Real analysis starts by looking carefully at the real number system R. R is an ordered field that also has the completeness property: every non-empty set bounded above has a least upper bound. From this follow the Archimedean property and the density of rationals. These ideas are the base for sequences, series and integration.
Key points
R is an ordered field: the usual rules of +, × hold, and for a < b we have a + c < b + c, and ac < bc when c > 0.
Absolute value: |x| = x if x ≥ 0, −x if x < 0. Triangle inequality: |x + y| ≤ |x| + |y|; also ||x| − |y|| ≤ |x − y|.
A set S is bounded above if some M satisfies x ≤ M for all x ∈ S (M is an upper bound). Bounded below similarly. Bounded = both.
Supremum (lub): the least upper bound of S. Infimum (glb): the greatest lower bound. They may or may not belong to S.
Characterisation: u = sup S iff (i) x ≤ u for all x ∈ S and (ii) for every ε > 0 there is x ∈ S with x > u − ε.
Completeness axiom: every non-empty subset of R that is bounded above has a supremum in R. Q does not have this property (the set {x ∈ Q : x² < 2} has no rational supremum).
Archimedean property: for any x ∈ R there is n ∈ N with n > x. Equivalently, for any ε > 0 there is n with 1/n < ε.
Density: between any two real numbers there is a rational and also an irrational number.
A neighbourhood of a is any set containing an open interval (a − ε, a + ε).
Important terms
Upper bound (उच्च सीमा) – a number not less than any element of the set.
Supremum – the smallest upper bound.
Maximum – a supremum that belongs to the set.
Ordered field – a field with an order compatible with + and ×.
Formulas
|xy| = |x||y|; |x − a| < ε ⇔ a − ε < x < a + ε
inf S = −sup(−S)
sup(A + B) = sup A + sup B
5 important questions
Q1. State the completeness property of R.
Every non-empty subset of R which is bounded above has a least upper bound (supremum) in R. Equivalently, every non-empty set bounded below has a greatest lower bound.
Q2. State the Archimedean property.
For every real number x there is a natural number n such that n > x. As a result, for every ε > 0 there is n ∈ N with 1/n < ε.
Q3. Find sup and inf of S = {1 − 1/n : n ∈ N}.
Elements: 0, 1/2, 2/3, 3/4, … increasing.
Smallest element is 0 (n = 1), so inf S = 0 = min S.
Every element is less than 1, so 1 is an upper bound.
For any ε > 0, choose n > 1/ε (Archimedean); then 1 − 1/n > 1 − ε. So no number below 1 is an upper bound.
sup S = 1, which is not attained (1 ∉ S).
Q4. Find sup and inf of A = {(−1)ⁿ/n : n ∈ N}.
Elements: −1, 1/2, −1/3, 1/4, −1/5, …
Positive terms 1/2, 1/4, … decrease, so the largest element is 1/2.
Negative terms −1, −1/3, … increase, so the smallest element is −1.
sup A = 1/2 (= max) and inf A = −1 (= min).
Q5. Prove that √2 is not rational.
Suppose √2 = p/q with p, q integers having no common factor. Then p² = 2q², so p² is even and hence p is even, say p = 2k. Then 4k² = 2q², so q² = 2k² and q is also even. Then p and q have the common factor 2, a contradiction. Hence √2 is irrational.
Exam tip
To prove u = sup S, always show two things: u is an upper bound, and u − ε is not an upper bound for any ε > 0. Use the Archimedean property for the second part.
A sequence is an ordered list of real numbers a₁, a₂, a₃, … following a rule. Real analysis studies when such a list approaches a fixed number (converges). This topic covers the ε–N definition of limit, bounded and monotonic sequences, the monotone convergence theorem, limit theorems, subsequences and Cauchy sequences.
Key points
A sequence is a function from natural numbers N to real numbers R, written (aₙ).
Convergence: (aₙ) → l if for every ε > 0 there is a natural number N such that |aₙ − l| < ε for all n ≥ N.
The limit of a convergent sequence is unique.
A sequence that does not converge either diverges (to +∞ or −∞) or oscillates (e.g. (−1)ⁿ).
Bounded: there is M with |aₙ| ≤ M for all n. Every convergent sequence is bounded, but a bounded sequence need not converge ((−1)ⁿ).
Monotonic: increasing (aₙ₊₁ ≥ aₙ) or decreasing (aₙ₊₁ ≤ aₙ).
Monotone convergence theorem: a bounded monotonic sequence converges. An increasing sequence bounded above converges to its supremum.
Algebra of limits: if aₙ → a and bₙ → b, then aₙ ± bₙ → a ± b, aₙbₙ → ab, aₙ/bₙ → a/b (b ≠ 0).
Sandwich (squeeze) theorem: if aₙ ≤ bₙ ≤ cₙ and aₙ, cₙ → l, then bₙ → l.
Subsequence: a sequence picked from (aₙ) keeping order. Every subsequence of a convergent sequence has the same limit.
Bolzano–Weierstrass theorem: every bounded sequence has a convergent subsequence.
Cauchy sequence: |aₘ − aₙ| < ε for all m, n ≥ N. In R, a sequence converges if and only if it is Cauchy (Cauchy's general principle of convergence).
Standard limits: 1/n → 0; rⁿ → 0 if |r| < 1; n^(1/n) → 1; (1 + 1/n)ⁿ → e.
Important terms
Supremum (least upper bound) – smallest number ≥ every term.
Infimum – greatest number ≤ every term.
Sequence (अनुक्रम) – ordered list indexed by natural numbers.
Completeness of R – every non-empty set bounded above has a supremum.
Q1. Define a convergent sequence. Give one example of a bounded sequence that does not converge.
(aₙ) converges to l if for each ε > 0 there is N with |aₙ − l| < ε for every n ≥ N. Example: aₙ = (−1)ⁿ is bounded by 1 but oscillates between −1 and 1, so it does not converge.
Q2. State the monotone convergence theorem.
Every monotonic sequence that is bounded converges. An increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum.
Q3. Using the definition, prove that (2n + 1)/(n + 3) → 2.
Bounded and increasing, so it converges to some l.
Take limits in aₙ₊₁² = 2 + aₙ: l² = 2 + l → l² − l − 2 = 0 → (l − 2)(l + 1) = 0.
Since all terms are positive, l = 2.
Exam tip
In ε–N proofs, first simplify |aₙ − l| to a simple expression in n, then solve for n. For recursive sequences, prove "bounded" and "monotonic" separately by induction before finding the limit.
An infinite series is the sum a₁ + a₂ + a₃ + … of the terms of a sequence. It converges if its partial sums approach a finite limit. This topic covers the basic results and the standard tests for convergence of positive-term series, and alternating series with absolute and conditional convergence.
Key points
Partial sum Sₙ = a₁ + a₂ + … + aₙ. Σaₙ converges to S if Sₙ → S.
Necessary condition: if Σaₙ converges then aₙ → 0. The converse is false: Σ1/n diverges though 1/n → 0. So if aₙ does not tend to 0, the series diverges (nth term test).
Geometric series Σ rⁿ⁻¹ converges to 1/(1 − r) if |r| < 1 and diverges if |r| ≥ 1.
p-series Σ1/nᵖ converges if p > 1 and diverges if p ≤ 1 (harmonic series p = 1 diverges).
