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/* libmath.b for GNU bc. */
/* This file is part of GNU bc.
Copyright (C) 1991, 1992, 1993, 1997 Free Software Foundation, Inc.
This program is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation; either version 2 of the License , or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program; see the file COPYING. If not, write to
the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA.
You may contact the author by:
e-mail: phil@cs.wwu.edu
us-mail: Philip A. Nelson
Computer Science Department, 9062
Western Washington University
Bellingham, WA 98226-9062
*************************************************************************/
scale = 20
/* Uses the fact that e^x = (e^(x/2))^2
When x is small enough, we use the series:
e^x = 1 + x + x^2/2! + x^3/3! + ...
*/
define e(x) {
auto a, d, e, f, i, m, n, v, z
/* a - holds x^y of x^y/y! */
/* d - holds y! */
/* e - is the value x^y/y! */
/* v - is the sum of the e's */
/* f - number of times x was divided by 2. */
/* m - is 1 if x was minus. */
/* i - iteration count. */
/* n - the scale to compute the sum. */
/* z - orignal scale. */
/* Check the sign of x. */
if (x<0) {
m = 1
x = -x
}
/* Precondition x. */
z = scale;
n = 6 + z + .44*x;
scale = scale(x)+1;
while (x > 1) {
f += 1;
x /= 2;
scale += 1;
}
/* Initialize the variables. */
scale = n;
v = 1+x
a = x
d = 1
for (i=2; 1; i++) {
e = (a *= x) / (d *= i)
if (e == 0) {
if (f>0) while (f--) v = v*v;
scale = z
if (m) return (1/v);
return (v/1);
}
v += e
}
}
/* Natural log. Uses the fact that ln(x^2) = 2*ln(x)
The series used is:
ln(x) = 2(a+a^3/3+a^5/5+...) where a=(x-1)/(x+1)
*/
define l(x) {
auto e, f, i, m, n, v, z
/* return something for the special case. */
if (x <= 0) return ((1 - 10^scale)/1)
/* Precondition x to make .5 < x < 2.0. */
z = scale;
scale = 6 + scale;
f = 2;
i=0
while (x >= 2) { /* for large numbers */
f *= 2;
x = sqrt(x);
}
while (x <= .5) { /* for small numbers */
f *= 2;
x = sqrt(x);
}
/* Set up the loop. */
v = n = (x-1)/(x+1)
m = n*n
/* Sum the series. */
for (i=3; 1; i+=2) {
e = (n *= m) / i
if (e == 0) {
v = f*v
scale = z
return (v/1)
}
v += e
}
}
/* Sin(x) uses the standard series:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! ... */
define s(x) {
auto e, i, m, n, s, v, z
/* precondition x. */
z = scale
scale = 1.1*z + 2;
v = a(1)
if (x < 0) {
m = 1;
x = -x;
}
scale = 0
n = (x / v + 2 )/4
x = x - 4*n*v
if (n%2) x = -x
/* Do the loop. */
scale = z + 2;
v = e = x
s = -x*x
for (i=3; 1; i+=2) {
e *= s/(i*(i-1))
if (e == 0) {
scale = z
if (m) return (-v/1);
return (v/1);
}
v += e
}
}
/* Cosine : cos(x) = sin(x+pi/2) */
define c(x) {
auto v;
scale += 1;
v = s(x+a(1)*2);
scale -= 1;
return (v/1);
}
/* Arctan: Using the formula:
atan(x) = atan(c) + atan((x-c)/(1+xc)) for a small c (.2 here)
For under .2, use the series:
atan(x) = x - x^3/3 + x^5/5 - x^7/7 + ... */
define a(x) {
auto a, e, f, i, m, n, s, v, z
/* a is the value of a(.2) if it is needed. */
/* f is the value to multiply by a in the return. */
/* e is the value of the current term in the series. */
/* v is the accumulated value of the series. */
/* m is 1 or -1 depending on x (-x -> -1). results are divided by m. */
/* i is the denominator value for series element. */
/* n is the numerator value for the series element. */
/* s is -x*x. */
/* z is the saved user's scale. */
/* Negative x? */
m = 1;
if (x<0) {
m = -1;
x = -x;
}
/* Special case and for fast answers */
if (x==1) {
if (scale <= 25) return (.7853981633974483096156608/m)
if (scale <= 40) return (.7853981633974483096156608458198757210492/m)
if (scale <= 60) \
return (.785398163397448309615660845819875721049292349843776455243736/m)
}
if (x==.2) {
if (scale <= 25) return (.1973955598498807583700497/m)
if (scale <= 40) return (.1973955598498807583700497651947902934475/m)
if (scale <= 60) \
return (.197395559849880758370049765194790293447585103787852101517688/m)
}
/* Save the scale. */
z = scale;
/* Note: a and f are known to be zero due to being auto vars. */
/* Calculate atan of a known number. */
if (x > .2) {
scale = z+5;
a = a(.2);
}
/* Precondition x. */
scale = z+3;
while (x > .2) {
f += 1;
x = (x-.2) / (1+x*.2);
}
/* Initialize the series. */
v = n = x;
s = -x*x;
/* Calculate the series. */
for (i=3; 1; i+=2) {
e = (n *= s) / i;
if (e == 0) {
scale = z;
return ((f*a+v)/m);
}
v += e
}
}
/* Bessel function of integer order. Uses the following:
j(-n,x) = (-1)^n*j(n,x)
j(n,x) = x^n/(2^n*n!) * (1 - x^2/(2^2*1!*(n+1)) + x^4/(2^4*2!*(n+1)*(n+2))
- x^6/(2^6*3!*(n+1)*(n+2)*(n+3)) .... )
*/
define j(n,x) {
auto a, d, e, f, i, m, s, v, z
/* Make n an integer and check for negative n. */
z = scale;
scale = 0;
n = n/1;
if (n<0) {
n = -n;
if (n%2 == 1) m = 1;
}
/* Compute the factor of x^n/(2^n*n!) */
f = 1;
for (i=2; i<=n; i++) f = f*i;
scale = 1.5*z;
f = x^n / 2^n / f;
/* Initialize the loop .*/
v = e = 1;
s = -x*x/4
scale = 1.5*z
/* The Loop.... */
for (i=1; 1; i++) {
e = e * s / i / (n+i);
if (e == 0) {
scale = z
if (m) return (-f*v/1);
return (f*v/1);
}
v += e;
}
}
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