Pure Mathematics 2
A-Level Mathematics Topic 2 12:32 English narration · English + 中文 subtitles burned in
Chapters
Transcript
In two thousand eleven, an earthquake off Japan measured nine on the Richter scale.
二〇一一年，日本外海的一次地震在里氏震级上测得九级。
A smaller quake might measure six.
一次较小的地震也许是六级。
Only three points apart — yet the first shook the ground a thousand times harder.
只相差三级——可第一次让地面震动的幅度却是后者的一千倍。
How can such vast numbers fit on one small scale?
如此巨大的数字，怎么能装进一把小小的标尺？
The answer is the logarithm.
答案就是对数。
On a log scale, every equal step multiplies by ten, folding the enormous into the manageable.
在对数刻度上，每走相等的一步，就乘以十，把庞大折叠成可掌控的大小。
It is one of the powerful new tools in Pure Mathematics, part two.
它正是纯数学第二部分里那些强大新工具之一。
Pure two picks up where pure one left off.
纯数学第二部分，接着第一部分继续讲。
We add new algebra — the modulus and polynomials — then logarithms and exponentials, a deeper layer of trigonometry, and sharper tools for calculus.
我们添上新的代数——绝对值与多项式—— 然后是对数与指数、更深一层的三角，以及更锋利的微积分工具。
Finally, when an equation has no exact answer, we learn to hunt the solution numerically.
最后，当一个方程没有精确解时，我们学会用数值方法去搜寻它的解。
Let's begin.
让我们开始吧。
First, new algebra.
先看新的代数。
The modulus of a number is its size, with any minus sign thrown away — so it is never negative.
一个数的绝对值，就是把正负号去掉后它的大小——所以它永远不为负。
Its graph is a sharp V that bounces off the axis.
它的图像是一个尖尖的 V，在横轴上弹起。
Two rules do most of the work.
两条规则就能完成大部分工作。
To compare two moduli, square both sides.
要比较两个绝对值，就把两边平方。
And a modulus less than a number means the inside lies within that distance of zero.
而绝对值小于某个数，意味着里面的量离零不超过那个距离。
So the inequality — mod three x plus eight, less than nine — unfolds into a simple range for x.
于是这个不等式——三 x 加八的绝对值小于九——就展开成 x 的一个简单范围。
Next, polynomials — sums of powers of x.
接下来是多项式——x 各次幂的和。
When you divide one by x minus a, you can read the remainder instantly, without dividing at all: it is just the polynomial evaluated at a.
当你用 x 减 a 去除它时，不必真的做除法， 就能立刻读出余数：它就是把多项式在 a 处求值。
That is the remainder theorem.
这就是余数定理。
And if that value comes out zero, then x minus a divides it exactly — the factor theorem.
而如果这个值恰好为零，那么 x 减 a 就能整除它——这就是因式定理。
So to test a suspected factor like x plus two, just substitute minus two and check for zero, which quickly pins down an unknown coefficient.
所以要检验一个可疑的因式，比如 x 加二，只需代入负二，看是否为零， 这就能很快定出一个未知系数。
Now logarithms and their partner, the exponential — logarithms and indices are inverses.
现在讲对数，以及它的搭档——指数。
Three laws turn multiplication into addition: the log of a product is the sum of the logs, division becomes subtraction, and a power comes down to the front.
三条定律把乘法变成加法：乘积的对数是各对数之和， 除法变成相减，而幂则落到前面成为系数。
The exponential and the natural logarithm are perfect inverses — each undoes the other.
指数与自然对数互为完美的反函数——彼此抵消。
And here is a favourite exam trick — linear form: a power law becomes a straight line the moment you take logs, so its gradient and intercept hand you the unknown constants.
还有一个考试常用的妙招：一旦取对数，幂律关系就变成一条直线， 于是它的斜率和截距就把未知常数交给了你。
Picture the two curves: e to the x climbing away, and the natural log rising slowly.
想象这两条曲线：e 的 x 次方一路上扬，自然对数缓缓上升。
They are exact mirrors across the line y equals x — each undoes the other.
它们关于直线 y 等于 x 互为精确镜像——彼此抵消。
So if you ever need to undo an exponential, take a natural log. If you need to undo a natural log, raise e to that power.
