(The analysis that follows is a light introduction. For more austere results, consult Paper 2).

We begin with some standard notation:

f(x) = c^{x}, 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.

h(x) = f^{(2)}(x) - x.

dh/dx = f^{(2)}(x)*f(x)*ln(c)^{2} - 1.

d^{2}h/dx^{2} = f^{(2)}(x)*(f(x))^{2}*ln(c)^{4}
+ f^{(2)}(x)*f(x)*ln(c)^{3}.

W(k, x) denotes the Lambert's W function branch corresponding to k.

__Lemma 1:__

If c > e^{(1/e)}, h(x) = 0 admits no real roots.

__Lemma 2:__

If c = e^{(1/e)}, h(x) = 0 admits exactly one real root x_{r} = e.

__Lemma 3:__

If 1 < c < e^{(1/e)}, h(x) = 0 admits exactly two real
roots {x_{r1}, x_{r2}}, given by: x_{rj} = e^{-W(j-2,-ln(c))},
j in {1,2}.

__Lemma 4:__

If e^{-e} < c < 1, h(x) = 0 admits exactly one real root
x_{r} = e^{-W(-ln(c))}.

__Lemma 5:__

If c = e^{-e}, h(x) = 0 admits exactly one real root x_{r}
= 1/e.

If 0 < c < e^{-e}, h(x) = 0 admits exactly three real
roots {x_{r}, x_{r1}, x_{r2}}, with x_{r}
= e^{-W(-ln(c))}, and x_{r1} < x_{r} < x_{r2}.

Lemmas 1-6 depend on the following lemmas:

__Lemma 7:__

If c > 1, then dh/dx = 0 admits exactly one real root: x_{crit} = ln(1/ln(c)*W(1/ln(c)))/ln(c).

__Lemma 8:__

If e^{-e} < c < 1, then dh/dx = 0 admits no real roots.

__Lemma 9:__

If c = e^{-e}, then dh/dx = 0 admits exactly one real root: x_{crit} = 1/e.

__Lemma 10:__

If 0 < c < e^{-e}, then dh/dx = 0 admits exactly two
real roots {x_{crit1}, x_{crit2}}:
x_{critj} = ln(1/ln(c)*W(j-2,1/ln(c)))/ln(c), for j in {1,2},
with x_{crit1} < x_{crit2}.

Lemma 10 depends on the following lemma:

__Lemma 11:__

W(-1,x) <= W(x) for all x in [-1/e, 0).

We begin with Lemma 11:

Follows from lemma 9 and lemma10 in a previous article, along with the fact that W(-1/e) = W(-1,-1/e) = -1. Here, we just exhibit the two graphs for x in [-1/e, 0):

>plot({W(x),W(-1,x)},x=-exp(-1)..0);

__Lemma 7.1:__

If c > 1, then dh/dx = 0 admits at most one real root.

__Proof:__

It suffices to show that dh/dx is strictly increasing throughout R.

c > 1, => ln(c) > 0. x1 < x2, => ln(c)*x1 < ln(c)*x2, => f(x1) < f(x2), => ln(c)*f(x1) < ln(c)*f(x2). Similarly: ln(c)*f^{(2)}(x1) < ln(c)*f^{(2)}(x2). Thus: f(x1)*f^{(2)}(x1)*ln(c)^{2} < f(x2)*f^{(2)}(x2)*ln(c)^{2},
=> f(x1)*f^{(2)}(x1)*ln(c)^{2} - 1 < f(x2)*f^{(2)}(x2)*ln(c)^{2}
- 1, => dh/dx|_{x1} < dh/dx|_{x2}, and the lemma follows.

__Proof of Lemma 7:__

c > 1, x_{crit} = ln(1/ln(c)*W(1/ln(c)))/ln(c), and verify that: x_{crit} is in R, dh/dx|_{x_crit} = 0, => dh/dx = 0 admits at least one real root, x_{crit}. By Lemma 7.1, dh/dx = 0 admits at most one real root. Thus: dh/dx = 0 admits exactly one real root: x_{crit}, and the lemma follows.

