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# Implementing Cordic Algorithms

OCT90: IMPLEMENTING CORDIC ALGORITHMS

# IMPLEMENTING CORDIC ALGORITHMS

## A single compact routine for computing transcendental functions

### Pitts Jarvis

Efficiently computing sines, cosines, and other transcendental functions is a process about which many programmers are blissfully ignorant. When these values are called for in a graphics or CAD program, we usually rely on a call to the compiler's run-time library. The library either derives the necessary values in some mysterious manner or calls the floating-point coprocessor to assist in the task.

The CORDIC (COordinate, Rotation DIgital Computer) family of algorithms is an elegant, efficient, and compact way to compute sines, cosines, exponentials, logarithms, and associated transcendental functions using one core routine. These truly remarkable algorithms compute these functions with n bits of accuracy in n iterations -- where each iteration requires only a small number of shifts and additions. Furthermore, these routines use only fixed-point arithmetic. Using these algorithms, you can cast your entire graphics application into fixed-point, and thus avoid the cost of run-time conversion from fixed- to floating-point representation and back.

Even if you don't plan on recasting your application into fixed-point, you just might be curious how your floating-point coprocessor works. The Intel numerics family (8087, 80287, and 80387) all use Cordic algorithms, in a form slightly different than described here, to compute circular functions. The Intel implementations are described by R. Nave1 and A. K. Yuen2.

The implementations may be contemporary, but the algorithms are not new. J. E. Volder3 coined the name in 1959. He applied these algorithms to build a special-purpose digital computer for real-time airborne navigation. D. S. Cochran4 identifies their use in the HP-35 calculator in 1972 to calculate the transcendental functions.

### Mathematical Manipulation

If we have a vector [x, y], we can rotate it through an angle a by multiplying it by the matrix Ra, defined in Example 1(a). Explicitly doing the matrix multiplication yields the equation in Example 1(b).

If we choose x = 1 and y = 0 and multiply that vector by Ra we are left with the vector [cos a, sin a].

Multiplying by two successive rotation matrices; Ra and Rb rotates the vector through the angle a + b, or more formally RaRb - Ra+b. If we choose to represent the angle a as a sum of angles ai for i = 0 through n (see Example 1(c) , then we can rotate the vector through the angle a by multiplying a series of rotation matrices Rao, Ra1, . . . Ran.

By picking the ai carefully, we can simplify the arithmetic. Notice that we can rewrite the rotation matrix by factoring out cos a as shown in Example 1(d). If we pick ai such that tan ai = 2-i for i = 0 through n, all of the multiplications by tan ai become right shifts by i bits.

Now we need to specify an algorithm so that we can represent a as the sum of the ai. Initialize a variable, z, to a. This z will be a residue quantity, which we are trying to drive to zero by adding or subtracting ai at the i-th step. At the first step, i= 0. At the i-th step, if z 0 then subtract ai from z. Otherwise add ai to z. At the last step i = n, and z is the error in our representation of a. Notice that in Example 1(e), for large i, each additional step yields one more bit of accuracy in our representation of a.

Figure 1 shows the relative magnitudes of the incremental angles, ai. Figure 2 gives an example of the convergence process with an initial angle of 0.65. Notice that successive iterations do not necessarily reduce the absolute error in the representation of the angle. Also notice that the error does not oscillate about zero.

At each step as we decompose a into the sum or difference of the ai. Figure 2 gives an example of the convergence pro's we could also multiply our vector [x, y] by the appropriate Rai. Figure 2 gives an example of the convergence pro or R-ai. Figure 2 gives an example of the convergence pro depending on whether we add or subtract ai. Figure 2 gives an example of the convergence pro. Remember, these multiplications are nothing more than shifts. We must also multiply in the still embarrassing factor cos ai. Figure 2 gives an example of the convergence pro. However, cosine is an even function and has the property that cos ai - cos (-a1). It does not matter whether we add or subtract the angle - we always multiply by the same factor! Because all of the cos ai can be factored out and grouped together, we can treat their product as a constant and compute it only once, along with all the ai = tan-12-i.

