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]

After some jugglery in Excel and a bit of Googleing

The float constant of squareroot(2) which has the value 1.4142135623 comes closest

[bonanova]

After some jugglery in Excel and a bit of Googleing

The float constant of squareroot(2) which has the value 1.4142135623 comes closest

Hi Supandi,

Nice going.

But you all know how much I love "showing your work" even tho sometimes I don't.

So let me expand the puzzle.

1. Prove it [elegantly]
   
2. Do any other values of x satisfy the equation?
   

[Yoruichi-san]

Lol...this is TANTALIZING!

I was totally reading the problem wrong at first...but now I think I get it...

I'm not sure how solid this is...I'm not sure how sound the assumption inf-1=inf is, but...

So using the notation in the OP, I call the rhs x^y -> x^y=2.

Now I take the log₂ of both sides, and using properties of log to simplify-> log₂(x^y)=log₂(2) -> ylog₂(x)=1

Rearranging -> log₂(x)=1/y

But since y is the infinite series (-1 term...that's what I'm not 100% sure about using), y=2

-> log₂(x)=1/2 ->x=2^1/2

[Guest]

Lets define a sequence y_i where

y₀=x

y₁=x^x=x^y₀

y₂=x^y₁

and in general y_i=x^y_i-1.

If y_i approaches a limit z as i increases, then y_i+1 also approaches z, in other words x^y_i approaches z.

At the limit we can substitute y_i for z to get x^z=z.

In this case z=2 so x^2=2.

That means there are only 2 solutions in this case (plus or minus sqrt(2)).

So if the sequence tended towards 3 there would be 1 solution (or 3 if you allow complex numbers), namely the cube root of 3.

[Guest]

Hi Supandi,

Nice going.

But you all know how much I love "showing your work" even tho sometimes I don't.

So let me expand the puzzle.

1. Prove it [elegantly]
   
2. Do any other values of x satisfy the equation?
   

If there is another value,

negative of square root of 2..... - sqrt(2)

[Guest]

Hi Supandi,

Nice going.

But you all know how much I love "showing your work" even tho sometimes I don't.

So let me expand the puzzle.

1. Prove it [elegantly]
   
2. Do any other values of x satisfy the equation?
   

Sure.

Its exactly what Octopuppy said. I couldn't have said it better.

Infact I couldn't have said it at all I'm kinda math illiterate

[Guest]

Lets define a sequence y_i where

y₀=x

y₁=x^x=x^y₀

y₂=x^y₁

and in general y_i=x^y_i-1.

If y_i approaches a limit z as i increases, then y_i+1 also approaches z, in other words x^y_i approaches z.

At the limit we can substitute y_i for z to get x^z=z.

In this case z=2 so x^2=2.

That means there are only 2 solutions in this case (plus or minus sqrt(2)).

So if the sequence tended towards 3 there would be 1 solution (or 3 if you allow complex numbers), namely the cube root of 3.

I don't think you need the "sort of" in there. Good work.

[bonanova]

Lol...this is TANTALIZING!

I was totally reading the problem wrong at first...but now I think I get it...

I'm not sure how solid this is...I'm not sure how sound the assumption inf-1=inf is, but...

So using the notation in the OP, I call the rhs x^y -> x^y=2.

Now I take the log₂ of both sides, and using properties of log to simplify-> log₂(x^y)=log₂(2) -> ylog₂(x)=1

Rearranging -> log₂(x)=1/y

But since y is the infinite series (-1 term...that's what I'm not 100% sure about using), y=2

-> log₂(x)=1/2 ->x=2^1/2

Very astute; and inf-1 is in fact inf, for the purposes of this riddle.

Lop off the leftmost x, and you haven't changed the right hand side.

So lop off the leftmost x of x^y = 2, and you get y = 2.

Thus x^y = x² = 2. And you have your result.

Before I declare Y-san, octopuppy and woon as solvers, however, and this is where it really becomes tantalizing:

Is there yet another value of x which satisfies the equation?

[Guest]

Very astute; and inf-1 is in fact inf, for the purposes of this riddle.

Is there yet another value of x which satisfies the equation?

Lop off the leftmost x, and you haven't changed the right hand side.

So lop off the leftmost x of x^y = 2, and you get y = 2.

Thus x^y = x² = 2. And you have your result.

Before I declare Y-san, octopuppy and woon as solvers, however, and this is where it really becomes tantalizing:

...other than -sqrt(2)?

?!? ?!?

There's another answer?

[Guest]

Take a variable x and raise it to the xth power: xx; call the result y.

Now raise x to the y power: xy; call that result y. i.e. replace the previous value of y with the value of xy.

Again, raise x to the y power: xy; and replace the previous value of y with the value of xy.

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?

