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Euler's Gamma function
*A Gamma function
Euler's Gamma function $Gamma(z)$ is defined for complex $z$ such that
$Re(z)>0$ by the integral
$$Gamma(z):=Integrate(z, 0, Infinity) Exp(t)*t^(z1)$$.
The Gamma function satisfies several identities that can be proved by
rearranging this integral; for example, $Gamma(z+1)=z*Gamma(z)$. This identity
defines $Gamma(z)$ for all complex $z$. The Gamma function is regular
everywhere except nonpositive integers ($0$, $1$, $2$, ...) where it diverges.
Special arguments
For real integers $n>0$, the Gamma function is the same as the factorial,
$$ Gamma(n+1) := n! $$,
so the factorial notation can be used for the Gamma function too. Some formulae
become a little simpler when written in factorials.
*A Gamma function!halfinteger arguments
The Gamma function is implemented as {Gamma(x)}. At integer values {n} of the
argument, {Gamma(n)} is computed exactly.
Because of overflow, it only makes sense to compute exact integer factorials for small numbers $n$.
Currently a warning message is printed if a factorial of $n>65535$ is requested.
For halfinteger arguments $Gamma(x)$ is
also computed exactly, using the following identities (here $n$ is a
nonnegative integer and we use the factorial notation):
$$ (+(2*n+1)/2)! = Sqrt(Pi)*(2*n+1)! / (2^(2*n+1)*n!) $$,
$$ ((2*n+1)/2)! = (1)^n*Sqrt(Pi) * (2^(2*n)*n!) / (2*n)! $$.
For efficiency, "double factorials" are used in this calculation.
The "double factorial" is defined as
$n!! = n*(n2)*...$ and satisfies the identities
$$ (2*n1)!! = (2*n)! /(2^n*n!) $$,
$$ (2*n)!! = 2^n*n! $$.
For convenience, one defines $0! = 1$, $0!! = 1$, $(1)!! = 1$.
If the factorial of a large integer or halfinteger $n$ needs to be computed not exactly but only with a certain floatingpoint precision, it is faster (for large enough $Abs(n)$) not to evaluate an exact integer product, but to use the floatingpoint numerical approximation.
This method is currently not implemented in Yacas.
*A Stirling's formula
There is also the famous Stirling's asymptotic formula for large factorials,
$$Ln(n!) <=> Ln(2*Pi*n)/2+n*Ln(n/e)+1/(12*n)1/(360*n^3)+...$$
An analogous formula for double factorials can be easily found.
Method 0: power series (rational arguments)
*A Gamma function!rational arguments
For "small" rational arguments, i.e. for numbers $s=p/q$ where $p$, $q$ are small integers, there is a faster method for computing $Gamma(s)$.
This method was used in [Smith 2001].
[Haible <i>et al.</i> 1998] give this method in conjunction with the binary splitting technique.
Repeated partial integration gives the expansion
$$ Gamma(s) = M^s*Exp(M)*Sum(n,0,Infinity,M^n/(s*(s+1)*...*(s+n))) $$
$$ + (Integrate(u,M,Infinity) u^(s1)*Exp(u)) $$.
Suppose that $1<=s<=2$.
Then the remainder is given by the last integral that is smaller than $M*Exp(M)$.
By choosing $M$ a large enough integer, the remainder can be made small enough to discard it.
Then we obtain a series of rational numbers that can be evaluated directly in $O(P^2)$ time or using the binary splitting technique in $O(M(P*Ln(P))*Ln(P))$ time.
This method is currently not implemented in Yacas.
Method 1: the LanczosSpouge formula
*A Gamma function!by LanczosSpouge method
For arbitrary complex arguments with nonnegative real part, the
function {Internal'GammaNum(x)} computes a uniform appoximation of Lanczos [Lanczos 1964] modified by Spouge [Spouge 1994].
(See also [Godfrey 2001] for some more explanations.) This function is also triggered when in numeric mode, eg. when
calling {N(Gamma(x))}, which is the preferred method for end users.
The method gives the $Gamma$function only for arguments with positive
real part; at negative values of the real part of the argument, the
$Gamma$function is computed via the identity
$$Gamma(x)*Gamma(1x) = Pi/Sin(Pi*x)$$.
The LanczosSpouge approximation formula depends on a parameter $a$,
$$Gamma(z) = (Sqrt(2*Pi)*(z+a)^(z+1/2)) / (z*e^(z+a)) * (1+e^(a1)/Sqrt(2*Pi) * Sum(k, 1, N, c[k]/(z+k)))$$,
with $N := Ceil(a)1$. The coefficients $c[k]$ are defined by
$$ c[k] = (1)^(k1)*(ak)^(k1/2)/(e^(k1)*(k1)!) $$.
The parameter $a$ is a free parameter of the approximation that determines also
the number of terms in the sum. Some choices of $a$ may lead to a slightly more
precise approximation, but larger $a$ is always better. The number of terms $N$
must be large enough to produce the required precision. The estimate of the relative error for
this formula is valid for all $z$ such that $Re(z)>0$ and is
$$error < (2*Pi)^(a)/Sqrt(2*Pi*a)*(a/(a+z))$$.
The lowest value of $a$ to produce $P$ correct digits is estimated as
$$a = (PLn(P)/Ln(10)) * (Ln(10)/Ln(2*Pi))1/2$$. In practical calculations, the integer logarithm routine {IntLog} is used and the constant $Ln(10)/Ln(2*Pi)$ is approximated from above by 659/526, so that $a$ is not underestimated.
The coefficients $c[k]$ and the parameter $a$ can be chosen to
achieve a greater precision of the approximation formula. However, the recipe
for the coefficients $c[k]$ given in the paper by Lanczos is too complicated
for practical calculations in arbitrary precision: the time it would take to
compute the array of $N$ coefficients $c[k]$ grows as $N^3$. Therefore it is
better to use less precise but much simpler formulae derived by Spouge.
Roundoff error in the Lanczos method
In the calculation of the sum $S:=Sum(k, 1, N, c[k]*(z+k)^(1))$, a certain roundoff error results from the changing signs in $c[k]$, and from the fact that some values $c[k]$ are much larger than the final value of the sum.
This leads to some cancellations and as a result to a certain loss of precision.
At version
*EVAL Version() : ","
{Yacas} is limited in its internal arbitrary precision facility that does not support
true floatingpoint computation but rather uses fixedpoint logic; this hinders precise calculations with
floatingpoint numbers. In the current version of the {Internal'GammaNum()} function, two workarounds
are implemented. First, a Horner scheme is used to compute the sum; this is somewhat
faster and leads to smaller roundoff errors. Second, intermediate calculations
are performed at 40{%} higher precision than requested. This is much slower but
allows to obtain results at desired precision.
If strict floatingpoint logic is used, the working precision necessary to compensate for the cancellations must be $1.1515*P$ digits for $P$ digits of the result.
This can be shown as follows.
The sum converges to a certain value $S$ which is related to the correct value of the Gamma function at $z$.
After some algebra we find that $S$ is of order $Sqrt(a)$ if $z>a$ and of order $a^(1/2z)$ if $a>z$.
Since $a$ is never a very large number, we can consider the value of $S$ to be roughly of order $1$, compared with exponentially large values of some of the terms $c[k]$ of this sum.
The magnitude of a coefficient $c[k]$ is estimated by Stirling's formula,
$$ Ln(Abs(c[k]))<=> (k1)*Ln((ak)/(k1)) $$.
(Here we have put approximately $k1<=>k1/2$ since $k$ is not small.)
This function has a maximum at $k<=>1+0.2178*(a1)$.
Then the largest magnitude of a coefficient $c[k]$ at this $k$ is approximately $Exp(0.2785*(a1))$.
The constant $0.2785=W(1/e)$ is the solution of $x+Ln(x)+1=0$.
Here $x=(k1)/(ak)$ and $W$ is Lambert's function.
Now, $a1<=>P*Ln(10)/Ln(2*Pi)$, so the maximum of $c[k]$ is $10^(0.1515*P)$.
Therefore we have a certain cancellation in the sum $S$: adding and subtracting some numbers of order $10^(0.1515*P)$ produces an answer of order $1$.
Therefore we need to have at least 15% more decimal digits to obtain $P$ digits of the final answer.
(The constant $W(1/e)/Ln(2*Pi)<=>0.1515$ is independent of the base 10.)
Other methods for the Gamma function
More traditional ways of calculating the Gamma function are the Stirling asymptotic series and the SweeneyBrent method of combined series.
Method 2: Stirling's asymptotic series
*A Gamma function!by asymptotic series
The Stirling asymptotic series for the Gamma function is
$$ Ln(Gamma(x)) <> (x1/2)*Ln(x)x+1/2*Ln(2*Pi) $$
$$ + Sum(n,1,Infinity, B[2*n]/(2*n*(2*n1)*x^(2*n1))) $$.
This series is valid asymptotically for large $Abs(x)$, also for complex values of $x$ (but excluding the negative real values).
Computation of this series up to $n=N$ requires finding the Bernoulli numbers $B[n]$ up to $N$.
For a given (large) value of $Abs(x)$, the terms of this series decrease at first, but then start to grow.
(Here $x$ can be a complex number.)
There exist estimates for the error term of the asymptotic series (see [Abramowitz <i>et al.</i> 1964], 6.1.42).
Roughly, the error is of the order of the first discarded term.
We can estimate the magnitude of the terms using the asymptotic formula for the Bernoulli numbers (see below).
After some algebra, we find that the value of $n$ at which the series starts to grow and diverge is
$n[0]<=>Pi*Abs(x)+2$.
Therefore at given $x$ we can only use the asymptotic series up to the $n[0]$th term.
For example, if we take $x=10$, then we find that the 32nd term of the asymptotic series has the smallest magnitude (about $10^(28)$) but the following terms start to grow.
To be on the safe side, we should drop a few more terms from the series.
Define the number of terms by $n[0]:=Pi*Abs(x)$.
Then the order of magnitude of the $n[0]$th term is $Exp(2*Pi*Abs(x))/(2*Pi^2*Abs(x))$.
This should be compared with the magnitude of the sum of the series which is of order $Abs(x*Ln(x))$.
We find that the relative precision of $P$ decimal digits or better is achieved if
$$ (2*Pi1)*x+(x+1)*Ln(x)+3.9 > P*Ln(10) $$.
If ($x$,$P$) do not satisfy this inequality, the asymptotic method does not provide enough precision.
For example, with $x=10$ we obtain $P<=35$.
So the asymptotic series gives the value of $Gamma(10)$ with not more than 35 decimal digits.
For very large $P$, the inequality is satisfied when roughly $x>P*Ln(10)/Ln(P)$.
Assuming that the Bernoulli numbers are precomputed, the complexity of this method is that of computing a Taylor series with $n[0]$ terms, which is roughly $O(Sqrt(P))*M(P)$.
What if $x$ is not large enough? Using the identity
$ x*Gamma(x) = Gamma(x+1) $,
we can reduce the computation of $Gamma(x)$ to $Gamma(x+M)$ for some integer $M$.
Then we can choose $M$ to be large enough so that the asymptotic series gives the required precision when evaluated at $x+M$.
We shall have to divide the result $M$ times by some long numbers to obtain $Gamma(x)$.
Therefore, the complexity of this method for given ($x$,$P$) is increased by
$M(P)*(P*Ln(10)/Ln(P)x)$.
For small $x$ this will be the dominant contribution to the complexity.
On the other hand, if the Bernoulli numbers are not available precomputed, then their calculation dominates the complexity of the algorithm.
Method 3: the SweeneyBrent trick
*A Gamma function!by SweeneyBrent method
The second method for the Gamma function $Gamma(x)$ was used in R. P. Brent's Fortran MP package [Brent 1978].
Brent refers to [Sweeney 1963] for the origin of this method.
Therefore, we called this method the "SweeneyBrent" method.
This method works well when $1<=x<2$ (other values of $x$ need to be reduced first).
The idea is to represent the Gamma function as a sum of two integrals,
$$ Gamma(x) = (Integrate(u,0,M) u^(x1)*Exp(u)) + (Integrate(u,M,Infinity) u^(x1)*Exp(u)) $$.
The above identity clearly holds for any $M$.
Both integrals in this equation can be approximated by power series (although we may do without the second integral altogether, for a small increase of computation time).
The parameter $M$ and the numbers of terms in the series must be chosen appropriately, as we shall see below.
The first integral in this equation can be found as a sum of the Taylor series (expanding $Exp(u)$ near $u=0$),
$$ (Integrate(u,0,M) u^(x1)*Exp(u)) = M^x*Sum(n,0,Infinity, ((1)^n*M^n)/((n+x)*n!)) $$.
This series absolutely converges for any finite $M$.
However, the series has alternating signs and incurs a certain roundoff error.
The second integral is smaller than $M^(x1)*Exp(M)$ and $M$ can be chosen large enough so that this is smaller than $10^(P)$.
This condition gives approximately $M>P*Ln(10)+Ln(P*Ln(10))$.
Now we can estimate the number of terms in the above series.
We know that the value of the Gamma function is of order $1$.
The condition that $n$th term of the series is smaller than $10^(P)$ gives $n*Ln(n/e*M)>P*Ln(10)$.
With the above value for $M$, we obtain $n=P*Ln(10)/W(1/e)$ where $W$ is Lambert's function; $W(1/e)<=>0.2785$.
The terms of the series are however not monotonic: first the terms grow and then they start to decrease, like in the Taylor series for the exponential function evaluated at a large argument.
The ratio of the ($k+1$)th term to the $k$th term is approximately $M/(k+1)$.
Therefore the terms with $k<=>M$ will be the largest and will have the magnitude of order $M^M/M! <=>Exp(M)<=>10^P$.
In other words, we will be adding and subtracting large numbers with $P$ digits before the decimal point, but we need to obtain a result with $P$ digits after the decimal point.
Therefore to avoid the roundoff error we need to increase the working precision to $2*P$ floatingpoint decimal digits.
It is quicker to compute this series if $x$ is a small rational number, because then the long multiplications can be avoided, or at high enough precision the binary splitting can be used.
Calculations are also somewhat faster if $M$ is chosen as an integer value.
If the second integral is approximated by an asymptotic series instead of a constant $Exp(M)$,
then it turns out that the smallest error of the series is $Exp(2*M)$.
Therefore we can choose a smaller value of $M$ and the roundoff error gets somewhat smaller.
According to [Brent 1978], we then need only $3/2*P$ digits of working precision, rather than $2*P$, for computing the first series
(and only $P/2$ digits for computing the second series).
However, this computational savings may not be significant enough to justify computing a second series.
Euler's constant $gamma$
*A Euler's constant $gamma$
Euler's constant $gamma$ is defined as
$$ gamma := Limit(n,Infinity) (Sum(k,1,n,1/k)Ln(n)) $$.
This constant is useful for various reasons, mostly when working with higher transcendental functions.
For example, $gamma$ is needed to obtain a Taylor series expansion of $Gamma(x)$.
Another useful relation is
$$ (D(x)Gamma(x)[x=1]) = gamma $$.
Approximately $gamma<=>0.577216$.
Method 1: Brent's decomposition
Computing $gamma$ by the series in the definition is extremely slow.
A much faster method can be used, based on some identities of Bessel functions.
(See [Brent <i>et al.</i> 1980] and [Gourdon <i>et al.</i> 2001].)
The basic formulae for the "fast" method (Brent's method "B1") are:
$$gamma <=> S(n) / V(n)  Ln(n) $$,
where $S(n)$ and $V(n)$ are some auxiliary functions, and $n$ is chosen to be high enough (precise estimates are given below).
First, the sequence $H[n]$ is defined as the partial sum of the harmonic series:
$$ H[n] := Sum(k,1,n,1/k) $$.
We also define $H[0]:=0$ for convenience.
The function $V(n)$ is the modified Bessel function $I0(2*n)$. It is computed as a Taylor series
$$ V(n) := Sum(k,0,Infinity, (n^k/k!)^2) $$.
