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Calculus

Limits, mean value theorems, integrals, maxima and minima, series

Calculus covers limits and continuity, the mean value theorems, Taylor series, definite and improper integrals, partial derivatives, maxima and minima of functions of one and two variables, multiple integrals and (for EE) Fourier series. These notes state every result with its exact hypotheses, because a theorem used where a hypothesis fails can give a wrong answer. They also collect the properties that shorten the work: symmetry, King's property, change of order of integration and the Jacobian.

39 previous-year questions · 38 solved · 7 practice questions

Limits, continuity and differentiability

The limit $lim_{x \to a} f (x) = L$ exists only when the left-hand limit $lim_{x \to a^{-}} f (x)$ and the right-hand limit $lim_{x \to a^{+}} f (x)$ both exist and equal the same finite $L$ . The value $f (a)$ plays no part in the limit. Throughout these notes angles are in radians, $ln$ is the natural logarithm, and $j$ is the imaginary unit, $j^{2} = - 1$ .

STANDARD LIMITS AT ZERO

x \to 0 lim \frac{sin x}{x} = 1, x \to 0 lim \frac{tan x}{x} = 1, x \to 0 lim \frac{1 - cos x}{x ^{2}} = \frac{1}{2} x \to 0 lim \frac{e ^{x} - 1}{x} = 1, x \to 0 lim \frac{a ^{x} - 1}{x} = ln a, x \to 0 lim \frac{ln ( 1 + x )}{x} = 1

$x$ in radians, $a > 0$ . Each follows from the Maclaurin series of the function.

EXPONENTIAL AND POWER LIMITS

x \to 0 lim (1 + x)^{1/ x} = e, x \to \infty lim (1 + \frac{a}{x})^{x} = e^{a}, x \to a lim \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}

The last holds for any real $n$ when $a > 0$ .

ORDERS OF GROWTH

x \to \infty lim \frac{ln x}{x ^{p}} = 0, x \to \infty lim \frac{x ^{p}}{e ^{a x}} = 0, x \to 0^{+} lim x^{p} ln x = 0

For $p > 0$ and $a > 0$ : a logarithm grows more slowly than any positive power, and a power more slowly than any growing exponential. Each follows from L'Hospital's rule. Consequence: $lim_{x \to 0^{+}} x^{x} = lim_{x \to 0^{+}} e^{x l n x} = e^{0} = 1$ .

THE ONE-TO-THE-POWER-INFINITY RULE

f (x) \to 1, ∣ g (x) ∣ \to \infty \Rightarrow lim f (x)^{g (x)} = e^{l i m g (x) [f (x) - 1]}

Valid whenever the limit in the exponent exists; if that limit is $+ \infty$ the answer is $\infty$ , and if it is $- \infty$ the answer is $0$ . It comes from $ln f \approx f - 1$ when $f \to 1$ .

L'HOSPITAL'S RULE

x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )}

Only for the forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , with $f, g$ differentiable and $g^{'} (x) \neq = 0$ near $a$ (except possibly at $a$ ), and only if $lim f^{'} (x) / g^{'} (x)$ exists (finite or infinite). Differentiate numerator and denominator separately, not by the quotient rule. Also valid for $a = \pm \infty$ and for one-sided limits.

Converting the other indeterminate forms

$0 \cdot \infty$ : write $f g = \frac{f}{1/ g}$ (or $\frac{g}{1/ f}$ ) to get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
$\infty - \infty$ : combine into a single fraction (common denominator, or rationalise a difference of roots).
$0^{0}, \infty^{0}, 1^{\infty}$ (with $f > 0$ ): put $y = f^{g}$ , so $ln y = g ln f$ is a $0 \cdot \infty$ form; find its limit $ℓ$ , then $lim y = e^{ℓ}$ .
For $x \to \infty$ with polynomials or roots, divide by the highest power first. For $x \to 0$ , substituting Maclaurin series is often faster than repeated L'Hospital.

CONTINUITY AND DIFFERENTIABILITY AT A POINT

x \to a lim f (x) = f (a), f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}

Continuity needs $f (a)$ defined, the limit to exist, and the two to be equal. Differentiability needs the left derivative ( $h \to 0^{-}$ ) and the right derivative ( $h \to 0^{+}$ ) to be finite and equal. Differentiable $\Rightarrow$ continuous; the converse is false.

Functions continuous on a closed interval [a, b]

Extreme values: a function continuous on the closed, bounded interval $[a, b]$ is bounded there and attains a greatest and a least value. On an open interval this can fail: $\frac{1}{x}$ is continuous on $(0, 1)$ but unbounded.
Intermediate value theorem: if $f$ is continuous on $[a, b]$ and $k$ lies between $f (a)$ and $f (b)$ , then $f (c) = k$ for some $c \in [a, b]$ . In particular, if $f (a)$ and $f (b)$ have opposite signs, $f$ has a root in $(a, b)$ .
Sums, products and compositions of continuous functions are continuous, and so is a quotient wherever its denominator is non-zero. Polynomials, $e^{x}$ , $sin x$ and $cos x$ are continuous for all $x$ , and $ln x$ for $x > 0$ .

Function (with $f (0) = 0$ )	Continuous at 0?	Differentiable at 0?	Reason
$∣ x ∣$	Yes	No	Left derivative $- 1$ , right derivative $+ 1$ : a corner
$x ∣ x ∣$	Yes	Yes, $f^{'} (0) = 0$	Difference quotient $∣ h ∣ \to 0$
$x^{1/3}$	Yes	No	Difference quotient $h^{- 2/3} \to \infty$ (vertical tangent)
$x sin (1/ x)$	Yes	No	Difference quotient $sin (1/ h)$ oscillates, no limit
$x^{2} sin (1/ x)$	Yes	Yes, $f^{'} (0) = 0$	But $f^{'} (x) = 2 x sin (1/ x) - cos (1/ x)$ has no limit at 0, so $f^{'}$ is not continuous there
$sgn (x)$	No	No	Jump from $- 1$ to $+ 1$

WORKED EXAMPLE

Evaluate $lim_{x \to 0} (cos x)^{1/ x^{2}}$ .

Solution

As $x \to 0$ , $cos x \to 1$ and $1/ x^{2} \to \infty$ : the form is $1^{\infty}$ .
By the rule, the limit is $e^{ℓ}$ with $ℓ = lim_{x \to 0} \frac{cos x - 1}{x ^{2}}$ .
From the standard limit $\frac{1 - cos x}{x ^{2}} \to \frac{1}{2}$ , so $ℓ = - \frac{1}{2}$ .
Check by L'Hospital on $ln y = \frac{ln cos x}{x ^{2}}$ (form $\frac{0}{0}$ ): $\frac{- tan x}{2 x} \to - \frac{1}{2}$ . Same result.

