Tower of power

[bonanova]

Take a variable x and raise it to the x^th power: x^x; call the result y.
Now raise x to the y power: x^y; call that result y. i.e. replace the previous value of y with the value of x^y.
Again, raise x to the y power: x^y; and replace the previous value of y with the value of x^y.

Repeat this an arbitrarily large number of times, and set the result equal to 2.
That creates the infinite exponential sequence

x to the x to the x to the x to the x to the x .... = 2.

What value of x, if any, satisfies the equation?

[Guest]

p(x) = lim_i→∞ a_i+1 = lim_i→∞ x^a_i = x ^{lim_i→∞ a_i} = x^p(x)(*** see edit)

*** edit2: actually, is my lim manipulation correct? can I swap powers and lim like that?

In answer to my own question, yes! Of course! Consider our sequence:

a₁, a₂, ...., a_i, ...., p(x)=n

and raise each term to p(x)=n

n^a₁, n^a₂, ...., n^a_i, ....

a₂, a₃, ...., a_i+1, ...., p(x)=n

So, as long as the original sequence converges, raising it to a power also converges to the same limit.

[Guest]

Same method solves p(x) = 100, yielding x = 100^1/100. Yet it is quite clear that P(100^1/100) does not converge to 100, but to some other much lower number.

If p(2^1/2) indeed converges to 2, then it must be some special significance/quality of the square root with respect to p(x), which we are yet to discover.

But as I said, there is no value of x which solves p(x)=100.

You need to think about it the other way around: if x (=y^1/y) <e^1/e (ie y<e) then p(x)=y . It cannot be used for solving p(x)=y for all y.

[Guest]

P(100^1/100) does not converge to 100, but to some other much lower number.

In fact, p(100^1/100) converges to y where y^1/y = 100^1/100 and y≠100, ie the 'paired' point on Chuck Rampart's graph<{POST_SNAPBACK}>.

For every y>e, p(y^1/y) converges to z where z^1/z = y^1/y, of which there is exactly one solution (z≠y) by my earlier proof that there is one maximum and the graph.

Bonanova cleverly confused us by choosing y = 2, so that x=2^1/2 and which also happens to equal 4^1/4, ie z=4

4^1/4 = 2^1/2

P(4^1/4) = P(2^1/2) = 2

Edited October 8, 2008 by foolonthehill

[Prime]

My complaint is that the method of solving p(x) = n by substituting p(x) with x^p(x) yields incorrect results where n is outside function's convergence (existence) range. I'd be more content if the method yielded some unreal solutions or something of the sort. Like in the case of x² = -1, for example.

Here is what I think p(x) graph looks like.

My other wild guess guess is that the portion of the graph between x=(0,1) is 90-degree rotation proportionally shrunk reflection of the graph for x=(1, e^1/e)

[EventHorizon]

Since my first post has mostly been ignored, I thought it would be good to explain it better. It answers questions such as why the "paired points" exist, which x's converge, why x>e^(1/e) does not converge, etc.

If you look at the graph of y=sqrt(2)^x and compare it to the graph of y=x, you'll notice that sqrt(2)^x > x over the ranges (-inf,2) and (4,inf).

sqrt(2)^x < x over the range (2,4).

This shows you that 2 is a stable equilibrium of the series (if you start somewhere in the range (-inf,4), you will end up at 2), and 4 is an unstable equilibrium (if you start near but not directly on it, you move away from it).

Here I used x and y as the independent and dependent variables for a graph (as opposed to the variable in the OP). The graph is used here to show that given ai = x, ai+1 = y. It essentially shows how the numbers move in the series given any and all starting points. If x = sqrt(2)^x, then x is an equilibrium point. This works for x=2,4. But then I show what the graph looks like around those ranges, and you can see that the series converges to 2 if initially started less than 4, to 4 if started at 4, and diverges to infinity when starting greater than 4.

I then discussed ranges for the x given in the OP.

if x is in the range (0,1), there is one stable equilibrium at y satisfying x = y^(1/y), where x < y < 1.

