This handout covers Topic 1: Pure Mathematics 纯数学 1. It is the algebra 代数 and calculus 微积分 core of the course. Each ## section is one syllabus subtopic.
A-Level Mathematics
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1 Pure Mathematics 1
1.1
Quadratics
Syllabus
- carry out the process of completing the square for a quadratic polynomial $ax^2 + bx + c$ and use a completed square form
- find the discriminant of a quadratic polynomial $ax^2 + bx + c$ and use the discriminant
- solve quadratic equations, and quadratic inequalities, in one unknown
- solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic
- recognise and solve equations in $x$ which are quadratic in some function of $x$
Source: Cambridge International syllabus
A suspension bridge: the main cable hangs in a parabola.A quadratic 二次式 is an expression of the form $ax^2 + bx + c$, where $a \neq 0$. The letters $a$, $b$, $c$ are the coefficients 系数 (the fixed numbers). Much of this section is about solving the equation $ax^2 + bx + c = 0$.
Completing the square
To complete the square 配方 means to write the quadratic in the form
$$a(x + p)^2 + q.$$This form is useful: it shows the vertex 顶点 (turning point) of the curve at $(-p,\ q)$, and it gives a quick way to solve the equation.Worked example. Write $9x^2 - 36x + 8$ in the form $p(x + q)^2 + r$.
Take the factor $9$ out of the first two terms, then complete the square inside:
$$\begin{aligned} 9x^2 - 36x + 8 &= 9\left(x^2 - 4x\right) + 8 \\ &= 9\left((x - 2)^2 - 4\right) + 8 \\ &= 9(x - 2)^2 - 36 + 8 = 9(x - 2)^2 - 28. \end{aligned}$$So $p = 9$, $q = -2$, $r = -28$.The discriminant
The discriminant 判别式 of $ax^2 + bx + c$ is
$$\Delta = b^2 - 4ac.$$It tells you how many real roots 实根 (real solutions) the equation $ax^2 + bx + c = 0$ has:Discriminant Roots $b^2 - 4ac > 0$ two distinct 相异 real roots $b^2 - 4ac = 0$ one repeated real root $b^2 - 4ac < 0$ no real roots
The sign of $b^2-4ac$ decides how many times the parabola meets the $x$-axis.Worked example. Find the values of the constant $k$ for which $3kx^2 + (k + 8)x + 3 = 0$ has two distinct real roots.
Here $a = 3k$, $b = k + 8$, $c = 3$. For two distinct real roots you need $b^2 - 4ac > 0$:
$$(k + 8)^2 - 4(3k)(3) > 0 \;\Rightarrow\; k^2 + 16k + 64 - 36k > 0 \;\Rightarrow\; k^2 - 20k + 64 > 0.$$Factorise: $(k - 4)(k - 16) > 0$, so $k < 4$ or $k > 16$. You also need $a \neq 0$, so $k \neq 0$.Quadratic equations and inequalities
To solve a quadratic equation 二次方程, factorise, complete the square, or use the formula
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$To solve a quadratic inequality 二次不等式 such as $(k - 4)(k - 16) > 0$, find the two roots, then decide which side of each root makes the statement true. A sketch of the parabola 抛物线 helps: the curve is above the $x$-axis (positive) outside the roots and below it (negative) between them.Simultaneous equations
To solve a pair of simultaneous equations 联立方程 where one is linear and one is quadratic, use substitution 代入: rearrange the linear equation for one letter, then put that into the quadratic. This gives a single quadratic to solve.
Equations that are quadratic in disguise
Some equations are quadratic in some function of $x$. For example $x^4 - 5x^2 + 4 = 0$ is quadratic in $x^2$: let $u = x^2$, solve $u^2 - 5u + 4 = 0$, then go back to $x$. You will use this idea again in trigonometry.
Vocabulary TrainEnglish Chinese Pinyin Pure Mathematics 纯数学 chún shù xué algebra 代数 dài shù calculus 微积分 wēi jī fēn quadratic 二次式 èr cì shì coefficient 系数 xì shù complete the square 配方 pèi fāng vertex 顶点 dǐng diǎn discriminant 判别式 pàn bié shì real roots 实根 shí gēn distinct 相异 xiāng yì quadratic equation 二次方程 èr cì fāng chéng quadratic inequality 二次不等式 èr cì bù děng shì parabola 抛物线 pāo wù xiàn simultaneous equations 联立方程 lián lì fāng chéng substitution 代入 dài rù 1.2
Functions
Syllabus
- understand the terms function, domain, range, one-one function, inverse function and composition of functions
- identify the range of a given function in simple cases, and find the composition of two given functions
- determine whether or not a given function is one-one, and find the inverse of a one-one function in simple cases
- illustrate in graphical terms the relation between a one-one function and its inverse
- understand and use the transformations of the graph of $y = f(x)$ given by $y = f(x) + a$, $y = f(x + a)$, $y = af(x)$, $y = f(ax)$ and simple combinations of these
Source: Cambridge International syllabus
A function 函数 is a rule that sends each input to exactly one output. Write it as $f(x)$.
- The domain 定义域 is the set of allowed inputs $x$.
- The range 值域 is the set of outputs the function actually produces.
A function is one-one 一一对应 if different inputs always give different outputs. (No output is repeated.) Only a one-one function has an inverse function 反函数 $f^{-1}$, which reverses the rule.
The composition 复合 of two functions means doing one after the other. $fg(x)$ means "do $g$ first, then $f$": $fg(x) = f(g(x))$. The composite function $fg$ exists only when the range of $g$ lies inside the domain of $f$.
Finding an inverse
To find $f^{-1}$: write $y = f(x)$, make $x$ the subject, then swap letters.
Worked example. The function $f(x) = (x + 3)^2 - 12$ is defined for $x \geqslant 0$. Find $f^{-1}(x)$.
Write $y = (x + 3)^2 - 12$ and solve for $x$:
$$(x + 3)^2 = y + 12 \;\Rightarrow\; x + 3 = \sqrt{y + 12} \;\Rightarrow\; x = \sqrt{y + 12} - 3.$$You take the positive square root because $x \geqslant 0$ means $x + 3 \geqslant 3 > 0$. So$$f^{-1}(x) = \sqrt{x + 12} - 3.$$Graphs of inverses and transformations
The graph of $y = f^{-1}(x)$ is the reflection 反射 of $y = f(x)$ in the line $y = x$.
Reflecting $y=f(x)$ in the line $y=x$ gives its inverse; a point $(a,b)$ becomes $(b,a)$.You should know these transformations 变换 of $y = f(x)$:
New equation Effect on the graph $y = f(x) + a$ translation 平移 up by $a$ $y = f(x + a)$ translation left by $a$ $y = a\,f(x)$ stretch 伸缩 in the $y$-direction, scale factor $a$ $y = f(ax)$ stretch in the $x$-direction, scale factor $\tfrac{1}{a}$ When two transformations are combined, the order can matter. State each one fully (type, direction, and amount).
Adding to the output or input slides the curve; a multiplier stretches it.Vocabulary TrainEnglish Chinese Pinyin function 函数 hán shù domain 定义域 dìng yì yù range 值域 zhí yù one-one 一一对应 yī yī duì yìng inverse function 反函数 fǎn hán shù composition 复合 fù hé reflection 反射 fǎn shè transformations 变换 biàn huàn translation 平移 píng yí stretch 伸缩 shēn suō 1.3
Coordinate geometry
Syllabus
- find the equation of a straight line given sufficient information
- interpret and use any of the forms $y = mx + c$, $y - y_1 = m(x - x_1)$, $ax + by + c = 0$ in solving problems
- understand that the equation $(x - a)^2 + (y - b)^2 = r^2$ represents the circle with centre $(a, b)$ and radius $r$
- use algebraic methods to solve problems involving lines and circles
- understand the relationship between a graph and its associated algebraic equation, and use the relationship between points of intersection of graphs and solutions of equations
Source: Cambridge International syllabus
Coordinate geometry 坐标几何 studies lines and circles using their equations.
Straight lines
The gradient 斜率 (steepness) of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is
$$m = \frac{y_2 - y_1}{x_2 - x_1}.$$You can write the equation of a straight line 直线方程 in any of these forms:$$y = mx + c, \qquad y - y_1 = m(x - x_1), \qquad ax + by + c = 0.$$Two lines are parallel 平行 when their gradients are equal, and perpendicular 垂直 (at right angles) when the product of their gradients is $-1$.Circles
The circle 圆 with centre 圆心 $(a, b)$ and radius 半径 $r$ has equation
$$(x - a)^2 + (y - b)^2 = r^2.$$An expanded form like $x^2 + y^2 - 6x + 10y - 27 = 0$ is the same circle: complete the square in $x$ and in $y$ to find the centre and radius.A tangent 切线 to a circle touches it at one point and is perpendicular to the radius at that point. This right-angle fact solves most circle problems.
A tangent touches the circle once and meets the radius at a right angle.Worked example. The points $P(1, 1)$ and $Q(7, 11)$ are the ends of a diameter 直径 of a circle. Find the equation of the circle.