Answer: $e^{- 1/2} \approx 0.6065$

EXAM TIP

L'Hospital's rule gives nonsense when the form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ : $lim_{x \to 0} \frac{c o s x}{x + 1} = 1$ , but differentiating top and bottom gives $\frac{- s i n 0}{1} = 0$ . And when $lim f^{'} / g^{'}$ does not exist, the original limit may still exist: $lim_{x \to \infty} \frac{x + s i n x}{x} = 1$ , although $\frac{1 + c o s x}{1}$ has no limit.

Mean value theorems and Taylor series

The mean value theorems guarantee a point $c$ strictly inside $(a, b)$ with a stated property. All three share two hypotheses: continuity on the closed interval $[a, b]$ and differentiability on the open interval $(a, b)$ . Rolle's theorem adds $f (a) = f (b)$ , and Cauchy's adds $g^{'} \neq = 0$ . When a hypothesis fails the conclusion is no longer guaranteed, though it may still happen to hold, so check every hypothesis before using a theorem.

ROLLE'S THEOREM

f continuous on [a, b], differentiable on (a, b), f (a) = f (b) \Rightarrow f^{'} (c) = 0 for some c \in (a, b)

Geometrically, a horizontal tangent between two equal heights. $f (x) = ∣ x ∣$ on $[- 1, 1]$ has $f (- 1) = f (1)$ but no such $c$ , because it is not differentiable at 0.

LAGRANGE'S MEAN VALUE THEOREM

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} for some c \in (a, b)

Needs $f$ continuous on $[a, b]$ and differentiable on $(a, b)$ . The tangent at $c$ is parallel to the chord. Rolle's theorem is the case $f (a) = f (b)$ . Equivalent form: $f (b) = f (a) + (b - a) f^{'} (c)$ . Consequences on an interval: $f^{'} = 0$ throughout $\Rightarrow$ $f$ is constant; $f^{'} > 0$ $\Rightarrow$ $f$ is strictly increasing; $f^{'} < 0$ $\Rightarrow$ strictly decreasing; $∣ f^{'} ∣ \leq M$ $\Rightarrow$ $∣ f (b) - f (a) ∣ \leq M (b - a)$ .

CAUCHY'S MEAN VALUE THEOREM

\frac{f ^{'} ( c )}{g ^{'} ( c )} = \frac{f ( b ) - f ( a )}{g ( b ) - g ( a )} for some c \in (a, b)

Needs $f, g$ continuous on $[a, b]$ , differentiable on $(a, b)$ , and $g^{'} (x) \neq = 0$ on $(a, b)$ (which also forces $g (b) \neq = g (a)$ ). Lagrange's theorem is the case $g (x) = x$ .

Theorem and function on $[a, b]$	Value of $c$
Lagrange, $f = p x^{2} + q x + r$ , $p \neq = 0$	$\frac{a + b}{2}$ (arithmetic mean)
Lagrange, $f = \frac{1}{x}$ , $0 < a < b$	$ab$ (geometric mean)
Lagrange, $f = ln x$ , $0 < a < b$	$\frac{b - a}{ln ( b / a )}$
Lagrange, $f = e^{x}$	$ln \frac{e ^{b} - e ^{a}}{b - a}$
Cauchy, $f = e^{x}$ , $g = e^{- x}$	$\frac{a + b}{2}$
Cauchy, $f = x$ , $g = \frac{1}{x}$ , $0 < a < b$	$ab$
Cauchy, $f = \frac{1}{x ^{2}}$ , $g = \frac{1}{x}$ , $0 < a < b$	$\frac{2 ab}{a + b}$ (harmonic mean)

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

Maclaurin series when $a = 0$ . Needs derivatives of every order at $a$ , and the series equals $f (x)$ exactly when the remainder $R_{n} (x) \to 0$ ; for the standard functions below this holds throughout the stated range. A convergent Taylor series can still differ from $f$ : $f (x) = e^{- 1/ x^{2}}$ with $f (0) = 0$ has every derivative zero at 0, so its Maclaurin series is 0. A power series that equals $f$ near $a$ is its Taylor series, so series may be built by substitution, multiplication, and term-by-term differentiation or integration inside the interval of convergence.

LAGRANGE FORM OF THE REMAINDER

R_{n} (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} (x - a)^{n + 1}, ξ between a and x

The error after the term in $(x - a)^{n}$ , valid when $f^{(n + 1)}$ exists on an interval containing $a$ and $x$ (with $f^{(n)}$ continuous there). With $n = 0$ it is exactly Lagrange's mean value theorem.

MACLAURIN SERIES VALID FOR ALL X

e^{x} sin x cos x sinh x = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots = x + \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} + \dots, cosh x = 1 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots

Odd functions have only odd powers, even functions only even powers.

MACLAURIN SERIES WITH A RESTRICTED RANGE

ln (1 + x) \frac{1}{1 - x} (1 + x)^{k} tan^{- 1} x tan x = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots = 1 + x + x^{2} + x^{3} + \dots = 1 + k x + \frac{k ( k - 1 )}{2 !} x^{2} + \frac{k ( k - 1 ) ( k - 2 )}{3 !} x^{3} + \dots = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \dots = x + \frac{x ^{3}}{3} + \frac{2 x ^{5}}{15} + \dots - 1 < x \leq 1 ∣ x ∣ < 1 ∣ x ∣ < 1 ∣ x ∣ \leq 1 ∣ x ∣ < \frac{π}{2}

For a non-negative integer $k$ the binomial series stops after $k + 1$ terms and holds for all $x$ . Replacing $x$ by $- x$ gives $ln (1 - x) = - x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3} - \dots$ for $- 1 \leq x < 1$ , and $\frac{1}{1 + x} = 1 - x + x^{2} - \dots$ for $∣ x ∣ < 1$ .

RADIUS OF CONVERGENCE OF A POWER SERIES

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

When this limit exists (it may be $\infty$ ). The series converges absolutely for $∣ x - a ∣ < R$ and diverges for $∣ x - a ∣ > R$ ; each end point $x = a \pm R$ must be tested separately. Example: $\sum_{n \geq 1} \frac{x ^{n}}{n}$ has $R = lim \frac{n + 1}{n} = 1$ ; it diverges at $x = 1$ (harmonic series) and converges at $x = - 1$ , so the interval is $- 1 \leq x < 1$ , the range of $- ln (1 - x)$ , which this series equals.