Comparison test: if 0 ≤ aₙ ≤ bₙ and Σbₙ converges, Σaₙ converges; if Σaₙ diverges, Σbₙ diverges.
Limit comparison test: if aₙ/bₙ → l (finite, non-zero), both series behave alike.
D'Alembert's ratio test: if aₙ₊₁/aₙ → l, then l < 1 converges, l > 1 diverges, l = 1 test fails.
Cauchy's root test: if (aₙ)^(1/n) → l, same conclusions.
Raabe's test is used when the ratio test fails: if n(aₙ/aₙ₊₁ − 1) → l, l > 1 converges, l < 1 diverges.
Cauchy's integral test: for positive decreasing f, Σf(n) and ∫₁^∞ f(x) dx converge or diverge together.
Leibniz test (alternating series Σ(−1)ⁿ⁺¹uₙ): converges if uₙ is decreasing and uₙ → 0.
Absolute convergence: Σ|aₙ| converges. Absolutely convergent ⇒ convergent. A series that converges but not absolutely is conditionally convergent (e.g. 1 − 1/2 + 1/3 − …).
Important terms
Series (श्रेणी) – sum of the terms of a sequence.
Positive-term series – all terms ≥ 0; its partial sums increase.
Alternating series – terms alternately positive and negative.
Formulas
Geometric: a + ar + ar² + … = a/(1 − r), |r| < 1
Ratio test: l = lim aₙ₊₁/aₙ
Root test: l = lim (aₙ)^(1/n)
Raabe: l = lim n(aₙ/aₙ₊₁ − 1)
5 important questions
Q1. Show by example that aₙ → 0 is not sufficient for convergence.
For the harmonic series Σ1/n, the terms 1/n → 0, but the series diverges (it is a p-series with p = 1). So aₙ → 0 is necessary but not sufficient.
Q2. Test the convergence of Σ1/(n² + 1).
1/(n² + 1) < 1/n² for all n. Σ1/n² converges (p = 2 > 1). By the comparison test, Σ1/(n² + 1) converges.
Q3. Test the convergence of Σ n/2ⁿ.
aₙ = n/2ⁿ, aₙ₊₁ = (n + 1)/2ⁿ⁺¹.
aₙ₊₁/aₙ = (n + 1)/(2n) = ½(1 + 1/n).
Limit as n → ∞: l = ½.
l < 1, so by D'Alembert's ratio test the series converges.
Q5. Show that 1 − 1/2 + 1/3 − 1/4 + … is conditionally convergent.
It is alternating with uₙ = 1/n.
uₙ₊₁ = 1/(n + 1) < 1/n = uₙ, so uₙ is decreasing.
uₙ → 0. By the Leibniz test, the series converges.
Σ|aₙ| = Σ1/n is the harmonic series, which diverges.
So the series converges but not absolutely: conditionally convergent.
Exam tip
Choose the test by the form: factorials and powers like 2ⁿ → ratio test; nth powers like (…)ⁿ → root test; rational functions of n → comparison with a p-series. If the ratio test gives 1, move to Raabe's test.
A group is a set with one operation that satisfies closure, associativity, identity and inverse. Groups capture the idea of symmetry and appear across mathematics and physics. This topic covers definitions and examples, abelian and cyclic groups, order of an element, subgroups and their tests, cosets and Lagrange's theorem.
Key points
A **group (G, *) satisfies: (1) closure**: a * b ∈ G; (2) associativity: (a * b) * c = a * (b * c); (3) identity e: a * e = e * a = a; (4) inverse: for each a there is a⁻¹ with a * a⁻¹ = e.
Abelian (commutative) group: a * b = b * a for all a, b.
Examples: (Z, +), (Q, +), (R − {0}, ×), {1, −1, i, −i} under multiplication, Zₙ under addition mod n.
Not groups: (N, +) (no inverses), (Z, ×) (2 has no inverse).
In a group, the identity is unique, each element has a unique inverse, (ab)⁻¹ = b⁻¹a⁻¹, and cancellation laws hold.
Order of a group |G| = number of elements. Order of an element a: the least positive n with aⁿ = e.
Cyclic group: G = ⟨a⟩, every element is a power of a generator a. Every cyclic group is abelian.
Subgroup: a subset H that is itself a group under the same operation.
One-step subgroup test: a non-empty H is a subgroup if a, b ∈ H ⇒ ab⁻¹ ∈ H. For a finite H, closure alone is enough.
Intersection of two subgroups is a subgroup; their union need not be.
Cosets: aH = {ah : h ∈ H}. Any two left cosets are equal or disjoint, and all have |H| elements.
Lagrange's theorem: for a finite group, the order of a subgroup divides the order of the group.
Corollaries: order of each element divides |G|; a group of prime order is cyclic.
Normal subgroup H: gH = Hg for all g ∈ G.
Important terms
Binary operation – rule combining two elements to give one element of the set.
Generator – element whose powers give the whole cyclic group.
Index [G : H] – number of distinct cosets of H in G.
Group (समूह) – set with an operation satisfying the four axioms.
Formulas
|G| = [G : H] × |H| (Lagrange)
(ab)⁻¹ = b⁻¹a⁻¹; (a⁻¹)⁻¹ = a
In a cyclic group of order n, aᵏ is a generator iff gcd(k, n) = 1
5 important questions
Q1. Define an abelian group with an example.
A group (G, ) in which a b = b * a for all a, b ∈ G. Example: (Z, +) — integers under addition — is abelian.
Q2. Find the order of each element of G = {1, −1, i, −i} under multiplication.
o(1) = 1; o(−1) = 2 since (−1)² = 1; o(i) = 4 since i² = −1, i³ = −i, i⁴ = 1; similarly o(−i) = 4. G is cyclic with generators i and −i.
Q3. Prove that if every element of a group is its own inverse, the group is abelian.
Given a = a⁻¹ for all a ∈ G.
For a, b ∈ G, ab ∈ G, so ab = (ab)⁻¹.
(ab)⁻¹ = b⁻¹a⁻¹ = ba (since b⁻¹ = b, a⁻¹ = a).
So ab = ba for all a, b. Hence G is abelian.
Q4. Show that the intersection of two subgroups H and K of G is a subgroup.
e ∈ H and e ∈ K, so H ∩ K is non-empty.
Let a, b ∈ H ∩ K. Then a, b ∈ H and a, b ∈ K.
H is a subgroup ⇒ ab⁻¹ ∈ H; K is a subgroup ⇒ ab⁻¹ ∈ K.
So ab⁻¹ ∈ H ∩ K.
By the one-step test, H ∩ K is a subgroup.
Q5. State and prove Lagrange's theorem.
Statement: if H is a subgroup of a finite group G, then |H| divides |G|.
The left cosets of H partition G (any two are equal or disjoint, and every g lies in gH).
Each coset aH has exactly |H| elements, since h → ah is one-one.
If there are k distinct cosets, then |G| = k|H|.
Hence |H| divides |G|. Example: Z₆ can have subgroups only of orders 1, 2, 3, 6.
Exam tip
When checking whether a set is a group, test all four axioms in order and write each line. In proofs, quote the result you use (cancellation, (ab)⁻¹ = b⁻¹a⁻¹, one-step test) by name.
