§参数方程和极坐标
- 参数方程（parametric equations）、图像和切线（tangents）
- 笛卡尔坐标（Cartesian coordinates）和极坐标（polar coordinates）的转换
- 极坐标曲线（polar curve）的切线
- 极坐标曲线围成的面积
27.1 参数方程
…suppose that both $x$ and $y$ are functions of another variable $t$.
$x=3\cos(t)$ 和 $y=3\sin(t)$，其中 $0{\leq}t{\leq}2\pi$ 是圆 $x^2+y^2=9$ 的参数化（parametrization）
The variable $t$ is called a parameter, and the above equations are called parametric equations.
描点作图：
| $t$ | $0$ | $\pi/6$ | $\pi/4$ | $\pi/3$ | $\pi/2$ |
|---|---|---|---|---|---|
| $x$ | $3$ | $3\sqrt{3}/2$ | $3/\sqrt{2}$ | $3/2$ | $0$ |
| $y$ | $0$ | $3/2$ | $3/\sqrt{2}$ | $3\sqrt{3}/2$ | $3$ |
可以得到圆（$x^2+y^2=(3\cos(t))^2+(3\sin(t))^2=9$）的四等分弧（quarter-arc）
$t$ 的周期为 $2\pi$，也可以使 $t$ 从 $0$ 向 $t<0$ 的区间取点作图，得到完整的圆
Notice that if you pick a point $(x, y)$ on the circle, there isn’t just one value of $t$ which corresponds to that point!
其他相同的参数化：
$x=3\cos(2t)$ 和 $y=3\sin(2t)$，其中 $0{\leq}t{\leq}\pi$
$x=3\sin(t)$ 和 $y=3\cos(t)$，其中 $0{\leq}t{\leq}2\pi$ （从 $(0,3)$ 出发顺时针）
椭圆（ellipse）的参数化：如 $x^2+4y^2=9$
令 $Y=2y$，则 $x^2+Y^2=9$
根据前面的参数化，$Y=2y=3\sin(t)$，所以
$x=3\cos(t)$ 和 $y={3\over2}\sin(t)$，其中 $0{\leq}t{\leq}2\pi$
$x^6+y^6=64$ 的曲线像一个膨胀的圆（bloated circle），“半径”（radius）为 $64^{1/6}=2$ 单位（2 units）的圆
$x=2\cos^{1/3}(t)$ 和 $y=2\sin^{1/3}(t)$，其中 $0{\leq}t{\leq}2\pi$
27.1.1 参数方程的微分
$$ {\mathrm{d}y\over\mathrm{d}t}={\mathrm{d}y\over\mathrm{d}x}{\mathrm{d}x\over\mathrm{d}t} \implies {\mathrm{d}y\over\mathrm{d}x}={\mathrm{d}y/\mathrm{d}t\over\mathrm{d}x/\mathrm{d}t}
\frac{y’(t)}{x’(t)} $$
例，找出 $x=4t^2-4$ 和 $y=2t-2t^3$，$t$ 为所有实数，在原点的任意切线方程
解得 $t=\pm1$ 曲线经过原点
${\mathrm{d}y\over\mathrm{d}x}={1\over4t}-{3t\over4}$，$t=1$ 时，斜率为 $-1/2$；$t=-1$ 时，斜率为 $1/2$ $$ {\mathrm{d^2}y\over\mathrm{d}x^2}={\mathrm{d}y’\over\mathrm{d}x}={\mathrm{d}y’/\mathrm{d}t\over\mathrm{d}x/\mathrm{d}t} $$ ${\mathrm{d^2}y\over\mathrm{d}x^2}=-{1\over32t^3}-{3\over32t}$，$t=1$ 时，${\mathrm{d^2}y\over\mathrm{d}x^2}=-{1\over8}$，表示曲线下凹（concave down）
27.2 极坐标
27.2.1 极坐标的转换
极坐标 → 笛卡尔坐标
$x=r\cos(\theta)$ 和 $y=r\sin(\theta)$
笛卡尔坐标 → 极坐标
$r^2=x^2+y^2$ 且，如果 $x\neq0$，$\arctan(\frac{y}{x})=\theta$，根据原点所在的象限判断 $\theta$ 的大小
例如，笛卡尔坐标中的点 $(-1,-1)$，在极坐标中为 $(\sqrt{2},{5\pi\over4}+2n\pi)$，$(-\sqrt{2},{\pi\over4}+2n\pi)$ ，$n\in\Bbb{Z}$
当考虑的角度 $\theta$ 是 $\pi/2$ 的整数倍（$x=0$ 或 $y=0$），可以直接从图像上入手
27.2.2 绘制极坐标系上的曲线
Suppose you know that $r = f(\theta)$ for some function f, and you want to sketch the graph of all points $(r, \theta)$ in polar coordinates where $r = f(\theta)$ for $\theta$ in some given range.
