David Foster

+

4

5

4

5

4

I have heard of tetration before, but it always seemed to make larger numbers than were useful. I have not seen any formula I can think of, that used it either, which is a bit of a shame in a way.

Start a hangout

When you repeat addition you get multiplication. When you repeat multiplication you get exponentiation. And when you repeat that, you get **tetration**!

Let's follow Donald Knuth and write exponentiation as ↑. Then 4 cubed is

4↑3 = 4 × (4 × 4) = 64

We are multiplying 4 by itself 3 times. Next comes tetration, which we write as ↑↑. Here's how it works:

4↑↑3 = 4 ↑ (4 ↑ 4) = 4 ↑ 256 ≈ 1.3 × 10¹⁵⁴

We are raising 4 to itself 3 times. Before the parentheses didn't matter, but now they do: we put the parentheses all the way to the right, since (a ↑ b) ↑ c equals a ↑ (b × c) while a ↑ (b ↑ c) is something really new.

As you can see, tetration lets us describe quite large numbers. 10↑↑10 is much bigger than the number of atoms in the observable universe.

In fact, 10↑↑10 is so big that you couldn't write it down if you wrote one decimal digit on each atom in the observable universe!

In fact, 10↑↑10 is so big you couldn't write the*number of its digits* if you wrote one digit of *that* number on each atom in the observable universe!!

In fact, 10↑↑10 is so big you couldn't write down the*number of digits in its number of digits* if you wrote one digit of *that* number on each atom in the observable universe!!!

**Puzzle:** About how many times could I keep going on here? Let's say there are 10 ↑ 80 atoms in the observable universe; this seems roughly right.

We could march forwards and create notations for numbers so huge that 10↑↑10 looks pathetically small.*And we will!* We can even look at infinite tetration. *And we will - that's our main goal here!* But it's also fun to ask questions like:

**Puzzle:** What is 10↑↑10.2 ?

This makes no sense at first: you can't write down a tower of powers with 10.2 tens in it. But you couldn't add 10.2 tens together at first, either, so 10 × 10.2 didn't make sense until someone explained what it meant. Same for exponentiation. So, maybe tetration can also be defined for fractions in some nice way. And maybe real numbers too. And maybe even complex numbers.

Believe it or not, this seems to be an open question! It's phrased as a precise conjecture here:

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

and there's a good candidate for the answer. The picture here, drawn by Dmitrii Kouznetsov, shows a graph of the candidate for e↑↑z as a function on the complex plane.

#bigness

Let's follow Donald Knuth and write exponentiation as ↑. Then 4 cubed is

4↑3 = 4 × (4 × 4) = 64

We are multiplying 4 by itself 3 times. Next comes tetration, which we write as ↑↑. Here's how it works:

4↑↑3 = 4 ↑ (4 ↑ 4) = 4 ↑ 256 ≈ 1.3 × 10¹⁵⁴

We are raising 4 to itself 3 times. Before the parentheses didn't matter, but now they do: we put the parentheses all the way to the right, since (a ↑ b) ↑ c equals a ↑ (b × c) while a ↑ (b ↑ c) is something really new.

As you can see, tetration lets us describe quite large numbers. 10↑↑10 is much bigger than the number of atoms in the observable universe.

In fact, 10↑↑10 is so big that you couldn't write it down if you wrote one decimal digit on each atom in the observable universe!

In fact, 10↑↑10 is so big you couldn't write the

In fact, 10↑↑10 is so big you couldn't write down the

We could march forwards and create notations for numbers so huge that 10↑↑10 looks pathetically small.

This makes no sense at first: you can't write down a tower of powers with 10.2 tens in it. But you couldn't add 10.2 tens together at first, either, so 10 × 10.2 didn't make sense until someone explained what it meant. Same for exponentiation. So, maybe tetration can also be defined for fractions in some nice way. And maybe real numbers too. And maybe even complex numbers.

Believe it or not, this seems to be an open question! It's phrased as a precise conjecture here:

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

and there's a good candidate for the answer. The picture here, drawn by Dmitrii Kouznetsov, shows a graph of the candidate for e↑↑z as a function on the complex plane.

#bigness

483

255

94 comments

David Foster

+

4

5

4

5

4

I have heard of tetration before, but it always seemed to make larger numbers than were useful. I have not seen any formula I can think of, that used it either, which is a bit of a shame in a way.

As I recall such big number related to kolmogorov complexity...

Samuel Loup

+

6

7

6

7

6

your graph, looks like a fractal

Ray Cromwell

+

7

8

7

8

7

I have a feeling http://en.wikipedia.org/wiki/Grahams_number is going to come up soon.

Akira Bergman

+

2

3

2

3

2

+David Foster , why is it not used? Too difficult or not needed?

David Foster

+

1

2

1

2

1

Not sure, both reasons I suppose. I just know I have seen a lot of formulas, and I haven't seen this used to do anything.

Mikhail Kandel

+

2

3

2

3

2

Why is there madness in the graph past e↑↑e?

With e comes chaos?

Useful information for us 'plebs', +John Baez. Why the superfluous parentheses?

Madness x Chaos = €¥!£

Matt McIrvin

+

4

5

4

5

4

+Samuel Loup Indeed, it bears a strange resemblance to the piece of the Mandelbrot set near the negative real axis. I immediately started wondering if that's coincidental.

