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arXiv:1907.07204v1 [math.CV] 16 Jul 2019
On the enumeration of the roots of arbitrary
separable equations using HW hyper-Lambert
maps
Ioannis Galidakis
Department of Mathematics
Agricultural University of Athens
jgal@aua.gr
Ioannis Papadoperakis
Department of Mathematics
Agricultural University of Athens
papadoperakis@aua.gr
May 2019
1
Abstract
In this article we use the HW maps to solve arbitrary equations f= 0, by
providing an effective enumeration of the roots of f, as these project on and at
the branches of the HW maps. This is just an enumeration of the pro jection
points (roots) of a pin-line on the Riemann surface of fthrough HW.
1 Introduction
The HW maps have been used to determine the attractors of the infinite expo-
nential whenever it falls into a p-cycle in [3] and in [2] to solve certain transcen-
dental equations such as Kepler’s Equation. They have also been used in [4] to
solve in closed form the generalized Abel differential equation. Here we display
a simple algebraic scheme which can be used to utilize the solution of arbitrary
equations using the HW maps, by providing an effective enumeration of all the
roots of arbitrary equations f= 0, using the branches of the maps HW. Imagine
an arbitrary multivalued f, for which we want force f(x) = 0. We line-pin the
entire Riemann surface of ffrom top to bottom starting at the complex origin.
The local projection pin points ziwill be exactly the roots of f= 0. Because
the branches of HW can be enumerated starting at the origin, all the roots zi
of f= 0 can therefore be enumerated and referenced by approximating just an
ǫpin through the origin.
2 Definitions
Suppose fn(z) are non-vanishing identically complex functions, with n≤n0∈
N. We define Fn(z): N×C→Cas:
Definition 2.1
Fn(z) = (1, if n= 1,
efn−1(z)Fn−1(z), if n > 1.
Definition 2.2 G(f1, f2,...,fn;z) = z·Fn+1(z)
If n= 0, then G(z) = z. If n= 1 then G(f1;z) = zef1(z). If n= 2 then
G(f2, f1;z) = zef2ef1(z). When we write about the HW, we can use the termi-
nology G(...;z), meaning that the corresponding function includes meaningful
terms-parameters. The order of the functions is immaterial and we can re-order
them to get to the function of interest here, which is the inverse of G(...;z),
denoted by,
HW(f1, f2,...,fn;y) (1)
In other words Gand HW satisfy the functional relation:
G(...; HW(...;y)) = y(2)
2
by supposing always that the list of parameters is identical on both sides. These
maps have been called generalized hyper-Lambert HW functions and in general
they are multivalued. We note that when n= 1, HW(y) satisfies a more general
form which comes from the Lambert function W, i.e., zef1(z)=y. The Lambert
function satisfies zez=y. The existence of all the HW is guaranteed in all cases
by the Lagrange Inversion Theorem (see [5, 201-202]).
3 An indexing scheme for the HW maps
3.1 An algebraic scheme
For the complex maps log and W, their indexing scheme is the simplest possible,
that is log(k, z ) and W(k, z), k∈Z. There exists an indexing scheme which
indexes identically the mappings HW but it is not integral. Dubinov in [1] solves
Kepler’s equation, using the following algebraic inversion:
E−ǫ·sin(E) = M⇒
E1−ǫsin(E)
E=M⇒
E·elog (1−ǫ·sinc E)=M⇒
E= HW [log(1 −ǫ·sinc(x)); M]
(3)
The inversion above can be generalized producing a removable pole at z0of
multiplicity n. Setting w= (z−z0)n, with z0such that f(z0) = y, we have:
f(z) = y⇒
(z−z0)n·f(z)
(z−z0)n=y⇒
w·elog(f(z)
w)=y⇒
w= HW log f(z)
w;y⇒
(z−z0)n= HW log f(z)
(z−z0)n;y⇒
z= HW log f(z)
(z−z0)n;y1
n
+z0
(4)
The scheme above gives an index into the set of the HW functions, in the
form of a functional parameter as log f(z)
(z−z0)n. Now, if we know f(z), this
scheme can give identities which must hold identifying this way the correspond-
ing function.
