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From: Robert Dodier <robert_dodier@us...>  20050102 19:26:03

Update of /cvsroot/maxima/maxima/doc/info In directory sc8prcvs1.sourceforge.net:/tmp/cvsserv10315 Modified Files: Number.texi Log Message: Revise description of "bern". Minor revision of "bernpoly". Strike out description of obsolete, nonexistent function "bgzeta". Minor revision of "bfzeta". Minor revision of "bfhzeta". Minor revision of "binomial". Revision of "burn". Commented out statements about execution times. Strike out obsolete function "bzeta". Expand and revise "cf" with addt'l info and examples. Minor revision of "cfdisrep". Minor revision of "cfexpand". Revision of "cflength". Index: Number.texi =================================================================== RCS file: /cvsroot/maxima/maxima/doc/info/Number.texi,v retrieving revision 1.8 retrieving revision 1.9 diff u d r1.8 r1.9  Number.texi 29 Dec 2004 08:01:39 0000 1.8 +++ Number.texi 2 Jan 2005 19:25:52 0000 1.9 @@ 6,125 +6,226 @@ @node Definitions for Number Theory, , Number Theory, Number Theory @section Definitions for Number Theory @... bern (x) gives the Xth Bernoulli number for integer X. ZEROBERN[TRUE] if set to FALSE excludes the zero BERNOULLI numbers. (See also BURN). +@defun bern (n) +Returns the @var{n}'th Bernoulli number for integer @code{n}. +@c WELL, ACTUALLY bern SIMPLIFIES, LIKE FACTORIAL  DO WE WANT TO GET INTO THAT ??? +@c OR JUST PRETEND IT'S "RETURNED" ??? +Bernoulli numbers equal to zero are suppressed if @code{zerobern} is @code{false}. @... defun +See also @code{burn}. @... bernpoly (v, n) generates the nth Bernoulli polynomial in the variable v. +@example +(%i1) zerobern: true$ +(%i2) map (bern, [0, 1, 2, 3, 4, 5, 6, 7, 8]); + 1 1 1 1 1 +(%o2) [1,  , , 0,  , 0, , 0,  ] + 2 6 30 42 30 +(%i3) zerobern: false$ +(%i4) map (bern, [0, 1, 2, 3, 4, 5, 6, 7, 8]); + 1 1 1 5 691 7 3617 43867 +(%o4) [1,  , ,  , ,  , ,  , ] + 2 6 30 66 2730 6 510 798 +@end example @end defun @... bfzeta (expr, n) BFLOAT version of the Riemann Zeta function. The 2nd argument is how many digits to retain and return, it's a good idea to request a couple of extra. This function is available by doing LOAD(BFFAC); . +@defun bernpoly (v, n) +Returns the @var{n}'th Bernoulli polynomial in the +variable @var{v}. @end defun @... bgzeta (s, fpprec) BGZETA is like BZETA, but avoids arithmetic overflow errors on large arguments, is faster on medium size arguments (say S=55, FPPREC=69), and is slightly slower on small arguments. It may eventually replace BZETA. BGZETA is available by doing LOAD(BFAC);. +@defun bfzeta (s, n) +Returns the Riemann zeta function for the argument @var{s}. +The return value is a big float (bfloat); +@var{n} is the number of digits in the return value. + +@code{load ("bffac")} loads this function. @end defun @... bhzeta (s, h, fpprec) gives FPPREC digits of +@defun bfhzeta (s, h, n) +Returns the Hurwitz zeta function for the arguments @var{s} and @var{h}. +The return value is a big float (bfloat); +@var{n} is the number of digits in the return value. + +The Hurwitz zeta function is defined as @example SUM((K+H)^S,K,0,INF) +sum ((k+h)^s, k, 0, inf) @end example This is available by doing LOAD(BFFAC);. + +@code{load ("bffac")} loads this function. @end defun @defun binomial (x, y) the binomial coefficient X*(X1)*...