A permutation of a set is a one-one onto mapping of the set to itself. All permutations of n symbols form the symmetric group Sₙ of order n!. We write permutations in two-row or cycle form, multiply them, find their order and decide whether they are even or odd. The even permutations form the alternating group Aₙ.
Key points
Sₙ: the group of all permutations of {1, 2, …, n} under composition; |Sₙ| = n!. Sₙ is non-abelian for n ≥ 3.
Product: we compose from right to left: (αβ)(x) = α(β(x)). (Some books use left to right; state your convention.)
Cycle (a₁ a₂ … aᵣ): a₁ → a₂ → … → aᵣ → a₁, other symbols fixed. A cycle of length 2 is a transposition.
Every permutation is a product of disjoint cycles, uniquely apart from order. Disjoint cycles commute.
Order of a permutation = lcm of the lengths of its disjoint cycles. A cycle of length r has order r.
Every cycle is a product of transpositions: (a₁ a₂ … aᵣ) = (a₁ aᵣ)(a₁ aᵣ₋₁)…(a₁ a₂), i.e. r − 1 transpositions.
A permutation is even or odd according as it is a product of an even or odd number of transpositions; this parity is fixed.
Aₙ, the set of even permutations, is a subgroup of Sₙ of order n!/2 (n ≥ 2).
Cayley's theorem: every group is isomorphic to a group of permutations.
Important terms
Permutation (क्रमचय) – a bijection of a set onto itself.
Disjoint cycles – cycles with no symbol in common.
Identity permutation – fixes every symbol.
Formulas
|Sₙ| = n!, |Aₙ| = n!/2
Order of σ = lcm(r₁, r₂, …) for disjoint cycle lengths rᵢ
A cycle of length r is even if r is odd, and odd if r is even
(a₁ a₂ … aᵣ)⁻¹ = (aᵣ … a₂ a₁)
5 important questions
Q1. What is the order of S₄ and of A₄?
|S₄| = 4! = 24 and |A₄| = 24/2 = 12.
Q2. Is the cycle (1 2 3 4) even or odd?
(1 2 3 4) = (1 4)(1 3)(1 2), a product of 3 transpositions, so it is odd.
Q3. Write σ as disjoint cycles, and find its order, parity and inverse.
Aₙ is non-empty since the identity is even. If α, β are even, β⁻¹ is also even (reverse the same transpositions), and αβ⁻¹ is a product of an even number of transpositions, so it is even. By the subgroup test, Aₙ is a subgroup of Sₙ.
Exam tip
Always state the multiplication convention (right to left) at the start. For order questions, first break into disjoint cycles — the lcm rule works only for disjoint cycles.
A normal subgroup is a subgroup whose left and right cosets are the same. Only for normal subgroups can we multiply cosets and get a new group, called the quotient (factor) group G/H. This topic covers the definition and tests for normality, standard examples, and the construction and order of quotient groups.
Key points
A subgroup H of G is normal (H ⊴ G) if gH = Hg for every g ∈ G.
Equivalent test: gHg⁻¹ ⊆ H for all g ∈ G, i.e. ghg⁻¹ ∈ H for all g ∈ G, h ∈ H.
In an abelian group every subgroup is normal.
{e} and G are always normal (trivial normal subgroups). A group with no other normal subgroup is called simple.
Every subgroup of index 2 is normal (the two left cosets and the two right cosets must both be H and G − H).
The centre Z(G) = {z : zg = gz for all g} is a normal subgroup.
The intersection of two normal subgroups is normal.
Quotient group G/H = {aH : a ∈ G} with product (aH)(bH) = (ab)H. This is well-defined only when H is normal.
Identity of G/H is H itself; inverse of aH is a⁻¹H; |G/H| = |G|/|H| = [G : H].
If G is abelian (or cyclic), G/H is abelian (or cyclic).
Important terms
Normal subgroup (सामान्य उपसमूह) – a subgroup invariant under conjugation.
Conjugate of h by g – the element ghg⁻¹.
Factor group – another name for the quotient group.
Formulas
H ⊴ G ⇔ gHg⁻¹ = H for all g
|G/H| = |G|/|H|
(aH)(bH) = abH, (aH)⁻¹ = a⁻¹H
5 important questions
Q1. Define a normal subgroup and give an example.
A subgroup H of G is normal if gH = Hg (equivalently gHg⁻¹ ⊆ H) for all g ∈ G. Example: A₃ = {e, (1 2 3), (1 3 2)} is normal in S₃ because it has index 2.
Q2. Prove that every subgroup of an abelian group is normal.
If G is abelian, then for g ∈ G and h ∈ H, ghg⁻¹ = hgg⁻¹ = h ∈ H. So gHg⁻¹ ⊆ H for all g, and H is normal.
Q3. Show that H = {e, (1 2)} is not normal in S₃.
Take g = (1 3), so g⁻¹ = (1 3).
Compute g(1 2)g⁻¹ = (1 3)(1 2)(1 3), working right to left.
1 → 3 → 3 → 1; 2 → 2 → 1 → 3; 3 → 1 → 2 → 2. So the result is (2 3).
(2 3) ∉ H, so gHg⁻¹ ⊄ H. Hence H is not normal.
Q4. Let G = Z₁₂ (addition mod 12) and H = ⟨4⟩. List G/H and find the order of 2 + H.
H = {0, 4, 8}, so |G/H| = 12/3 = 4.
Cosets: H = {0, 4, 8}, 1 + H = {1, 5, 9}, 2 + H = {2, 6, 10}, 3 + H = {3, 7, 11}.
(2 + H) + (2 + H) = 4 + H = H, the identity.
So the order of 2 + H is 2. (G/H is cyclic of order 4, generated by 1 + H.)
Q5. Prove that a subgroup of index 2 is normal.
Let [G : H] = 2. The left cosets are H and G − H; the right cosets are also H and G − H. If g ∈ H then gH = H = Hg. If g ∉ H then gH = G − H = Hg. So gH = Hg for every g, and H is normal.
Exam tip
To prove normality, the shortest route is often "abelian group" or "index 2". To disprove it, find one g and one h with ghg⁻¹ outside H and show the calculation clearly.
A homomorphism is a map between two groups that respects the group operation. Its kernel is a normal subgroup, and the fundamental (first isomorphism) theorem says G/ker f is isomorphic to the image of f. An isomorphism is a one-one, onto homomorphism, meaning two groups have exactly the same structure.
Key points
f : G → G′ is a homomorphism if f(ab) = f(a)f(b) for all a, b ∈ G (using each group's own operation).
Properties: f(e) = e′, f(a⁻¹) = f(a)⁻¹, f(aⁿ) = f(a)ⁿ, and the order of f(a) divides the order of a.
Kernel: ker f = {a ∈ G : f(a) = e′}. It is a normal subgroup of G.
Image f(G) is a subgroup of G′.
f is one-one (a monomorphism) iff ker f = {e}.
An isomorphism is a homomorphism that is one-one and onto. We write G ≅ G′.
Isomorphic groups have the same order, the same number of elements of each order, and both are abelian (or cyclic) or neither.
First isomorphism (fundamental) theorem: if f : G → G′ is an onto homomorphism, then G/ker f ≅ G′.
Every cyclic group of order n is isomorphic to Zₙ, and every infinite cyclic group to (Z, +).
An isomorphism of G onto itself is an automorphism.