例如，在极坐标系上绘制 $r=3\sin(\theta)$ 在 $0\leq\theta\leq\pi$ 的曲线
在笛卡尔坐标系上，以 $\theta$ 为横轴，$r$ 为纵轴绘制区间内的曲线：在 $[0,\pi]$ 上，距离 $r$ 从 $0$ 增加到 $3$，然后回到 $0$
由 $y=r\sin(\theta)$ 和 $r=3\sin(\theta)$，可得 $r^2=3y$
$$ x^2+y^2=3y \implies x^2+y^2-3y+({3\over2})^2=({3\over2})^2 \implies x^2+(y-{3\over2})^2=({3\over2})^2 $$
曲线是以 $(0,3/2)$ 为圆心，半径为 $3/2$ 的圆
极坐标曲线
- The curve given by $r = 1 + \cos(\theta)$ is called a cardioid. The curve $r = 1 + {3\over4} \cos(\theta)$ is an example of a limacon, of which the cardioid is a special case.
- In the above graph of $r = \sin(3\theta)$, the angle $\theta$ only goes from $0$ to $\pi$. As $\theta$ goes from $\pi$ to $2\pi$, the graph is retraced, just as in the case of the circle $r = \sin(\theta)$.
- The curve given by $r = \theta/\pi$ is an example of a spiral of Archimedes. This is not periodic: as $\theta$ increases, the spiral gets bigger and bigger.
- The curve given by $r = 2=(1 + \sin(\theta))$ looks like a parabola. In fact, you should try to show that the above equation becomes $x^2 = 4 - 4y$ in Cartesian coordinates.
27.2.3 极坐标曲线的切线
Luckily, finding tangents to polar curves is just a special case of finding tangents to curves given by parametric equations.
因为 $r=f(\theta)$，参数方程可以写作 $x=f(\theta)\cos(\theta)$ 和 $y=f(\theta)\sin(\theta)$，故所以切线斜率为： $$ {\mathrm{d}y\over\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta} $$
27.2.4 极坐标曲线围成的面积
扇形的面积：半径平方的二分之一乘上角的弧度
在 $r=f(\theta)$ 之内，介于 $\theta=\theta_0$ 和 $\theta=\theta_1$ 之间的面积为 $$ \int_{\theta_0}^{\theta_1}{1\over2}r^2\mathrm{d}\theta $$
例 $r=1+2\cos(\theta)$ 在区间 $[0,2\pi]$ 内围成的面积
在 27.2.2 中，绘制了曲线 $r=1+2\cos(\theta)$ 的图像，当 $r=2\pi/3$ 和 $r=4\pi/3$ 时，曲线经过原点，且在该区间内 $r<0$ （图像折回）
当使用 $\int_{\theta_0}^{\theta_1}{1\over2}r^2\mathrm{d}\theta$ 计算时，没有考虑 $r<0$ 的情况，实际计算了曲线 $r=|1+2\cos(\theta)|$ 围成的面积
需要将 $r<0$ 的区间内曲线围成的面积的二倍，从积分的计算结果中减去，即 $$ S=\int_{0}^{2\pi}{1\over2}(1+2\cos(\theta))^2\mathrm{d}\theta-2\int_{4\pi/3}^{2\pi/3}{1\over2}(1+2\cos(\theta))^2\mathrm{d}\theta $$
§复数
- 基本运算（basic manipulations）及解二次方程（solving quadratic equations）
- 复平面（complex plane）及复数的笛卡尔和极坐标形式
- 解形如 $z^n=w$ 的方程
- 解形如 $e^x=w$ 的方程
- 利用幂级数和复数的一些技巧求解一些级数问题
28.1 基础
$-1$ 的平方根（square roots）：$i^2=-1$ $$ (-i)^2=(-1)^2(i^2)=1(-1)=-1 $$
…, when we say that a number is imaginary, we mean that its square is a negative number. The only imaginary numbers are of the form $yi$ where $y$ is a real number not equal to $0$. You can also write $iy$ instead of $yi$.
…
Notice that all imaginary numbers are complex numbers; for example, $2i = 0 + 2i$. All real numbers are also complex numbers; for example, $-13 = -13 + 0i$. Every complex number has a real and an imaginary part. If $z = x + iy$, then the real part is $x$ and the imaginary part is $y$. These are written as $Re(z)$ and $Im(z)$, respectively.
