Further Pure Mathematics 1
A-Level Further Mathematics Topic 1 14:28 English narration · English + 中文 subtitles burned in
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Here is a cubic equation, crossing the axis at three roots.
这是一个三次方程，在横轴上交于三个根。
Finding those roots exactly can be hard — or even impossible by hand.
要精确求出这些根可能很难——甚至手算根本做不到。
But here is a beautiful secret of Further Maths: you can know their sum, and their product, instantly, just by reading the coefficients, without ever solving the equation.
但进阶数学有一个美妙的秘密：你可以立刻知道它们的和，以及它们的积， 只需读一读系数，而完全不用解方程。
That is the spirit of this whole topic — finding the deep structure hidden inside the mathematics.
这正是整个主题的精神所在—— 去发现数学内部隐藏的深层结构。
Further Pure Mathematics builds a richer toolkit.
进阶纯数学搭建起一套更丰富的工具箱。
We will connect the roots of an equation to its coefficients, sketch more daring graphs, sum series exactly, command matrices and the transformations they perform, chart curves in polar coordinates, sharpen our vectors, and finally, prove statements true for all of infinity.
我们会把方程的根和它的系数联系起来， 画出更大胆的图像，精确地求和级数，驾驭矩阵以及它们所施行的变换， 用极坐标描画曲线，磨利我们的向量，最后，证明对整个无穷都成立的命题。
Let's begin.
让我们开始吧。
Start with the roots of a polynomial.
从多项式的根开始。
For a quadratic, the sum of the two roots is minus b over a, and their product is c over a — straight from the coefficients.
对一个二次式，两个根之和是负 b 比 a，两根之积是 c 比 a—— 直接从系数读出来。
For a cubic, the pattern extends: the sum of the roots, the sum of their pairs, and the product, each read from a coefficient.
对一个三次式，这个规律延伸下去：根之和、根两两之积的和、以及三根之积， 各自从一个系数读出。
These are symmetric functions.
这些都是对称函数。
And with a clever substitution, you can build a brand new equation whose roots are shifted — say, each one larger by one — without ever finding the originals.
而通过一个巧妙的代换，你可以构造出一个全新的方程， 它的根被平移了——比如每个都大一——却完全不必求出原来的根。
Here is the exam's favourite move.
这是考试最爱的一招。
Take a cubic.
取一个三次方程。
From its coefficients you already know the sum of the roots, the sum of their pairs, and their product — without solving anything.
从它的系数，你已经知道根之和、根两两之积的和、 以及三根之积——完全不用解方程。
Now suppose the question asks for the sum of the squares of the roots.
现在假设题目要求根的平方和。
Never find the roots.
千万别去求根。
Instead, use an identity: the sum of the squares equals the square of the sum, minus twice the sum of the pairs.
而是用一个恒等式：平方和等于和的平方，减去两倍的两两之积之和。
Put the numbers in — six squared, minus two times eleven — and the answer is fourteen.
代入数字——六的平方，减去二乘十一——答案就是十四。
The second root-coefficient technique builds a whole new equation, and it is worth working in full.
第二种根与系数的技巧是造出一个全新的方程，值得完整做一遍。
Take x squared minus five x plus six equals zero, which has roots alpha and beta.
取 x 平方减五 x 加六等于零，它的两根是 alpha 和 beta。
From the coefficients, the sum is five and the product is six.
由系数可知，和是五，积是六。
Now find the equation whose roots are alpha plus one and beta plus one.
现在要求以 alpha 加一和 beta 加一为根的方程。
There are two routes.
有两条路。
First, transform the sum and the product directly.
第一，直接变换和与积。
The new sum is alpha plus one, plus beta plus one — that is five plus two, so seven.
新的和是 alpha 加一，加上 beta 加一——也就是五加二，等于七。
The new product is alpha plus one, times beta plus one, which expands to alpha beta plus alpha plus beta plus one — six plus five plus one, so twelve.
新的积是 alpha 加一乘以 beta 加一，展开就是 alpha beta 加 alpha 加 beta 加一—— 六加五加一，等于十二。
Or substitute instead.
或者改用代换。
If w is x plus one, then x is w minus one, so put w minus one into the original equation and expand: w squared minus seven w plus twelve equals zero.