Not all angles can be represented as the sum of ai. There is a domain of convergence outside of which we cannot reduce the angle to within an-1 of zero. See Example 1(f). For the algorithm to work, we must start with a such that |a| amax 1.74. This conveniently falls just outside the first quadrant. If we are given an angle outside the first quadrant, we can scale it by dividing by /2 obtaining a quotient Q and a remainder D where |D| < /2 < a?? Since the algorithm computes both sine and cosine of D, we pick the appropriate value and sign depending on the value of Q.

What about angles within the domain of convergence? It's not obvious that the strange set we've picked (see Example 1(g)) can represent all angles within the domain of convergence to within a~~ But, using mathematical induction. Walther proves that the scheme works.

### The Circular Functions

First, start with [x,y,z] The x and y are as before z is the quantity that we drive to zero with an initial value of angle a. The first step in the loop decides whether to add or subtract ai from the residue z. The variable s is correspondingly positive or negative The second step reduces the magnitude of z and effects the multiplications by the tan a~ The expression ~~~~ means shift y right by ~ bits.

#### Example 2: (a) The basic algorithm; (b) the inverse algorithm

```  (a)
for i from 0 to n do
{
if (z <img src="http://twimgs.com/ddj/ddj/images/ddj9010f/gteq.gif"> 0) then s:= 1 else s:= -1;
[x,y,z] := [x - s y>>i, y + s x>>i, z - s a<SUB>1</SUB>]
}

(b)
for i from 0 to n do
{
if (y <img src="http://twimgs.com/ddj/ddj/images/ddj9010f/gteq.gif"> 0) then s:= 1 else s:= -1;
[x,y,z] := [x + s y>>i, y - s x>>i, z - s a<SUB>1</SUB>]
}
```

When you start the algorithm with [x,y,z] and then drive z to zero as specified by Example 2(a). We are left with the quantities in Example 3(a). Where K is a constant. It is just the product of the cos ai as in Example 3(b).

For the curious, K 0.607. The value of K can be precomputed by setting [x,y,z] to [1,0,0] and running the algorithm as before. The result is shown in Example 3(c). Take the reciprocal of the final x and we have K. Therefore to compute sin a and cos a, set [x,y,z] to [K, O, ~] and run the algorithm, Example 3(d) shows the result. In effect, we start with a vector, [x,y] and rotate it through a given ~~ a my driving z to zero. Running the algorithm with the special case where the vector initially lies along the x axis and is of length K, rotates the vector by angle a and leaves behind cos a and sin a. This relationship is shown in Figure 3.

To compute tan 1a instead of z, we could choose to drive y to zero. Driv. ing y to zero rotates the vector through the angle a, the angle subtended by the vector and the x axis, leaving the vector lying along the x axis. Start with the vector anywhere in the first or fourth quadrant and an initial value of zero in z. The first or fourth quadrant is used because almost all vectors in the second or third quadrant will not converge. At the i-th step, if y >- 0, the vector lies in the first quadrant, subtract ai from z. Move the vector closer to the x axis; rotate it by negative ai by multiplying by the rotation matrix R-ai. If y < 0, the vector lies in the fourth quadrant, add ai to z and multiply the vector by Rai. At the end, z has the negative of the angle of the original vector [x, y], tan-1 y/x = tan-1a.

Changing the sign of ai has no effect on the computed values of x and y and leaves the original angle a in z rather than its negative. With this change, the inverse algorithm to drive y to zero becomes the expression shown as in the algorithm in Example 2(b).

Starting with [x, y, z] and then driving y to zero using the inverse algorithm leaves behind the quantities in Example 3(e).

### Hyperbolic Functions

#### Table 1: Hyperbolic functions

```Hyperbolic Function               Circular Function
---------------------------------------------------

e<SUP>x</SUP>+e<SUP>-x</SUP>                    e<SUP>ix</SUP> + e<SUP>-ix</SUP>
cosh x = ----------         cos x = --------------
2                            2
e<SUP>x</SUP>-e<SUP>-x</SUP>                    e<SUP>ix</SUP> - e<SUP>-ix</SUP>
sinh x = ----------         sin x = --------------
2                       2i
[cosh a sinh a]           [cosh a -sin a]
H<SUB>a</SUB> = [sinh a cosh a]      R<SUB>a</SUB> = [sin a cos a]

H<SUB>a</SUB>H<SUB>b</SUB> = H<SUB>a+b</SUB>           R<SUB>a</SUB>R<SUB>b</SUB> = R<SUB>a+b</SUB>

e<SUP>x</SUP> = cosh x + sinh x

x-1
In x = 2 tanh<SUP>-1</SUP> ---
x+1
```

By analogy, use Ha as the rotation matrix and represent a using the set ai = tanh-1 2-i for i = 1 to n. Notice that for hyperbolics, a0 is infinity.

