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AngouriMath.Functions.Algebra

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Classes within the AngouriMath.Functions.Algebra namespace

• AntiderivativeBreaks

  Summary

  The points strictly inside a numeric range where an antiderivative can stop being
  continuous, so that a definite integral is taken through it piece by piece.
  

  Remarks

  F(b) - F(a) is the integral only where F is continuous on [a, b]. A pole
  of the integrand between the bounds is a point where F is not even defined: the
  integral of 1/x^2 over [-1, 1] came back as -1/x at 1 less -1/x at -1, which is -2, a negative answer for a positive integrand whose integral
  diverges. An antiderivative can also jump where the integrand is continuous, and the
  answer through it is then wrong by the jump: arccot(x) jumps at 0 in this library's
  convention, and a half-angle substitution writes tan(x/2).
  The places read are where a node of the antiderivative stops being continuous: the zeros
  of a denominator, of a base raised to a negative power, of a logarithm's argument, of the
  argument of sgn, arccot, arcsec and arccsc, and the zeros of the
  cosine or the sine under a tangent, secant, cotangent or cosecant of a linear argument.
  Where one of them cannot be listed -- a root with a symbol in it, a periodic argument
  that is not linear, more breaks than MostBreaks -- the caller is told so,
  and integrates as it did before.
  A break is carried as its exact point, where the one-sided limits are taken, and its
  decimal value, which only orders the breaks and places them in the range.
  https://github.com/asc-community/AngouriMath/issues/1508
  

• AsymptoticSeries

  Summary

  A power series in w around w -> 0+, truncated at a known exponent, whose
  coefficients are arbitrary expressions not containing w.
  

  Remarks

  This is what Gruntz's algorithm expands in, and it is not a Taylor series: the
  exponents may be negative and fractional, and the coefficients are symbolic rather
  than numeric. What is wanted from it in the end is only the leading term, but the
  leading terms of a sum can cancel -- which is the whole point of the algorithm, since
  e^(x + e^-x) - e^x cancels to every order when its two parts are expanded separately
  -- so the series has to be carried far enough to see past the cancellation, and how
  far that is is not known in advance. The caller raises the order until a leading term
  survives.
  The value of log(w) is supplied from outside rather than left as a logarithm of
  a symbol. Gruntz chooses w as an exponential, so its logarithm is an ordinary
  expression in x, and expanding a logarithm needs it: without it the constant term of
  log(c * w^e) cannot be told from zero.
  

• BreakpointIntegration

  Summary

  A definite integral whose integrand can jump inside the range, split at the jumps. A
  piecewise breaks where a case's condition changes truth; floor(x) and
  ceil(x) break at every whole number. Between two breakpoints the piecewise is one
  of its cases and the step is one constant, so the pieces integrate through an
  antiderivative and add.
  

  Remarks

  F(b) - F(a) is the integral only where F is continuous on [a, b], and
  an antiderivative found with the piecewise's cases or the floor taken as constants is
  continuous between two jumps and not across one. Taken across one it answered
  integral(x - floor(x), x, 0, 3) with 0 and the tent map's integral over
  [0, 1] with 1, where they are 3/2 and 1/2; with a symbolic
  bound it answered (n - floor(n))^3 / 3, and with a piecewise it handed back a
  piecewise that still mentioned the integration variable. None of that happens here: an
  integrand with a break never goes through an antiderivative between two bounds.
  Finitely many breakpoints -- a piecewise whose conditions compare the variable with
  numbers -- are collected, the range is cut at the ones inside it, and on each piece the
  piecewise is replaced by the case that holds at the piece's midpoint, which is exact
  because the condition has no other place to change. A condition that compares the
  variable with something symbolic, or a bound that is symbolic, is declined: the pieces
  depend on where the jumps fall, and answering across them is what this exists to stop.
  A piecewise whose conditions do not mention the variable is a constant here and goes
  through the antiderivative as one.
  Infinitely many, evenly spaced -- a floor or a ceiling of the variable -- are the
  same split written as a sum: on [n, n + 1) the floor is n, the ceiling
  n + 1, x is n + t with t over [0, 1), and the integral
  over whole bounds is a sum over n of an integral over t with no step in it,
  which the summation's closed forms answer; +oo is allowed as the upper bound. A
  numeric bound that is not whole contributes the piece up to the nearest whole number,
  on which the step is one constant. The fractional part x - floor(x) is substituted
  as a unit, since written out it arrives as n + t - n, and a term subtracted from
  itself simplifies to a conditional zero rather than to nothing
  (#1174).
  Offered only where every piece then resolves; an integral rewritten as an unevaluated sum
  of unevaluated integrals is not an answer. Question I.2 of
  #1212, and the
  review of #1215,
  which asked for the general mechanism rather than the floor's special case.
  

• CubeRootLogarithmAnsatz

  Summary

  A rational function of x and one cube root y = P(x)^(1/3) of a polynomial,
  integrated by an ansatz over logarithms of L - y for linear L, their
  conjugates, and a rational part in y and y^2.
  