The centre is the midpoint of $PQ$:
$$\left(\frac{1 + 7}{2},\ \frac{1 + 11}{2}\right) = (4, 6).$$The radius is half the length of $PQ$:$$r = \tfrac12\sqrt{(7 - 1)^2 + (11 - 1)^2} = \tfrac12\sqrt{36 + 100} = \tfrac12\sqrt{136} = \sqrt{34}.$$So the circle is $(x - 4)^2 + (y - 6)^2 = 34$.Vocabulary TrainEnglish Chinese Pinyin coordinate geometry 坐标几何 zuò biāo jǐ hé gradient 斜率 xié lǜ equation of a straight line 直线方程 zhí xiàn fāng chéng parallel 平行 píng xíng perpendicular 垂直 chuí zhí circle 圆 yuán centre 圆心 yuán xīn radius 半径 bàn jìng tangent 切线 qiè xiàn diameter 直径 zhí jìng 1.4
Circular measure
Syllabus
- understand the definition of a radian, and use the relationship between radians and degrees
- use the formulae $s = r\theta$ and $A = \tfrac{1}{2}r^2\theta$ in solving problems concerning the arc length and sector area of a circle
Source: Cambridge International syllabus
Radians
A radian 弧度 is another way to measure angles. One radian is the angle at the centre of a circle that cuts off an arc 弧 equal in length to the radius. The link between radians and degrees 度 is
$$\pi \text{ radians} = 180^\circ.$$So to change degrees to radians, multiply by $\dfrac{\pi}{180}$; to change radians to degrees, multiply by $\dfrac{180}{\pi}$.Arc length and sector area
For a sector 扇形 with radius $r$ and angle $\theta$ in radians:
$$\text{arc length} = s = r\theta, \qquad \text{sector area} = A = \tfrac12 r^2 \theta.$$A chord 弦 cuts the sector into a triangle and a segment 弓形. The segment area is the sector minus the triangle:$$\text{segment} = \tfrac12 r^2 \theta - \tfrac12 r^2 \sin\theta = \tfrac12 r^2(\theta - \sin\theta).$$
The shaded segment is the part of the sector between the chord and the arc.Worked example. A sector has centre $O$ and the angle at $O$ is $\tfrac{2}{3}\pi$ radians. Show that the segment cut off by the chord has area about $0.614 r^2$.
$$\text{segment} = \tfrac12 r^2\left(\tfrac{2}{3}\pi - \sin\tfrac{2}{3}\pi\right) = \tfrac12 r^2(2.0944 - 0.8660) = \tfrac12 r^2(1.2284) \approx 0.614 r^2.$$Vocabulary TrainEnglish Chinese Pinyin radian 弧度 hú dù arc 弧 hú degrees 度 dù sector 扇形 shàn xíng chord 弦 xián segment 弓形 gōng xíng 1.5
Trigonometry
Syllabus
- sketch and use graphs of the sine, cosine and tangent functions (for angles of any size, and using either degrees or radians)
- use the exact values of the sine, cosine and tangent of $30°$, $45°$, $60°$, and related angles
- use the notations $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$ to denote the principal values of the inverse trigonometric relations
- use the identities $\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta \equiv 1$
- find all the solutions of simple trigonometrical equations lying in a specified interval
Source: Cambridge International syllabus
A Ferris wheel: a point on the rim rises and falls like a sine curve.Graphs and exact values
You must know the shape of the graphs of the sine 正弦, cosine 余弦 and tangent function 正切 (written $\sin$, $\cos$, $\tan$). The sine and cosine graphs wave between $-1$ and $1$ and repeat every $360^\circ$ ($2\pi$). Learn these exact values:
$\theta$ $30^\circ$ $45^\circ$ $60^\circ$ $\sin\theta$ $\tfrac12$ $\tfrac{1}{\sqrt2}$ $\tfrac{\sqrt3}{2}$ $\cos\theta$ $\tfrac{\sqrt3}{2}$ $\tfrac{1}{\sqrt2}$ $\tfrac12$ $\tan\theta$ $\tfrac{1}{\sqrt3}$ $1$ $\sqrt3$
Over one turn $\sin$ and $\cos$ stay between $-1$ and $1$; $\tan$ shoots off at $90^\circ$ and $270^\circ$.The notations $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$ mean the inverse angle (the principal value 主值).
Identities
An identity 恒等式 is true for every value of the angle. The two you must know are
$$\tan\theta \equiv \frac{\sin\theta}{\cos\theta}, \qquad \sin^2\theta + \cos^2\theta \equiv 1.$$Use them to rewrite an equation so it contains only one trig function.Solving trigonometric equations
To solve a trigonometric equation 三角方程, first reduce it to one function, then find every solution in the given interval.
Worked example. Solve $6\sin\theta = 1 + \dfrac{2}{\sin\theta}$ for $-180^\circ < \theta < 180^\circ$.
Multiply through by $\sin\theta$ to clear the fraction. This makes a quadratic in $\sin\theta$:
$$6\sin^2\theta - \sin\theta - 2 = 0 \;\Rightarrow\; (3\sin\theta - 2)(2\sin\theta + 1) = 0.$$So $\sin\theta = \tfrac23$ or $\sin\theta = -\tfrac12$.- $\sin\theta = \tfrac23$: $\theta = 41.8^\circ$ or $\theta = 180^\circ - 41.8^\circ = 138.2^\circ$.
- $\sin\theta = -\tfrac12$: $\theta = -30^\circ$ or $\theta = -150^\circ$.
The four solutions are $\theta = -150^\circ,\ -30^\circ,\ 41.8^\circ,\ 138.2^\circ$.
Vocabulary TrainEnglish Chinese Pinyin sine 正弦 zhèng xián cosine 余弦 yú xián tangent function 正切 zhèng qiē principal value 主值 zhǔ zhí identity 恒等式 héng děng shì trigonometric equation 三角方程 sān jiǎo fāng chéng 1.6
Series
Syllabus
- use the expansion of $(a + b)^n$, where $n$ is a positive integer
- recognise arithmetic and geometric progressions
- use the formulae for the $n$th term and for the sum of the first $n$ terms to solve problems involving arithmetic or geometric progressions
- use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression
Source: Cambridge International syllabus
The binomial expansion
For a positive integer $n$, the binomial expansion 二项展开式 is
$$(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + b^n,$$where $\binom{n}{r} = \dfrac{n!}{r!\,(n - r)!}$ is a binomial coefficient 二项式系数.Worked example. Find the first three terms, in ascending powers of $x$, of $(2 - px)^5$.
$$(2 - px)^5 = 2^5 + \binom{5}{1}2^4(-px) + \binom{5}{2}2^3(-px)^2 + \cdots = 32 - 80px + 80p^2x^2 + \cdots$$Arithmetic and geometric progressions
A progression 数列 (sequence) is a list of terms following a rule.
- An arithmetic progression 等差数列 (AP) adds a fixed common difference 公差 $d$ each step. The $n$th term 项 is $u_n = a + (n - 1)d$, and the sum of the first $n$ terms is $S_n = \tfrac{n}{2}\big(2a + (n - 1)d\big)$.
- A geometric progression 等比数列 (GP) multiplies by a fixed common ratio 公比 $r$ each step. The $n$th term is $u_n = ar^{\,n-1}$, and $S_n = \dfrac{a(1 - r^n)}{1 - r}$.
A GP converges 收敛 (settles to a limit) when $|r| < 1$. Then it has a sum to infinity 无穷和
$$S_\infty = \frac{a}{1 - r}.$$Worked example. The third term of a GP is $18$ and the sum of the first three terms is $26$. The common ratio is negative. Find the sum to infinity.
From $ar^2 = 18$ you get $a = \dfrac{18}{r^2}$. Put this into $a(1 + r + r^2) = 26$:
$$18(1 + r + r^2) = 26r^2 \;\Rightarrow\; 8r^2 - 18r - 18 = 0 \;\Rightarrow\; (4r + 3)(r - 3) = 0.$$The ratio is negative, so $r = -\tfrac34$ and $a = \dfrac{18}{(3/4)^2} = 32$. Then$$S_\infty = \frac{32}{1 - (-\tfrac34)} = \frac{32}{\tfrac74} = \frac{128}{7}.$$Vocabulary TrainEnglish Chinese Pinyin binomial expansion 二项展开式 èr xiàng zhǎn kāi shì binomial coefficient 二项式系数 èr xiàng shì xì shù progression 数列 shù liè arithmetic progression 等差数列 děng chā shù liè common difference 公差 gōng chāi term 项 xiàng geometric progression 等比数列 děng bǐ shù liè common ratio 公比 gōng bǐ converges 收敛 shōu liǎn sum to infinity 无穷和 wú qióng hé 1.7
Differentiation
Syllabus
- understand the gradient of a curve at a point as the limit of the gradients of a suitable sequence of chords, and use the notations $f'(x)$, $f''(x)$, $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ for first and second derivatives
- use the derivative of $x^n$ (for any rational $n$), together with constant multiples, sums and differences of functions, and of composite functions using the chain rule
- apply differentiation to gradients, tangents and normals, increasing and decreasing functions and rates of change
- locate stationary points and determine their nature, and use information about stationary points in sketching graphs
Source: Cambridge International syllabus
Differentiation 微分 finds the gradient of a curve 曲线斜率 at each point. The gradient is the limit 极限 of the gradients of shorter and shorter chords, called the derivative 导数.
The rules
Write the derivative as $f'(x)$ or $\dfrac{dy}{dx}$. The basic rule is
$$\frac{d}{dx}\left(x^n\right) = n x^{n-1} \quad \text{for any rational } n.$$Differentiate sums term by term, and use the chain rule 链式法则 for a function inside a function:$$\frac{d}{dx}\,f(g(x)) = f'(g(x)) \cdot g'(x).$$Differentiating again gives the second derivative 二阶导数 $f''(x)$ or $\dfrac{d^2y}{dx^2}$.Using the derivative
- Tangent and normal. The gradient of the curve at a point is the gradient of the tangent there. The normal 法线 is perpendicular to the tangent, so its gradient is $-\dfrac{1}{\text{(tangent gradient)}}$.
- Increasing or decreasing. The function is an increasing function 增函数 where $\dfrac{dy}{dx} > 0$, and a decreasing function 减函数 where $\dfrac{dy}{dx} < 0$.
- Rate of change. A derivative is a rate of change 变化率. Linked rates use the chain rule, e.g. $\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$.
Stationary points
A stationary point 驻点 is where $\dfrac{dy}{dx} = 0$. Test its nature with the second derivative: $f''(x) > 0$ gives a minimum point 极小值点, and $f''(x) < 0$ gives a maximum point 极大值点.
