE it turns hard limits, coefficient-finding, approximation and local-extremum questions into short polynomial arithmetic, provided you know the standard series and where each is valid.

2 previous-year questions · 2 solved · 6 practice questions

1. Taylor and Maclaurin series

Convention: all functions here are real-valued functions of one real variable, angles are in radians, and $ln$ is the natural logarithm. If $f$ has derivatives of every order at $x = a$ , its Taylor series about $a$ is built from the values $f^{(n)} (a)$ . The special case $a = 0$ is the Maclaurin series.

TAYLOR SERIES ABOUT X = A

f (x) = n = 0 \sum \infty \frac{f ^{(n)} ( a )}{n !} (x - a)^{n} = f (a) + f^{'} (a) (x - a) + \frac{f ^{''} ( a )}{2 !} (x - a)^{2} + \dots

Valid for those $x$ where the series converges to $f (x)$ ; convergence is not automatic, so state the range of $x$ .

MACLAURIN SERIES (A = 0)

f (x) = n = 0 \sum \infty \frac{f ^{(n)} ( 0 )}{n !} x^{n}

The coefficient of $x^{n}$ is $f^{(n)} (0) / n!$ , so $f^{(n)} (0) = n! \times (coefficient of x^{n})$ .

The degree- $n$ Taylor polynomial $P_{n} (x) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$ keeps the terms up to $(x - a)^{n}$ ; it matches $f$ and its first $n$ derivatives at $a$ . Uniqueness: if a power series $\sum c_{n} (x - a)^{n}$ equals $f (x)$ on an open interval containing $a$ , then $c_{n} = f^{(n)} (a) / n!$ for every $n$ , so that power series is the Taylor series of $f$ about $a$ . This is why a series obtained from the standard series by substitution, multiplication or term-by-term integration (inside their ranges) can be used to read off coefficients and derivatives.

2. Standard Maclaurin series and their ranges

EXPONENTIAL

e^{x} = n = 0 \sum \infty \frac{x ^{n}}{n !} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots

Converges for all real $x$ .

SINE

sin x = n = 0 \sum \infty \frac{( - 1 ) ^{n} x ^{2 n + 1}}{( 2 n + 1 )!} = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots

Converges for all real $x$ ; only odd powers appear.

COSINE

cos x = n = 0 \sum \infty \frac{( - 1 ) ^{n} x ^{2 n}}{( 2 n )!} = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

Converges for all real $x$ ; only even powers appear.

NATURAL LOGARITHM

ln (1 + x) = n = 1 \sum \infty \frac{( - 1 ) ^{n + 1} x ^{n}}{n} = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots

Valid for $- 1 < x \leq 1$ . At $x = 1$ it gives $ln 2$ ; at $x = - 1$ it diverges.

BINOMIAL SERIES

(1 + x)^{m} = k = 0 \sum \infty (k m) x^{k} = 1 + m x + \frac{m ( m - 1 )}{2 !} x^{2} + \dots, (k m) = \frac{m ( m - 1 ) \dots ( m - k + 1 )}{k !}

For any real $m$ it is valid for $∣ x ∣ < 1$ and diverges for $∣ x ∣ > 1$ (radius 1) unless $m$ is a non-negative integer, in which case the series stops and is exact for all $x$ . At $x = \pm 1$ convergence depends on $m$ (for example, for $m > 0$ it also converges at both endpoints).

GEOMETRIC SERIES

\frac{1}{1 - x} = n = 0 \sum \infty x^{n} = 1 + x + x^{2} + \dots

Valid for $∣ x ∣ < 1$ . Replace $x$ by $- x$ for $1/ (1 + x) = 1 - x + x^{2} - \dots$ .

INVERSE TANGENT

tan^{- 1} x = n = 0 \sum \infty \frac{( - 1 ) ^{n} x ^{2 n + 1}}{2 n + 1} = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \dots

Valid for $- 1 \leq x \leq 1$ ; obtained by integrating $1/ (1 + t^{2}) = 1 - t^{2} + t^{4} - \dots$ from $0$ to $x$ . At $x = 1$ it gives $π /4$ .

TANGENT (FIRST TERMS)

tan x = x + \frac{x ^{3}}{3} + \frac{2 x ^{5}}{15} + \dots

Valid for $∣ x ∣ < π /2$ ; only odd powers appear. The first two terms are usually enough for limits as $x \to 0$ .