I could find the paper, but it also describes more than 2 up-arrows... There could be 5, for instance…

See http://www.tetration.org/Tetration/index.html

See http://www.tetration.org/Tetration/index.html

Haw Bawheed

+

1

2

1

2

1

Try teach this to someone with adha

Ziyad Khamis

+

1

2

1

2

1

Interesting in a seemingly complex way..

Timothy Gugerty

+

1

2

1

2

1

I don't trust intelligence that cannot make observation

Oliy Bowman

+

1

2

1

2

1

dude its maths no one likes it

Alfredo Ramos

+

1

2

1

2

1

This is what I've been saying for years now!!

Martin Smith

+

1

2

1

2

1

Like? How many fingers you got, dude!? +Oliy Bowman

Peter Luschny

+

4

5

4

5

4

"The copyright holder of this file [the plot of the graph], Dmitrii Kouznetsov, allows anyone to use it for any purpose, provided that the copyright holder is properly attributed." I do not see an attribution to the copyright holder in the post.

Pete Hopkins

+

3

4

3

4

3

That is cool! Bet u cant explain the properties of the female reproductive organ though.

Ryan Ng

+

1

2

1

2

1

Out of this world.

John Baez

+

3

4

3

4

3

+David Foster - Yes, tetration seems *vastly* less useful than addition, multiplication and exponentiation. I'm not sure why. Perhaps it's because these operations get a bit 'worse' each time. For example, exponentiation isn't commutative and associative the way addition and multiplication are. Still, exponentiation is important because

(d/dx) e ↑ x = e ↑ x

What differential equation do we get from tetration?

Another possibility is that people haven't thought of many good uses for tetration because not enough people know about it!

(d/dx) e ↑ x = e ↑ x

What differential equation do we get from tetration?

Another possibility is that people haven't thought of many good uses for tetration because not enough people know about it!

John Baez

+

3

4

3

4

3

+Ray Cromwell - Yes, Graham's number is coming up soon! People who want to dive in now should try this:

http://www.mrob.com/pub/math/largenum.html

What I'm really interested in now is how describing*really really* large numbers - numbers so large they make Graham's number look small - uses the theory of infinities. There's something spooky about this.

http://www.mrob.com/pub/math/largenum.html

What I'm really interested in now is how describing

+John Baez or because complexity of it (in any sense) is very high.

John Baez

+

4

5

4

5

4

+Martin Smith wrote: "Why the superfluous parentheses?"

Addition and multiplication are associative but exponentiation and all the higher 'hyperoperations' aren't, so I wanted to emphasize that they're defined in a way where the parentheses are pushed to the right:

4 × 3 = 4 + (4 + 4) = (4 + 4) + 4

4↑3 = 4 × (4 × 4) = (4 × 4) × 4

4↑↑3 = 4 ↑ (4 ↑ 4) ≠ (4 ↑ 4) ↑ 4

4↑↑↑3 = 4 ↑↑ (4 ↑↑ 4) ≠ (4 ↑↑ 4) ↑↑ 4

But, once we get used to this, we can say that 4 ↑ 4 ↑ 4 is defined to be 4 ↑ (4 ↑ 4), and so on. People usually do this for exponentiation and I guess they do it for tetration and the rest too.

http://en.wikipedia.org/wiki/Hyperoperation

Addition and multiplication are associative but exponentiation and all the higher 'hyperoperations' aren't, so I wanted to emphasize that they're defined in a way where the parentheses are pushed to the right:

4 × 3 = 4 + (4 + 4) = (4 + 4) + 4

4↑3 = 4 × (4 × 4) = (4 × 4) × 4

4↑↑3 = 4 ↑ (4 ↑ 4) ≠ (4 ↑ 4) ↑ 4

4↑↑↑3 = 4 ↑↑ (4 ↑↑ 4) ≠ (4 ↑↑ 4) ↑↑ 4

But, once we get used to this, we can say that 4 ↑ 4 ↑ 4 is defined to be 4 ↑ (4 ↑ 4), and so on. People usually do this for exponentiation and I guess they do it for tetration and the rest too.

http://en.wikipedia.org/wiki/Hyperoperation

John Baez

+

1

2

1

2

1

+Peter Luschny - Thanks. I added the attribution.

I only learnt subtraction and addition never heard of all of u saying cause I after holiday thean pri3 hope to learn at pri3 or pri4 but I think is working of division and multiplication 😀😀😀

John Baez

+

4

5

4

5

4

+Toby Bartels wrote: "With e comes chaos?"

I bet something like that is right. The infinite tetration

x ↑ (x ↑ (x ↑ (x ↑ (x ↑ (x ↑ (x ↑ (x ↑ ( ...

converges for e ↑ (1/e) ≤ x ≤ e ↑ (e - 1). This was shown by Euler!

http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights

I bet something like that is right. The infinite tetration

x ↑ (x ↑ (x ↑ (x ↑ (x ↑ (x ↑ (x ↑ (x ↑ ( ...

converges for e ↑ (1/e) ≤ x ≤ e ↑ (e - 1). This was shown by Euler!

http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights

John Baez

+

1

2

1

2

1

So over 200 people +1ed this and nobody tried the puzzle??? What wimps! :-)

I'm hoping the mathematicians that see how to do this realize it's so easy that they want to leave it for other people.

