Flashcards · H240 · Topic 1

Pure mathematics flashcards

88 cards on Topic 1, subtopics 1.01 to 1.10. Pick a subtopic or study them all.

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1.01 Proof

Proof by deduction: Start from known facts and apply valid logical steps to reach the result. Example: (2m + 1) + 2n = 2(m + n) + 1, which is odd because m + n is an integer.
Proof by exhaustion: Split the statement into a finite list of cases, prove each one, and say that the cases cover every possibility. Example: checking n = 0, 1, 2, 3, 4 shows n⁴ leaves remainder 0 or 1 on division by 5.
Disproof by counter-example: Find one value that satisfies the condition but makes the conclusion false, and check it. Example: "if a > b then 1/a < 1/b" fails for a = 2, b = −1, since 1/2 > −1.
Proof by contradiction: Assume the statement is false, then deduce something impossible. So the assumption is wrong and the statement is true. Used for "is irrational" and "there is no …" results, such as √2 is irrational.
What do ⇒, ⇐ and ⇔ mean?: P ⇒ Q: if P then Q (P implies Q). P ⇐ Q: Q implies P. P ⇔ Q: P ⇒ Q and Q ⇒ P (if and only if). To prove ⇔, prove both directions.
The converse and the contrapositive of P ⇒ Q: Converse: Q ⇒ P. It may be false even when P ⇒ Q is true. Contrapositive: not Q ⇒ not P. It is equivalent to P ⇒ Q.
How do you write an even integer, an odd integer and a rational number in a proof?: Even: 2n. Odd: 2n + 1 (n an integer). Rational: a/b with a and b integers, b ≠ 0. Always define the letters, and finish with a sentence stating what has been proved.

1.02 Algebra and functions

What does x^(m/n) mean? Evaluate 16^(−3/4).: x^(m/n) = (ⁿ√x)ᵐ = ⁿ√(xᵐ). A negative index means the reciprocal. 16^(−3/4): ⁴√16 = 2, 2³ = 8, so the value is 1/8.
Rationalise 6/(√7 − 1).: Multiply top and bottom by the conjugate √7 + 1: 6(√7 + 1)/(7 − 1) = √7 + 1.
Write x² + 8x + 3 in completed-square form and state the vertex.: (x + 4)² − 16 + 3 = (x + 4)² − 13. Vertex (−4, −13), a minimum because the x² coefficient is positive.
Factor theorem (including factors of the form ax − b): (x − a) is a factor of f(x) ⇔ f(a) = 0. (ax − b) is a factor of f(x) ⇔ f(b/a) = 0. For (2x + 1), substitute x = −1/2.
Find the remainder when 3x³ − 2x + 1 is divided by (x + 2).: f(−2) = 3(−8) − 2(−2) + 1 = −24 + 4 + 1 = −19. (Long division gives quotient 3x² − 6x + 10, remainder −19.)
Partial fractions: the forms for (x + p)(x + q), (x + p)² and (x + p)(x + q)²: A/(x + p) + B/(x + q) A/(x + p) + B/(x + p)² A/(x + p) + B/(x + q) + C/(x + q)² If the top degree is not less than the bottom degree, divide first.
Rewrite |x − c| < d and |x − c| > d without the modulus (d > 0).: |x − c| < d ⇔ c − d < x < c + d. |x − c| > d ⇔ x < c − d or x > c + d (two regions joined by "or").
Solve |3x + 5| = x + 1.: Case 1: 3x + 5 = x + 1 gives x = −2, but then x + 1 = −1 < 0: reject. Case 2: −(3x + 5) = x + 1 gives x = −3/2, but x + 1 = −1/2 < 0: reject. No solutions.
Function, one-one and many-one: Function: every x in the domain has exactly one image f(x). One-one: each output comes from exactly one input. Many-one: two or more inputs share an output (still a function). State a range in terms of f(x), not x.
The inverse function f⁻¹: when it exists, its domain and range, and its graph: It exists only if f is one-one. Domain of f⁻¹ = range of f; range of f⁻¹ = domain of f. The graph of y = f⁻¹(x) is the reflection of y = f(x) in y = x.
Describe y = f(x) + a, y = f(x + a), y = af(x) and y = f(ax).: f(x) + a: translation by (0, a). f(x + a): translation by (−a, 0). af(x): stretch parallel to the y-axis, scale factor a. f(ax): stretch parallel to the x-axis, scale factor 1/a.
Sketching a factorised polynomial: what happens at a root with an odd power, and with an even power?: Odd power: the curve crosses the x-axis (flattening as it crosses if the power is 3 or more). Even power: the curve touches the axis and turns back. The sign of the highest-power term fixes the ends.
Asymptotes and shape of y = a/x and y = a/x² for a > 0: Both have asymptotes x = 0 and y = 0. y = a/x: in the first and third quadrants. y = a/x²: above the x-axis on both sides, symmetrical in the y-axis.

