Although the `gp`

calculator is programmable, it is useful to have
a number of preprogrammed loops, including sums, products, and a certain
number of recursions. Also, a number of functions from numerical analysis
like numerical integration and summation of series will be described here.

One of the parameters in these loops must be the control variable, hence a simple variable name. In the descriptions, the letter X will always denote any simple variable name, and represents the formal parameter used in the function. The expression to be summed, integrated, etc. is any legal PARI expression, including of course expressions using loops.

**Library mode.**
Since it is easier to program directly the loops in library mode, these
functions are mainly useful for GP programming. On the other hand, numerical
routines code a function (to be integrated, summed, etc.) with two parameters
named

GEN (*eval)(void*,GEN) void *E; \\ context: eval(E, x) must evaluate your function at x.

see the Libpari manual for details.

**Numerical integration.**
Starting with version 2.2.9 the "double exponential" univariate
integration method is implemented in `intnum`

and its variants. Romberg
integration is still available under the name `intnumromb`

, but
superseded. It is possible to compute numerically integrals to thousands of
decimal places in reasonable time, as long as the integrand is regular. It is
also reasonable to compute numerically integrals in several variables,
although more than two becomes lengthy. The integration domain may be
non-compact, and the integrand may have reasonable singularities at
endpoints. To use `intnum`

, you must split the integral into a sum
of subintegrals where the function has no singularities except at the
endpoints. Polynomials in logarithms are not considered singular, and
neglecting these logs, singularities are assumed to be algebraic (asymptotic
to C(x-a)^{-α} for some α > -1 when x is
close to a), or to correspond to simple discontinuities of some (higher)
derivative of the function. For instance, the point 0 is a singularity of
abs(x).

See also the discrete summation methods below, sharing the prefix `sum`

.

Asymptotic expansion of *expr*, corresponding to a sequence u(n),
assuming it has the shape
u(n) ~ ∑_{i ≥ 0} a_i n^{-iα}
with rational coefficients a_i with reasonable height; the algorithm
is heuristic and performs repeated calls to limitnum, with
`k`

and `alpha`

are as in `limitnum`

? f(n) = n! / (n^n*exp(-n)*sqrt(n)); ? asympnum(f) %2 = [] \\ failure ! ? l = limitnum(f) %3 = 2.5066282746310005024157652848110452530 ? asympnum(n->f(n)/l) \\ normalize %4 = [1, 1/12, 1/288, -139/51840]

and we indeed get a few terms of Stirling's expansion. Note that it helps to normalize with a limit computed to higher accuracy:

? \p100 ? L = limitnum(f) ? \p38 ? asympnum(n->f(n)/L) \\ we get more terms! %6 = [1, 1/12, 1/288, -139/51840, -571/2488320, 163879/209018880,\ 5246819/75246796800, -534703531/902961561600]

If `alpha`

is not an integer, loss of accuracy is
expected, so it should be precomputed to double accuracy, say:

? \p38 ? asympnum(n->-log(1-1/n^Pi),,Pi) %1 = [0, 1, 1/2, 1/3] ? asympnum(n->-log(1-1/sqrt(n)),,1/2) %2 = [0, 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12, \ 1/13, 1/14, 1/15, 1/16, 1/17, 1/18, 1/19, 1/20, 1/21, 1/22] ? localprec(100); a = Pi; ? asympnum(n->-log(1-1/n^a),,a) \\ better ! %4 = [0, 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12]

The library syntax is

, where **asympnum**(void *E, GEN (*u)(void *,GEN,long), long muli, GEN alpha, long prec)`u(E, n, prec)`

must return u(n) in precision `prec`

.
Also available is
`GEN `

, where u
must be a vector of sufficient length as above.**asympnum0**(GEN u, long muli, GEN alpha, long prec)

Given a continued fraction `CF`

output by `contfracinit`

, evaluate
the first `lim`

terms of the continued fraction at `t`

(all
terms if `lim`

is negative or omitted; if positive, `lim`

must be
less than or equal to the length of `CF`

.

The library syntax is `GEN `

.**contfraceval**(GEN CF, GEN t, long lim)

Given M representing the power series S = ∑_{n ≥ 0} M[n+1]z^n,
transform it into a continued fraction; restrict to n ≤ `lim`

if latter is non-negative. M can be a vector, a power
series, a polynomial, or a rational function.
The result is a 2-component vector [A,B] such that
S = M[1] / (1+A[1]z+B[1]z^2/(1+A[2]z+B[2]z^2/(1+...1/(1+A[lim/2]z)))).
Does not work if any coefficient of M vanishes, nor for series for
which certain partial denominators vanish.

The library syntax is `GEN `

.**contfracinit**(GEN M, long lim)

Numerical derivation of *expr* with respect to X at X = a.

? derivnum(x=0,sin(exp(x))) - cos(1) %1 = -1.262177448 E-29

A clumsier approach, which would not work in library mode, is

? f(x) = sin(exp(x)) ? f'(0) - cos(1) %1 = -1.262177448 E-29

When a is a power series, compute `derivnum(t = a,f)`

as f'(a) =
(f(a))'/a'.

The library syntax is

. Also
available is **derivnum**(void *E, GEN (*eval)(void*,GEN), GEN a, long prec)`GEN `

, which also allows power series for a.**derivfun**(void *E, GEN (*eval)(void *, GEN), GEN a, long prec)

Numerical
integration of (2iπ)^{-1}*expr* with respect to X on the circle
|X-a |= R.
In other words, when *expr* is a meromorphic
function, sum of the residues in the corresponding disk; *tab* is as in
`intnum`

, except that if computed with `intnuminit`

it should be with
the endpoints `[-1, 1]`

.

? \p105 ? intcirc(s=1, 0.5, zeta(s)) - 1 time = 496 ms. %1 = 1.2883911040127271720 E-101 + 0.E-118*I

The library syntax is

.**intcirc**(void *E, GEN (*eval)(void*,GEN), GEN a,GEN R,GEN tab, long prec)

Initialize tables for use with integral transforms such Fourier,
Laplace or Mellin transforms, in order to compute
∫_a^b f(t) k(t,z) dt
for some kernel k(t,z).
The endpoints a and b are coded as in `intnum`

, f is the
function to which the integral transform is to be applied and the
non-negative integer m is as in `intnum`

: multiply the number of
sampling points roughly by 2^m, hopefully increasing the accuracy. This
function is particularly useful when the function f is hard to compute,
such as a gamma product.

