A-Level Further Mathematics · ⁨מתמטיקה מתקדמת A-Level⁩

For a polynomial equation, the roots 根 are linked to the coefficients 系数. For $ax^2 + bx + c = 0$ with roots $\alpha, \beta$:

For a cubic $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$: Sums like $\sum\alpha$ and $\sum\alpha\beta$ are symmetric functions 对称函数 of the roots (unchanged if the roots are swapped). To find an equation whose roots are changed in a simple way, use a substitution 代换 (for example, put $w = \alpha + 1$).

Worked example. The equation $x^2 - 5x + 6 = 0$ has roots $\alpha, \beta$. Find the equation with roots $\alpha + 1, \beta + 1$.

Here $\alpha + \beta = 5$ and $\alpha\beta = 6$. The new sum is $(\alpha + 1) + (\beta + 1) = 7$, and the new product is $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 6 + 5 + 1 = 12$. So the new equation is

A rational function 有理函数 is a fraction of two polynomials. When the top has degree exactly one higher than the bottom, the graph has an oblique asymptote 斜渐近线 (a slanted line the curve approaches); find it by dividing out. You should also be able to relate the graph of $y = f(x)$ to those of $y^2 = f(x)$, $y = \dfrac{1}{f(x)}$, $y = |f(x)|$ and $y = f(|x|)$. Find the set of values the function can take using the discriminant 判别式, and locate its turning points 驻点.

Worked example. Find the oblique asymptote of $y = \dfrac{x^2 + x + 1}{x + 1}$.

Divide out the fraction: $\dfrac{x^2 + x + 1}{x + 1} = x + \dfrac{1}{x + 1}$. As $x \to \pm\infty$ the remainder $\dfrac{1}{x+1} \to 0$, so the curve approaches the line $y = x$ — the oblique asymptote.

Learn the standard results:

The method of differences 差分法 sums a series by cancelling middle terms: if partial fractions 部分分式 write each term as $f(r) - f(r+1)$, almost everything cancels. From the sum to $n$ terms you can see whether a series is convergent 收敛 and, if so, find its sum to infinity 无穷和.

Worked example. Find $\displaystyle\sum_{r=1}^{n} (2r + 1)$.

A matrix 矩阵 is a rectangular block of numbers. Matrix addition adds entries in matching positions; you can also subtract and multiply matrices (multiplication is row $\times$ column). The zero matrix 零矩阵 has every entry $0$, and the identity matrix 单位矩阵 (or unit matrix) $I$ leaves any matrix unchanged under multiplication.

For a $2\times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant 行列式 is $\det A = ad - bc$. If $\det A \neq 0$ the matrix is non-singular (otherwise it is a singular matrix 奇异矩阵), and the inverse matrix 逆矩阵 is

For a product, $(AB)^{-1} = B^{-1}A^{-1}$. A $2\times 2$ matrix can represent a geometric transformation 几何变换 of the plane — a rotation 旋转, reflection 反射, enlargement 放大, stretch 拉伸 or shear 切变: the determinant gives the area scale factor, and a product $AB$ means "do $B$, then $A$". Points or lines that do not move are called invariant points 不变点 and invariant lines 不变直线.

Worked example. Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$.

$\det A = 3\times 4 - 1\times 2 = 10$, so

Polar coordinates 极坐标 give a point by its distance $r$ from the origin and its angle $\theta$. They link to Cartesian coordinates 直角坐标 by

You should sketch simple polar curves 极坐标曲线, and find the area of a sector with

Worked example. Convert the polar equation $r = 4\cos\theta$ to Cartesian form.

Multiply both sides by $r$: $r^2 = 4r\cos\theta$, so $x^2 + y^2 = 4x$. Completing the square gives $(x - 2)^2 + y^2 = 4$, a circle of radius $2$ centred at $(2, 0)$.

In three dimensions a plane 平面 can be written as $ax + by + cz = d$, or $\mathbf{r}\cdot\mathbf{n} = p$, or $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$. Besides the scalar product, there is the vector product 向量积 of two vectors 向量:

which gives a vector at right angles to both. With scalar and vector products you can find distances, angles, and where lines and planes meet — including the shortest distance between skew lines 异面直线.

