arctan 反正切 即知道一个角的正切值而求这个角的度数
换句话说 知道一个点在直角坐标系中的坐标，将其转化为极坐标系的坐标
再换句话说 已知P(x,y)求ρ和θ

从二分查找法说起
先从一个例子说起，知道平面上一点在直角坐标系下的坐标（X,Y）=（100，200），如何求的在极坐标系下的坐标（ρ,θ）。用计算器计算一下可知答案是（223.61，63.435）。
为了突出重点，这里我们只讨论X和Y都为正数的情况。这时θ=atan(y/x)。求θ的过程也就是求atan 函数的过程。Cordic算法采用的想法很直接，将（X，Y）旋转一定的度数，如果旋转完纵坐标变为了0，那么旋转的度数就是θ。坐标旋转的公式可能大家都忘了，这里把公式列出了。设（x,y）是原始坐标点，将其以原点为中心，顺时针旋转θ之后的坐标记为（x’,y’）,则有如下公式：

也可以写为矩阵形式：

如何旋转呢，可以借鉴二分查找法的思想。我们知道θ的范围是0到90度。那么就先旋转45度试试。

旋转之后纵坐标为70.71，还是大于0，说明旋转的度数不够，接着再旋转22.5度（45度的一半）。

这时总共旋转了45+22.5=67.5度。结果纵坐标变为了负数，说明θ<67.5度，这时就要往回转，还是二分查找法的思想，这次转11.25度。

这时总共旋转了45+22.5-11.25=56.25度。又转过头了，接着旋转，这次顺时针转5.625度。

这时总共旋转了45+22.5-11.25+5.625=61.875度。这时纵坐标已经很接近0了。我们只是说明算法的思想，因此就不接着往下计算了。计算到这里我们给的答案是 61.875±5.625。二分查找法本质上查找的是一个区间，因此我们给出的是θ值的一个范围。同时，坐标到原点的距离ρ也求出来了，ρ=223.52。与标准答案比较一下计算的结果还是可以的。旋转的过程图示如下。

可能有读者会问，计算中用到了 sin 函数和 cos 函数，这些值又是怎么计算呢。很简单，我们只用到很少的几个特殊点的sin 函数和 cos 函数的值，提前计算好存起来，用时查表。

下面给出上面方法的C 语言实现。
#include <stdio.h>
#include <stdlib.h>
double my_atan2(double x, double y);
int main(void)
{
double z = my_atan2(100.0, 200.0);
printf("\n z = %f \n", z);
return 0;
}
double my_atan2(double x, double y)
{
const double sine[] = {0.7071067811865,0.3826834323651,0.1950903220161,0.09801714032956,
0.04906767432742,0.02454122852291,0.01227153828572,0.006135884649154,0.003067956762966
,0.001533980186285,7.669903187427045e-4,3.834951875713956e-4,1.917475973107033e-4,
9.587379909597735e-5,4.793689960306688e-5,2.396844980841822e-5
};
const double cosine[] = {0.7071067811865,0.9238795325113,0.9807852804032,0.9951847266722,
0.9987954562052,0.9996988186962,0.9999247018391,0.9999811752826,0.9999952938096,
0.9999988234517,0.9999997058629,0.9999999264657,0.9999999816164,0.9999999954041,
0.999999998851,0.9999999997128
};
int i = 0;
double x_new, y_new;
double angleSum = 0.0;
double angle = 45.0;
for(i = 0; i < 15; i++)
{
if(y > 0)
{
x_new = x * cosine[i] + y * sine[i];
y_new = y * cosine[i] - x * sine[i];
x = x_new;
y = y_new;
angleSum += angle;
}
else
{
x_new = x * cosine[i] - y * sine[i];
y_new = y * cosine[i] + x * sine[i];
x = x_new;
y = y_new;
angleSum -= angle;
}
printf("Debug: i = %d angleSum = %f, angle = %f\n", i, angleSum, angle);
angle /= 2;
}
return angleSum;
}
程序运行的输出结果如下：
Debug: i = 0 angleSum = 45.000000, angle = 45.000000
Debug: i = 1 angleSum = 67.500000, angle = 22.500000
Debug: i = 2 angleSum = 56.250000, angle = 11.250000
Debug: i = 3 angleSum = 61.875000, angle = 5.625000
Debug: i = 4 angleSum = 64.687500, angle = 2.812500
Debug: i = 5 angleSum = 63.281250, angle = 1.406250
Debug: i = 6 angleSum = 63.984375, angle = 0.703125
Debug: i = 7 angleSum = 63.632813, angle = 0.351563
Debug: i = 8 angleSum = 63.457031, angle = 0.175781
Debug: i = 9 angleSum = 63.369141, angle = 0.087891
Debug: i = 10 angleSum = 63.413086, angle = 0.043945
Debug: i = 11 angleSum = 63.435059, angle = 0.021973
Debug: i = 12 angleSum = 63.424072, angle = 0.010986
Debug: i = 13 angleSum = 63.429565, angle = 0.005493
Debug: i = 14 angleSum = 63.432312, angle = 0.002747
z = 63.432312