**Limitation.** the endpoints a and b must be at infinity,
with the same asymptotic behaviour. Oscillating types are not supported.
This is easily overcome by integrating vectors of functions, see example
below.

**Examples.**

***** numerical Fourier transform
F(z) = ∫_{- oo }^{+ oo } f(t)e^{-2iπ z t} dt.
First the easy case, assume that f decrease exponentially:

f(t) = exp(-t^2); A = [-oo,1]; B = [+oo,1]; \p200 T = intfuncinit(t = A,B , f(t)); F(z) = { my(a = -2*I*Pi*z); intnum(t = A,B, exp(a*t), T); } ? F(1) - sqrt(Pi)*exp(-Pi^2) %1 = -1.3... E-212

Now the harder case, f decrease slowly: we must specify the oscillating behaviour. Thus, we cannot precompute usefully since everything depends on the point we evaluate at:

f(t) = 1 / (1+ abs(t)); \p200 \\ Fourier cosine transform FC(z) = { my(a = 2*Pi*z); intnum(t = [-oo, a*I], [+oo, a*I], cos(a*t)*f(t)); } FC(1)

***** Fourier coefficients: we must integrate over a period, but
`intfuncinit`

does not support finite endpoints.
The solution is to integrate a vector of functions !

FourierSin(f, T, k) = \\ first k sine Fourier coeffs { my (w = 2*Pi/T); my (v = vector(k+1)); intnum(t = -T/2, T/2, my (z = exp(I*w*t)); v[1] = z; for (j = 2, k, v[j] = v[j-1]*z); f(t) * imag(v)) * 2/T; } FourierSin(t->sin(2*t), 2*Pi, 10)

The same technique can be used instead of `intfuncinit`

to integrate f(t) k(t,z) whenever the list of z-values is known
beforehand.

Note that the above code includes an unrelated optimization: the sin(j w t) are computed as imaginary parts of exp(i j w t) and the latter by successive multiplications.

***** numerical Mellin inversion
F(z) = (2iπ)^{-1} ∫_{c -i oo }^{c+i oo } f(s)z^{-s} ds
= (2π)^{-1} ∫_{- oo }^{+ oo }
f(c + i t)e^{-log z(c + it)} dt.
We take c = 2 in the program below:

f(s) = gamma(s)^3; \\ f(c+it) decrease as exp(-3Pi|t|/2) c = 2; \\ arbitrary A = [-oo,3*Pi/2]; B = [+oo,3*Pi/2]; T = intfuncinit(t=A,B, f(c + I*t)); F(z) = { my (a = -log(z)); intnum(t=A,B, exp(a*I*t), T)*exp(a*c) / (2*Pi); }

The library syntax is

.**intfuncinit**(void *E, GEN (*eval)(void*,GEN), GEN a,GEN b,long m, long prec)

Numerical integration
of *expr* on ]a,b[ with respect to X, using the
double-exponential method, and thus O(Dlog D) evaluation of
the integrand in precision D. The integrand may have values
belonging to a vector space over the real numbers; in particular, it can be
complex-valued or vector-valued. But it is assumed that the function is
regular on ]a,b[. If the endpoints a and b are finite and the
function is regular there, the situation is simple:

? intnum(x = 0,1, x^2) %1 = 0.3333333333333333333333333333 ? intnum(x = 0,Pi/2, [cos(x), sin(x)]) %2 = [1.000000000000000000000000000, 1.000000000000000000000000000]

An endpoint equal to ± oo is coded as `+oo`

or `-oo`

, as
expected:

? intnum(x = 1,+oo, 1/x^2) %3 = 1.000000000000000000000000000

In basic usage, it is assumed that the function does not decrease exponentially fast at infinity:

? intnum(x=0,+oo, exp(-x)) *** at top-level: intnum(x=0,+oo,exp(- *** ^-------------------- *** exp: overflow in expo().

We shall see in a moment how to avoid that last problem, after describing
the last *optional* argument *tab*.

**The tab.**

***** a non-negative integer m, we multiply the number of sampling points
by 2^m, hopefully increasing accuracy. Note that the running time
increases roughly by a factor 2^m. One may try consecutive values of m
until they give the same value up to an accepted error.

***** a set of integration tables containing precomputed x_i and w_i
as output by `intnuminit`

. This is useful if several integrations of
the same type are performed (on the same kind of interval and functions,
for a given accuracy): we skip a precomputation of O(Dlog D)
elementary functions in accuracy D, whose running time has the same order
of magnitude as the evaluation of the integrand. This is in particular
usefule for multivariate integrals.

**Specifying the behavior at endpoints.**
This is done as follows. An endpoint a is either given as such (a scalar,
real or complex, `oo`

or `-oo`

for ± oo ), or as a two
component vector [a,α], to indicate the behavior of the integrand in a
neighborhood of a.

If a is finite, the code [a,α] means the function has a singularity of the form (x-a)^{α}, up to logarithms. (If α \ge 0, we only assume the function is regular, which is the default assumption.) If a wrong singularity exponent is used, the result will lose a catastrophic number of decimals:

? intnum(x=0, 1, x^(-1/2)) \\ assume x^{-1/2} is regular at 0 %1 = 1.9999999999999999999999999999827931660 ? intnum(x=[0,-1/2], 1, x^(-1/2)) \\ no, it's not %2 = 2.0000000000000000000000000000000000000 ? intnum(x=[0,-1/10], 1, x^(-1/2)) \\ using a wrong exponent is bad %3 = 1.9999999999999999999999999999999901912

If a is ± oo , which is coded as `+oo`

or `-oo`

,
the situation is more complicated, and [±`oo`

,α] means:

***** α = 0 (or no α at all, i.e. simply ±`oo`

)
assumes that the integrand tends to zero moderately quickly, at least as
O(x^{-2}) but not exponentially fast.

***** α > 0 assumes that the function tends to zero exponentially fast
approximately as exp(-α x). This includes oscillating but quickly
decreasing functions such as exp(-x)sin(x).