1.03 Coordinate geometry in the x-y plane

Gradient of the line ax + by + c = 0: −a/b. For example 4x + 6y − 9 = 0 gives y = −(2/3)x + 3/2: gradient −2/3, y-intercept 3/2.
Conditions for parallel and perpendicular lines: Parallel: m₁ = m₂. Perpendicular: m₁ × m₂ = −1, so the perpendicular gradient is −1/m.
Find the centre and radius of x² + y² + 10x − 2y + 10 = 0.: Complete the square: (x + 5)² − 25 + (y − 1)² − 1 + 10 = 0, so (x + 5)² + (y − 1)² = 16. Centre (−5, 1), radius 4.
Three circle properties used in coordinate geometry: The angle in a semicircle is 90°. The perpendicular from the centre to a chord bisects the chord. A tangent is perpendicular to the radius at the point of contact.
How do you convert parametric equations to a Cartesian equation?: 1. Rearrange the simpler equation for the parameter, or use an identity such as sin²θ + cos²θ ≡ 1. 2. Eliminate the parameter. 3. State any restriction on x or y that the parameter range causes.
Parametric equations of the circle with centre (a, b) and radius r: x = a + r cos θ, y = b + r sin θ, for 0 ≤ θ < 2π. Parametric forms are not unique.
x = 3 − 2 cos θ. What values can x take?: Since −1 ≤ cos θ ≤ 1, 1 ≤ x ≤ 5. The parameter range restricts the part of the Cartesian curve that exists.
Convert x = 2 − 5t, y = t² + 1 to a Cartesian equation.: t = (2 − x)/5, so y = (2 − x)²/25 + 1. If t ≥ 0 is stated, then x ≤ 2.

1.04 Sequences and series

Find the coefficient of x² in (2 − x)⁶.: ⁶C₂ × 2⁴ × (−1)² = 15 × 16 = 240.
Binomial expansion of (1 + x)ⁿ for rational n, and its validity: (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + … Valid for |x| < 1. For (a + bx)ⁿ, take out aⁿ first; valid for |bx/a| < 1.
Expand (1 + 2x)⁻³ up to x³ and state the validity.: 1 + (−3)(2x) + (−3)(−4)(2x)²/2 + (−3)(−4)(−5)(2x)³/6 + … = 1 − 6x + 24x² − 80x³ + …, valid for |x| < 1/2.
Arithmetic series: nth term and sum of n terms: uₙ = a + (n − 1)d Sₙ = (n/2)(2a + (n − 1)d) = (n/2)(a + l), where l is the last term.
Geometric series: nth term, sum of n terms and sum to infinity: uₙ = arⁿ⁻¹ Sₙ = a(1 − rⁿ)/(1 − r) S∞ = a/(1 − r), only when |r| < 1.
How many terms are in Σ from r = m to k, and what is Σ c over those terms?: There are k − m + 1 terms, so Σ c = c(k − m + 1), not c.
xₙ₊₁ = 0.4xₙ + 9 converges. Find its limit.: If xₙ → L, then L = 0.4L + 9, so 0.6L = 9 and L = 15.
Increasing, decreasing and periodic sequences: Increasing: uₙ₊₁ > uₙ for all n. Decreasing: uₙ₊₁ < uₙ for all n. Periodic: the terms repeat in a cycle; the order is the cycle length.