We can now list how the most important categories of complex functions are
solved based on this index.
3
3.2 Polynomial functions
Suppose then that f(z) =
N
Q
k=1
(z−zk)nk. Keeping kfixed and setting w=
(z−zk)nk, we have,
w= HW log f(z)
w;y⇒
z= HW log f(z)
(z−zk)nk;y1
nk+zk⇒
z= HW [log(f(z)) −nklog (z−zk) ; y]1
nk+zk
(5)
Theorem 3.1 If f(z) =
N
Q
k=1
(z−zk)nkis a complex polynomial function, then
the inverse of f(z)relative to yis given by the function HW and the last equation
of (5), whose Riemann surface has at most m=
N
P
k=1
nkbranches, indexed by
m, with k∈N.
Proof: The last expression of (5) is true for any k∈ {1,2,...,N}, therefore the
multiplicity is at least mbecause for each kthe multiplicity is at least nkand
each nkmay give different branches. This means that the expression can index
fully all the branches of the corresponding HW using only an integral index k.
Theorem 3.2 If f(z)is a complex polynomial function, the roots of f(z) = y
are given directly by a suitable HW function.
Proof: Using equation (1) of Definition (2.2), follows that for each HW,
HW(...; 0) = 0, therefore calculating the corresponding HW of the last equa-
tions in (5) at y= 0, forces z=zkand these are the roots of f(z) = y.
Therefore, we can extract all the roots of equation f(z) = y, manually. The
first root, suppose z1, is extracted as,
z1= HW log f(z)
z;y
g1(z) = f(z)−y
z−z1
Having the root z1, the rest of the roots can be extracted recursively for
1≤k≤N−1 as,
zk+1 = lim
ǫ→0+HW log gk(z)
z;ǫ
gk+1(z) = gk(z)
z−zk+1
4
and the Theorem follows.
3.3 Rational functions
We suppose that f(z) = P(z)/Q(z), with P(z), Q(z) polynomial functions. We
have similar results here.
Theorem 3.3 If f(z) = P(z)/Q(z)is a complex rational function such that
N= max {deg(P),deg(Q)}, then the inverse of f(z)relative to yis given by:
z= HW log P(z)−y·Q(z)
(z−zk)nk;y1
nk+zk
whose Riemann surface has at most m=
N
P
k=1
nkbranches, indexed by m,
with k∈N.
Proof: If F(z) = P(z)−y·Q(z), then F(z) is a polynomial of degree N, in
which case the Theorem follows similarly, with f(z) replaced by F(z).
Theorem 3.4 If f(z)is a complex rational function, the roots of f(z) = ycan
be given by a suitable HW function.
Proof: Similarly, if F(z) = P(z)−y·Q(z), then F(z) is polynomial map of
degree N, therefore we can extract its roots as:
z1= lim
ǫ→0+HW log F(z)
z;ǫ
g1(z) = F(z)
z−z1
Having z1, the rest of the roots can be extracted recursively for 1 ≤k≤N−1
as,
zk+1 = lim
ǫ→0+HW log gk(z)
z;ǫ
gk+1(z) = gk(z)
z−zk+1
and the Theorem follows.
We observe that when Q(z) = 1, the case of a polynomial function arises.
3.4 Analytic functions
For an analytic function f(z) =
∞
P
n=0
αn·(z−z0)nin some region D⊆C, with
z0∈D, we have similar results.
5
Theorem 3.5 If f(z) =
∞
P
n=0
αn·(z−z0)nis a complex analytic function, then
the inverse of f(z)relative to yis given by a suitable HW function:
z= HW log f(z)
(z−zk)nk;y1
nk+zk
whose Riemann surface has infinitely many branches given by n∈N.
Proof: Suppose TN(z) =
N
P
n=0
αn·(z−z0)n, is the corresponding Taylor poly-
nomial of degree N. Then TN(z) is obviously a polynomial function, therefore
the inverse of TN(z) relative to yis given again by Theorem (3.1).
z= HW log TN(z)
(z−zk)nk;y1
nk+zk(6)
TN(z)→f(z) uniformly in compact subsets and the HW are analytic ([3]),
therefore (6) implies that the inverse is given by:
z= lim
N→∞
HW log TN(z)
(z−zk)nk;y1
nk+zk⇒
z= HW "log lim
N→∞
TN(z)
(z−zk)nk!;y#
1
nk
+zk⇒
z= HW log f(z)
(z−zk)nk;y1
nk+zk
(7)
and the Theorem follows.