WORKED EXAMPLE

Find the value of $c$ given by Lagrange's mean value theorem for $f (x) = x^{3} - 3 x$ on $[0, 2]$ .

Solution

$f$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$ : the theorem applies.
Chord slope: $\frac{f ( 2 ) - f ( 0 )}{2 - 0} = \frac{( 8 - 6 ) - 0}{2} = 1$ .
$f^{'} (c) = 3 c^{2} - 3 = 1 \Rightarrow c^{2} = \frac{4}{3} \Rightarrow c = \pm \frac{2}{3}$ .
Only a root inside $(0, 2)$ counts: $c = \frac{2}{3} \approx 1.155$ .

Answer: $c = 2/ 3 \approx 1.155$

WORKED EXAMPLE

Find the coefficient of $x^{4}$ in the Maclaurin series of $e^{x} cos x$ .

Solution

Multiply the series $e^{x} = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \frac{x ^{4}}{24} + \dots$ and $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ .
Products giving $x^{4}$ : $1 \cdot \frac{x ^{4}}{24}$ , $\frac{x ^{2}}{2} \cdot (- \frac{x ^{2}}{2})$ and $\frac{x ^{4}}{24} \cdot 1$ . The terms $x$ and $\frac{x ^{3}}{6}$ would need odd powers from $cos x$ , which has none.
Coefficient: $\frac{1}{24} - \frac{1}{4} + \frac{1}{24} = - \frac{1}{6}$ .
Check with $j$ : $e^{x} cos x = Re e^{(1 + j) x}$ , so the coefficient of $x^{n}$ is $Re (1 + j)^{n} / n!$ . Here $(1 + j)^{4} = (2 j)^{2} = - 4$ , giving $- 4/24 = - 1/6$ .

Answer: $- \frac{1}{6}$

EXAM TIP

For a limit such as $lim_{x \to 0} \frac{x - s i n x}{x ^{3}}$ , substitute the series: $x - sin x = \frac{x ^{3}}{6} - \frac{x ^{5}}{120} + \dots$ , so the limit is $\frac{1}{6}$ at once, where L'Hospital needs three rounds. Keep terms up to the power in the denominator.

EXAM TIP

In a mean value question, reject any root $c$ that is not strictly inside $(a, b)$ , and check the hypotheses before solving: a function with a corner or a vertical tangent inside the interval (such as $∣ x - 1∣$ on an interval containing 1, or $x^{2/3}$ on one containing 0) does not satisfy them.

Definite and improper integrals

FUNDAMENTAL THEOREM OF CALCULUS

\int_{a}^{b} f (x) d x = F (b) - F (a) where F^{'} = f, \frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)

Needs $f$ continuous on $[a, b]$ . Evaluating an antiderivative across a point where $f$ is unbounded is not justified and can give a wrong answer (see the tip at the end of this section): split the integral there and treat each piece as improper.

INTEGRATION BY PARTS

\int_{a}^{b} u v^{'} d x = [u v]_{a}^{b} - \int_{a}^{b} u^{'} v d x \int e^{a x} sin b x d x = \frac{e ^{a x} ( a sin b x - b cos b x )}{a ^{2} + b ^{2}} + C \int e^{a x} cos b x d x = \frac{e ^{a x} ( a cos b x + b sin b x )}{a ^{2} + b ^{2}} + C

$u$ and $v$ with continuous derivatives on $[a, b]$ . Choose as $u$ the factor that simplifies when differentiated (a power of $x$ , a logarithm, an inverse trigonometric function). The last two lines come from integrating by parts twice; $a^{2} + b^{2} \neq = 0$ and $C$ is an arbitrary constant.

MEAN VALUE THEOREM FOR INTEGRALS

\int_{a}^{b} f (x) d x = f (c) (b - a) for some c \in [a, b]

Needs $f$ continuous on $[a, b]$ . $f (c)$ is the average value of $f$ over the interval.

KING'S PROPERTY

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

Substitute $x \to a + b - x$ . Adding the two forms often cancels the awkward part: $2 I = \int_{a}^{b} [f (x) + f (a + b - x)] d x$ .

EVEN AND ODD INTEGRANDS

\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x (f even), \int_{- a}^{a} f (x) d x = 0 (f odd)

Even: $f (- x) = f (x)$ ; odd: $f (- x) = - f (x)$ . Any $f$ is its even part $\frac{f ( x ) + f ( - x )}{2}$ plus its odd part $\frac{f ( x ) - f ( - x )}{2}$ , and only the even part survives a symmetric integral.

REFLECTION ABOUT THE MIDPOINT

\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x if f (2 a - x) = f (x); \int_{0}^{2 a} f (x) d x = 0 if f (2 a - x) = - f (x)

Example: $\int_{0}^{π} sin^{n} x d x = 2 \int_{0}^{π /2} sin^{n} x d x$ , and $\int_{0}^{π} cos^{n} x d x = 0$ for odd $n$ .

PERIODIC INTEGRANDS

\int_{0}^{n T} f (x) d x = n \int_{0}^{T} f (x) d x, \int_{a}^{a + T} f (x) d x = \int_{0}^{T} f (x) d x

For $f (x + T) = f (x)$ and a positive integer $n$ : every full period gives the same integral. Example: $∣ sin x ∣$ has period $π$ , so $\int_{0}^{10 π} ∣ sin x ∣ d x = 10 \int_{0}^{π} sin x d x = 20$ .

REDUCTION FORMULA FOR POWERS OF SINE AND COSINE

\int_{0}^{π /2} sin^{n} x d x = \int_{0}^{π /2} cos^{n} x d x = \frac{( n - 1 ) ( n - 3 ) \dots}{n ( n - 2 ) \dots} \times {\frac{π}{2} 1 n even n odd

$n$ a positive integer. Each product keeps only its positive factors, so it runs down to 1 or 2; a product with no factors is 1, so $n = 1$ gives 1. Examples: $n = 4$ gives $\frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{π}{2} = \frac{3 π}{16}$ ; $n = 5$ gives $\frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}$ .

