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zerobern is false.
See also burn.
(%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
load ("bffac") loads this function.
The Hurwitz zeta function is defined as
sum ((k+h)^-s, k, 0, inf)
load ("bffac") loads this function.
(x + y)!/(x! y!).
If x and y are integers, then the numerical value of the binomial
coefficient is computed.
If y, or x - y, is an integer,
the binomial coefficient is expressed as a polynomial.
burn may be more efficient than bern for large, isolated n
(perhaps n greater than 105 or so),
as bern computes all the Bernoulli numbers up to index n before returning.
burn exploits the observation that (rational) Bernoulli numbers can be
approximated by (transcendental) zetas with tolerable efficiency.
load ("bffac") loads this function.
cf.
cf evaluates its arguments after binding listarith to false.
cf returns a continued fraction, represented as a list.
A continued fraction a + 1/(b + 1/(c + ...))
is represented by the list [a, b, c, ...].
The list elements a, b, c, ... must evaluate to integers.
expr may also contain sqrt (n) where n is an integer.
In this case cf will give as many
terms of the continued fraction as the value of the variable
cflength times the period.
A continued fraction can be evaluated to a number
by evaluating the arithmetic representation
returned by cfdisrep.
See also cfexpand for another way to evaluate a continued fraction.
See also cfdisrep, cfexpand, and cflength.
Examples:
(%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]
cflength controls how many periods of the continued fraction
are computed for algebraic, irrational numbers.
(%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]
cfdisrep.
(%i1) cflength: 3$ (%i2) cfdisrep (cf (sqrt (3)))$ (%i3) ev (%, numer); (%o3) 1.731707317073171
cf.
(%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]
a + 1/(b + 1/(c + ...))
from the list representation of a continued fraction [a, b, c, ...].
(%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 (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
cflength controls the number of terms of the continued
fraction the function cf will give, as the value cflength times the
period. Thus the default is to give one period.
(%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]
divsum (n, k) returns the sum of the divisors of n
raised to the k'th power.
divsum (n) returns the sum of the divisors of n.
(%i1) divsum (12); (%o1) 28 (%i2) 1 + 2 + 3 + 4 + 6 + 12; (%o2) 28 (%i3) divsum (12, 2); (%o3) 210 (%i4) 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 12^2; (%o4) 210
For the Euler-Mascheroni constant, see %gamma.
(%i1) map (euler, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]); (%o1) [1, 0, - 1, 0, 5, 0, - 61, 0, 1385, 0, - 50521]
factorial (x) the same as x!.
See !.
fib(0) equal to 0 and fib(1) equal to 1,
and
fib (-n) equal to (-1)^(n + 1) * fib(n).
After calling fib,
prevfib is equal to fib (x - 1),
the Fibonacci number preceding the last one computed.
(%i1) map (fib, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]); (%o1) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
%phi,
which is (1 + sqrt(5))/2, approximately 1.61803399.
By default, Maxima does not know about %phi.
After executing tellrat (%phi^2 - %phi - 1) and algebraic: true,
ratsimp can simplify some expressions containing %phi.
(%i1) fibtophi (fib (n));
n n
%phi - (1 - %phi)
(%o1) -------------------
2 %phi - 1
(%i2) fib (n-1) + fib (n) - fib (n+1);
(%o2) - fib(n + 1) + fib(n) + fib(n - 1)
(%i3) ratsimp (fibtophi (%));
(%o3) 0
(%i1) l: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]$ (%i2) map (lambda ([a], inrt (10^a, 3)), l); (%o2) [2, 4, 10, 21, 46, 100, 215, 464, 1000, 2154, 4641, 10000]
(%i1) l: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]$ (%i2) map (lambda ([a], jacobi (a, 9)), l); (%o2) [1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0]
load ("functs") loads this function.
minfactorial then turns one into a polynomial times the other.
(%i1) n!/(n+2)!;
n!
(%o1) --------
(n + 2)!
(%i2) minfactorial (%);
1
(%o2) ---------------
(n + 1) (n + 2)
partfrac does a complete
partial fraction decomposition. The algorithm employed is based on
the fact that the denominators of the partial fraction expansion (the
factors of the original denominator) are relatively prime. The
numerators can be written as linear combinations of denominators, and
the expansion falls out.
(%i1) 1/(1+x)^2 - 2/(1+x) + 2/(2+x);
2 2 1
(%o1) ----- - ----- + --------
x + 2 x + 1 2
(x + 1)
(%i2) ratsimp (%);
x
(%o2) - -------------------
3 2
x + 4 x + 5 x + 2
(%i3) partfrac (%, x);
2 2 1
(%o3) ----- - ----- + --------
x + 2 x + 1 2
(x + 1)
true if n is a prime, false if not.
sqrt (n) where n is an integer,
i.e., the element whose norm is unity.
This amounts to solving Pell's equation a^2 - n b^2 = 1.
(%i1) qunit (17); (%o1) sqrt(17) + 4 (%i2) expand (% * (sqrt(17) - 4)); (%o2) 1
true
When zerobern is false,
bern excludes the Bernoulli numbers which are equal to zero.
See bern.
zeta (n) for all other arguments,
including rational noninteger, floating point, and complex arguments.
See also bfzeta and zeta%pi.
(%i1) map (zeta, [-4, -3, -2, -1, 0, 1, 2, 3, 4, 5]);
2 4
1 1 1 %pi %pi
(%o1) [0, ---, 0, - --, - -, inf, ----, zeta(3), ----, zeta(5)]
120 12 2 6 90
true
When zeta%pi is true, zeta returns an expression
proportional to %pi^n for even integer n.
Otherwise, zeta returns a noun form zeta (n)
for even integer n.
(%i1) zeta%pi: true$
(%i2) zeta (4);
4
%pi
(%o2) ----
90
(%i3) zeta%pi: false$
(%i4) zeta (4);
(%o4) zeta(4)
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