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