? intnum(x=0, +oo, exp(-2*x)) *** at top-level: intnum(x=0,+oo,exp(- *** ^-------------------- *** exp: exponent (expo) overflow ? intnum(x=0, [+oo, 2], exp(-2*x)) \\ OK! %1 = 0.50000000000000000000000000000000000000 ? intnum(x=0, [+oo, 3], exp(-2*x)) \\ imprecise exponent, still OK ! %2 = 0.50000000000000000000000000000000000000 ? intnum(x=0, [+oo, 10], exp(-2*x)) \\ wrong exponent ==> disaster %3 = 0.49999999999952372962457451698256707393

As the last exemple shows, the exponential decrease rate
*must* be indicated to avoid overflow, but the method is robust enough
for a rough guess to be acceptable.

***** α < -1 assumes that the function tends to 0 slowly, like
x^{α}. Here the algorithm is less robust and it is essential to give a
sharp α, unless α ≤ -2 in which case we use
the default algorithm as if α were missing (or equal to 0).

? intnum(x=1, +oo, x^(-3/2)) \\ default %1 = 1.9999999999999999999999999999646391207 ? intnum(x=1, [+oo,-3/2], x^(-3/2)) \\ precise decrease rate %2 = 2.0000000000000000000000000000000000000 ? intnum(x=1, [+oo,-11/10], x^(-3/2)) \\ worse than default %3 = 2.0000000000000000000000000089298011973

The last two codes are reserved for oscillating functions. Let k > 0 real, and g(x) a non-oscillating function tending slowly to 0 (e.g. like a negative power of x), then

***** α = k * I assumes that the function behaves like cos(kx)g(x).

***** α = -k* I assumes that the function behaves like sin(kx)g(x).

Here it is critical to give the exact value of k. If the oscillating part is not a pure sine or cosine, one must expand it into a Fourier series, use the above codings, and sum the resulting contributions. Otherwise you will get nonsense. Note that cos(kx), and similarly sin(kx), means that very function, and not a translated version such as cos(kx+a).

**Note.** If f(x) = cos(kx)g(x) where g(x) tends to zero
exponentially fast as exp(-α x), it is up to the user to choose
between [±`oo`

,α] and [±`oo`

,k* I], but a good rule of
thumb is that
if the oscillations are weaker than the exponential decrease, choose
[±`oo`

,α], otherwise choose [±`oo`

,k*I], although the
latter can
reasonably be used in all cases, while the former cannot. To take a specific
example, in the inverse Mellin transform, the integrand is almost always a
product of an exponentially decreasing and an oscillating factor. If we
choose the oscillating type of integral we perhaps obtain the best results,
at the expense of having to recompute our functions for a different value of
the variable z giving the transform, preventing us to use a function such
as `intfuncinit`

. On the other hand using the exponential type of
integral, we obtain less accurate results, but we skip expensive
recomputations. See `intfuncinit`

for more explanations.

We shall now see many examples to get a feeling for what the various parameters achieve. All examples below assume precision is set to 115 decimal digits. We first type

? \p 115

**Apparent singularities.** In many cases, apparent singularities
can be ignored. For instance, if f(x) = 1
/(exp(x)-1) - exp(-x)/x, then ∫_0^ oo f(x)dx = γ, Euler's
constant `Euler`

. But

? f(x) = 1/(exp(x)-1) - exp(-x)/x ? intnum(x = 0, [oo,1], f(x)) - Euler %1 = 0.E-115

But close to 0 the function f is computed with an enormous loss of accuracy, and we are in fact lucky that it get multiplied by weights which are sufficiently close to 0 to hide this:

? f(1e-200) %2 = -3.885337784451458142 E84

A more robust solution is to define the function differently near special points, e.g. by a Taylor expansion

? F = truncate( f(t + O(t^10)) ); \\ expansion around t = 0 ? poldegree(F) %4 = 7 ? g(x) = if (x > 1e-18, f(x), subst(F,t,x)); \\ note that 7.18 > 105 ? intnum(x = 0, [oo,1], g(x)) - Euler %2 = 0.E-115

It is up to the user to determine constants such as the 10^{-18} and 10 used above.

**True singularities.** With true singularities the result is worse.
For instance

? intnum(x = 0, 1, x^(-1/2)) - 2 %1 = -3.5... E-68 \\ only 68 correct decimals ? intnum(x = [0,-1/2], 1, x^(-1/2)) - 2 %2 = 0.E-114 \\ better

**Oscillating functions.**

? intnum(x = 0, oo, sin(x) / x) - Pi/2 %1 = 16.19.. \\ nonsense ? intnum(x = 0, [oo,1], sin(x)/x) - Pi/2 %2 = -0.006.. \\ bad ? intnum(x = 0, [oo,-I], sin(x)/x) - Pi/2 %3 = 0.E-115 \\ perfect ? intnum(x = 0, [oo,-I], sin(2*x)/x) - Pi/2 \\ oops, wrong k %4 = 0.06... ? intnum(x = 0, [oo,-2*I], sin(2*x)/x) - Pi/2 %5 = 0.E-115 \\ perfect ? intnum(x = 0, [oo,-I], sin(x)^3/x) - Pi/4 %6 = -0.0008... \\ bad ? sin(x)^3 - (3*sin(x)-sin(3*x))/4 %7 = O(x^17)

We may use the above linearization and compute two oscillating integrals with
endpoints `[oo, -I]`

and `[oo, -3*I]`

respectively, or
notice the obvious change of variable, and reduce to the single integral
(1/2)∫_0^ oo sin(x)/xdx. We finish with some more complicated
examples:

? intnum(x = 0, [oo,-I], (1-cos(x))/x^2) - Pi/2 %1 = -0.0003... \\ bad ? intnum(x = 0, 1, (1-cos(x))/x^2) \ + intnum(x = 1, oo, 1/x^2) - intnum(x = 1, [oo,I], cos(x)/x^2) - Pi/2 %2 = 0.E-115 \\ perfect ? intnum(x = 0, [oo, 1], sin(x)^3*exp(-x)) - 0.3 %3 = -7.34... E-55 \\ bad ? intnum(x = 0, [oo,-I], sin(x)^3*exp(-x)) - 0.3 %4 = 8.9... E-103 \\ better. Try higher m ? tab = intnuminit(0,[oo,-I], 1); \\ double number of sampling points ? intnum(x = 0, oo, sin(x)^3*exp(-x), tab) - 0.3 %6 = 0.E-115 \\ perfect

**Warning.** Like `sumalt`

, `intnum`

often assigns a
reasonable value to diverging integrals. Use these values at your own risk!
For example:

? intnum(x = 0, [oo, -I], x^2*sin(x)) %1 = -2.0000000000...