所以要解开指数，就取自然对数；要解开自然对数， 就把 e 升到那个幂。
That single fact unlocks growth models, decay, and every exam equation with the unknown stuck in the exponent.
这一条事实就能打开增长模型、衰减，以及所有未知数卡在指数上的考题。
Worked example.
例题。
Solve four to the x is less than zero point zero five.
解四的 x 次方小于零点零五。
First take natural logs of both sides: x times the natural log of four is less than the natural log of zero point zero five.
先对两边取自然对数：x 乘四的自然对数， 小于零点零五的自然对数。
Then divide by the natural log of four — it is positive, so the inequality keeps its direction.
再除以四的自然对数——它是正的，所以不等号方向不变。
The right-hand side is about minus two point one six, to three significant figures.
右边约等于负二点一六，保留三位有效数字。
So x is less than that value.
因此 x 小于这个值。
Trigonometry gains three new functions, each the reciprocal of one you already know: secant, cosecant, and cotangent — one over cosine, one over sine, one over tangent.
三角这里新增了三个函数，每一个都是你已经认识的某个函数的倒数：正割、余割、余切—— 也就是一比余弦、一比正弦、一比正切。
They arrive with two new trigonometric identities, close cousins of the Pythagorean one: secant squared equals one plus tangent squared, and cosecant squared equals one plus cotangent squared.
它们还带来两条新的恒等式，是勾股恒等式的近亲： 正割的平方等于一加正切的平方，余割的平方等于一加余切的平方。
Reach for the identity that leaves you with a single function.
要挑那条能让你只剩下一个函数的恒等式来用。
Look at the secant graph beside ordinary cosine.
看正割图像，对照普通的余弦。
Wherever cosine is zero, the secant shoots up to an asymptote and never touches the axis there.
余弦为零的地方，正割冲向渐近线，在那里永远不碰到横轴。
That is why equations in secant often flip back to cosine at the end: one over cosine makes the vertical walls visible, but cosine itself is friendlier to solve.
这就是为什么正割方程常常最后又改回余弦：一比余弦让竖直的墙显出来， 但余弦本身更好解。
Always state the domain — secant is undefined at those same zeros of cosine.
一定要写清定义域——正割在余弦的那些零点处没有定义。
Three more families of formulas.
还有三族公式。
The compound angle formulas expand the sine or the cosine of a sum or a difference.
复合角公式，把和角或差角的正弦、余弦展开。
The double angle formulas rewrite the sine and cosine of twice an angle in terms of the single angle — priceless for simplifying, and for integrating.
二倍角公式，把二倍角的正弦和余弦，改写成单个角的形式—— 这对化简和积分都极为宝贵。
And the R-formula is a small piece of magic: it collapses a sine plus a cosine into one single shifted wave.
而辅助角公式则是一个小小的魔法： 它把一个正弦加一个余弦，合并成单独一个平移过的波。
That instantly reveals the maximum, the minimum, and every solution of the equation.
这一下子就揭示出最大值、最小值，以及方程的每一个解。
Worked example.
例题。
Solve two tangent squared theta plus three secant theta equals eighteen, for theta between minus one hundred eighty and one hundred eighty degrees.
解二倍正切平方 theta 加三倍正割 theta 等于十八，theta 在负一百八十度到 一百八十度之间。
First replace tangent squared with secant squared minus one, so only one function remains.
先把正切平方换成正割平方减一，这样只剩一个函数。
That rearranges to a quadratic in secant: two secant squared plus three secant minus twenty equals zero.
整理成正割的 二次方程：二正割平方加三正割减二十等于零。
Factor it — two secant minus five, times secant plus four.
因式分解——二正割减五，乘正割加四。
So secant is five over two, or secant is minus four.
所以正割是五比二，或正割是负四。
Then cosine is two fifths or minus one quarter.
于是余弦是五分之二或负四分之一。
That gives four solutions in the given range: about plus or minus sixty six point four degrees, and plus or minus one hundred four point five degrees.
在给定范围内 有四个解：大约正负六十六点四度，以及正负一百零四点五度。
Differentiation now needs a short list of standard results you must know cold.
现在微分需要一张你必须倒背如流的标准结果短表。
The exponential differentiates to itself.
指数求导还是它自己。
The natural log of x differentiates to one over x.
x 的自然对数求导是一比 x。
Sine and cosine trade places — the derivative of sine is cosine, and the derivative of cosine is minus sine.
