纯数学3

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:

First check the top is lower degree than the bottom. If it is not ("top-heavy"), divide the polynomial first: dividing gives a quotient 商 plus a remainder 余数 over the original bottom, and you split only that remaining proper fraction.

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

Worked example: a repeated factor. Express $\dfrac{5x+4}{(x-1)(x+2)^2}$ in partial fractions.

A squared factor needs two fractions, one over $(x+2)$ and one over $(x+2)^2$:

so $5x+4 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)$. Put $x = 1$: $9 = 9A$, so $A = 1$. Put $x = -2$: $-6 = -3C$, so $C = 2$. No value of $x$ removes both $A$ and $C$, so compare the $x^2$ terms instead: $0 = A + B$, so $B = -1$. Hence

Check with $x = 0$: the left side is $\dfrac{4}{(-1)(4)} = -1$ and the right side is $-1 - \tfrac12 + \tfrac12 = -1$.

Common mistakes. Leaving out the $\dfrac{B}{x+2}$ term for a squared factor, and splitting a top-heavy fraction without dividing first.

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 $|x| < 1$:

Worked example. Expand $(4 - x)^{-1/2}$ as far as the term in $x^2$, and state the values of $x$ for which it is valid.

The formula needs a $1$ in front, so take out the $4$ first, raised to the same power:

Now use $n = -\tfrac12$ with $-\tfrac{x}{4}$ in place of $x$:

Multiply by $\tfrac12$: $(4 - x)^{-1/2} = \dfrac12 + \dfrac{x}{16} + \dfrac{3x^2}{256} + \cdots$. It is valid when $\left|\tfrac{x}{4}\right| < 1$, that is $|x| < 4$.

Common mistakes. Writing $4(1 - \tfrac{x}{4})^{-1/2}$ (the $4$ must also be raised to the power $-\tfrac12$); losing the minus sign inside $\left(-\tfrac{x}{4}\right)^2$ in later terms; and giving the validity as $|x| < 1$ instead of $|x| < 4$.

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.

The three new trig functions are the reciprocals 倒数 of the familiar ones: 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 $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$ for $\sin\theta\ne0$. It equals $1/\tan\theta$ wherever both expressions are defined. (Memory aid: match the third letter — sec goes with cosine.) The two Pythagorean identities above come straight from dividing $\sin^2\theta + \cos^2\theta \equiv 1$ by $\cos^2\theta$ or $\sin^2\theta$. For a numerical method, an iteration $x_{n+1}=F(x_n)$ converges 收敛 if its values approach a finite limit. Smaller successive changes alone do not prove convergence; when $F$ is continuous at that limit, the limit is a fixed point. Check it in the original equation.

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:

Worked example. Differentiate $y = \tan^{-1}(3x)$.

Use the chain rule with the result above: $\dfrac{dy}{dx} = \dfrac{1}{1 + (3x)^2} \times 3 = \dfrac{3}{1 + 9x^2}.$

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 $v = \sin x$:

Worked example: substitution. Use the substitution $u = x + 1$ to find $\displaystyle\int_0^3 \frac{x}{\sqrt{x+1}}\,dx$.

Change all three parts: $x = u - 1$, $dx = du$, and the limits $x = 0$ and $x = 3$ become $u = 1$ and $u = 4$:

The common error is to keep the old limits $0$ and $3$ after changing to $u$.

Worked example: partial fractions. Show that $\displaystyle\int_3^4 \frac{x+4}{(x+1)(x-2)}\,dx = \ln\frac{16}{5}$.

Use the partial fractions found earlier, then integrate each piece as a logarithm:

This is $2\ln 2 + 2\ln 2 - \ln 5 = \ln 16 - \ln 5 = \ln\dfrac{16}{5}$. Because the answer is given, show the log laws step by step.

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 nonzero 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).

The scalar product 数量积 (dot product) of $\mathbf{a}$ and $\mathbf{b}$ is

Worked example. Find the angle between $\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}$.

Worked example: do two lines meet? Line $l$ is $\mathbf{r} = \mathbf{i} + 3\mathbf{j} + t(\mathbf{i} - \mathbf{j} + 2\mathbf{k})$ and line $m$ is $\mathbf{r} = -\mathbf{j} + \mathbf{k} + s(3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})$. Show that they intersect and find the point.