I'm not sure there isn't something wrong with all these proofs...

root(2)^root(2) = 1.632

root(2)^root(2)^root(2) = 2

BUT

root(2)^root(2)^root(2)^root(2) = 2.6651

and as we do enven more powers, the number gets ever higher.

I think the problem is:

that you've assumed that there is an answer to the question, and so that (incorrect) assumption has led to the proofs above.

If we start with x<1 then x^....^x tends to 0.

If x = 1, then x^...^x = 1

If x > 1, then x^...^x tends to infinity.

So... we can't get to = 2, and the problem has no solutions.

Or have I misunderstood something?

Edited October 6, 2008 by armcie

[bonanova]

I'm not sure there isn't something wrong with all these proofs...

root(2)^root(2) = 1.632

root(2)^root(2)^root(2) = 2

BUT

root(2)^root(2)^root(2)^root(2) = 2.6651

and as we do enven more powers, the number gets ever higher.

I think the problem is:

that you've assumed that there is an answer to the question, and so that (incorrect) assumption has led to the proofs above.

If we start with x<1 then x^....^x tends to 0.

If x = 1, then x^...^x = 1

If x > 1, then x^...^x tends to infinity.

So... we can't get to = 2, and the problem has no solutions.

Or have I misunderstood something?

Nope, you're spot on.

Convergence of the infinite exponent stack was not addressed.

It was only asserted, in violation of bonanova's prime directive.

The equation LHS = 2 is meaningless if LHS is undefined.

To validate the puzzle, I offer the following proof that LHS converges.

If the proof is valid, then we can discuss solutions to the equation LSH = 2.

If the proof is flawed, then the puzzle is solved by identifying the flaw.

Note: LHS = left hand side of the above equation

LHS is the limit [if a limit exists] of the sequence sqrt(2), sqrt(2) to the sqrt(2), ....

Write the sequence as a₁, a₂, a₃, ... where i-1 is the height of the exponent stack of term a_i.

Since a_i+2 = sqrt(2) ^a_i+1 > sqrt(2) ^a_i = a_i+1 . then 1 < a_i < a_i+1 for each i >= 1.

If the topmost sqrt(2) in any a_i is replaced by the larger value 2, a_i itself becomes 2.

So the sequence is increasing and bounded above; therefore it has a limit and converges.

[bonanova]

...other than -sqrt(2)?

?!? ?!?

There's another answer?

Sorry, I misspoke.

The additional question, rather, is ... let x to the x to the x to the x ... = 4. instead of = 2.

Then is not x = sqrt(2) a solution to this equation as well?

i.e., x^y = 4 where y = 4 leads to x = 4th root of 4 = sqrt(2) again.

Therefore, what kind of creature is sqrt(2) to the sqrt(2) to the .... - that it can equal both 2 and 4.

[Guest]

root(2)^root(2) = 1.632

root(2)^root(2)^root(2) = 2

BUT

root(2)^root(2)^root(2)^root(2) = 2.6651

and as we do enven more powers, the number gets ever higher.

(x^x)^x is not the same as x^(x^x). Your answer uses the former, while the OP clearly describes the latter.

D.

[Guest]

Sorry, I misspoke.

The additional question, rather, is ... let x to the x to the x to the x ... = 4. instead of = 2.

Then is not x = sqrt(2) a solution to this equation as well?

i.e., x^y = 4 where y = 4 leads to x = 4th root of 4 = sqrt(2) again.

Therefore, what kind of creature is sqrt(2) to the sqrt(2) to the .... - that it can equal both 2 and 4.

An interesting beasty indeed. Experimentation with Excel shows that the series does indeed converge on 2. If you take your first "result" to be 4 instead of sqrt(2) the series stays on 4. But otherwise it either converges to 2 (1st result<4) or goes flying off into infinity (1st result>4). Though I'm sure if you had stated in the question that it converged on 4 I would have given you the wrong answer and not spotted the fact that it doesn't converge there at all. That would have made a much more cheeky question!

[Prime]

Here is my inductive reasoning. I don't know if it could be considered as a proof, though.

(2^1/2)² = 2.

For the first step, let y = (2^1/2)^(2^1/2). Then y < 2.

For any y, if y < 2 then (2^1/2)^y < 2.

And since we substitute the above exp​ression for y on every step, y shall forever be less than 2.

[Guest]

Sorry, I misspoke.

The additional question, rather, is ... let x to the x to the x to the x ... = 4. instead of = 2.

Then is not x = sqrt(2) a solution to this equation as well?

i.e., x^y = 4 where y = 4 leads to x = 4th root of 4 = sqrt(2) again.

Therefore, what kind of creature is sqrt(2) to the sqrt(2) to the .... - that it can equal both 2 and 4.

To quote Keanu Reeves: "Whoa"

I have no idea how to resolve that.

[Guest]

Sorry, I misspoke.

The additional question, rather, is ... let x to the x to the x to the x ... = 4. instead of = 2.