These series actually alternates between above p(x) and below it. It looks like I didn't try a small enough x initially because it looks like somewhere between .05 and .1 it starts to tend to stabilize on alternating between two values. So in the range [.1,1) it converges as I said before, but in the range (0,.1) I haven't quite figured out the exact value at which it stabilizes on two alternating values. This should have been apparent due to x=0. 0^1=0 => 0^0 => 1 (according to google)....so it will alternate between 0 and 1.

if x is 1, there is one stable equilibrium at 1.

Obviously if you have 1^z = 1, 1^1=1. So for any starting point, it converges on 1 quickly.

if x is in the range (1,e^(1/e)), there is one stable equilibrium (somewhere in the range (x,e)) and one unstable equilibrium (greater than e) both satisfying x = y^(1/y).

These follow the same pattern as my first example with x=sqrt(2). The graph between the two points is less than the graph of y=x, which makes the lesser of the two points stable and the upper unstable.

if x = e^(1/e), there is one unstable equilibrium at e.

I was incorrect saying that it was a stable equilibrium here. It is stable from below, but diverges from above. You can see this because (e^(1/e))^z > z for all z except z=e where it is equal. It asymptotically approaches e from the bottom, but if you start just above e it starts out moving slowly away and then only increases the speed at which it diverges.

if x is in the range (e^(1/e),inf), there is no equilibrium.... the graph compared to the graph of y=x shows that the number always increases.

A simple proof of this is the following (let the x from the original post be z to allow x to be the independent variable in the graph). Look at the graph of y = z^x compared to y=x where z is in the range (e^(1/e),inf). The graphs do not touch or cross. You can determine the minimum of the function y=(z^x)-x, which will be a positive number (call it alpha) which represents the smallest change in the series. This shows that as the series continues, the next element in the series will be at least alpha greater than the previous value. Since the series will always be increasing by at least a given positive number, it cannot converge.

The series essentially start with x (from the OP) (or equivalently the power/multiplicative identity, 1), which shows that you cannot get to p(x) equilibria values greater than e without starting directly on them.

Edited October 8, 2008 by EventHorizon

[Guest]

My complaint is that the method of solving p(x) = n by substituting p(x) with x^p(x) yields incorrect results where n is outside function's convergence (existence) range. I'd be more content if the method yielded some unreal solutions or something of the sort. Like in the case of x² = -1, for example.

I see your point that unreal solutions to xⁿ=n would make it obvious that p(x) is undefined, but can't give you any more answer than "it isn't"!

Here is what I think p(x) graph looks like.

My other wild guess guess is that the portion of the graph between x=(0,1) is 90-degree rotation proportionally shrunk reflection of the graph for x=(1, e^1/e)

I agree with your graph (except that x=1.5^2/3 and x=2^1/2 are the wrong way round) and perhaps if we could show that p(x) is continuous in at least (1,e^1/e), then we might show that dp(x)/dx -> inf as x-> e^1/e?

Edited October 8, 2008 by foolonthehill

[Guest]

Since my first post has mostly been ignored, I thought it would be good to explain it better. It answers questions such as why the "paired points" exist, which x's converge, why x>e^(1/e) does not converge, etc.

If you look at the graph of y=sqrt(2)^x and compare it to the graph of y=x, you'll notice that sqrt(2)^x > x over the ranges (-inf,2) and (4,inf).

sqrt(2)^x < x over the range (2,4).

Apologies for not understanding your point first time round, but I think you show a very clear explanation. We have been focussing on the series defined iteratively for a₁=x. ; a_i+1=x^a_i. Your point is that it actually doesn't matter what a_i (let's call it d) is, we can take any number and by continually raising x^a_i, we will still converge on the same (stable equilibrium) p(x) for a certain set of starting numbers, d.

This explains Prime's problem with the idea that x=2^1/2 solves xⁿ=n for n=4: if we start with a_i=4, then a_i+1 = (2^1/2)⁴ = 4. So it seems that, if we modify the OP to say that a¹=d for some d, then p(x)=4 does has a solution x=2^1/2, d=4.

Looks like your suspicions were well founded Prime. Apologies for doubting you!