At a maximum or a minimum the tangent is flat, so $\frac{dy}{dx}=0$.Worked example. The curve $y = 4x^{1/2} - x$ has a maximum point at $x = a$. Find $a$.
$$\frac{dy}{dx} = 2x^{-1/2} - 1 = 0 \;\Rightarrow\; \frac{2}{\sqrt{x}} = 1 \;\Rightarrow\; \sqrt{x} = 2 \;\Rightarrow\; x = 4.$$So $a = 4$.Vocabulary TrainEnglish Chinese Pinyin differentiation 微分 wēi fēn gradient of a curve 曲线斜率 qū xiàn xié lǜ limit 极限 jí xiàn derivative 导数 dǎo shù chain rule 链式法则 liàn shì fǎ zé second derivative 二阶导数 èr jiē dǎo shù normal 法线 fǎ xiàn increasing function 增函数 zēng hán shù decreasing function 减函数 jiǎn hán shù rate of change 变化率 biàn huà lǜ stationary point 驻点 zhù diǎn minimum point 极小值点 jí xiǎo zhí diǎn maximum point 极大值点 jí dà zhí diǎn 1.8
Integration
Syllabus
- understand integration as the reverse process of differentiation, and integrate $(ax + b)^n$ (for any rational $n$ except $-1$), together with constant multiples, sums and differences
- solve problems involving the evaluation of a constant of integration
- evaluate definite integrals
- use definite integration to find the area of a region bounded by a curve and lines parallel to the axes, or between a curve and a line or between two curves, and a volume of revolution about one of the axes
Source: Cambridge International syllabus
Integration 积分 is the reverse of differentiation. Reversing the power rule gives
$$\int (ax + b)^n \, dx = \frac{(ax + b)^{n+1}}{a(n + 1)} + C \quad (n \neq -1).$$The $+\,C$ is the constant of integration 积分常数. If you know one point on the curve, substitute it to find $C$.Definite integrals and area
A definite integral 定积分 has limits and gives a number:
$$\int_p^q f(x)\, dx = \big[F(x)\big]_p^q = F(q) - F(p).$$The area of the region 区域 between a curve and the $x$-axis, from $x = p$ to $x = q$, is $\displaystyle\int_p^q y \, dx$. For the area between two curves, integrate (top curve $-$ bottom curve).
The definite integral $\int_0^4 y\,dx$ is the shaded area under the curve.Worked example. The curve $y = 4x^{1/2} - x$ meets the $x$-axis again at $x = 16$. Find the area between the curve and the $x$-axis from $x = 0$ to $x = 4$.
$$\int_0^4 \left(4x^{1/2} - x\right) dx = \left[\frac{8}{3}x^{3/2} - \frac{x^2}{2}\right]_0^4 = \frac{8}{3}(8) - \frac{16}{2} = \frac{64}{3} - 8 = \frac{40}{3}.$$Volume of revolution
When a region is turned all the way around an axis it sweeps out a solid. The volume of revolution 旋转体体积 about the $x$-axis is
$$V = \pi \int_p^q y^2 \, dx,$$and about the $y$-axis it is $V = \pi \displaystyle\int x^2 \, dy$. For example, the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$, turned about the $x$-axis, has volume $\pi\displaystyle\int_0^4 x\, dx = \pi\big[\tfrac{x^2}{2}\big]_0^4 = 8\pi$.Vocabulary TrainEnglish Chinese Pinyin integration 积分 jī fēn constant of integration 积分常数 jī fēn cháng shù definite integral 定积分 dìng jī fēn region 区域 qū yù volume of revolution 旋转体体积 xuán zhuǎn tǐ tǐ jī -
2 Pure Mathematics 2
This handout covers Topic 2: Pure Mathematics 纯数学 2. It adds the modulus and polynomial algebra, logarithms 对数 and the exponential function 指数函数, more trigonometry, and new ways to differentiate and integrate.
2.1
Algebra
Syllabus
- understand the meaning of $|x|$, sketch the graph of $y = |ax + b|$ and use relations such as $|a| = |b| \Leftrightarrow a^2 = b^2$ and $|x - a| < b \Leftrightarrow a - b < x < a + b$ when solving equations and inequalities
- divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, and identify the quotient and remainder (which may be zero)
- use the factor theorem and the remainder theorem
Source: Cambridge International syllabus
The modulus
The modulus 绝对值 $|x|$ is the size of a number with its sign removed, so $|x| \geqslant 0$ always. The graph of $y = |ax + b|$ is a "V" shape that bounces off the $x$-axis. Two useful rules for solving equations and inequalities are
$$|a| = |b| \;\Leftrightarrow\; a^2 = b^2, \qquad |x - a| < b \;\Leftrightarrow\; a - b < x < a + b.$$
Taking the modulus folds the part of the line below the axis upward into a V.Worked example. Solve $|3x + 8| < 9$.
Using the second rule with the inequality written as $-9 < 3x + 8 < 9$:
$$-9 < 3x + 8 < 9 \;\Rightarrow\; -17 < 3x < 1 \;\Rightarrow\; -\tfrac{17}{3} < x < \tfrac13.$$Polynomial division and the factor and remainder theorems
A polynomial 多项式 is a sum of powers of $x$, such as $2x^4 + 3x^2 - 5$. Its degree 次数 is the highest power. When you divide one polynomial by another you get a quotient 商 and a remainder 余数.
- Remainder theorem 余数定理: the remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$.
- Factor theorem 因式定理: $(x - a)$ is a factor of $p(x)$ exactly when $p(a) = 0$.
Worked example. The polynomial $p(x) = 2x^4 + kx^3 + kx^2 + 17x + 18$ has factor $(x + 2)$. Find $k$.
By the factor theorem $p(-2) = 0$:
$$2(16) + k(-8) + k(4) + 17(-2) + 18 = 0 \;\Rightarrow\; 16 - 4k = 0 \;\Rightarrow\; k = 4.$$Vocabulary TrainEnglish Chinese Pinyin modulus 绝对值 jué duì zhí polynomial 多项式 duō xiàng shì degree 次数 cì shù quotient 商 shāng remainder 余数 yú shù remainder theorem 余数定理 yú shù dìng lǐ factor theorem 因式定理 yīn shì dìng lǐ 2.2
Logarithmic and exponential functions
Syllabus
- understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base)
- understand the definition and properties of $e^x$ and $\ln x$, including their relationship as inverse functions and their graphs
- use logarithms to solve equations and inequalities in which the unknown appears in indices
- use logarithms to transform a given relationship to linear form, and hence determine unknown constants by considering the gradient and/or intercept
Source: Cambridge International syllabus
Earthquake strength is measured on the logarithmic Richter scale.
Populations can grow exponentially when resources are plentiful.A logarithm answers the question "what power?". If $a^x = y$ then $x = \log_a y$. Logarithms and indices 指数 (powers) are reverse ideas. The laws of logarithms 对数定律 are
$$\log(mn) = \log m + \log n, \qquad \log\!\frac{m}{n} = \log m - \log n, \qquad \log(m^k) = k\log m.$$The exponential function $e^x$ and the natural logarithm 自然对数 $\ln x$ are inverse functions, so $\ln(e^x) = x$ and $e^{\ln x} = x$. When the unknown is in the power, take logs of both sides.
$e^x$ and $\ln x$ undo each other, so each is the other reflected in $y=x$.Worked example. Solve $4^x < 0.05$.
$$4^x < 0.05 \;\Rightarrow\; x\ln 4 < \ln 0.05 \;\Rightarrow\; x < \frac{\ln 0.05}{\ln 4} = -2.16 \ (\text{3 s.f.}).$$Linear form 线性形式: a relationship like $y = Ax^n$ becomes a straight line if you take logs: $\ln y = \ln A + n\ln x$. Plotting $\ln y$ against $\ln x$ gives a line with gradient $n$ and intercept $\ln A$, so you can find the unknown constants.
Vocabulary TrainEnglish Chinese Pinyin Pure Mathematics 纯数学 chún shù xué logarithms 对数 duì shù exponential function 指数函数 zhǐ shù hán shù indices 指数 zhǐ shù laws of logarithms 对数定律 duì shù dìng lǜ natural logarithm 自然对数 zì rán duì shù linear form 线性形式 xiàn xìng xíng shì 2.3
Trigonometry
Syllabus
- understand the relationship of the secant, cosecant and cotangent functions to cosine, sine and tangent, and use properties and graphs of all six trigonometric functions for angles of any magnitude
- use trigonometrical identities for the simplification and exact evaluation of expressions and in the course of solving equations, selecting an appropriate identity, showing familiarity in particular with $\sec^2\theta \equiv 1 + \tan^2\theta$ and $\csc^2\theta \equiv 1 + \cot^2\theta$, the expansions of $\sin(A \pm B)$, $\cos(A \pm B)$ and $\tan(A \pm B)$, the formulae for $\sin 2A$, $\cos 2A$ and $\tan 2A$, and the expression of $a\sin\theta + b\cos\theta$ in the forms $R\sin(\theta \pm \alpha)$ and $R\cos(\theta \pm \alpha)$
Source: Cambridge International syllabus
There are three more functions, each the reciprocal of one you know: the secant 正割 $\sec\theta = \dfrac{1}{\cos\theta}$, the cosecant 余割 $\csc\theta = \dfrac{1}{\sin\theta}$, and the cotangent 余切 $\cot\theta = \dfrac{1}{\tan\theta}$.
$\sec\theta=1/\cos\theta$ rises to an asymptote wherever $\cos\theta=0$.You must know these trigonometric identities 三角恒等式 and choose the right one for each problem:
$$\sec^2\theta \equiv 1 + \tan^2\theta, \qquad \csc^2\theta \equiv 1 + \cot^2\theta.$$You also use the compound angle 复合角 formulae for $\sin(A \pm B)$, $\cos(A \pm B)$, $\tan(A \pm B)$, the double angle 二倍角 formulae$$\sin 2A = 2\sin A\cos A, \quad \cos 2A = 2\cos^2 A - 1, \quad \tan 2A = \frac{2\tan A}{1 - \tan^2 A},$$and the R-formula 辅助角公式 $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \dfrac{b}{a}$.Worked example. Solve $2\tan^2\theta + 3\sec\theta = 18$ for $-180^\circ < \theta < 180^\circ$.