PRODUCTS OF POWERS OF SINE AND COSINE

\int_{0}^{π /2} sin^{m} x cos^{n} x d x = \frac{[ ( m - 1 ) ( m - 3 ) \dots ] [ ( n - 1 ) ( n - 3 ) \dots ]}{( m + n ) ( m + n - 2 ) \dots} \times {\frac{π}{2} 1 m and n both even otherwise

$m, n$ non-negative integers. Each product keeps only its positive factors, running down to 1 or 2, and a product with no factors (for $m$ or $n$ equal to 0 or 1) is 1. With $m = 0$ it is the formula above. Examples: $m = n = 2$ gives $\frac{1 \cdot 1}{4 \cdot 2} \cdot \frac{π}{2} = \frac{π}{16}$ ; $m = 3, n = 2$ gives $\frac{2 \cdot 1}{5 \cdot 3 \cdot 1} = \frac{2}{15}$ . The Beta function below gives the same values.

LEIBNIZ RULE

\frac{d}{d x} \int_{a (x)}^{b (x)} f (x, t) d t = f (x, b (x)) b^{'} (x) - f (x, a (x)) a^{'} (x) + \int_{a (x)}^{b (x)} \frac{\partial f}{\partial x} (x, t) d t

Needs $f$ and $\partial f / \partial x$ continuous and $a (x), b (x)$ differentiable. If the integrand does not contain $x$ the last term is zero; if the limits are constants only the last term remains (differentiation under the integral sign).

An integral is improper when a limit of integration is infinite (first kind) or the integrand is unbounded at a point of $[a, b]$ (second kind). It is defined as a limit, for example $\int_{a}^{\infty} f d x = lim_{R \to \infty} \int_{a}^{R} f d x$ , and it converges only if that limit is finite. An integral over $(- \infty, \infty)$ , or across an interior point where $f$ blows up, must converge on each piece separately: $\int_{- \infty}^{\infty} x d x$ diverges even though $\int_{- R}^{R} x d x = 0$ for every $R$ (that symmetric value is only the Cauchy principal value).

P-TEST FOR IMPROPER INTEGRALS

\int_{1}^{\infty} \frac{d x}{x ^{p}} converges ⟺ p > 1 (value \frac{1}{p - 1}), \int_{0}^{1} \frac{d x}{x ^{p}} converges ⟺ p < 1 (value \frac{1}{1 - p})

Both diverge at $p = 1$ (logarithm). For non-negative $f$ and $g$ : comparison test, if $0 \leq f \leq g$ then $\int g$ convergent $\Rightarrow$ $\int f$ convergent, and $\int f$ divergent $\Rightarrow$ $\int g$ divergent; limit comparison, if $f / g$ tends to a finite non-zero limit, both converge or both diverge. For any $f$ : if $\int ∣ f ∣$ converges, so does $\int f$ (absolute convergence).

STANDARD IMPROPER INTEGRALS

\int_{0}^{\infty} e^{- a x} d x = \frac{1}{a}, \int_{0}^{\infty} x^{n} e^{- a x} d x = \frac{n !}{a ^{n + 1}}, \int_{- \infty}^{\infty} e^{- a x^{2}} d x = \frac{π}{a} \int_{0}^{\infty} e^{- a x} cos b x d x = \frac{a}{a ^{2} + b ^{2}}, \int_{0}^{\infty} e^{- a x} sin b x d x = \frac{b}{a ^{2} + b ^{2}}, \int_{0}^{\infty} \frac{sin x}{x} d x = \frac{π}{2}

$a > 0$ , $b$ real and $n$ a non-negative integer; for real $n > - 1$ the second result is $\frac{Γ ( n + 1 )}{a ^{n + 1}}$ . The last integral converges, but not absolutely.

GAMMA FUNCTION

Γ (n) = \int_{0}^{\infty} x^{n - 1} e^{- x} d x (n > 0), Γ (n + 1) = n Γ (n), Γ (\frac{1}{2}) = π

$Γ (n + 1) = n!$ for non-negative integers $n$ , so $Γ (1) = 1$ . Example: $Γ (\frac{5}{2}) = \frac{3}{2} \cdot \frac{1}{2} π = \frac{3 π}{4}$ .

BETA FUNCTION

B (m, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x = \frac{Γ ( m ) Γ ( n )}{Γ ( m + n )} = 2 \int_{0}^{π /2} sin^{2 m - 1} θ cos^{2 n - 1} θ d θ

$m, n > 0$ and $B (m, n) = B (n, m)$ . The trigonometric form gives $\int_{0}^{π /2} sin^{p} θ cos^{q} θ d θ = \frac{1}{2} B (\frac{p + 1}{2}, \frac{q + 1}{2})$ for $p, q > - 1$ .

AREA, ARC LENGTH AND VOLUME

A L V = \int_{a}^{b} ∣ f (x) - g (x) ∣ d x, A = \frac{1}{2} \int_{α}^{β} r^{2} d θ = \int_{a}^{b} 1 + (\frac{d y}{d x})^{2} d x = \int_{t_{1}}^{t_{2}} (\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} d t = \int_{α}^{β} r^{2} + (\frac{d r}{d θ})^{2} d θ = π \int_{a}^{b} y^{2} d x, V_{y} = 2 π \int_{a}^{b} x f (x) d x

Area between two curves and area swept by a polar curve $r (θ)$ ; arc length for $y = f (x)$ , for a parametric curve, and for a polar curve. $V$ is the volume of the solid formed by revolving the region under $y = f (x)$ , $a \leq x \leq b$ , about the $x$ -axis; $V_{y}$ (shell method, for $f \geq 0$ and $0 \leq a < b$ ) is the volume when the same region is revolved about the $y$ -axis.

WORKED EXAMPLE

Evaluate $I = \int_{2}^{5} \frac{x}{x + 7 - x} d x$ .

Solution

Here $a + b = 7$ , so replace $x$ by $7 - x$ : $I = \int_{2}^{5} \frac{7 - x}{7 - x + x} d x$ .
Add the two forms: $2 I = \int_{2}^{5} \frac{x + 7 - x}{x + 7 - x} d x = \int_{2}^{5} 1 d x = 3$ .
Hence $I = \frac{3}{2}$ .

Answer: $I = 1.5$

WORKED EXAMPLE

If $F (x) = \int_{x}^{x^{2}} ln t d t$ for $x > 1$ , find $F^{'} (2)$ .

Solution

The integrand does not contain $x$ , so only the two boundary terms of the Leibniz rule remain.
$F^{'} (x) = ln (x^{2}) \cdot \frac{d}{d x} (x^{2}) - ln x \cdot \frac{d}{d x} (x) = 2 ln x \cdot 2 x - ln x = (4 x - 1) ln x$ .
$F^{'} (2) = 7 ln 2 \approx 4.852$ .