Then is not x = sqrt(2) a solution to this equation as well?

i.e., x^y = 4 where y = 4 leads to x = 4th root of 4 = sqrt(2) again.

Therefore, what kind of creature is sqrt(2) to the sqrt(2) to the .... - that it can equal both 2 and 4.

I think you'll find it's a consequence of the fact that 2*2 = 2^2.

[Prime]

Sorry, I misspoke.

The additional question, rather, is ... let x to the x to the x to the x ... = 4. instead of = 2.

Then is not x = sqrt(2) a solution to this equation as well?

i.e., x^y = 4 where y = 4 leads to x = 4th root of 4 = sqrt(2) again.

Therefore, what kind of creature is sqrt(2) to the sqrt(2) to the .... - that it can equal both 2 and 4.

I don't see 2^1/2 as an answer here. To me it seems more like 2^(2^1/2).

Inductive reasoning could be the same as in my previous post.

[Guest]

I don't see 2^1/2 as an answer here. To me it seems more like 2^(2^1/2).

Inductive reasoning could be the same as in my previous post.

Help...I must be way off base here but:

if x to the x to the x to the x .... equals 2

then xlog (x to the x to the x to the x.... ) equals log2

thus x log2 = log 2

and x = 1

I only see sqrt(2) to the sqrt(2) to the sqrt(2) = 2

Obviously my logic is very faulty!

[Guest]

Help...I must be way off base here but:

if x to the x to the x to the x .... equals 2

then xlog (x to the x to the x to the x.... ) equals log2

thus x log2 = log 2

and x = 1

I only see sqrt(2) to the sqrt(2) to the sqrt(2) = 2

Obviously my logic is very faulty!

I think the problem is in the second line. It should be

(x to the x to the x to the x...)logx=log2

Which then gives you

2logx=log2

logx²=log2

x²=2

x=sqrt(2)

However, you can just as easily do it with 4:

(x to the x to the x to the x...)logx=log4

4logx=log4

logx⁴=log4

x⁴=4

x=sqrt(2)

So, I'm still totally stumped.

[Prime]

Help...I must be way off base here but:

if x to the x to the x to the x .... equals 2

then xlog (x to the x to the x to the x.... ) equals log2

thus x log2 = log 2

and x = 1

I only see sqrt(2) to the sqrt(2) to the sqrt(2) = 2

Obviously my logic is very faulty!

What is the base for the logarithm? It makes sense to consider base x. Let us also designate the infinite string of x exponents as p(x). The condition is that p(x) =2. Then

log_xp(x) = log_x2

p(x)log_xx = log_x2

p(x) = log_x2

2 = log_x2

And thus x = 2^1/2

Having that, I am not sure my answer for the p(x) = 4 in my previous post is correct.

[Guest]

I think the problem is in the second line. It should be

(x to the x to the x to the x...)logx=log2

Which then gives you

2logx=log2

logx²=log2

x²=2

x=sqrt(2)

However, you can just as easily do it with 4:

(x to the x to the x to the x...)logx=log4

4logx=log4

logx⁴=log4

x⁴=4

x=sqrt(2)

So, I'm still totally stumped.

And still confused here! More help required....thanks

If I take a simple finite example and use my math:

2 ^ 2 ^ 2 ^ 2 ^ 2 = 65,536

2log 2 ^ 2 ^ 2 ^ 2 = log 65,536

2log 256 = log 65,536

4.82 = 4.82

If I understand your correction would I get?

2 ^ 2 ^ 2 ^ 2 ^ 2 = 65,536

2 ^ 2 ^ 2 ^2 log 2 = log 65,536

256log 2 = log 65,536

77.06 not = 4.82

[Prime]

And still confused here! More help required....thanks

If I take a simple finite example and use my math:

2 ^ 2 ^ 2 ^ 2 ^ 2 = 65,536

...

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.

[Guest]

I think you'll find it's a consequence of the fact that 2*2 = 2^2.

d3k3 has the key here I think - another way to express it is that √2⁴=4 (ie 2 x 2 = √2.√2 x √2.√2 = 2²). This can also be expanded infinitely to √2^{√2^4...}=4

For the problem at hand, if you define a₁=4, then a_n=4 (because a_n=√2^a_n-1). I am not sure of the exact flaw in bona's suggested solution, but I think we are somehow looking at a different series, as he has already shown us that a_n is bounded above by 2 for all n>=1.

hmmm....not quite the elegant solution I had in mind, (or that I am sure bona has tucked away!)

[Guest]

Alright, this is driving me crazy. Let's go over this one more time:

p(x)=x^(x^(x^(...

p(sqrt(2))=2

P(sqrt(2))=4

2!=4

The only way I can see to resolve this is to just say that p(x) does not exist. Certainly not very satisfying, but what else can you say?