Important terms
Homomorphism (समाकारिता) – an operation-preserving map.
Natural homomorphism – the map G → G/H, a ↦ aH; its kernel is H.
Endomorphism – a homomorphism of G into itself.
Formulas
f(ab) = f(a)f(b); f(e) = e′; f(a⁻¹) = f(a)⁻¹
G/ker f ≅ f(G)
|G| = |ker f| × |f(G)| for finite G
5 important questions
Q1. Define the kernel of a homomorphism. Is it a subgroup?
ker f = {a ∈ G : f(a) = e′}. Yes: it contains e, and if a, b ∈ ker f then f(ab⁻¹) = f(a)f(b)⁻¹ = e′, so ab⁻¹ ∈ ker f. It is in fact a normal subgroup.
Q2. Show that f : (R, +) → (R⁺, ×), f(x) = eˣ, is an isomorphism.
Homomorphism: f(x + y) = e^(x+y) = eˣ·eʸ = f(x)f(y). One-one: eˣ = eʸ ⇒ x = y. Onto: any t > 0 equals f(log t). So f is an isomorphism and (R, +) ≅ (R⁺, ×).
Q3. Prove that f is one-one iff ker f = {e}.
If f is one-one and a ∈ ker f, then f(a) = e′ = f(e), so a = e.
Conversely, let ker f = {e} and f(a) = f(b).
Then f(ab⁻¹) = f(a)f(b)⁻¹ = e′, so ab⁻¹ ∈ ker f = {e}.
Hence ab⁻¹ = e, i.e. a = b. So f is one-one.
Q4. Use the fundamental theorem to show Z/nZ ≅ Zₙ.
Define f : Z → Zₙ by f(a) = a mod n (the remainder of a on division by n).
f(a + b) = (a + b) mod n = f(a) +ₙ f(b), so f is a homomorphism.
f is onto, since f(r) = r for r = 0, 1, …, n − 1.
ker f = {a : a mod n = 0} = nZ.
By the fundamental theorem, Z/nZ ≅ Zₙ.
Q5. Show that the determinant map GL(2, R) → R* is a homomorphism and find its kernel.
GL(2, R) = invertible 2 × 2 real matrices under multiplication; R* = non-zero reals under multiplication.
det(AB) = det A × det B, so det is a homomorphism. It is onto: diag(t, 1) has determinant t.
Kernel = {A : det A = 1} = SL(2, R), the special linear group.
Hence SL(2, R) is normal in GL(2, R), and GL(2, R)/SL(2, R) ≅ R*.
Exam tip
For any isomorphism question, write three headings: homomorphism, one-one, onto — and prove each in one or two lines. To show two groups are NOT isomorphic, find a property one has and the other lacks (order, abelian, number of elements of order 2).
A ring is a set with two operations, addition and multiplication, that behave like those of the integers. A field is a ring in which every non-zero element has a multiplicative inverse, like the rational or real numbers. This topic covers rings, integral domains, fields, subrings, ideals and ring homomorphisms.
Key points
Ring (R, +, ·): (R, +) is an abelian group; multiplication is closed and associative; distributive laws hold: a(b + c) = ab + ac and (a + b)c = ac + bc.
Commutative ring: ab = ba. Ring with unity: has 1 with 1·a = a·1 = a.
Examples: Z, Q, R, C, Zₙ (mod n), the set of 2 × 2 real matrices (non-commutative), 2Z (no unity).
Zero divisors: non-zero a, b with ab = 0. Example: in Z₆, 2·3 = 0.
Integral domain: a commutative ring with unity (1 ≠ 0) and no zero divisors. Example: Z.
In an integral domain, cancellation holds: ab = ac and a ≠ 0 ⇒ b = c.
Field: a commutative ring with unity in which every non-zero element has a multiplicative inverse. Examples: Q, R, C, Zₚ (p prime).
Every field is an integral domain. Every finite integral domain is a field.
Zₙ is a field if and only if n is prime.
Subring: a non-empty S with a − b ∈ S and ab ∈ S for all a, b ∈ S.
Ideal I: a subring with ra ∈ I and ar ∈ I for every r ∈ R, a ∈ I. Example: nZ is an ideal of Z.
Ring homomorphism f: f(a + b) = f(a) + f(b), f(ab) = f(a)f(b). Its kernel is an ideal.
Characteristic of a ring: least positive n with n·a = 0 for all a (0 if none). The characteristic of an integral domain is 0 or a prime.
Important terms
Unit – element with a multiplicative inverse.
Quotient ring R/I – ring of cosets of an ideal I.
Division ring (skew field) – every non-zero element has an inverse, but multiplication may not commute.
Field (क्षेत्र) – commutative division ring.
Formulas
a·0 = 0; (−a)(−b) = ab
Kernel: ker f = {a ∈ R : f(a) = 0}
Subring test: a − b ∈ S and ab ∈ S
Ideal test: a − b ∈ I and ra, ar ∈ I
5 important questions
Q1. Define an integral domain. Is Z₆ an integral domain?
An integral domain is a commutative ring with unity and no zero divisors. Z₆ is not, because 2 ≠ 0 and 3 ≠ 0 but 2·3 = 6 ≡ 0 (mod 6).
Q2. Prove that in any ring, a·0 = 0.
a·0 = a(0 + 0) = a·0 + a·0. Adding the additive inverse of a·0 to both sides gives 0 = a·0.
Q3. Prove that every field is an integral domain.
A field F is commutative with unity; we only need: no zero divisors.
Suppose ab = 0 with a ≠ 0.
Since F is a field, a⁻¹ exists.
b = 1·b = (a⁻¹a)b = a⁻¹(ab) = a⁻¹·0 = 0.
So ab = 0 ⇒ a = 0 or b = 0. F has no zero divisors and is an integral domain.
Q4. Show that Z₅ is a field by finding the inverse of each non-zero element.
Z₅ = {0, 1, 2, 3, 4} is a commutative ring with unity 1 under addition and multiplication mod 5.
1 × 1 = 1, so 1⁻¹ = 1.
2 × 3 = 6 ≡ 1, so 2⁻¹ = 3 and 3⁻¹ = 2.
4 × 4 = 16 ≡ 1, so 4⁻¹ = 4.
Every non-zero element has an inverse, so Z₅ is a field. (In general Zₚ is a field for prime p.)
Q5. Show that the kernel of a ring homomorphism f: R → R' is an ideal of R.
f(0) = 0, so 0 ∈ ker f; it is non-empty.
If a, b ∈ ker f: f(a − b) = f(a) − f(b) = 0 − 0 = 0, so a − b ∈ ker f.
If a ∈ ker f and r ∈ R: f(ra) = f(r)f(a) = f(r)·0 = 0, so ra ∈ ker f.
Similarly f(ar) = 0, so ar ∈ ker f.
Hence ker f is an ideal of R.
Exam tip
Learn the chain: field ⇒ integral domain ⇒ commutative ring with unity, and the counterexamples (Z is an integral domain but not a field; Z₆ has zero divisors). Writing tables for Zₙ makes inverse questions quick.
Ideals play the same role in rings that normal subgroups play in groups. An ideal lets us form a quotient ring R/I. We study ideals, principal, prime and maximal ideals, ring homomorphisms and their kernels, and the key results connecting prime ideals with integral domains and maximal ideals with fields.