RADIUS OF CONVERGENCE (RATIO TEST)

R = n \to \infty lim \frac{c _{n}}{c _{n + 1}} for n = 0 \sum \infty c_{n} (x - a)^{n}

When this limit exists (it may be $\infty$ ), the series converges absolutely for $∣ x - a ∣ < R$ and diverges for $∣ x - a ∣ > R$ ; the endpoints $∣ x - a ∣ = R$ must be checked separately. For a series with only odd or only even powers, apply the ratio test to consecutive non-zero terms instead.

Function	Range of validity	Terms present
$e^{x}$	all real $x$	all powers
$sin x$	all real $x$	odd powers
$cos x$	all real $x$	even powers
$ln (1 + x)$	$- 1 < x \leq 1$	all powers $\geq 1$
$(1 + x)^{m}$ , $m$ not a non-negative integer	$∣ x ∣ < 1$ (endpoints depend on $m$ )	all powers
$1/ (1 - x)$	$∣ x ∣ < 1$	all powers
$tan^{- 1} x$	$- 1 \leq x \leq 1$	odd powers
$tan x$	$∣ x ∣ < π /2$	odd powers

3. Building new series and finding coefficients

Procedure to get the coefficient of a given power

Do not differentiate repeatedly unless forced; rewrite the function using the standard series.
Substitute: replace $x$ by the inner expression, e.g. $e^{2 x} = \sum (2 x)^{n} / n!$ and $ln (1 + x^{2}) = x^{2} - x^{4} /2 + \dots$ .
For a product, multiply the two truncated series and collect only the terms of the wanted power; keep terms up to that power and no further.
For a quotient, write $1/ g$ as a series (often geometric or binomial) first, then multiply.
Inside the open interval of convergence $∣ x - a ∣ < R$ , a power series may be differentiated or integrated term by term, and the radius stays the same (this is how $ln (1 + x)$ follows from $1/ (1 + x)$ , and $tan^{- 1} x$ from $1/ (1 + x^{2})$ ).
If $c_{n}$ is the coefficient of $(x - a)^{n}$ , then $f^{(n)} (a) = n! c_{n}$ .

To expand about a point $a \neq = 0$ , set $t = x - a$ and expand in $t$ . For example, about $a = 1$ , $ln x = ln (1 + t) = t - t^{2} /2 + t^{3} /3 - \dots$ , valid for $0 < x \leq 2$ .

WORKED EXAMPLE

Find the coefficient of $x^{3}$ in the Maclaurin series of $e^{x} sin x$ .

Solution

$e^{x} = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \dots$ and $sin x = x - \frac{x ^{3}}{6} + \dots$ .
Terms giving $x^{3}$ : $1 \cdot (- \frac{x ^{3}}{6})$ and $\frac{x ^{2}}{2} \cdot x$ .
Sum of coefficients: $- \frac{1}{6} + \frac{1}{2} = \frac{1}{3}$ .

Answer: $\frac{1}{3}$

4. Truncation and the size of the error

TAYLOR'S THEOREM WITH LAGRANGE REMAINDER

f (x) = P_{n} (x) + R_{n} (x), R_{n} (x) = \frac{f ^{(n + 1)} ( c )}{( n + 1 )!} (x - a)^{n + 1} for some c strictly between a and x

Needs $f^{(n)}$ continuous on the closed interval between $a$ and $x$ and $f^{(n + 1)}$ to exist on the open interval (having $n + 1$ continuous derivatives is enough). Hence $∣ R_{n} (x) ∣ \leq \frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}$ , where $M$ is the largest value of $∣ f^{(n + 1)} ∣$ on that interval.

ALTERNATING SERIES ERROR BOUND

∣ error after n terms ∣ \leq ∣ first omitted term ∣

Applies when, from the first omitted term on, the terms alternate in sign and decrease in magnitude to 0; the error then has the same sign as the first omitted term. Examples: $ln (1 + x)$ for $0 < x \leq 1$ ; $sin x$ , $cos x$ and $tan^{- 1} x$ for $∣ x ∣ \leq 1$ .