Except for the repeated terms and some changes of sign, the algorithms for hyperbolic functions are identical to the circular functions. Listing One, page 157, shows this in detail.

For hyperbolic functions, we start with [x,y,z] and then drive z to zero. This yields the quantities in Example 4(b). Starting with x,y,z and then driving y to zero gives the quantity shown in Example 4(c). For hyperbolics, K 1.21.

Some interesting special cases include the exponential, square root, and natural logarithm. The exponential case is in Example 4(d) while the square root and logarithm cases are in Example 4(e).

### Calculating the Constants

The program, written in C, uses fixed point arithmetic for all calculations. All constants and variables used to calculate functions are declared as the type long. The code assumes that a long is at least 32 bits. I have decided to represent numbers in the range - 4 <= x < 4; this lets me represent e as a fixed point number. The high order bit is for the sign. The low order fractionBits (a constant defined as 29) bits hold the fractional part of the number. The remainder of the bits between the sign bit and the fractional part hold the integer part of the number. Figure 4 shows the fixed point format in graphic form.

I use power series to calculate the incremental angles ai, as shown in Example 5(c) and Example 5(d), respectively. How do we know the number of terms necessary to evaluate tan-1 and tanh-1 to 32 bits of precision? First consider the value of x for which tan-1 x = x to 32 bits of precision. A theorem of numerical analysis states that for an alternating sum where the absolute values of the terms decrease monotonically, the error is less than the absolute value of the first neglected term. Solving the equation x3/3 = 2-32 for x yields x = 3(6 * 2-11); therefore for i 11, tan-1 2-i = 2-i with 32 bits of precision.

For the higher powers of two, we need to solve the relation 2-in/n = 2-32 for n for each of the cases i = 1 to 10. We do not even attempt the calculation for i = 0. The series for tan-1 1 converges very slowly, even after 500 terms the third digit is still changing. Fortunately, we know that the answer is /4. Computing the rest is not as much work. The array terms has the gory details.

As usual, tanh-1 is more perverse. It is not an alternating sum and does not meet the conditions of the theorem used above. Consider the second neglected term of tanh-1 1/2. It is less than 1/4 of the first neglected term because the series includes only every other power of two. All of the other neglected terms can have no effect on the 33rd bit. The series for the other arguments, 1/4, 1/8, ..., converges even faster. Therefore, the number of terms calculated for tan-1 works just as well for tanh-1 for 32-bit accuracy.

Before computing the power series, we still need to compute the coefficients, 1/k, for each term k = 1, 3, 5, ... 27. We fill the coefficient array long a[28] with odd indices by calling the routine Reciprocal, which takes two arguments and returns a long. The first argument is the integer for the desired reciprocal. The second specifies the desired precision for the fractional part of the result. Reciprocal uses a simple as can be restoring division; it is the algorithm we all learned in grade school for long division. The elements of the array a with even indices get OL because there are no terms in the power series with even exponents.

Everything is ready to fill the arrays atan[fractionBits+ 1] and atanh[fractionBits+1].

The routine Poly2 evaluates the power series for the specified number of terms for the specified power of two using Horner's rule. The coefficients come from the array a, which we just carefully filled. Horner's rule is the recommended method for evaluating polynomials. A polynomial as in Example 5(e) can be rewritten as in Example 5(f). This simple recursive formula evaluates the polynomial with n multiplications and n additions. We compute the prescaling constants K by using the method explained above; in the program we call these XOC and XOH, for the circular and hyperbolic constants, respectively. Program output to this point is shown in Listing Two, page 158.

The routines Circular, InvertCircular, Hyperbolic, and InvertHyperbolic are the C implementations of the algorithms described above. They all take as arguments the initial values for [x,y,z]; they leave their results in the global variables X, Y, and Z. Considering their versatility and the wide range of functions they compute, these routines are compact and elegant!