Key points
A non-empty subset I of a ring R is an ideal if (i) a − b ∈ I for all a, b ∈ I, and (ii) ra ∈ I and ar ∈ I for all r ∈ R, a ∈ I.
Every ring has the trivial ideals {0} and R. A field has only these two ideals.
Principal ideal ⟨a⟩: the smallest ideal containing a. In Z, every ideal is principal: nZ.
Quotient ring R/I = {a + I} with (a + I) + (b + I) = (a + b) + I and (a + I)(b + I) = ab + I.
A ring homomorphism f : R → R′ satisfies f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b).
Kernel of a ring homomorphism is an ideal, and R/ker f ≅ f(R) (fundamental theorem).
Prime ideal P ≠ R: ab ∈ P ⇒ a ∈ P or b ∈ P.
Maximal ideal M ≠ R: there is no ideal strictly between M and R.
For a commutative ring R with unity: R/P is an integral domain iff P is prime; R/M is a field iff M is maximal. So every maximal ideal is prime.
In Z: nZ is prime iff n is prime (or n = 0), and for a prime p, pZ is maximal and Z/pZ = Zₚ is a field.
Important terms
Ideal (गुणजावली) – a subring that absorbs multiplication by ring elements.
Integral domain – commutative ring with unity and no zero divisors.
Principal ideal domain (PID) – an integral domain in which every ideal is principal, e.g. Z.
Formulas
I ideal ⇔ a − b ∈ I and ra, ar ∈ I
R/ker f ≅ f(R)
Z/nZ ≅ Zₙ; Zₙ is a field ⇔ n is prime
5 important questions
Q1. Define an ideal and give an example.
A non-empty subset I of a ring R closed under subtraction and such that ra, ar ∈ I for all r ∈ R and a ∈ I. Example: 3Z = {…, −3, 0, 3, 6, …} is an ideal of Z.
Q2. Show that 6Z is not a prime ideal of Z.
2 × 3 = 6 ∈ 6Z, but neither 2 nor 3 belongs to 6Z. So 6Z is not prime.
Q3. Prove that a field has no ideals other than {0} and itself.
Let I be an ideal of a field F with I ≠ {0}. Take a ≠ 0 in I.
Since F is a field, a⁻¹ ∈ F, so a⁻¹a = 1 ∈ I.
Then for every x ∈ F, x = x·1 ∈ I. So I = F.
Q4. Prove that the kernel of a ring homomorphism is an ideal.
Let f : R → R′ be a homomorphism and K = ker f. K is non-empty since f(0) = 0′.
If a, b ∈ K: f(a − b) = f(a) − f(b) = 0′, so a − b ∈ K.
If r ∈ R and a ∈ K: f(ra) = f(r)f(a) = f(r)·0′ = 0′, so ra ∈ K; similarly ar ∈ K.
Hence K is an ideal of R.
Q5. In Z₁₂, find the ideal ⟨4⟩ and the quotient ring Z₁₂/⟨4⟩.
(2 + ⟨4⟩)(2 + ⟨4⟩) = 4 + ⟨4⟩ = ⟨4⟩, the zero element, while 2 + ⟨4⟩ is not zero.
So the quotient has a zero divisor and is not an integral domain; it behaves like Z₄. Hence ⟨4⟩ is not a prime ideal of Z₁₂.
Exam tip
Remember the pair of key theorems: prime ideal ↔ integral domain, maximal ideal ↔ field (for commutative rings with unity). Use Z and nZ for quick examples and counter-examples.
Linear Differential Equations with Constant Coefficients
Quick summary
These are equations like a₀ dⁿy/dxⁿ + a₁ dⁿ⁻¹y/dxⁿ⁻¹ + … + aₙy = X, where a₀, a₁, … are constants. Using the operator D = d/dx we write them as f(D)y = X. The complete solution is the complementary function (CF) plus a particular integral (PI). This topic gives the rules for both.
Key points
D notation: D = d/dx, D² = d²/dx², and 1/D means integration.
Complete solution: y = CF + PI.
CF is the solution of f(D)y = 0. Write the auxiliary equation (AE) f(m) = 0 and find its roots.
Roots real and distinct m₁, m₂: CF = c₁e^(m₁x) + c₂e^(m₂x).
Roots real and equal m, m: CF = (c₁ + c₂x)e^(mx).
Complex roots α ± iβ: CF = e^(αx)(c₁ cos βx + c₂ sin βx).
PI = (1/f(D)) X, found by short-cut rules depending on X.
X = e^(ax): PI = e^(ax)/f(a) if f(a) ≠ 0. If f(a) = 0, multiply by x and differentiate f: PI = x e^(ax)/f'(a).
X = sin ax or cos ax: replace D² by −a² (if the denominator is not zero).
X = xᵐ: write 1/f(D) as [1 ± φ(D)]⁻¹, expand by the binomial series up to Dᵐ.
X = e^(ax)V: PI = e^(ax) (1/f(D + a)) V (shifting rule).
Cauchy–Euler equation x²y'' + axy' + by = X becomes one with constant coefficients by putting x = eᶻ.
Important terms
Complementary function (पूरक फल) – general solution of the homogeneous equation; has n arbitrary constants.
Particular integral – any one solution of the full equation; no arbitrary constants.
Auxiliary equation – f(m) = 0 obtained by replacing D by m.
Formulas
1/f(D) e^(ax) = e^(ax)/f(a), f(a) ≠ 0
1/(D² + a²) sin ax = −(x/(2a)) cos ax; 1/(D² + a²) cos ax = (x/(2a)) sin ax
AE: m² + 4 = 0 → m = ±2i. CF = c₁ cos 2x + c₂ sin 2x.
PI for sin x: replace D² by −1: sin x/(−1 + 4) = (sin x)/3.
PI for sin 2x: D² → −4 makes the denominator zero (failure case). Use 1/(D² + a²) sin ax = −(x/(2a)) cos ax with a = 2: −(x/4) cos 2x.
y = c₁ cos 2x + c₂ sin 2x + (sin x)/3 − (x cos 2x)/4.
Q5. Solve (D² + 1)y = x².
AE: m² + 1 = 0 → m = ±i. CF = c₁ cos x + c₂ sin x.
PI = (1 + D²)⁻¹ x² = (1 − D² + D⁴ − …) x².
D²(x²) = 2, D⁴(x²) = 0. So PI = x² − 2.
y = c₁ cos x + c₂ sin x + x² − 2.
Check: y_PI'' + y_PI = 2 + x² − 2 = x² ✓.
Exam tip
Always write CF first, then PI, then y = CF + PI. Watch for the failure case (denominator zero) — it is asked almost every year and needs the "multiply by x" rule.
Cauchy–Euler Equations and Variation of Parameters
Quick summary
Some linear differential equations have variable coefficients. The Cauchy–Euler (homogeneous linear) equation, with terms like x²y″ and xy′, becomes an equation with constant coefficients when we put x = eᶻ. The method of variation of parameters gives a particular integral for any right-hand side once the complementary function is known.
Key points
Cauchy–Euler equation: x²y″ + a₁xy′ + a₂y = X(x), where a₁, a₂ are constants.
Put x = eᶻ (z = log x) and D = d/dz. Then xy′ = Dy, x²y″ = D(D − 1)y, x³y‴ = D(D − 1)(D − 2)y.
