(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.

d²h/dx² = 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.

Lemma 6:

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:

Proof of 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). d²h/dx² = 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 => d²h/dx²  <  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, d²h/dx²|_{y_crit} = 0, d³h/dx³|_{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 = -e², 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->-oodh/dx = -1, lim_x->+oodh/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, d²h/dx²|_{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: d³h/dx³|_{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, => d²h/dx² = 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, => d²h/dx² < 0, and the lemma follows.

Lemma 6.2:

dh/dx is strictly positive in (x_crit1, x_crit2).

Proof:

d²h/dx² = 0, => y_crit = ln(-1/ln(c))/ln(c), and verify: d³h/dx³|_{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: d²h/dx²|_xcrit1 = e^{W(-1,1/ln(c))}*W(-1,1/ln(c))*ln(c)²*(W(-1,1/ln(c))+1) > 0,
d²h/dx²|_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)} = c⁰ = 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}