减少乘法运算
现在已经有点 Cordic 算法的样子了，但是我们看到没次循环都要计算 4 次浮点数的乘法运算，运算量还是太大了。还需要进一步的改进。改进的切入点当然还是坐标变换的过程。我们将坐标变换公式变一下形。

可以看出 cos(θ)可以从矩阵运算中提出来。我们可以做的再彻底些，直接把 cos(θ) 给省略掉。

省略cos(θ)后发生了什么呢，每次旋转后的新坐标点到原点的距离都变长了，放缩的系数是1/cos(θ)。不过没有关系，我们求的是θ，不关心ρ的改变。这样的变形非常的简单，但是每次循环的运算量一下就从4次乘法降到了2次乘法了。

还是给出 C 语言的实现：
double my_atan3(double x, double y)
{
const double tangent[] = {1.0,0.4142135623731,0.1989123673797,0.09849140335716,0.04912684976947,
0.02454862210893,0.01227246237957,0.006136000157623,0.003067971201423,
0.001533981991089,7.669905443430926e-4,3.83495215771441e-4,1.917476008357089e-4,
9.587379953660303e-5,4.79368996581451e-5,2.3968449815303e-5
};
int i = 0;
double x_new, y_new;
double angleSum = 0.0;
double angle = 45.0;
for(i = 0; i < 15; i++)
{
if(y > 0)
{
x_new = x + y * tangent[i];
y_new = y - x * tangent[i];
x = x_new;
y = y_new;
angleSum += angle;
}
else
{
x_new = x - y * tangent[i];
y_new = y + x * tangent[i];
x = x_new;
y = y_new;
angleSum -= angle;
}
printf("Debug: i = %d angleSum = %f, angle = %f, ρ = %f\n", i, angleSum, angle, hypot(x,y));
angle /= 2;
}
return angleSum;
}
计算的结果是：
Debug: i = 0 angleSum = 45.000000, angle = 45.000000, ρ = 316.227766
Debug: i = 1 angleSum = 67.500000, angle = 22.500000, ρ = 342.282467
Debug: i = 2 angleSum = 56.250000, angle = 11.250000, ρ = 348.988177
Debug: i = 3 angleSum = 61.875000, angle = 5.625000, ρ = 350.676782
Debug: i = 4 angleSum = 64.687500, angle = 2.812500, ρ = 351.099697
Debug: i = 5 angleSum = 63.281250, angle = 1.406250, ρ = 351.205473
Debug: i = 6 angleSum = 63.984375, angle = 0.703125, ρ = 351.231921
Debug: i = 7 angleSum = 63.632813, angle = 0.351563, ρ = 351.238533
Debug: i = 8 angleSum = 63.457031, angle = 0.175781, ρ = 351.240186
Debug: i = 9 angleSum = 63.369141, angle = 0.087891, ρ = 351.240599
Debug: i = 10 angleSum = 63.413086, angle = 0.043945, ρ = 351.240702
Debug: i = 11 angleSum = 63.435059, angle = 0.021973, ρ = 351.240728
Debug: i = 12 angleSum = 63.424072, angle = 0.010986, ρ = 351.240734
Debug: i = 13 angleSum = 63.429565, angle = 0.005493, ρ = 351.240736
Debug: i = 14 angleSum = 63.432312, angle = 0.002747, ρ = 351.240736
z = 63.432312