*(XY+1)/Y!. If X and Y are integers, then the numerical value of the binomial coefficient is computed. If Y, or the value XY, is an integer, the binomial coefficient is expressed as a polynomial. +The binomial coefficient @code{(x + y)!/(x! y!)}. +If @code{x} and @code{y} are integers, then the numerical value of the binomial +coefficient is computed. +If @code{y}, or @code{x  y}, is an integer, +the binomial coefficient is expressed as a polynomial. @end defun @defun burn (n) is like BERN(N), but without computing all of the uncomputed Bernoullis of smaller index. So BURN works efficiently for large, isolated N. (BERN(402) takes about 645 secs vs 13.5 secs for BURN(402). BERN's time growth seems to be exponential, while BURN's is about cubic. But if next you do BERN(404), it only takes 12 secs, since BERN remembers all in an array, whereas BURN(404) will take maybe 14 secs or maybe 25, depending on whether Maxima needs to BFLOAT a better value of %PI.) BURN is available by doing LOAD(BFFAC);. BURN uses an observation of WGD that (rational) Bernoulli numbers can be approximated by (transcendental) zetas with tolerable efficiency. +Returns the @var{n}'th Bernoulli number for integer @code{n}. +@code{burn} may be more efficient than @code{bern} for large, isolated @code{n}, +as @code{bern} computes all the Bernoulli numbers up to index @code{n} before returning. @... defun +@c STATEMENTS ABOUT TIMING NEED VERIFICATION !!! +@c (BERN(402) takes about 645 secs vs 13.5 secs for BURN(402). +@c The time to compute @code{bern} is approximately exponential, +@c while the time to compute @code{burn} is approximately cubic. +@c But if next you do BERN(404), it only takes 12 secs, +@c since BERN remembers all in an array, whereas BURN(404) will take +@c maybe 14 secs or maybe 25, depending on whether Maxima needs to +@c BFLOAT a better value of %PI.) @... bzeta   This function is obsolete, see BFZETA. +@code{burn} exploits the observation that (rational) Bernoulli numbers can be +approximated by (transcendental) zetas with tolerable efficiency. + +@code{load ("bffac")} loads this function. @end defun @defun cf (expr) converts exp into a continued fraction. exp is an expression composed of arithmetic operators and lists which represent continued fractions. A continued fraction a+1/(b+1/(c+...)) is represented by the list [a,b,c,...]. a,b,c,.. must be integers. Exp may also involve SQRT(n) where n is an integer. In this case CF will give as many +Converts @var{expr} into a continued fraction. +@var{expr} is an expression +comprising continued fractions and square roots of integers. +Operands in the expression may be combined with arithmetic operators. +Aside from continued fractions and square roots, +factors in the expression must be integer or rational numbers. +Maxima does not know about operations on continued fractions outside of @code{cf}. + +@code{cf} evaluates its arguments after binding @code{listarith} to @code{false}. +@code{cf} returns a continued fraction, represented as a list. + +A continued fraction @code{a + 1/(b + 1/(c + ...))} +is represented by the list @code{[a, b, c, ...]