I'm hoping the mathematicians that see how to do this realize it's so easy that they want to leave it for other people.

Brian Jones

+

2

3

2

3

2

I'm thinking of a number between 1 and 10 ^^ 10. What number am I thinking of?

Matt McIrvin

+

4

5

4

5

4

+John Baez The other reason exponentiation is useful has specifically to do with small powers and the physics of our universe. Powers of 2, 3 and sometimes 4 come up a lot because we live in a space with a small number of dimensions that has symmetries mixing its dimensions, and also in which Pythagorean-style distance metrics are important. Is there anything analogous for tetration?

It doesn't make sense

awesome dude

nice

John Baez

+

1

2

1

2

1

+Rwitwik Sinha - those "maybes" are all explained in the Wikipedia article I linked to:

http://en.wikipedia.org/wiki/Tetration#Extension_to_real_heights

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

However, there really is an open question here. We need a good mathematician to solve it!

http://en.wikipedia.org/wiki/Tetration#Extension_to_real_heights

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

However, there really is an open question here. We need a good mathematician to solve it!

Romain Brasselet

+

2

3

2

3

2

For the first puzzle, 10↑↑n has 10↑↑(n-1)+1 digits.

Assume that 10^80 is roughly 10^10 (i.e. 10↑↑2) to make it easier. It is anyway much smaller than 10↑↑3.

We can take the number of digits (nod) of the previous number 8 times (10-2). So we can write the nod of the nod of the nod of the nod of the nod of the nod of the nod of the nod of 10↑↑10 and write a single digit of that number on each particle in the universe. That should fit!

Assume that 10^80 is roughly 10^10 (i.e. 10↑↑2) to make it easier. It is anyway much smaller than 10↑↑3.

We can take the number of digits (nod) of the previous number 8 times (10-2). So we can write the nod of the nod of the nod of the nod of the nod of the nod of the nod of the nod of 10↑↑10 and write a single digit of that number on each particle in the universe. That should fit!

Greetings.

Excellent.

OL.

Excellent.

OL.

Easily sorted out with some Excel equations =)

Matt McIrvin

+

1

2

1

2

1

My previous comment is maybe related to the observation that exponentiation is not commutative. When an operation is commutative, the symmetric case where you plug in two or more of the same thing is probably especially important. (Anticommutative too, but there the important observation is just that you get nothing.) And that's how you go up a step in this sequence.

When it's not commutative, that might well reduce the significance of the symmetric case, at least for finite numbers.

When it's not commutative, that might well reduce the significance of the symmetric case, at least for finite numbers.

For the second puzzle, I can only give a naive suggestion.

The same way that 10^0.5=x can be solved by looking for x^2=10, to solve 10↑↑0.5=x, I would look for x such that x↑↑2=10, which should be around 2.5.

The same way that 10^0.5=x can be solved by looking for x^2=10, to solve 10↑↑0.5=x, I would look for x such that x↑↑2=10, which should be around 2.5.

I'll try my hand at the puzzle, even though I have no math training beyond high school, and without reading the wikipedia article.

10↑↑10 = 10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ 10))))))))

10^10 = 10 billion

That means 10↑↑4 is already greater than a googolplex (10^(10^100)), which already can't fit into the universe. However, that's not a bad starting point...

With that in mind, I can safely say that you can't write "the digits of the digits" as far as 10↑↑7, as each progressive ↑↑ is unwritable.

However, since 10^80 is a fraction of 10^100, the math to figure out exactly how many times you would need to regress "the digits of the digits" eludes me. I know it'd be at least 7 times, but I'd wager on it being significantly more.

10↑↑10 = 10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ 10))))))))

10^10 = 10 billion

That means 10↑↑4 is already greater than a googolplex (10^(10^100)), which already can't fit into the universe. However, that's not a bad starting point...

With that in mind, I can safely say that you can't write "the digits of the digits" as far as 10↑↑7, as each progressive ↑↑ is unwritable.

However, since 10^80 is a fraction of 10^100, the math to figure out exactly how many times you would need to regress "the digits of the digits" eludes me. I know it'd be at least 7 times, but I'd wager on it being significantly more.

Timothy Gowers

+

2

3

2

3

2

I'm going to write T(2,k) for a tower of 2s of height k. One of the difficulties we have here is in finding nice algebraic rules satisfied by this function. That's important because the way we extend exponentiation to non-integer powers is by insisting on the rule a^{x+y}=a^xa^y plus something else such as continuity or monotonicity. Can we say anything nice about T(2,r+s)? It doesn't seem so.

A closely related question is to find a nice function f such that f(f(x))=log x. I wondered about approximating log x by (x^c-1)/c for a very small constant c, but found that even finding a nice function f such that f(f(x))=x^2-1 seems to be annoyingly challenging.

A closely related question is to find a nice function f such that f(f(x))=log x. I wondered about approximating log x by (x^c-1)/c for a very small constant c, but found that even finding a nice function f such that f(f(x))=x^2-1 seems to be annoyingly challenging.