Worked example. Find $\mathbf{a}\times\mathbf{b}$ for $\mathbf{a} = \mathbf{i} + \mathbf{j}$ and $\mathbf{b} = \mathbf{j} + \mathbf{k}$.

Mathematical induction 数学归纳法 proves a result for every positive integer $n$ in two steps:

1. Base case: show the result is true for $n = 1$.
2. Inductive step: assume it is true for $n = k$, then prove it for $n = k + 1$.

If both steps work, the result is true for all $n$. Often you first make a conjecture 猜想 (a sensible guess) from a few cases, then confirm it by inductive proof 归纳证明.

Worked example. Prove that $\displaystyle\sum_{r=1}^{n} r = \tfrac12 n(n+1)$.

Base case $n = 1$: the left side is $1$ and the right side is $\tfrac12(1)(2) = 1$. True. Inductive step: assume $\displaystyle\sum_{r=1}^{k} r = \tfrac12 k(k+1)$. Then

This is the formula with $n = k + 1$, so by induction it holds for all $n$.

• Use the relationships between roots and coefficients (sum, sum of pairs, product) instead of solving the polynomial.
• For proof by induction, set out the base case, assume the result for $n = k$, prove it for $n = k+1$, and write a clear closing statement.
• Sketch rational-function graphs with their asymptotes and turning points; use the method of differences for a summation.
• For matrices, know the determinant, inverse and invariant lines, and interpret each transformation geometrically.

The hyperbolic functions 双曲函数 are built from the exponential function:

They obey identities much like the trigonometric ones, the main one being $\cosh^2 x - \sinh^2 x = 1$. The three reciprocals complete the set: $\operatorname{sech} x = \dfrac{1}{\cosh x}$, $\operatorname{cosech} x = \dfrac{1}{\sinh x}$, and $\coth x = \dfrac{1}{\tanh x} = \dfrac{\cosh x}{\sinh x}$ – all built from $e^x$ too, and worth recognising in integrals and mark schemes. The inverse hyperbolic functions 反双曲函数 have a logarithmic form 对数形式, for example $\sinh^{-1} x = \ln\!\left(x + \sqrt{x^2 + 1}\right)$.

Worked example. Show that $\cosh^2 x - \sinh^2 x = 1$.

You can write three linear equations in three unknowns as a single matrix equation 矩阵方程 $A\mathbf{x} = \mathbf{b}$. If $A$ is non-singular there is one solution; if $A$ is singular the equations are either inconsistent (no solution) or have infinitely many.

For a square matrix $A$, the characteristic equation 特征方程 is $\det(A - \lambda I) = 0$. Its solutions are the eigenvalues 特征值 $\lambda$, and for each one the vector $\mathbf{v}$ with $A\mathbf{v} = \lambda\mathbf{v}$ is an eigenvector 特征向量. You can then write $A = QDQ^{-1}$, where $D$ is a diagonal matrix 对角矩阵 of eigenvalues and the columns of $Q$ are the eigenvectors.

Worked example. Find the eigenvalues of $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.

The characteristic equation is

So $\lambda = 1$ or $\lambda = 3$. (The eigenvector for $\lambda = 3$ is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and for $\lambda = 1$ is $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.)

You can now differentiate the hyperbolic functions and the inverse functions $\sin^{-1}x$, $\tan^{-1}x$, $\sinh^{-1}x$ and so on. You can also find $\dfrac{d^2y}{dx^2}$ for curves given implicitly or parametrically.

A Maclaurin's series 麦克劳林级数 writes a function as a power series:

For example $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$

Worked example. Find the Maclaurin series of $f(x) = \ln(1+x)$ up to the term in $x^3$.

$f(0) = 0$; $f'(x) = \dfrac{1}{1+x}$ so $f'(0) = 1$; $f''(x) = \dfrac{-1}{(1+x)^2}$ so $f''(0) = -1$; $f'''(x) = \dfrac{2}{(1+x)^3}$ so $f'''(0) = 2$. Substituting into the series:

Learn these standard integrals:

A trigonometric substitution, a hyperbolic substitution, or completing the square 配方法 in the denominator handles related forms. A reduction formula 递推公式 links an integral $I_n$ to $I_{n-1}$, so you can work down step by step. Integration also gives the arc length 弧长 of a curve and the surface area of revolution 旋转曲面面积 when a curve is turned about an axis.