Note the formula ∫_0^ oo sin(x)/x^sdx = cos(π s/2) Γ(1-s) , a priori valid only for 0 < Re(s) < 2, but the right hand side provides an analytic continuation which may be evaluated at s = -2...

**Multivariate integration.**
Using successive univariate integration with respect to different formal
parameters, it is immediate to do naive multivariate integration. But it is
important to use a suitable `intnuminit`

to precompute data for the
*internal* integrations at least!

For example, to compute the double integral on the unit disc x^2+y^2 ≤ 1 of the function x^2+y^2, we can write

? tab = intnuminit(-1,1); ? intnum(x=-1,1, intnum(y=-sqrt(1-x^2),sqrt(1-x^2), x^2+y^2, tab),tab) - Pi/2 %2 = -7.1... E-115 \\ OK

The first *tab* is essential, the second optional. Compare:

? tab = intnuminit(-1,1); time = 4 ms. ? intnum(x=-1,1, intnum(y=-sqrt(1-x^2),sqrt(1-x^2), x^2+y^2)); time = 3,092 ms. \\ slow ? intnum(x=-1,1, intnum(y=-sqrt(1-x^2),sqrt(1-x^2), x^2+y^2, tab), tab); time = 252 ms. \\ faster ? intnum(x=-1,1, intnum(y=-sqrt(1-x^2),sqrt(1-x^2), x^2+y^2, tab)); time = 261 ms. \\ theinternalintegral matters most

The library syntax is

,
where an omitted **intnum**(void *E, GEN (*eval)(void*,GEN), GEN a,GEN b,GEN tab, long prec)*tab* is coded as `NULL`

.

Numerical integration of *expr* on the compact interval [a,b] with
respect to X using Gauss-Legendre quadrature; `tab`

is either omitted
or precomputed with `intnumgaussinit`

. As a convenience, it can be an
integer n in which case we call
`intnumgaussinit`

(n) and use n-point quadrature.

? test(n, b = 1) = T=intnumgaussinit(n);\ intnumgauss(x=-b,b, 1/(1+x^2),T) - 2*atan(b); ? test(0) \\ default %1 = -9.490148553624725335 E-22 ? test(40) %2 = -6.186629001816965717 E-31 ? test(50) %3 = -1.1754943508222875080 E-38 ? test(50, 2) \\ double interval length %4 = -4.891779568527713636 E-21 ? test(90, 2) \\ n must almost be doubled as well! %5 = -9.403954806578300064 E-38

On the other hand, we recommend to split the integral and change variables rather than increasing n too much:

? f(x) = 1/(1+x^2); ? b = 100; ? intnumgauss(x=0,1, f(x)) + intnumgauss(x=1,1/b, f(1/x)*(-1/x^2)) - atan(b) %3 = -1.0579449157400587572 E-37

The library syntax is `GEN `

.**intnumgauss0**(GEN X, GEN b, GEN expr, GEN tab = NULL, long prec)

Initialize tables for n-point Gauss-Legendre integration of
a smooth function f lon a compact
interval [a,b] at current `realprecision`

. If n is omitted, make a
default choice n ~ `realprecision`

, suitable for analytic
functions on [-1,1]. The error is bounded by

((b-a)^{2n+1} (n!)^4)/((2n+1)[(2n)!]^3) f^{(2n)} (ξ) , a < ξ < b

so, if the interval length increases, n should be increased as well.

? T = intnumgaussinit(); ? intnumgauss(t=-1,1,exp(t), T) - exp(1)+exp(-1) %1 = -5.877471754111437540 E-39 ? intnumgauss(t=-10,10,exp(t), T) - exp(10)+exp(-10) %2 = -8.358367809712546836 E-35 ? intnumgauss(t=-1,1,1/(1+t^2), T) - Pi/2 %3 = -9.490148553624725335 E-22 ? T = intnumgaussinit(50); ? intnumgauss(t=-1,1,1/(1+t^2), T) - Pi/2 %5 = -1.1754943508222875080 E-38 ? intnumgauss(t=-5,5,1/(1+t^2), T) - 2*atan(5) %6 = -1.2[...]E-8

On the other hand, we recommend to split the integral and change variables
rather than increasing n too much, see `intnumgauss`

.

The library syntax is `GEN `

.**intnumgaussinit**(long n, long prec)

Initialize tables for integration from
a to b, where a and b are coded as in `intnum`

. Only the
compactness, the possible existence of singularities, the speed of decrease
or the oscillations at infinity are taken into account, and not the values.
For instance intnuminit(-1,1) is equivalent to intnuminit(0,Pi),
and intnuminit([0,-1/2],oo) is equivalent to
intnuminit([-1,-1/2], -oo); on the other hand, the order matters
and
intnuminit([0,-1/2], [1,-1/3]) is *not* equivalent to
intnuminit([0,-1/3], [1,-1/2]) !

If m is present, it must be non-negative and we multiply the default number of sampling points by 2^m (increasing the running time by a similar factor).

The result is technical and liable to change in the future, but we document
it here for completeness. Let x = φ(t), t ∈ ]- oo , oo [ be an
internally chosen change of variable, achieving double exponential decrease of
the integrand at infinity. The integrator `intnum`

will compute
h ∑_{|n| < N} φ'(nh) F(φ(nh))
for some integration step h and truncation parameter N.
In basic use, let

[h, x0, w0, xp, wp, xm, wm] = intnuminit(a,b);

***** h is the integration step

***** x_0 = φ(0) and w_0 = φ'(0),

***** *xp* contains the φ(nh), 0 < n < N,

***** *xm* contains the φ(nh), 0 < -n < N, or is empty.

***** *wp* contains the φ'(nh), 0 < n < N,

***** *wm* contains the φ'(nh), 0 < -n < N, or is empty.

The arrays *xm* and *wm* are left empty when φ is an odd
function. In complicated situations when non-default behaviour is specified at
end points, `intnuminit`

may return up to 3 such arrays, corresponding
to a splitting of up to 3 integrals of basic type.

If the functions to be integrated later are of the form F = f(t) k(t,z)
for some kernel k (e.g. Fourier, Laplace, Mellin,...), it is
useful to also precompute the values of f(φ(nh)), which is accomplished
by `intfuncinit`

. The hard part is to determine the behaviour
of F at endpoints, depending on z.

The library syntax is `GEN `

.**intnuminit**(GEN a, GEN b, long m, long prec)

Numerical integration of *expr* (smooth in ]a,b[), with respect to
X. Suitable for low accuracy; if *expr* is very regular (e.g. analytic
in a large region) and high accuracy is desired, try `intnum`

first.

Set *flag* = 0 (or omit it altogether) when a and b are not too large, the
function is smooth, and can be evaluated exactly everywhere on the interval
[a,b].

If *flag* = 1, uses a general driver routine for doing numerical integration,
making no particular assumption (slow).

*flag* = 2 is tailored for being used when a or b are infinite using the
change of variable t = 1/X. One *must* have ab > 0, and in fact if
for example b = + oo , then it is preferable to have a as large as
possible, at least a ≥ 1.

If *flag* = 3, the function is allowed to be undefined
at a (but right continuous) or b (left continuous),
for example the function sin(x)/x between x = 0 and 1.

The user should not require too much accuracy: `realprecision`

about
30 decimal digits (`realbitprecision`

about 100 bits) is OK,
but not much more. In addition, analytical cleanup of the integral must have
been done: there must be no singularities in the interval or at the
boundaries. In practice this can be accomplished with a change of
variable. Furthermore, for improper integrals, where one or both of the
limits of integration are plus or minus infinity, the function must decrease
sufficiently rapidly at infinity, which can often be accomplished through
integration by parts. Finally, the function to be integrated should not be
very small (compared to the current precision) on the entire interval. This
can of course be accomplished by just multiplying by an appropriate constant.

Note that infinity can be represented with essentially no loss of
accuracy by an appropriate huge number. However beware of real underflow
when dealing with rapidly decreasing functions. For example, in order to
compute the ∫_0^ oo e^{-x^2}dx to 28 decimal digits, then one can
set infinity equal to 10 for example, and certainly not to `1e1000`

.

The library syntax is

,
where **intnumromb_bitprec**(void *E, GEN (*eval)(void*,GEN), GEN a, GEN b, long flag, long bitprec)`eval`

(x, E) returns the value of the function at x.
You may store any additional information required by `eval`

in E, or set
it to `NULL`

. The historical variant
The library syntax is

, where **intnumromb**(..., long prec)`prec`

is expressed in words,
not bits, is obsolete and should no longer be used.

Lagrange-Zagier numerical extrapolation of *expr*, corresponding to a
sequence
u_n, either given by a closure `n- > u(n)`

or by a vector of values
I.e., assuming that u_n tends to a finite limit ℓ, try to determine
ℓ. This routine is purely numerical and heuristic, thus may or may not
work on your examples; k is ignored if u is given by a vector,
and otherwise is a multiplier such that we extrapolate from u(kn).

Assume that u_n has an asymptotic expansion in n^{-α} : u_n = ℓ + ∑_{i ≥ 1} a_i n^{-iα} for some a_i.

? limitnum(n -> n*sin(1/n)) %1 = 1.0000000000000000000000000000000000000 ? limitnum(n -> (1+1/n)^n) - exp(1) %2 = 0.E-37 ? limitnum(n -> 2^(4*n+1)*(n!)^4 / (2*n)! /(2*n+1)! ) %3 = 3.1415926535897932384626433832795028842 ? Pi %4 = 3.1415926535897932384626433832795028842

If u_n is given by a vector, it must be long enough for the extrapolation
to make sense: at least k times the current `realprecision`

. The
preferred format is thus a closure, although it becomes inconvenient
when u_n cannot be directly computed in time polynomial in log n,
for instance if it is defined as a sum or by induction. In that case,
passing a vector of values is the best option. It usually pays off to
interpolate u(kn) for some k > 1:

? limitnum(vector(10,n,(1+1/n)^n)) *** ^-------------------- *** limitnum: non-existent component in limitnum: index < 20 \\ at this accuracy, we must have at least 20 values ? limitnum(vector(20,n,(1+1/n)^n)) - exp(1) %5 = -2.05... E-20 ? limitnum(vector(20,n, m=10*n;(1+1/m)^m)) - exp(1) \\ better accuracy %6 = 0.E-37 ? v = vector(20); s = 0; ? for(i=1,#v, s += 1/i; v[i]= s - log(i)); ? limitnum(v) - Euler %9 = -1.6... E-19 ? V = vector(200); s = 0; ? for(i=1,#V, s += 1/i; V[i]= s); ? v = vector(#V \ 10, i, V[10*i] - log(10*i)); ? limitnum(v) - Euler %13 = 6.43... E-29

The library syntax is

, where **limitnum**(void *E, GEN (*u)(void *,GEN,long), long muli, GEN alpha, long prec)`u(E, n, prec)`

must return u(n) in precision `prec`

.
Also available is
`GEN `

, where u
must be a vector of sufficient length as above.**limitnum0**(GEN u, long muli, GEN alpha, long prec)

Product of expression
*expr*, initialized at x, the formal parameter X going from a to
b. As for `sum`

, the main purpose of the initialization parameter x
is to force the type of the operations being performed. For example if it is
set equal to the integer 1, operations will start being done exactly. If it
is set equal to the real 1., they will be done using real numbers having
the default precision. If it is set equal to the power series 1+O(X^k) for
a certain k, they will be done using power series of precision at most k.
These are the three most common initializations.

As an extreme example, compare

? prod(i=1, 100, 1 - X^i); \\ this has degree 5050 !! time = 128 ms. ? prod(i=1, 100, 1 - X^i, 1 + O(X^101)) time = 8 ms. %2 = 1 - X - X^2 + X^5 + X^7 - X^12 - X^15 + X^22 + X^26 - X^35 - X^40 + \ X^51 + X^57 - X^70 - X^77 + X^92 + X^100 + O(X^101)

Of course, in this specific case, it is faster to use `eta`

,
which is computed using Euler's formula.

? prod(i=1, 1000, 1 - X^i, 1 + O(X^1001)); time = 589 ms. ? \ps1000 seriesprecision = 1000 significant terms ? eta(X) - % time = 8ms. %4 = O(X^1001)

The library syntax is

.**produit**(GEN a, GEN b, char *expr, GEN x)

Product of expression *expr*,
initialized at 1. (i.e. to a *real* number equal to 1 to the current
`realprecision`

), the formal parameter X ranging over the prime numbers
between a and b.