Answer: $F^{'} (2) = 7 ln 2 \approx 4.852$

EXAM TIP

$\int_{- 1}^{1} \frac{d x}{x ^{2}}$ is not $[- \frac{1}{x}]_{- 1}^{1} = - 2$ ; a negative value for a positive integrand is the warning sign. The integrand is unbounded at $x = 0$ and $\int_{0}^{1} x^{- 2} d x$ diverges ( $p = 2 \geq 1$ ), so the whole integral diverges.

Several variables: limits, partial derivatives and Euler's theorem

For a function of two variables, $lim_{(x, y) \to (a, b)} f (x, y) = L$ means that $f (x, y) \to L$ however $(x, y)$ approaches $(a, b)$ : along every path, not only along straight lines. Two paths that give different values prove that the limit does not exist; agreement along any number of paths proves nothing. To show that a limit at the origin exists, put $x = r cos θ$ , $y = r sin θ$ and bound $∣ f - L ∣$ by a function of $r$ alone that tends to 0. $f$ is continuous at $(a, b)$ when this limit exists and equals $f (a, b)$ .

WORKED EXAMPLE

Decide whether each limit exists: (i) $lim_{(x, y) \to (0, 0)} \frac{x ^{2} y}{x ^{2} + y ^{2}}$ ; (ii) $lim_{(x, y) \to (0, 0)} \frac{x y}{x ^{2} + y ^{2}}$ .

Solution

(i) In polar form the function is $\frac{r ^{3} cos ^{2} θ sin θ}{r ^{2}} = r cos^{2} θ sin θ$ , and $∣ r cos^{2} θ sin θ ∣ \leq r \to 0$ whatever $θ$ does. The limit is $0$ .
(ii) Along the line $y = m x$ with $x \neq = 0$ : $\frac{m x ^{2}}{x ^{2} + m ^{2} x ^{2}} = \frac{m}{1 + m ^{2}}$ . This is $0$ along the $x$ -axis ( $m = 0$ ) and $\frac{1}{2}$ along $y = x$ ( $m = 1$ ).
Two paths give two different values, so the limit in (ii) does not exist.

Answer: (i) exists and equals $0$ ; (ii) does not exist.

EXAM TIP

Straight lines are not enough. For $\frac{x ^{2} y}{x ^{4} + y ^{2}}$ , every line $y = m x$ gives $\frac{m x}{x ^{2} + m ^{2}} \to 0$ and both axes give 0, but the parabola $y = x^{2}$ gives $\frac{x ^{4}}{2 x ^{4}} = \frac{1}{2}$ , so there is no limit at the origin. Also, unlike the one-variable case, the existence of $f_{x}$ and $f_{y}$ at a point does not make $f$ continuous there: $\frac{x y}{x ^{2} + y ^{2}}$ with $f (0, 0) = 0$ has $f_{x} (0, 0) = f_{y} (0, 0) = 0$ but no limit at the origin.

For $z = f (x, y)$ the partial derivative $f_{x} = \partial f / \partial x$ is the ordinary derivative in $x$ with $y$ held constant. Second derivatives are written $f_{xx}$ , $f_{y y}$ and the mixed $f_{x y} = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x})$ . If $f_{x y}$ and $f_{y x}$ are continuous near a point, they are equal there, so the order of differentiation does not matter for the functions met in practice.

TOTAL DIFFERENTIAL AND CHAIN RULE

d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y, \frac{d z}{d t} = \frac{\partial z}{\partial x} \frac{d x}{d t} + \frac{\partial z}{\partial y} \frac{d y}{d t}

For $z = f (x, y)$ with $x = x (t)$ , $y = y (t)$ . For $x = x (u, v)$ , $y = y (u, v)$ : $z_{u} = z_{x} x_{u} + z_{y} y_{u}$ and $z_{v} = z_{x} x_{v} + z_{y} y_{v}$ . The total differential also estimates small changes: $Δ z \approx z_{x} Δ x + z_{y} Δ y$ .

IMPLICIT DIFFERENTIATION

f (x, y) = 0 \Rightarrow \frac{d y}{d x} = - \frac{f _{x}}{f _{y}}

Valid where $f_{y} \neq = 0$ . Note the minus sign.

TAYLOR SERIES IN TWO VARIABLES

f (a + h, b + k) = f (a, b) + (h f_{x} + k f_{y}) + \frac{1}{2 !} (h^{2} f_{xx} + 2 hk f_{x y} + k^{2} f_{y y}) + \dots

All derivatives are evaluated at $(a, b)$ , and $f$ needs continuous partial derivatives of every order used. Stopping after the first bracket gives the linear approximation $f (a + h, b + k) \approx f (a, b) + h f_{x} + k f_{y}$ , the total differential again. At a stationary point the first bracket is zero and the second decides the nature of the point, which is where the $r t - s^{2}$ test comes from.

EULER'S THEOREM ON HOMOGENEOUS FUNCTIONS

x f_{x} + y f_{y} = n f, x^{2} f_{xx} + 2 x y f_{x y} + y^{2} f_{y y} = n (n - 1) f

For $f$ differentiable and homogeneous of degree $n$ , meaning $f (t x, t y) = t^{n} f (x, y)$ for all $t > 0$ (continuous second partial derivatives are needed for the second identity). $n$ may be negative or fractional. With more variables: $\sum_{i} x_{i} \partial f / \partial x_{i} = n f$ .

EULER'S THEOREM FOR A FUNCTION OF A HOMOGENEOUS FUNCTION

F (u) = f (x, y) homogeneous of degree n \Rightarrow x u_{x} + y u_{y} = n \frac{F ( u )}{F ^{'} ( u )}

Use when $u$ itself is not homogeneous but some $F (u)$ is; needs $F^{'} (u) \neq = 0$ . Writing $g (u) = n \frac{F ( u )}{F ^{'} ( u )}$ , the second-order form is $x^{2} u_{xx} + 2 x y u_{x y} + y^{2} u_{y y} = g (u) [g^{'} (u) - 1]$ . Example: for $u = sin^{- 1} \frac{x ^{2} + y ^{2}}{x + y}$ , $sin u$ is homogeneous of degree 1, so $x u_{x} + y u_{y} = \frac{s i n u}{c o s u} = tan u$ , and the second-order sum is $tan u (sec^{2} u - 1) = tan^{3} u$ .

WORKED EXAMPLE

For $u = ln \frac{x ^{4} + y ^{4}}{x + y}$ with $x, y > 0$ , find $x u_{x} + y u_{y}$ .