消除乘法运算
我们已经成功的将乘法的次数减少了一半，还有没有可能进一步降低运算量呢？还要从计算式入手。

第一次循环时，tan(45)=1，所以第一次循环实际上是不需要乘法运算的。第二次运算呢？
Tan(22.5)=0.4142135623731,很不幸，第二次循环乘数是个很不整的小数。是否能对其改造一下呢？答案是肯定的。第二次选择22.5度是因为二分查找法的查找效率最高。如果选用个在22.5到45度之间的值，查找的效率会降低一些。如果稍微降低一点查找的效率能让我们有效的减少乘法的次数，使最终的计算速度提高了，那么这种改进就是值得的。
我们发现tan(26.565051177078)=0.5，如果我们第二次旋转采用26.565051177078度，那么乘数变为0.5，如果我们采用定点数运算的话（没有浮点协处理器时为了加速计算我们会大量的采用定点数算法）乘以0.5就相当于将乘数右移一位。右移运算是很快的，这样第二次循环中的乘法运算也被消除了。
类似的方法，第三次循环中不用11.25度，而采用 14.0362434679265 度。
Tan(14.0362434679265)= 1/4
乘数右移两位就可以了。剩下的都以此类推。
tan(45)= 1
tan(26.565051177078)= 1/2
tan(14.0362434679265)= 1/4
tan(7.1250163489018)= 1/8
tan(3.57633437499735)= 1/16
tan(1.78991060824607)= 1/32
tan(0.8951737102111)= 1/64
tan(0.4476141708606)= 1/128
tan(0.2238105003685)= 1/256

还是给出C语言的实现代码，我们采用循序渐进的方法，先给出浮点数的实现（因为用到了浮点数，所以并没有减少乘法运算量，查找的效率也比二分查找法要低，理论上说这个算法实现很低效。不过这个代码的目的在于给出算法实现的示意性说明，还是有意义的）。
double my_atan4(double x, double y)
{
const double tangent[] = {1.0, 1 / 2.0, 1 / 4.0, 1 / 8.0, 1 / 16.0,
1 / 32.0, 1 / 64.0, 1 / 128.0, 1 / 256.0, 1 / 512.0,
1 / 1024.0, 1 / 2048.0, 1 / 4096.0, 1 / 8192.0, 1 / 16384.0
};
const double angle[] = {45.0, 26.565051177078, 14.0362434679265, 7.1250163489018, 3.57633437499735,
1.78991060824607, 0.8951737102111, 0.4476141708606, 0.2238105003685, 0.1119056770662,
0.0559528918938, 0.027976452617, 0.01398822714227, 0.006994113675353, 0.003497056850704
};
int i = 0;
double x_new, y_new;
double angleSum = 0.0;
for(i = 0; i < 15; i++)
{
if(y > 0)
{
x_new = x + y * tangent[i];
y_new = y - x * tangent[i];
x = x_new;
y = y_new;
angleSum += angle[i];
}
else
{
x_new = x - y * tangent[i];
y_new = y + x * tangent[i];
x = x_new;
y = y_new;
angleSum -= angle[i];
}
printf("Debug: i = %d angleSum = %f, angle = %f, ρ = %f\n", i, angleSum, angle[i], hypot(x, y));
}
return angleSum;
}
程序运行的输出结果如下：
Debug: i = 0 angleSum = 45.000000, angle = 45.000000, ρ = 316.227766
Debug: i = 1 angleSum = 71.565051, angle = 26.565051, ρ = 353.553391
Debug: i = 2 angleSum = 57.528808, angle = 14.036243, ρ = 364.434493
Debug: i = 3 angleSum = 64.653824, angle = 7.125016, ρ = 367.270602
Debug: i = 4 angleSum = 61.077490, angle = 3.576334, ρ = 367.987229
Debug: i = 5 angleSum = 62.867400, angle = 1.789911, ρ = 368.166866
Debug: i = 6 angleSum = 63.762574, angle = 0.895174, ρ = 368.211805
Debug: i = 7 angleSum = 63.314960, angle = 0.447614, ρ = 368.223042
Debug: i = 8 angleSum = 63.538770, angle = 0.223811, ρ = 368.225852
Debug: i = 9 angleSum = 63.426865, angle = 0.111906, ρ = 368.226554
Debug: i = 10 angleSum = 63.482818, angle = 0.055953, ρ = 368.226729
Debug: i = 11 angleSum = 63.454841, angle = 0.027976, ρ = 368.226773
Debug: i = 12 angleSum = 63.440853, angle = 0.013988, ρ = 368.226784
Debug: i = 13 angleSum = 63.433859, angle = 0.006994, ρ = 368.226787
Debug: i = 14 angleSum = 63.437356, angle = 0.003497, ρ = 368.226788
z = 63.437356