[EventHorizon]

I just found out the smallest x such that a stable p(x) exists!

Instead of x, I'll use z (and p(z)) to leave x as an independent parameter for graphs etc.

The reason z^z^z^z^z.....diverges with small values of z is that the slope of y=z^x is less than -1 when it intersects the line y=x. This means that the intersection is an unstable equilibrium because z^x will be further from the equilibrium than x.

To find the specific value of z such that the slope at the point of intersection is -1, I did the following.

First I took the derivative of y=z^x with respect to x.

dy/dx = (ln z)(z^x)dx

Since the intersection is where y=x, this leaves me with a system of 2 equations and 2 unknowns. Let w = p(z).

-1=(ln z)z^w

w=z^w

-1=(ln z)w

-1/w = ln z

e^(-1/w) = z

(e^(-1/w))^w=z^w

e^-1=w

w = 1/e

-1/(1/e) = ln z

-e = ln z

e^-e = z

So the lower bound for x (from OP) that converges is e^-e ~ .066

So the range for x for which the series can converge is [1/(e^e),e^(1/e)], and their corresponding p(x) values are 1/e and e respectively.....seems fitting, don't you think?

So, for the record, here are the ranges for x and what equilibria they produce...

Range, Equilibria

(-inf,0), no idea...

0, diverges by alternating 1's and 0's

(0,1/(e^e)), one unstable equilibrium in the range (x,1/e), but ends up alternating between two values (which themselves converge) if not started on the equilibrium

[1/(e^e),1], one stable equilibrium in the range [1/e,1]

(1,e^(1/e)), one stable equilibrium (somewhere in the range (x,e)) and one unstable equilibrium (greater than e)

e^(1/e), an equilibrium at p(x)=e that is stable from below and unstable from above

((e^(1/e)),inf), no equilibria (diverges)

Edited October 9, 2008 by EventHorizon

[Guest]

So, for the record, here are the ranges for x and what equilibria they produce...

Range, Equilibria

(-inf,0), no idea...

0, diverges by alternating 1's and 0's

(0,1/(e^e)), one unstable equilibrium in the range (x,1/e), but ends up alternating between two values (which themselves converge) if not started on the equilibrium

[1/(e^e),1], one stable equilibrium in the range [1/e,1]

(1,e^(1/e)), one stable equilibrium (somewhere in the range (x,e)) and one unstable equilibrium (greater than e)

e^(1/e), an equilibrium at p(x)=e that is stable from below and unstable from above

((e^(1/e)),inf), no equilibria (diverges)

Alright, I follow you on all this and I have no arguments.

The one thing I still can't get past is: what value does the sequence converge to (when it does)? If the x=5^1/5 sequence doesn't converge to 5, what does it converge to? When does y^1/y=5^1/5 when y!=5?

There has to be an exact expression, right?

[bonanova]

The one thing I still can't get past is: what value does the sequence converge to (when it does)? If the x=5^1/5 sequence doesn't converge to 5, what does it converge to? When does y^1/y=5^1/5 when y!=5?

There has to be an exact expression, right?

Well, we know that p(x) is bounded above by e, so p will never reach an integer value greater than 2.

What are the integer values? Here are a few.

I evaluated the exponent stack to 10 million terms, and compared to the 10-million-and-one^th term to be sure of convergence.

... n n^(1/n) ... p[n^(1/n)]

... 1 1.000000000 1.000000000

... 2 1.414213562 2.000000000

... 3 1.442249570 2.47805268

... 4 1.414213562 2.000000000

... 5 1.379729661 1.764921915

... 6 1.348006155 1.624243846

... 7 1.320469248 1.530140119

... 8 1.296839555 1.462501432

... 9 1.276518007 1.411381742

.. 10 1.258925412 1.371288574

. 100 1.047128548 1.049519190

.1000 1.006931669 1.006980222

10000 1.000921458 1.000922309

[bonanova]

I just found out the smallest x such that a stable p(x) exists!

......