正弦和余弦互换位置——正弦的导数是余弦， 余弦的导数是负正弦。
And the derivative of tangent is secant squared.
而正切的导数是正割的平方。
Learn them once; every chain rule and product rule call depends on them.
记牢一次；之后每次链式法则 和乘积法则都靠它们。
Calculus grows new muscles too.
微积分也长出了新的肌肉。
When two functions are multiplied together, you cannot just differentiate each part on its own.
当两个函数相乘时，你不能只是各自分别求导。
The product rule handles it: differentiate the first times the second, plus the first times the derivative of the second.
乘积法则来处理它：第一个的导数乘以第二个，加上第一个乘以第二个的导数。
Watch the growing rectangle — its expanding area is exactly this rule made visible.
看这个不断长大的长方形——它扩张的面积，正是这条法则的可视化。
When one function is divided by another, the quotient rule takes over.
当一个函数被另一个除时，就轮到商法则出场。
It looks fiddly, but the pattern is fixed: the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.
它看着繁琐，但套路是固定的： 分子的导数乘以分母，减去分子乘以分母的导数，全部除以分母的平方。
Learn the pattern once, and every fraction yields.
把这个套路记熟一次，之后每个分式都会乖乖就范。
Worked example.
例题。
Differentiate y equals six x times cosine of x squared plus one.
对 y 等于六 x 乘以余弦 x 平方加一 求导。
It is a product of six x and a cosine of a bracket.
这是六 x 与一个括号余弦的乘积。
Let u be six x, so u dash is six.
令 u 为六 x，则 u 撇是六。
For v, use the chain rule for the cosine: the outside gives minus sine of the bracket, times the inside derivative two x.
对 v 用链式法则处理余弦：外层给出负正弦括号， 再乘以内层导数二 x。
So the product rule gives six times cosine of the bracket, plus six x times minus two x sine of the bracket.
于是乘积法则给出六乘括号的余弦，加上六 x 乘负二 x 乘括号的正弦。
That simplifies to six cosine of the bracket, minus twelve x squared sine of the bracket.
化简为六乘括号的余弦，减去十二 x 平方乘括号的正弦。
Sometimes a curve is not written as y equals a function of x.
有时曲线并不是写成 y 等于 x 的某个函数。
It might be given by parametric equations, with x and y each depending on a third variable, t.
它可能以参数形式给出， x 和 y 各自依赖第三个变量 t。
Then the gradient is simply the rate of y over the rate of x.
这时斜率就是 y 的变化率除以 x 的变化率。
Or the equation might be written implicitly, all tangled together.
又或者方程是隐式的，全都纠缠在一起。
Then you differentiate every term, applying the chain rule to the y parts, and solve for the gradient.
那就对每一项求导，对 y 的部分用链式法则， 再解出斜率。
Every new derivative can be run in reverse to give a new integral — integration by substitution when a chain-rule pattern appears.
每一个新的导数，都能反过来运行，给出一个新的积分。
So the exponential integrates back to itself, one over the bracket gives a logarithm, and sine and cosine trade places.
于是指数积分回它自己， 一比那个括号给出对数，正弦和余弦互换位置。
Just remember to divide by the number in front of x.
只要记得除以 x 前面的那个数。
And when you meet a power of sine or cosine, you cannot integrate it directly — first use a double angle identity to flatten the power, then integrate term by term.
而当你遇到正弦或余弦的幂时，不能直接积分——先用二倍角恒等式把幂压平，再逐项积分。
Worked example.
例题。
Find the integral of six sine squared x.
求六倍正弦平方 x 的积分。
You cannot integrate sine squared directly.
不能直接积分正弦平方。
First replace it with a double angle identity: sine squared x equals half of one minus cosine two x.
先用二倍角恒等式替换： 正弦平方 x 等于二分之一乘以一减余弦二 x。
Then six times that identity becomes three minus three cosine two x.
再乘六，变成三减三余弦二 x。
Integrate term by term.
逐项积分。
The integral is three x minus three over two sine two x, plus a constant.
结果是三 x 减二分之三正弦二 x，再加一个常数。
But some integrals have no exact answer at all.
但有些积分根本没有精确解。
When that happens, we estimate.
遇到这种情况，我们就估算。
The trapezium rule slices the area into thin strips, and treats the top of each as a straight line — a trapezium.