### References

1. R. Nave. "Implementation of Transcendental Functions on a Numerics Processor." Microprocessing and Microprogramming, vol. 11, num. 3 - 4, pp. 221 - 225, March - April 1983.

2. A. K. Yuen. "Intel's Floating-Point Processors." Electro/88 Conference Record, pp. 48/5/1 - 7, 1988.

3. J. E. Volder. "The Cordic TrigonometricComputing Technique." IRE Transactions Electronic Computers, vol. EC - 8, pp. 330 - 334, September 1959.

4. D. S. Cochran. "Algorithms and Accuracy in the HP-35." HewlettPackard Journal, pp. 10 - 11, June 1972.

5. J. S. Walther. "A Unified Algorithm for Elementary Functions." In 1971 Proceedings of the Joint Spring Computer Conference, pp. 379 - 385, 1971.

_IMPLEMENTING CORDIC ALGORITHMS_ by Pitts Jarvis

```<a name="0210_000c">

/* cordicC.c -- J. Pitts Jarvis, III
* cordicC.c computes CORDIC constants and exercises the basic algorithms.
* Represents all numbers in fixed point notation.  1 bit sign,
* longBits-1-n bit integral part, and n bit fractional part.  n=29 lets us
* represent numbers in the interval [-4, 4) in 32 bit long.  Two's
* complement arithmetic is operative here.
*/

#define fractionBits 29
#define longBits 32
#define One    (010000000000>>1)
#define HalfPi (014441766521>>1)

/* cordic algorithm identities for circular functions, starting with [x, y, z]
*  and then
*  driving z to 0 gives: [P*(x*cos(z)-y*sin(z)), P*(y*cos(z)+x*sin(z)), 0]
*  driving y to 0 gives: [P*sqrt(x^2+y^2), 0, z+atan(y/x)]
*  where K = 1/P = sqrt(1+1)* . . . *sqrt(1+(2^(-2*i)))
*  special cases which compute interesting functions
*  sin, cos      [K, 0, a] -> [cos(a), sin(a), 0]
*  atan          [1, a, 0] -> [sqrt(1+a^2)/K, 0, atan(a)]
*      [x, y, 0] -> [sqrt(x^2+y^2)/K, 0, atan(y/x)]
* for hyperbolic functions, starting with [x, y, z] and then
* driving z to 0 gives: [P*(x*cosh(z)+y*sinh(z)), P*(y*cosh(z)+x*sinh(z)), 0]
* driving y to 0 gives: [P*sqrt(x^2-y^2), 0, z+atanh(y/x)]
* where K = 1/P = sqrt(1-(1/2)^2)* . . . *sqrt(1-(2^(-2*i)))
*  sinh, cosh    [K, 0, a] -> [cosh(a), sinh(a), 0]
*  exponential   [K, K, a] -> [e^a, e^a, 0]
*  atanh         [1, a, 0] -> [sqrt(1-a^2)/K, 0, atanh(a)]
*         [x, y, 0] -> [sqrt(x^2-y^2)/K, 0, atanh(y/x)]
*  ln            [a+1, a-1, 0] -> [2*sqrt(a)/K, 0, ln(a)/2]
*  sqrt          [a+(K/2)^2, a-(K/2)^2, 0] -> [sqrt(a), 0, ln(a*(2/K)^2)/2]
*  sqrt, ln      [a+(K/2)^2, a-(K/2)^2, -ln(K/2)] -> [sqrt(a), 0, ln(a)/2]
* for linear functions, starting with [x, y, z] and then
*  driving z to 0 gives: [x, y+x*z, 0]
*  driving y to 0 gives: [x, 0, z+y/x]
*/

long X0C, X0H, X0R; /* seed for circular, hyperbolic, and square root */
long OneOverE, E;   /* the base of natural logarithms */
long HalfLnX0R;       /* constant used in simultanous sqrt, ln computation */

/* compute atan(x) and atanh(x) using infinite series
*    atan(x) =  x - x^3/3 + x^5/5 - x^7/7 + . . . for x^2 < 1
*     atanh(x) = x + x^3/3 + x^5/5 + x^7/7 + . . . for x^2 < 1