  Remarks

  The curve y^3 = P(x) has no rational parametrisation for a cubic P, so no
  substitution makes these rational; the ones that are elementary at all -- Welz's
  x/((1 + x)(1 - x^3)^(1/3)), 1/(x^3 - 3x^2 + 7x - 5)^(1/3),
  (x - 1)/((x + 1)(2 + x^3)^(1/3)) -- are integrated by logarithms of
  L(x) - y, with L linear, and of their conjugates L - w y for the
  cube roots of unity w, which pair into ln(L^2 + L y + y^2) and
  arctan((2L + y)/(sqrt(3) y)). The norm (L - y)(L - wy)(L - w^2 y) = L^3 - P is where the logarithm has its poles, so the L that can occur are the ones
  whose cube agrees with P at the places the integrand has poles: the linear
  function through two points of the curve, or tangent to it at one. Each of those is
  a candidate here -- the tangent at a rational pole, at infinity, and at a root of
  P; the line through a pole and a root, or two poles -- and for an irreducible
  quadratic factor Q of the denominator with P constant modulo Q,
  the cube roots of that constant times a cube root of unity of Q[x]/(Q), which
  is where ln(2^(1/3) - y), ln(2^(1/3) x + y) and
  ln(2^(1/3)(x - 1) - y) in Welz's answers over 1 - x + x^2 come from.
  The constants those bring in are cube roots of rationals, 2^(1/3) most often,
  so the coefficients live in Q(c) with c^3 rational; the linear system is
  solved in that field exactly, as polynomials in c reduced modulo c^3 - m,
  rather than numerically -- a symbol for c would not be told from zero at
  c^3 - 2, and a floating c would give floating coefficients. One such
  constant per integrand; two independent ones would need their compositum and are
  declined. With symbols in the coefficients and every constant rational, the general
  symbolic solver is used instead.
  The identity sum c_k g_k' = f is an identity in the function field
  Q(x)[y]/(y^3 - P): everything is brought over one denominator in x, the
  numerators reduced by y^3 = P to A + B y + C y^2, and the three
  coefficient polynomials in x matched term by term. The answer is checked against
  the integrand at sampled points before it is returned.
  https://github.com/asc-community/AngouriMath/issues/718
  

• EquationSolver

• Gruntz

  Summary

  Gruntz's algorithm for the limit of an expression as x tends to positive infinity,
  after D. Gruntz, "On Computing Limits in a Symbolic Manipulation System", ETH 1996.
  

  Remarks

  The idea is to work out which subexpressions grow fastest, rewrite the expression in
  terms of a single one of them, and read the answer off the leading term of the power
  series that leaves. Two functions are in the same comparability class when
  log|f| / log|g| tends to something finite and non-zero, and the set of subexpressions
  in the fastest class present is the mrv set. Every member of it can be written as a
  power of any other member times something slower, so one member is picked -- turned
  upside down if need be, so that it tends to zero -- and everything is written in terms
  of it. What is left is a series in that one quantity, and the sign of the leading
  exponent says whether the limit is zero, infinite, or the limit of the leading
  coefficient, which is a smaller problem of the same kind.
  This answers the limits that defeat term-by-term expansion because the terms cancel to
  every order. lim x -> +oo e^(x + e^(-x)) - e^x is the standard one: expanding the two
  exponentials separately gives two divergent series whose difference cancels entirely,
  while rewriting the whole expression in w = e^(-x) gives (e^w - 1)/w, whose leading
  term is 1.
  The scope here is the exp-log functions: what can be built from x and the rationals
  with the four operations, exp and log. That is the class the algorithm is proved for,
  because those functions are eventually monotone and so comparable at all. Anything
  else -- a sine, an unknown function, a factorial -- makes this decline rather than
  guess, since sin(x) has no limit at infinity and no comparability class either.
  

• IndefiniteIntegralSolver

• IntegralPatterns

• Integration

• LimitFunctional

• LimitSolvers

• OrdinaryDifferentialEquation

  Summary

  The first-order linear ordinary differential equation, solved by its integrating factor.
  #241

  Remarks

  For y' + f(x) y = g(x), multiplying through by m = e^(int f dx) makes the left
  side the derivative of m y — which is the whole method, since integrating both sides
  then gives m y = int (m g) dx. Nothing here is a heuristic: where the two integrals
  come out, the answer is exact, and where either does not this declines.
  The unknown is an Application. A bare Variable cannot
  stand for a function of x: derivative(y, x) is 0, because y does
  not contain x and the library is right about that. Written apply(y, x) it is
  an application that does not reduce, and its derivative stays symbolic — which is exactly
  what an equation about an unknown function needs.
  Linearity is checked, not assumed. The coefficients are read off by differentiating
  with respect to the unknown and its derivative, which is only valid if the equation really
  is linear in them — so the reading is put back together and compared with what it came
  from. An equation that is not linear fails that comparison and is declined rather than
  answered wrongly.
  

• RothsteinTrager

  Summary

  The integral of a rational function with rational coefficients, whatever its
  denominator: the Hermite reduction for the rational part of the answer, and the
  Rothstein–Trager resultant for the logarithmic part, written in real terms after Rioboo.
  