令 w 等于 x 加一，那么 x 就是 w 减一， 把 w 减一代回原方程并展开：w 平方减七 w 加十二等于零。
Both routes give the same equation, x squared minus seven x plus twelve equals zero — which is why the substitution is worth knowing as a check on your algebra, not just as an alternative.
两条路给出同一个方程，x 平方减七 x 加十二等于零—— 这正是为什么代换值得掌握：它是对你代数运算的一个检验，而不只是另一种做法。
Next, rational functions — one polynomial over another.
接下来是有理函数——一个多项式除以另一个多项式。
When the top is one degree higher than the bottom, the graph has an oblique asymptote, a slanting line it hugs far out.
当分子的次数正好比分母高一次时， 图像就有一条斜渐近线，一条它在远处紧贴的斜线。
To find it, simply divide out the fraction.
要找到它，只需把分式作除法。
The whole-number part is the asymptote; the remainder fades away as x grows large, so the curve approaches that slant line on both sides.
那个整式部分就是渐近线；余项在 x 变大时逐渐消失，于是曲线在两侧都逼近那条斜线。
When the top of a rational function has degree exactly one HIGHER than the bottom, the graph has an oblique asymptote — a slanted line the curve approaches.
当有理函数分子的次数恰好比分母高一次时，图像就有一条斜渐近线—— 曲线所趋近的一条斜直线。
Find it by dividing out.
用多项式除法把它找出来。
Take y equals x squared plus x plus one, all over x plus one.
取 y 等于 x 平方加 x 加一，除以 x 加一。
Division gives x plus one over x plus one.
作除法得到 x 加上一除以 x 加一。
As x goes to plus or minus infinity that remainder goes to zero, so the curve approaches the line y equals x.
当 x 趋于正负无穷时， 余项趋于零，所以曲线趋近直线 y 等于 x。
Look at the picture: far from the origin the curve hugs that slant line, and near x equals minus one it runs off along the vertical asymptote instead.
看这幅图：在远离原点的地方曲线紧贴那条斜线， 而在 x 等于负一附近，它转而沿着竖直渐近线跑掉。
From the graph of y equals f of x you must be able to sketch four relatives.
由 y 等于 f x 的图像，你必须能画出四个相关的图。
y squared equals f of x exists only where f is positive or zero, and it is symmetric about the x-axis.
y 平方等于 f x 只在 f 非负的地方存在，而且关于 x 轴对称。
One over f of x turns zeros into vertical asymptotes and asymptotes into zeros, and keeps the sign.
一除以 f x 把零点变成竖直渐近线，把渐近线变成零点，并保持符号不变。
The modulus of f of x reflects everything below the axis UP above it.
f x 的绝对值把横轴以下的部分全部翻折到上方。
And f of the modulus of x throws away the left half and replaces it with a mirror image of the right — so it is always symmetric about the y-axis.
而 f 关于 x 绝对值的函数，把左半边丢掉， 换成右半边的镜像——所以它一定关于 y 轴对称。
Two more tools go with this section: use the discriminant to find the set of values the function can take, and differentiate to locate turning points.
这一节还配套两件工具：用判别式求函数的值域， 以及求导来定出驻点。
Now, summing series exactly.
现在，精确地求和级数。
Memorize three standard results: the sum of the first n whole numbers, the sum of their squares, and the sum of their cubes.
记住三个标准结果：前 n 个自然数之和、它们平方之和、它们立方之和。
For trickier sums, use the elegant method of differences.
对更棘手的求和，用优雅的差分法。
If partial fractions let you write each term as a difference, then almost every term cancels its neighbour — a telescoping collapse — leaving just the first and the last.
如果部分分式能让你把每一项写成一个差， 那么几乎每一项都会和它的邻项相消——像望远镜一样收拢——只剩下第一项和最后一项。
Three results you must know by heart.
有三个结果你必须背下来。
The sum of r from one to n is a half n n plus one.
r 从一到 n 的和是二分之一 n 乘以 n 加一。
The sum of r squared is a sixth n n plus one two n plus one.
r 平方的和是六分之一 n 乘以 n 加一乘以二 n 加一。
The sum of r cubed is a quarter n squared n plus one squared — and notice that is the square of the first one.
r 三次方的和是四分之一 n 平方乘以 n 加一的平方—— 注意它正是第一个结果的平方。
Now use them: to find the sum of two r plus one, split the sum term by term into two times the sum of r, plus the sum of one.