We observe that in this case the inverse function has infinitely many branches,
since Nis not bounded.
Theorem 3.6 If f(z)is a complex analytic function, the roots of f(z) = yare
given again by a HW function.
Proof: We can extract the roots as:
z1= HW log f(z)
z;y
g1(z) = f(z)−y
z−z1
(8)
The rest of the roots can be again extracted recursively for 1 ≤kas,
6
zk+1 = lim
ǫ→0+HW log gk(z)
z;ǫ
gk+1(z) = gk(z)
z−zk+1
(9)
and the Theorem follows.
4HW functional index
An open problem set in [2, 1114-1115] is whether there is a way to effectively
index the numbering of the branches of the HW functions. With the following
Theorem we show that the answer is affirmative.
Theorem 4.1 If f(z)is a complex function and zk∈C,k∈N, such that
f(zk) = yand suppose gk(z)follows as in equations (8) -(9). Then, if HW is
the inverse of f(z)relative to y, the following scheme covers all the branches of
this inverse of f(z):
zk+1 =
HW h(0,) log f(z)
z;yi, if k= 0,
lim
ǫ→0+HW hk, log gk(z)
z;ǫi, if k > 0.
Proof: The proof follows from (3) along with Theorems 3.1, 3.3 and 3.5.
Note that for a specific analytic fexpanded around zk, we define F(z) =
log f(z)
(z−zk)nk. The map Fcreates a Laurent series with residue exp(ank),
which is gotten from the HW through the Residue Theorem of Cauchy for f,
with winding number ankaround zk. Consequently, the repeated application of
F(via gk) for z=zk, extracts recursively all the roots zkof the inverse and as
such it can be used as an index for the corresponding Riemann surface.
5 Conclusions
The HW maps can solve any equation f= 0, provided it can be brought into
a separable form with all z’s on the left and one won the right. Further, the
enumeration of the roots is origin consistent relative to the force of f.
6 Appendix: programming with the HW maps
Code for the HW maps is given below. Arguments are HW(functional index,y,n)
restart;
Digits:=40;
HW := proc ()
7
local y, n, c, s, p, sol, i, aprx, dy, dist, r, newr, oldr, fun, dfun,eps;
if nargs < 2 then ERROR("At least two arguments required") end if;
n := args[-1]; y := args[-2]; c := [args[1 .. -3]];
if y = 0 then 0 else dist := infinity;
eps:=1e-10; fun := 1; for i from 1 to nargs-2 do fun := exp(c[-i]*fun) end do;
fun := z*fun-y; dfun := diff(fun, z);
s := series(fun, z, n); p := convert(s, polynom);
sol := {fsolve(p = 0, z, complex)};
for i from 1 to nops(sol) do aprx := evalf(subs(z = op(i, sol), fun));
dy := evalf(abs(aprx)); if dy <= dist then r := op(i, sol);
dist := dy end if end do; oldr := r;
newr := r-evalf(subs(z = r, fun)/subs(z = r, dfun));
for i from 1 to 1000 while abs((oldr-newr)/oldr)>eps do oldr := newr;
newr := newr-evalf(subs(z = newr, fun)/subs(z = newr, dfun)) end do;
newr end if end proc:
Example: Using the program with five decimal digits accuracy to solve the
equation (z−2)(z−3)(z−5) = 2,
y:=2;
f:=z->(z-2)*(z-3)*(z-5);
z1:=HW(log(f(z)/z),y,10);
g1:=z->(f(z)-y)/(z-z1);
z2:=HW(log(g1(z)/z),1e-20,10);
g2:=z->g1(z)/(z-z2);
z3:=HW(log(g2(z)/z),1e-20,10);
gives:
z1≃2.36523 −0.69160i
z2≃2.36523 + 0.69160i
z3≃5.26953
Using the program for an approximate solution with Maple,
solve(f(z)=y,z);
evalf(%);
gives:
5.26953, 2.36523+0.69160i, 2.36523-0.69160i.