The function $S(n)$ is defined by a series like $V(n)$ but with each term multiplied by $H[k]$:
$$ S(n) := Sum(k,0,Infinity, (n^k/k!)^2*H[k]) $$.
Note that we need to compute $S(n)$ and $V(n)$ with enough precision, so the sum over $k$ will have to be performed up to a large enough $k$.
(In practice, we do not really need to know the limit $k[max]$ beforehand.
Instead, we can simply add terms to the series for $V(n)$ and $S(n)$ until desired precision is achieved.
Knowing $k[max]$ in advance would help only if we needed to compare this method for computing $gamma$ with some other method, or if we would use the rectangular method of evaluating the Taylor series.)
According to [Brent <i>et al.</i> 1980], the error of this approximation of $gamma$, assuming that $S(n)$ and $V(n)$ are computed exactly, is
$$ Abs(gammaS(n)/V(n)) < Pi*Exp(4*n) $$.
Therefore the parameter $n$ is proportional to the number of digits we need.
If we need $P$ decimal digits (of absolute, not relative, precision), then we have to choose
$$ n>(P*Ln(10)+Ln(Pi))/4$$.
The required number of terms $k[max]$ in the summation over $k$ to get $S(n)$ and $V(n)$ with this precision can be approximated as usual via Stirling's formula.
It turns out that $k[max]$ is also proportional to the number of digits,
$k[max]<=>2.07*P$.
Therefore, this method of computing $gamma$ has "linear convergence", i.e. the number of iterations is linear in the number of correct digits we need in the result.
Of course, all calculations need to be performed with the working precision.
The working precision must be a few digits more than $P$ because we accumulate about $Ln(k[max])/Ln(10)$ digits of roundoff error by performing $k[max]$ arithmetic operations.
Brent mentions a small improvement on this method (his method "B3").
It consists of estimating the error of the approximation of $gamma$ by an asymptotic series.
Denote $W(n)$ the function
$$ W(n):=(1/(4*n))*Sum(k,0,2*n, (2*k)! ^3/((k!)^4)*(16*n)^(2*k)) $$.
This function is basically the asymptotic series for $I0(2*n)*K0(2*n)$, where $I0$ and $K0$ are modified Bessel functions.
The sum in this series has to be evaluated until about $k=2*n$ to get a good precision.
Then a more precise approximation for $gamma$ is
$$ gamma <=> U(n)/V(n)W(n)/V(n)^2 $$.
The precision of this approximation is of order $O(Exp(8*n))$ instead of $O(Exp(4*n))$ in Brent's method "B1".
However, this is not such a great savings in time, because almost as much additional work has to be done to compute $W(n)$.
Brent estimated that the method "B3" is about 20{%} faster than "B1".
Method 2: Brent's summation trick
Computation of $S(n)$ seems to need a running computation of $H[k]$ and a long multiplication by $H[k]$ at each term.
To compute $S(n)$ and $V(n)$ faster and more accurately, Brent suggested the following trick that avoids this long multiplication and computes $H[k]$ simultaneously with the series.
Define the function $U(n):=S(n)V(n)*Ln(n)$.
Then $gamma<=>U(n)/V(n)$.
The series for $U(n)$ is
$ U(n) := Sum(k,0,Infinity, A[k]) $,
with
$$ A[k] := (n^k/k!)^2*(H[k]Ln(n)) $$.
If we denote
$$ B[k] := (n^k/k!)^2 $$
the $k$th term of the series for $V(n)$, then we can compute $A[k]$ and $B[k]$ simultaneously using the recurrence relations
$A[0]= Ln(n)$, $B[0]=1$,
$$ B[k]=B[k1]*n^2/k^2 $$,
$$ A[k]=1/k*(A[k1]*n^2/k+B[k]) $$.
This trick can be formulated for any sequence $A[k]$ of the form $A[k]=B[k]*C[k]$, where the sequences $B[k]$ and $C[k]$ are given by the recurrences
$ B[k]=p(k)*B[k1]$ and $C[k]=q(k)+C[k1]$.
Here we assume that $p(k)$ and $q(k)$ are known functions of $k$ that can be computed to $P$ digits using $O(P)$ operations, e.g. rational functions with short constant coefficients.
Instead of evaluating $B[k]$ and $C[k]$ separately and multiplying them using a long multiplication, we note that $p(k)*A[k1] = B[k]*C[k1]$.
This allows to compute $A[k]$ by using the following two recurrences:
$$ B[k] = p(k)*B[k1] $$,
$$ A[k]=p(k)*A[k1]+q(k)*B[k] $$.
All multiplications and divisions in these recurrence relations are performed with short integers, so the long multiplications are avoided.
The time complexity of this method is $O(P^2)$ where $P$ is the required number of digits.
A variant of binary splitting method can be used to reduce the complexity to $O(M(P*Ln(P))*Ln(P))$ which gives an asymptotic advantage at very high precision.
Also, it turns out that we can use a variant of the fast "rectangular method" to evaluate the series for $U(n)$ and $V(n)$ simultaneously.
(We can consider these series as Taylor series in $n^2$.)
This however does not speed up the evaluation of $gamma$.
This happens because the rectangular method requires long multiplications and leads in this case to increased roundoff errors.
The rectangular method for computing a power series in $x$ is less efficient than a straightforward computation when $x$ is a "short" rational or integer number.
The "rectangular method" for computing $Sum(k,0,N,x^k*A[k])$ needs to be able to convert a coefficient of the Taylor series into the next coefficient $A[k+1]$ by "short" operations, more precisely, by some multiplications and divisions by integers of order $k$.
The $j$th column of the rectangle ($j=0$, 1, ...) consists of numbers $x^(r*j)*A[r*j]$, $x^(r*j)*A[r*j+1]$, ..., $x^r*A[r*j+r1]$.
The numbers of this column are computed sequentially by short operations, starting from the $x^(r*j)*A[j*r]$ which is known from the end of the previous column.
The recurrence relation for $A[k]$ is not just some multiplication by rational numbers, but also contains an addition of $B[k]$.
However, if we also use the rectangular method for $V(n)$, the number $x^(r*j)*B[r*j]$ will be known and so we will be able to use the recurrence relation to get $x^(r*j)*A[r*j+1]$ and all following numbers of the column.
Derivation of method 1
Brent's "B1" method can be derived from the Taylor series for the modified Bessel function $BesselI(nu,z)$,
$$ BesselI(nu,z) = Sum(k,0,Infinity, z^(nu+2*k)/(2^(nu+2*k)*Gamma(nu+k+1)*k!)) $$,
and the definition of the modified Bessel function $K0(z)$,
$$ K0(z) :=  D(nu)BesselI(nu,z)[nu=0]$$.
Here the derivative wrt $nu$ is taken at $nu=0$.
This derivative can be evaluated using the above Taylor series for $BesselI(nu,z)$ and expressed through the series for $S(n)$.
To compute this, we need the derivative of the Gamma function at integer arguments $n$:
$$ (D(x) Gamma(x)[x=n+1]) = n! *(H[n]gamma) $$.
The resulting identity in the way it is used here is
$$ gamma+Ln(n) = (S0(2*n)K0(2*n))/I0(2*n) $$.
Since $K0(2*n)$ decays very quickly at large $n$, we obtain the approximation
$$ gamma <=> S0(2*n)/I0(2*n)Ln(n)+O(Exp(4*n)) $$.
Catalan's constant $G$
Catalan's constant $G$ is defined by
$$ G = Sum(n,0,Infinity,(1)^n/(2*n+1)^2) $$.
This series converges very slowly.
There are several alternative methods.
Best method so far
A fast series to evaluate this constant is:
$$ G = Pi/8*Ln(2+Sqrt(3)) + 3/8*Sum(n,0,Infinity,(n!)^2/((2*n)! *(2*n+1)^2))$$
[Bailey <i>et al.</i> 1997].
To obtain $P$ decimal digits of relative precision, we need to take at most $P*Ln(10)/Ln(4)$ terms of the series.
The sum can be efficiently evaluated using Horner's scheme, for example
$$ 1+1/(2*3)*(1/3+2/(2*5)*(1/5+3/(2*7)*(1/7+ ...) ) ) $$.