Solution

$u$ is not homogeneous, but $e^{u} = \frac{x ^{4} + y ^{4}}{x + y}$ is homogeneous of degree $4 - 1 = 3$ .
With $F (u) = e^{u}$ : $x u_{x} + y u_{y} = 3 \frac{F ( u )}{F ^{'} ( u )} = 3 \frac{e ^{u}}{e ^{u}} = 3$ .
Direct check: $u = ln (x^{4} + y^{4}) - ln (x + y)$ , so $x u_{x} + y u_{y} = \frac{4 x ^{4} + 4 y ^{4}}{x ^{4} + y ^{4}} - \frac{x + y}{x + y} = 4 - 1 = 3$ .

Answer: $3$

EXAM TIP

Euler's theorem needs genuine homogeneity. Test by replacing $x, y$ with $t x, t y$ and factoring out $t^{n}$ : $x^{2} + y$ fails the test, and indeed $x f_{x} + y f_{y} = 2 x^{2} + y$ is not a constant multiple of $f$ . $\frac{1}{x ^{2} + y ^{2}}$ passes with degree $- 1$ .

Maxima and minima

One variable: procedure

Find the critical points: interior points where $f^{'} (x) = 0$ and interior points where $f^{'}$ does not exist.
Second derivative test at a stationary point $c$ ( $f^{'} (c) = 0$ ): $f^{''} (c) < 0$ gives a local maximum, $f^{''} (c) > 0$ a local minimum, and $f^{''} (c) = 0$ is inconclusive.
If inconclusive, find the first non-zero derivative $f^{(k)} (c)$ . For even $k$ there is an extremum (maximum if $f^{(k)} (c) < 0$ , minimum if $f^{(k)} (c) > 0$ ); for odd $k$ there is none. Examples: $x^{3}$ at 0 ( $k = 3$ , no extremum), $x^{4}$ at 0 ( $k = 4$ , minimum).
Where $f^{'}$ does not exist (such as $∣ x ∣$ at 0), use the sign change of $f^{'}$ : from $+$ to $-$ is a maximum, from $-$ to $+$ a minimum.
A point of inflection is where $f^{''}$ changes sign (the curve changes between concave up and concave down). $f^{''} (c) = 0$ alone is not enough: $x^{4}$ has $f^{''} (0) = 0$ , but $f^{''} = 12 x^{2} \geq 0$ on both sides and $x = 0$ is a minimum.
Absolute extrema of a continuous $f$ on a closed interval $[a, b]$ : compare $f$ at every critical point inside and at both end points; the largest value is the maximum and the smallest the minimum.

TWO VARIABLES: SECOND DERIVATIVE TEST

r = f_{xx}, s = f_{x y}, t = f_{y y}, D = r t - s^{2} at a point where f_{x} = f_{y} = 0

Classify with the table below; $f$ needs continuous second partial derivatives near the point. When $D > 0$ , $r$ and $t$ have the same sign, so either may be used.

Condition at a stationary point	Nature
$r t - s^{2} > 0$ and $r < 0$	Local maximum
$r t - s^{2} > 0$ and $r > 0$	Local minimum
$r t - s^{2} < 0$	Saddle point (neither maximum nor minimum)
$r t - s^{2} = 0$	Test fails: examine $f$ near the point directly

LAGRANGE MULTIPLIERS

\nabla f = λ \nabla g, g = 0 ⟺ f_{x} = λ g_{x}, f_{y} = λ g_{y}, f_{z} = λ g_{z}, g (x, y, z) = 0

Extremise $f (x, y, z)$ subject to the constraint $g (x, y, z) = 0$ , at points where $\nabla g \neq = 0$ . Solve the four equations for $x, y, z, λ$ , evaluate $f$ at each candidate and compare. The method finds candidates; it does not by itself say which is a maximum.

WORKED EXAMPLE

Find and classify the stationary points of $f (x, y) = x^{3} + y^{3} - 6 x y$ .

Solution

$f_{x} = 3 x^{2} - 6 y = 0 \Rightarrow y = \frac{x ^{2}}{2}$ ; $f_{y} = 3 y^{2} - 6 x = 0 \Rightarrow x = \frac{y ^{2}}{2}$ .
Substitute: $x = \frac{1}{2} (\frac{x ^{2}}{2})^{2} = \frac{x ^{4}}{8} \Rightarrow x (x^{3} - 8) = 0 \Rightarrow x = 0$ or $x = 2$ . Stationary points: $(0, 0)$ and $(2, 2)$ .
$r = 6 x$ , $s = - 6$ , $t = 6 y$ .
At $(0, 0)$ : $r t - s^{2} = 0 - 36 < 0$ , a saddle point.
At $(2, 2)$ : $r t - s^{2} = 144 - 36 = 108 > 0$ and $r = 12 > 0$ , a local minimum with $f (2, 2) = 8 + 8 - 24 = - 8$ .

Answer: Saddle point at $(0, 0)$ ; local minimum $- 8$ at $(2, 2)$ ; no local maximum.

WORKED EXAMPLE

Find the minimum of $x^{2} + y^{2} + z^{2}$ subject to $x + 2 y + 2 z = 9$ .

Solution

$f = x^{2} + y^{2} + z^{2}$ , $g = x + 2 y + 2 z - 9$ . Conditions: $2 x = λ$ , $2 y = 2 λ$ , $2 z = 2 λ$ .
So $y = z = λ = 2 x$ . The constraint gives $x + 4 x + 4 x = 9 \Rightarrow x = 1$ , $y = z = 2$ .
$f (1, 2, 2) = 1 + 4 + 4 = 9$ . On the plane $f \to \infty$ as $∣ (x, y, z) ∣ \to \infty$ , so $f$ attains a least value there (and has no greatest); that least value must occur at a candidate, and $(1, 2, 2)$ is the only one.
Check: $f$ is the squared distance from the origin to the plane, $(\frac{9}{1 + 4 + 4})^{2} = 9$ .

Answer: Minimum value $9$ , at $(1, 2, 2)$ .

EXAM TIP

A stationary point need not be an extremum, and a local extremum need not be absolute: $x^{3} + y^{3} - 6 x y$ has a local minimum $- 8$ but no absolute minimum, since $f (x, 0) = x^{3} \to - \infty$ as $x \to - \infty$ . On a closed, bounded region, also check the boundary and the corners.