现在来用它们： 要求二 r 加一的和，就把求和逐项拆开， 变成二乘以 r 的和，再加上一的和。
That is n n plus one plus n, which factorises to n times n plus two.
也就是 n 乘以 n 加一，再加 n，因式分解后得到 n 乘以 n 加二。
The other technique here is the method of differences: if partial fractions write each term as f of r minus f of r plus one, almost everything cancels, and what survives tells you whether the series is convergent.
这一节的另一种方法是差分法：如果用部分分式把每一项写成 f r 减 f r 加一，那么几乎所有项都会相消， 剩下的部分就告诉你这个级数是否收敛。
Now something visual: matrices.
现在来点直观的：矩阵。
A matrix is a block of numbers.
矩阵是一块数字。
Matrix addition adds matching entries, and multiplication composes transformations.
矩阵加法把对应位置的元素相加， 而矩阵乘法则是把变换复合起来。
Its real power is geometric — it transforms the whole plane.
它真正的威力是几何的——它变换整个平面。
Watch a rotation, a reflection, an enlargement, a stretch, and a shear.
看它做旋转、反射、放大、拉伸和切变。
Points and lines that do not move are invariant points and invariant lines.
变换中不动的点和直线，叫做不变点和不变直线。
The determinant tells you the area scale factor.
行列式告诉你面积的缩放倍数。
And every non-singular matrix has an inverse, which undoes the transformation and lets you solve equations.
而每一个非奇异矩阵都有逆矩阵，它把变换还原回去，还能帮你解方程。
Let's make a matrix concrete.
我们把矩阵讲具体些。
Take this two by two matrix.
取这个二乘二矩阵。
Its determinant is the leading diagonal product minus the other diagonal product — three times four, minus one times two, which is ten.
它的行列式，是主对角线之积减去另一条对角线之积—— 三乘四，减一乘二，等于十。
That determinant is also the area scale factor: the transformation makes every area ten times larger.
这个行列式也是面积的缩放倍数：变换让每个面积放大十倍。
Because it is not zero, the matrix has an inverse.
因为它不为零，这个矩阵有逆矩阵。
Swap the leading diagonal, change the sign of the other two, and divide by the determinant.
把主对角线互换，把另外两个变号，再除以行列式。
Some curves are far easier to describe not by x and y, but by a distance and an angle — polar coordinates.
有些曲线，用 x 和 y 来描述很费劲，用一个距离和一个角度反而容易得多——这就是极坐标。
As the angle sweeps around, the radius changes, tracing shapes a normal graph struggles with.
当角度扫过一圈时，半径随之改变，画出普通图像难以描绘的形状。
Watch this beautiful cardioid appear from one simple equation.
看这条美丽的心形线，从一个简单的方程里浮现出来。
And the area swept out is found by integrating one half r squared.
而它扫过的面积，用二分之一 r 平方来积分求得。
A polar equation can hide a familiar shape.
极坐标方程能把熟悉的形状藏起来。
Take the equation: r equals four cosine theta.
取这个方程：r 等于四倍余弦西塔。
Multiply both sides by r.
两边同时乘以 r。
On the left, r squared becomes x squared plus y squared.
左边，r 的平方变成 x 平方加 y 平方。
On the right, r cosine theta is simply x.
右边，r 乘余弦西塔就等于 x。
So the equation becomes x squared plus y squared equals four x.
于是方程变成 x 平方加 y 平方等于四 x。
Complete the square, and out comes a circle — radius two, centred at the point two, zero.
配方之后，一个圆就出现了——半径二，圆心在点二、零。
Polar coordinates locate a point by its distance r from the origin and its angle theta, instead of by x and y.
极坐标用点到原点的距离 r 和角度 theta 来确定位置， 而不是用 x 和 y。
The picture shows why the conversions are just right-angled trigonometry: x equals r cos theta, y equals r sin theta, and r squared equals x squared plus y squared.
