Title: InkMaths / Further Pure 2: groups, matrices and number theory
On every note of this deck, unless a note says otherwise:
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Front: STANDARD RESULT — LEARN IT
Arc length:
cartesian, then polar = ?\(\text{cartesian, then polar} = ?\)cartesian, then polar = ?
Back: ∫ √1 + (dydx)^2 dx, ∫ √r^2 + (drdθ)^2 dθ\(\text{∫} \sqrt{1 + (\frac{dy}{dx})^{2}} \, dx, \text{∫} \sqrt{r^{2} + (\frac{dr}{dθ})^{2}} \, dθ\)∫ √(1 + (dy/dx)^2) dx, ∫ √(r^2 + (dr/dθ)^2) dθ

Front: CHECK YOURSELF
A group has order 8. Explain why it cannot contain an element of order 3.
Back: The powers of such an element would form a subgroup of order 3, and 3 does not divide 8, contradicting Lagrange's theorem.

Front: DEFINE
Define: Bezout's identity.
Back: The highest common factor of a and b can be written as ax + by for integers x and y, found by unwinding the Euclidean algorithm. When that factor is 1, x is the inverse of a modulo b.

Front: SPOT THE ERROR
Say what is wrong with this claim:
The letters of BANANA can be arranged in 6! ways.
Back: 6! counts the arrangements as though every letter were distinguishable. Three As and two Ns are not, so divide by the repeats: 6!/(3!2!) = 60, not 720. Any arrangement question with a repeated object needs that denominator, and forgetting it is the single most common combinatorics error.

Front: STANDARD RESULT — LEARN IT
Inversion:
w = 1z: image of a line missing 0?\(w = \frac{1}{z}: \text{image of a line missing 0?}\)w = (1/z): image of a line missing 0?
Back: a circle through 0, and the map sends that circle back to a line\(\text{a circle through 0, and the map sends that circle back to a line}\)a circle through 0, and the map sends that circle back to a line

Front: DEFINE
Define: Cayley-Hamilton theorem.
Back: Every square matrix satisfies its own characteristic equation, which turns any power of it into a linear combination of lower powers and, when det M ≠ 0, hands over the inverse without cofactors.

Front: DEFINE
Define: group.
Back: A set with a binary operation satisfying closure, associativity, the existence of an identity, and an inverse for every element. Commutativity is not required.

Front: STANDARD RESULT — LEARN IT
Surface of revolution:
about the x-axis, S = ?\(\text{about the x-axis, } S = ?\)about the x-axis, S = ?
Back: 2π ∫ y ds = 2π ∫ y √1 + (dydx)^2 dx\(2π \text{∫} y \, ds = 2π \text{∫} y \sqrt{1 + (\frac{dy}{dx})^{2}} \, dx\)2π ∫ y ds = 2π ∫ y √(1 + (dy/dx)^2) dx

Front: STANDARD RESULT — LEARN IT
Reduction formula:
I_n = ∫_0^π/2 sin ^n x dx: n I_n = ?\(I_{n} = \text{∫}_{0}^{π/2} \sin^{n} x \, dx: n I_{n} = ?\)I_n = ∫_0^(π/2) sin^n x dx: n I_n = ?
Back: (n - 1) I_n-2, with I_0 = π2, I_1 = 1\((n - 1) I_{n-2}, \text{with } I_{0} = \frac{π}{2}, \, I_{1} = 1\)(n - 1) I_(n - 2), with I_0 = (π/2), I_1 = 1

Front: DEFINE
Define: Lagrange's theorem.
Back: The order of any subgroup of a finite group divides the order of the group, so the order of every element divides it too.

Front: STANDARD RESULT — LEARN IT
Diagonalisation:
M^n = ?\(M^{n} = ?\)M^n = ?
Back: P D^n P^-1, P of eigenvectors, D of eigenvalues in matching order\(P D^{n} P^{-1}, \text{P of eigenvectors, D of eigenvalues in matching order}\)P D^n P^(-1), P of eigenvectors, D of eigenvalues in matching order

Front: STANDARD RESULT — LEARN IT
Eigenvalues:
2 by 2 characteristic equation = ?\(\text{2 by 2 characteristic equation} = ?\)2 by 2 characteristic equation = ?
Back: λ^2 - (trace)λ + det = 0\(λ^{2} - (\text{trace})λ + \text{det} = 0\)λ^2 - (trace)λ + det = 0

Front: STANDARD RESULT — LEARN IT
Fermat's little theorem:
p prime, a not a multiple of p: a^p-1 ≡ ?\(p \text{ prime, a not a multiple of p}: a^{p-1} \equiv ?\)p prime, a not a multiple of p: a^(p - 1) ≡ ?
Back: 1 (mod p), so exponents reduce modulo p - 1\(1 \, (\text{mod } p), \text{so exponents reduce modulo } p - 1\)1 (mod p), so exponents reduce modulo p - 1

Front: STANDARD RESULT — LEARN IT
Recurrence relations:
u_n+2 + b u_n+1 + c u_n = 0: u_n = ?\(u_{n+2} + b u_{n+1} + c u_{n} = 0: u_{n} = ?\)u_(n + 2) + b u_(n + 1) + c u_n = 0: u_n = ?
Back: A m_1^n + B m_2^n, repeated root needs (A + Bn)m^n\(A m_{1}^{n} + B m_{2}^{n}, \text{repeated root needs } (A + Bn)m^{n}\)A m_1^n + B m_2^n, repeated root needs (A + Bn)m^n

Front: DEFINE
Define: Apollonius locus.
Back: The set of points whose distances from two fixed points are in a fixed ratio other than 1. It is a circle, not a line; equal distances give the perpendicular bisector instead.