注意：虚部不包括 $i$，即 $Im(2-3i)=-3$
Just add (or subtract) the real parts, and then do the imaginary parts.
Multiplication isn’t much harder|you just expand, but remember to change $i^2$ into $-1$ whenever you see it. In fact, because $i^4 = 1$, we can see that the powers of $i$ keep on cycling through $1, i, -1, -i$. For example, $i^{101} = i$ since $i^{100} = 1$ (remembering that $100$ is divisible by $4$).
- $i^3=i^2{\times}i=-1{\times}i=-i$
- $i^4=i^3{\times}i=-i{\times}i=1$
- $i^5=i^4{\times}i=1{\times}i=i$
- ……
复数 $x+iy$ 和 $x-iy$ 相乘可以得到一个实数： $$ (x+iy)(x-iy)=x^2-(iy)^2=x^2-i^2y^2=x^2+y^2 $$
If $z = x + iy$, the related number $x - iy$ is so important that it has a name: it is called the complex conjugate of $x + iy$ and denoted $\bar{z}$.
Note that the complex conjugate of a real number is the same number.
…given a complex number $z = x + iy$, let’s define the modulus of $z$ to be $\sqrt{x^2 + y^2}$. We write the modulus of $z$ as $|z|$.
$$ |x+iy|=\sqrt{x^2+y^2} $$
28.1.1 复指数函数
$e^x=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$ 中当实数 $x$ 替换为复数 $z$ 时，可以用比式判别法（ratio test）证明级数收敛
$$ e^x=\sum_{n=0}^{\infty}\frac{z^n}{n!} $$
在实数的指数函数中 $e^xe^y=e^{x+y}$，用麦克劳林级数表示为 $$ \sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{\infty}\frac{y^m}{m!}=\sum_{k=0}^{\infty}\frac{(x+y)^k}{k!} $$ 等式左右 $x^ny^m$ 的系数相等，当实数 $x$ 和 $y$ 替换为复数 $z$ 和 $w$ 时，同样成立
28.2 复平面
在复平面上用极坐标表示复数：$z=x+iy=r\cos(\theta)+ir\sin(\theta)$
欧拉公式（Euler’s identity）： $$ e^{i\theta}=\cos(\theta)+i\sin(\theta) $$
This means that the complex number $e^{i\theta}$, as defined in the previous section, has polar coordinates $(1, \theta)$ when you plot it on the complex number plane.
等式右边为 $r=1$ 时，在复平面上极坐标和复数的转换
所以 $e^{i\theta}$ 在以 $x$ 正半轴为 $\theta$ 起点的单位圆上
Let’s say that a complex number like $re^{i\theta}$ is in polar form, (as opposed to $x + iy$, which is in Cartesian form).
如果 $(x,y)$ 和 $(r,\theta)$ 表示复平面上的同一点，那么 $x+iy=re^{i\theta}$
Let’s agree that when we’re dealing with complex numbers, we’ll never let $r$ be negative.
在复平面上用极坐标表示复数时，必须使 $r\geq0$（一点在极坐标系上有无穷多种表示方式）
So $e^{i\theta}$ is periodic in $\theta$ with period $2\pi$.
例如，点 $(1,3\pi/2)$ 又可以表示为 $(1,-\pi/2)$，对于复数，意味着 $e^{i(3\pi/2)}=e^{i(-\pi/2)}$
28.2.1 极坐标形式的转换
极坐标 → 笛卡尔坐标
欧拉公式：$e^{i\theta}=\cos(\theta)+i\sin(\theta)$
笛卡尔坐标 → 极坐标
参考27.2.1
$r=\sqrt{x^2+y^2}=|z|$ 且 $\tan(\theta)=\frac{y}{x}$ （注意：$\tan(\pi/2)$ 没有意义）
因为 $r\geq0$，所以不考虑 $r=-\sqrt{x^2+y^2}$ 的情况
The modulus $|z|$ is therefore the distance from the point $z$ to the origin (in the complex number plane). The angle $\theta$ is called the argument of $z$ and is written $arg(z)$. (Normally one requires that $0 \leq arg(z) \leq 2\pi$ so that there’s no ambiguity.)
In fact, sometimes the polar form of $z$ is referred to as mod-arg form.
28.3 复数的高次幂
…it’s really easy to multiply and take powers in polar form.