MuChawn Golden

+

1

2

1

2

1

I would love if my money reproduced this way!

Puzzle 1 makes me wonder about repeated logarithms (log log... repeated k times... log log x) . Do they exist and do they have any interesting applications?

dude .. awesome stuff.. ↑↑

tatration seems intersting.. ;-)

tatration seems intersting.. ;-)

Matt Prosper

+

1

2

1

2

1

My brain is all melted and hurty.

It looks like you've had a seizure whilst using spyragraph.

A little Excel can sort this out. Here is a good start http://www.supersoftbox.com/macro_spreadsheet_example.php

Boris Borcic

+

1

2

1

2

1

I think you people don't see how great this really is. We have a *painted portrait* of a sexy lady identified as "the extension to complex heights of tetration". The picture is of her, but it deserves the name of "painted portrait" rather than just picture, because the text tells us her existence is just conjecture. Therefore, the picture is by definition what's usually called "Artist's vision" in popular science channels. OTOH, Wikipedia tells us that her avatar is available online:)

Brillantly confuse to my puny brain cells. But it is excellant math skills

Did anybody read Conway on the sensuality of quadratic forms? The two conjugated fixed points suggest a sensual link to things really quadratic, dont they?

Rereading the comments in the thread, I find myself wanting to be told about special triplets such that (x^y)^z = x^(y^z) when ^ is in general non-associative. Lazy, I agree :)

John Baez

+

1

2

1

2

1

+Timothy Gowers wrote: "Can we say anything nice about T(2,r+s)? It doesn't seem so."

I'll stick to up-arrow notation. We have rules like

a × (b + c) = (a × b) + (a × c)

a ↑ (b + c) = (a ↑ b) × (a ↑ c)

so continuing this pattern we'd naturally hope for

a ↑↑ (b + c) = (a ↑↑ b) ↑ (a ↑↑c)

**Puzzle:** is this rule true or not?

Another approach to determining a ↑↑ b for real numbers b is to look for a 'nice' analytic function that agrees with a ↑↑ b for all natural numbers b. This is how we define a! for lots of real numbers a: people call it the gamma function and, just out of perversity, set it up so that Γ(a) = (a-1)! for natural numbers a. Here 'nice' refers to the fact that there are lots of analytic functions that vanish at all natural numbers a, so we can't expect a unique extension unless we impose some extra conditions. And this is what the Wikipedia article suggests doing:

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

I'll stick to up-arrow notation. We have rules like

a × (b + c) = (a × b) + (a × c)

a ↑ (b + c) = (a ↑ b) × (a ↑ c)

so continuing this pattern we'd naturally hope for

a ↑↑ (b + c) = (a ↑↑ b) ↑ (a ↑↑c)

Another approach to determining a ↑↑ b for real numbers b is to look for a 'nice' analytic function that agrees with a ↑↑ b for all natural numbers b. This is how we define a! for lots of real numbers a: people call it the gamma function and, just out of perversity, set it up so that Γ(a) = (a-1)! for natural numbers a. Here 'nice' refers to the fact that there are lots of analytic functions that vanish at all natural numbers a, so we can't expect a unique extension unless we impose some extra conditions. And this is what the Wikipedia article suggests doing:

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

John Baez

+

1

2

1

2

1

+Boris Borcic wrote: "the text tells us her existence is just conjecture."

As far as I can tell it's not the*existence* of this function that's just a conjecture. As far as I can tell, it exists. The main point seems to be that there's a *unique* function obeying some conditions. Wikipedia says:

"There is a conjecture that there exists a unique function F which is a solution of the equation F(z+1)=exp(F(z)) and satisfies the additional conditions that F(0)=1 and F(z) approaches the fixed points of the logarithm (roughly 0.31813150520476413531 ± 1.33723570143068940890i) as z approaches ±i∞ and that F is holomorphic in the whole complex z-plane, except the part of the real axis at z≤−2. The complex double precision approximation of this function is available online."

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

That's not completely clear, but it gives me the feeling that*existence* is known, or easy, while uniqueness is the main open question.

As far as I can tell it's not the

"There is a conjecture that there exists a unique function F which is a solution of the equation F(z+1)=exp(F(z)) and satisfies the additional conditions that F(0)=1 and F(z) approaches the fixed points of the logarithm (roughly 0.31813150520476413531 ± 1.33723570143068940890i) as z approaches ±i∞ and that F is holomorphic in the whole complex z-plane, except the part of the real axis at z≤−2. The complex double precision approximation of this function is available online."

http://en.wikipedia.org/wiki/Tetration#Extension_to_complex_heights

That's not completely clear, but it gives me the feeling that

+John Baez Aren't you euphemistic? :) You are certainly right, but it boils down the same: the lady can't be specially sexy if she is run-of-the-mill :)

John Baez

+

2

3

2

3

2

Thanks for being the first ones with the guts to tackle this puzzle, +Romain Brasselet and +Davis Centis! I write puzzles so people can have fun talking about them, so I feel sad if nobody tries them.

Romain wisely defined**nod** to mean 'number of digits' and wrote approximately:

"So we can write the nod of the nod of the nod of the nod of the nod of the nod of the nod of the nod of 10↑↑10 and write a single digit of that number on each atom in the universe."