* To calcuate these functions to 32 bits of precision, pick
* terms[i] s.t. ((2^-i)^(terms[i]))/(terms[i]) < 2^-32
* For x <= 2^(-11), atan(x) = atanh(x) = x with 32 bits of accuracy  */
unsigned terms[11]= {0, 27, 14, 9, 7, 5, 4, 4, 3, 3, 3};static long a[28], atan[fractionBits+1], atanh[fractionBits+1], X, Y, Z;
#include <stdio.h>   /* putchar is a marco for some */

/* Delta is inefficient but pedagogical */
#define Delta(n, Z) (Z>=0) ? (n) : -(n)
#define abs(n) (n>=0) ? (n) : -(n)

/* Reciprocal, calculate reciprocol of n to k bits of precision
*  a and r form integer and fractional parts of the dividend respectively  */
long
Reciprocal(n, k) unsigned n, k;
{
unsigned i, a= 1; long r= 0;
for (i= 0; i<=k; ++i) {r += r; if (a>=n) {r += 1; a -= n;}; a += a;}
return(a>=n? r+1 : r); /* round result */
}

/* ScaledReciprocal, n comes in funny fixed point fraction representation */
long
ScaledReciprocal(n, k) long n; unsigned k;
{
long a, r=0; unsigned i;
a= 1L<<k;
for (i=0; i<=k; ++i) {r += r; if (a>=n) {r += 1; a -= n;}; a += a;};
return(a>=n? r+1 : r); /* round result */
}

/* Poly2 calculates polynomial where the variable is an integral power of 2,
*   log is the power of 2 of the variable
*   n is the order of the polynomial
*   coefficients are in the array a[]   */
long
Poly2(log, n) int log; unsigned n;
{
long r=0; int i;
for (i=n; i>=0; --i) r= (log<0? r>>-log : r<<log)+a[i];
return(r);
}
WriteFraction(n) long n;
{
unsigned short i, low, digit; unsigned long k;
putchar(n < 0 ? '-' : ' '); n = abs(n);
putchar((n>>fractionBits) + '0'); putchar('.');
low = k = n << (longBits-fractionBits); /* align octal point at left */
k   >>=  4;   /* shift to make room for a decimal digit */
for (i=1; i<=8; ++i)
{
digit = (k *= 10L) >> (longBits-4);
low = (low & 0xf) * 10;
k += ((unsigned long) (low>>4)) - ((unsigned long) digit << (longBits-4));
putchar(digit+'0');
}
}
WriteRegisters()
{  printf("  X: "); WriteVarious(X);
printf("  Y: "); WriteVarious(Y);
printf("  Z: "); WriteVarious(Z);
}
WriteVarious(n) long n;
{
WriteFraction(n); printf("  0x%08lx 0%011lo\n", n, n);
}
Circular(x, y, z) long x, y, z;
{
int i;
X = x; Y = y; Z = z;
for (i=0; i<=fractionBits; ++i)
{
x= X>>i; y= Y>>i; z= atan[i];
X -= Delta(y, Z);
Y += Delta(x, Z);
Z -= Delta(z, Z);
}
}
InvertCircular(x, y, z) long x, y, z;
{
int i;
X = x; Y = y; Z = z;
for (i=0; i<=fractionBits; ++i)
{
x= X>>i; y= Y>>i; z= atan[i];
X -= Delta(y, -Y);
Z -= Delta(z, -Y);
Y += Delta(x, -Y);
}
}
Hyperbolic(x, y, z) long x, y, z;
{
int i;
X = x; Y = y; Z = z;
for (i=1; i<=fractionBits; ++i)
{
x= X>>i; y= Y>>i; z= atanh[i];
X += Delta(y, Z);
Y += Delta(x, Z);
Z -= Delta(z, Z);
if ((i==4)||(i==13))
{
x= X>>i; y= Y>>i; z= atanh[i];
X  +=  Delta(y, Z);
Y += Delta(x, Z);
Z  -=  Delta(z, Z);
}
}
}
InvertHyperbolic(x, y, z) long x, y, z;
{
int i;
X = x; Y = y; Z = z;  for (i=1; i<=fractionBits; ++i)
{
x= X>>i; y= Y>>i; z= atanh[i];
X += Delta(y, -Y);
Z -= Delta(z, -Y);
Y += Delta(x, -Y);