极坐标形式让乘法的幂的计算更加简单！
28.4 求解 $z^n=w$
在关于复数 $z$ 方程 $z^n=w$ 中，$n$ 为整数，$w$ 为复数
令 $z=re^{i\theta}$，$w=Re^{i\phi}$
例如，$z^5=-\sqrt{3}+i$
$R=\sqrt{(-\sqrt{3})^2+(1)^2}=2$，$\tan(\phi)=-1/\sqrt{3}\implies\phi=5\pi/6$ （位于第二象限）
$r^5e^{i(5\theta)}=2e^{i(5\pi/6)}$，其中 $r$ 为非负实数，$r=2^{1/5}$，$\theta={\pi\over6}+{2k\pi\over5}$
如果对实数 $A$ 和实数 $B$，等式 $e^{iA}=e^{iB}$ 成立， 那么 $A=B+2k\pi,k\in\Bbb{Z}$
…since $n = 5$, you only need to use the first five values for $k$, namely, $k = 0, 1, 2, 3, 4$.
…unless $w = 0$, the equation $z^n = w$ has n different solutions, which occur when $k = 0. 1,\dots , n - 1$.
方程的解均匀分布（evenly spaced around the circle）在半径为 $2^{1/5}$ 单位（modulus）的圆上，连续解的幅角（arguments）之差为 $2\pi/5$（即圆周的五分之一），形成规则的五边形（pentagon）
In general, there are $n$ solutions to the equation $z^n = w$, which when plotted form the vertices of a regular $n$-sided polygon. (The exception is if $w = 0$, in which case $z = 0$ is the only solution, but it is of multiplicity $n$.)
- Write $z = re^{i\theta}$ in polar coordinates. Then $z^n = r^ne^{in\theta}$.
- Convert $w$ to polar coordinates. Let’s say that $w = Re^{i\phi}$.
- Since $z^n = w$, we can write the original equation as $r^ne^{in\theta} = Re^{i\phi}$. Here, the values of $n$, $R$, and $\phi$ should be filled in with your values, but $r$ and $\theta$ are always what we need to find (so they appear as variables).
- Decompose into two equations: $r^n = R$ and $e^{in\theta} = e^{i\phi}$.
- The first is simple to solve: take $n$th roots to get $r = R^{1/n}$.
- For the second, use the above triple-boxed principle to get $n\theta = \phi+2\pi k$, where $k$ is an integer.
- Divide this by $n$, then write out all the different values for $\theta$ when $k = 0, 1, 2,\dots, n - 1$.
- Substitute the value of $r$ and the different values of $\theta$ into $z= re^{i\theta}$ to get $n$ different values for $z$, which are the solutions.
- If necessary, change each and every one of those solutions into Cartesian coordinates.
28.4.1 一些变式
例 1
求解 $(z-2)^3=i$ 时，令 $Z=z-2$，计算 $Z^3=i$
例 2
求解 $z^2+\frac{1}{\sqrt{2}}z-\frac{\sqrt{3}}{8}i=0$
根据二次方程的求根公式 $z=\frac{-\frac{1}{\sqrt{2}}\pm\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{2}i}}{2}$
令 $Z^2=\frac{1}{2}+\frac{\sqrt{3}}{2}i$，解得 $Z=\pm(\frac{\sqrt{3}}{2}+\frac{1}{2}i)$
所以 $z=\frac{\frac{-1}{\sqrt{2}}\pm(\frac{\sqrt{3}}{2}+\frac{1}{2}i)}{2}$ 化简即可
例 3
将 $z^4-z^2+1$ 因式分解为复数（factor $z^4-z^2+1$ over the complex numbers）
令 $Z=z^2$，根据二次方程的求根公式 $Z=\frac{1\pm\sqrt{-3}}{2}={1\over2}\pm{\sqrt{3}\over2}i$
计算 $z^2={1\over2}\pm{\sqrt{3}\over2}i$
$z^2={1\over2}+{\sqrt{3}\over2}i$，$\theta={\pi\over6},{7\pi\over6}$
$z^2={1\over2}-{\sqrt{3}\over2}i$，$\theta={5\pi\over6},{11\pi\over6}$
求得 $z$ 的 $4$ 个解
$$ z^4-z^2+1=(z-\frac{\sqrt{3}+i}{2})(z-\frac{\sqrt{3}-i}{2})(z-\frac{-\sqrt{3}+i}{2})(z-\frac{-\sqrt{3}-i}{2}) $$
To get the real factorization, we need to use a nice fact: if $w$ is any complex number, then $(z - w)(z - \bar w)$ has real coefficients when you multiply it out. Indeed, you get $z^2 - (w + \bar w)z + w \bar w$, but it’s easy enough to see that $w + \bar w = 2Re(w)$ (which is real), and we’ve already seen that $w \bar w = |w|^2$, which is also real.