That looks like 8 nods to me. Davis Centis said 'at least 7'. So does this mean that you guys agree: you can't take the nod^7 of 10↑↑10 and write a single digit of that number on each atom in the universe, but you can with the nod^8?

By the way, Romain said 'particle' where I said 'atom'. I don't feel like estimating the number of photons or neutrinos in the universe. It could be big, and with photons it's hard to even define.

Romain wisely defined

"So we can write the nod of the nod of the nod of the nod of the nod of the nod of the nod of the nod of 10↑↑10 and write a single digit of that number on each atom in the universe."

That looks like 8 nods to me. Davis Centis said 'at least 7'. So does this mean that you guys agree: you can't take the nod^7 of 10↑↑10 and write a single digit of that number on each atom in the universe, but you can with the nod^8?

By the way, Romain said 'particle' where I said 'atom'. I don't feel like estimating the number of photons or neutrinos in the universe. It could be big, and with photons it's hard to even define.

John Baez

+

2

3

2

3

2

Later we'll get to numbers X that are so big that if you take the biggest true sentence

**We can't write the number of digits of the number of digits of the number of digits of.... X in this universe with just one digit on each atom in the observable universe.**

we can't write*this* sentence in the universe! And so on.

we can't write

Well, it could very well be more than 8... a lot more. There's a big gap between the *nod* of a googolplex (10^(10^100)) and 10^80. We then start to exacerbate this difference with each successive 10^. Thus, while I said at least 7, my intuitive understanding of math suggests that it would actually be much higher. It's tempting to say that it could be 8 **nods**, but with each successive 10^, a small fraction might inflate to the point that you need dozens of **nods**!

Thanks for the puzzle +John Baez !

Thanks for the puzzle +John Baez !

You're welcome! I approached this puzzle by thinking about some easier ones first:

1) What's the number of digits of

10^10 ?

2) What's the number of digits of the number of digits of

10^(10^10) ?

3) What's the number of digits of the number of digits of the number of digits of

10^(10^(10^10)) ?

and so on.

1) What's the number of digits of

10^10 ?

2) What's the number of digits of the number of digits of

10^(10^10) ?

3) What's the number of digits of the number of digits of the number of digits of

10^(10^(10^10)) ?

and so on.

John Baez

+

3

4

3

4

3

I'm really happy this post has gotten over four hundred +1s so far, because I'm trying to get lots of people excited about math. But I'll only feel satisfied when I start seeing things like +10↑↑1.5 next to my posts.

David Foster

+

1

2

1

2

1

I wonder if logs, or even repeated logs, might make this problem easier.

Maybe... or maybe not.

Maybe... or maybe not.

Cool new word, don't understand anything beyond that.

+David Foster - it's all about repeated logs, since the number of digits in a natural number x is approximately log(x). (To be precise, it's 1 more than the biggest integer that's ≤ log(x).)

Different

Well, to add something yet on the more felt than reasoned or want-to-feel-smart side, to me the infinite's first function is to police asymptotics, and what this seems to be leading at, is perhaps then a "hyperbolic rotation" of "wheels within wheels", *asymptotics within asymptotics?*

math is hot

COOL

Aw...i think i will just have bologna sandwich.

Mind Blown!

stefan jeffers

+

1

2

1

2

1

After reading that Wiki article, my brain feels like that graph looks.

Boris Borcic

+

1

2

1

2

1

+stefan jeffers You are aware, of course, that this is an ambiguous declaration?

Wait...wut?

Toby Bartels

+

1

2

1

2

1

+John Baez wrote in part:

>a ↑↑ (b ↑ c) = (a ↑↑ b) ↑ (a ↑↑c)

Surely you mean

a ↑↑ (b + c) = (a ↑↑ b) ↑ (a ↑↑c)

which is still false because ↑ is nonassociative.

>a ↑↑ (b ↑ c) = (a ↑↑ b) ↑ (a ↑↑c)

Surely you mean

a ↑↑ (b + c) = (a ↑↑ b) ↑ (a ↑↑c)

which is still false because ↑ is nonassociative.

+Toby Bartels - Whoops, I'll fix that typo for future people who encounter that puzzle before your answer. Btw, didn't you and James Dolan work on these iterated operations and their algebraic properties at some point?

Thanks, +John Baez! Great attention! Both to your work & those that inquire of you. <3

John Baez

+

1

2

1

2

1

+Matt McIrvin - Interesting comments. Yes, it seems the main uses of exponentiation in everyday scientific life are x^2, x^3, x^4 and e^(cx). These probably account for over 95% of them. I know of *no uses* of tetration and the higher-order hyperoperations other than to name insanely large numbers. Wikipedia lists some real-world applications of the Lambert W function:

http://en.wikipedia.org/wiki/Lambert_W_function#Example_5

which is defined by z = W(z) exp(W(z)), and which is closely related to infinite tetration:

z^(z^(z^(z^(z^(..... = W(-ln(z)) / -ln(z)

But that's all I've got.

http://en.wikipedia.org/wiki/Lambert_W_function#Example_5

which is defined by z = W(z) exp(W(z)), and which is closely related to infinite tetration:

z^(z^(z^(z^(z^(..... = W(-ln(z)) / -ln(z)

But that's all I've got.