Replace $\tan^2\theta$ with $\sec^2\theta - 1$ to get one function:
$$2(\sec^2\theta - 1) + 3\sec\theta = 18 \;\Rightarrow\; 2\sec^2\theta + 3\sec\theta - 20 = 0 \;\Rightarrow\; (2\sec\theta - 5)(\sec\theta + 4) = 0.$$So $\sec\theta = \tfrac52$ or $\sec\theta = -4$, giving $\cos\theta = \tfrac25$ or $\cos\theta = -\tfrac14$. The solutions are $\theta = \pm 66.4^\circ$ and $\theta = \pm 104.5^\circ$.Vocabulary TrainEnglish Chinese Pinyin secant 正割 zhèng gē cosecant 余割 yú gē cotangent 余切 yú qiē trigonometric identities 三角恒等式 sān jiǎo héng děng shì compound angle 复合角 fù hé jiǎo double angle 二倍角 èr bèi jiǎo R-formula 辅助角公式 fǔ zhù jiǎo gōng shì 2.4
Differentiation
Syllabus
- use the derivatives of $e^x$, $\ln x$, $\sin x$, $\cos x$, $\tan x$, together with constant multiples, sums, differences and composites
- differentiate products and quotients
- find and use the first derivative of a function which is defined parametrically or implicitly
Source: Cambridge International syllabus
Learn these standard derivatives:
$$\frac{d}{dx}e^x = e^x, \quad \frac{d}{dx}\ln x = \frac{1}{x}, \quad \frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x, \quad \frac{d}{dx}\tan x = \sec^2 x.$$For a product or a quotient of two functions, use:
- the product rule 乘积法则: $(uv)' = u'v + uv'$;
- the quotient rule 商法则: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
When a curve is given by parametric equations 参数方程 $x = x(t)$, $y = y(t)$, the gradient is $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$. When $y$ is defined implicitly 隐式 (not made the subject), differentiate every term with respect to $x$, using the chain rule on the $y$ terms, then solve for $\dfrac{dy}{dx}$.
Worked example. Given $y = 6x\cos(x^2 + 1)$, find $\dfrac{dy}{dx}$.
Use the product rule with $u = 6x$ and $v = \cos(x^2 + 1)$ (and the chain rule for $v$):
$$\frac{dy}{dx} = 6\cos(x^2 + 1) + 6x\cdot\big(-2x\sin(x^2 + 1)\big) = 6\cos(x^2 + 1) - 12x^2\sin(x^2 + 1).$$Vocabulary TrainEnglish Chinese Pinyin product rule 乘积法则 chéng jī fǎ zé quotient rule 商法则 shāng fǎ zé parametric equations 参数方程 cān shù fāng chéng implicitly 隐式 yǐn shì 2.5
Integration
Syllabus
- extend the idea of 'reverse differentiation' to include the integration of $e^{ax + b}$, $\dfrac{1}{ax + b}$, $\sin(ax + b)$, $\cos(ax + b)$ and $\sec^2(ax + b)$
- use trigonometrical relationships in carrying out integration
- understand and use the trapezium rule to estimate the value of a definite integral
Source: Cambridge International syllabus
Reverse each new derivative. For a linear inside function $(ax + b)$:
$$\int e^{ax+b}\,dx = \frac{1}{a}e^{ax+b} + C, \qquad \int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax + b| + C,$$$$\int \sin(ax+b)\,dx = -\frac{1}{a}\cos(ax+b) + C, \qquad \int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b) + C, \qquad \int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b) + C.$$To integrate a power of $\sin$ or $\cos$, first use an identity to remove the power. When you cannot integrate exactly, the trapezium rule 梯形法则 estimates a definite integral:$$\int_a^b y\,dx \approx \tfrac{h}{2}\big[y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1})\big].$$
Each strip of width $h$ is a trapezium; their areas add up to estimate the integral.Worked example. Find $\displaystyle\int 6\sin^2 x\,dx$.
Use $\sin^2 x = \tfrac12(1 - \cos 2x)$:
$$\int 6\sin^2 x\,dx = \int (3 - 3\cos 2x)\,dx = 3x - \tfrac32\sin 2x + C.$$Vocabulary TrainEnglish Chinese Pinyin trapezium rule 梯形法则 tī xíng fǎ zé 2.6
Numerical solution of equations
Syllabus
- locate approximately a root of an equation, by means of graphical considerations and/or searching for a sign change
- understand the idea of, and use the notation for, a sequence of approximations which converges to a root of an equation
- understand how a given simple iterative formula of the form $x_{n+1} = F(x_n)$ relates to the equation being solved, and use a given iteration, or an iteration based on a given rearrangement of an equation, to determine a root to a prescribed degree of accuracy
Source: Cambridge International syllabus
Many equations cannot be solved exactly. Two ideas help you find a root 根 (a solution).
- Sign change 变号: if $f(a)$ and $f(b)$ have opposite signs (and the graph has no break between them), a root lies between $a$ and $b$.
- Iteration 迭代: rearrange the equation into the form $x = F(x)$, then use the iterative formula 迭代公式 $x_{n+1} = F(x_n)$. Start from a first guess $x_0$ and repeat. If the values settle down, they converge 收敛 to a root. Keep going until the answer is steady to the accuracy asked for.
$f(a)$ and $f(b)$ have opposite signs, so a root is trapped between $a$ and $b$.
Each step goes up to $y=F(x)$ then across to $y=x$; the staircase closes in on the root.Worked example. A root $\beta$ of an equation satisfies $x = \sqrt[3]{-2x - 4.5}$, and $-1.4 < \beta < -1.0$. Use the iteration $x_{n+1} = \sqrt[3]{-2x_n - 4.5}$ with $x_0 = -1.2$.
$$x_1 = \sqrt[3]{-2(-1.2) - 4.5} = \sqrt[3]{-2.1} = -1.281, \qquad x_2 = \sqrt[3]{-2(-1.281) - 4.5} = -1.247, \quad \ldots$$The values settle near $-1.26$, so $\beta = -1.26$ (3 s.f.).Vocabulary TrainEnglish Chinese Pinyin root 根 gēn sign change 变号 biàn hào iteration 迭代 dié dài iterative formula 迭代公式 dié dài gōng shì converge 收敛 shōu liǎn -
3 Pure Mathematics 3
This handout covers Topic 3: Pure Mathematics 纯数学 3. Subtopics 3.1–3.6 build on the algebra, logarithms, trigonometry, differentiation, integration and numerical methods of Pure Mathematics 2, so this handout explains what is new in Pure 3 and then covers the three big new areas: vectors, differential equations and complex numbers.
3.1
Algebra
Syllabus
- understand the meaning of $|x|$, sketch the graph of $y = |ax + b|$ and use relations such as $|a| = |b| \Leftrightarrow a^2 = b^2$ and $|x - a| < b \Leftrightarrow a - b < x < a + b$ when solving equations and inequalities
- divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, and identify the quotient and remainder (which may be zero)
- use the factor theorem and the remainder theorem
- recall an appropriate form for expressing rational functions in partial fractions, and carry out the decomposition, in cases where the denominator is no more complicated than $(ax + b)(cx + d)(ex + f)$, $(ax + b)(cx + d)^2$ or $(ax + b)(cx^2 + d)$
- use the expansion of $(1 + x)^n$, where $n$ is a rational number and $|x| < 1$
Source: Cambridge International syllabus
Two new tools join the algebra from Pure 2.
Partial fractions
A single fraction with a factorised bottom can be split into a sum of simpler fractions. This is called writing it in partial fractions 部分分式, and it makes a rational function 有理函数 (a fraction of polynomials) easy to integrate or expand. Match the form to the bottom:
$$\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}, \qquad \frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}.$$Worked example. Express $\dfrac{x+4}{(x+1)(x-2)}$ in partial fractions.
Write $\dfrac{x+4}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$, so $x + 4 = A(x-2) + B(x+1)$. Put $x = 2$: $6 = 3B$, so $B = 2$. Put $x = -1$: $3 = -3A$, so $A = -1$. Hence
$$\frac{x+4}{(x+1)(x-2)} = \frac{2}{x-2} - \frac{1}{x+1}.$$The binomial expansion for a rational power
The binomial expansion 二项展开式 also works when the power is a fraction or is negative, as long as $|x| < 1$:
$$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$For example $(1 + x)^{1/2} = 1 + \tfrac12 x - \tfrac18 x^2 + \cdots$ for $|x| < 1$.Vocabulary TrainEnglish Chinese Pinyin Pure Mathematics 纯数学 chún shù xué partial fractions 部分分式 bù fèn fēn shì rational function 有理函数 yǒu lǐ hán shù binomial expansion 二项展开式 èr xiàng zhǎn kāi shì 3.2 3.3 3.6
Logarithms, trigonometry and numerical methods (from Pure 2)
Syllabus
- understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base)
- understand the definition and properties of $e^x$ and $\ln x$, including their relationship as inverse functions and their graphs
- use logarithms to solve equations and inequalities in which the unknown appears in indices
- use logarithms to transform a given relationship to linear form, and hence determine unknown constants by considering the gradient and/or intercept
- understand the relationship of the secant, cosecant and cotangent functions to cosine, sine and tangent, and use properties and graphs of all six trigonometric functions for angles of any magnitude
- use trigonometrical identities for the simplification and exact evaluation of expressions and in the course of solving equations, selecting an appropriate identity, showing familiarity in particular with $\sec^2\theta \equiv 1 + \tan^2\theta$ and $\csc^2\theta \equiv 1 + \cot^2\theta$, the expansions of $\sin(A \pm B)$, $\cos(A \pm B)$ and $\tan(A \pm B)$, the formulae for $\sin 2A$, $\cos 2A$ and $\tan 2A$, and the expression of $a\sin\theta + b\cos\theta$ in the forms $R\sin(\theta \pm \alpha)$ and $R\cos(\theta \pm \alpha)$
- locate approximately a root of an equation, by means of graphical considerations and/or searching for a sign change
- understand the idea of, and use the notation for, a sequence of approximations which converges to a root of an equation
- understand how a given simple iterative formula of the form $x_{n+1} = F(x_n)$ relates to the equation being solved, and use a given iteration, or an iteration based on a given rearrangement of an equation, to determine a root to a prescribed degree of accuracy
Source: Cambridge International syllabus
Subtopics 3.2, 3.3 and 3.6 are the same skills you met in Pure 2: the laws of logarithms with $e^x$ and $\ln x$; the identities $\sec^2\theta \equiv 1 + \tan^2\theta$ and $\csc^2\theta \equiv 1 + \cot^2\theta$, the compound- and double-angle formulae, and the $R$-form of $a\sin\theta + b\cos\theta$; and solving an equation numerically by a sign change 变号 and an iterative formula 迭代公式 $x_{n+1} = F(x_n)$. Use them exactly as before.
Vocabulary TrainEnglish Chinese Pinyin sign change 变号 biàn hào iterative formula 迭代公式 dié dài gōng shì 3.4
Differentiation
Syllabus
- use the derivatives of $e^x$, $\ln x$, $\sin x$, $\cos x$, $\tan x$, $\tan^{-1} x$, together with constant multiples, sums, differences and composites
- differentiate products and quotients
- find and use the first derivative of a function which is defined parametrically or implicitly
Source: Cambridge International syllabus
The methods are those of Pure 2 (the product, quotient and chain rules, with parametric and implicit curves). One derivative is added — the inverse tangent 反正切:
$$\frac{d}{dx}\tan^{-1} x = \frac{1}{1 + x^2}.$$Vocabulary TrainEnglish Chinese Pinyin inverse tangent 反正切 fǎn zhèng qiè 3.5
Integration
Syllabus
- extend the idea of 'reverse differentiation' to include the integration of $e^{ax + b}$, $\dfrac{1}{ax + b}$, $\sin(ax + b)$, $\cos(ax + b)$, $\sec^2(ax + b)$ and $\dfrac{1}{x^2 + a^2}$
- use trigonometrical relationships in carrying out integration
- integrate rational functions by means of decomposition into partial fractions
- recognise an integrand of the form $\dfrac{k\,f'(x)}{f(x)}$, and integrate such functions
- recognise when an integrand can usefully be regarded as a product, and use integration by parts
- use a given substitution to simplify and evaluate either a definite or an indefinite integral
Source: Cambridge International syllabus
Pure 3 adds several powerful methods.
- A new standard integral: $\displaystyle\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$.
- Partial fractions: split a rational function first, then integrate each piece as a logarithm.
- The pattern $\dfrac{k\,f'(x)}{f(x)}$: this integrates to $k\ln|f(x)| + C$. For example $\displaystyle\int \frac{2x}{x^2 + 1}\,dx = \ln(x^2 + 1) + C$.
- Integration by parts 分部积分, used for a product: $\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$.
- Integration by substitution 换元积分: a given change of variable turns a hard integral into an easy one.
Worked example. Find $\displaystyle\int x\cos x\,dx$.
Use integration by parts with $u = x$ and $\dfrac{dv}{dx} = \cos x$, so $\dfrac{du}{dx} = 1$ and $v = \sin x$:
$$\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C.$$Vocabulary TrainEnglish Chinese Pinyin integration by parts 分部积分 fēn bù jī fēn integration by substitution 换元积分 huàn yuán jī fēn 3.7
Vectors
Syllabus
- use standard notations for vectors (column vectors, $x\mathbf{i} + y\mathbf{j}$, $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, $\overrightarrow{AB}$, $\mathbf{a}$)
- carry out addition and subtraction of vectors and multiplication of a vector by a scalar, and interpret these operations in geometrical terms
- calculate the magnitude of a vector, and use unit vectors, displacement vectors and position vectors
- understand the significance of all the symbols used when the equation of a straight line is expressed in the form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$, and find the equation of a line, given sufficient information
- determine whether two lines are parallel, intersect or are skew, and find the point of intersection of two lines when it exists
- use formulae to calculate the scalar product of two vectors, and use scalar products in problems involving lines and points
Source: Cambridge International syllabus
Forces like wind and water are vectors — they have both size and direction.A vector 向量 has both size and direction. Write it as a column, or as $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, or as $\overrightarrow{AB}$.
- The magnitude 模长 (length) of $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ is $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$.
- A unit vector 单位向量 has magnitude $1$; divide a vector by its magnitude to make one.
- A position vector 位置向量 gives a point's place from the origin; a displacement vector 位移向量 $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ goes from one point to another. Multiplying by a scalar 标量 (a plain number) stretches a vector.
Lines and the scalar product
A straight line through point $\mathbf{a}$ in direction $\mathbf{b}$ has vector equation $\mathbf{r} = \mathbf{a} + t\mathbf{b}$. Two lines may be parallel 平行, may intersect 相交 at a point, or may be skew lines 异面直线 (not parallel and never meeting).
Start at the point $\mathbf{a}$, then add $t$ copies of the direction $\mathbf{b}$ to reach any point on the line.The scalar product 数量积 (dot product) of $\mathbf{a}$ and $\mathbf{b}$ is
$$\mathbf{a}\cdot\mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta,$$where $\theta$ is the angle between them. So $\mathbf{a}\cdot\mathbf{b} = 0$ means the vectors are perpendicular.
The scalar product picks out $|\mathbf{b}|\cos\theta$, how far $\mathbf{b}$ reaches along $\mathbf{a}$.Worked example. Find the angle between $\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}$.
$$\mathbf{a}\cdot\mathbf{b} = (1)(2) + (2)(2) + (2)(1) = 8, \qquad |\mathbf{a}| = |\mathbf{b}| = 3.$$So $\cos\theta = \dfrac{8}{3\times 3} = \dfrac{8}{9}$, giving $\theta = 27.3^\circ$.Vocabulary TrainEnglish Chinese Pinyin vector 向量 xiàng liàng magnitude 模长 mó zhǎng unit vector 单位向量 dān wèi xiàng liàng position vector 位置向量 wèi zhì xiàng liàng displacement vector 位移向量 wèi yí xiàng liàng scalar 标量 biāo liàng parallel 平行 píng xíng intersect 相交 xiāng jiāo skew lines 异面直线 yì miàn zhí xiàn scalar product 数量积 shù liàng jī 3.8
Differential equations
Syllabus
- formulate a simple statement involving a rate of change as a differential equation
- find by integration a general form of solution for a first order differential equation in which the variables are separable
- use an initial condition to find a particular solution
- interpret the solution of a differential equation in the context of a problem being modelled by the equation
Source: Cambridge International syllabus
A differential equation 微分方程 links a quantity to its rate of change. To solve a first-order equation whose variables are separable 可分离变量, put all the $y$ terms on one side and all the $x$ terms on the other, then integrate both sides. This gives the general solution 通解, which contains a constant. An initial condition 初始条件 (a known value) fixes the constant and gives the particular solution 特解.
Worked example. Solve $\dfrac{dy}{dx} = xy$, given that $y = 1$ when $x = 0$.
Separate the variables and integrate:
$$\int \frac{1}{y}\,dy = \int x\,dx \;\Rightarrow\; \ln y = \tfrac12 x^2 + c \;\Rightarrow\; y = A e^{x^2/2}.$$Using $y = 1$ at $x = 0$ gives $A = 1$, so $y = e^{x^2/2}$.
The constant $A$ gives a whole family of curves; the condition $y=1$ at $x=0$ selects $y=e^{x^2/2}$.Vocabulary TrainEnglish Chinese Pinyin differential equation 微分方程 wēi fēn fāng chéng separable 可分离变量 kě fēn lí biàn liàng general solution 通解 tōng jiě initial condition 初始条件 chū shǐ tiáo jiàn particular solution 特解 tè jiě 3.9
Complex numbers
Syllabus
- understand the idea of a complex number, recall the meaning of the terms real part, imaginary part, modulus, argument, conjugate, and use the fact that two complex numbers are equal if and only if both real and imaginary parts are equal
- carry out operations of addition, subtraction, multiplication and division of two complex numbers expressed in Cartesian form $x + iy$
- use the result that, for a polynomial equation with real coefficients, any non-real roots occur in conjugate pairs
- represent complex numbers geometrically by means of an Argand diagram
- carry out operations of multiplication and division of two complex numbers expressed in polar form $r(\cos\theta + i\sin\theta)$ or $re^{i\theta}$
- find the two square roots of a complex number
- understand in simple terms the geometrical effects of conjugating a complex number and of adding, subtracting, multiplying and dividing two complex numbers
- illustrate simple equations and inequalities involving complex numbers by means of loci in an Argand diagram
Source: Cambridge International syllabus
Self-similar patterns like Romanesco arise from iterating functions in the complex plane.A complex number 复数 has the form $z = x + iy$, where $i^2 = -1$. Here $x$ is the real part 实部 and $y$ is the imaginary part 虚部. This $x + iy$ is the Cartesian form 直角坐标形式. Two complex numbers are equal only when their real parts match and their imaginary parts match.
- The conjugate 共轭 of $z = x + iy$ is $z^* = x - iy$. For a polynomial with real coefficients, any non-real roots come in conjugate pairs.
- The modulus 模 is $|z| = \sqrt{x^2 + y^2}$ (its distance from the origin) and the argument 辐角 is the angle the point makes, measured from the positive real axis.
- You can plot $z$ as a point on an Argand diagram 阿干图 (the complex plane).
- The polar form 极坐标形式 is $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, where $r = |z|$ and $\theta$ is the argument. Multiplying multiplies the moduli and adds the arguments.
On the Argand diagram $|z|$ is the distance from $O$, $\arg z$ the angle, and $z^{*}$ the reflection in the real axis.To divide, multiply top and bottom by the conjugate of the bottom.
Worked example. Write $\dfrac{3 + i}{1 - i}$ in the form $x + iy$.
$$\frac{3 + i}{1 - i} = \frac{(3 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{3 + 3i + i + i^2}{1 + 1} = \frac{2 + 4i}{2} = 1 + 2i.$$An equation or inequality in $z$ describes a locus 轨迹 (a path or region) on the Argand diagram. For example $|z - a| = r$ is a circle of radius $r$ centred at $a$.