This takes $O(P^2)$ operations because there are $O(P)$ short multiplications and divisions of $P$digit numbers by small integers.
At very high precision, the binary splitting technique can be used instead of Horner's scheme to further reduce the computation time to $O(M(P)*Ln(P)^2)$.
A drawback of this scheme is that it requires a separate highprecision computation of $Pi$, $Sqrt(3)$ and of the logarithm.
Other methods
Another method is based on an identity by Ramanujan.
The formula is
$$ G = Sum(k,0,Infinity,(k!)^2/(2*k+1)! *2^(k1)*Sum(j,0,k,1/(2*j+1))) $$.
The $k$th term of the outer sum is of order $2^(k)$ and so we need to take slightly more than $P*Ln(10)/Ln(2)$ terms for $P$ decimal digits.
But each term is more complicated than in the first method's series.
The running time of this formula seems to be a few times slower than the previous method.
This method combined with Brent's summation trick (see the section on the Euler constant) was used in [Fee 1990].
Brent's trick allows to avoid a separate computation of the harmonic sum and all long multiplications.
Catalan's constant is obtained as a limit of $G[k]$ where $G[0]=B[0]=1/2$ and
$$ B[k]=k/(2*k+1)*B[k1] $$,
$$ G[k]=1/(2*k+1)*(k*G[k1]+B[k]) $$.
The running time is $O(P^2)$.
Since only rational numbers are involved, the binary splitting technique can be used at high precision.
A third formula is more complicated but the convergence is much faster and there is no need to evaluate any other transcendental functions.
This formula is called "Broadhurst's series".
$$ G = 3/2*Sum(k,0,Infinity, 1/16^k*Sum(i,0,7,a[i]/(8*k+i)^2))  1/4*Sum(k,0,Infinity, 1/4096^k*Sum(i,0,7,a[i]/(8*k+i)^2)) $$.
Here the arrays of coefficients are defined as $a[i]=[0;1;1;1/2;0;1/4;1/4;1/8;]$ and $b[i]=[0;1;1/2;1/8;0;1/64;1/128;1/512;]$.
We need to take only $P*Ln(10)/Ln(16)$ terms of the first series and $P*Ln(10)/Ln(4096)$ terms of the second series.
However, each term is about six times more complicated than one term of the first method's series.
So there are no computational savings (unless $Ln(x)$ is excessively slow).
Riemann's Zeta function
*A Riemann's Zeta function
*A Bernoulli numbers
Riemann's Zeta function $Zeta(s)$ is defined for complex $s$ such that $Re(s)>0$ as a sum of inverse powers of integers:
$$Zeta(s) := Sum(n,0,Infinity, 1/n^s)$$.
This function can be analytically continued to the entire complex plane except
the point $s=1$ where it diverges. It satisfies several identities, for
example, a formula useful for negative $Re(s)$,
$$Zeta(1s) = (2*Gamma(s))/(2*Pi)^s*Cos((Pi*s)/2)*Zeta(s)$$,
and a formula for even integers that helps in numerical testing,
$$Zeta(2*n) = ((1)^(n+1)*(2*Pi)^(2*n))/(2*(2*n)!)*B[2*n]$$,
where $B[n]$ are Bernoulli numbers.
(The values at negative integer $n$ are given by $Zeta(n)= B[n+1]/(n+1)$.)
The classic book [Bateman <i>et al.</i> 1953], vol. 1, describes many results concerning the properties of $Zeta(s)$.
For the numerical evaluation of Riemann's Zeta function with arbitrary
precision to become feasible, one needs special algorithms. Recently P. Borwein [Borwein 1995]
gave a simple and quick approximation algorithm for $Re(s)>0$. See
also [Borwein <i>et al.</i> 1999] for a review of methods.
*A Riemann's Zeta function!by Borwein's algorithm
It is the "third" algorithm (the simplest one) from P. Borwein's paper which is implemented in
Yacas. The approximation formula valid for $Re(s) > (n1)$ is
$$ Zeta(s) = 1/(2^n*(12^(1s)))*Sum(j,0,2*n1, e[j]/(j+1)^s) $$,
where the coefficients $e[j]$ for $j=0$, ..., $2*n1$ are defined by
$$ e[j] := (1)^(j1)*(Sum(k,0,jn, n! /(k! *(nk)!))2^n) $$,
and the empty sum (for $j<n$) is taken to be zero. The parameter $n$ must be
chosen high enough to achieve the desired precision. The error estimate for this
formula is approximately
$$error < 8^(n)$$
for the relative precision, which means that to achieve $P$ correct digits we
must have $n>P*Ln(10)$/$Ln(8)$.
This method requires to compute $n$ times the exponential and the logarithm to find the power $(j+1)^(s)$.
This power can be computed in asymptotic time $O(M(P)*Ln(P))$, unless $s$ is an integer, in which case this computation is $O(M(P))$, the cost of one division by an integer $(j+1)^s$.
Therefore the complexity of this method is at most $O(P*M(P)*Ln(P))$.
The function {Zeta(s)} calls {Internal'ZetaNum(s)} to compute this approximation formula
for $Re(s)>1/2$ and uses the identity above to get the value for other $s$.
*A Riemann's Zeta function!by direct summation
For very large values of $s$, it is faster to use more direct methods implemented in the routines {Internal'ZetaNum1(s,N)} and {Internal'ZetaNum2(s,N)}.
If the required precision is $P$ digits and
$s > 1 + Ln(10)/Ln(P)*P$,
then it is more efficient to compute the defining series for $Zeta(s)$,
$$Zeta(s) <=> Sum(k,1,N, 1/k^s)$$,
up to a certain number of terms $N$, than to use Borwein's method.
The required number of terms $N$ is given by $$N = 10^(P/(s1))$$.
This is implemented in the routine
{Internal'ZetaNum1(s)}.
For example, at 100 digits of precision it is advisable to use
{Internal'ZetaNum1(s)} only for $s>50$, since it would require $N<110$ terms in
the series, whereas the expression used in {Internal'ZetaNum(s)} uses
$n=Ln(10)/Ln(8)*P$ terms (of a different series).
The complexity of this method is $O(N)*M(P)$.
*A Riemann's Zeta function!by product over primes
Alternatively, one can use {Internal'ZetaNum2(n,N)} which computes the infinite product over prime numbers $p[i]$
$$1/Zeta(n) <=> Factorize(k,1,M,(11/p[k]^s))$$.
Here $M$ must be chosen such that the $M$th prime number $p[M]>N$.
To obtain $P$ digits of precision, $N$ must be chosen as above, i.e.
$N > 10^(P/(s1))$.
Since only prime numbers $p[i]$ are used, this formula is asymptotically faster than {Internal'ZetaNum1}.
(Prime numbers have to be obtained and tested but this is quick for numbers of size $n$, compared with other operations.)
The number of primes up to $N$ is asymptotically
$pi(N) <> N/Ln(N)$ and therefore this procedure is faster by a factor $O(Ln(N)) <> O(Ln(n))$. However, for $n<250$ it is better (with Yacas internal math) to use the {Internal'ZetaNum1} routine because it involves fewer multiplications.
Zeta function at special arguments
Recently some identities dating back to S. Ramanujan have been studied in relation to Riemann's Zeta function.
A side result was the development of fast methods for computing the Zeta function at odd integer arguments
(see [Borwein <i>et al.</i> 1999] for a review)
and at small rational arguments ([Kanemitsu <i>et al.</i> 2001]).
*A Riemann's Zeta function!integer arguments
The value $Zeta(3)$, also known as the Apery's constant, can be computed using the following geometrically convergent series:
$$ Zeta(3) = 5/2*Sum(k,1,Infinity, (1)^(k1)/(k^3*Bin(2*k,k))) $$.
For other odd integers $n$ there is no general analogous formula.
