Re: Bringing back an old tetration curiosity?

On Sep 17, 8:12 am, theronr...@xxxxxxxxx wrote:
The proposed method in this thread is flawed. The generated "curve"
has branches, so it is not a function.

I've seen that thread, and I was curious about
the method myself.

By "branches," do you mean in the sense of a
complex function such as log(z) or sqrt(z)
having multiple branches? I'm wondering how
that could be, since we dealing with:

f(x + 1) = sqrt(2)^f(x) (2)

a real-valued function. If we need a branch
of the exponential, then we want the branch
that maps reals to positive reals. And of
course by "sqrt(2)" we mean the positive
value of the square root.

But I see that your graph does reveal an
important question -- is f well-defined on
the entire domain [-2,inf]? In other words
suppose we define S_n as:

S_0 = N union {-2,-1,0}
S_n+1 = S_n union (the set of all reals that
are the sum of a nonnegative integer and
an element of the image of the negation
function, restricted to the image of S_n)

Intuitively, S_n is the domain of the
function after n iterations, i.e.,

S_0 contains 0, 1, 2, etc.
S_1 contains -sqrt(2), -sqrt(2)^sqrt(2),
1-sqrt(2)^sqrt(2), etc.
S_2 contains -1/2, 1/2, etc.
(see original 2004 thread for reason)

Then so far, f is only well-defined on the
union of all S_n's. Let S be this union,
and clearly S is a proper subset of [-2,inf]
since S is the countable union of countable
sets and thus countable. The question is,
is S a _dense_ subset of S_n?

The graph suggests that it may not be. And
if it isn't, then Theron is correct, and
the attempt fails.

Of course, a graph is not a proof. I'm not
quite sure how one may make a rigorous
proof of the existence or nonexistence of a
function satisfying (1)-(3).

Here's a similar example. Suppose we were to
try to define a function from the nonnegative
reals to the nonnegative reals as follows:

(a) g(0) = 0
(b) g(x+1) = g(x)^2 + 1
(c) g(1/x) = 1/g(x)

(Notice that using continued fractions, g is
well-defined for all nonnegative rationals.)

Now we try to prove that no C^1 function can
possibly satisfy (a)-(c). Clearly g(1) = 1,
but what if we tried to calculate the
derivative at x = 1? We apply the Chain Rule
to rule (b):

g(x+1) = g(x)^2 + 1
g'(x+1) = 2g(x)g'(x)

then we plug in x = 0:

g'(0+1) = 2g(0)g'(0) (*)
g'(1) = 0 since g(0) = 0

Now we find g'(2) and g'(3):

g'(1+1) = 2g(1)g'(1)
g'(2) = 0 since g'(1) = 0

g'(2+1) = 2g(2)g'(2)
g'(3) = 0 since g'(2) = 0

And obviously we end up proving that g'
vanishes at all nonnegative rationals.

Then we apply the Chain Rule to (c):

g(1/x) = 1/g(x)
-1/x^2 g'(1/x) = -g'(x)/g(x)^2
g'(1/x) = g'(x) x^2/g(x)^2
g'(1/2) = g'(2) 2^2/g(2)^2
= 0 since g'(2) = 0

and so we show that g' vanishes for all
reciprocals of nonnegative integers --
and ultimately at all nonnegative
rationals (via continued fractions). And
since g' is continuous (as g is C^1), we
have that g' must vanish for all
nonnegative _reals_!

So we have a blatant contradiction, for
g' is zero everywhere, and the only C^1
functions whose derivative vanishes
everywhere are constant function. Yet
g(0) = 0 and g(1) = 1!

(I'm not sure if the proof is completely
rigorous, though. In line (*) I referred
to g'(0), but g is only defined for x>0,
so only a one-sided derivative exists. I
don't know how to apply the Chain Rule
to one-sided derivatives. Perhaps we've
only proved the right-hand derivative of
g'(1) to be zero, but since g is C^1,
the left-hand derivative is also 0? Or
else this proof fails.)

I'd like to see something similar above
for the OP's function.

.

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