__Lemma 8.1:__

If 0 < c < 1 and y_{crit} = ln(-1/ln(c))/ln(c), then
if x < y_{crit}, dh/dx is strictly increasing and if x > y_{crit},
dh/dx is strictly decreasing.

__Proof:__

0 < c < 1, => ln(c) < 0, and verify that y_{crit} is in R. x < y_{crit} => ln(c)*x > ln(-1/ln(c)), => f(x) > -1/ln(c). d^{2}h/dx^{2} = f^{(2)}(x)*f(x)*[f(x)*ln(c)^{4}+ln(c)^{3}]
> f^{(2)}(x)*(-1/ln(c))*[(-1/ln(c)*ln(c)^{4}+ln(c)^{3}]
= 0. Similarly, x > y_{crit} => d^{2}h/dx^{2} < 0, and the lemma follows.

__Corollary:__

If 0 < c < 1, then dh/dx possesses a global maximum M at y_{crit}.

__Proof:__

0 < c < 1 => y_{crit} is in R as before, and verify that: dh/dx|_{y_crit} = M = -ln(c)/e - 1, d^{2}h/dx^{2}|_{y_crit} = 0, d^{3}h/dx^{3}|_{y_crit} = ln(c)^{3}/e
= N < 0. and the lemma follows from the above and lemma 8.1.

__Proof of Lemma 8:__

It suffices to show that for the M of the above corollary, M < 0.

If e^{-e} < c < 1, verify that: M = -ln(c)/e - 1 < 0, and the lemma follows from the corollary.

__Proof of Lemma 9:__

It suffices to show that dh/dx < 0, for all x in R - {1/e}. Apply the proof of lemma 8, with c = e^{-e}, y_{crit}
= 1/e, M = 0 and N = -e^{2}, and the lemma follows from lemma 8.1.

__Proof of Lemma 10:__

If 0 < c < e^{-e}, verify that: M = -ln(c)/e - 1 > 0, lim_{x->-oo}dh/dx = -1, lim_{x->+oo}dh/dx = -1. Note that 1/ln(c) is in (-1/e, 0), so by Lemma
9 and Lemma 10,
both W(-1,1/ln(c)) and W(1/ln(c)) are in R, => x_{critj} = ln(1/ln(c)*W(j-2,1/ln(c)))/ln(c), for j in {1,2},
are in R and satisfy: dh/dx|_{x_critj} = 0. and the lemma follows the above and lemma 8.1.

__Lemma 1.1:__

If c > 1 and x_{crit} = ln(W(1/ln(c))/ln(c))/ln(c), then if
x > x_{crit}, h(x) is strictly increasing and if x < x_{crit},
h(x) is strictly decreasing.

__Proof:__

If c > 1, verify that x_{crit} is in R. x > x_{crit} => x*ln(c) > ln(W(1/ln(c))/ln(c)), => f(x) > W(1/ln(c))/ln(c), => ln(c)*f(x) > W(1/ln(c)), => f^{(2)}(x) > e^{W(1/ln(c))}. dh/dx = f^{(2)}(x)*f(x)*ln(c)^{2} - 1 > e^{W(1/ln(c))}*W(1/ln(c))/ln(c)*ln(c)^{2}
- 1 = 1/ln(c)*1/ln(c)*ln(c)^{2} - 1 = 0. Similarly: x < x_{crit} => dh/dx < 0, and the lemma follows.

__Corollary:__

If c > 1, then h(x) possesses a global minimum m at x_{crit}.

__Proof:__

c > 1 => x_{crit} is in R, and verify that: h(x_{crit}) = m = f^{(2)}(x_{crit}) - x_{crit}, dh/dx|_{x_crit} = 0, d^{2}h/dx^{2}|_{x_crit} = ln(c)*[W(1/ln(c))
+ 1] = N > 0, and the lemma follows from lemma 1.1.