梯形法则把面积切成一条条细窄的长条，并把每一条的顶部当作一条直线——一个梯形。
Add the strip areas together, and you get a good numerical estimate.
把这些长条的面积加起来，就得到一个不错的数值估计。
More strips, a better answer.
长条越多，答案越准。
Write the formula carefully.
仔细写出公式。
The width of each strip is h.
每条长条的宽度是 h。
The estimate is h over two, times the first ordinate plus the last, plus twice the sum of all the ordinates in between.
估计值是 h 除以二，乘以第一个纵坐标加最后一个， 再加上中间所有纵坐标之和的两倍。
End strips count once; every middle strip is shared by two trapezia, so those ordinates get the factor two.
两端各算一次；中间每条被两个梯形共用， 所以那些纵坐标乘二。
Use more strips when the curve bends sharply — the straight tops fit better when h is small.
曲线弯得厉害时多用几条——h 越小，直线顶边越贴合。
The same spirit solves equations we cannot crack by hand.
同样的思路，也能解出我们手算不出来的方程。
First, trap a root: if the function changes sign between two points, a solution must lie between them.
第一步，锁住一个根： 如果函数在两点之间变号，那么它们之间必有一个解。
Then close in on it.
然后逐步逼近它。
One powerful method rides the tangent, sliding down to the axis again and again, each step landing nearer the root — until the answer stops moving.
有一种强大的方法沿着切线滑动，一次又一次滑到横轴，每一步都落得离根更近—— 直到答案不再移动。
Before any iteration, trap the root with a sign change.
在任何迭代之前，先用变号锁住根。
Evaluate the function at two trial points.
在两个试探点求函数值。
If one value is positive and the other is negative, and the graph has no break between them, a root must sit in that interval.
如果一个正、一个负， 且图像中间不断裂，那么根必在这个区间里。
In the exam, write both function values with their signs — that single line earns the location mark even before you start iterating.
考试里把两个函数值连同符号写出来—— 这一行在开始迭代之前就能拿到定位分。
Iteration rearranges the equation into x equals F of x, then uses the iterative formula x n plus one equals F of x n.
迭代把方程改写成 x 等于 F 的 x。
Start from a first guess, then feed each answer back in.
从一个初值开始，再把每个答案代回去： x 下标 n 加一 等于 F 的 x 下标 n。
On the graph, each step goes up to the curve y equals F of x, then across to the line y equals x — a staircase that closes in on their crossing when the process is convergent.
在图上，每一步先升到曲线 y 等于 F 的 x， 再横到直线 y 等于 x——若过程收敛，这道阶梯就逼近它们的交点。
Keep going until the digits stop changing at the accuracy the question asks for.
一直做到题目要求的精度下数字不再变化为止。
Worked example.
例题。
A root beta satisfies x equals the cube root of minus two x minus four point five, and it sits between minus one point four and minus one.
根 beta 满足 x 等于负二 x 减四点五的立方根，并且落在负一点四与负一之间。
Use the iteration x n plus one equals the cube root of minus two x n minus four point five.
用迭代：x 下标 n 加一 等于负二 x 下标 n 减四点五的立方根。
Start with x nought equal to minus one point two.
从 x 零等于负一点二开始。
The next value is about minus one point two eight one, then about minus one point two four seven.
下一个值约负一点二八一，再下一个约负一点二四七。
The sequence settles near minus one point two six, so beta is minus one point two six to three significant figures.
数列稳定在负一点二六附近， 所以 beta 是负一点二六，保留三位有效数字。
Before you go, four ways to keep your marks.
结束之前，四个保住分数的办法。
First, use the log laws to solve any equation with the unknown in the power, and remember e and the natural log are inverses.
第一，凡是未知数在幂上的方程，就用对数定律来解， 并记住 e 与自然对数互为反函数。
Second, for a product or a quotient, pick the correct rule and name it.
第二，遇到乘积或商，选对法则并写出它的名字。
Third, in numerical work, always show a sign change to trap the root, and give the accuracy asked for.
第三，做数值题时，一定要展示一次变号来锁住根，并按要求的精度作答。
Fourth, for a modulus equation, consider both the positive and the negative cases, then sketch to check.
第四，做绝对值方程时，把正、负两种情形都考虑，再画图检验。