1.05 Trigonometry

Sine rule, cosine rule and area of a triangle: a/sin A = b/sin B = c/sin C a² = b² + c² − 2bc cos A Area = ½ab sin C (C is the angle between a and b)
Small-angle approximations (θ in radians): sin θ ≈ θ cos θ ≈ 1 − θ²/2 tan θ ≈ θ
Find the exact value of tan 300°.: Fourth quadrant, reference angle 60°, tangent negative there: tan 300° = −√3.
Domains and ranges (principal values) of arcsin x, arccos x and arctan x: arcsin: −1 ≤ x ≤ 1, −π/2 ≤ y ≤ π/2. arccos: −1 ≤ x ≤ 1, 0 ≤ y ≤ π. arctan: all real x, −π/2 < y < π/2.
sec θ, cosec θ, cot θ and the two identities that use them: sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = 1/tan θ. sec²θ ≡ 1 + tan²θ cosec²θ ≡ 1 + cot²θ
Find the exact value of tan 15°.: tan(45° − 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1) = 2 − √3.
sin A = 3/5 with A acute. Find sin 2A and cos 2A.: cos A = 4/5. sin 2A = 2(3/5)(4/5) = 24/25. cos 2A = 1 − 2(9/25) = 7/25.
Write a cos θ + b sin θ as R cos(θ − α).: Expand: R cos θ cos α + R sin θ sin α. Match R cos α = a and R sin α = b. R = √(a² + b²), tan α = b/a. Greatest value R, least −R.
Solve sin 2x = sin x for 0 ≤ x ≤ 2π.: 2 sin x cos x − sin x = 0, so sin x(2 cos x − 1) = 0. sin x = 0: x = 0, π, 2π. cos x = 1/2: x = π/3, 5π/3. Five solutions; dividing by sin x would lose three.

1.06 Exponentials and logarithms

Definition of a logarithm: y = aˣ ⇔ x = logₐ y (a > 0, a ≠ 1). Also logₐ a = 1 and logₐ 1 = 0.
Laws of logarithms: logₐ x + logₐ y = logₐ(xy) logₐ x − logₐ y = logₐ(x/y) k logₐ x = logₐ(xᵏ) log(x + y) cannot be split.
Graph of y = aˣ (a > 0), and the effect of adding c: Passes through (0, 1), always positive, asymptote y = 0. Increasing if a > 1, decreasing if 0 < a < 1. y = aˣ + c has asymptote y = c and y-intercept 1 + c.
Gradient of y = e^(kx), and why it matters in models: dy/dx = k e^(kx). The rate of change is proportional to the current value: k > 0 gives growth, k < 0 gives decay.
Solve aˣ = b.: Take logs: x ln a = ln b, so x = ln b/ln a. Isolate the power first and round only at the end.
Which graph gives a straight line for y = axⁿ, and which for y = kbˣ?: y = axⁿ: plot log y against log x; gradient n, intercept log a. y = kbˣ: plot log y against x; gradient log b, intercept log k.
N = N₀e^(kt): what do the constants mean, and what is the doubling time or half-life?: N₀ is the initial value (t = 0); k > 0 is growth, k < 0 is decay. Doubling time or half-life = ln 2/|k|.
Is (log x)² equal to 2 log x?: No. The power law applies only to the input of the log: log(x²) = 2 log x, but (log x)² is log x multiplied by itself.

1.07 Differentiation

Differentiation from first principles: f′(x) = lim (h → 0) [f(x + h) − f(x)]/h: the limit of the chord gradient as the chord shrinks to a tangent.
Derivatives of e^(kx), a^(kx), ln x, sin kx, cos kx and tan kx: k e^(kx) k ln a × a^(kx) 1/x k cos kx −k sin kx k sec² kx (radians)
Convex, concave and point of inflection: Convex on an interval: f″(x) > 0 there. Concave: f″(x) < 0. Point of inflection: f″(x) = 0 and the curve changes between concave and convex (f″ changes sign).
How do you classify a stationary point?: Solve dy/dx = 0. Then d²y/dx² > 0: minimum; d²y/dx² < 0: maximum. If d²y/dx² = 0, check the sign of dy/dx either side (− to + is a minimum, + to − a maximum).
Product rule and quotient rule: y = uv: dy/dx = uv′ + vu′. y = u/v: dy/dx = (vu′ − uv′)/v².
Chain rule, connected rates, and dy/dx from dx/dy: dy/dx = (dy/du) × (du/dx). Connected rates: dV/dt = (dV/dr) × (dr/dt). dy/dx = 1/(dx/dy).
Implicit and parametric differentiation: Implicit: differentiate each term with respect to x; d/dx(yⁿ) = nyⁿ⁻¹ dy/dx. Parametric: dy/dx = (dy/dt) ÷ (dx/dt).
Differentiate y = (2x + 1)e^(−3x).: Product rule: 2e^(−3x) − 3(2x + 1)e^(−3x) = −(6x + 1)e^(−3x).
Increasing and decreasing functions: Increasing on an interval where dy/dx > 0; decreasing where dy/dx < 0. Factorise the derivative and use a sign table or sketch.