__Proof of Lemma 1:__

It suffices to show that if c > e^{(1/e)}, m > 0. We will show: x_{crit} = ln(W(1/ln(c))/ln(c))/ln(c) < -ln(ln(c))/ln(c),
<=> ln(W(1/ln(c))/ln(c)) < -ln(ln(c)), <=> W(1/ln(c))/ln(c) < 1/ln(c), <=> W(1/ln(c)) < 1, which holds by virtue of: c > e^{(1/e)}, <=> ln(c) > 1/e, <=> 1/ln(c) < e, <=> W(1/ln(c)) < W(e) = 1, so: g(x_{crit}) > 0, from Lemma
11 in the previous article, and therefore: f(x_{crit}) - x_{crit} > 0, => f(x_{crit}) > x_{crit}, => f^{(2)}(x_{crit}) > f(x_{crit}) > x_{crit},
=> f^{(2)}(x_{crit}) > x_{crit}, => f^{(2)}(x_{crit}) - x_{crit} = m > 0, => and the lemma follows.

__Proof of Lemma 2:__

It suffices to show that if c = e^{(1/e)}, m = 0. c = e^{(1/e)}, => x_{crit} = e, h(x_{crit}) = m = 0, N = 2/e, and the lemma follows from lemma 1.1.

__Proof of Lemma 3:__

Similar to the proof of lemma 1, and note that in this case: x_{crit} = ln(W(1/ln(c))/ln(c))/ln(c) > -ln(ln(c))/ln(c),
and verify that: and lim_{x->-∞}h(x) = +∞ and lim_{x->+∞}h(x) =
+∞,
the aforementioned expressions are roots, and the lemma follows from
lemma 1.1.

__Proof of Lemma 4:__

It suffices to show that dh/dx < 0, for all x, lim_{x->-∞}h(x)
= +∞, and lim_{x->+∞}h(x) = -∞. e^{-e} < c < 1, => -e < ln(c) < 0, => -ln(c)
<
e, => -ln(c)/e < 1. dh/dx = f^{(2)}(x)*f(x)*ln(c)^{2} - 1. Now verify that the point: y_{crit} = ln(-1/ln(c))/ln(c) is a critical point of the
function
dh/dx and furthermore: d^{3}h/dx^{3}|_{y_crit} = ln(c)^{3}/e
< 0, so y_{crit} is a maximum of dh/dx. Furthermore: x < y_{crit}, => x < ln(-1/ln(c))/ln(c), => x*ln(c) > ln(-1/ln(c)), => f(x) > -1/ln(c), => f(x)*ln(c) < -1, => f^{(2)}(x) < 1/e, => f^{(2)}(x)*ln(c) < ln(c)/e, => dh/dx = f^{(2)}(x)*f(x)*ln(c)^{2} - 1 < -ln(c)/e
- 1 < 0. Similarly, x > y_{crit}, => dh/dx < 0. Next verify that x_{r} is a root. The last two limits follow easily, and the lemma follows.

__Proof of Lemma 5:__

if c = e^{-e}, verify that: x_{crit} = ln(W(1/ln(c))/ln(c))/ln(c) = ln(W(-1,
1/ln(c))/ln(c))/ln(c)
= 1/e satisfies: dh/dx|_{x_crit} = 0, h(x_{crit}) = 0. One then verifies that dh/dx < 0, for all x in R - {1/e}, which
is similar to the proof of lemma 4, and the lemma follows.

__Lemma 6.1:__

Let y_{crit} = ln(-1/ln(c))/ln(c). Then dh/dx is strictly
increasing
in (x_{crit1}, y_{crit}] and strictly decreasing in [y_{crit},
x_{crit2}).