John Baez

+

2

3

2

3

2

That sounds interesting, +John Lato! I should learn about that. On the other extreme, I guess that some rapidly growing functions show up as lower bounds for the complexity of some very hard problems. So it seems these functions are more about logic and computation than physics and differential equations.

There's also x^(4-epsilon), for dealing with singularities in quantum field theory...

corinne osh

+

1

2

1

2

1

Brilliant!

Toby Bartels

+

1

2

1

2

1

+John Baez wrote ‘didn't you and James Dolan work on these iterated operations and their algebraic properties at some point?’

Jim was thinking about something like rings with three (or more) operations, corresponding to addition, multiplication, and exponentiation (and maybe tetration, which he of course called ‘eka-exponentiation’), each of which distributed over the previous one (but on one side only).

Jim was thinking about something like rings with three (or more) operations, corresponding to addition, multiplication, and exponentiation (and maybe tetration, which he of course called ‘eka-exponentiation’), each of which distributed over the previous one (but on one side only).

Toby Bartels

+

3

4

3

4

3

+John Baez & +Matt McIrvin

>x^2, x^3, x^4, e^(cx), x^(4-epsilon)

More generally, exponentation is almost always used when either the base or the exponent is constant, or at least when there's a big difference in how constant they are (on the continuum from explicit constant through unspecified constant through parameter through independent variable to dependent variable). So no x^x, and certainly no tetration.

This is why, in calculus class, they almost always teach the sum rule d(u + v) = du + dv and product rule d(uv) = v du + u dv in full glory, but almost never do the same for the exponentiation rule d(u^v) = u^(v-1) (v du + u ln u dv), preferring to teach only the special cases in which u or v is constant.

>x^2, x^3, x^4, e^(cx), x^(4-epsilon)

More generally, exponentation is almost always used when either the base or the exponent is constant, or at least when there's a big difference in how constant they are (on the continuum from explicit constant through unspecified constant through parameter through independent variable to dependent variable). So no x^x, and certainly no tetration.

This is why, in calculus class, they almost always teach the sum rule d(u + v) = du + dv and product rule d(uv) = v du + u dv in full glory, but almost never do the same for the exponentiation rule d(u^v) = u^(v-1) (v du + u ln u dv), preferring to teach only the special cases in which u or v is constant.

stefan jeffers

+

2

3

2

3

2

OK, +Boris Borcic , after a few hours to recover, I would say my brain felt stirred, scattered, splattered, and not at all holomorphic on the part of the real axis at z ≤ −2. :-)

John Baez

+

3

4

3

4

3

Shaken, not stirred.

Boris Borcic

+

2

3

2

3

2

The graph isn't obvious to decipher in practice. The legend says that lines are traced where ln|f| and arg(f) are integer values, but what do the line colors mean - aside making the result prettier? What does the symmetric white *shape* extending around the positive real axis correspond to? Its border exhibits intriguing self-similarities... makes one wish for an explorer applet like those for the Mandelbrot or Julia sets.

Romain Brasselet

+

1

2

1

2

1

Back to the puzzle, 10↑↑2 is a 1 followed by ten 0s, so its nod is clearly 11.

10↑↑3 is a 1 followed by 10^10 0s, so its nod is 10↑↑2 +1. Then the nod of 10↑↑2 +1, following the previous result, is 11.

10↑↑4 is a 1 followed by 10^10^10 0s, so its nod is 10↑↑3+1. So the nod of the nod of the nod of 10↑↑4 is 11.

In general, 10↑↑n=10^(10↑↑(n-1)). Also, 10^n has n+1 digits. These considerations yield that the nod of 10↑↑n is 10↑↑(n-1) +1. In a simple language, taking the nod of a tetration of 10 decreases the "tetration exponent" by 1.

Now, we want to find how many times we have to take the nod of 10↑↑10 to reach a number smaller than 10^80 (which is comprised between 10↑↑2 and 10↑↑3).

Based on the previous results, again I find 8.

However, this is the first I play with tetration and I may have made a huge mistake somewhere...

10↑↑3 is a 1 followed by 10^10 0s, so its nod is 10↑↑2 +1. Then the nod of 10↑↑2 +1, following the previous result, is 11.

10↑↑4 is a 1 followed by 10^10^10 0s, so its nod is 10↑↑3+1. So the nod of the nod of the nod of 10↑↑4 is 11.

In general, 10↑↑n=10^(10↑↑(n-1)). Also, 10^n has n+1 digits. These considerations yield that the nod of 10↑↑n is 10↑↑(n-1) +1. In a simple language, taking the nod of a tetration of 10 decreases the "tetration exponent" by 1.

Now, we want to find how many times we have to take the nod of 10↑↑10 to reach a number smaller than 10^80 (which is comprised between 10↑↑2 and 10↑↑3).

Based on the previous results, again I find 8.

However, this is the first I play with tetration and I may have made a huge mistake somewhere...

Your solution looks correct to me, +Romain Brasselet! Congratulations!

Great! Now, I really want to understand what 10↑↑10.2 is... even more now that you said you'll be satisfied when your post reaches 10↑↑1.5 +1s.

Again, by writing 10↑↑x=10^(10↑↑(x-1)), we see that we only have to care about solving 10↑↑x where 0<x<1. The rest will follow.