$|z-a|=r$ is the set of points a fixed distance $r$ from $a$ — a circle.Vocabulary TrainEnglish Chinese Pinyin complex number 复数 fù shù real part 实部 shí bù imaginary part 虚部 xū bù Cartesian form 直角坐标形式 zhí jiǎo zuò biāo xíng shì conjugate 共轭 gòng è complex modulus 模 mó argument 辐角 fú jiǎo Argand diagram 阿干图 ā gàn tú polar form 极坐标形式 jí zuò biāo xíng shì locus 轨迹 guǐ jì -
4 Mechanics
This handout covers Topic 4: Mechanics 力学. It studies how forces make objects move. Throughout this topic, take the acceleration of free fall as $g = 10\ \text{m s}^{-2}$.
4.1
Forces and equilibrium
Syllabus
- identify the forces acting in a given situation
- understand the vector nature of force, and find and use components and resultants
- use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, that the sum of the components in any direction is zero
- understand that a contact force between two surfaces can be represented by two components, the normal component and the frictional component
- use the model of a 'smooth' contact, and understand the limitations of this model
- understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship $F = \mu R$ or $F \leqslant \mu R$, as appropriate
- use Newton's third law
Source: Cambridge International syllabus
A force 力 is a push or a pull. It is a vector, so it has size and direction. Because it is a vector, you can split a force into components 分量 (usually horizontal and vertical), and you can add several forces into one resultant 合力.
A force at angle $\theta$ has a horizontal part $F\cos\theta$ and a vertical part $F\sin\theta$.A particle is in equilibrium 平衡 when the forces are balanced: the vector sum of the forces is zero. In practice this means the components in any direction add to zero.
Friction
When two surfaces touch, the contact force 接触力 between them has two parts: the normal reaction 法向反作用力 $R$, at right angles to the surface, and the friction 摩擦力 $F$, along the surface, which opposes sliding. A "smooth" surface is a model with no friction.
Friction can only grow up to a maximum. At that maximum the body is about to slip — this is limiting friction 最大静摩擦力. The maximum is set by the coefficient of friction 摩擦系数 $\mu$:
$$F \leqslant \mu R, \qquad \text{with } F = \mu R \text{ at the point of slipping}.$$
On a rough surface the contact force splits into the normal reaction $R$ and friction $F$ (at most $\mu R$).By Newton's third law 牛顿第三定律, the two surfaces push on each other with equal and opposite forces.
Vocabulary TrainEnglish Chinese Pinyin Mechanics 力学 lì xué force 力 lì components 分量 fèn liàng resultant 合力 hé lì equilibrium 平衡 píng héng contact force 接触力 jiē chù lì normal reaction 法向反作用力 fǎ xiàng fǎn zuò yòng lì friction 摩擦力 mó cā lì limiting friction 最大静摩擦力 zuì dà jìng mó cā lì coefficient of friction 摩擦系数 mó cā xì shù Newton's third law 牛顿第三定律 niú dùn dì sān dìng lǜ 4.2
Kinematics of motion in a straight line
Syllabus
- understand the concepts of distance and speed as scalar quantities, and of displacement, velocity and acceleration as vector quantities
- sketch and interpret displacement-time graphs and velocity-time graphs, and in particular appreciate that the area under a velocity-time graph represents displacement, the gradient of a displacement-time graph represents velocity, and the gradient of a velocity-time graph represents acceleration
- use differentiation and integration with respect to time to solve simple problems concerning displacement, velocity and acceleration
- use appropriate formulae for motion with constant acceleration in a straight line
Source: Cambridge International syllabus
Distance 距离 and speed 速率 are scalars (size only). Displacement 位移, velocity 速度 and acceleration 加速度 are vectors (size and direction).
On a velocity-time graph 速度时间图, the area under the graph is the displacement and the gradient is the acceleration. On a displacement-time graph, the gradient is the velocity. More generally, differentiate with respect to time to go from displacement to velocity to acceleration, and integrate to go back.
The shaded area gives the distance travelled; the slope of the line gives the acceleration.For motion with constant acceleration 匀加速, use these formulae (the "suvat" equations):
$$v = u + at, \qquad s = ut + \tfrac12 at^2, \qquad v^2 = u^2 + 2as, \qquad s = \tfrac12(u + v)t.$$Worked example. A car starts from rest and accelerates at $2.5\ \text{m s}^{-2}$ for $4\ \text{s}$. Find its speed and the distance travelled.
$$v = 0 + 2.5\times 4 = 10\ \text{m s}^{-1}, \qquad s = 0 + \tfrac12(2.5)(4^2) = 20\ \text{m}.$$Vocabulary TrainEnglish Chinese Pinyin distance 距离 jù lí speed 速率 sù lǜ displacement 位移 wèi yí velocity 速度 sù dù acceleration 加速度 jiā sù dù velocity-time graph 速度时间图 sù dù shí jiān tú constant acceleration 匀加速 yún jiā sù 4.3
Momentum
Syllabus
- use the definition of linear momentum and show understanding of its vector nature
- use conservation of linear momentum to solve problems that may be modelled as the direct impact of two bodies
Source: Cambridge International syllabus
A Newton's cradle demonstrates conservation of momentum in collisions.The momentum 动量 of a body is $\text{mass} \times \text{velocity}$. It is a vector. In a direct collision of two bodies, the total momentum is unchanged. This is the conservation of momentum 动量守恒:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$$
The total momentum is the same before and after the collision.Worked example. A body of mass $2\ \text{kg}$ moving at $3\ \text{m s}^{-1}$ hits a stationary body of mass $1\ \text{kg}$, and they stick together. Find their common speed afterwards.
$$2(3) + 1(0) = (2 + 1)v \;\Rightarrow\; v = \frac{6}{3} = 2\ \text{m s}^{-1}.$$Vocabulary TrainEnglish Chinese Pinyin momentum 动量 dòng liàng conservation of momentum 动量守恒 dòng liàng shǒu héng 4.4
Newton's laws of motion
Syllabus
- apply Newton's laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod
- use the relationship between mass and weight
- solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration
- solve simple problems which may be modelled as the motion of connected particles
Source: Cambridge International syllabus
Newton's laws of motion 牛顿运动定律 connect force and acceleration. The key one is: resultant force $=$ mass 质量 $\times$ acceleration,
$$F = ma.$$The weight 重力 of a body is the force of gravity on it: $W = mg$. Forces in a problem may include weight, friction, tension 张力 in a string, and the push in a rod.For motion on an inclined plane 斜面, split each force into a part along the slope and a part at right angles to it, then apply $F = ma$ along the slope. For connected particles 连接质点 (joined by a string), apply $F = ma$ to each body, or to the whole system.
On a slope, resolve the forces along the slope and at right angles to it.Worked example. A block of mass $12\ \text{kg}$ is pulled up a rough plane by a rope parallel to the slope. The plane is at $20^\circ$ to the horizontal, the coefficient of friction is $0.4$, and the acceleration is $2\ \text{m s}^{-2}$. Find the tension in the rope.
The normal reaction is $R = mg\cos 20^\circ = 120\cos 20^\circ = 112.8\ \text{N}$, so the friction is $F = \mu R = 0.4 \times 112.8 = 45.1\ \text{N}$. Along the slope, $T - mg\sin 20^\circ - F = ma$:
$$T = ma + mg\sin 20^\circ + F = 12(2) + 120\sin 20^\circ + 45.1 = 24 + 41.0 + 45.1 = 110\ \text{N (3 s.f.)}.$$Vocabulary TrainEnglish Chinese Pinyin Newton's laws of motion 牛顿运动定律 niú dùn yùn dòng dìng lǜ mass 质量 zhì liàng weight 重力 zhòng lì tension 张力 zhāng lì inclined plane 斜面 xié miàn connected particles 连接质点 lián jiē zhì diǎn 4.5
Energy, work and power
Syllabus
- understand the concept of the work done by a force, and calculate the work done by a constant force when its point of application undergoes a displacement not necessarily parallel to the force
- understand the concepts of gravitational potential energy and kinetic energy, and use appropriate formulae
- understand and use the relationship between the change in energy of a system and the work done by the external forces, and use in appropriate cases the principle of conservation of energy
- use the definition of power as the rate at which a force does work, and use the relationship between power, force and velocity for a force acting in the direction of motion
- solve problems involving, for example, the instantaneous acceleration of a car moving on a hill against a resistance
Source: Cambridge International syllabus
A roller coaster trades potential energy for kinetic energy as it rises and falls.The work done 功 by a constant force is the force times the distance moved in the direction of the force: $W = Fd\cos\theta$, where $\theta$ is the angle between the force and the motion. Work is measured in joules (J).
Energy comes in forms you can calculate:
- Kinetic energy 动能 (energy of movement): $\text{KE} = \tfrac12 mv^2$.
- Gravitational potential energy 重力势能 (energy of height): $\text{PE} = mgh$.
The work done by the outside forces equals the change in the total energy. When no friction acts, the total energy stays the same — the conservation of energy 能量守恒.
Power 功率 is the rate of doing work. For a force pulling in the direction of motion, $P = Fv$ (power $=$ force $\times$ velocity). Power is measured in watts (W).
Worked example. A car engine works at $12\ \text{kW}$ while the car moves at $20\ \text{m s}^{-1}$ on a level road. Find the driving force.
$$P = Fv \;\Rightarrow\; F = \frac{P}{v} = \frac{12000}{20} = 600\ \text{N}.$$Vocabulary TrainEnglish Chinese Pinyin work done 功 gōng kinetic energy 动能 dòng néng gravitational potential energy 重力势能 zhòng lì shì néng conservation of energy 能量守恒 néng liàng shǒu héng power 功率 gōng lǜ -
5 Probability & Statistics 1
This handout covers Topic 5: Probability & Statistics 概率统计 1. It is about describing data, counting choices, and working out the chance of events.