The corresponding expressions for $Zeta(5)$ and $Zeta(7)$ are
$$ Zeta(5) = 2*Sum(k,1,Infinity, (1)^(k1)/(k^5*Bin(2*k,k)))  5/2*Sum(k,1,Infinity, (1)^(k1)/(k^3*Bin(2*k,k))*Sum(j,1,k1,1/j^2)) $$;
$$ Zeta(7) = 5/2*Sum(k,1,Infinity, (1)^(k1)/(k^7*Bin(2*k,k))) + 25/2*Sum(k,1,Infinity, (1)^(k1)/(k^3*Bin(2*k,k))*Sum(j,1,k1,1/j^4)) $$.
In these series the term $Bin(2*k,k)$ grows approximately as $4^k$ and therefore one can take no more than $P*Ln(10)/Ln(4)$ terms in the series to get $P$ decimal digits of relative precision.
For odd integer $n$ there are the following special relations: for $n:=Mod(3,4)$,
$$ Zeta(n) = (2*Pi)^n/(2*(n+1)!)*Sum(k,0,(n+1)/2, (1)^(k1)*Bin(n+1,2*k)*B[n+12*k]*B[2*k]) $$
$$  2*Sum(k,1,Infinity,1/(k^n*(Exp(2*Pi*k)1))) $$,
and for $n:=Mod(1,4)$,
$$ Zeta(n) = (2*Pi)^n/((n+1)! *(n1)) $$
$$ ... *Sum(k,0,(n+1)/4,(1)^k*(n+14*k)*Bin(n+1,2*k)*B[n+12*k]*B[2*k]) $$
$$ 2*Sum(k,1,Infinity, (Exp(2*Pi*k)*(1+(4*Pi*k)/(k1))1) / (k^n*(Exp(2*Pi*k)1)^2)) $$.
These relations contain geometrically convergent series,
and it suffices to take $P*Ln(10)/(2*Pi)$ terms to obtain $P$ decimal digits of relative precision.
Finally, [Kanemitsu <i>et al.</i> 2001] gave a curious formula for the values of Riemann's Zeta function at rational values between 0 and 2 (their "Corollary 1").
This formula is very complicated but contains a geometrically convergent series.
We shall have to define several auxiliary functions to make the formula more understandable.
We shall be interested in the values of $Zeta(p/N)$ where $p$, $N$ are integers.
For integer $h$, $N$ such that $0<=h<=N$,
and for arbitrary real $x$,
$$ Zeta((2*h1)/N) = (N*x)/Pi*((2*Pi)/x)^((2*h1)/N) * Sin((Pi*(2*h1))/(2*N)) * L(x,N,h) $$,
$$ Zeta((N2*h+1)/N) = N/Gamma((N2*h+1)/N)*x^((N2*h+1)/N)*L(x,N,h) $$.
Here the function $L(x,N,h)$ (containing the "Lambert series") is defined by
$$ L(x,N,h) := Sum(n,1,Infinity, n^(N2*h)/(Exp(n^N*x)1))S(x,N,h)$$
$$+1/2*Zeta(N+2*h)Zeta(2*h)/x $$,
and the function $S(x,N,h)$ is defined by
$$ S(x,N,h) := (1)^(h+1)/N*((2*Pi)/x)^((N2*h+1)/N) $$
$$ ... * Sum(n,0,Infinity, 1/n^((2*h1)/N) * (Sum(k,1,N/2,f(2*k1,x,n,N,h)))) $$
for even $N$, and by
$$ S(x,N,h) := (1)^(h+1)/N*((2*Pi)/x)^((N2*h+1)/N) $$
$$ ... * Sum(n,0,Infinity, 1/n^((2*h1)/N) * (f(0,x,n,N) + Sum(k,1,(N1)/2,f(2*k,x,n,N,h)))) $$
for odd $N$.
The auxiliary function $f(j,x,n,N,h)$ is defined by
$$ f(0,x,n,N) := Exp(A)/(2*Sinh(A)) $$,
$$ f(j,x,n,N,h):=(Cos(2*B+C)  Exp(2*A)*Cos(C)) / (Cosh(2*A)Cos(2*B)) $$
for integer $j>=1$, and in these expressions
$$ A:=Pi*((2*Pi*n)/x)^(1/N)*Cos((Pi*j)/(2*N)) $$,
$$ B:=Pi*((2*Pi*n)/x)^(1/N)*Sin((Pi*j)/(2*N)) $$,
and $C:=(Pi*j)/2*(2*h1)/N$.
Practical calculations using this formula are of the same asymptotic complexity as Borwein's method above. (It is not clear whether this method has a significant computational advantage.)
The value of $x$ can be chosen at will, so we should find such $x$ as to minimize the cost of computation.
There are two series to be computed: the terms in the first one decay as $Exp(n^N*x)$ while the terms in the second one (containing $f$) decay only as
$$ Exp(2*A)<>Exp(2*Pi*((2*Pi*n)/x)^(1/N)) $$.
Also, there are about $N$ copies of the second series.
Clearly, a very small value should be chosen for $x$ so that the second series converges somewhat faster.
For a target precision of $P$ decimal digits, the required numbers of terms $n[1]$, $n[2]$ for the first and the second series can be estimated as
$ n[1]<=>((P*Ln(10))/x)^(1/N) $,
$ n[2]<=>x/(2*Pi)*((P*Ln(10))/(2*Pi))^N $.
(Here we assume that $N$, the denominator of the fraction $p/N$, is at least $10$.
This scheme is impractical for very large $N$ because it requires to add $O(N)$ slightly different variants of the second series.)
The total cost is proportional to the time it takes to compute $Exp(x)$ or $Cos(x)$ and to roughly $n[1]+N*n[2]$.
The value of $x$ that minimizes this cost function is approximately
$$ x<=>(2*Pi)/N^2*((2*Pi)/(P*Ln(10)))^(N1) $$.
With this choice, the asymptotic total cost is $O(P*M(P))$.
Lambert's $W$ function
Lambert's $W$ function is (a multiplevalued, complex function) defined for any (complex) $z$ by
$ W(z) * Exp(W(z)) = z$.
This function is sometimes useful to represent solutions of transcendental equations or integrals.
*A Lambert's $W$ function!asymptotics
Asymptotics of Lambert's $W$ function are
$$W((z^22)/2/e) = 1 + z  z^3/3 + 11/72*z^4  ...$$ (this is only good really close to z=0);
$W(x) = xx^2+3/2*x^3...$ (this is good very near x=0);
$$W(x) = Ln(x)  Ln(Ln(x)) + Ln(Ln(x))/Ln(x) + 1/2*(Ln(Ln(x))/Ln(x))^2 + ...$$ (good for very large $x$ but it is not straightforward to find higher terms!).
Here are some inequalities to help estimate $W(x)$ at large $x$ (more exactly, for $x>e$):
$$ Ln(x/Ln(x))<W(x)<Ln(x) $$,
$$ Ln(x/(Ln(x)1))<W(x)<Ln(x/(Ln(x)1))+0.13$$.
*A Lambert's $W$ function!uniform approximations
One can also find uniform rational approximations, e.g.:
$$W(x) <=> Ln(1+x)*(1Ln(1+Ln(1+x))/(2+Ln(1+x)))$$ (uniformly good for $x>0$, relative error not more than $10^(2)$); and
$$W(x) <=> (x*e) / (1+( (2*e*x+2)^(1/2) 1/Sqrt(2) +1/(e1))^(1) )$$
(uniformly good for $1/e<x<0$, relative error not more than $10^(3)$).
There exists a uniform approximation of the form
$$ W(x) <=> (a+L)*(fLn(1+c*Ln(1+d*Y))+2*L)/(a+1/2+L) $$,
where $a=2.3436$, $b=0.8842$, $c=0.9294$, $d=0.5106$, $f= 1.2133$ are constants, $Y:=Sqrt(2*e*x+2)$ and $L:=Ln(1+b*Y)$.
(The coefficients $a$, ..., $f$ are
found numerically by matching the asymptotics at three points
$x= 1/e$, $x=0$, $x=Infinity$.)
This approximation miraculously works over the whole complex plane within relative error at most $0.008$.
The behavior of this formula at the branching points $x= 1/e$ and $x=Infinity$ correctly mimics $W(x)$ if the branch cuts of the square root and of the logarithm are chosen appropriately (e.g. the common branch cut along the negative real semiaxis).