The equation becomes linear with constant coefficients in z; solve as usual and then put z = log x, eᶻ = x.
Legendre's linear equation: (a + bx)²y″ + a₁(a + bx)y′ + a₂y = X. Put a + bx = eᶻ; then (a + bx)y′ = bDy and (a + bx)²y″ = b²D(D − 1)y.
Variation of parameters for y″ + P(x)y′ + Q(x)y = R(x): if y₁, y₂ are independent solutions of the homogeneous equation, take y = u y₁ + v y₂.
Wronskian W = y₁y₂′ − y₂y₁′ (non-zero for independent solutions).
u = −∫ y₂R/W dx and v = ∫ y₁R/W dx. The PI is u y₁ + v y₂.
Variation of parameters works even for R(x) = sec x, tan x, cosec x, where the short operator methods fail.
Important terms
Complementary function (CF) – general solution of the equation with right side 0.
Particular integral (PI) – any one solution of the full equation.
Wronskian (रॉन्स्कियन) – the determinant used to test independence of solutions.
Formulas
x = eᶻ: x dy/dx = Dy, x² d²y/dx² = D(D − 1)y
W = y₁y₂′ − y₂y₁′
PI = −y₁∫(y₂R/W) dx + y₂∫(y₁R/W) dx
5 important questions
Q1. What substitution reduces a Cauchy–Euler equation to constant coefficients?
Put x = eᶻ, i.e. z = log x, and write D = d/dz. Then x dy/dx = Dy and x² d²y/dx² = D(D − 1)y.
Q2. Define the Wronskian of y₁ and y₂.
W(y₁, y₂) = y₁y₂′ − y₂y₁′. If W ≠ 0, the two solutions are linearly independent. For y₁ = cos x, y₂ = sin x, W = cos²x + sin²x = 1.
Q3. Solve x²y″ − 3xy′ + 4y = 0.
Put x = eᶻ: D(D − 1)y − 3Dy + 4y = 0, i.e. (D² − 4D + 4)y = 0.
Auxiliary equation m² − 4m + 4 = 0 gives m = 2, 2 (repeated).
y = (c₁ + c₂z)e²ᶻ.
Back to x: y = (c₁ + c₂ log x)x².
Check y = x²: x²(2) − 3x(2x) + 4x² = 0 ✓.
Q4. Solve x²y″ + xy′ − y = x².
Put x = eᶻ: D(D − 1)y + Dy − y = e²ᶻ, i.e. (D² − 1)y = e²ᶻ.
CF: m = ±1, so c₁eᶻ + c₂e⁻ᶻ = c₁x + c₂/x.
PI = e²ᶻ/(2² − 1) = e²ᶻ/3 = x²/3.
y = c₁x + c₂/x + x²/3.
Check the PI: x²(2/3) + x(2x/3) − x²/3 = x² ✓.
Q5. Solve y″ + y = sec x by variation of parameters.
CF: y₁ = cos x, y₂ = sin x; W = 1. R = sec x.
u = −∫ sin x sec x dx = −∫ tan x dx = log(cos x).
v = ∫ cos x sec x dx = ∫ 1 dx = x.
PI = cos x log(cos x) + x sin x.
y = c₁cos x + c₂sin x + cos x log(cos x) + x sin x.
Exam tip
After solving in z, never forget to change back to x (eᶻ = x, z = log x) — leaving the answer in z loses marks. In variation of parameters, compute W first and write u and v formulas before integrating.
Some problems involve more than one unknown function or three variables at once. Simultaneous equations of the form dx/P = dy/Q = dz/R are solved by grouping terms or by using multipliers to find two independent integrals. A total differential equation P dx + Q dy + R dz = 0 can be solved when it satisfies an integrability condition. Systems of linear equations in t are solved by eliminating one variable.
Key points
Form: dx/P = dy/Q = dz/R, where P, Q, R are functions of x, y, z. The solution is two independent relations u = c₁, v = c₂.
Method of grouping: if two of the fractions involve only two variables, solve them directly to get one integral.
Method of multipliers: if l, m, n are chosen so that lP + mQ + nR = 0, then l dx + m dy + n dz = 0; if this is exact, integrate it. Each fraction also equals (l dx + m dy + n dz)/(lP + mQ + nR).
Common multipliers: (1, 1, 1) and (x, y, z).
Total DE: P dx + Q dy + R dz = 0 is integrable iff P(∂Q/∂z − ∂R/∂y) + Q(∂R/∂x − ∂P/∂z) + R(∂P/∂y − ∂Q/∂x) = 0.
If the left side is an exact differential, the solution is found directly: e.g. y dx + x dy = d(xy).
Linear systems such as dx/dt = ax + by, dy/dt = cx + dy are solved by eliminating y (or x) to get one second-order equation with constant coefficients.
Important terms
Integral of a system – a relation u(x, y, z) = c satisfied along every solution.
Condition of integrability (समाकलनक्षमता अट) – the condition that P dx + Q dy + R dz = 0 has a solution f(x, y, z) = c.
Exact differential – an expression that equals du for some function u.
Formulas
dx/P = dy/Q = dz/R = (l dx + m dy + n dz)/(lP + mQ + nR)
Many important differential equations with variable coefficients (Legendre, Bessel, Hermite) cannot be solved in closed form. We assume the solution is a power series y = Σ aₙxⁿ, substitute, and find a recurrence relation for the coefficients. Near a regular singular point we use the Frobenius method, y = Σ aₙx^(n+r), where r comes from the indicial equation.
Key points
Write the equation as y″ + P(x)y′ + Q(x)y = 0.
x = x₀ is an ordinary point if P and Q are analytic there (have Taylor series). Otherwise it is a singular point.
A singular point x₀ is regular if (x − x₀)P and (x − x₀)²Q are analytic there; otherwise irregular.
Near an ordinary point: put y = Σ aₙxⁿ, y′ = Σ naₙxⁿ⁻¹, y″ = Σ n(n − 1)aₙxⁿ⁻². Shift indices so every sum has xⁿ, and equate coefficients to zero.
This gives a recurrence relation for aₙ; a₀ and a₁ stay arbitrary, giving two independent solutions.
Frobenius method (regular singular point at 0): put y = Σ aₙx^(n+r), a₀ ≠ 0. The lowest power gives the indicial equation for r.
If the indicial roots differ by a non-integer, both give independent series solutions. Equal roots or roots differing by an integer need extra work (a log term may appear).
Important terms
Ordinary point (सामान्य बिंदू) – a point where the coefficients are analytic.
Recurrence relation – a formula giving aₙ₊₂ (or aₙ₊₁) in terms of earlier coefficients.
Indicial equation – the quadratic in r from the lowest power of x.
Q1. Define an ordinary point and a regular singular point.
For y″ + Py′ + Qy = 0, x₀ is ordinary if P and Q are analytic at x₀. It is a regular singular point if it is not ordinary but (x − x₀)P and (x − x₀)²Q are analytic there.
Q2. Solve y′ = y by the power series method.
Put y = Σ aₙxⁿ: Σ (n + 1)aₙ₊₁xⁿ = Σ aₙxⁿ, so aₙ₊₁ = aₙ/(n + 1). Then aₙ = a₀/n!, and y = a₀(1 + x + x²/2! + …) = a₀eˣ.