The library syntax is

.**prodeuler**(void *E, GEN (*eval)(void*,GEN), GEN a,GEN b, long prec)

infinite product of
expression *expr*, the formal parameter X starting at a. The evaluation
stops when the relative error of the expression minus 1 is less than the
default precision. In particular, non-convergent products result in infinite
loops. The expressions must always evaluate to an element of ℂ.

If *flag* = 1, do the product of the (1+*expr*) instead.

The library syntax is

(**prodinf**(void *E, GEN (*eval)(void*,GEN), GEN a, long prec)*flag* = 0), or `prodinf1`

with the same arguments (*flag* = 1).

Find a real root of expression
*expr* between a and b, under the condition
*expr*(X = a) * *expr*(X = b) ≤ 0. (You will get an error message
`roots must be bracketed in solve`

if this does not hold.)
This routine uses Brent's method and can fail miserably if *expr* is
not defined in the whole of [a,b] (try `solve(x = 1, 2, tan(x))`

).

The library syntax is

.**zbrent**(void *E,GEN (*eval)(void*,GEN),GEN a,GEN b,long prec)

Find zeros of a continuous function in the real interval [a,b] by naive
interval splitting. This function is heuristic and may or may not find the
intended zeros. Binary digits of *flag* mean

***** 1: return as soon as one zero is found, otherwise return all
zeros found;

***** 2: refine the splitting until at least one zero is found
(may loop indefinitely if there are no zeros);

***** 4: do a multiplicative search (we must have a > 0 and *step* >
1), otherwise an additive search; *step* is the multiplicative or
additive step.

***** 8: refine the splitting until at least one zero is very close to an
integer.

? solvestep(X=0,10,1,sin(X^2),1) %1 = 1.7724538509055160272981674833411451828 ? solvestep(X=1,12,2,besselj(4,X),4) %2 = [7.588342434..., 11.064709488...]

The library syntax is

.**solvestep**(void *E, GEN (*eval)(void*,GEN), GEN a,GEN b, GEN step,long flag,long prec)

Sum of expression *expr*,
initialized at x, the formal parameter going from a to b. As for
`prod`

, the initialization parameter x may be given to force the type
of the operations being performed.

As an extreme example, compare

? sum(i=1, 10^4, 1/i); \\ rational number: denominator has 4345 digits. time = 236 ms. ? sum(i=1, 5000, 1/i, 0.) time = 8 ms. %2 = 9.787606036044382264178477904

The library syntax is

.**somme**(GEN a, GEN b, char *expr, GEN x)

Numerical summation of the series *expr*, which should be an
alternating series (-1)^k a_k, the formal variable X starting at
a. Use an algorithm of Cohen, Villegas and Zagier (*Experiment. Math.*
**9** (2000), no. 1, 3--12).

If *flag* = 0, assuming that the a_k are the moments of a positive
measure on [0,1], the relative error is O(3+sqrt8)^{-n} after using
a_k for k ≤ n. If `realprecision`

is p, we thus set
n = log(10)p/log(3+sqrt8) ~ 1.3 p; besides the time needed to
compute the a_k, k ≤ n, the algorithm overhead is negligible: time
O(p^2) and space O(p).

If *flag* = 1, use a variant with more complicated polynomials, see
`polzagier`

. If the a_k are the moments of w(x)dx where w
(or only xw(x^2)) is a smooth function extending analytically to the whole
complex plane, convergence is in O(14.4^{-n}). If xw(x^2) extends
analytically to a smaller region, we still have exponential convergence,
with worse constants. Usually faster when the computation of a_k is
expensive. If `realprecision`

is p, we thus set
n = log(10)p/log(14.4) ~ 0.86 p; besides the time needed to
compute the a_k, k ≤ n, the algorithm overhead is *not*
negligible: time O(p^3) and space O(p^2). Thus, even if the analytic
conditions for rigorous use are met, this variant is only worthwile if the
a_k are hard to compute, at least O(p^2) individually on average:
otherwise we gain a small constant factor (1.5, say) in the number of
needed a_k at the expense of a large overhead.

The conditions for rigorous use are hard to check but the routine is best used heuristically: even divergent alternating series can sometimes be summed by this method, as well as series which are not exactly alternating (see for example Section se:user_defined). It should be used to try and guess the value of an infinite sum. (However, see the example at the end of Section se:userfundef.)

If the series already converges geometrically,
`suminf`

is often a better choice:

```
? \p28
? sumalt(i = 1, -(-1)^i / i) - log(2)
time = 0 ms.
%1 = -2.524354897 E-29
? suminf(i = 1, -(-1)^i / i) \\ Had to hit
````C-C`

*** at top-level: suminf(i=1,-(-1)^i/i)
*** ^------
*** suminf: user interrupt after 10min, 20,100 ms.
? \p1000
? sumalt(i = 1, -(-1)^i / i) - log(2)
time = 90 ms.
%2 = 4.459597722 E-1002
? sumalt(i = 0, (-1)^i / i!) - exp(-1)
time = 670 ms.
%3 = -4.03698781490633483156497361352190615794353338591897830587 E-944
? suminf(i = 0, (-1)^i / i!) - exp(-1)
time = 110 ms.
%4 = -8.39147638 E-1000 \\ faster and more accurate

The library syntax is

. Also
available is **sumalt**(void *E, GEN (*eval)(void*,GEN),GEN a,long prec)`sumalt2`

with the same arguments (*flag* = 1).

Sum of expression *expr* over the positive divisors of n.
This function is a trivial wrapper essentially equivalent to

D = divisors(n); for (i = 1, #D, X = D[i]; eval(expr))

(except that `X`

is lexically scoped to the `sumdiv`

loop). If *expr* is a multiplicative function, use `sumdivmult`

.

Sum of *multiplicative* expression *expr* over the positive
divisors d of n. Assume that *expr* evaluates to f(d)
where f is multiplicative: f(1) = 1 and f(ab) = f(a)f(b) for coprime
a and b.

infinite sum of expression
*expr*, the formal parameter X starting at a. The evaluation stops
when the relative error of the expression is less than the default precision
for 3 consecutive evaluations. The expressions must always evaluate to a
complex number.