*A Lambert's $W$ function!by Halley's method
The numerical procedure uses Halley's method.
Halley's iteration for the equation $W*Exp(W)=x$ can be written as
$$ W'=W(Wx*Exp(W))/(W+1(W+2)/(W+1)*(Wx*Exp(W))/2) $$. It has cubic convergence for any initial value $W> Exp(1)$.
The initial value is computed using one of the uniform approximation formulae.
The good precision of the uniform approximation guarantees rapid convergence of the iteration scheme to the correct root of the equation, even for complex arguments $x$.
Bessel functions
*A Bessel functions
Bessel functions are a family of special functions solving the equation
$$ (Deriv(x,2)w(x))+1/x*(D(x)w(x))+(1n^2/x^2)*w(x)=0 $$. There are two linearly independent solutions which can be taken as the pair of Hankel functions $H1(n,x)$, $H2(n,x)$, or as the pair of BesselWeber functions $J[n]$, $Y[n]$.
These pairs are linearly related,
$J[n]=1/2*(H1(n,x)+H2(n,x))$, $J[n]=1/(2*I)*(H1(n,x)H2(n,x))$. The function $H2(n,x)$ is the complex conjugate of $H1(n,x)$. This arrangement of four functions is very similar to the relation between $Sin(x)$, $Cos(x)$ and $Exp(I*x)$, $Exp(I*x)$, which are all solutions of $(Deriv(x,2)f(x))+f(x)=0$.
*A Bessel functions!asymptotics
For large values of $Abs(x)$, there is the following asymptotic series:
$$ H1(n,x) <> Sqrt(2/(Pi*x))*Exp(I*zeta)*Sum(k,0,Infinity, I^k*A(k,n)/x^k) $$,
where
$zeta:=x1/2*n*Pi1/4*Pi$ and
$$ A(k, n) := ( (4*n^21^2)*(4*n^23^2)*...*(4*n^2(2*k1)^2))/(k! *8^k) $$.
From this one can find the asymptotic series for
$ J[n] <> Sqrt(2/(Pi*x))*Cos(zeta)*Sum(k,0,Infinity, (1)^k*A(2*k,n)*x^(2*k))  Sqrt(2/(Pi*x))*Sin(zeta)*Sum(k,0,Infinity, (1)^k*A(2*k+1,n)*x^(2*k1)) $ and
$ Y[n] <> Sqrt(2/(Pi*x))*Sin(zeta)*Sum(k,0,Infinity, (1)^k*A(2*k,n)*x^(2*k)) + Sqrt(2/(Pi*x))*Cos(zeta)*Sum(k,0,Infinity, (1)^k*A(2*k+1,n)*x^(2*k1)) $.
The error of a truncated asymptotic series is not larger than the first discarded term if the number of terms is larger than $n1/2$.
(See the book [Olver 1974] for derivations.
It seems that each asymptotic series requires special treatment and yet in all cases the error is about the same as the first discarded term.)
Currently Yacas can compute {BesselJ(n,x)} for all $x$ where $n$ is an integer and for $Abs(x)<= 2*Gamma(n)$ when
$n$ is a real number. Yacas currently uses the Taylor series
when $Abs(x)<= 2*Gamma(n)$ to compute the numerical value:
$$ BesselJ(n,x) := Sum(k,0,Infinity,(1)^k*x^(2*k+n)/(2^(2*k+n)* k! * Gamma(k+n+1))) $$.
*A Bessel functions!by recurrence relation
If $Abs(x) > 2*Gamma(n)$ and $n$ is an integer, then Yacas uses the forward recurrence
relation:
$$ BesselJ(n,x) := (2*(n+1)/x)*BesselJ(n+1,x)  BesselJ(n+2,x) $$
until the given {BesselJ} function is represented in terms of higher order terms which
all satisfy $Abs(x) <= 2*Gamma(n)$. Note that when $n$ is much smaller than $x$, this
algorithm is quite slow because the number of Bessel function evaluations grows like $2^i$, where $i$ is
the number of times the recurrence identity is used.
We see from the definition that when $Abs(x)<=2*Gamma(n)$, the absolute value
of each term is always decreasing (which is called absolutely monotonely
decreasing). From this we know that if we stop after $i$ iterations, the error
will be bounded by the absolute value of the next term. So given a set
precision, turn this into a value $epsilon$, so that we can check if the current
term will contribute to the sum at the prescribed precision. Before doing this,
Yacas currently increases the precision by 20{%} to do interim calculations.
This is a heuristic that works, it is not backed by theory. The value $epsilon$
is given by $epsilon:=5*10^(prec)$, where $prec$ was the previous precision. This
is directly from the definition of floating point number which is correct to
$prec$ digits: A number correct to $prec$ digits has a rounding error no
greater than $5*10^(prec)$. Beware that some books incorrectly have $.5$
instead of $5$.
Bug: Something is not right with complex numbers, but pure imaginary are OK.
Bernoulli numbers and polynomials
*A Bernoulli numbers!definition
*A Bernoulli polynomials!definition
The Bernoulli numbers $B[n]$ are rational numbers that are frequently used in applications.
The first few numbers are $B[0]=1$, $B[1]= 1/2$, $B[2]=4$, $B[3]=0$, $B[4]= 1/30$.
The numbers $B[n]$ can be defined
by the series expansion of the following generating function,
$$z/(e^z1) = Sum(n, 0, Infinity, B[n]*z^n/n!)$$.
The Bernoulli polynomials $B(x)[n]$ are defined similarly by
$$(z*Exp(z*x))/(e^z1) = Sum(n, 0, Infinity, B(x)[n]*z^n/n!)$$.
The Bernoulli polynomials are related to Bernoulli numbers by
$$B(x)[n] = Sum(k,0,n,Bin(n,k)*x^k*B[nk])$$,
where $Bin(n,k)$ are binomial coefficients.
*A sums of integer powers
Bernoulli numbers and polynomials are used in various Taylor series expansions, in the EulerMaclauren series resummation formula, in Riemann's Zeta function and so on. For example, the sum of (integer) $p$th powers of consecutive integers is given by
$$Sum(k, 0, n1, k^p) = (B(n)[p+1]  B[p+1])/(p+1)$$.
The Bernoulli polynomials $B(x)[n]$ can be found by first computing an array of Bernoulli numbers up to $B[n]$ and then applying the above formula for the coefficients.
We consider two distinct computational tasks: evaluate a Bernoulli number exactly as
a rational, or find it approximately to a specified floatingpoint precision.
There are also two possible problem settings: either we need to
evaluate <i>all</i> Bernoulli numbers $B[n]$ up to some $n$ (this situation occurs most often in practice), or we only
need one isolated value $B[n]$ for some large $n$. Depending on how large $n$
is, different algorithms can be chosen in these cases.
Exact evaluation of Bernoulli numbers
*A Bernoulli numbers!by recurrence
In the {Bernoulli()} routine,
Bernoulli numbers are evaluated exactly (as rational numbers) via one of the two algorithms. The first, simpler algorithm ({BernoullliArray()}) uses the recurrence relation,
$$B[n] = 1/(n+1)*Sum(k,0,n1,B[k]*Bin(n+1,k))$$.
This formula requires to know the entire set of $B[k]$ with $k$ up to a given
$n$ to compute $B[n]$. Therefore at large $n$ this algorithm
is a very slow way to compute $B[n]$ if we do not need all other $B[k]$.
*A Bernoulli numbers!by recurrence!cost estimate
Here is an estimate of the cost of {BernoullliArray}.
Suppose $M(P)$ is the time needed to multiply $P$digit integers.
The required number of digits $P$ to store the numerator of $B[n]$ is asymptotically $P <> n*Ln(n)$. At each of the $n$ iterations we need to multiply $O(n)$ large rational numbers by large coefficients and take a GCD to simplify the resulting fractions. The time for GCD is logarithmic in $P$. So the complexity of this algorithm is $O(n^2*M(P)*Ln(P))$ with $P <> n*Ln(n)$.