The first series is e^(x²/2), since x^(2k)/(2ᵏk!) gives 1, ½, ⅛, 1/48 ✓.
Q5. Find the indicial equation and its roots for 2xy″ + y′ + y = 0.
x = 0 is a regular singular point. Put y = Σ aₙx^(n+r).
2xy″ gives Σ 2(n + r)(n + r − 1)aₙx^(n+r−1); y′ gives Σ (n + r)aₙx^(n+r−1); y gives higher powers.
The lowest power x^(r−1): a₀[2r(r − 1) + r] = a₀ r(2r − 1) = 0.
Indicial equation r(2r − 1) = 0, roots r = 0 and r = ½. They differ by a non-integer, so there are two independent Frobenius series.
Exam tip
Before equating coefficients, shift every sum so it has the same power xⁿ and the same starting index. Write a few coefficients and then spot the pattern.
Legendre and Bessel functions are special functions that come from solving Legendre's and Bessel's differential equations by series. They appear in physics problems with spherical and cylindrical symmetry. We study their definitions, Rodrigues' formula, generating functions, recurrence relations and orthogonality.
Key points
Legendre's equation: (1 − x²)y″ − 2xy′ + n(n + 1)y = 0. For integer n, one solution is the polynomial Pₙ(x).
Q5. Given J_(1/2) = √(2/(πx)) sin x and J_(−1/2) = √(2/(πx)) cos x, find J_(3/2).
Use Jₙ₋₁ + Jₙ₊₁ = (2n/x)Jₙ with n = ½: J_(−1/2) + J_(3/2) = (1/x)J_(1/2).
J_(3/2) = (1/x)J_(1/2) − J_(−1/2).
= √(2/(πx)) (sin x/x − cos x).
Exam tip
Learn the first four Legendre polynomials and the two main Bessel recurrences by heart — most short questions use only these. In "express in Legendre polynomials" problems, start from the highest power.
The Laplace transform changes a function of time f(t) into a function of s, F(s). It turns differential equations into algebraic equations, which are easier to solve. This topic covers the definition, standard transforms, properties (linearity, shifting, derivatives), inverse transforms and solving initial value problems.
Key points
Definition: L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt, for those s where the integral converges.
Existence: f should be piecewise continuous and of exponential order (|f(t)| ≤ Me^(at)).
Linearity: L{af + bg} = aL{f} + bL{g}.
First shifting theorem: L{e^(at) f(t)} = F(s − a).
Second shifting theorem: L{f(t − a) u(t − a)} = e^(−as)F(s), where u is the unit step function.
Learn the table of seven standard transforms and the derivative formula with initial values. In inverse problems, partial fractions plus the first shifting theorem solve almost every question.
The inverse Laplace transform takes a function F(s) back to the function f(t) whose transform it is. We use standard tables, partial fractions, the shifting theorem and the convolution theorem. The main application is solving linear differential equations with initial conditions: the equation becomes algebraic in s, we solve for Y(s), and invert.
Key points
If L{f(t)} = F(s), then f(t) = L⁻¹{F(s)}. The inverse is linear: L⁻¹{aF + bG} = aL⁻¹{F} + bL⁻¹{G}.
If L⁻¹{F(s)} = f(t) and L⁻¹{G(s)} = g(t), then L⁻¹{F(s)G(s)} = ∫₀ᵗ f(u)g(t − u) du, the convolution f ∗ g. Also f ∗ g = g ∗ f.
Exam tip
Always check your partial-fraction constants by the "cover-up" method and verify the final y(t) with the initial conditions — it takes 30 seconds and catches most mistakes.
A Fourier series represents a periodic function as a sum of a constant and sine and cosine terms. This topic covers Euler's formulas for the coefficients, Dirichlet's conditions, series over (−π, π) and (0, 2π), even and odd functions, half-range sine and cosine series, and change of interval to (−l, l).
Key points
f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx), n = 1 to ∞, for a function of period 2π.
The coefficients come from Euler's formulas, using orthogonality of sin nx and cos nx over a period.
Dirichlet's conditions: in one period f is bounded, single-valued, has finitely many discontinuities and finitely many maxima and minima. Then the series converges to f(x) at points of continuity and to ½[f(x⁺) + f(x⁻)] at a jump.
Even function on (−π, π): bₙ = 0 (cosine series). Odd function: a₀ = aₙ = 0 (sine series).
Half-range series: a function given only on (0, π) can be expanded as a pure sine series (odd extension) or a pure cosine series (even extension).
For period 2l (interval (−l, l)), replace nx by nπx/l.
Putting a special value of x in a series gives sums of numerical series, e.g. Σ1/n² = π²/6.
Useful integrals: sin nπ = 0, cos nπ = (−1)ⁿ.
Important terms
Periodic function (आवर्ती फलन) – f(x + T) = f(x).
Fourier coefficients – a₀, aₙ, bₙ.
Half-range expansion – series using only sines or only cosines on half the period.
Orthogonality – integral of the product of two different basis functions over a period is zero.
Q1. Write Euler's formulas for the Fourier coefficients in (−π, π).
a₀ = (1/π)∫₋π^π f(x) dx, aₙ = (1/π)∫₋π^π f(x) cos nx dx, bₙ = (1/π)∫₋π^π f(x) sin nx dx, and f(x) = a₀/2 + Σ(aₙ cos nx + bₙ sin nx).
Q2. What happens to the Fourier coefficients of an even function in (−π, π)?
For even f, f(x) sin nx is odd, so bₙ = 0. Also a₀ = (2/π)∫₀^π f dx and aₙ = (2/π)∫₀^π f cos nx dx. The series contains only the constant and cosine terms.
Q3. Find the Fourier series of f(x) = x² in (−π, π) and deduce that Σ1/n² = π²/6.
f is even, so bₙ = 0.
a₀ = (2/π)∫₀^π x² dx = (2/π)(π³/3) = 2π²/3.
∫₀^π x² cos nx dx = [x² sin nx/n]₀^π − (2/n)∫₀^π x sin nx dx = 0 − (2/n)(π(−1)ⁿ⁺¹/n) = 2π(−1)ⁿ/n².
aₙ = (2/π) × 2π(−1)ⁿ/n² = 4(−1)ⁿ/n².
x² = π²/3 + 4Σ(−1)ⁿ cos nx/n² = π²/3 − 4[cos x − cos 2x/4 + cos 3x/9 − …].
Put x = π (cos nπ = (−1)ⁿ, so (−1)ⁿ(−1)ⁿ = 1): π² = π²/3 + 4Σ1/n² → Σ1/n² = π²/6.
Q4. Find the half-range cosine series of f(x) = x in (0, π).
a₀ = (2/π)∫₀^π x dx = (2/π)(π²/2) = π.
∫₀^π x cos nx dx = [x sin nx/n + cos nx/n²]₀^π = ((−1)ⁿ − 1)/n².
aₙ = (2/π)((−1)ⁿ − 1)/n²: zero for even n, −4/(πn²) for odd n.
x = π/2 − (4/π)[cos x + cos 3x/3² + cos 5x/5² + …].
(Putting x = 0 gives 1 + 1/9 + 1/25 + … = π²/8.)
Q5. Find the half-range sine series of f(x) = 1 in (0, π).
