Mathematics Specialised Level 4 flashcards

40 cards. Read the term and the definition here. The deck at the top of this page is optional.

Remainder theorem

When a polynomial P(x) is divided by (x − a), the remainder equals P(a). If the remainder is zero, then (x − a) is a factor of P(x).

Quadratic discriminant

For ax² + bx + c = 0 with real coefficients, Δ = b² − 4ac. Positive gives two distinct real roots, zero gives one repeated root, negative gives no real roots.

Partial fractions decomposition

A rational function with distinct linear factors is split as A/(x−a) + B/(x−b). Repeated factors require one term per power, and irreducible quadratics take a linear numerator.

Logarithm product law

For positive x and y and base a > 0 with a ≠ 1, logₐ(xy) = logₐx + logₐy. The corresponding quotient law subtracts logarithms, and the power law multiplies by the exponent.

Perpendicular gradients

Two non-vertical lines are perpendicular exactly when the product of their gradients equals −1. A line with gradient m therefore has perpendicular gradient −1/m.

Circle equation

The circle with centre (h, k) and radius r has equation (x−h)² + (y−k)² = r². Expanding gives x² + y² + 2gx + 2fy + c = 0 with centre (−g, −f).

Distance point to line

The perpendicular distance from point (x₀, y₀) to line ax + by + c = 0 is |ax₀ + by₀ + c| divided by √(a² + b²).

Circle tangent property

A tangent to a circle at point P is perpendicular to the radius drawn to P. Consequently the normal at P passes through the circle's centre.

Arithmetic series sum

For an arithmetic sequence with first term a and common difference d, the sum of n terms is Sₙ = n/2 [2a + (n−1)d], equivalently n/2 (a + l) where l is the last term.

Geometric series sum

For first term a and common ratio r ≠ 1, Sₙ = a(1 − rⁿ)/(1 − r). If |r| < 1 the infinite sum converges to a/(1 − r).

Binomial theorem

For non-negative integer n, (a + b)ⁿ = Σ C(n,k) aⁿ⁻ᵏ bᵏ from k = 0 to n. The general term is C(n,k) aⁿ⁻ᵏ bᵏ.

Method of differences

If a general term can be written uₙ = f(n) − f(n+1), then the sum from n = 1 to N telescopes to f(1) − f(N+1).

Product and quotient rules

For differentiable u and v, (uv)′ = u′v + uv′ and, where v ≠ 0, (u/v)′ = (u′v − uv′)/v².

Chain rule

If y = f(u) and u = g(x), then dy/dx = dy/du · du/dx. Equivalently, the derivative of f(g(x)) is f′(g(x))·g′(x).

Implicit differentiation

When y is defined implicitly by an equation in x and y, differentiate both sides with respect to x, applying the chain rule to y terms, then rearrange to solve for dy/dx.

Second derivative test

At a stationary point where dy/dx = 0, if d²y/dx² > 0 the point is a local minimum; if d²y/dx² < 0 it is a local maximum; if zero the test is inconclusive.

Integration by parts

For differentiable u and v, ∫u dv = uv − ∫v du. Choosing u according to the LIATE order often simplifies the remaining integral.

Integration by substitution

Replace a composite integrand by setting u = g(x), du = g′(x) dx. For definite integrals, change the limits to the corresponding u-values before evaluating.

Volume of revolution

Rotating y = f(x) about the x-axis from x = a to x = b gives volume V = π∫ₐᵇ y² dx, provided the curve does not cross the axis.

Trapezium rule

With h = (b−a)/n and equally spaced ordinates f₀ to fₙ, the integral is approximately h/2 [f₀ + 2(f₁ + … + fₙ₋₁) + fₙ].

Double angle identities

sin2A = 2 sinA cosA; cos2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1; tan2A = 2tanA/(1 − tan²A) where defined.

R-form of a sinθ + b cosθ

a sinθ + b cosθ can be written R sin(θ + α) with R = √(a² + b²) and tanα = b/a. The maximum value is R and the minimum is −R.

Cosine rule

In a triangle with sides a, b, c opposite angles A, B, C respectively, a² = b² + c² − 2bc cosA. The rule finds a side from two sides and the included angle.

General solution of sinθ = k

For |k| ≤ 1, sinθ = k has solutions θ = (−1)ⁿ arcsin k + nπ for integer n, where arcsin k lies in [−π/2, π/2].

Dot product

For vectors a and b, a·b = |a||b|cosθ, where θ is the angle between them. In components a·b = a₁b₁ + a₂b₂ + a₃b₃.

Inverse of 2×2 matrix

For M = [[a,b],[c,d]] with determinant ad − bc ≠ 0, M⁻¹ = (1/(ad−bc))[[d,−b],[−c,a]].

Cross product area

The magnitude |a × b| equals |a||b|sinθ and gives the area of the parallelogram spanned by a and b. The direction is perpendicular to both vectors.

Eigenvalue equation

A scalar λ is an eigenvalue of square matrix A if det(A − λI) = 0. A corresponding eigenvector v ≠ 0 satisfies (A − λI)v = 0.

De Moivre's theorem

For real θ and integer n, [r(cosθ + i sinθ)]ⁿ = rⁿ(cos nθ + i sin nθ). It underlies finding powers and roots of complex numbers.

Roots of a complex number

If w = ρe^{iφ}, the n distinct roots of zⁿ = w are z = ρ^{1/n} e^{i(φ+2kπ)/n} for k = 0, 1, …, n−1, equally spaced on a circle.

Euler's formula

For real θ, e^{iθ} = cosθ + i sinθ. It gives cosθ = (e^{iθ} + e^{−iθ})/2 and sinθ = (e^{iθ} − e^{−iθ})/(2i).

Locus |z − a| = r

The set of points z in the complex plane satisfying |z − a| = r is a circle centred at a with radius r, because modulus measures distance.

Bayes' theorem

P(A|B) = P(B|A)P(A)/P(B). The denominator is often computed by the law of total probability over a partition of the sample space.

Binomial distribution mean and variance

If X ~ B(n, p), then E(X) = np and Var(X) = np(1 − p). The probability mass function is P(X = k) = C(n,k) pᵏ(1−p)ⁿ⁻ᵏ.

Standardising a normal variable

If X ~ N(μ, σ²), then Z = (X − μ)/σ follows N(0, 1). Probabilities for X are found using the standard normal cumulative distribution Φ.

Variance of a random variable

Var(X) = E(X²) − [E(X)]². For constants a and b, E(aX + b) = aE(X) + b and Var(aX + b) = a² Var(X).

Proof by contradiction

To prove statement P, assume ¬P and derive a logical contradiction. The contradiction establishes that ¬P is false, so P must be true.

Mathematical induction

Prove the base case P(1), assume P(k) as the inductive hypothesis, then prove P(k+1). This establishes P(n) for all integers n ≥ 1.

Proof by contrapositive

To prove P ⇒ Q, prove the logically equivalent ¬Q ⇒ ¬P. This is often easier when the negations have simpler algebraic forms.

Counterexample

A single case that satisfies the hypothesis of a universal statement but fails its conclusion disproves that statement. It is the standard method of disproof.