Worked example. Find $\displaystyle\int \frac{1}{\sqrt{4 - x^2}}\,dx$.

Here $a^2 = 4$, so $a = 2$ and

Bounding a sum by an integral. Draw rectangles of unit width under, or over, a decreasing curve such as $y=\frac1x$. Each rectangle of height $\frac1r$ is trapped between two integral strips, so summing gives

Shrinking the rectangle width to $\frac1n\to 0$ turns such a sum into a Riemann sum 黎曼和 that equals the integral in the limit, e.g. $\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^{n}f\!\left(\frac{r}{n}\right)=\int_0^1 f(x)\,dx$.

A complex number 复数 in polar form is $z = r(\cos\theta + i\sin\theta)$. De Moivre's theorem 棣莫弗定理 says that for any integer $n$,

It is used to expand $\cos n\theta$ and $\sin n\theta$ in powers, to sum series, and to find the $n$ roots of unity 单位根 (the solutions of $z^n = 1$, equally spaced around the unit circle).

Worked example. Use de Moivre's theorem to express $\cos 3\theta$ in terms of $\cos\theta$.

Take the real part of $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$:

For a first order linear equation $\dfrac{dy}{dx} + P(x)\,y = Q(x)$, multiply by the integrating factor 积分因子 $\mu = e^{\int P\,dx}$. The left side then becomes $\dfrac{d}{dx}(\mu y)$, so you can integrate directly.

For a linear equation with constant coefficients, the general solution 通解 is the sum of two parts: the complementary function 余函数 (the solution of the equation with the right side set to $0$, found from the auxiliary equation) and a particular integral 特积分 (any one solution of the full equation). Initial conditions then fix the constants to give the particular solution 特解; simpler equations with separable variables 可分离变量 are integrated directly.

Worked example. Solve $\dfrac{dy}{dx} + 2y = e^x$.

The integrating factor is $\mu = e^{\int 2\,dx} = e^{2x}$. Multiplying through gives $\dfrac{d}{dx}\!\left(y\,e^{2x}\right) = e^{3x}$, so

Second order, constant coefficients. For $\dfrac{d^2y}{dx^2}+b\dfrac{dy}{dx}+cy=f(x)$, solve the auxiliary equation 辅助方程 $m^2+bm+c=0$. The complementary function depends on its roots:

• two distinct real roots $m_1,m_2$: $y=Ae^{m_1x}+Be^{m_2x}$;
• a repeated root $m$: $y=(A+Bx)e^{mx}$;
• complex roots $p\pm qi$: $y=e^{px}(A\cos qx+B\sin qx)$.

Then add a particular integral by trying a form matching $f(x)$ (a polynomial, $ae^{bx}$, or $a\cos px+b\sin px$); if that trial already appears in the complementary function, multiply it by $x$.

Worked example. Solve $\dfrac{d^2y}{dx^2}-4\dfrac{dy}{dx}+4y=0$. The auxiliary equation $m^2-4m+4=(m-2)^2=0$ has a repeated root $m=2$, so $y=(A+Bx)e^{2x}$.

• Learn the hyperbolic identities and derivatives, and use the inverse hyperbolic functions in integration.
• Solve a linear differential equation as complementary function $+$ particular integral, then apply the boundary conditions.
• Use de Moivre's theorem for powers and roots of complex numbers and to derive trigonometric identities.
• Reduce an integral to a standard form by a stated substitution.

A projectile 抛射体 moves freely under gravity, so it has constant acceleration 匀加速 $g$ downwards and no horizontal acceleration. Treat the horizontal and vertical motions separately. If it is launched at speed $u$ and angle $\alpha$:

Eliminating $t$ gives the Cartesian equation 直角坐标方程 of the trajectory 轨迹 (the path), which is a parabola. For a fixed launch speed, all the trajectories lie under a bounding parabola 包络抛物线 (the envelope of reachable points). The range on level ground is $\dfrac{u^2\sin 2\alpha}{g}$ and the greatest height is $\dfrac{u^2\sin^2\alpha}{2g}$.

Worked example. A ball is thrown at $u = 20\ \text{m s}^{-1}$ at $30^\circ$ to the horizontal. Find the range and greatest height.