__Proof:__

if x in (x_{crit}, y_{crit}] then x < y_{crit}, => x < ln(-1/ln(c))/ln(c), => x*ln(c) > ln(-1/ln(c)), => f(x) > -1/ln(c), => f(x)*ln(c) < -1, => f^{(2)}(x) < 1/e, => f^{(2)}(x)*ln(c) > ln(c)/e, => d^{2}h/dx^{2} = f^{(2)}(x)*f(x)*[f(x)*ln(c)^{3}+ln(c)^{4}]
=

f^{(2)}(x)*ln(c)*f(x)*[f(x)*ln(c)^{2}+ln(c)^{3}]
> ln(c)/e*(-1/ln(c))*[(-1/ln(c)*ln(c)^{2}+ln(c)^{3}]
= ln(c)^{2}/e - ln(c)^{3}/e > 0. Similarly: x > y_{crit}, => d^{2}h/dx^{2} < 0, and the lemma follows.

__Lemma 6.2:__

dh/dx is strictly positive in (x_{crit1}, x_{crit2}).

__Proof:__

d^{2}h/dx^{2} = 0, => y_{crit} =
ln(-1/ln(c))/ln(c),
and verify: d^{3}h/dx^{3}|_{y_crit} = ln(c)^{3}/e
< 0, so y_{crit} gives rise to a (global) maximum for dh/dx. If dh/dx|_{x'} = 0 for some x' not in {x_{crit1}, x_{crit2}},
x' in (x_{crit1}, x_{crit2}), then either x' in (x_{crit1},
y_{crit}) or x' in (y_{crit}, x_{crit2}). But
then,
since dh/dx|_{x_crit1} = dh/dx|_{x_crit2} = 0, dh/dx
would
have to be both increasing and decreasing in that corresponding
interval.
This would mean that either dh/dx is a constant in that interval or
that
lemma 6.1 is violated in that interval. Both are contradictions, and
the
lemma follows.

__Proof of Lemma 6:__

If 0 < c < e^{-e}, => ln(c) < -e, => 1/ln(c)
> -1/e,
=> W(1/ln(c)) > -1, and W(-1,1/ln(c)) < -1, by Lemma
11. Now verify that: x_{r} is a root. x_{crit1} = ln(W(-1, 1/ln(c))/ln(c))/ln(c), and x_{crit2} = ln(W(1/ln(c))/ln(c))/ln(c), satisfy: dh/dx|_{x_critj} = 0, j in {1, 2}, and furthermore: d^{2}h/dx^{2}|_{xcrit1} = e^{W(-1,1/ln(c))}*W(-1,1/ln(c))*ln(c)^{2}*(W(-1,1/ln(c))+1)
> 0,

d^{2}h/dx^{2}|_{xcrit2} = e^{W(1/ln(c))}*W(1/ln(c))*ln(c)^{2}*(W(1/ln(c))+1)
< 0, so x_{crit1} gives rise to a (local) minimum and x_{crit2}
gives rise to a (local) maximum. It suffices to show that h(x) is strictly decreasing if x < x_{crit1}
and if x > x_{crit2}, lim_{x->-∞}h(x) = +∞, lim_{x->+∞}h(x)
= -∞, and that there is exactly one root between x_{crit1} and
x_{crit2}. x < x_{crit1}, => x < ln(W(-1,1/ln(c))/ln(c))/ln(c), => x*ln(c) > ln(W(-1,1/ln(c))/ln(c)), => f(x) > W(-1,1/ln(c))/ln(c), => f(x)*ln(c) < W(-1, 1/ln(c)) < -1, and f^{(2)}(x) < e^{W(-1, 1/ln(c))} < 1/e, dh/dx = f^{(2)}(x)*f(x)*ln(c)^{2} - 1 < -ln(c)/e
- 1 < -(-e)/e - 1 = 0. Similarly: x > x_{crit2}, => dh/dx < 0. Next, the two limits: c < 1, so lim_{x->+∞}f^{(2)}(x) = 0, => lim_{x->+∞}h(x)
= -∞. lim_{x->-∞}f^{(2)}(x) = lim_{x->+∞}f^{(2)}(-x)
= lim_{x->+∞}c^{(1/c^x)} = c^{0} = 1, => lim_{x->-∞}h(x) = +∞. Next verify that: h(x_{crit1}) < 0 and h(x_{crit2}) > 0: For this it suffices to show that: x_{crit1} < r < x_{crit2}, where r is the single
root of g(x) = 0 (in the range 0 < c < e^{-e}), for then,
since by Lemma 11 in
a previous article g(x) is strictly decreasing everywhere: g(x_{crit1}) > 0 and g(x_{crit2}) < 0, and
therefore: f^{(2)}(x_{crit1}) > f(x_{crit1}) > x_{crit1},
and f^{(2)}(x_{crit2}) < f(x_{crit2}) < x_{crit2}, and the desired result follows.