Multiple integrals

A double integral $\iint_{R} f (x, y) d A$ is evaluated as an iterated integral: the inner integral is taken first with the outer variable held constant, and its limits may depend on the outer variable; the outer limits must be constants. For a continuous $f$ on a bounded region the order of integration may be changed, but the limits must be re-derived from the region, never simply swapped. A triple integral $∭_{E} f d V$ is handled the same way, one variable at a time.

Changing the order of integration

Read the region from the given limits: the inner limits are the bounding curves, the outer limits the extreme values of the outer variable.
Sketch the region and mark where its boundary curves intersect.
For the new order, fix the new outer variable and draw a strip across the region in the new inner direction: the curves where the strip enters and leaves are the new inner limits.
The new outer limits are the least and greatest values of the new outer variable over the region. If the entry or exit curve changes part-way, split the region into pieces.

AREA AND VOLUME

A = \iint_{R} d A, V = \iint_{R} f (x, y) d A (f \geq 0), V = ∭_{E} d V

Area of a plane region $R$ , volume under the surface $z = f (x, y)$ over $R$ , and volume of a solid $E$ . Replacing 1 by a density gives mass.

CHANGE OF VARIABLES AND THE JACOBIAN

\iint_{R} f (x, y) d x d y = \iint_{R^{'}} f (x (u, v), y (u, v)) \frac{\partial ( x , y )}{\partial ( u , v )} d u d v, \frac{\partial ( x , y )}{\partial ( u , v )} = x_{u} y_{u} x_{v} y_{v}

Valid when the map $(u, v) \mapsto (x, y)$ takes $R^{'}$ one-to-one onto $R$ (except possibly on the boundary), with continuous partial derivatives and a non-zero Jacobian inside. The absolute value of the Jacobian is the area scale factor. The Jacobians of a transformation and its inverse multiply to 1: $\frac{\partial ( x , y )}{\partial ( u , v )} \cdot \frac{\partial ( u , v )}{\partial ( x , y )} = 1$ .

System	Transformation	Element	Ranges
Polar (plane)	$x = r cos θ, y = r sin θ$	$d A = r d r d θ$	$r \geq 0, 0 \leq θ < 2 π$
Cylindrical	$x = r cos θ, y = r sin θ, z = z$	$d V = r d r d θ d z$	$r \geq 0, 0 \leq θ < 2 π$
Spherical	$x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ$	$d V = ρ^{2} sin ϕ d ρ d ϕ d θ$	$ρ \geq 0, 0 \leq ϕ \leq π, 0 \leq θ < 2 π$
Scaled polar (ellipse)	$x = a r cos θ, y = b r sin θ$	$d A = ab r d r d θ$	$0 \leq r \leq 1$ covers $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} \leq 1$

EXAM TIP

Convention used here: in spherical coordinates $ϕ$ is the angle from the positive $z$ -axis ( $0 \leq ϕ \leq π$ ) and $θ$ is the angle in the $x y$ -plane from the positive $x$ -axis. Some texts swap the two letters; the volume element is $ρ^{2} sin (angle from the z -axis)$ either way. In polar work do not drop the factor $r$ in $d A = r d r d θ$ .

STANDARD RESULTS

\int_{- \infty}^{\infty} e^{- x^{2}} d x = π, ellipse area = π ab, sphere volume = \frac{4}{3} π a^{3}, tetrahedron \frac{x}{a} + \frac{y}{b} + \frac{z}{c} \leq 1, x, y, z \geq 0 : V = \frac{ab c}{6}

The first follows by squaring the integral and switching to polar coordinates: $(\int_{- \infty}^{\infty} e^{- x^{2}} d x)^{2} = \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = π$ (with $a, b, c > 0$ in the others).

WORKED EXAMPLE

Evaluate $\int_{0}^{2} \int_{x /2}^{1} e^{y^{2}} d y d x$ .

Solution

$e^{y^{2}}$ has no elementary antiderivative, so change the order.
Region: $0 \leq x \leq 2$ , $\frac{x}{2} \leq y \leq 1$ , the triangle with vertices $(0, 0)$ , $(0, 1)$ and $(2, 1)$ .
For fixed $y \in [0, 1]$ , $x$ runs from $0$ to $2 y$ : $I = \int_{0}^{1} \int_{0}^{2 y} e^{y^{2}} d x d y = \int_{0}^{1} 2 y e^{y^{2}} d y$ .
With $u = y^{2}$ : $I = [e^{y^{2}}]_{0}^{1} = e - 1$ .

Answer: $e - 1 \approx 1.718$

WORKED EXAMPLE

Evaluate $\iint_{R} e^{- (x^{2} + y^{2})} d A$ over the disc $R : x^{2} + y^{2} \leq 4$ .

Solution

Polar coordinates: $x^{2} + y^{2} = r^{2}$ , $d A = r d r d θ$ , $0 \leq r \leq 2$ , $0 \leq θ \leq 2 π$ .
$I = \int_{0}^{2 π} \int_{0}^{2} e^{- r^{2}} r d r d θ = 2 π [- \frac{1}{2} e^{- r^{2}}]_{0}^{2}$ .
$I = 2 π \cdot \frac{1}{2} (1 - e^{- 4}) = π (1 - e^{- 4})$ .

Answer: $π (1 - e^{- 4}) \approx 3.084$

Fourier series (listed by EE)

A function $f$ given on $(- L, L)$ and extended periodically with period $2 L$ can be written as a sum of cosines and sines. Convention used here: the constant term is written $\frac{a _{0}}{2}$ , so one formula gives both $a_{0}$ and $a_{n}$ , and $\frac{a _{0}}{2}$ is the average value of $f$ over a period. Some texts write the constant term as $a_{0}$ with $a_{0} = \frac{1}{2 L} \int_{- L}^{L} f d x$ ; the series itself is the same.

FOURIER SERIES ON (-L, L)

f (x) = \frac{a _{0}}{2} + n = 1 \sum \infty [a_{n} cos \frac{nπ x}{L} + b_{n} sin \frac{nπ x}{L}]

Period $2 L$ . For period $2 π$ put $L = π$ , giving $a_{n} cos n x + b_{n} sin n x$ .

FOURIER COEFFICIENTS

a_{n} = \frac{1}{L} \int_{- L}^{L} f (x) cos \frac{nπ x}{L} d x (n = 0, 1, 2, \dots), b_{n} = \frac{1}{L} \int_{- L}^{L} f (x) sin \frac{nπ x}{L} d x (n = 1, 2, \dots)

Any interval of length $2 L$ , such as $(0, 2 L)$ , may be used. They follow from orthogonality: over one period, $cos \frac{mπ x}{L} cos \frac{nπ x}{L}$ and $sin \frac{mπ x}{L} sin \frac{nπ x}{L}$ integrate to $L$ if $m = n \geq 1$ and to $0$ if $m \neq = n$ , and every sine-cosine product integrates to $0$ .