So the lower bound for x (from OP) that converges is e^-e ~ .066

So the range for x for which the series can converge is [1/(e^e),e^(1/e)],

and their corresponding p(x) values are 1/e and e respectively.....

seems fitting, don't you think?

Nice analysis.

It's interesting that when Euler did this analysis in 1778, he determined

the convergence range as 1 <= x <= e^(1/e).

Why he didn't see convergence for any values less than unity is not clear.

Anyway,

I started to play with the derivative but switched to brute force calculation.

I calculated p(x) for x = unity downward, to see where the calculated values began to diverge.

I verified that alternating terms of the sequence converged either to the same or different stable values.

Again, 10 million exponents compared to 10 million + 1 exponents.

Then I successively subdivided the range where divergence set in.

These graphs show the last two ranges of x that I looked at.

The result is interesting.

I calculate 1/e^e to be 0.06598803585....

The calculated point of divergence is slightly larger : 0.0659928... - a difference of 0.000004 ...

The calculated 1/e^e value lies to the left of the second graph entirely.

Here are the plots of p[x] vs x - the second plot is a blowup of the indicated region on the first plot.

[EventHorizon]

Alright, I follow you on all this and I have no arguments.

The one thing I still can't get past is: what value does the sequence converge to (when it does)? If the x=5^1/5 sequence doesn't converge to 5, what does it converge to? When does y^1/y=5^1/5 when y!=5?

There has to be an exact expression, right?

Possibly. I've tried for a little while to find a way to calculate it exactly and looked around on the net, but can't find it as of yet. I may try again a little later. It seems that X^(1/X) is just a hard function to work with.

[Guest]

It's interesting that when Euler did this analysis in 1778, he determined

the convergence range as 1 <= x <= e^(1/e).

Why he didn't see convergence for any values less than unity is not clear.

So you're telling me we're all smarter than Euler. That's going on my resume for sure.

[EventHorizon]

Nice analysis.

Thanks.

The result is interesting.

I calculate 1/e^e to be 0.06598803585....

The calculated point of divergence is slightly larger : 0.0659928... - a difference of 0.000004 ...

The calculated 1/e^e value lies to the left of the second graph entirely.

Here are the plots of p[x] vs x - the second plot is a blowup of the indicated region on the first plot.

I was a little unsure that 1/(e^e) would converge....if the slope at the intersection is really -1, then there isn't much pushing the series towards the intersection point when very close to it. But barely beyond it, it seems like it should converge. I'll think about it more and perform similar calculations and post any interesting results I stumble on.

[bonanova]

Thanks.

I was a little unsure that 1/(e^e) would converge....if the slope at the intersection is really -1, then there isn't much pushing the series towards the intersection point when very close to it. But barely beyond it, it seems like it should converge. I'll think about it more and perform similar calculations and post any interesting results I stumble on.

I wouldn't worry. I think the analysis "has to be" right.

I suspect possibly doing 10 million successive exponentiations - even with 32 bits of accuracy - could result in a systematic drift of the order seen here.

Altho I scaled back to 10 thousand and the numbers were the same. [No doubt of convergence.] *still scratching head*

I'd take the calculations [as well as the exquisite simplicity / symmetry] of your result as verification, not contradiction.

[Guest]

I wouldn't worry. I think the analysis "has to be" right.

I suspect possibly doing 10 million successive exponentiations - even with 32 bits of accuracy - could result in a systematic drift of the order seen here.

Altho I scaled back to 10 thousand and the numbers were the same. [No doubt of convergence.] *still scratching head*

I'd take the calculations [as well as the exquisite simplicity / symmetry] of your result as verification, not contradiction.

A good test might be to see exactly how close sqrt(2) comes to 2, so you have an idea what the roundoff error might be.

[bonanova]

A good test might be to see exactly how close sqrt(2) comes to 2, so you have an idea what the roundoff error might be.

The program returned an integer 2.

That means it converged to a floating point value indistinguishable from 2.

But it's not the computed p(x) value that's differing here, rather the value of the independent variable x for which the computed p(x) begins to oscillate between two different values.