Indeed: x_{r} < x_{crit2}, <=> -W(-ln(c))/ln(c) < ln(W(1/ln(c))/ln(c))/ln(c), <=> -W(-ln(c)) > ln(W(1/ln(c))/ln(c)), <=> e^{-W(-ln(c))} > W(1/ln(c))/ln(c), <=> -W(-ln(c))/ln(c) > W(1/ln(c)), <=> -W(-ln(c)) < W(1/ln(c)), <=> W(-ln(c)) > -W(1/ln(c)). But: ln(c) < -e, => -ln(c) > e, => W(-ln(c)) > 1, and ln(c) < -e, => 1/ln(c) > -1/e, => W(1/ln(c)) > -1,
=> -W(1/ln(c))
< 1, and the inequality above holds.

Similar for: x_{crit1} < x_{r} and the desired result follows.
It follows that h(x) has at least two roots. By the Mean Value Theorem and the following two inequalities: h(x_{crit1})
< 0 and h(x_{crit2}) > 0, h(x) has at least three roots,
(one
more between x_{crit1} and x_{crit2}). If h(x) had more
than three roots, the fourth root would have to lie in (x_{crit1},
x_{crit2}). But this is impossible, for then: dh/dx would have
to be negative somewhere inside (x_{crit1}, x_{crit2})
and this contradicts lemma 6.2.

To order the roots, it suffices to show that x_{crit1} <
x_{r} < x_{crit2}. But this has been shown already
above
(since x_{r} = r = e^{-W(-ln(c))}), and the lemma
follows.

We can now modify our Maple code for "solveAuxR", to take care of this case as well:

> solveAux2R:=proc(c)

> local fc,fb1,fb2,xcf1,xcf2,r1,r2,r3;

> fc:=evalf(c); #Turn c into a float

> fb1:=evalf(exp(exp(-1)));fb2:=evalf(exp(-exp(1)));
#Turn
e^(1/e) into a float.

> if fb1=fc then #If c is equal to e^(1/e) then

> r1:= evalf(exp(1)); #one real root=e.

> RETURN(r1);

> elif fc>1 and fc<fb1 then #If c is in (1,e^(1/e)), then
calculate
real roots.

> r1:=evalf(exp(-W(-log(fc)))); #First root
always
given by W.

> r2:=evalf(exp(-W(-1,-log(fc)))); #Second root given by
W(-1,x)

> RETURN({r1,r2});

> elif c=1 then #If c=1, then equation is degenerate: h(1,x) = 1-x

> r1:=1;

> RETURN(r1);

> elif fc>=fb2 and fc<1 then

> r1:= evalf(exp(-W(-log(fc))));

> RETURN(r1);

> elif fc>0 and fc<fb2 then

> xcf1:=evalf(log(1/log(fc)*W(-1,1/log(fc)))/log(fc));

> xcf2:=evalf(log(1/log(fc)*W(1/log(fc)))/log(fc));

> r1:=evalf(exp(-W(-log(fc))));

> r2:=fsolve(fc^x=log(x)/log(fc),x,x=0..xcf1);

> r3:=fsolve(fc^x=log(x)/log(fc),x,x=xcf2..infinity);

> RETURN({r1,r2,r3});

> elif c=0 then

> r1:=1;

> RETURN(r1);

> else

> RETURN(`No Real Roots.`);

> fi;

> end:

Let's now try our code:

> c:=1.4;

> solveAux2R(c);

{1.886663306, 4.410292793}

> c:=0.0142;

> solveAux2R(c);

{.2905280101, .9178938293, .2013709754e-1}