*A Bernoulli numbers!from Zeta function
For large (even) values of the index $n$, a single Bernoulli number $B[n]$ can be
computed by a more efficient procedure: the integer part and the fractional
part of $B[n]$ are found separately
(this method is also well explained in [Gourdon <i>et al.</i> 2001]).
*A Clausen  von Staudt, theorem of
*A Bernoulli numbers!fractional part of
First, by the theorem of Clausen  von
Staudt, the fractional part of ($B[n]$) is the same as the fractional part of
the sum of all inverse prime numbers $p$ such that $n$ is divisible by $p1$.
To illustrate the theorem, take $n=10$ with
$B[10]=5/66$. The number $n=10$ is divisible only by 1, 2, 5, and 10; this
corresponds to $p=2$, $3$, $6$ and $11$. Of these, $6$ is not a prime.
Therefore, we exclude 6 and take the sum $1/2 + 1/3 + 1/11 = 61/66$. The
theorem now says that $61/66$ has the same fractional part as $B[10]$; in
other words, $B[10]=i+f$ where $i$ is some unknown integer and the fractional
part $f$ is a nonnegative rational number, $0<=f<1$, which is now known to be
$61/66$. Indeed $B[10]= 1+61/66$. So one can find the fractional part of the Bernoulli number relatively quickly by
just checking the numbers that might divide $n$.
*A Stirling's formula
*A Bernoulli numbers!asymptotic
Now one needs to obtain the
integer part of $B[n]$. The number $B[n]$ is positive if $Mod(n,4)=2$ and negative if
$Mod(n,4)=0$. One can use Riemann's Zeta function identity for even integer
values of the argument and compute the value $zeta(n)$ precisely enough so that
the integer part of the Bernoulli number is determined. The required precision
is found by estimating the Bernoulli number from the same identity
in which
one approximates $zeta(n)=1$, i.e.
$$ Abs(B[2*n]) <=> 2*(2*n)! /(2*Pi)^(2*n) $$.
To estimate the factorial of large numbers, we can use Stirling's asymptotic formula
$$Ln(n!) <=> Ln(2*Pi*n)/2+n*Ln(n/e)$$. The result is that for large $n$,
$$ Abs(B[2*n]) <> 2*(n/(Pi*e))^(2*n)$$.
At such
large values of the argument $n$, it is feasible to use the routines {Internal'ZetaNum1(n, N)} or {Internal'ZetaNum2(n,N)} to compute the zeta function.
These routines approximate $zeta(n)$ by the defining series
$$zeta(n) <=> Sum(k,1,N, 1/k^n)$$.
The remainder of the
sum is of order $N^(n)$. By straightforward algebra one obtains a lower bound on $N$,
$$N > n/(2*Pi*e)$$,
for this sum to give enough precision to compute the integer part of the
Bernoulli number $B[n]$.
*A Bernoulli numbers!from Zeta function!example
For example, let us compute $B[20]$ using this method.
* 1. We find the fractional part, using the fact that 20 is divided only by 1, 2, 4, 5, 10, and 20 and only 2, 3, 5, 11 are prime:
In> 1/2 + 1/3 + 1/5 + 1/11;
Out> 371/330;
This number has fractional part equal to 41/330 and it's the same as the fractional part of $B[20]$. Since $B[20]$ must be negative,
this means that $B[20]$ is of the form $(X+41/330)$ where $X$ is some positive integer which we now have to find.
* 2. We estimate the magnitude of $Abs(B[20])$ to find the required precision to compute its integer part. We use the identity
$$Abs(B[20])=(2*20!)/(2*Pi)^20*zeta(20)$$,
Stirling's formula
$$Ln(20!) = (Ln(2*Pi)+Ln(20))/2+20*Ln(20/e) <=> 42.3$$,
and also the fact that $zeta(20)=1+2^(20)+...$ is approximately equal to 1.
We find that the number $B[20]$ has approximately
$$ 1+Ln(Abs(B[20]))/Ln(10) = 1+(Ln(2)+Ln(20!)20*Ln(2*Pi))/Ln(10) <=> 3.72$$
decimal digits before the decimal point; so a precision of 3 or 4 mantissa digits would surely be enough to compute its integer part.
* 3. We now need to compute $zeta(20)$ to 4 decimal digits. The series
$$zeta(20) <=> Sum(k,1,N, 1/k^20)$$
will have an error of order $N^(20)$. We need this error to be less than
$10^(4)$ and this is achieved with $N>2$. Therefore it is enough to compute
the sum up to $N=2$:
In> N(1+1/2^20)
Out> 1.0000009536;
* 4. Now we find the integer part of $Abs(B[20])$. This time we do not use Stirling's formula for the factorial but compute the exact factorial.
In> N( 2*20! /(2*Pi)^20*1.0000009536 )
Out> 529.1242423667;
(Note the space after the factorial sign  it is needed for the syntax to parse correctly.) Therefore we know that the integer part of $Abs(B[20])$ is 529.
* 5. Since $B[20] = (X+41/330)$ and we found that $X=529$, we obtain
In> (529+41/330);
Out> 174611/330;
This is exactly the Bernoulli number $B[20]$.
*A {Bernoulli1}
All these steps are implemented in the routine {Bernoulli1}. The function {Bernoulli1Threshold()} returns the smallest $n$ for which $B[n]$ is to be computed via this routine instead of the recursion relation. Its current default value is 20. This value can be set with {SetBernoulli1Threshold(threshold)}.
The complexity of {Bernoulli1} is estimated as the complexity of finding all primes up to $n$ plus the complexity of computing the factorial, the power and the Zeta function. Finding the prime numbers up to $n$ by checking all potential divisors up to $Sqrt(n)$ requires $O(n^(3/2)*M(Ln(n)))$ operations with precision $O(Ln(n))$ digits. For the second step we need to evaluate $n!$, $Pi^n$ and $zeta(n)$ with precision of $P=O(n*Ln(n))$ digits. The factorial is found in $n$ short multiplications with $P$digit numbers (giving $O(n*P)$), the power of $pi$ in $Ln(n)$ long multiplications (giving $O(M(P)*Ln(n))$), and {Internal'ZetaNum2(n)} (the asymptotically faster algorithm) requires $O(n*M(P))$ operations. The Zeta function calculation dominates the total cost because $M(P)$ is slower than $O(P)$. So the total complexity of {Bernoulli1} is $O(n*M(P))$ with $P <> n*Ln(n)$.
Note that this is the cost of finding just one Bernoulli number, as opposed to the $O(n^2*M(P)*Ln(P))$ cost of finding all Bernoulli numbers up to $B[n]$ using the first algorithm {Internal'BernoulliArray}. If we need a complete table of Bernoulli numbers, then {Internal'BernoulliArray} is only marginally (logarithmically) slower. So for finding complete Bernoulli tables, {Bernoulli1} is better only for very large $n$.
Approximate calculation of Bernoulli numbers
If Bernoulli numbers do not have to be found exactly but only to a certain floatingpoint precision $P$ (this is usually the case for most numerical applications), then the situation is rather different. First, all calculations can be performed using floatingpoint numbers instead of exact rationals. This significantly speeds up the recurrencebased algorithms.
*A Bernoulli numbers!by recurrence!numerical stability
However, the recurrence relation used in {Internal'BernoulliArray} turns out to be numerically unstable and needs to be replaced by another [Brent 1978].
Brent's algorithm computes the Bernoulli numbers divided by factorials, $C[n]:=B[2*n]/(2*n)!$ using a (numerically stable) recurrence relation
$$2*C[k]*(14^(k)) = (2*k1)/(4^k*(2*k)!)  Sum(j, 1, k1, C[kj]/(4^j*(2*j)!))$$.
The numerical instability of the usual recurrence relation
$$ Sum(j,0,k1,C[kj]/((2*j+1)!)) = (k1/2)/((2*k+1)!) $$
and the numerical stability of Brent's recurrence are not obvious. Here is one way to demonstrate them.