Dirichlet conditions and convergence

$f$ is periodic with period $2 L$ and absolutely integrable over a period.
In one period $f$ has finitely many discontinuities, all finite jumps, and finitely many maxima and minima.
Then the series converges to $f (x)$ wherever $f$ is continuous, and to the midpoint $\frac{f ( x ^{+} ) + f ( x ^{-} )}{2}$ at a jump. This includes the end points: at $x = \pm L$ it converges to $\frac{f ( - L ^{+} ) + f ( L ^{-} )}{2}$ .

Symmetry of $f$	Series contains	Coefficients
Even, $f (- x) = f (x)$	Constant and cosines only	$b_{n} = 0$ , $a_{n} = \frac{2}{L} \int_{0}^{L} f (x) cos \frac{nπ x}{L} d x$
Odd, $f (- x) = - f (x)$	Sines only	$a_{n} = 0$ (including $a_{0}$ ), $b_{n} = \frac{2}{L} \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x$
Half-wave, $f (x + L) = - f (x)$	Odd harmonics only	$a_{0} = 0$ , and $a_{n} = b_{n} = 0$ for even $n$

HALF-RANGE SERIES ON (0, L)

cosine: a_{n} = \frac{2}{L} \int_{0}^{L} f (x) cos \frac{nπ x}{L} d x, sine: b_{n} = \frac{2}{L} \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x

A function given only on $(0, L)$ is extended evenly (cosine series) or oddly (sine series), each with period $2 L$ . Both converge to $f$ at every point of $(0, L)$ where $f$ is continuous. They differ outside $(0, L)$ and can differ at its end points: the sine series is $0$ at $x = 0$ and $x = L$ , while the cosine series converges to $f (0^{+})$ at $x = 0$ and to $f (L^{-})$ at $x = L$ .

PARSEVAL'S IDENTITY

\frac{1}{L} \int_{- L}^{L} f^{2} (x) d x = \frac{a _{0}^{2}}{2} + n = 1 \sum \infty (a_{n}^{2} + b_{n}^{2})

Written for the $\frac{a _{0}}{2}$ convention; holds whenever $\int f^{2}$ over a period is finite, in particular under the Dirichlet conditions above. Halving both sides: mean square value $= (\frac{a _{0}}{2})^{2} + \frac{1}{2} \sum_{n \geq 1} (a_{n}^{2} + b_{n}^{2})$ .

COMPLEX EXPONENTIAL FORM

f (x) = n = - \infty \sum \infty c_{n} e^{j nπ x / L}, c_{n} = \frac{1}{2 L} \int_{- L}^{L} f (x) e^{- j nπ x / L} d x

$j = - 1$ . $c_{0} = \frac{a _{0}}{2}$ , $c_{n} = \frac{a _{n} - j b _{n}}{2}$ for $n \geq 1$ , and for real $f$ , $c_{- n} = \overline{c_{n}}$ (the complex conjugate).

WORKED EXAMPLE

Find the Fourier series of $f (x) = x^{2}$ on $(- 1, 1)$ , extended with period 2, and use it to find $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}}$ .

Solution

$L = 1$ and $f$ is even, so $b_{n} = 0$ .
$a_{0} = \frac{2}{1} \int_{0}^{1} x^{2} d x = \frac{2}{3}$ , so the constant term is $\frac{a _{0}}{2} = \frac{1}{3}$ .
$a_{n} = 2 \int_{0}^{1} x^{2} cos nπ x d x$ . Integrating by parts twice (the $sin nπ$ terms vanish): $\int_{0}^{1} x^{2} cos nπ x d x = \frac{2 c o s nπ}{n ^{2} π ^{2}} = \frac{2 ( - 1 ) ^{n}}{n ^{2} π ^{2}}$ , so $a_{n} = \frac{4 ( - 1 ) ^{n}}{n ^{2} π ^{2}}$ .
Series: $x^{2} = \frac{1}{3} + \frac{4}{π ^{2}} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n ^{2}} cos nπ x$ for $- 1 \leq x \leq 1$ . The periodic extension is continuous, so equality holds at the end points too.
Put $x = 1$ , where $cos nπ = (- 1)^{n}$ : $1 = \frac{1}{3} + \frac{4}{π ^{2}} \sum_{n = 1}^{\infty} \frac{1}{n ^{2}}$ , so $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ .

Answer: $\sum_{n = 1}^{\infty} 1/ n^{2} = π^{2} /6 \approx 1.645$

EXAM TIP

For a piecewise-smooth $f$ , the decay of the coefficients shows how smooth the periodic extension is: with a jump they fall off like $\frac{1}{n}$ ; if it is continuous but has corners (jumps in $f^{'}$ ), like $\frac{1}{n ^{2}}$ . Near a jump the partial sums overshoot by about 9% of the jump however many terms are taken (Gibbs phenomenon), while at the jump itself the series converges to the midpoint.

EXAM TIP

Before integrating, check symmetry about the origin and half-wave symmetry: they can remove whole families of coefficients. Remember $cos nπ = (- 1)^{n}$ and $sin nπ = 0$ for integer $n$ .

PREVIOUS-YEAR QUESTIONS

39 questions from the EC, EE and IN papers

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

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GATE 2026 EC Q.36

GATE 2026 · ECQ.36 · 2 marks · MCQ

Consider the two series, $S_{A}$ and $S_{B}$ , where

S_{A} = n = 1 \sum \infty \frac{n ^{2}}{2 ^{n}}

S_{B} = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \dots

Which of the following statements is correct for the two given series?

(A)
Both $S_{A}$ and $S_{B}$ converge.
(B)
Neither $S_{A}$ nor $S_{B}$ converges.
(C)
$S_{A}$ converges but $S_{B}$ does not converge.
(D)
$S_{B}$ converges but $S_{A}$ does not converge.

Answer and solution

GATE 2026 EC Q.36: answer and solution

See the question in the paper

Calculus

Limits, continuity and differentiability

Mean value theorems and Taylor series

Definite and improper integrals

Several variables: limits, partial derivatives and Euler's theorem

Maxima and minima

Multiple integrals

Fourier series (listed by EE)

39 questions from the EC, EE and IN papers

7 questions to test yourself