A first-order approximation is $f (a) + f^{'} (a) (x - a)$ and a second-order one adds $\frac{1}{2} f^{''} (a) (x - a)^{2}$ . For small $x$ : $e^{x} \approx 1 + x$ , $sin x \approx x$ , $cos x \approx 1 - x^{2} /2$ , $ln (1 + x) \approx x$ , $(1 + x)^{m} \approx 1 + m x$ . When $f^{(n + 1)}$ is bounded near $a$ , the Lagrange form shows the error of the degree- $n$ approximation is of order $∣ x - a ∣^{n + 1}$ , so it shrinks fast as $x \to a$ .

EXAM TIP

For $sin x$ the error after the term in $x^{2 k - 1}$ is at most $∣ x ∣^{2 k + 1} / (2 k + 1)!$ for every real $x$ , because $∣ sin^{(n)} ∣ \leq 1$ (the $x^{2 k}$ term is zero, so the remainder starts at order $2 k + 1$ ). For $e^{x}$ on $[0, h]$ use $max e^{c} = e^{h}$ in the Lagrange remainder.

5. Using series to evaluate limits

For a limit as $x \to 0$ of a $0/0$ form, expand numerator and denominator until the first non-cancelling terms appear, then compare lowest powers. This is often shorter than repeated L'Hôpital steps.

Limit procedure

Expand each function with the standard series, writing explicit terms up to a safe order.
Cancel the terms that match; let the lowest surviving term of the numerator be $c x^{p}$ and that of the denominator $d x^{q}$ , with $c, d \neq = 0$ .
If $p = q$ the limit is $c / d$ ; if $p > q$ it is $0$ ; if $p < q$ the limit is not finite (the one-sided limits are infinite, with opposite signs when $q - p$ is odd).
Expand the numerator at least up to $x^{q}$ ; if every term up to $x^{q}$ cancels, the limit is $0$ . Stopping before the cancellation ends gives a wrong answer.

Limit as $x \to 0$	Value
$\frac{sin x}{x}$	$1$
$\frac{1 - cos x}{x ^{2}}$	$\frac{1}{2}$
$\frac{e ^{x} - 1}{x}$	$1$
$\frac{ln ( 1 + x )}{x}$	$1$
$\frac{x - sin x}{x ^{3}}$	$\frac{1}{6}$
$\frac{e ^{x} - 1 - x}{x ^{2}}$	$\frac{1}{2}$

WORKED EXAMPLE

Evaluate $x \to 0 lim \frac{e ^{2 x} - 1 - 2 x}{x ^{2}}$ .

Solution

$e^{2 x} = 1 + 2 x + \frac{( 2 x ) ^{2}}{2 !} + \frac{( 2 x ) ^{3}}{3 !} + \dots = 1 + 2 x + 2 x^{2} + \frac{4 x ^{3}}{3} + \dots$ .
Numerator: $e^{2 x} - 1 - 2 x = 2 x^{2} + \frac{4 x ^{3}}{3} + \dots$ .
Divide by $x^{2}$ : $2 + \frac{4 x}{3} + \dots \to 2$ .

Answer: $2$

6. Traps and quick checks

EXAM TIP

Do not quote the range of $ln (1 + x)$ as all real $x$ : it is $- 1 < x \leq 1$ . The series of $1/ (1 - x)$ needs $∣ x ∣ < 1$ , and a binomial series $(1 + x)^{m}$ with $m$ not a non-negative integer has radius 1 (it diverges for $∣ x ∣ > 1$ ), while $e^{x}$ , $sin x$ , $cos x$ converge for every real $x$ .

EXAM TIP

When substituting, the range changes too: $ln (1 + 2 x)$ is valid for $- \frac{1}{2} < x \leq \frac{1}{2}$ , and the series of $\frac{1}{1 + x ^{2}}$ only for $∣ x ∣ < 1$ , even though the function itself is defined for every real $x$ .

EXAM TIP

Parity check: $sin x$ is odd so only odd powers appear; $cos x$ is even so only even powers appear. A product of an odd and an even function has only odd powers.

EXAM TIP

The coefficient of $x^{n}$ is $f^{(n)} (0) / n!$ , not $f^{(n)} (0)$ . Forgetting the $n!$ is a common slip in coefficient questions.

EXAM TIP

A function can have a Taylor series that converges but not to the function (for example $f (x) = e^{- 1/ x^{2}}$ for $x \neq = 0$ , $f (0) = 0$ , has all derivatives 0 at 0, so its Maclaurin series is identically 0). The standard series in Section 2 do converge to their functions in the ranges stated there.