The moment 力矩 of a force about a point is force $\times$ perpendicular distance; it measures turning effect. The weight of a body acts at its centre of mass 质心, which you can find using symmetry 对称 for a uniform flat shape (a lamina 薄片), or by treating a composite body as a set of particles.

A rigid body under coplanar forces 共面力 is in equilibrium 平衡 when two conditions both hold: the vector sum of the forces is zero, and the sum of the moments about any point is zero. A body may also be on the edge of toppling 翻倒 (turning over) or sliding 滑动 (slipping).

Worked example. A uniform beam $AB$ of length $4\ \text{m}$ and weight $100\ \text{N}$ rests on a pivot at $A$. A vertical force $F$ at $B$ keeps it horizontal. Find $F$.

Take moments about $A$ (the weight acts at the centre, $2\ \text{m}$ from $A$):

For a particle moving in a circle of radius $r$, the angular speed 角速度 $\omega$ links to the speed by $v = r\omega$. The acceleration points towards the centre — the centripetal acceleration 向心加速度 — with size

In a horizontal circle 水平圆 the speed is constant — as in a conical pendulum 圆锥摆 (a mass swung on a string). In a vertical circle 竖直圆 use energy conservation, because the speed changes with height; the normal contact force 法向接触力 or string tension provides the centripetal force.

Worked example. A particle moves in a horizontal circle of radius $2\ \text{m}$ with angular speed $3\ \text{rad s}^{-1}$. Find its speed and acceleration.

Hooke's law 胡克定律 says the tension in an elastic string or spring is proportional to its extension $x$:

where $L$ is the natural length and $\lambda$ is the modulus of elasticity 弹性模量. Stretching stores elastic potential energy 弹性势能:

Worked example. An elastic string of natural length $2\ \text{m}$ and modulus $50\ \text{N}$ is stretched by $0.5\ \text{m}$. Find the tension and the stored energy.

When the force depends on position $x$, use acceleration in the form $a = v\dfrac{dv}{dx}$, which turns Newton's law into a differential equation 微分方程 relating $v$ and $x$.

Worked example. A particle of mass $8\ \text{kg}$ moves along a line under a variable force 变力 of size $(x^3 + 4x)\ \text{N}$ acting in the direction of motion. When $x = 0$, $v = 1$. Find $v$ in terms of $x$.

Newton's law gives $8v\dfrac{dv}{dx} = x^3 + 4x$. Separating and integrating:

At $x = 0$, $v = 1$ gives $C = 4$, so $4v^2 = \tfrac14 x^4 + 2x^2 + 4 = 4\left(\tfrac{x^2}{4} + 1\right)^2$. Hence $v = \tfrac14 x^2 + 1$.

In a collision, total linear momentum is conserved — the conservation of linear momentum 动量守恒. The bounciness is measured by Newton's experimental law 牛顿实验定律, which defines the coefficient of restitution 恢复系数 $e$:

Here $e = 1$ is perfectly elastic (no energy lost) and $e = 0$ is inelastic (the bodies stay together). In an oblique impact 斜碰撞, resolve the velocities along and perpendicular to the line of impact, applying restitution along it.

Worked example. A sphere $A$ of mass $2\ \text{kg}$ moving at $5\ \text{m s}^{-1}$ hits a stationary sphere $B$ of mass $3\ \text{kg}$, with $e = 0.5$. Find the speeds afterwards.

Momentum: $2(5) = 2v_A + 3v_B$, so $2v_A + 3v_B = 10$. Restitution: $v_B - v_A = 0.5(5) = 2.5$. Solving together gives $v_A = 0.5\ \text{m s}^{-1}$ and $v_B = 3.0\ \text{m s}^{-1}$.

• For projectiles, resolve into horizontal (constant velocity) and vertical ($a = g$) motion, linked by the same time.
• For a rigid body in equilibrium, take moments about a point that removes an unknown force.
• For circular motion, use $F = mv^2/r = m\omega^2 r$ towards the centre; in a vertical circle check the minimum speed at the top.
• With a variable force, use $a = v\,\frac{dv}{dx}$ and integrate; elastic PE $= \frac{\lambda x^2}{2L}$.