有了上面的准备，我们可以来讨论定点数算法了。所谓定点数运算，其实就是整数运算。我们用256 表示1度。这样的话我们就可以精确到1/256=0.00390625 度了，这对于大多数的情况都是足够精确的了。256 表示1度，那么45度就是 45*256 = 115200。其他的度数以此类推。
序号 度数 ×256 取整
1 45.0 11520 11520
2 26.565051177078 6800.65310133196 6801
3 14.0362434679265 3593.27832778918 3593
4 7.1250163489018 1824.00418531886 1824
5 3.57633437499735 915.541599999322 916
6 1.78991060824607 458.217115710994 458
7 0.8951737102111 229.164469814035 229
8 0.4476141708606 114.589227740302 115
9 0.2238105003685 57.2954880943458 57
10 0.1119056770662 28.647853328949 29
11 0.0559528918938 14.3239403248137 14
12 0.027976452617 7.16197186995294 7
13 0.01398822714227 3.58098614841984 4
14 0.006994113675353 1.79049310089035 2
15 0.003497056850704 0.8952465537802 1

C 代码如下：
int my_atan5(int x, int y)
{
const int angle[] = {11520, 6801, 3593, 1824, 916, 458, 229, 115, 57, 29, 14, 7, 4, 2, 1};
int i = 0;
int x_new, y_new;
int angleSum = 0;
x *= 1024;// 将 X Y 放大一些，结果会更准确
y *= 1024;
for(i = 0; i < 15; i++)
{
if(y > 0)
{
x_new = x + (y >> i);
y_new = y - (x >> i);
x = x_new;
y = y_new;
angleSum += angle[i];
}
else
{
x_new = x - (y >> i);
y_new = y + (x >> i);
x = x_new;
y = y_new;
angleSum -= angle[i];
}
printf("Debug: i = %d angleSum = %d, angle = %d\n", i, angleSum, angle[i]);
}
return angleSum;
}
计算结果如下：
Debug: i = 0 angleSum = 11520, angle = 11520
Debug: i = 1 angleSum = 18321, angle = 6801
Debug: i = 2 angleSum = 14728, angle = 3593
Debug: i = 3 angleSum = 16552, angle = 1824
Debug: i = 4 angleSum = 15636, angle = 916
Debug: i = 5 angleSum = 16094, angle = 458
Debug: i = 6 angleSum = 16323, angle = 229
Debug: i = 7 angleSum = 16208, angle = 115
Debug: i = 8 angleSum = 16265, angle = 57
Debug: i = 9 angleSum = 16236, angle = 29
Debug: i = 10 angleSum = 16250, angle = 14
Debug: i = 11 angleSum = 16243, angle = 7
Debug: i = 12 angleSum = 16239, angle = 4
Debug: i = 13 angleSum = 16237, angle = 2
Debug: i = 14 angleSum = 16238, angle = 1
z = 16238
16238/256=63.4296875度，精确的结果是63.4349499度，两个结果的差为0.00526，还是很精确的。
到这里 CORDIC 算法的最核心的思想就介绍完了。当然，这里介绍的只是CORDIC算法最基本的内容，实际上，利用CORDIC 算法不光可以计算 atan 函数，其他的像 Sin，Cos，Sinh，Cosh 等一系列的函数都可以计算，不过那些都不在本文的讨论范围内了。另外，每次旋转时到原点的距离都会发生变化，而这个变化是确定的，因此可以在循环运算结束后以此补偿回来，这样的话我们就同时将（ρ,θ）都计算出来了。

＠_＠

请问一下，用cordic计算e的指数运算，是不是有一个范围