Based on my first intuition, 10↑↑0.5=x could be solved by taking x↑↑2=10, which gives x=2.5. However 10^2.5 is roughly 300, which your post has already reached, so it cannot be the answer...

Again, by writing 10↑↑x=10^(10↑↑(x-1)), we see that we only have to care about solving 10↑↑x where 0<x<1. The rest will follow.

Based on my first intuition, 10↑↑0.5=x could be solved by taking x↑↑2=10, which gives x=2.5. However 10^2.5 is roughly 300, which your post has already reached, so it cannot be the answer...

+Romain Brasselet - the Wikipedia article is pretty helpful if you're trying to figure out what 10↑↑10.2 is:

http://en.wikipedia.org/wiki/Tetration#Extension_to_real_heights

It says there's no generally agreed-on answer, but it has a lot of useful facts and references, and it presents a candidate that seems good to me. When it gets to complex heights, it gives a link to Mathematica code that calculates what*may be* the unique best version of x↑↑z when z is complex. I haven't had the energy to make any progress on these questions myself....

http://en.wikipedia.org/wiki/Tetration#Extension_to_real_heights

It says there's no generally agreed-on answer, but it has a lot of useful facts and references, and it presents a candidate that seems good to me. When it gets to complex heights, it gives a link to Mathematica code that calculates what

Toby Bartels

+

1

2

1

2

1

>the nod of 10↑↑n is 10↑↑(n-1) +1

Or, the log of 10↑↑n is 10↑↑(n-1), period, which is slightly easier to calculate with.

Here, by ‘log’ I mean the floor of the common logarithm, which is always 1 less than the number of decimal digits of a positive integer.

Or, the log of 10↑↑n is 10↑↑(n-1), period, which is slightly easier to calculate with.

Here, by ‘log’ I mean the floor of the common logarithm, which is always 1 less than the number of decimal digits of a positive integer.

Glenn Phillips

+

2

3

2

3

2

I've been looking at the for a week now and I still have no idea how to read it... but its bloody beautiful

Daniel Geisler

+

1

2

1

2

1

+John Baez - As the author of http://www.tetration.org I have been deeply involved in extending tetration to the complex numbers as well as matrices. You may be interested to know that Philippe Flajolet's research and your writing have had the biggest impact on my research since the turn of the millennia. Back in 2006 I noticed an article while searching on “tetration” with Google - it was the Wikipedia article on tetration. I made what I felt were significant improvements to the article, but after struggling for a year with people who didn’t follow Wikipedia’s policy of no unpublished personal research I didn’t want my name associated with the article.

Here are some things to consider regarding extending tetration. For links on who has done what regarding tetration see http://tetration.org/Links/index.html.

Extending tetration is just an example of a much broader and important problem, that of fractional iteration. If you conquer fractional iteration you can not only extend tetration, but also pentation and all the higher operators of the Ackermann function. I met with Stephen Wolfram in 1986 several times and discussed my work with extending tetration. Wolfram was very interested because at the time he was trying to unify different dynamical systems into a single system that could mathematically model any dynamical system. There are two dynamical systems deemed capable of representing the dynamical systems in physics - PDEs and iterated functions. The problem is that iterated functions are discrete while physics given every appearance of being continuous. Therefore a theory of fractional iteration would provide a new inroad into understanding physics. See

R. Aldrovandi and L. P. Freitas,

Continuous iteration of dynamical maps,

J. Math. Phys. 39, 5324 (1998)

for a great paper on continuous/fractional iteration with an application to the Naiver Stokes equation. My derivation of fractionally iterated functions is experimentally consistent with Aldrovandi and Freitas, but I haven’t proven their mathematical equivalence.

What combinatorial structure is associated with iterated functions. Look at a simpler problem and ask what combinatorial structure is associated with composite functions? The answer is set partitions or Bell numbers. Aldrovandi and Freitas paper follows this line of thought and use Bell matrices in their paper. Riordan’s book Combinatorial Identities and the chapter on Bell polynomials gives further insights. Bell polynomials provide a way of representing arbitrary functions in terms of the derivitives of composite functions. The combinatorial structure associated with Bell polynomials has a number of names: Schroeder’s Fourth Problem, hierarchies total partitions and phylogenetic trees. See http://oeis.org/A000311 for more information. My own work focuses on the derivatives of iterated functions which are Bell polynomials because iterated functions are trivially composite functions. See http://tetration.org/Combinatorics/index.html to view how Schroeder’s Fourth Problem is recovered from the coefficients of the derivatives of iterated functions.

