Solving the First
Auxiliary Real Exponential Equation: cx = x.
(The analysis that follows is a light introduction. For more
austere results, consult Paper 2).
(Lambert's W function in these articles will be denoted as W).
Preliminaries:
We have seen how the W function readily solves the (complex)
exponential
equation:
cz=z,
providing all the solutions (which are infinite, since the W
is multi-valued), as:
z = -W(k,-log(c))/log(c), for k in Z. (1) (where k denotes
the
appropriate W branch, and where k=0 denotes the principal branch).
(In effect, the points z above, are all the fixed points of the
function
f(z) = cz. For details on this, please consult the two
articles
on infinite exponentials).
We have also seen in the analysis of the complex
case, that the iterates of f(z) = cz often become a
p-cycle,
when we perturb z away from the function's fixed points.
Whenever such p-cycles are formed, we are interested in solving what
some authors have denoted as the auxiliary exponential equations
f(n)(z) = z, n >= 1. (Please check the references).
For example, the first auxiliary exponential equation, is: cz=z.
The second auxiliary exponential equation, is: ccz
=
z. The third auxiliary exponential equation is: cccz
= z, and so on. In practice, this amounts to figuring the fixed points
of iterates of f(n)(z), for n >=1. In this article, we
will
address the problem of solving the first auxiliary real exponential
equation:
cx = x, for c > 0, x in R.
We begin with some standard notation:
f(x) = cx, c > 0, x in R. f(n)(x) = {f(x), iff n = 1, f(f(n-1)(x)) iff
n > 1}. g(x) = f(x) - x. dg/dx = f(x)*ln(c) - 1. d2g/dx2 = f(x)*ln(c)2. W(k, x) denotes Lambert's W branch corresponding to k.
Continuity follows from the above definitions and basic analysis, so
it is implicitly assumed throughout this article. Note that for c >
0 all
the involved functions are analytic everywhere, and derivatives of all
orders exist everywhere. The continuity of the LW(-1,x) for x in
[-1/e,0),
will be addressed in a separate article.
We will approach the problem, using the following 6 lemmas:
Lemma 1:
If c > e(1/e), g(x) = 0 admits no real roots.
Lemma 2:
If c = e(1/e), g(x) = 0 admits exactly one real root, xr
= e.
Lemma 3:
If 1 < c < e(1/e), g(x) = 0 admits exactly two real
roots {xr1, xr2}, given by: xrj = e-W(j-2,-ln(c)),
j in {1,2}.
Lemma 4:
If e-e < c < 1, g(x) = 0 admits exactly one real root,
xr = e-W(-ln(c)).
Lemma 5:
If c = e-e, g(x) = 0 admits exactly one real root xr
= 1/e.
Lemma 6:
If 0 < c < e-e, g(x) = 0 admits exactly one real root,
xr = e-W(-ln(c)).
Lemmas 1-6 depend on the following lemmas:
Lemma 7:
If c > 1, then dg/dx = 0 admits exactly one real root:
xcrit = -ln(ln(c))/ln(c).
Lemma 8:
If c < 1, then dg/dx = 0 admits no real roots.
Lemma 3 depends on the following lemmas:
Lemma 9:
The function W(x) is real valued, continuous and strictly increasing
for x in [-1/e, +oo).
Lemma 10:
The function W(-1,x) is real valued, continuous and strictly
decreasing
for x in [-1/e, 0).
We begin with Lemmas 9 and 10:
Proof of Lemma 9:
Please see lemmas on previous articles on the Lambert's W function,
particularly
the infinite exponentials article, the complex
case, W's expansion and approximating
the principal branch of the W or check the
references
particularly [31].
Proof of Lemma 10:
We will take this up in another article, which will be devoted to
calculating
the W branch corresponding to -1.
Proof of lemmas 7 & 8:
dg/dx = 0, =>
f(x)*ln(c) - 1 = 0, =>
cx*ln(x) - 1 = 0, =>
e(ln(c)*x)*ln(x) = 1, =>
e(ln(c)*x) = 1/ln(c), =>
ln(c)*x = ln(1/ln(c)), =>
xcrit = -ln(ln(c))/ln(c).
Note that if c > 1, => ln(c) > 0, => ln(ln(c)) in R,
=> xcrit
in R, while if c < 1, => ln(c) < 0, => ln(ln(c)) not in R,
and Lemmas
7 and 8 follow.