这幅图说明了为什么换算就是直角三角形里的三角函数： x 等于 r cos theta，y 等于 r sin theta， 而 r 平方等于 x 平方加 y 平方。
Going the other way, theta comes from the arctangent of y over x — and you must check the QUADRANT, because the arctangent alone cannot tell the third quadrant from the first.
反过来算时，theta 由 y 除以 x 的反正切得到—— 而且你必须检查象限，因为单凭反正切分不清第三象限和第一象限。
To sketch a polar curve, plot r straight from r as a function of theta — no conversion needed.
要画极坐标曲线，直接把 r 作为 theta 的函数描点即可——不需要换算。
This is the cardioid, r equals one plus cos theta.
这是心形线，r 等于一加 cos theta。
At theta zero, r is two, the far right point.
theta 为零时 r 等于二，也就是最右边那一点。
At theta equals pi over two, r is one.
theta 等于二分之 pi 时 r 等于一。
At theta equals pi, r is zero, which is the dimple at the origin.
theta 等于 pi 时 r 等于零，那就是原点处的凹口。
Join those and the heart shape appears.
把这些点连起来，心形就出现了。
And to find the area of a sector, integrate a half r squared with respect to theta.
而要求扇形面积，就对二分之一 r 平方关于 theta 积分。
Note that carefully: it is r SQUARED, and the half belongs there — it is not the Cartesian area formula in disguise.
要仔细记住：是 r 的平方，而且那个二分之一是必需的—— 它并不是直角坐标面积公式的变形。
Convert the polar equation r equals four cos theta into Cartesian form.
把极坐标方程 r 等于四 cos theta 化成直角坐标形式。
The trick is worth remembering: multiply both sides by r.
这个技巧值得记住：两边同乘以 r。
That gives r squared equals four r cos theta.
于是得到 r 平方等于四 r cos theta。
Now both pieces convert directly — r squared is x squared plus y squared, and r cos theta is x.
现在两部分都能直接换算——r 平方就是 x 平方加 y 平方， 而 r cos theta 就是 x。
So x squared plus y squared equals four x.
所以 x 平方加 y 平方等于四 x。
Completing the square in x gives x minus two, all squared, plus y squared equals four.
对 x 配方，得到 x 减二的平方，加上 y 平方，等于四。
That is a circle of radius two centred at the point two, zero — a circle through the origin, which is what r equals four cos theta traces as theta sweeps round.
这是一个半径为二、圆心在二零点的圆—— 一个过原点的圆，而这正是 theta 转一圈时 r 等于四 cos theta 所描出的轨迹。
Vectors gain a powerful new operation.
向量获得了一种强大的新运算。
Take two vectors in space.
取空间中的两个向量。
Their vector product is itself a vector — one that points at right angles to both of them, perpendicular to the plane they span.
它们的向量积本身也是一个向量—— 一个同时垂直于它们两者的向量，垂直于它们张成的平面。
Its length equals the area of the parallelogram they form.
它的长度， 等于它们所围成的平行四边形的面积。
With it, you can describe a plane, find angles, and even the shortest distance between two lines that never meet.
有了它，你能描述一个平面、求夹角， 甚至求出两条永不相交的直线之间的最短距离。
Besides the scalar product there is the vector product: a cross b equals the magnitude of a, times the magnitude of b, times sin theta, times the unit normal.
除了数量积，还有向量积：a 叉乘 b 等于 a 的模， 乘以 b 的模，乘以 sin theta，再乘以单位法向量。
Two things make it useful and both are in the picture.
有两点让它有用，而且图上都画出来了。
It gives a vector at RIGHT ANGLES to both — which is how you get a normal to a plane from two vectors lying in it.
它给出一个同时垂直于两者的向量—— 这正是你如何由平面内的两个向量得到平面法向量的方法。
And its LENGTH equals the area of the parallelogram the two vectors span.
而它的长度等于这两个向量张成的平行四边形的面积。
Note the contrast with the scalar product: that uses cosine and returns a number, this uses sine and returns a vector.
注意它和数量积的对照：数量积用余弦，返回一个数； 向量积用正弦，返回一个向量。
Find a cross b where a is i plus j, and b is j plus k.