1.08 Integration

Indefinite vs definite integral: Indefinite: no limits, answer is a function plus c. Definite: limits, answer is a number.
∫ e^(kx) dx, ∫ 1/x dx, ∫ sin kx dx and ∫ cos kx dx: (1/k)e^(kx) + c ln|x| + c −(1/k) cos kx + c (1/k) sin kx + c
Area between two curves: ∫ₐᵇ (upper − lower) dx, where a and b are the x-coordinates of the intersections. Regions below the x-axis give negative integrals, so split at the roots for an area.
Find ∫ 5x²/(x³ + 2) dx.: The derivative of x³ + 2 is 3x², so the top is (5/3) × 3x². Answer: (5/3) ln|x³ + 2| + c.
∫ k f′(x)/f(x) dx and ∫ f′(x)[f(x)]ⁿ dx: ∫ k f′(x)/f(x) dx = k ln|f(x)| + c ∫ f′(x)[f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n + 1) + c, n ≠ −1
Integration by parts: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. Choose u to be the part that simplifies when differentiated (for example x in x eˣ, or ln x).
Integration by substitution u = g(x): what must you replace?: Replace every x, replace dx using du/dx, and change the limits to u-values (or substitute back before using x-limits).
How do you solve a separable differential equation dy/dx = f(x)g(y)?: 1. Separate: ∫ (1/g(y)) dy = ∫ f(x) dx. 2. Integrate, adding one constant. 3. Use the initial condition to find it. 4. Make y the subject if asked: ln|y − a| = g(x) + c gives y − a = Ae^(g(x)).
Solve dy/dx = 2xy + 6x with y = 1 when x = 0.: dy/dx = 2x(y + 3), so ∫ 1/(y + 3) dy = ∫ 2x dx and ln|y + 3| = x² + c. y + 3 = Ae^(x²); at x = 0, A = 4. y = 4e^(x²) − 3.
Area as the limit of a sum: Area = lim (δx → 0) Σ f(x)δx = ∫ f(x) dx. Each strip has area about f(x)δx.

1.09 Numerical methods

Locating a root by a change of sign: If f is continuous on [a, b] and f(a), f(b) have opposite signs, then f(x) = 0 has a root between a and b. Quote both values and state the continuity.
How do you show that a root is 1.76 to 2 decimal places?: Show a sign change of f between 1.755 and 1.765 (half a unit either side of 1.76).
When can the change-of-sign method fail?: When the curve touches the axis (a repeated root), when there is an even number of roots in the interval, or when there is a vertical asymptote (a sign change with no root).
Fixed-point iteration xₙ₊₁ = g(xₙ): when does it converge, and how is it drawn?: It converges to α if |g′(α)| < 1 and x₁ is close enough to α. Diagram: draw y = x and y = g(x); go vertically to the curve, then horizontally to the line, repeatedly (a staircase or a cobweb).
Newton–Raphson formula: xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ)
Use Newton–Raphson on f(x) = x² − 2 with x₁ = 1 to find x₂ and x₃.: f′(x) = 2x. x₂ = 1 − (−1)/2 = 1.5. x₃ = 1.5 − 0.25/3 = 1.4167 (4 d.p.), approaching √2.
When does Newton–Raphson fail?: When f′(xₙ) = 0 (division by zero). A starting value near a stationary point can send the next value far from the root.
Trapezium rule: when is it an overestimate, and when an underestimate?: (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)], h = (b − a)/n. Convex curve (f″ > 0): overestimate. Concave curve (f″ < 0): underestimate.
Bounding an integral with rectangles for an increasing function: Left-edge heights give a lower bound; right-edge heights give an upper bound. For a decreasing function they swap.

1.10 Vectors

Scalar vs vector: Scalar: size only (mass, time, speed, distance). Vector: size and direction (displacement, velocity, acceleration, force).
Magnitude of xi + yj and of xi + yj + zk: |xi + yj| = √(x² + y²) |xi + yj + zk| = √(x² + y² + z²)
Position vectors a and b of A and B: find →AB, the distance AB and the midpoint.: →AB = b − a. AB = |b − a|. Midpoint: (a + b)/2.
When are two vectors parallel?: When b = ka for some non-zero scalar k.
Find the components of a vector with magnitude r at angle θ to the positive x-axis.: x = r cos θ, y = r sin θ, so the vector is r cos θ i + r sin θ j.
Find the magnitude and direction of 3i − 4j.: Magnitude √(9 + 16) = 5. It points into the fourth quadrant, at tan⁻¹(4/3) = 53.1° below the positive x-axis (306.9° measured anticlockwise).
Unit vector in the direction of a: â = a/|a|. For 3i − 4j it is (3/5)i − (4/5)j.