Consider the usual recurrence (above). For large $k$, the number $C[k]$ is approximately $C[k] <=> 2*(1)^k*(2*Pi)^(2*k)$. Suppose we use this recurrence to compute $C[k]$ from previously found values $C[k1]$, $C[k2]$, etc. and suppose that we have small relative errors $e[k]$ of finding $C[k]$. Then instead of the correct $C[k]$ we use $C[k]*(1+e[k])$ in the recurrence. Now we can derive a relation for the error sequence $e[k]$ using the approximate values of $C[k]$. It will be a linear recurrence of the form
$$ Sum(j,0,k1,(1)^(kj)*e[kj]*((2*Pi)^(2*j)/((2*j+1)!)) ) = (k1/2)/((2*k+1)!) * (2*Pi)^(2*k) $$.
Note that the coefficients for $j>5$ are very small but the coefficients for $0<=j<=5$ are of order 1. This means that we have a cancellation in the first 5 or so terms that produces a very small number $C[k]$ and this may lead to a loss of numerical precision. To investigate this loss, we find eigenvalues of the sequence $e[k]$, i.e. we assume that $e[k] = lambda^k$ and find $lambda$. If $Abs(lambda)>1$, then a small initial error $e[1]$ will grow by a power of $lambda$ on each iteration and it would indicate a numerical instability.
The eigenvalue of the sequence $e[k]$ can be found approximately for large $k$ if we notice that the recurrence relation for $e[k]$ is similar to the truncated Taylor series for $Sin(x)$. Substituting $e[k] = lambda^k$ into it and disregarding a very small number $(2*Pi)^(2*k)$ on the right hand side, we find
$$ Sum(j,0,k1,(lambda)^(kj)*((2*Pi)^(2*j)/((2*j+1)!)) ) <=> lambda^k*Sin((2*Pi)/Sqrt(lambda)) <=> 0 $$,
which means that $lambda=4$ is a solution. Therefore the recurrence is unstable.
By a very similar calculation one finds that the inverse powers of $4$ in Brent's recurrence make the largest eigenvalue of the error sequence $e[k]$ almost equal to $1$ and therefore the recurrence is stable. Brent gives the relative error in the computed $C[k]$ as $O(k^2)$ times the roundoff error in the last digit of precision.
*A Bernoulli numbers!by recurrence!cost estimate
The complexity of Brent's method is given as $O(n^2*P+n*M(P))$ for finding all Bernoulli numbers up to $B[n]$ with precision $P$ digits. This computation time can be achieved if we compute the inverse factorials and powers of $4$ approximately by floatingpoint routines that know how much precision is needed for each term in the recurrence relation. The final long multiplication by $(2*k)!$ computed to precision $P$ adds $M(P)$ to each Bernoulli number.
The noniterative method using the Zeta function does not perform much better if a Bernoulli number $B[n]$ has to be computed with significantly fewer digits $P$ than the full $O(n*Ln(n))$ digits needed to represent the integer part of $B[n]$. (The fractional part of $B[n]$ can always be computed relatively quickly.) The Zeta function needs $10^(P/n)$ terms, so its complexity is $O(10^(P/n)*M(P))$ (here by assumption $P$ is not very large so $10^(P/n)<n/(2*Pi*e)$; if $n>P$ we can disregard the power of $10$ in the complexity formula). We should also add $O(Ln(n)*M(P))$ needed to compute the power of $2*Pi$. The total complexity of {Bernoulli1} is therefore $O(Ln(n)*M(P) + 10^(P/n)*M(P))$.
If only one Bernoulli number is required, then {Bernoulli1} is always faster. If all Bernoulli numbers up to a given $n$ are required, then Brent's recurrence is faster for certain (small enough) $n$.
Currently Brent's recurrence is implemented as {Internal'BernoulliArray1()} but it is not used by {Bernoulli} because the internal arithmetic is not yet able to correctly compute with floatingpoint precision.
Error function $Erf(x)$ and related functions
The error function $Erf(z)$ is defined for any (complex) $z$ by
$$ Erf(z) := (2/Sqrt(Pi))*Integrate(t,0,z)Exp(t^2) $$.
The complementary error function $Erfc(x)$ is defined for real $x$ as
$$ Erfc(x) := (2/Sqrt(Pi))*(Integrate(t,x,Infinity)Exp(t^2)) = 1Erf(x) $$.
The imaginary error function $Erfi(x)$ is defined for real $x$ as
$$ Erfi(x) := (2/Sqrt(Pi))*Integrate(t,x,Infinity)Exp(t^2) $$.
Numerical computation of the error function $Erf(z)$ needs to be performed by
different methods depending on the value of $z$ and its position in the complex
plane, and on the required precision. We follow the book [Tsimring 1988] and the paper [Thacher 1963].
(These texts, however, do not describe arbitraryprecision computations.)
*A error function $Erf(x)$!summary of methods
The function $Erf(z)$ has the following approximations that are useful for its
numerical computation:
* 1. The Taylor series at $z=0$,
$$ (Sqrt(Pi)/(2*z))*Erf(z) = 1z^2/3+z^4/(2! *5) ... + (z^2)^n/((2*n+1)*n!)+O(z^(2*n+1))$$
* 2. The Taylor series for a slightly modified function,
$$ (Sqrt(Pi)/2)*(e^(z^2)/z)*Erf(z) = 1+(2*z^2)/3+... + (2*z^2)^n/(2*n+1)!! + O(z^(2*n+1))$$.
* 3. Asymptotic expansion of $Erfc(z)$ at $z=Infinity$:
$$ Erfc(z) = e^(z^2)/(z*Sqrt(Pi))*(11/(2*z^2) + ... + (2*n1)!! /(2*z^2)^n+...) $$.
* 4. Continued fraction expansion due to Laplace:
$$ (Sqrt(Pi)/2)*z*e^(z^2)*Erfc(z) = z/(z+1/2/(z+1/(z+3/2/(z+...)))) $$
The continued fraction in the RHS of the above equation can be rewritten equivalently as
$$ 1/(1+v/(1+(2*v)/(1+(3*v)/(1+...)))) $$
if we define $v:=(2*z^2)^(1)$.
Here we shall analyze the convergence and precision of these methods.
We need to choose a good method to compute $Erf(z)$ with (relative) precision $P$
decimal digits for a given (complex) number $z$, and to obtain estimates for the
necessary number of terms to take.
Both Taylor series converge absolutely for all $z$, but they do not converge uniformly fast; in fact these series are not very useful for large $z$ because a very large number of slowly decreasing terms gives a significant contribution to the result, and the roundoff error (especially for the first series with the alternating signs) becomes too high. Both series converge well for $Abs(z)<1$.
*A error function $Erf(x)$!by Taylor series
Consider method 1 (the first Taylor series). We shall use the method 1 only for $Abs(z)<=1$. If the absolute error of the truncated Taylor series is estimated as the first discarded term, the precision after taking all terms up to and including $z^(2*n)$ is approximately $z^(2*n+2)/(n+2)!$. The factorial can be approximated by Stirling's formula,
$ n! <=> n^n * e^(n)$. The value of $Erf(z)$ at small $z$ is of order 1, so we can take the absolute error to be equal to the relative error of the series that starts with $1$. Therefore, to obtain $P$ decimal digits of precision, we need the number of terms $n$ that satisfies the inequality
$$ Abs(e*z^2/(n+1))^(n+1) < 10^(P)$$.
(We have substituted $n+1$ instead of $n+2$ which made the inequality stronger.)
The error will be greatest when $Abs(z)=1$. For these values of $z$, the above inequality is satisfied when $n>1+Exp(1+W(P*Ln(10)/e))$ where $W$ is Lambert's $W$ function.
*A error function $Erf(x)$!by asymptotic series
Consider method 3 (the asymptotic series).
Due to limitations of the asymptotic series, we shall use the method 3 only for
large enough values of $z$ and small enough precision.
There are two important cases when calculating $Erf(z)$ for large (complex) $z$: the case of $z^2>0$ and the case of $z^2<0$. In the first case (e.g. a real $z$), the function $Erf(z)$ is approximately $1$ for large $Abs(z)$ (if $Re(z)>0$, and approximately $1$ if $Re(z)<0$). In the second case (e.g. pure imaginary $z=I*t$) the function $Erf(z)$ rapidly grows as $Exp(z^2)/z$ at large $Abs(z)$.
*REM FIXME need to fill in on the details