5.1
Representation of data
Syllabus
- select a suitable way of presenting raw statistical data, and discuss advantages and/or disadvantages that particular representations may have
- draw and interpret stem-and-leaf diagrams, box-and-whisker plots, histograms and cumulative frequency graphs
- understand and use different measures of central tendency (mean, median, mode) and variation (range, interquartile range, standard deviation)
- use a cumulative frequency graph
- calculate and use the mean and standard deviation of a set of data (including grouped data) either from the data itself or from given totals $\sum x$ and $\sum x^2$, or coded totals $\sum(x - a)$ and $\sum(x - a)^2$, and use such totals in solving problems which may involve up to two data sets
Source: Cambridge International syllabus
Choose a diagram that suits the data. You should be able to draw and read:
- a stem-and-leaf diagram 茎叶图 (keeps the original values and shows the shape);
- a box-and-whisker plot 箱线图 (shows the lowest value, the three quartiles, and the highest value);
- a histogram 直方图 (for grouped data, where the area of each bar shows the frequency);
- a cumulative frequency 累积频数 graph (running totals, used to estimate the median and quartiles).
The box spans the quartiles $Q_1$ to $Q_3$; the whiskers reach the lowest and highest values.Averages and spread
A measure of central tendency 集中趋势 is a single "middle" value:
- the mean 平均数 $\bar{x} = \dfrac{\sum x}{n}$ (the average);
- the median 中位数 (the middle value when the data is in order);
- the mode 众数 (the most common value).
A measure of variation 离散程度 shows how spread out the data is:
- the range (of data) 极差 (highest $-$ lowest);
- the interquartile range 四分位距 (upper quartile $-$ lower quartile);
- the standard deviation 标准差 $\sigma = \sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2}$.
You often work from the totals $\sum x$ and $\sum x^2$. The square of the standard deviation is the variance 方差.
Worked example. For $10$ values, $\sum x = 50$ and $\sum x^2 = 300$. Find the mean and standard deviation.
$$\bar{x} = \frac{50}{10} = 5, \qquad \sigma = \sqrt{\frac{300}{10} - 5^2} = \sqrt{30 - 25} = \sqrt{5} = 2.24.$$Vocabulary TrainEnglish Chinese Pinyin Probability & Statistics 概率统计 gài lǜ tǒng jì stem-and-leaf diagram 茎叶图 jīng yè tú box-and-whisker plot 箱线图 xiāng xiàn tú histogram 直方图 zhí fāng tú cumulative frequency 累积频数 lěi jī pín shuò central tendency 集中趋势 jí zhōng qū shì mean 平均数 píng jūn shù median 中位数 zhōng wèi shù mode 众数 zhòng shù variation 离散程度 lí sàn chéng dù range (of data) 极差 jí chà interquartile range 四分位距 sì fēn wèi jù standard deviation 标准差 biāo zhǔn chà variance 方差 fāng chà 5.2
Permutations and combinations
Syllabus
- understand the terms permutation and combination, and solve simple problems involving selections
- solve problems about arrangements of objects in a line, including those involving repetition (e.g. the number of ways of arranging the letters of the word 'NEEDLESS') and restriction (e.g. the number of ways several people can stand in a line if two particular people must, or must not, stand next to each other)
Source: Cambridge International syllabus
A permutation 排列 is an arrangement where order matters; a combination 组合 is a selection where order does not matter. The numbers are
$${}^nP_r = \frac{n!}{(n-r)!}, \qquad {}^nC_r = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}.$$To arrange objects in a line when some are repeated, divide by the factorial of each repeat count.Worked example. How many different arrangements are there of the letters of the word NEEDLESS?
There are $8$ letters, with E repeated $3$ times and S repeated $2$ times:
$$\frac{8!}{3!\,2!} = \frac{40320}{6 \times 2} = 3360.$$Vocabulary TrainEnglish Chinese Pinyin permutation 排列 pái liè combination 组合 zǔ hé 5.3
Probability
Syllabus
- evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations
- use addition and multiplication of probabilities, as appropriate, in simple cases
- understand the meaning of exclusive and independent events, including determination of whether events $A$ and $B$ are independent by comparing the values of $P(A \cap B)$ and $P(A) \times P(B)$
- calculate and use conditional probabilities in simple cases
Source: Cambridge International syllabus
Dice: a familiar starting point for probability.Find a probability by counting equally likely outcomes, or by using permutations and combinations. Combine probabilities with these rules:
- addition for "or": $P(A \cup B) = P(A) + P(B) - P(A \cap B)$;
- multiplication for "and" when events are independent: $P(A \cap B) = P(A)\,P(B)$.
The overlap of the two circles is $A\cap B$; the addition rule subtracts it once so it is not counted twice.Two events are mutually exclusive events 互斥事件 if they cannot both happen, and independent events 独立事件 if one happening does not change the chance of the other. To test independence, check whether $P(A \cap B) = P(A)\times P(B)$. A conditional probability 条件概率 is the chance of $A$ given that $B$ has happened: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.
Worked example. Events have $P(A) = 0.5$, $P(B) = 0.4$ and $P(A \cap B) = 0.2$. Are $A$ and $B$ independent?
Test: $P(A)\times P(B) = 0.5 \times 0.4 = 0.2 = P(A \cap B)$. The values are equal, so $A$ and $B$ are independent.
Vocabulary TrainEnglish Chinese Pinyin mutually exclusive events 互斥事件 hù chì shì jiàn independent events 独立事件 dú lì shì jiàn conditional probability 条件概率 tiáo jiàn gài lǜ 5.4
Discrete random variables
Syllabus
- draw up a probability distribution table relating to a given situation involving a discrete random variable $X$, and calculate $E(X)$ and $\mathrm{Var}(X)$
- use formulae for probabilities for the binomial and geometric distributions, and recognise practical situations where these distributions are suitable models
- use formulae for the expectation and variance of the binomial distribution and for the expectation of the geometric distribution
Source: Cambridge International syllabus
A discrete random variable 离散型随机变量 $X$ takes separate values, each with a probability. List them in a probability distribution table 概率分布表; the probabilities must add to $1$. Then the expectation 期望 (mean) and variance are
$$E(X) = \sum x\,P(X = x), \qquad \mathrm{Var}(X) = \sum x^2\,P(X = x) - \big(E(X)\big)^2.$$Two special models:
- the binomial distribution 二项分布 $X \sim B(n, p)$, for the number of successes in $n$ independent trials: $P(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$, with $E(X) = np$ and $\mathrm{Var}(X) = np(1-p)$;
- the geometric distribution 几何分布, for the trial on which the first success happens: $P(X = r) = (1-p)^{r-1}p$, with $E(X) = \dfrac{1}{p}$.
Worked example. $X \sim B(10, 0.3)$. Find $P(X = 2)$ and $E(X)$.
$$P(X = 2) = \binom{10}{2}(0.3)^2(0.7)^8 = 45 \times 0.09 \times 0.05765 = 0.233, \qquad E(X) = 10 \times 0.3 = 3.$$
The distribution $B(10,0.3)$: each bar is $P(X=r)$, clustered around the mean $np=3$.Vocabulary TrainEnglish Chinese Pinyin discrete random variable 离散型随机变量 lí sàn xíng suí jī biàn liàng probability distribution table 概率分布表 gài lǜ fēn bù biǎo expectation 期望 qī wàng binomial distribution 二项分布 èr xiàng fēn bù geometric distribution 几何分布 jǐ hé fēn bù 5.5
The normal distribution
Syllabus
- understand the use of a normal distribution to model a continuous random variable, and use normal distribution tables
- solve problems concerning a variable $X$, where $X \sim N(\mu, \sigma^2)$, including finding the value of $P(X > x_1)$, or a related probability, given the values of $x_1$, $\mu$, $\sigma$, and finding a relationship between $x_1$, $\mu$ and $\sigma$ given the value of $P(X > x_1)$ or a related probability
- recall conditions under which the normal distribution can be used as an approximation to the binomial distribution, and use this approximation, with a continuity correction, in solving problems
Source: Cambridge International syllabus
A Galton board: balls falling through pins pile up into the bell-shaped normal distribution.The normal distribution 正态分布 models a continuous random variable 连续型随机变量 with a symmetric bell shape. Write $X \sim N(\mu, \sigma^2)$, where $\mu$ is the mean and $\sigma$ is the standard deviation. To use the tables, standardize 标准化 to the variable $Z \sim N(0, 1)$:
$$Z = \frac{X - \mu}{\sigma}.$$
A Galton board: balls fall through rows of pegs and pile up into the bell-shaped normal distribution
Then $P(X < x) = P\!\left(Z < \dfrac{x - \mu}{\sigma}\right)$, which you read from the normal table $\Phi$.
A normal probability is an area under the bell curve; standardizing rescales it to $Z\sim N(0,1)$.The normal distribution is also a good approximation 近似 to the binomial when $n$ is large. Because you replace a discrete variable by a continuous one, apply a continuity correction 连续性校正 (adjust by $0.5$).
When $n$ is large the binomial bars follow a normal curve of the same mean and variance.Worked example. Bags of rice have mass $X \sim N(\mu, 0.14^2)$. Given that $P(X < 1.48) = 0.22$, find $\mu$.
From the table, $P(Z < z) = 0.22$ gives $z = -0.772$. So
$$\frac{1.48 - \mu}{0.14} = -0.772 \;\Rightarrow\; \mu = 1.48 + 0.772 \times 0.14 = 1.59\ \text{kg (3 s.f.)}.$$Vocabulary TrainEnglish Chinese Pinyin normal distribution 正态分布 zhèng tài fēn bù continuous random variable 连续型随机变量 lián xù xíng suí jī biàn liàng standardize 标准化 biāo zhǔn huà approximation 近似 jìn sì continuity correction 连续性校正 lián xù xìng jiào zhèng -
6 Probability & Statistics 2
This handout covers Topic 6: Probability & Statistics 概率统计 2. It adds the Poisson model, combining random variables, continuous distributions, and the ideas of estimation and testing.