if ((i==4)||(i==13))
{
x= X>>i; y= Y>>i; z= atanh[i];
X += Delta(y, -Y);
Z -= Delta(z, -Y);
Y += Delta(x, -Y);
}
}
}
Linear(x, y, z) long x, y, z;
{
int i;
X = x; Y = y; Z = z; z= One;
for (i=1; i<=fractionBits; ++i)
{
x  >>= 1; z  >>= 1; Y += Delta(x, Z); Z -= Delta(z, Z);
}
}
InvertLinear(x, y, z) long x, y, z;
{
int i;
X = x; Y = y; Z = z; z= One;
for (i=1; i<=fractionBits; ++i)
{
Z -= Delta(z  >>= 1, -Y); Y += Delta(x  >>= 1, -Y);
}
}

/*********************************************************/
main()
{
int i; long r;
/*system("date");*//* time stamp the log for UNIX systems */
for (i=0; i<=13; ++i)
{
a[2*i]= 0; a[2*i+1]= Reciprocal(2*i+1, fractionBits);
}
for (i=0; i<=10; ++i) atanh[i]= Poly2(-i, terms[i]);
atan[0]= HalfPi/2; /* atan(2^0)= pi/4 */
for (i=1; i<=7; ++i) a[4*i-1]= -a[4*i-1];
for (i=1; i<=10; ++i) atan[i]= Poly2(-i, terms[i]);
for (i=11; i<=fractionBits; ++i) atan[i]= atanh[i]= 1L<<(fractionBits-i);
printf("\natanh(2^-n)\n");
for (i=1; i<=10; ++i){printf("%2d ", i); WriteVarious(atanh[i]);}
r= 0;
for (i=1; i<=fractionBits; ++i)
r +=  atanh[i];
r +=  atanh[4]+atanh[13];
for (i=0; i<=10; ++i){printf("%2d ", i); WriteVarious(atan[i]);}
r= 0; for (i=0; i<=fractionBits; ++i) r +=  atan[i];

/* all the results reported in the printfs are calculated with my HP-41C */
printf("\n\n--------------------circular functions--------------------\n");
printf("Grinding on [1, 0, 0]\n");
Circular(One, 0L, 0L); WriteRegisters();
printf("\n  K: "); WriteVarious(X0C= ScaledReciprocal(X, fractionBits));
printf("\nGrinding on [K, 0, 0]\n");
Circular(X0C, 0L, 0L); WriteRegisters();
printf("\nGrinding on [K, 0, pi/6] -> [0.86602540, 0.50000000, 0]\n");
Circular(X0C, 0L, HalfPi/3L); WriteRegisters();
printf("\nGrinding on [K, 0, pi/4] -> [0.70710678, 0.70710678, 0]\n");
Circular(X0C, 0L, HalfPi/2L); WriteRegisters();
printf("\nGrinding on [K, 0, pi/3] -> [0.50000000, 0.86602540, 0]\n");
Circular(X0C, 0L, 2L*(HalfPi/3L)); WriteRegisters();
printf("\n------Inverse functions------\n");
printf("Grinding on [1, 0, 0]\n");
InvertCircular(One, 0L, 0L); WriteRegisters();
printf("\nGrinding on [1, 1/2, 0] -> [1.84113394, 0, 0.46364761]\n");
InvertCircular(One, One/2L, 0L); WriteRegisters();
printf("\nGrinding on [2, 1, 0] -> [3.68226788, 0, 0.46364761]\n");
InvertCircular(One*2L, One, 0L); WriteRegisters();
printf("\nGrinding on [1, 5/8, 0] -> [1.94193815, 0, 0.55859932]\n");
InvertCircular(One, 5L*(One/8L), 0L); WriteRegisters();
printf("\nGrinding on [1, 1, 0] -> [2.32887069, 0, 0.78539816]\n");
InvertCircular(One, One, 0L); WriteRegisters();
printf("\n--------------------hyperbolic functions--------------------\n");
printf("Grinding on [1, 0, 0]\n");
Hyperbolic(One, 0L, 0L); WriteRegisters();
printf("\n  K: "); WriteVarious(X0H= ScaledReciprocal(X, fractionBits));
printf("  R: "); X0R= X0H>>1; Linear(X0R, 0L, X0R); WriteVarious(X0R= Y);
printf("\nGrinding on [K, 0, 0]\n");
Hyperbolic(X0H, 0L, 0L); WriteRegisters();