Check even/odd first; it saves half the work. In integration by parts, use the shortcut ∫uv = uv₁ − u'v₂ + u''v₃ − … and write cos nπ = (−1)ⁿ, sin nπ = 0 straight away.
A partial differential equation (PDE) involves partial derivatives of a function of two or more variables. This topic covers forming PDEs by eliminating arbitrary constants or functions, Lagrange's linear equation Pp + Qq = R, standard non-linear forms, classification of second-order equations, and the method of separation of variables.
Key points
Notation: z = f(x, y); p = ∂z/∂x, q = ∂z/∂y, r = ∂²z/∂x², s = ∂²z/∂x∂y, t = ∂²z/∂y².
Order = order of the highest derivative; degree = power of that derivative.
Formation by eliminating arbitrary constants: differentiate partially w.r.t. x and y and eliminate the constants.
Formation by eliminating an arbitrary function: differentiate and eliminate the function's derivative.
Lagrange's linear equation: Pp + Qq = R (P, Q, R functions of x, y, z). Auxiliary (subsidiary) equations: dx/P = dy/Q = dz/R. If u = c₁ and v = c₂ are two independent solutions, the general solution is φ(u, v) = 0.
Methods for the auxiliary equations: grouping, and using multipliers (l, m, n) with lP + mQ + nR = 0.
Non-linear first order standard forms: f(p, q) = 0; z = px + qy + f(p, q) (Clairaut's form, solution z = ax + by + f(a, b)); f(z, p, q) = 0; f₁(x, p) = f₂(y, q). Charpit's method is the general method.
Classification of A uₓₓ + B uₓᵧ + C uᵧᵧ + … = 0: B² − 4AC < 0 elliptic; = 0 parabolic; > 0 hyperbolic.
Laplace's equation uₓₓ + uᵧᵧ = 0 is elliptic; heat equation uₜ = c²uₓₓ is parabolic; wave equation uₜₜ = c²uₓₓ is hyperbolic.
Separation of variables: assume u = X(x)T(t), separate into ordinary differential equations equal to a constant, solve, then apply boundary conditions.
Important terms
Complete integral – solution containing as many arbitrary constants as independent variables.
General solution – solution containing an arbitrary function.
Boundary conditions (सीमा अटी) – values of the solution prescribed at the edges.
Formulas
Lagrange: dx/P = dy/Q = dz/R → φ(u, v) = 0
Clairaut: z = px + qy + f(p, q) → z = ax + by + f(a, b)
For Lagrange's equation, always write the auxiliary equations first and find two independent integrals; the answer is φ(u, v) = 0. In separation of variables, show clearly where the separation constant is used to match the given condition.
Calculus of variations finds the function y(x) that makes an integral (a functional) as small or large as possible, such as the shortest curve between two points or the curve of quickest descent. The necessary condition is Euler's equation (the Euler–Lagrange equation). Its solutions are called extremals.
Key points
A functional I[y] = ∫ₐᵇ F(x, y, y′) dx assigns a number to each function y(x) with fixed end values y(a) = A, y(b) = B.
Euler's equation: ∂F/∂y − d/dx (∂F/∂y′) = 0. Any y that makes I stationary must satisfy it.
Solutions of Euler's equation satisfying the end conditions are extremals.
Special case 1: F does not contain y. Then ∂F/∂y′ = constant.
Special case 2: F does not contain x. Then F − y′ ∂F/∂y′ = constant (Beltrami identity).
Special case 3: F does not contain y′. Then ∂F/∂y = 0, an ordinary equation.
Shortest distance between two points in a plane: F = √(1 + y′²) gives a straight line.
Brachistochrone (quickest descent under gravity): the extremal is a cycloid.
Minimum surface of revolution: the extremal is a catenary y = c cosh((x − d)/c).
For isoperimetric problems (an extra integral condition), use F + λG in Euler's equation.
Important terms
Functional (फलनक) – a rule assigning a number to a function.
Extremal – a curve satisfying Euler's equation.
Variation δy – a small change in the function y.
Formulas
∂F/∂y − d/dx (∂F/∂y′) = 0
F free of y: ∂F/∂y′ = c
F free of x: F − y′(∂F/∂y′) = c
5 important questions
Q1. Write Euler's equation for the functional ∫ F(x, y, y′) dx.
∂F/∂y − (d/dx)(∂F/∂y′) = 0.
Q2. Show that the shortest curve between two points in a plane is a straight line.
Length = ∫ √(1 + y′²) dx. F has no y, so ∂F/∂y′ = y′/√(1 + y′²) = c. This gives y′ = constant = m, so y = mx + k, a straight line.
Q3. Find the extremal of ∫₀¹ (y′² + 12xy) dx with y(0) = 0, y(1) = 1.
By Euler's equation the bracket is 0, so d/dx (F − y′F_y′) = 0.
Hence F − y′ ∂F/∂y′ = constant.
Exam tip
First check which variable is missing from F — it decides which short form to use and saves time. Always apply the end conditions at the last step to get the particular extremal.
Numerical methods give approximate answers when exact methods are hard or impossible. This topic covers finding roots of equations (bisection, regula falsi, Newton–Raphson), finite differences and interpolation (Newton's and Lagrange's formulas), numerical integration (trapezoidal and Simpson's rules), and solving ODEs (Euler and Runge–Kutta methods).
Key points
Bisection method: if f(a) and f(b) have opposite signs, a root lies in (a, b). Take the midpoint and keep the half where the sign changes. Always converges, but slowly.
Regula falsi (false position): uses the chord joining (a, f(a)) and (b, f(b)) instead of the midpoint.
Newton–Raphson method: fast (quadratic convergence) when the starting value is close to the root; fails if f'(x) is near 0.
Finite differences: forward Δyᵢ = yᵢ₊₁ − yᵢ; backward ∇yᵢ = yᵢ − yᵢ₋₁. For a polynomial of degree n, the n-th differences are constant.
Newton's forward interpolation is used near the beginning of a table; backward near the end. Both need equal intervals.
Lagrange's interpolation works for unequal intervals.
Trapezoidal rule: joins points by straight lines. Simpson's 1/3 rule: fits parabolas; needs an even number of intervals; more accurate. Simpson's 3/8 rule needs intervals in multiples of 3.
The interval must be divided into an even number of equal sub-intervals (an odd number of ordinates), because the rule fits a parabola through each set of three points.
Q2. Write the Newton–Raphson formula and one limitation.
xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ). It fails or converges slowly if f'(x) is zero or very small near the root, or if the starting value is far from the root.
Q3. Find √2 by the Newton–Raphson method, starting from x₀ = 1.5.
Q5. Use Euler's method with h = 0.1 to find y(0.2) for dy/dx = x + y, y(0) = 1.
f(x, y) = x + y; x₀ = 0, y₀ = 1.
y₁ = y₀ + h f(x₀, y₀) = 1 + 0.1(0 + 1) = 1.1 at x = 0.1.
y₂ = y₁ + h f(x₁, y₁) = 1.1 + 0.1(0.1 + 1.1) = 1.1 + 0.12 = 1.22 at x = 0.2.
Exact solution y = 2eˣ − x − 1 gives y(0.2) ≈ 1.2428; Euler's error shows why RK4 is preferred.
Exam tip
Make neat tables of x and y values in integration and interpolation questions — it reduces errors and earns method marks. Keep at least four or five decimal places in working and round only the final answer.