If the series converges slowly, make sure `realprecision`

is low (even 28
digits may be too much). In this case, if the series is alternating or the
terms have a constant sign, `sumalt`

and `sumpos`

should be used
instead.

```
? \p28
? suminf(i = 1, -(-1)^i / i) \\ Had to hit
````C-C`

*** at top-level: suminf(i=1,-(-1)^i/i)
*** ^------
*** suminf: user interrupt after 10min, 20,100 ms.
? sumalt(i = 1, -(-1)^i / i) - log(2)
time = 0 ms.
%1 = -2.524354897 E-29

The library syntax is

.**suminf**(void *E, GEN (*eval)(void*,GEN), GEN a, long prec)

Numerical summation of f(n) at high accuracy using Euler-MacLaurin,
the variable n taking values from a to + oo , where f is assumed to
have positive values and is a C^ oo function; `a`

must be an integer
and `tab`

, if given, is the output of `sumnuminit`

. The latter
precomputes abcissas and weights, speeding up the computation; it also allows
to specify the behaviour at infinity via `sumnuminit([+oo, asymp])`

.

? \p500 ? z3 = zeta(3); ? sumpos(n = 1, n^-3) - z3 time = 2,332 ms. %2 = 2.438468843 E-501 ? sumnum(n = 1, n^-3) - z3 \\ here slower than sumpos time = 2,752 ms. %3 = 0.E-500

**Complexity.**
The function f will be evaluated at O(D log D) real arguments,
where D ~ `realprecision`

.log(10). The routine is geared
towards slowly decreasing functions: if f decreases exponentially fast,
then one of `suminf`

or `sumpos`

should be preferred.
If f satisfies the stronger hypotheses required for Monien summation,
i.e. if f(1/z) is holomorphic in a complex neighbourhood of [0,1],
then `sumnummonien`

will be faster since it only requires O(D/log D)
evaluations:

? sumnummonien(n = 1, 1/n^3) - z3 time = 1,985 ms. %3 = 0.E-500

The `tab`

argument precomputes technical data
not depending on the expression being summed and valid for a given accuracy,
speeding up immensely later calls:

? tab = sumnuminit(); time = 2,709 ms. ? sumnum(n = 1, 1/n^3, tab) - z3 \\ now much faster than sumpos time = 40 ms. %5 = 0.E-500 ? tabmon = sumnummonieninit(); \\ Monien summation allows precomputations too time = 1,781 ms. ? sumnummonien(n = 1, 1/n^3, tabmon) - z3 time = 2 ms. %7 = 0.E-500

The speedup due to precomputations becomes less impressive when the function f is expensive to evaluate, though:

? sumnum(n = 1, lngamma(1+1/n)/n, tab); time = 14,180 ms. ? sumnummonien(n = 1, lngamma(1+1/n)/n, tabmon); \\ fewer evaluations time = 717 ms.

**Behaviour at infinity.**
By default, `sumnum`

assumes that *expr* decreases slowly at infinity,
but at least like O(n^{-2}). If the function decreases like n^{α}
for some -2 < α < -1, then it must be indicated via

tab = sumnuminit([+oo, alpha]); /* alpha < 0 slow decrease */

otherwise loss of accuracy is expected. If the functions decreases quickly, like exp(-α n) for some α > 0, then it must be indicated via

tab = sumnuminit([+oo, alpha]); /* alpha > 0 exponential decrease */

otherwise exponent overflow will occur.

? sumnum(n=1,2^-n) *** at top-level: sumnum(n=1,2^-n) *** ^---- *** _^_: overflow in expo(). ? tab = sumnuminit([+oo,log(2)]); sumnum(n=1,2^-n, tab) %1 = 1.000[...]

As a shortcut, one can also input

sumnum(n = [a, asymp], f)

instead of

tab = sumnuminit(asymp); sumnum(n = a, f, tab)

**Further examples.**

? \p200 ? sumnum(n = 1, n^(-2)) - zeta(2) \\ accurate, fast time = 200 ms. %1 = -2.376364457868949779 E-212 ? sumpos(n = 1, n^(-2)) - zeta(2) \\ even faster time = 96 ms. %2 = 0.E-211 ? sumpos(n=1,n^(-4/3)) - zeta(4/3) \\ now much slower time = 13,045 ms. %3 = -9.980730723049589073 E-210 ? sumnum(n=1,n^(-4/3)) - zeta(4/3) \\ fast but inaccurate time = 365 ms. %4 = -9.85[...]E-85 ? sumnum(n=[1,-4/3],n^(-4/3)) - zeta(4/3) \\ with decrease rate, now accurate time = 416 ms. %5 = -4.134874156691972616 E-210 ? tab = sumnuminit([+oo,-4/3]); time = 196 ms. ? sumnum(n=1, n^(-4/3), tab) - zeta(4/3) \\ faster with precomputations time = 216 ms. %5 = -4.134874156691972616 E-210 ? sumnum(n=1,-log(n)*n^(-4/3), tab) - zeta'(4/3) time = 321 ms. %7 = 7.224147951921607329 E-210

Note that in the case of slow decrease (α < 0), the exact
decrease rate must be indicated, while in the case of exponential decrease,
a rough value will do. In fact, for exponentially decreasing functions,
`sumnum`

is given for completeness and comparison purposes only: one
of `suminf`

or `sumpos`

should always be preferred.

? sumnum(n=[1, 1], 2^-n) \\ pretend we decrease as exp(-n) time = 240 ms. %8 = 1.000[...] \\ perfect ? sumpos(n=1, 2^-n) %9 = 1.000[...] \\ perfect and instantaneous

The library syntax is

where an omitted **sumnum**((void *E, GEN (*eval)(void*, GEN), GEN a, GEN tab, long prec))*tab* is coded as `NULL`

.

Initialize tables for Euler--MacLaurin delta summation of a series with
positive terms. If given, `asymp`

is of the form [`+oo`

, α],
as in `intnum`

and indicates the decrease rate at infinity of functions
to be summed. A positive
α > 0 encodes an exponential decrease of type exp(-α n) and
a negative -2 < α < -1 encodes a slow polynomial decrease of type
n^{α}.

? \p200 ? sumnum(n=1, n^-2); time = 200 ms. ? tab = sumnuminit(); time = 188 ms. ? sumnum(n=1, n^-2, tab); \\ faster time = 8 ms. ? tab = sumnuminit([+oo, log(2)]); \\ decrease like 2^-n time = 200 ms. ? sumnum(n=1, 2^-n, tab) time = 44 ms. ? tab = sumnuminit([+oo, -4/3]); \\ decrease like n^(-4/3) time = 200 ms. ? sumnum(n=1, n^(-4/3), tab); time = 221 ms.