7. Taylor series and local extrema

Convention: $f$ is a function of one real variable and $a$ is an interior point of its domain; a local maximum or minimum compares $f (a)$ only with values on some open interval around $a$ . At a critical point ( $f^{'} (a) = 0$ ) the expansion $f (a + h) = f (a) + \frac{1}{2} f^{''} (a) h^{2} + \dots$ shows the shape of $f$ near $a$ : the first non-zero term decides whether $f (a + h) - f (a)$ keeps one sign for small $h$ .

HIGHER-ORDER DERIVATIVE TEST

f^{'} (a) = f^{''} (a) = \dots = f^{(n - 1)} (a) = 0, f^{(n)} (a) \neq = 0 (n \geq 2)

Assume $f^{(n)}$ is continuous near $a$ . If $n$ is even: local minimum when $f^{(n)} (a) > 0$ , local maximum when $f^{(n)} (a) < 0$ . If $n$ is odd: no local extremum at $a$ . For $n = 2$ this is the second-derivative test.

WORKED EXAMPLE

Does $f (x) = e^{x} - 1 - x - \frac{x ^{2}}{2}$ have a local extremum at $x = 0$ ?

Solution

From the exponential series, $f (x) = \frac{x ^{3}}{6} + \frac{x ^{4}}{24} + \dots$ .
So $f^{'} (0) = f^{''} (0) = 0$ and $f^{'''} (0) = 3! \times \frac{1}{6} = 1 \neq = 0$ .
The first non-zero derivative has odd order 3, so there is no local extremum: $f (x) < 0$ for small $x < 0$ and $f (x) > 0$ for small $x > 0$ , while $f (0) = 0$ .

Answer: No local extremum at $x = 0$ .

EXAM TIP

$f^{''} (a) = 0$ alone decides nothing: $x^{4}$ (local minimum), $- x^{4}$ (local maximum) and $x^{3}$ (neither) all have $f^{'} (0) = f^{''} (0) = 0$ . Go to the first non-zero derivative.

PREVIOUS-YEAR QUESTIONS

2 questions from the DA papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE DA 2026 · GATE DA 2025

GATE 2026 DA Q.35

GATE 2026 · DAQ.35 · 1 mark · NAT

The value of $\sum_{i = 0}^{\infty} \sum_{j = 1}^{\infty} 2^{- i} 3^{- j}$ is ______________. (Answer in integer)

Answer and solution

GATE 2026 DA Q.35: answer and solution

See the question in the paper

Tap to open full size

GATE 2026 DA Q.35: answer and solution

GATE 2025 DA Q.14

GATE 2025 · DAQ.14 · 1 mark · MCQ

Let $f (x) = \frac{e ^{x} - e ^{- x}}{2}$ , $x \in R$ . Let $f^{(k)} (a)$ denote the $k^{t h}$ derivative of $f$ evaluated at $a$ . What is the value of $f^{(10)} (0)$ ? (Note: ! denotes factorial)

(A)
$0$
(B)
$1$
(C)
$\frac{1}{10 !}$
(D)
$\frac{2}{10 !}$

Answer and solution

GATE 2025 DA Q.14: answer and solution

See the question in the paper

Tap to open full size

GATE 2025 DA Q.14: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · NAT

Evaluate $x \to 0 lim \frac{sin x - x cos x}{x ^{3}}$ . Give your answer rounded to three decimal places.

Answer and solution

Answer: $0.3333$ (accepted range $\pm 0.001$ )

$sin x = x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots$ and $x cos x = x - \frac{x ^{3}}{2} + \frac{x ^{5}}{24} - \dots$ .
$sin x - x cos x = (- \frac{1}{6} + \frac{1}{2}) x^{3} + (\frac{1}{120} - \frac{1}{24}) x^{5} + \dots = \frac{x ^{3}}{3} - \frac{x ^{5}}{30} + \dots$ .
Divide by $x^{3}$ : $\frac{1}{3} - \frac{x ^{2}}{30} + \dots \to \frac{1}{3} \approx 0.333$ .

