纯数学2

The modulus

The modulus 绝对值 $|x|$ is the size of a number with its sign removed, so $|x| \geqslant 0$ always. For $a\ne0$, 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

Worked example. Solve $|3x + 8| < 9$.

Using the second rule with the inequality written as $-9 < 3x + 8 < 9$:

Worked example: modulus on both sides. Solve $|x - 2| > |2x + 1|$.

Both sides are never negative, so squaring keeps the inequality the right way round:

The quadratic is negative between its roots, so $-3 < x < \tfrac13$. Check $x = 0$: $|{-2}| = 2 > 1 = |1|$, true.

Common mistake. Solving only $x - 2 > 2x + 1$ (giving $x < -3$) misses the other cases. Squaring, or sketching both V graphs, catches them all.

Polynomial division and the factor and remainder theorems

A polynomial 多项式 is a finite sum of terms with non-negative integer 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 $p(-2) = 0$:

Worked example: dividing by a quadratic. Find the quotient and remainder when $2x^3 - 3x^2 + 4x - 5$ is divided by $x^2 + 1$.

Divide the leading terms: $2x^3 \div x^2 = 2x$. Subtract $2x(x^2 + 1) = 2x^3 + 2x$ to leave $-3x^2 + 2x - 5$. Next, $-3x^2 \div x^2 = -3$; subtract $-3(x^2 + 1) = -3x^2 - 3$ to leave $2x - 2$. That has lower degree than $x^2 + 1$, so stop:

The quotient is $2x - 3$ and the remainder is $2x - 2$. Dividing by a quadratic can leave a linear remainder, not just a number.

A logarithm answers the question "what power?". If $a^x = y$ then $x = \log_a y$. For real logarithms, the base is positive and not 1, and the argument is positive. Logarithms and indices 指数 (powers) are reverse ideas. The laws of logarithms 对数定律 are

The exponential function $e^x$ and the natural logarithm 自然对数 $\ln x$ (for $x>0$) 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.

Worked example. Solve $4^x < 0.05$.

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.

Worked example: unknowns in two powers. Solve $2^{x+1} = 3^x$.

Take logs of both sides and bring the powers down: $(x + 1)\ln 2 = x\ln 3$. Collect the $x$ terms: $\ln 2 = x(\ln 3 - \ln 2)$, so

Worked example: finding constants from a line. Two variables satisfy $y = Ax^n$. The graph of $\ln y$ against $\ln x$ is a straight line through $(0.5, 2.1)$ and $(2.5, 5.1)$. Find $A$ and $n$.

The gradient is $n = \dfrac{5.1 - 2.1}{2.5 - 0.5} = 1.5$. Then $\ln A = 2.1 - 1.5(0.5) = 1.35$, so $A = e^{1.35} = 3.86$. Remember to undo the log: giving $A = 1.35$ is a common slip.

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 $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$ where $\sin\theta\ne0$. It equals $1/\tan\theta$ where both expressions are defined.

You must know these trigonometric identities 三角恒等式 and choose the right one for each problem:

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 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 $R\cos\alpha=a$, $R\sin\alpha=b$. Choose the quadrant from both signs; $\tan\alpha=b/a$ alone is insufficient, and if $a=b=0$ the expression is zero.

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:

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 $\theta = \pm 104.5^\circ$.

Worked example: the R-formula. Express $3\sin\theta + 4\cos\theta$ as $R\sin(\theta + \alpha)$, and solve $3\sin\theta + 4\cos\theta = 2$ for $0^\circ < \theta < 360^\circ$.

Expand: $R\sin(\theta + \alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta$. Compare: $R\cos\alpha = 3$ and $R\sin\alpha = 4$. So $R = \sqrt{3^2 + 4^2} = 5$ and $\tan\alpha = \tfrac43$, giving $\alpha = 53.13^\circ$.

Now $5\sin(\theta + 53.13^\circ) = 2$, so $\sin(\theta + 53.13^\circ) = 0.4$. The angle $\theta + 53.13^\circ$ lies between $53.13^\circ$ and $413.13^\circ$, so its values are $180^\circ - 23.58^\circ = 156.42^\circ$ and $360^\circ + 23.58^\circ = 383.58^\circ$ (the basic value $23.58^\circ$ is too small). Subtract $53.13^\circ$:

Shifting the interval before listing the angles is the step most often missed.

Learn these standard derivatives:

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 $v$):

Worked example: a parametric tangent. A curve has $x = t^2$ and $y = t^3 - 3t$. Find the equation of the tangent at the point where $t = 2$.

Differentiate each with respect to $t$: $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2 - 3$, so

At $t = 2$ the point is $(4, 2)$, so the tangent is $y - 2 = \tfrac94(x - 4)$, that is $4y = 9x - 28$. Use the point $(x, y) = (4, 2)$, not $(t, \ldots) = (2, \ldots)$.

Integration is reverse differentiation — reverse each new derivative; harder integrals may need integration by substitution 换元积分 (developed in Pure 3). For a linear inside function $(ax + b)$ with $a\ne0$:

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:

Worked example. Find $\displaystyle\int 6\sin^2 x\,dx$.

Use $\sin^2 x = \tfrac12(1 - \cos 2x)$:

Worked example: the trapezium rule. Use the trapezium rule with $2$ intervals to estimate $\displaystyle\int_0^1 \sqrt{1 + x^3}\,dx$, and say whether it is an overestimate or an underestimate.

With $h = 0.5$ the ordinates are $y_0 = 1$, $y_1 = \sqrt{1.125} = 1.0607$ and $y_2 = \sqrt2 = 1.4142$:

The curve bends upward (it is convex) on $[0, 1]$, so each straight top lies above the curve, and $1.13$ is an overestimate. (The true value is $1.11$.) Remember: the number of ordinates is one more than the number of intervals.

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 are convergent they settle down and converge 收敛 to a root. Keep going until the answer is steady to the accuracy asked for.

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 $x_0 = -1.2$.

The values settle near $-1.26$, so $\beta = -1.26$ (3 s.f.).

• Use the laws of logarithms to solve equations; remember $\ln$ and $e^x$ are inverses.
• For numerical methods, show a sign change to locate a root, set out the iteration clearly, and give the answer to the stated accuracy.
• Learn the chain, product and quotient rules and identify which the function needs.
• Solve modulus equations $|f(x)| = g(x)$ by considering both the positive and negative cases, and sketch to check.