But your suggestion still demonstrates that the precision limitation involved in computing 2^.5 and 4^.25 did not deter finding the [exact] condition p(x) = 2 for those cases. I didn't think of that.

The mystery [albeit a small one] persists.

[bonanova]

I'm going to add my 2 cents at this point.

I think that within the perimeters of the OP, all the salient analysis has been done, and the relevant conclusions have been reached. Congratulations to all.

Peter Winkler found this gem of a paradox, surprisingly, from a question on an American High School Mathematics Exam back in the 1960s. The question simply asked, "If x^x^x^x^x^x .... = 2, what is x?" Questions of this type invite the student to show insight more than computational prowess. And in this case s/he was asked to see that removing the first x wouldn't change anything, so that x^2 = 2 could be written instead.

All well and good.

Until some genius kid noted the a similar substitution in x^x^x^x^x^x^x .... = 4 would produce x^4 = 4, and the same value of x namely 2^.5 produced different results.

Translating this for BrainDen, I decided to ask it in two parts. the "=2" question in the OP, to get thinking locked into the notion of substituting by removing the first x, and in later post introduce the "=4" question to induce some head scratching. Which was sadistically satisfying to watch.

I'm crediting armcie with raising the question of convergence. But I didn't declare it solved at that point. That would be cruel - to me of course - prolonging the scratching seemed like a good idea. So I glossed over the convergence issue [or tried to] by showing convergence [for x = 2^.5] to the value 2. But I didn't show [because it's not true] that it converged to the value 4. That was the part of the puzzle still to be solved - to find the range of x for which p(x) converged.

And we finally got there, with contributions by many, you know who you are, and finally now very nicely by eventhorizon.

The red herring in this delicious little gem of course is the equation x^y = y. Or x = y*(1/y). Thanks to Chuck Rampart for plotting it, and thereby fixing attention to it. And the search [hope?] for other magical integer solutions, given the nice result for 2. The imponderable was that plots of y(x) or x(y) can go far outside the region of p(x)'s convergence.

Outside that region is meaningless for the purpose of this puzzle. Just as the student's notion that x^x^x^x^x^x^x .... =4 has a "solution." As we found here, the infinite exponential stack can never have a value greater than e. Yup, e's our man. The fact that x^4 = 4 does in fact have a solution [the red herring again] kept the discussion alive.

The puzzle is solved, however, as soon as the range of x for which p(x) converges has been found, and anything outside that range is rejected. Everything outside that range is fair play in discussing the properties of x^y = y, but it's irrelevant for discussions of p(x).

Anyway, kudos to all, and I hope it was fun.

Here's Winkler's summary, and I quote:

"The expression x^x^x^x^x ... is meaningful and equal to the lower root of x = y^(1/y) as long as 1 <= x <= e^(1/e). [sic] For x = e^(1/e) the expression is equal to e, but as soon as x exceeds e^(1/e) the sequence diverges to infinity.

"This observation was first made by Leonhard Euler in 1778!"

[Prime]

Since my first post has mostly been ignored, I thought it would be good to explain it better. It answers questions such as why the "paired points" exist, which x's converge, why x>e^(1/e) does not converge, etc.

If you look at the graph of y=sqrt(2)^x and compare it to the graph of y=x, you'll notice that sqrt(2)^x > x over the ranges (-inf,2) and (4,inf).

sqrt(2)^x < x over the range (2,4).

This shows you that 2 is a stable equilibrium of the series (if you start somewhere in the range (-inf,4), you will end up at 2), and 4 is an unstable equilibrium (if you start near but not directly on it, you move away from it).

Here I used x and y as the independent and dependent variables for a graph (as opposed to the variable in the OP). The graph is used here to show that given ai = x, ai+1 = y. It essentially shows how the numbers move in the series given any and all starting points. If x = sqrt(2)^x, then x is an equilibrium point. This works for x=2,4. But then I show what the graph looks like around those ranges, and you can see that the series converges to 2 if initially started less than 4, to 4 if started at 4, and diverges to infinity when starting greater than 4.

...

Fascinating research! I feel a graphical illustration for 1 <= x <= e^1/e is in order here.

So for the function r^x, where r^k < k, raising r into the resulting power will lower the result repeatedly. (With each step we’d be raising r to a smaller power). Where for r^k = k, repeated operations will keep the result the same (k). Such power is found at the equilibrium point k where r=k^1/k.

Therefore, the substitution method of solution where we substitute p(x) for x^p(x) does not solve p(x) = c equation, it solves x^c = c instead. However, the solution for the latter happens to coincide with the solution for the former, where c happened to be in the convergence range for the p(x).

To bring it to science for the interval (1, e^1/e):

1. Do we have a proof that x^1/x has its maximum at x=e?

2. What about proof that there may not be any other convergence points within the convergence interval? I mean other than the points of equilibrium (intersection of y=x with y=r^x). It is possible that repeated steps of r^k would lower the result by a smaller interval every time never reaching some point C_k above the lower equilibrium point. (Even though inspection does not support that.)

[EventHorizon]

Fascinating research! I feel a graphical illustration for 1 <= x <= e^1/e is in order here.

...

To bring it to science for the interval (1, e^1/e):

1. Do we have a proof that x^1/x has its maximum at x=e?

2. What about proof that there may not be any other convergence points within the convergence interval? I mean other than the points of equilibrium (intersection of y=x with y=r^x). It is possible that repeated steps of r^k would lower the result by a smaller interval every time never reaching some point C_k above the lower equilibrium point. (Even though inspection does not support that.)

Proof of (1) is fairly simple. Set the first derivative to zero.

As for (2), it seems you would need the graph to approach and only touch the line y=x once but not cross it. The only one that does this is x=e^(1/e), which was shown to converge to e (from below). So...no other convergence points.

(I thought I would add in a little more explanation now that I reread (2). The slope of the graph of r^x at the point of convergence determines how quickly it will converge. Which is why it would need to not cross if the series was to slow too much to converge.)

Edited October 9, 2008 by EventHorizon

[Prime]

Proof of (1) is fairly simple. Set the first derivative to zero.

As for (2), it seems you would need the graph to approach and only touch the line y=x once but not cross it. The only one that does this is x=e^(1/e), which was shown to converge to e (from below). So...no other convergence points.

(I thought I would add in a little more explanation now that I reread (2). The slope of the graph of r^x at the point of convergence determines how quickly it will converge. Which is why it would need to not cross if the series was to slow too much to converge.)

OK, since it is simple, I’ll go through the motions:

y = x^1/x

lny = lnx^1/x

lny = (1/x)*lnx

Differentiating both sides:

y'*1/y = -x^-2*lnx + x^-2

y’ = y*x^-2*(1 - lnx)

y’ = x^1/x*x^-2*(1 - lnx)

The only way the above equation yields zero is lnx = 1 --> x = e.

For the second point the illustration is as following:

Each vertical line is the difference between y = x and y = r^x.

Corresponding horizontal line plots the next x-value for consecutive step (raising to a resulting power.) The lines form isosceles triangles. Thus the x-step is converging onto zero at the same rate as y-step and both coordinates will converge at the same point – the point of equilibrium.

So the problem p(x) = c is solved and proven here for 1 <= c <= e.

For any given x=a where 1 <= a <= e^1/e, we don’t have a neat finite formula to find c, such that a = c^1/c. But we can find an approximation using spreadsheet.

[EventHorizon]

I wouldn't worry. I think the analysis "has to be" right.

I suspect possibly doing 10 million successive exponentiations - even with 32 bits of accuracy - could result in a systematic drift of the order seen here.

Altho I scaled back to 10 thousand and the numbers were the same. [No doubt of convergence.] *still scratching head*

I'd take the calculations [as well as the exquisite simplicity / symmetry] of your result as verification, not contradiction.

I thought I would look at the following equation (again using z and p(z) as opposed to x's to leave x for graphing),

y=(z^(z^x)-x)/p(z)-x, (obviously undefined at p(z))

This equation represents the percent change in the error (distance to equilibrium) after two more exponents. If this equation has a minimum greater than 0, then the remaining distance will be at most 1-min(y) (=less than 1) times the previous distance after two more exponents. The error after 4 more exponents will be (1-min(y))^2 times the previous error, etc.

Obviously, we can get the error as small as desired because the multiplier is less than 1.

I've checked the minimum of the equation using a calculator (yeah, I'm lazy.....and the math was getting a little ugly) for a number of z values between 1/(e^e) and .0659928, and it seems to me that the only time it reaches 0 is at 1/(e^e). So it seems you may be right in thinking it was some sort of precision error/drift. I may try the math again later to try and get a more rigorous proof....

Edited October 10, 2008 by EventHorizon

[Prime]

I just found out the smallest x such that a stable p(x) exists!

Instead of x, I'll use z (and p(z)) to leave x as an independent parameter for graphs etc.

The reason z^z^z^z^z.....diverges with small values of z is that the slope of y=z^x is less than -1 when it intersects the line y=x. This means that the intersection is an unstable equilibrium because z^x will be further from the equilibrium than x.

...

I couldn’t quite follow the logic of the derivative at the point of intersection of y=z^x with y=x. Why is the derivative a deciding factor for the convergence of those oscillating values?

Anyway, here is how I envision the behavior of z^(z^(z^(… for 0 < z < 1.

The spiraling lines plot the value for the exponent on each next step.

Notice the first turn of the spiral when the z^z - z > z^z - z^(z^z)). Or z^(z^z)) > z^z. That should always be the case as long as z < c, where c is the convergence point, or the point of intersection of y=x with y=z^x. (z^c = c). If the spiral did not get narrower with each turn, y=z^x would have a local minimum somewhere in that interval and we know that’s not the case.

Why would such spiral never rich the conversion point c? With every turn it appears to encounter a steeper rise of the y=z^x and should converge faster…

[Prime]

And I just noted a mental hick-up I had. I said "as long as z < c (the convergence point). But that is always the case.

[EventHorizon]

I couldn’t quite follow the logic of the derivative at the point of intersection of y=z^x with y=x. Why is the derivative a deciding factor for the convergence of those oscillating values?

Anyway, here is how I envision the behavior of z^(z^(z^(… for 0 < z < 1.

...

The spiraling lines plot the value for the exponent on each next step.

Notice the first turn of the spiral when the z^z - z > z^z - z^(z^z)). Or z^(z^z)) > z^z. That should always be the case as long as z < c, where c is the convergence point, or the point of intersection of y=x with y=z^x. (z^c = c). If the spiral did not get narrower with each turn, y=z^x would have a local minimum somewhere in that interval and we know that’s not the case.

Why would such spiral never rich the conversion point c? With every turn it appears to encounter a steeper rise of the y=z^x and should converge faster…

Why is the derivative a deciding factor for the convergence of those oscillating values?

Consider the following...

Let the slope at p(x) be -1+e = -(1-e).

Let a1 = p(x)-d.

When very close to the intersection/equilibrium, they are basically straight lines crossing.

a2 = p(x) + (-d*-(1-e)) = d(1-e).

What I'm doing here is simply treating them as straight lines and seeing what the next value will be.

a3 = p(x) + d(1-e)*-(1-e) = p(x)-d(1-e)^2

...

an = p(x) + ((-1)^n) * d * (1-e)^(n-1)

if e is greater than 0, (1-e) is less than 1 and the error is decreasing.

If e is 0, there will be no movement.

If e is negative (e.g., the slope is less than -1), then (1-e) is greater than 1 and the error is increasing.

What this is saying is that the series will converge if the slope at the equilibrium is in the range (-1,1). Not only that, but the closer the slope is to 0, the faster the convergence (alluded to in a previous post).

Therefore, if the derivative at the equilibrium is less than -1, the series will not converge to p(x). When x = 1/(e^e), the slope at the equilibrium is exactly -1. With a little more math work (described in my last post) we may be able to show that it will converge for all x in (1/(e^e),e^(1/e)]. Consequently.....the slopes at the equilibrium for x values of 1/(e^e) and e^(1/e) are -1 and 1 respectively.