The library syntax is `GEN `

.**sumnuminit**(GEN asymp = NULL, long prec)

Numerical summation ∑_{n ≥ a} f(n) at high accuracy, the variable n taking values from the integer a to + oo using Monien summation, which assumes that f(1/z) has a complex analytic continuation in a (complex) neighbourhood of the segment [0,1].

The function f is evaluated at O(D / log D) real arguments,
where D ~ `realprecision`

.log(10).
By default, assume that f(n) = O(n^{-2}) and has a non-zero asymptotic
expansion
f(n) = ∑_{i ≥ 2} a_i n^{-i}
at infinity. To handle more complicated behaviours and allow time-saving
precomputations (for a given `realprecision`

), see `sumnummonieninit`

.

The library syntax is `GEN `

.**sumnummonien0**(GEN n, GEN f, GEN tab = NULL, long prec)

Initialize tables for Monien summation of a series ∑_{n ≥ n_0} f(n) where f(1/z) has a complex analytic continuation in a (complex) neighbourhood of the segment [0,1].

By default, assume that f(n) = O(n^{-2}) and has a non-zero asymptotic
expansion
f(n) = ∑_{i ≥ 2} a_i / n^i
at infinity. Note that the sum starts at i = 2! The argument `asymp`

allows to specify different expansions:

***** a real number α > 1 means
f(n) = ∑_{i ≥ 1} a_i / n^{α i}
(Now the summation starts at 1.)

***** a vector [α,β] of reals, where we must have α > 0
and α + β > 1 to ensure convergence, means that
f(n) = ∑_{i ≥ 1} a_i / n^{α i + β}
Note that `asymp`

= [α, α] is equivalent to
`asymp`

= α.

? \p38 ? s = sumnum(n = 1, sin(1/sqrt(n)) / n) %1 = 2.3979771206715998375659850036324914714 ? sumnummonien(n = 1, sin(1/sqrt(n)) / n) - s %2 = -0.001[...] \\ completely wrong ! ? t = sumnummonieninit([1/2,1]); \\ f(n) = sum_i 1 / n^(i/2+1) ? sumnummonien(n = 1, sin(1/sqrt(n)) / n, t) - s %3 = 0.E-37 \\ now correct

The argument w is used to sum expressions of the form
∑_{n ≥ n_0} f(n) w(n),
for varying f *as above*, and fixed weight function w, where we
further assume that the auxiliary sums
g_w(m) = ∑_{n ≥ n_0} w(n) / n^{α m + β}
converge for all m ≥ 1. Note that for non-negative integers k,
and weight w(n) = (log n)^k, the function g_w(m) = ζ^{(k)}(α m +
β) has a simple expression; for general weights, g_w is
computed using `sumnum`

. The following variants are available

***** an integer k ≥ 0, to code w(n) = (log n)^k;
only the cases k = 0,1 are presently implemented; due to a poor
implementation of ζ derivatives, it is not currently worth it
to exploit the special shape of g_w when k > 0;

***** a `t_CLOSURE`

computing the values w(n), where we
assume that w(n) = O(n^ε) for all ε > 0;

***** a vector [w, `fast`

], where w is a closure as above
and `fast`

is a scalar;
we assume that w(n) = O(n^{`fast`

+ε}); note that
`w`

= [w, 0] is equivalent to `w`

= w.

***** a vector [w, `oo`

], where w is a closure as above;
we assume that w(n) decreases exponentially. Note that in this case,
`sumnummonien`

is provided for completeness and comparison purposes only:
one of `suminf`

or `sumpos`

should be preferred in practice.

The cases where w is a closure or w(n) = log n are the only ones where
n_0 is taken into account and stored in the result. The subsequent call to
`sumnummonien`

*must* use the same value.

? \p300 ? sumnummonien(n = 1, n^-2*log(n)) + zeta'(2) time = 536 ms. %1 = -1.323[...]E-6 \\ completely wrong, f does not satisfy hypotheses ! ? tab = sumnummonieninit(, 1); \\ codes w(n) = log(n) time = 18,316 ms. ? sumnummonien(n = 1, n^-2, tab) + zeta'(2) time = 44 ms. %3 = -5.562684646268003458 E-309 \\ now perfect ? tab = sumnummonieninit(, n->log(n)); \\ generic, about as fast time = 18,693 ms. ? sumnummonien(n = 1, n^-2, tab) + zeta'(2) time = 40 ms. %5 = -5.562684646268003458 E-309 \\ identical result

The library syntax is `GEN `

.**sumnummonieninit**(GEN asymp = NULL, GEN w = NULL, GEN n0 = NULL, long prec)

Numerical summation of the series *expr*, which must be a series of
terms having the same sign, the formal variable X starting at a. The
algorithm used is Van Wijngaarden's trick for converting such a series into
an alternating one, then we use `sumalt`

. For regular functions, the
function `sumnum`

is in general much faster once the initializations
have been made using `sumnuminit`

.

The routine is heuristic and assumes that *expr* is more or less a
decreasing function of X. In particular, the result will be completely
wrong if *expr* is 0 too often. We do not check either that all terms
have the same sign. As `sumalt`

, this function should be used to
try and guess the value of an infinite sum.

If *flag* = 1, use `sumalt`

(,1) instead of `sumalt`

(,0), see
Section se:sumalt. Requiring more stringent analytic properties for
rigorous use, but allowing to compute fewer series terms.

To reach accuracy 10^{-p}, both algorithms require O(p^2) space;
furthermore, assuming the terms decrease polynomially (in O(n^{-C})), both
need to compute O(p^2) terms. The `sumpos`

(,1) variant has a smaller
implied constant (roughly 1.5 times smaller). Since the `sumalt`

(,1)
overhead is now small compared to the time needed to compute series terms,
this last variant should be about 1.5 faster. On the other hand, the
achieved accuracy may be much worse: as for `sumalt`

, since
conditions for rigorous use are hard to check, the routine is best used
heuristically.

The library syntax is

. Also
available is **sumpos**(void *E, GEN (*eval)(void*,GEN),GEN a,long prec)`sumpos2`

with the same arguments (*flag* = 1).