printf("\nGrinding on [K, 0, 1] -> [1.54308064, 1.17520119, 0]\n");
Hyperbolic(X0H, 0L, One); WriteRegisters();
printf("\nGrinding on [K, K, -1] -> [0.36787944, 0.36787944, 0]\n");
Hyperbolic(X0H, X0H, -One); WriteRegisters();
OneOverE = X; /* save value ln(1/e) = -1 */
printf("\nGrinding on [K, K, 1] -> [2.71828183, 2.71828183, 0]\n");
Hyperbolic(X0H, X0H, One); WriteRegisters();
E = X; /* save value ln(e) = 1 */
printf("\n------Inverse functions------\n");
printf("Grinding on [1, 0, 0]\n");
InvertHyperbolic(One, 0L, 0L); WriteRegisters();
printf("\nGrinding on [1/e + 1, 1/e - 1, 0] -> [1.00460806, 0,
-0.50000000]\n");
InvertHyperbolic(OneOverE+One,OneOverE-One, 0L); WriteRegisters();
printf("\nGrinding on [e + 1, e - 1, 0] -> [2.73080784, 0, 0.50000000]\n");
InvertHyperbolic(E+One, E-One, 0L); WriteRegisters();
printf("\nGrinding on (1/2)*ln(3) -> [0.71720703, 0, 0.54930614]\n");
InvertHyperbolic(One, One/2L, 0L); WriteRegisters();
printf("\nGrinding on [3/2, -1/2, 0] -> [1.17119417, 0, -0.34657359]\n");    InvertHyperbolic(One+(One/2L), -(One/2L), 0L); WriteRegisters();
printf("\nGrinding on sqrt(1/2) -> [0.70710678, 0, 0.15802389]\n");
InvertHyperbolic(One/2L+X0R, One/2L-X0R, 0L); WriteRegisters();
printf("\nGrinding on sqrt(1) -> [1.00000000, 0, 0.50449748]\n");
InvertHyperbolic(One+X0R, One-X0R, 0L); WriteRegisters();
HalfLnX0R = Z;
printf("\nGrinding on sqrt(2) -> [1.41421356, 0, 0.85117107]\n");
InvertHyperbolic(One*2L+X0R, One*2L-X0R, 0L); WriteRegisters();
printf("\nGrinding on sqrt(1/2), ln(1/2)/2 -> [0.70710678, 0,
-0.34657359]\n");
InvertHyperbolic(One/2L+X0R, One/2L-X0R, -HalfLnX0R); WriteRegisters();
printf("\nGrinding on sqrt(3)/2, ln(3/4)/2 -> [0.86602540, 0,
-0.14384104]\n");
InvertHyperbolic((3L*One/4L)+X0R, (3L*One/4L)-X0R, -HalfLnX0R);
WriteRegisters();
printf("\nGrinding on sqrt(2), ln(2)/2 -> [1.41421356, 0, 0.34657359]\n");
InvertHyperbolic(One*2L+X0R, One*2L-X0R, -HalfLnX0R);
WriteRegisters();
exit(0);
}

<a name="0210_000d"><a name="0210_000d">
<a name="0210_000e">```
[LISTING TWO]
```<a name="0210_000e">

atanh (2^-n)

1   0.54930614   0x1193ea7a 002144765172
2   0.25541281   0x082c577d 001013053575
3   0.12565721   0x04056247 000401261107
4   0.06258157   0x0200ab11 000200125421
5   0.03126017   0x01001558 000100012530
6   0.01562627   0x008002aa 000040001252
7   0.00781265   0x00400055 000020000125
8   0.00390626   0x0020000a 000010000012
9   0.00195312   0x00100001 000004000001
10  0.00097656   0x00080000 000002000000

atan (26-n)

0   0.78539816   0x1921fb54 003110375524
1   0.46364760   0x0ed63382 001665431602
2   0.24497866   0x07d6dd7e 000765556576
3   0.12435499   0x03fab753 000376533523
4   0.06241880   0x01ff55bb 000177652673
6   0.01562372   0x007ffd55 000037776525
7   0.00781233   0x003fffaa 000017777652
8   0.00390622   0x001ffff5 000007777765
9   0.00195312   0x000ffffe 000003777776
10  0.00097656   0x0007ffff 000001777777

```

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