Lemma 11:
If c > 1, then: if x < xcrit, then g(x) is strictly
decreasing
and if x > xcrit, then g(x) is strictly increasing.
Proof:
It suffices to show that dg/dx > 0, if x > xcrit, and
dg/dx
< 0, if x < xcrit.
c > 1, => ln(c) > 0.
x < xcrit, => x < -ln(ln(c))/ln(c), => x*ln(c)
<
-ln(ln(c)), => f(x) < 1/ln(c), => f(x)*ln(c) < 1, =>
f(x)*ln(c) -
1 < 0, => dg/dx < 0.
Similarly, x > xcrit, => dg/dx > 0,
and the lemma follows.
Lemma 12:
If c < 1, then g(x) is strictly decreasing everywhere.
Proof:
It suffices to show dg/dx < 0, everywhere.
c < 1, => ln(c) < 0, and f(x) is strictly increasing
everywhere.
x < xcrit, => x < -ln(ln(c))/ln(c), => x*ln(c)
> -ln(ln(c)),
=> f(x) > 1/ln(c), => f(x)*ln(x) < 1, => f(x)*ln(x) - 1
< 0, => dg/dx
< 0.
x > xcrit, => x > -ln(ln(c))/ln(c), => x*ln(c)
< -ln(ln(c)),
=> f(x) < 1/ln(c), => f(x)*ln(x) < 1, => f(x)*ln(x) - 1
< 0, =>
dg/dx < 0,
and the lemma follows.
Lemma 13:
If c > 1, then limx->+oog(x)=+oo, and limx->-oog(x)=+oo.
Proof:
The first limit follows trivially from the fact that for sufficiently
large positive x, f(x) > x.
The second limit follows trivially from the fact that for sufficiently
large negative x, -x > f(-x).
Proof of Lemma 1:
If c > 1 then, =>
d2g/dx2|x_crit = ln(c) > 0, => m
=
g(xcrit) is a minimum.
It suffices to show that m > 0, for all x in R.
Note:
g(xcrit) = 1/ln(c)+ln(ln(c))/ln(c).
But c > e(1/e), => ln(c) > 1/e => ln(ln(c)) >
-1, => g(xcrit)
> 0, and the lemma follows from lemma 11.
Proof of Lemma 2:
As above, c = e(1/e), => g(xcrit) = 0, and
note
that xcrit = xr = e, and the lemma follows from
lemma
11.
Proof of Lemma 3:
As above, 1 < c < e(1/e), => g(xcrit)
<
0, verify that the aforementioned expressions are roots, and the lemma
follows from lemmas 11 and 13.
Proof of Lemma 4:
e-e < c < 1, => ln(c) < 0.
dg/dx = f(x)*ln(c) - 1 = e(ln(c)*x)*ln(c) - 1. No matter
what x is, e(ln(c)*x) > 0, and ln(c) is always < 0, so
dg/dx
< 0, and g(x) is strictly decreasing everywhere. => It has at
most one
real root. But (1) gives a real root, so it has at least one real root,
=> g(x) has exactly one root, that given by (1).
Proof of Lemma 5:
Similar to the proof of Lemma 4 and note that:
g(1/e) = 0.
Proof of Lemma 6:
Similar to the proof of Lemma 4.
Now we are ready for some Maple code to solve our equation:
> solveAux1R:=proc(c)
> local fc,fb1,fb2,r1,r2;
> fc:=evalf(c); #turn c into a float
> fb1:=evalf(exp(exp(-1)));fb2:=evalf(exp(-exp(1)));
> if fc>fb1 then
> RETURN(`Ro Real Roots`);
> elif fc=fb1 then
> r1:=evalf(exp(1));
> RETURN(r1);
> elif 1<fc and fc<fb1 then
> r1:=evalf(exp(-W(-1,-ln(c))));
> r2:=evalf(exp(-W(0,-ln(c))));
> RETURN({r1,r2});
> elif c=1 then #If c=1, then equation is degenerate: h(1,x) = 1-x
> r1:=1;
> RETURN(r1);
> elif fb2<fc and fc<1 then
> r1:=evalf(exp(-W(-ln(c))));
> RETURN(r1);
> elif fc=fb2 then
> r1:=evalf(exp(-1));
> RETURN(r1);
> elif 0<fc and fc<fb2 then
> r1:=evalf(exp(-W(-ln(c))));
> RETURN(r1);
> elif c=0 then
> r1:=1;
> RETURN(r1);
> else
> RETURN(`No Real Roots`);
> fi;
> end:
