numeric::polysysroots
--
numerical roots of a system of polynomial equationsnumeric::polysysroots
(eqs, ..)
returns
numerical approximations of all real and complex roots of the
polynomial system of equations eqs
.
numeric::polysysroots(eqs)
numeric::polysysroots(eqs, vars)
eqs |
- | a polynomial equation or a list or a set of such
equations. Also expressions and polynomials of domain type DOM_POLY are accepted wherever an
equation is expected. They are interpreted as homogeneous
equations. |
vars |
- | an unknown or a list or set of unknowns. Unknowns may be identifiers or indexed identifiers. |
a set of lists of equations. The set {[]}
containing an
empty list is returned, if no solutions can be computed.
The function is sensitive to the environment variable DIGITS
, which determines the
numerical working precision.
linsolve
, numeric::fsolve
, numeric::linsolve
, numeric::polyroots
, numeric::realroot
, numeric::realroots
, polylib::realroots
, solve
vars
, then numeric::indets
(eqs)
is
used in place of vars
.[x1=value1,x2=value2,..] ,where x1,x2,.. are the unknowns. These simplified equations should be regarded as constraints on the unknowns. E.g., if an unknown x[1], say, does not turn up in the form x[1]=.. in the solution, then there is no constraint on this unknown and it is an arbitrary parameter. This holds true in general for all unknowns that do not turn up on the left hand side of the solved equations. Cf. example 2.
vars
determines the
ordering of the solved equations. If a set vars
is used, then an internal ordering is used.If the solution set of eqs
is not
finite, then numeric::polysysroots
may return solutions
with some of the unknowns remaining as free parameters. In this case
the representation of the solution depends on the ordering of the
unknowns! Cf. example 3. Further, if higher
degree polynomials are involved, then
numeric::polysysroots
may fail to compute roots. Cf.
example 5. The same may happen, when
eqs
contains symbolic parameters.
numeric::fsolve
to compute a single
numerical root, if numeric::polysysroots
cannot compute
all roots of the system. Note, however, that numeric::fsolve
does not accept
symbolic parameters in the equations.numeric::polyroots
to compute all
roots of a single univariate polynomial with numerical
coefficients.numeric::polysysroots
is a hybrid routine: it calls
the symbolic solver
solve(eqs, vars, BackSubstitution = FALSE)and processes its symbolic result numerically via
solvelib::allvalues
. Cf. example 4.Equations, expressions as well as DOM_POLY
objects may be used to specify the
polynomials:
>> numeric::polysysroots(x^2 = PI^2, x)
{[x = -3.141592654], [x = 3.141592654]}
>> numeric::polysysroots({x^2 + y^2 - 1, x^2 - y^2 = 1/2}, [x, y])
{[x = -0.8660254038, y = -0.5], [x = -0.8660254038, y = 0.5], [x = 0.8660254038, y = -0.5], [x = 0.8660254038, y = 0.5]}
>> numeric::polysysroots({poly(x^2 + y + 1), y^2 + x = 1}, [x, y])
{[x = -0.4533976515, y = -1.20556943], [x = 0.0, y = -1.0], [x = 0.2266988258 - 1.467711509 I, y = 1.102784715 + 0.6654569512 I], [x = 0.2266988258 + 1.467711509 I, y = 1.102784715 - 0.6654569512 I]}
Symbolic parameters are accepted:
>> numeric::polysysroots(x^2 + y + exp(z), [x, y])
1/2 {[x = (- 1.0 y - 1.0 exp(z)) ], 1/2 [x = - 1.0 (- 1.0 y - 1.0 exp(z)) ]}
The returned solutions may contain some of the unknowns remaining as free parameters:
>> numeric::polysysroots({x^2 + y^2 = z}, [x, y, z])
2 1/2 2 1/2 {[x = (z - 1.0 y ) ], [x = - 1.0 (z - 1.0 y ) ]}
The ordering of the unknowns determines the
representation of the solution, if the solution set is not finite.
First, the following equation is solved for x
leaving
y
as a free parameter:
>> numeric::polysysroots({x^3 = y^2}, [x, y])
2 1/3 2 1/3 {[x = (y ) ], [x = - (0.5 + 0.8660254038 I) (y ) ], 2 1/3 [x = - (0.5 - 0.8660254038 I) (y ) ]}
Reordering the unknowns leads to a representation with
x
as a free parameter:
>> numeric::polysysroots({x^3 = y^2}, [y, x])
3 1/2 3 1/2 {[y = (x ) ], [y = - 1.0 (x ) ]}
The symbolic solver produces a RootOf
solution of the following
system:
>> eqs := {y^2 - y = x, x^3 = y^3 + x}:
>> solve(eqs, BackSubstitution = FALSE)
2 {[x = 0, y = 0], [x = y - y, y = 3 2 4 5 RootOf(3 X3 - 2 X3 - X3 - 3 X3 + X3 + 1, X3)]}
Internally, numeric::polysysroots
calls
solve
and processes this
result numerically:
>> numeric::polysysroots(eqs, [x, y])
{[x = -0.238328984, y = 0.6080324762], [x = 0.0, y = 0.0], [x = 0.8911259621, y = -0.5682349751], [x = 2.237302267, y = 2.077118343], [x = - 1.445049623 + 0.1279930535 I, y = 0.441542078 - 1.094745154 I], [x = - 1.445049623 - 0.1279930535 I, y = 0.441542078 + 1.094745154 I]}
>> delete eqs:
The following equation is solved for the first of the specified unknowns:
>> eqs := y^5 - PI*y = x: solve(eqs, [x, y])
5 {[x = y - y PI]}
numeric::polysysroots
processes this output
numerically:
>> numeric::polysysroots(eqs, [x, y])
5 {[x = y - 3.141592654 y]}
The equation is solved for y
when the
unknowns are reordered. However, no simple representation of the
solution exists, so a RootOf
object is returned:
>> solve(eqs, [y, x])
5 {[y = RootOf(x + X5 PI - X5 , X5)]}
The roots represented by the RootOf
expression cannot be computed
numerically, because the symbolic parameter x
is
involved:
>> numeric::polysysroots(eqs, [y, x])
5 {[y = RootOf(x + 3.141592654 X6 - 1.0 X6 , X6)]}
>> delete eqs: