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?

[Yoruichi-san]

Wow...that's really interesting!

I think it may have to do with...

When we deal with decimal exponents, there can be more than one value. An obvious example is x^(0.5), i.e. the square root has both a positive and a negative value. Fourth roots have positive, negative, positive imaginary, and negative imaginary values. More complicated roots have more complex values, such as a+bi,a-bi, etc...

So here the exponent is sqrt(2), which is 1.414.....

x^(1.414....) = x^(1)*x^(0.414....)

so each time we take it to a power of sqrt(2), there is actually more than one possible value it returns, which could explain why the series also equals to 4. Let me think some way to prove whether these are the only values...

[Yoruichi-san]

Wow...that's really interesting!

I think it may have to do with...

When we deal with decimal exponents, there can be more than one value. An obvious example is x^(0.5), i.e. the square root has both a positive and a negative value. Fourth roots have positive, negative, positive imaginary, and negative imaginary values. More complicated roots have more complex values, such as a+bi,a-bi, etc...

So here the exponent is sqrt(2), which is 1.414.....

x^(1.414....) = x^(1)*x^(0.414....)

so each time we take it to a power of sqrt(2), there is actually more than one possible value it returns, which could explain why the series also equals to 4. Let me think some way to prove whether these are the only values...

And now that I think about it, all our proofs have been kind of going backwards, based off assuming that the series converges to a certain value, i.e. 2 or 4, but there is no forwards proof of what value the series converges to... And I don't know how well Excel deals with complex numbers...;P

Edited October 7, 2008 by Yoruichi-san

[Guest]

The question:

x^x^x...^x=2 in this equation if you take the first x aside,

you will have

x^(x^x^x..^x)=2 and from the first equation, we know that the part in the paranthesis is equal to 2. then x^2 = 2 and thus result is x=sqrt(2)

But, if you leave first x in place and take the last one aside : (x^x^x...x^x)^x = 2 then as we know that that inside of the paranthesis =2,

2^x = 2 . Thus x=1. This result is incorrect, it is obvious. Sqrt(2) result is obviously false because if you use a calculater after 4-5 operations you will exceed 2.

Here, we have two proved results, but they are also proved to be false. Because it is just a paradox. And the trick of it is "infinite numbers".

[bonanova]

Has anyone calculated the values of x that give convergence?

Step 1 to some answers. Consider:

x to the x to the x to the x to the x to the x .... = e.

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

e lies between 2 and 4.

If the sequence converges then x^y = y.

[Prime]

I'd venture to say for any x = n^1/n, where n is a positive integer, the function x^(x^(x^... converges.

The function x^1/x has its maximum somewhere between x = 2 and x = 3. Perhaps, it is the ubiquitous e.

The logarithmic solution/proof shown here before must be wrong, since it seems to give incorrect results for numbers other than 2.

The proof that I offered in one of the previous posts just shows that for x=2^1/2 the expression is less than 2. But that is not necessarily the lowest limit.

Just a wild guess: the function x^(x^(x^... is convergent for numbers up to the maximum of the function y=x^1/x.

[Guest]

The function x^1/x has its maximum somewhere between x = 2 and x = 3. Perhaps, it is the ubiquitous e.

Yep, looks like e's your man

[Guest]

2^(2^(2^(2^2))) = 2^65536

Huge number, it would take an entire notebook just to write it down.

Raising exponent into some power is not the same as raising the result of exponentiation into some power.

Ah, I see the error of my ways...thanks.

But if the problem was the way I envisioned x^x^x^x,.... = 2 (no brackets), then how could the answer be 1?

[Guest]

Alright, so I'm getting that x^1/x has a maximum at x=e. What this means (I think) is that there should be more pairs like 2 and 4 which appear to have the same solution. This is because the graph for y=x^1/x looks like this:

So, for instance, 5 should have a duplicate solution somewhere between 1 and 2. How can we find that?

Let's go over this again: Let p(x) be the infinite exponential sequence.

p(x)=5

x^p(x)=5

x⁵=5

x=5^1/5

But x=5^1/5 should be the solution for some other value of p(x).

5^1/5=y^1/y where y!=5

Just guessing and checking I found y is approx. 1.76495, but I can't find an exact expression that would make sense with that, and I can't figure out another way to solve that equation.

So, I'm stuck again. Would finding a Tayloy expansion for x^1/x help?

[Prime]

...

Let's go over this again: Let p(x) be the infinite exponential sequence.

p(x)=5

x^p(x)=5

x⁵=5

x=5^1/5

...

I think it is quite evident at this point that the solution x=n^1/n for p(x) = n is wrong. Why -- is a different question.

Your graph of the x^1/x function shows maximum at e, after which the function diminishes and tends to 1. So if the maximum is at e and the maximum for p(x) is at x=e^1/e where p(x) <= e, after which the values for p(x) must diminish. Therefore, p(x) = 5 should not have any solution. Really, if we start with a smaller value when constructing the infinite string p(x) we should not expect p(x) to be larger.

At this point I am compelled to remind that x^1/x as a domain for the values for which p(x) converges was a wild guess. We need to prove the top limit where p(x) still has a finite result (converges). Bonanova has asked for that already.

We know that p(x) converges for some x. We proved that for x=n^1/n, when p(x) < n (which is an off the mark upper limit), and we know that starting at some point p(x) runs off to infinity. For example, it is quite evidently the case for x = 2.

Now we suspect that the maximum for p(x) is at x = e^1/e and p(x) <= e at that point.

Also note that for x < 1, p(x) behavior is interesting too. Basically, p(x) tends to 1 there, but it may have a local minimum somewhere between x=0 and 1.

[Guest]

I think it is quite evident at this point that the solution x=n^1/n for p(x) = n is wrong. Why -- is a different question.

Your graph of the x^1/x function shows maximum at e, after which the function diminishes and tends to 1. So if the maximum is at e and the maximum for p(x) is at x=e^1/e where p(x) <= e, after which the values for p(x) must diminish. Therefore, p(x) = 5 should not have any solution. Really, if we start with a smaller value when constructing the infinite string p(x) we should not expect p(x) to be larger.

At this point I am compelled to remind that x^1/x as a domain for the values for which p(x) converges was a wild guess. We need to prove the top limit where p(x) still has a finite result (converges). Bonanova has asked for that already.

We know that p(x) converges for some x. We proved that for x=n^1/n, when p(x) < n (which is an off the mark upper limit), and we know that starting at some point p(x) runs off to infinity. For example, it is quite evidently the case for x = 2.

Now we suspect that the maximum for p(x) is at x = e^1/e and p(x) <= e at that point.

Also note that for x < 1, p(x) behavior is interesting too. Basically, p(x) tends to 1 there, but it may have a local minimum somewhere between x=0 and 1.

Saying p(x) converges for x in (0,e^1/e) based on the x^1/x formula is better than a wild guess. We got there backwards - we solved p(x)=2 for x=2^1/2, and it seemed to make sense. I can't see any reason why a number in that window wouldn't converge, and I can't see how a number outside that window would converge, or what it would converge to.

After going through numerically, it definitely looks like p(5^1/5) converges to that 1.76495 value, not 5. I feel pretty confident in saying p(x)<e for any x where it converges, but I can't prove that.

What would probably go along way towards finding that proof would be to find an exact formula for that 1.76495 value (or whatever the case may be). I've tried everything I can think of that makes sense, and none of it fits.

Edited October 7, 2008 by Chuck Rampart

[Guest]

edit: scratch what I said - I think some of it's wrong so I'll repost in a minute!

Edited October 7, 2008 by foolonthehill

[Prime]

Saying p(x) converges for x in (0,e^1/e) based on the x^1/x formula is better than a wild guess. We got there backwards - we solved p(x)=2 for x=2^1/2, and it seemed to make sense. I can't see any reason why a number in that window wouldn't converge, and I can't see how a number outside that window would converge, or what it would converge to.

...

I have already posted an inductive proof that for x inside that window (n^1/n), p(x) converges. But the other side has not been proven here yet. That is how a number outside that window would not converge. It would be very interesting to come up with a proof, or at least some demonstration of that point.

Here is my understanding of how logarithmic solution for p(x) = n yielding x = n^1/n is wrong.

1. Given p(x) = n

2. As p(x) = x^(x^(x^(x^ ... then p(x) = x^p(x) = x^(x^(p(x)) , and so on. Furthermore, we can break an infinite string into two infinite strings. And thus p(x) = p(x)^p(x).

3. log_xp(x) = log_xn

Substituting p(x) from (2):

4. log_xx^p(x) = log_xn --> p(x)log_xx = log_xn --> p(x) = log_xn --> x = n^1/n.

However, we could have substituted p(x) with p(x)^p(x), and then the equation would become:

p(x)log_xp(x) = log_xn --> n*log_xp(x) = log_xn, which would yield a different solution for x.

(As we have seen with the example of p(x)=4).

Therefore, I think logarithmic solution is illegal.

Yoruichi-san already mentioned that we don’t know the range for the p(x) -- what kind of solutions it yields, complex numbers, or what.

Here is an interesting possibility:

What if p(x) excludes rational values from its range? In other words, the true (minimal) upper limit for converging p(x) is always non-rational number?

Edited October 7, 2008 by Prime

[EventHorizon]

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).

As for the range of values that converge (where x is the x stated 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.

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

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).

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

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.

if x is negative, complex numbers make things difficult.....

[Guest]

To summarise where I think the ideas are at the moment

for given m, I define a_i+1 = x^a_i where a₁=x

and, if it exists, p(x) = lim_i->∞ a_i

Bonanova's original problem

bonanova asked 'What value of x gives p(x)=2?' and it is agreed that x = √2 gives a solution because

2 = p(x) = x^p(x)

2 = p(x) = x²

x = √2

but

armcie pointed out that this is only valid if p(x) exists, so bonanova showed that for x=√2, the series is increasing and bounded above.

so on to the next problem

so bonanova asked to explain why x=√2 also appears to be a solution for p(x)=4:

4 = p(x) = x^p(x)

4 = p(x) = x⁴

x = √2

However, we should be careful as this is still only valid if p(x) is well defined. If the series is divergent, p(x) ≠ x^p(x) so we cannot solve for x this way. It is still true that √2⁴ = 4, but that is of no consequence to the problem!

Covergence

By inspection and prime's reasoning, we seem to agree that p(x) exists (ie a_i converges) if x<e^1/e, but here's (half) a formal proof

(I assume that x>1 for everything here)

strictly increasing:

a₁ = x < x^x = a₂ (because x≠1) and

a_i+1 = x^a_i < x^a_i+1 = a_i+2 (because x>1)

so by induction a_i is strictly increasing.

bounded above

if we can find a y such that x = y^y, then

aⁱ⁺¹ = y^a_i/y < y iff a_i < y since a₁ = y^1/y < y if y>1

Therefore, as long as we can find a y>1 that solves x=y^y, then a_i is convergent.

the e^1/e limit

So we ask what values of x can be described as x = y^1/y

x = y^1/y

log x = 1/y . log y

x' . 1/x = 1/y² - 1/y² . log y (where x = dx/dy)

x' = y^1/y/_y² (1-log y)

which is positive in the range (0,e) and negative in the range (e,∞). At y=e, x'=0, and give a maximum x=e^1/e

Therefore, x can only be written in this form if x<=e^1/e, and is therefore convergent in the range (0,e^1/e).

The bit left is to prove that it is divergent for x>e^1/e and I haven't found that yet....

edit: reformat

Edited October 7, 2008 by foolonthehill

[EventHorizon]

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

Oops....it looks like it is stable from below, but unstable from above....a saddle point?

[Yoruichi-san]

I have already posted an inductive proof that for x inside that window (n^1/n), p(x) converges. But the other side has not been proven here yet. That is how a number outside that window would not converge. It would be very interesting to come up with a proof, or at least some demonstration of that point.

Here is my understanding of how logarithmic solution for p(x) = n yielding x = n^1/n is wrong.

1. Given p(x) = n

2. As p(x) = x^(x^(x^(x^ ... then p(x) = x^p(x) = x^(x^(p(x)) , and so on. Furthermore, we can break an infinite string into two infinite strings. And thus p(x) = p(x)^p(x).

3. log_xp(x) = log_xn

Substituting p(x) from (2):

4. log_xx^p(x) = log_xn --> p(x)log_xx = log_xn --> p(x) = log_xn --> x = n^1/n.

However, we could have substituted p(x) with p(x)^p(x), and then the equation would become:

p(x)log_xp(x) = log_xn --> n*log_xp(x) = log_xn, which would yield a different solution for x.

(As we have seen with the example of p(x)=4).

Therefore, I think logarithmic solution is illegal.

Umm...I don't think we can use this substitution. This series is created by adding a base each time, and it can't be split in half that way along any exponent in the middle...i.e. (3^3)^(3^3) is not equivalent to 3^(3^(3^3)). But I myself am still iffy on using inf=inf-1...

[unreality]

1 + -1 + 1 + -1 + 1 + -1 .... = ?

(1 + -1) + (1 + -1) + (1 + -1) .... = 0

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) .... = 1

1 = 0

infinity minus one

On a more serious note, think of one representation of the Golden Ratio:

p = √(1+√(1+√(1+√(1+√(1+√(1+√(1+√(1+...

square both sides

p² = 1+√(1+√(1+√(1+√(1+√(1+√(1+√(1+...

replace the thing after "1+" with the original equation:

p² = 1 + p

Which is of course the defining property of the Golden Ratio, which, with some basic algebra, arises its amazing properties deeply connected with the Fibonacci numbers

What we did, though, was reduce one part of the infinite series, yet it was infinite, so there was still the infinite series afterwards

[Yoruichi-san]

1 + -1 + 1 + -1 + 1 + -1 .... = ?

(1 + -1) + (1 + -1) + (1 + -1) .... = 0

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) .... = 1

1 = 0

infinity minus one

On a more serious note, think of one representation of the Golden Ratio:

p = √(1+√(1+√(1+√(1+√(1+√(1+√(1+√(1+...

square both sides

p² = 1+√(1+√(1+√(1+√(1+√(1+√(1+√(1+...

replace the thing after "1+" with the original equation:

p² = 1 + p

Which is of course the defining property of the Golden Ratio, which, with some basic algebra, arises its amazing properties deeply connected with the Fibonacci numbers

What we did, though, was reduce one part of the infinite series, yet it was infinite, so there was still the infinite series afterwards

Well, I think we're using the assumption, as Octupuppy said, that if a series converges to some value at infinity then the infinity-1 series also converges to that value, i.e....

if the series converges then the differences between adjacent values in the series become smaller until they stop changing. For a series to converge, the slope, i.e. the change in the values, of the series has to approach 0, so that y_inf/y_inf-1 ->1.

Your values in your example series do not converge, they just change between 0 and 1. The differences between adjacent values never gets smaller and never approaches 0.

Anyways, thank you for helping me convince myself that my earlier work actually works ;P.

[Prime]

Umm...I don't think we can use this substitution. This series is created by adding a base each time, and it can't be split in half that way along any exponent in the middle...i.e. (3^3)^(3^3) is not equivalent to 3^(3^(3^3)). But I myself am still iffy on using inf=inf-1...

That is not how I split it. I split the infinite x^(x^(x... into two infinite x^(x^(x^... (x^(x^(x...

Anyhow, we can see that the following solution is wrong.

p(x) = n

x^p(x) = n

xⁿ = n

x = n^1/n

We can tell that just by observation. The solution suggests that for p(x) = 100, x = 100^1/100. Whereas we solve that for p(x) = 2, x = 2^1/2. On the other hand, 2^1/2 > 100^1/100.

Thus this solution suggests that if we start the infinite exponent string with a smaller number, with correspondingly smaller incremental steps we are going to arrive to a larger result than if we started from larger number with larger incremental steps. That makes no sense.

Never mind that for 2^1/2 the p(x) seems to converge somewhere close to 2. That is coincedental.

My conclusion: the solution by splitting the infinite string p(x) whichever way, whether into several infinite strings, or just one term and the infinite string -- does not work!

[Yoruichi-san]

That is not how I split it. I split the infinite x^(x^(x... into two infinite x^(x^(x^... (x^(x^(x...

Anyhow, we can see that the following solution is wrong.

p(x) = n

x^p(x) = n

xⁿ = n

x = n^1/n

We can tell that just by observation. The solution suggests that for p(x) = 100, x = 100^1/100. Whereas we solve that for p(x) = 2, x = 2^1/2. On the other hand, 2^1/2 > 100^1/100.

Thus this solution suggests that if we start the infinite exponent string with a smaller number, with correspondingly smaller incremental steps we are going to arrive to a larger result than if we started from larger number with larger incremental steps. That makes no sense.

Never mind that for 2^1/2 the p(x) seems to converge somewhere close to 2. That is coincedental.

My conclusion: the solution by splitting the infinite string p(x) whichever way, whether into several infinite strings, or just one term and the infinite string -- does not work!

Then I must be misunderstanding your notation, because it seems to me that p(x)^p(x) is splitting it (x^(x^(x^...)^(x^(x^(x...), i.e. putting a parenthesis between the two infinite series whereas the original only puts parentheses around higher exponents...

But I agree that a condition is that the series actually converges for a particular x. However, I don't necessarily agree that 2^(0.5)>100^(0.01), it depends on whether you're using a positive/negative/complex root for each...100^(0.01) has 100 roots

[Guest]

Anyhow, we can see that the following solution is wrong.

p(x) = n

x^p(x) = n

xⁿ = n

x = n^1/n

We can tell that just by observation. The solution suggests that for p(x) = 100, x = 100^1/100. Whereas we solve that for p(x) = 2, x = 2^1/2. On the other hand, 2^1/2 > 100^1/100.

Thus this solution suggests that if we start the infinite exponent string with a smaller number, with correspondingly smaller incremental steps we are going to arrive to a larger result than if we started from larger number with larger incremental steps. That makes no sense.

Never mind that for 2^1/2 the p(x) seems to converge somewhere close to 2. That is coincedental.

My conclusion: the solution by splitting the infinite string p(x) whichever way, whether into several infinite strings, or just one term and the infinite string -- does not work!

Why is that solution wrong? Surley, it is true iff p(x) exists, ie. a_i converges.

Both you and I have shown that a_idoes converge for x <= e^1/e because x can be expressed as x=y^1/y. Therefore, the solution x = n^1/n solves p(x)=n iff n<p(e^1/e)=p(e)

It is not possible to solve p(x)=100 and get x=100^1/100, because p(100^1/100) converges to some number smaller than e. There is no series which converge to 100.

[Guest]

I realised my last summary post #40<{POST_SNAPBACK}> is incomplete - we also need, after proving convergence for x in (0,e^1/e), to find solution for x, given p(x)=n.

n = p(x) = x^p(x)

n = y^n/y

n1/n = y1/y

which as we have seen from Chuck Rampart's chart could have two solutions: y=p and another one which has not yet been well defined. Let's call them y1 and y2. Then, x = y1^1/y1 = y2^1/y2. There is still only one value of x that solves p(x)=n

remaining problems

The remaining parts to the problem I see that are left are:

prove that a_i is convergent for x in (0,1)

prove that a_i is divergent for x in (e^1/_e,∞)

clarification

Reading some replies here, I think we need to be clear that a given x does not produce two solutions for p(x). That is, x=2^1/2=√2 does not produce p(x)=2 and p(x)=4. p(x) can only have one value and it is demonstrably p(x)=2. p(x)=4 is undefined - there is no such p(x).

Edited October 8, 2008 by foolonthehill

[Prime]

All I can see so far, is that we found that the expression x^(x^(x^... which we designate as p(x) converges for x = n^1/n. However, we don't know what it converges to. Solving p(x) = n by substituting p(x) with x^p(x) must be an invalid method. The fact that 2^1/2^(2^1/2^(2^1/2^(2^1/2^( ... tends to 2 is a coincedence. It may have a limit somewhere lower than 2.

The function x^1/x was proposed as a domain for x values for which p(x) converges. The only point on x^1/x which is significant for our problem is its maximum, which we suspect at x=e where function value is e^1/e. Assuming that is the maximum, if we found that there is x > e^1/e for which p(x) converges, then the function x^1/x loses its significance.

[Guest]

All I can see so far, is that we found that the expression x^(x^(x^... which we designate as p(x) converges for x = n^1/n. However, we don't know what it converges to. Solving p(x) = n by substituting p(x) with x^p(x) must be an invalid method.

It is not invalid if the sequence converges. If and only if a_i converges, then p(x) = n = x^n because

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

The fact that 2^1/2^(2^1/2^(2^1/2^(2^1/2^( ... tends to 2 is a coincedence. It may have a limit somewhere lower than 2.

???? For x=2^1/2, simply doing the calculations in Excel shows that it gets to 2 pretty quickly and my proof above proves it.

The function x^1/x was proposed as a domain for x values for which p(x) converges. The only point on x^1/x which is significant for our problem is its maximum, which we suspect at x=e where function value is e^1/e. Assuming that is the maximum, if we found that there is x > e^1/e for which p(x) converges, then the function x^1/x loses its significance.

You are right: if we can find x > e^1/e for which a_i converges, then the funciton loses its significance. However, by experiment it is pretty clear that any value above this does not converge, so I will continue to look for proof of divergence.

edit:bbcode - again!

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

Edited October 8, 2008 by foolonthehill

[Prime]

...

???? For x=2^1/2, simply doing the calculations in Excel shows that it gets to 2 pretty quickly and my proof above proves it.

...

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.