6.1
The Poisson distribution
Syllabus
- use formulae to calculate probabilities for the distribution $\mathrm{Po}(\lambda)$
- use the fact that if $X \sim \mathrm{Po}(\lambda)$ then the mean and variance of $X$ are each equal to $\lambda$
- understand the relevance of the Poisson distribution to the distribution of random events, and use the Poisson distribution as a model
- use the Poisson distribution as an approximation to the binomial distribution where appropriate
- use the normal distribution, with continuity correction, as an approximation to the Poisson distribution where appropriate
Source: Cambridge International syllabus
People arriving at random in a queue follow a Poisson distribution.The Poisson distribution 泊松分布 $X \sim \mathrm{Po}(\lambda)$ models the number of random events in a fixed interval, when events happen at a steady average rate $\lambda$:
$$P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}.$$For a Poisson variable the mean and the variance are both equal to $\lambda$. The Poisson distribution is a good approximation 近似 to the binomial when $n$ is large and $p$ is small. The normal distribution (with continuity correction) approximates the Poisson when $\lambda$ is large.Worked example. $X \sim \mathrm{Po}(3)$. Find $P(X = 2)$.
$$P(X = 2) = e^{-3}\frac{3^2}{2!} = e^{-3}\times 4.5 = 0.224.$$
The Poisson distribution $\mathrm{Po}(3)$: for a Poisson variable the mean and variance both equal $\lambda$.Vocabulary TrainEnglish Chinese Pinyin Probability & Statistics 概率统计 gài lǜ tǒng jì Poisson distribution 泊松分布 pō sōng fēn bù approximation 近似 jìn sì 6.2
Linear combinations of random variables
Syllabus
- use, when solving problems, the results that $E(aX + b) = aE(X) + b$ and $\mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X)$
- use the results $E(aX + bY) = aE(X) + bE(Y)$ and $\mathrm{Var}(aX + bY) = a^2\,\mathrm{Var}(X) + b^2\,\mathrm{Var}(Y)$ for independent $X$ and $Y$
- use the result that if $X$ has a normal distribution then so does $aX + b$, and that if $X$ and $Y$ have independent normal distributions then $aX + bY$ has a normal distribution
- use the result that if $X$ and $Y$ have independent Poisson distributions then $X + Y$ has a Poisson distribution
Source: Cambridge International syllabus
When you change a variable by a linear rule, the expectation 期望 (mean) and variance 方差 follow these rules:
$$E(aX + b) = aE(X) + b, \qquad \mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X).$$For two independent variables $X$ and $Y$:$$E(aX + bY) = aE(X) + bE(Y), \qquad \mathrm{Var}(aX + bY) = a^2\,\mathrm{Var}(X) + b^2\,\mathrm{Var}(Y).$$Two useful facts: if $X$ has a normal distribution 正态分布 then so does $aX + b$; and the sum of independent Poisson variables is again Poisson.Worked example. $X$ has mean $5$ and variance $4$. Find $E(3X - 1)$ and $\mathrm{Var}(3X - 1)$.
$$E(3X - 1) = 3(5) - 1 = 14, \qquad \mathrm{Var}(3X - 1) = 3^2 \times 4 = 36.$$Vocabulary TrainEnglish Chinese Pinyin expectation 期望 qī wàng variance 方差 fāng chà normal distribution 正态分布 zhèng tài fēn bù 6.3
Continuous random variables
Syllabus
- understand the concept of a continuous random variable, and recall and use properties of a probability density function
- use a probability density function to solve problems involving probabilities, and to calculate the mean and variance of a distribution
Source: Cambridge International syllabus
A continuous random variable 连续型随机变量 can take any value in a range. Its probabilities come from a probability density function 概率密度函数 $f(x)$, with two key properties:
$$f(x) \geqslant 0, \qquad \int_{-\infty}^{\infty} f(x)\,dx = 1.$$A probability is the area under $f$, and the mean is found by integration:$$P(a < X < b) = \int_a^b f(x)\,dx, \qquad E(X) = \int_{-\infty}^{\infty} x\,f(x)\,dx.$$
For a continuous variable, the probability $P(ais the area under $f(x)$ between $a$ and $b$. Worked example. A continuous variable has $f(x) = \tfrac12 x$ for $0 \leqslant x \leqslant 2$ (and $0$ elsewhere). Find $E(X)$.
$$E(X) = \int_0^2 x\cdot\tfrac12 x\,dx = \int_0^2 \tfrac12 x^2\,dx = \left[\tfrac{x^3}{6}\right]_0^2 = \frac{8}{6} = \frac{4}{3}.$$Vocabulary TrainEnglish Chinese Pinyin continuous random variable 连续型随机变量 lián xù xíng suí jī biàn liàng probability density function 概率密度函数 gài lǜ mì dù hán shù 6.4
Sampling and estimation
Syllabus
- understand the distinction between a sample and a population, and appreciate the necessity for randomness in choosing samples
- explain in simple terms why a given sampling method may be unsatisfactory
- recognise that a sample mean can be regarded as a random variable, and use the facts that $E(\bar{X}) = \mu$ and that $\mathrm{Var}(\bar{X}) = \dfrac{\sigma^2}{n}$
- use the fact that $\bar{X}$ has a normal distribution if $X$ has a normal distribution
- use the Central Limit Theorem where appropriate
- calculate unbiased estimates of the population mean and variance from a sample, using either raw or summarised data
- determine and interpret a confidence interval for a population mean in cases where the population is normally distributed with known variance or where a large sample is used
- determine, from a large sample, an approximate confidence interval for a population proportion
Source: Cambridge International syllabus
Statistics studies a sample to learn about a whole population.A sample 样本 is a small group chosen from the whole population 总体. Good sampling needs randomness 随机性, so that every member has a fair chance of being chosen.
The sample mean $\bar{X}$ is itself a random variable, with
$$E(\bar{X}) = \mu, \qquad \mathrm{Var}(\bar{X}) = \frac{\sigma^2}{n}.$$By the Central Limit Theorem 中心极限定理, for a large sample $\bar{X}$ is approximately normal, whatever the shape of the population.
Whatever the population's shape, the sample mean $\bar{X}$ has a narrow, near-normal distribution centred on $\mu$.From a sample you can find unbiased estimates 无偏估计 of the population mean and variance. A confidence interval 置信区间 gives a range that probably contains the true mean. When the population is normal with known $\sigma$ (or the sample is large), a $95\%$ interval is
$$\bar{x} \pm 1.96\,\frac{\sigma}{\sqrt{n}}.$$
A $95\%$ confidence interval stretches $1.96$ standard errors each side of the sample mean.You can also find a confidence interval for a population proportion 总体比例 from a large sample.
Worked example. A sample of $n = 64$ has mean $\bar{x} = 50$, from a population with $\sigma = 8$. Find a $95\%$ confidence interval for the population mean.
$$50 \pm 1.96\times\frac{8}{\sqrt{64}} = 50 \pm 1.96 \;\Rightarrow\; (48.0,\ 52.0).$$Vocabulary TrainEnglish Chinese Pinyin sample 样本 yàng běn population 总体 zǒng tǐ randomness 随机性 suí jī xìng Central Limit Theorem 中心极限定理 zhōng xīn jí xiàn dìng lǐ unbiased estimates 无偏估计 wú piān gū jì confidence interval 置信区间 zhì xìn qū jiān population proportion 总体比例 zǒng tǐ bǐ lì 6.5
Hypothesis tests
Syllabus
- understand the nature of a hypothesis test, the difference between one-tailed and two-tailed tests, and the terms null hypothesis, alternative hypothesis, significance level, rejection region (or critical region), acceptance region and test statistic
- formulate hypotheses and carry out a hypothesis test in the context of a single observation from a population which has a binomial or Poisson distribution, using direct evaluation of probabilities or a normal approximation to the binomial or the Poisson distribution, where appropriate
- formulate hypotheses and carry out a hypothesis test concerning the population mean in cases where the population is normally distributed with known variance or where a large sample is used
- understand the terms Type I error and Type II error in relation to hypothesis tests
- calculate the probabilities of making Type I and Type II errors in specific situations involving tests based on a normal distribution or direct evaluation of binomial or Poisson probabilities
Source: Cambridge International syllabus
A hypothesis test 假设检验 uses sample data to judge a claim. You set up two statements: the null hypothesis 原假设 $H_0$ (the claim being tested, usually "no change") and the alternative hypothesis 备择假设 $H_1$ (what you suspect instead). The test is one-tailed 单尾 if $H_1$ points one way (e.g. $\mu > 50$) and two-tailed 双尾 if it allows both ways ($\mu \neq 50$).
You fix a significance level 显著性水平 (often $5\%$), work out a test statistic 检验统计量 from the data, and see whether it lands in the rejection region 拒绝域. If it does, you reject $H_0$.
A two-tailed test at $5\%$ rejects $H_0$ only if the test statistic falls in a shaded tail beyond $\pm1.96$.Two mistakes are possible: a Type I error 第一类错误 is rejecting $H_0$ when it is actually true; a Type II error 第二类错误 is accepting $H_0$ when it is actually false.
Worked example. A population is claimed to have mean $50$, with $\sigma = 8$. A sample of $n = 64$ gives $\bar{x} = 52$. Test at the $5\%$ level whether the mean has changed.
$H_0\!: \mu = 50$ and $H_1\!: \mu \neq 50$ (two-tailed). The test statistic is
$$z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{52 - 50}{8/8} = 2.$$The critical value at $5\%$ (two-tailed) is $1.96$. Since $2 > 1.96$, you reject $H_0$: there is evidence the mean has changed.Vocabulary TrainEnglish Chinese Pinyin hypothesis test 假设检验 jiǎ shè jiǎn yàn null hypothesis 原假设 yuán jiǎ shè alternative hypothesis 备择假设 bèi zé jiǎ shè one-tailed 单尾 dān wěi two-tailed 双尾 shuāng wěi significance level 显著性水平 xiǎn zhù xìng shuǐ píng test statistic 检验统计量 jiǎn yàn tǒng jì liàng rejection region 拒绝域 jù jué yù Type I error 第一类错误 dì yī lèi cuò wù Type II error 第二类错误 dì èr lèi cuò wù