$z^4-z^2+1=(z^2-\sqrt{3}z+1)(z^2+\sqrt{3}z+1)$
28.5 求解 $e^z=w$
$$ e^z=e^{x+iy}=e^xe^{iy} $$
The modulus is $e^x$ and the argument is $y$.
例 1
求解 $e^z=-\sqrt{3}+i$
解：
$e^xe^y=2e^{i(5\pi/6)}$
$x=2,y={5\pi\over6}+2\pi k$
$z=\ln(2)+i({5\pi\over6}+2\pi k)$
So the solutions are equally spaced on the vertical line $x = \ln(2)$. Incidentally, this means that they form an arithmetic progression of complex numbers.
例 2
求解 $e^{2iz+3}=i$
解：
$2iz+3=2i(x+yi)+3=(-2y+3)+i(2x)$
$e^{-2y+3}e^{i(2x)}=1e^{i\pi/2}$
$-2y+3=\ln(1)\implies y={3\over2}$
$2x={\pi\over2}+2\pi k\implies x={\pi\over4}+\pi k$
$z={\pi\over4}+\pi k+{3\over2}i$
28.6 一些三角级数
三角级数（trigonometric series） $$ \sum_{n=0}^{\infty}(a_n\cos(n\theta)+b_n\sin(n\theta)) $$ 三角级数转换成幂级数
$\sum\limits_{n=0}^{\infty}\frac{\sin(n\theta)}{n!}$，$\sum\limits_{n=0}^{\infty}\frac{\cos(n\theta)}{n!}$ $$ \sum_{n=0}^{\infty}\frac{\cos(n\theta)}{n!}+i\sum_{n=0}^{\infty}\frac{\sin(n\theta)}{n!} $$ 由欧拉等式得 $$ \sum_{n=0}^{\infty}\frac{\cos(n\theta)+i\sin(n\theta)}{n!}=\sum_{n=0}^{\infty}\frac{e^{in\theta}}{n!}=\sum_{n=0}^{\infty}\frac{(e^{i\theta})^n}{n!} $$ 对任意复数 $z$ 有 $\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}=e^z$ $$ \sum_{n=0}^{\infty}\frac{(e^{i\theta})^n}{n!}=e^{e^{i\theta}} $$ $e^{e^{i\theta}}=e^{\cos(\theta)+i\sin(\theta)}=e^{\cos(\theta)}e^{i\sin(\theta)}$
由欧拉等式，$e^{i\sin(\theta)}=\cos(\sin(\theta))+i\sin(\sin(\theta))$
$$ \begin{aligned} \sum\limits_{n=0}^{\infty}\frac{\cos(n\theta)}{n!}+i\sum_{n=0}^{\infty}\frac{\sin(n\theta)}{n!}&=e^{\cos(\theta)}[\cos(\sin(\theta))+i\sin(\sin(\theta))] \&=e^{\cos(\theta)}\cos(\sin(\theta))+ie^{\cos(\theta)}\sin(\sin(\theta)) \end{aligned} $$
Now, if two complex numbers are equal, then their real parts must be equal, and also their imaginary parts must be equal.
相等的两个复数，它们的实部相等，虚部也相等
$\sum\limits_{n=0}^{\infty}\frac{\cos(n\theta)}{n!}=e^{\cos(\theta)}\cos(\sin(\theta))$，$\sum\limits_{n=0}^{\infty}\frac{\sin(n\theta)}{n!}=e^{\cos(\theta)}\sin(\sin(\theta))$ 对所有实数 $\theta$ 成立
28.7 欧拉公式和幂级数
$$ \begin{aligned} e^{i\theta}&=1+(i\theta)+\frac{(i\theta)^2}{2!}+{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}+\frac{(i\theta)^6}{6!}\cdots \&=1+i\theta-\frac{\theta^2}{2!}-\frac{\theta^3}{3!}i+\frac{\theta^4}{4!}+\frac{\theta^5}{5!}i-\frac{\theta^6}{6!}-\frac{\theta^7}{7!}i+\cdots \end{aligned} $$
实部（real part）： $$ 1+i\theta-{\theta^2\over2!}+{\theta^4\over4!}-{\theta^6\over6!}+\dots=\cos(\theta) $$ 虚部（imaginary part）： $$ \theta-{\theta^3\over3!}+{\theta^5\over5!}-{\theta^7\over7!}+\dots=\sin(\theta) $$