The directions are not multiples of each other, so the lines are not parallel. Use a different letter for each line's parameter and set the components equal:

Add the $x$ and $y$ equations: $4 = -1 + 5s$, so $s = 1$ and then $t = 2$. Now test the third equation: $2(2) = 4$ and $1 + 3(1) = 4$. It holds, so the lines meet, at $\mathbf{i} + 3\mathbf{j} + 2(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 3\mathbf{i} + \mathbf{j} + 4\mathbf{k}$, the point $(3, 1, 4)$.

If the third equation had failed, the lines would be skew. Always use the equation you did not solve as the check.

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:

Worked example: forming the equation. Water leaks from a tank. The depth $h$ cm decreases at a rate proportional to $\sqrt h$. At $t = 0$ minutes the depth is $16$ cm, and at $t = 10$ it is $9$ cm. When is the tank empty?

"Decreases at a rate proportional to $\sqrt h$" gives $\dfrac{dh}{dt} = -k\sqrt h$, with $k > 0$. Separate and integrate:

At $t = 0$: $2\sqrt{16} = c$, so $c = 8$. At $t = 10$: $2\sqrt9 = -10k + 8$, so $k = 0.2$. The tank is empty when $h = 0$: $0 = -0.2t + 8$, so $t = 40$ minutes.

Common mistakes. Leaving out the minus sign for a decrease, and adding the constant only after rearranging (so writing $y = e^{x^2/2} + c$ instead of $y = Ae^{x^2/2}$ in the example above).

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, for $z e0$, the argument 辐角 is the angle the point makes, measured from the positive real axis. Zero has no argument.
• 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.
• The loci of points satisfying a condition on $z$ are drawn on the Argand diagram — e.g. $|z - a| = r$ is a circle, $|z - a| = |z - b|$ a perpendicular bisector, and $\arg(z - a) = \theta$ a half-line. The square roots of a complex number come from solving $w^2 = z$.

To divide, multiply top and bottom by the conjugate of the bottom.

Worked example. Write $\dfrac{3 + i}{1 - i}$ in the form $x + iy$.

Worked example: modulus and argument. Find the modulus and argument of $z = -1 + \sqrt3\,i$.

$|z| = \sqrt{(-1)^2 + (\sqrt3)^2} = 2$. The point $(-1, \sqrt3)$ is in the second quadrant, so draw it before using a calculator. The angle with the negative real axis is $\tan^{-1}\sqrt3 = \tfrac{\pi}{3}$, so $\arg z = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$. Hence $z = 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right)$.

A calculator's $\tan^{-1}\!\left(\tfrac{\sqrt3}{-1}\right)$ gives $-\tfrac{\pi}{3}$, which is the wrong quadrant. In the same way $\arg(-1 - i) = -\tfrac{3\pi}{4}$, not $\tfrac{\pi}{4}$.

Worked example: a square root. Find the square roots of $3 + 4i$.

Let $(a + bi)^2 = 3 + 4i$ with $a$ and $b$ real. Then $a^2 - b^2 + 2abi = 3 + 4i$, so compare real and imaginary parts:

Substitute $b = \dfrac{2}{a}$: $a^2 - \dfrac{4}{a^2} = 3$, so $a^4 - 3a^2 - 4 = 0$, giving $(a^2 - 4)(a^2 + 1) = 0$. Since $a$ is real, $a^2 = 4$, so $a = 2$ with $b = 1$, or $a = -2$ with $b = -1$. The square roots are $\pm(2 + i)$. Check: $(2 + i)^2 = 4 + 4i - 1 = 3 + 4i$.

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 $a$.

• Split a rational function into partial fractions before integrating or expanding.
• Choose the right integration technique (substitution, by parts, or partial fractions) from the form of the integrand.
• Use the scalar (dot) product for the angle between vectors and to test for perpendicularity; write a line as $\mathbf{r} = \mathbf{a} + t\mathbf{b}$.
• Give complex numbers in the form asked for (Cartesian or modulus-argument) and show them on an Argand diagram.