}. +The list elements @code{a}, @code{b}, @code{c}, ... must evaluate to integers. +@var{expr} may also contain @code{sqrt (n)} where @code{n} is an integer. +In this case @code{cf} will give as many terms of the continued fraction as the value of the variable CFLENGTH[1] times the period. Thus the default is to give one period. (CF binds LISTARITH to FALSE so that it may carry out its function.) +@code{cflength} times the period. @... defun +A continued fraction can be evaluated to a number +by evaluating the arithmetic representation +returned by @code{cfdisrep}. +See also @code{cfexpand} for another way to evaluate a continued fraction. + +See also @code{cfdisrep}, @code{cfexpand}, and @code{cflength}. + +Examples: + +@itemize @bullet +@item +@var{expr} is an expression comprising continued fractions and square roots of integers. @... cfdisrep (list) converts the continued fraction represented by list into general representation. @example (%i1) CF([1,2,3]+[1,2,1]); (%o1) [1, 1, 1, 2] (%i2) CFDISREP(%);  1 (%o2) 1 +   1  1 +   1  1 +   2 +(%i1) cf ([5, 3, 1]*[11, 9, 7] + [3, 7]/[4, 3, 2]); +(%o1) [59, 17, 2, 1, 1, 1, 27] +(%i2) cf ((3/17)*[1, 2, 5]/sqrt(11) + (8/13)); +(%o2) [0, 1, 1, 1, 3, 2, 1, 4, 1, 9, 1, 9, 2] +@end example + +@item +@code{cflength} controls how many periods of the continued fraction +are computed for algebraic, irrational numbers. + +@example +(%i1) cflength: 1$ +(%i2) cf ((1 + sqrt(5))/2); +(%o2) [1, 1, 1, 1, 2] +(%i3) cflength: 2$ +(%i4) cf ((1 + sqrt(5))/2); +(%o4) [1, 1, 1, 1, 1, 1, 1, 2] +(%i5) cflength: 3$ +(%i6) cf ((1 + sqrt(5))/2); +(%o6) [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] +@end example + +@item +A continued fraction can be evaluated by evaluating the arithmetic representation +returned by @code{cfdisrep}. + +@example +(%i1) cflength: 3$ +(%i2) cfdisrep (cf (sqrt (3)))$ +(%i3) ev (%, numer); +(%o3) 1.731707317073171 +@end example + +@item +Maxima does not know about operations on continued fractions outside of @code{cf}. + +@example +(%i1) cf ([1,1,1,1,1,2] * 3); +(%o1) [4, 1, 5, 2] +(%i2) cf ([1,1,1,1,1,2]) * 3; +(%o2) [3, 3, 3, 3, 3, 6] +@end example + +@end itemize +@end defun +@defun cfdisrep (list) +Constructs and returns an ordinary arithmetic expression +of the form @code{a + 1/(b + 1/(c + ...))} +from the list representation of a continued fraction @code{[a, b, c, ...]}. +@example +(%i1) cf ([1, 2, 3] + [1, 2, 1]); +(%o1) [1, 1, 1, 2] +(%i2) cfdisrep (%); + 1 +(%o2) 1 +  + 1 + 1 +  + 1 + 1 +  + 2 @end example + @end defun @defun cfexpand (x) gives a matrix of the numerators and denominators of the nexttolast and last convergents of the continued fraction x. @... (%i1) CF(SQRT(3)); (%o1) [1, 1, 2, 1, 2, 1, 2, 1] (%i2) CFEXPAND(%);  [71 97] (%o2) [ ]  [41 56] (%i3) %o2[1,2]/%o2[2,2],NUMER; (%o3) 1.7321429 +Returns a matrix of the numerators and denominators of the +last (column 1) and nexttolast (column 2) convergents of the continued fraction @code{x}. +@example +(%i1) cf (rat (ev (%pi, numer))); +`rat' replaced 3.141592653589793 by 103993//33102 = 3.141592653011902 +(%o1) [3, 7, 15, 1, 292] +(%i2) cfexpand (%); + [ 103993 355 ] +(%o2) [ ] + [ 33102 113 ] +(%i3) %[1,1]/%[2,1], numer; +(%o3) 3.141592653011902 @end example + @end defun @defvar cflength  default: [1] controls the number of terms of the continued fraction the function CF will give, as the value CFLENGTH[1] times the +Default value: 1 + +@code{cflength} controls the number of terms of the continued +fraction the function @code{cf} will give, as the value @code{cflength} times the period. Thus the default is to give one period. +@example +(%i1) cflength: 1$ +(%i2) cf ((1 + sqrt(5))/2); +(%o2) [1, 1, 1, 1, 2] +(%i3) cflength: 2$ +(%i4) cf ((1 + sqrt(5))/2); +(%o4) [1, 1, 1, 1, 1, 1, 1, 2] +(%i5) cflength: 3$ +(%i6) cf ((1 + sqrt(5))/2); +(%o6) [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] +@end example + @end defvar @defun divsum (n, k) 