  Remarks

  The partial-fraction rules answer a denominator that factors over the rationals, and
  a biquadratic one over the reals, and stop there: Bronstein's
  (6 - 3x^2 + x^4)/(4 + 5x^2 - 5x^4 + x^6) has a denominator irreducible over
  Q whose real factors carry the roots of a cubic, and its antiderivative is
  arctan(x (1 - 3x^2 + x^4)/2) and two more arctangents of linears, with no root
  of that cubic anywhere in it. The logarithmic part of the integral of A/D, with
  D squarefree, is sum c ln(gcd(D, A - c D')) over the roots c of the
  resultant R(t) = res_x(D, A - t D'), and the residues c are what decide the
  field the answer needs, not the roots of D: here they are ±i/2 and
  ±i, and the answer is over Q with an i that pairs into arctangents.
  This takes the rational part first by the Hermite reduction, then the resultant, its
  squarefree decomposition -- a root of multiplicity i is a residue shared by
  i roots of D, and the gcd for it has degree i -- and the
  irreducible factors of each squarefree part over Q. A linear factor is a
  rational residue and a logarithm; a quadratic one is a pair of conjugate residues
  a ± w with w^2 rational, and the gcd is computed in Q(w): for
  w real the pair is two logarithms with a root in them, and for w = i b it
  is a ln(P^2 + b^2 Q^2) + b LogToAtan(P, b Q), Rioboo's continuous arctangent
  form. A factor of higher degree puts the residues in a field this does no arithmetic in,
  and those logarithms are written as a sum over the roots of the factor of the denominator
  they belong to, sum(A(r)/D'(r) ln(x - r), r in { r : E(r) = 0 }) (https://github.com/asc-community/AngouriMath/issues/1285).
  Bronstein, Symbolic Integration I, §2.2 (HermiteReduce), §2.5
  (IntRationalLogPart) and §2.8 (LogToReal, LogToAtan). The resultant is taken as a
  Sylvester determinant rather than from a subresultant remainder sequence, so the gcd
  is computed for each residue in its own field instead of being read off the
  sequence -- that is the classical Rothstein–Trager rather than Lazard–Rioboo–Trager,
  affordable here because the fields are of degree at most two. The answer is checked
  against the integrand at sampled points before it is returned.
  https://github.com/asc-community/AngouriMath/issues/718
  

• SquareRootLogarithmAnsatz

  Summary

  A rational function of x and one square root y = sqrt(P(x)) of a quadratic,
  a cubic or a quartic, integrated by an ansatz over logarithms of A - B y and
  arctangents of A/(B y) for polynomials A and B, and a rational
  part in y.
  

  Remarks

  The curve y^2 = P(x) is elliptic for a squarefree cubic or quartic P, so no
  substitution makes these rational; the ones that are elementary at all -- Welz's
  (1 + x)/((x - 2) sqrt(1 + x^3)), which is -(2/3) atanh((1 + x)^2/(3 sqrt(1 + x^3))),
  Bronstein's x/sqrt(x^4 + 10x^2 - 96x - 71) -- are integrated by logarithms of
  A - B y. The norm (A - By)(A + By) = A^2 - B^2 P is where such a logarithm
  has its poles, so the A and B that can occur are the ones for which
  A - By vanishes at the places the integrand has poles: at a rational pole
  a with P(a) a square, A/B agrees with a branch of y to some
  order there, which is a Padé approximant of the branch's series -- Welz's
  (1 + x)^2/3 is the second-order Taylor polynomial of sqrt(1 + x^3) at
  2, where the norm (1 + x)(2 - x)^3 has its triple root -- and at infinity
  likewise for a quartic with a square leading coefficient, which is Bronstein's. Where
  P(a) is the negative of a square the branch is imaginary, the conjugate pair of
  logarithms is one arctangent of A/(B y), and the norm is A^2 + B^2 P.
  Each Padé approximant at each place, for a few orders, is a candidate; the identity
  sum c_k g_k' = f is one in Q(x)[y]/(y^2 - P), brought over one denominator
  and matched coefficient by coefficient, and solved exactly over the rationals.
  An irreducible quadratic factor of the denominator is a pair of conjugate places, at
  which the branch begins with a line alpha + beta x over the rationals or a
  Q(sqrt(d)) of its own, lifted to any power of the factor; what such a candidate
  contributes to the system is the conjugate difference of its logarithms over
  sqrt(d), which is rational, so that candidates over different fields sit in one
  system over the rationals. For a quadratic P the curve is rational and Euler's
  substitutions answer in principle, but through a rational function whose residues can
  lie in a field of degree four -- Timofeev's (3 + x)/((1 + x^2) sqrt(1 + x + x^2)) -- and here it is a logarithm and an arctangent of (1 +- x)/(sqrt(2) y). A
  rational pole where P is neither a square nor the negative of one wants the field
  of its square root, and is left for now. The answer is checked against the integrand at
  sampled points before it is returned.
  https://github.com/asc-community/AngouriMath/issues/718