Example 2: Using the program to five digits of accuracy to solve the equa-
tion (z−2)(z−3)/(z−5)/(z−1) = 2,
y:=2;
f:=z->(z-2)*(z-3)/(z-5)/(z-1);
P:=unapply(numer(f(z)),z);
Q:=unapply(denom(f(z)),z);
8
F:=unapply(P(z)-y*Q(z),z);
z1:=HW(log(F(z)/z),1e-10,10);
g1:=z->F(z)/(z-z1);
z2:=HW(log(g1(z)/z),1e-10,10);
gives:
z1≃6.37228
z2≃0.62771
Using Maple approximation code,
solve(F(z)=y,z);
evalf(%);
gives:
6.37228, 0.62771.
Example 3: Using the program to five decimals of accuracy to solve the
equation sin(z) = 1/2,
y:=1/2;
f:=z->sin(z);
z1:=HW(log(f(z)/z),y,10);
g1:=z->(f(z)-y)/(z-z1);
z2:=HW(log(g1(z)/z),1e-20,10);
g2:=z->g1(z)/(z-z2);
z3:=HW(log(g2(z)/z),1e-20,10);
gives:
z1≃0.52359
z2≃2.61799
z3≃ −3.66519
z4≃ · · ·
The results are approximations of the numbers π/6, 5π/6, −7π/6,..., which
are the roots of sin(z) = 1/2. Many more complex equations can be solved here,
provided they are separable and the terms are analytic, like
Example 5: Using the code with five decimal accuracy to solve the equation
sin(z) + exp(sin(z))/p1 + tanh(z),
y=1/2;
f:=sin(z)+exp(sin(z))/sqrt(1+tanh(z));
z1:=HW(log(f(z)/z), y, 10);
g1 := z->(f(z)-y)/(z-z1);
z2:=HW(log(f1(z)/z),1e-20,10);
9
g2:=z->g1(z)/(z-z2);
z3:=HW(log(g2(z)/z),1e-20,10);
g3:=z->g2(z)/(z-z3);
z4:=HW(log(g3(z)/z), 0.1e-19, 10);
gives:
z1≃ −0.37435
z2≃ −1.71811
z3≃3.26659
z4≃6.15846
While using Maple approximation code,
solve(f(z)=y,z);
gives an open answer in terms of “RootOf ”, i.e. it cannot relay the roots
directly.
Example 6: Using the code with five decimal accuracy to solve the equation
z3−4z2+ 5z,
y:=2;
f:=z->z^3-4*z^2+5*z;
z1:=HW(log(f(z)/z),y,10);
g1:=z->(f(z)-y)/(z-z1);
z2:=HW(log(f1(z)/z),1e-20,10);
g2:=z->g1(z)/(z-z2);
z3:=HW(log(g2(z)/z),1e-20,10);
gives:
z1≃2
z2≃1.00005
z3≃1.00000
The description calculates correctly roots with multiplicity greater than 1.
The example is f(z)−y= (z−1)2(z−2), therefore the multiplicity of the
root 1 is indeed 2.
10
References
[1] A.E. Dubinov and I.N. Galidakis. An explicit solution to Kepler’s equation.
Physics of Particles and Nuclei. Letters, 4(3):213–216, May 2007.
[2] I.N. Galidakis. On some applications of the generalized hyper-Lambert func-
tions. Complex Variables and Elliptic Equations, 52(12):1101–1119, Dec.
2007.
[3] I.N. Galidakis. On solving the p-th complex auxiliary equation f(p)(z) = z.
Complex Variables, 50(13):977–997, Oct. 2005.
[4] P. Nastou, Y. Stamatiou, and A. Tsiakalos. Solving a class of odes arising in
the analysis of a computer security process using generalized hyper-lambert
functions. International Journal of Applied Mathematics and Computation
(IJAMC), 4(3):67–76, 2012.
[5] S. Saks and A. Sygmund. Analytic Functions. Hafner Publishing Company,
New York, 1952.
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