PRACTICE 21 mark · MCQ

The coefficient of $x^{3}$ in the Maclaurin series of $ln (1 + 2 x)$ is

(A)
$\frac{8}{3}$
(B)
$\frac{2}{3}$
(C)
$- \frac{8}{3}$
(D)
$\frac{4}{3}$

Answer and solution

Answer: (A)

$ln (1 + u) = u - \frac{u ^{2}}{2} + \frac{u ^{3}}{3} - \dots$ with $u = 2 x$ .
The $x^{3}$ term is $\frac{( 2 x ) ^{3}}{3} = \frac{8 x ^{3}}{3}$ .
The coefficient is $\frac{8}{3}$ (positive, since the $u^{3}$ term has sign $+$ ).

PRACTICE 32 marks · MSQ

Consider real $x$ and the Maclaurin series of the functions below. Which statements are correct?

(A)
The series of $ln (1 + x)$ converges at $x = 1$ .
(B)
The series of $sin x$ converges for every real $x$ .
(C)
The series of $(1 + x)^{1/2}$ converges for every real $x$ .
(D)
The series of $ln (1 + x)$ converges at $x = - 1$ .

Answer and solution

Answer: (A), (B)

A is true: at $x = 1$ the series is the alternating series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ , which converges (to $ln 2$ ).
B is true: the sine series converges for all real $x$ .
C is false: the binomial series with non-integer exponent $\frac{1}{2}$ has radius of convergence 1, so it diverges for $∣ x ∣ > 1$ (for example at $x = 2$ ).
D is false: at $x = - 1$ the series becomes $- (1 + \frac{1}{2} + \frac{1}{3} + \dots)$ , the negative of the divergent harmonic series.

PRACTICE 42 marks · NAT

Let $f (x) = e^{x^{2}} cos x$ . Find the coefficient of $x^{4}$ in its Maclaurin series. Give your answer as a decimal rounded to four places.

Answer and solution

Answer: $0.041667$ (accepted range $\pm 0.001$ )

$e^{x^{2}} = 1 + x^{2} + \frac{x ^{4}}{2} + \dots$ and $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ .
Terms giving $x^{4}$ : $1 \cdot \frac{x ^{4}}{24}$ , $x^{2} \cdot (- \frac{x ^{2}}{2})$ and $\frac{x ^{4}}{2} \cdot 1$ .
Sum: $\frac{1}{24} - \frac{1}{2} + \frac{1}{2} = \frac{1}{24} \approx 0.0417$ .

PRACTICE 52 marks · NAT

Using the Taylor polynomial of degree 2 of $f (x) = x$ about $a = 4$ , approximate $4.4$ . Give your answer to four decimal places.

Answer and solution

Answer: $2.0975$ (accepted range $\pm 0.00005$ )

$f (4) = 2$ , $f^{'} (x) = \frac{1}{2 x}$ so $f^{'} (4) = \frac{1}{4}$ .
$f^{''} (x) = - \frac{1}{4 x ^{3/2}}$ so $f^{''} (4) = - \frac{1}{32}$ .
With $h = 0.4$ : $P_{2} = 2 + \frac{1}{4} (0.4) + \frac{1}{2} (- \frac{1}{32}) (0.4)^{2}$ .
$= 2 + 0.1 - 0.0025 = 2.0975$ .
Check: $f^{'''} (x) = \frac{3}{8 x ^{5/2}} \leq \frac{3}{256}$ on $[4, 4.4]$ , so the error is at most $\frac{3}{256} \cdot \frac{( 0.4 ) ^{3}}{3 !} = 0.000125$ ; indeed $4.4 = 2.09762 \dots$ , about $0.00012$ above the approximation.

PRACTICE 61 mark · MCQ

For $f (x) = cos x - 1 + \frac{x ^{2}}{2}$ , the point $x = 0$ is

(A)
a local minimum of $f$
(B)
a local maximum of $f$
(C)
neither a local maximum nor a local minimum of $f$
(D)
not a critical point of $f$

Answer and solution

Answer: (A)

$cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , so $f (x) = \frac{x ^{4}}{24} - \frac{x ^{6}}{720} + \dots$ .
Hence $f^{'} (0) = f^{''} (0) = f^{'''} (0) = 0$ and $f^{(4)} (0) = 4! \times \frac{1}{24} = 1 > 0$ .
The first non-zero derivative has even order and is positive, so $x = 0$ is a local minimum.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE DA

Taylor series

Preparing in Kolkata? IAE Academy's course for GATE DA.