Tetration is based on iterated functions which are based on complex dynamics. Working alone for decades it has become very important for me to find flaws in my extension of tetration. The research community mostly consists of I. N. Galidakis, and myself working as individuals and researchers associated with the Tetration Forum including Kouznetsov http://math.eretrandre.org/tetrationforum/ . My point of difference with the Tetration Forum and Kouznetsov is that they are proposing solutions to the extension of tetration that are easily seen as being inconsistent with complex dynamics. In fact they claim that fixed points are irrelevant to extending tetration. According to complex dynamics -

Lennart Carleson, Theodore W. Gamelin

Complex Dynamics

Springer-Verlag

In reviewing Chapter 2 - the Classification of Fixed Points, one can see that because of the conjugacy of fixed points that there can be no “simple” single equation to extending tetration. In the classification of fixed points there are hyperbolic, super-attracting, parabolic rationally neutral, rationally neutral and irrationally neutral fixed points. Each type of fixed point leads to a different type of tetration. For example e ↑ (1/e) is the sole parabolic rationally neutral fixed point for tetration, so e ↑ (1/e) must have it’s own separate equation for extending tetration. Not only is my work consistent with the Classification of Fixed Points, it provides an algebraic explanation of the Classification of Fixed Points that is much simpler than the topologically based proofs in Complex Dynamics.

While I have been able to fractionally iterate any complex function and therefore tetration, there is a major problem everyone researching tetration faces, that of convergence. In my mind all the real action in tetration is focused on people showing that they have a convergent solution. A couple of years ago the issue was quite active on MathOverflow. See http://mathoverflow.net/questions/tagged/fractional-iteration .

Here are some things to consider regarding extending tetration. For links on who has done what regarding tetration see http://tetration.org/Links/index.html.

Extending tetration is just an example of a much broader and important problem, that of fractional iteration. If you conquer fractional iteration you can not only extend tetration, but also pentation and all the higher operators of the Ackermann function. I met with Stephen Wolfram in 1986 several times and discussed my work with extending tetration. Wolfram was very interested because at the time he was trying to unify different dynamical systems into a single system that could mathematically model any dynamical system. There are two dynamical systems deemed capable of representing the dynamical systems in physics - PDEs and iterated functions. The problem is that iterated functions are discrete while physics given every appearance of being continuous. Therefore a theory of fractional iteration would provide a new inroad into understanding physics. See

R. Aldrovandi and L. P. Freitas,

Continuous iteration of dynamical maps,

J. Math. Phys. 39, 5324 (1998)

for a great paper on continuous/fractional iteration with an application to the Naiver Stokes equation. My derivation of fractionally iterated functions is experimentally consistent with Aldrovandi and Freitas, but I haven’t proven their mathematical equivalence.

What combinatorial structure is associated with iterated functions. Look at a simpler problem and ask what combinatorial structure is associated with composite functions? The answer is set partitions or Bell numbers. Aldrovandi and Freitas paper follows this line of thought and use Bell matrices in their paper. Riordan’s book Combinatorial Identities and the chapter on Bell polynomials gives further insights. Bell polynomials provide a way of representing arbitrary functions in terms of the derivitives of composite functions. The combinatorial structure associated with Bell polynomials has a number of names: Schroeder’s Fourth Problem, hierarchies total partitions and phylogenetic trees. See http://oeis.org/A000311 for more information. My own work focuses on the derivatives of iterated functions which are Bell polynomials because iterated functions are trivially composite functions. See http://tetration.org/Combinatorics/index.html to view how Schroeder’s Fourth Problem is recovered from the coefficients of the derivatives of iterated functions.

Tetration is based on iterated functions which are based on complex dynamics. Working alone for decades it has become very important for me to find flaws in my extension of tetration. The research community mostly consists of I. N. Galidakis, and myself working as individuals and researchers associated with the Tetration Forum including Kouznetsov http://math.eretrandre.org/tetrationforum/ . My point of difference with the Tetration Forum and Kouznetsov is that they are proposing solutions to the extension of tetration that are easily seen as being inconsistent with complex dynamics. In fact they claim that fixed points are irrelevant to extending tetration. According to complex dynamics -

Lennart Carleson, Theodore W. Gamelin

Complex Dynamics

Springer-Verlag

In reviewing Chapter 2 - the Classification of Fixed Points, one can see that because of the conjugacy of fixed points that there can be no “simple” single equation to extending tetration. In the classification of fixed points there are hyperbolic, super-attracting, parabolic rationally neutral, rationally neutral and irrationally neutral fixed points. Each type of fixed point leads to a different type of tetration. For example e ↑ (1/e) is the sole parabolic rationally neutral fixed point for tetration, so e ↑ (1/e) must have it’s own separate equation for extending tetration. Not only is my work consistent with the Classification of Fixed Points, it provides an algebraic explanation of the Classification of Fixed Points that is much simpler than the topologically based proofs in Complex Dynamics.

While I have been able to fractionally iterate any complex function and therefore tetration, there is a major problem everyone researching tetration faces, that of convergence. In my mind all the real action in tetration is focused on people showing that they have a convergent solution. A couple of years ago the issue was quite active on MathOverflow. See http://mathoverflow.net/questions/tagged/fractional-iteration .

I think it'd be better if the "correct" ordering of brackets shown above was considered just one case and if the theory introduced a new concept similar to commutativity, except for tetration, governing in which order the powers are applied. For example Googol = (10↑10)↑10 whereas 10↑(10↑10) is ten times that. In "super tetration" any tetration would require a set with modulus equal to the 2nd term of the tetration minus 1, governing the order in which powers are applied. I imagine that would be much more versatile. If you really wanted to make it interesting you could then ask what if the "ordering" set contained non-integers, you could see what happens if the two (or more) powers are applied at the same time instead of sequentially! Actually doing it... that's something for a better man than me!

Add a comment...