求 a 叉乘 b，其中 a 是 i 加 j，b 是 j 加 k。
Write the determinant with i, j, k across the top, a's components on the second row, and b's on the third.
写出行列式，第一行是 i、j、k，第二行是 a 的分量，第三行是 b 的分量。
Expanding along the top row gives i times one, minus j times one, plus k times one.
沿第一行展开，得到 i 乘以一，减去 j 乘以一，加上 k 乘以一。
Watch the minus sign on the j term — it is part of the expansion, not a mistake, and forgetting it is the single most common error here.
注意 j 那一项前面的负号——它是展开式的一部分，不是写错了， 而漏掉它正是这里最常见的错误。
So the answer is i minus j plus k.
所以答案是 i 减 j 加 k。
Check it if you like: dot that with either original vector and you get zero, confirming it is perpendicular to both.
你可以验算一下： 把它与原来任一个向量作数量积都得零，这就确认了它同时垂直于两者。
In three dimensions a plane can be written three ways, and questions switch between them freely: as a x plus b y plus c z equals d, as r dot n equals p, or in parametric form as a plus lambda b plus mu c.
在三维空间中，平面有三种写法，而题目会在它们之间自由切换： 写成 a x 加 b y 加 c z 等于 d，写成 r 点乘 n 等于 p， 或者写成参数形式 a 加 lambda b 加 mu c。
With the scalar and vector products together you can find distances, angles and intersections — where a line meets a plane, where two planes meet.
把数量积和向量积配合起来，你就能求距离、角度和交点—— 直线与平面的交点、两个平面的交线。
The hardest standard case is skew lines: lines in three dimensions that are not parallel and never meet.
最难的标准情形是异面直线：三维空间中既不平行又永不相交的两条直线。
Their shortest distance runs along the common perpendicular, and you get that direction from the vector product of the two direction vectors.
它们之间的最短距离沿着公垂线， 而那个方向由两条直线方向向量的向量积给出。
Finally, mathematical induction — a way to prove a statement true for every positive whole number, all at once.
最后，数学归纳法——一种一次性证明命题对每一个正整数都成立的方法。
It works like a line of dominoes.
它就像一排多米诺骨牌。
First, the base case: knock over the very first one.
第一步，基例：推倒最前面的那一张。
Then, the inductive step: show that each domino, if it falls, knocks over the next.
第二步，归纳步骤： 证明每一张骨牌只要倒下，就会撞倒下一张。
Together, those two steps topple the entire infinite line.
这两步合在一起，就推倒了整条无穷长的骨牌。
Now let's actually write the proof — the exam wants every part.
现在我们真正把证明写出来——考试要求每一部分都写。
Prove that the sum of the first n whole numbers is one half n times n plus one.
证明：前 n 个自然数之和， 等于二分之一 n 乘 n 加一。
Base case: for n equals one, both sides equal one, so it holds.
基例：当 n 等于一时，两边都等于一，所以成立。
Inductive step: assume the formula is true for n equals k.
归纳步骤：假设公式对 n 等于 k 成立。
For the next term, add k plus one to both sides, factor, and tidy up — and you reach exactly the formula with k plus one.
对下一项，两边都加上 k 加一，因式分解，整理—— 你正好得到把 k 换成 k 加一的公式。
So each true case forces the next.
于是每一个成立的情形都逼出下一个。
That completes the mathematical induction.
这就完成了归纳法。
Often you first make a conjecture from a few cases, then confirm it by inductive proof.
常常你先从几个情形做出一个猜想，再用归纳证明去确认它。
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 relationships between roots and coefficients, rather than solving the polynomial.
第一，用根与系数的关系，而不是去硬解多项式。
Second, for induction, set out the base case, the assumption, the step to k plus one, and a clear closing statement.
第二，做归纳法时，写清基例、假设、到 k 加一的步骤，以及一句清晰的结语。
Third, sketch rational graphs with their asymptotes — divide out to find the slant one.
第三，画有理函数图像要带上渐近线——作除法找出那条斜的。
Fourth, for matrices, know the determinant and inverse, and read every transformation geometrically.
第四，对矩阵，掌握行列式和逆矩阵，并从几何角度去解读每一个变换。