Is there a "largest function"?

Question

In one of my classes, the professor asked about what we think the largest function was. Many thought perhaps ${e^x}^{e^x}$, but I thought about $n!$

When I talk about a "largest function", I mean the function that increases the quickest.

The professor asked about a function larger than $n!$ to which I responded, $2n!$

Although snarky in nature, it is technically true.

So my question is this:

What is the "largest function" if we define "largest" as being "increases the quickest". A parent function is what we need, as it prevents someone like myself from putting a larger coefficient before the function.

You have to precise what a "function" should be. The busy-beaver function for example grows asymptotically faster than any computable function. — Peter, Jun 29 '18 at 17:27
If you take the functions $2^x,2^{(2^x)},2^{(2^{(2^x)})},\cdots$ you get a sequence of functions such that every function grows much faster than the previous one, not only somewhat faster. — Peter, Jun 29 '18 at 17:29
See Scott Aaronson's excellent essay, "Who Can Name the Bigger Number?" — Jair Taylor, Jun 29 '18 at 17:37
You should rule out the "obvious responses" with a more careful definition of a growth rate. Study https://en.wikipedia.org/wiki/Fast-growing_hierarchy to understand how this is meant. — Peter, Jun 29 '18 at 17:39
A very good survey of large numbers and super-fast growing functions is here : https://sites.google.com/site/largenumbers/home — Peter, Jun 29 '18 at 17:41
@Peter You have really given me some good stuff to read here. Seemingly, I don't see anyone posting a larger parent function than $n!$. Of course raising something to itself infinitely is going to be larger than $n!$ but I am asking mostly about the "largest" (as defined above) generalized form of a family of functions. You have been a great help to me with this article. — Prime, Jun 29 '18 at 17:43
@SincerelyPrime Asymptotically $n!$ only grows like $10^n$ which is not very fast. Don't misinterprete this. Of course, $n!$ eventually is much larger than $10^n$, but for very large $n$, it makes no significant difference. — Peter, Jun 29 '18 at 17:44
I'm a little surprised (well a lot surprised) a professor teaching this level of class would ask a naive question like this and expect his students not to be too sophisticated for it. At least you have the decency to recognize you need "parent" functions or classes of functions to have this answerable without the utterly obvious $2f'(x)>f'(x)$. But you need to have the concept of "parent" function much more precisely defined to have this question worth discussion. (The busy-beaver function is probably the most relevant comment on this thread.) — fleablood, Jun 29 '18 at 17:46
@fleablood I agree : Just saying "squaring a function gives a function growing faster" or similar statements misses the point. — Peter, Jun 29 '18 at 17:49
Hmm. The professor was asking this is passing in the middle of a lecture to help motivate one of his points. Perhaps I should repost with much more rigorous definitions and include a link to this question? @fleablood — Prime, Jun 29 '18 at 17:50
@SincerelyPrime Here : https://sites.google.com/site/largenumbers/home/3-2/andre_joyce you can read about the "battle" between Andre Joyce and Jonathan Bowers who can create the larger numbers. Very funny to read! — Peter, Jun 29 '18 at 17:56
I'm no mathematician but check out the busy Beaver function https://en.wikipedia.org/wiki/Busy_beaver — sudo rm -rf slash, Jun 29 '18 at 20:18
Why do you think that none of the suggested functions are "larger" (by which you mean "faster-growing") than $n!$? Even your classmates' suggestion of $e^x^{e^x}$ grows faster. And the busy beaver function clearly grows faster. — Kyle Strand, Jun 29 '18 at 22:06
@KyleStrand I am simply wondering if one exists. The prof says the one my classmates suggested is not a parent function as it is a function raised to itself. The scope of my question and its intent seems to have been missed by many, per fleabod’s and Peter’s suggestion. — Prime, Jun 29 '18 at 23:46
@SincerelyPrime To me, that seems to confirm that the idea of "parent function" is not well defined. Recursive exponentiation is, I think, a very distinct concept from simple exponentiation; I think what's what Peter's comment was getting at. — Kyle Strand, Jun 30 '18 at 00:12
@SincerelyPrime could you ask your professor to clarify parent functions and report back to us? — Prime, Jun 30 '18 at 03:29
I would think a "parent function" of the enumerations of the infinities (f(0)=ℵ₀, f(1)=ℵ₁, ...) would grow pretty darn fast! — fluffy, Jun 30 '18 at 04:31
@JairTaylor - The biggest number do exist. It's 40. Explanation — Egor Skriptunoff, Jun 30 '18 at 09:35
Since there is no number larger than infinity right? Howabout f(x) = infinity you cannot get bigger than that? — Stephen Quan, Jul 02 '18 at 04:30
"parent function" needs to be nailed down. Perhaps you mean something like big O? "$g$ is $O(f)$" means "there exists a positive constant $C$ such that, for all sufficiently large values of $x$, $g(x) < C f(x)$. So the notion of "parent function" you're looking for might be big-O equivalence class; that would prevent the simple "your function times 2" trick, as you wanted (but it would still allow other only slightly more complicated tricks). — Don Hatch, Jul 02 '18 at 20:03
@EgorSkriptunoff I thought you were going to say the largest number is "about 45 billion"! — Kyle Strand, Jul 09 '18 at 20:13

Kevin Meier · Answer 1 · 2018-06-30T14:49:15.033

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There are many large functions, e.g. $e^n$, $n!$ etc.

And you might know that $e^n$ grows faster than $n^k$ for any $k\geq 1$.

But there are other interesting functions, e.g. the Busy Beaver function. It asymtotically grows faster than any computable function. That means you cannot even write a computer program that produces a faster growing function.

The nice thing is: The busy beaver function is well-defined, but not computable:)! This function really gives an upper bound for the growth of computable functions (e.g., it grows much faster than any function that just contains hyperoperators or the TREE function).

edit: Of course, there are more and even faster growing functions.

edited Jun 30 '18 at 14:49

answered Jun 29 '18 at 17:43

Kevin Meier

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I remember proving this in my proofs class! Good times, for sure. Your response adheres moreso to my original line of thinking, I think. – Prime Jun 29 '18 at 17:45
Your answer got me thinking that there must even exist functions which grow faster than all well-defined functions. However when trying to prove it I started wondering whether "well-defined" is well-defined. – kasperd Jun 30 '18 at 22:31
@kasperd: You can't complete such a proof, because you might be living in a pointwise definable model of set theory where ervery function is well-defined. – hmakholm left over Monica Jul 01 '18 at 00:49

score 32 · Accepted Answer · answered Jun 29 '18 at 17:38

32

Look into hyperoperators.

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

This is a sequence of binary operators, each generating larger numbers than the previous. Define $f_n(x) = n \uparrow^n x$. You now have an infinite sequence of functions, each one in the sequence grows faster than the previous one. And they will grow MUCH faster than $n!$.

answered Jun 29 '18 at 17:38

SlipEternal

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To give you an idea how how big this gets: $$\begin{align}f_1(2) & = 1 \ f_2(2) & = 4 \ f_3(2) & = 3^{27} \ f_4(2) & = \text{Wolframalpha cannot calculate how many digits this value has}\end{align}$$ – SlipEternal Jun 29 '18 at 19:35
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Numberphile did a video that featured hyperoperators a few years ago. – Denis de Bernardy Jun 30 '18 at 12:44
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While this is true, this only demonstrates the existence of an infinite sequence of "larger" functions, rather than proving the nonexistence of a "largest" function. (You might think it's obvious how to extend this to a proof that there is no "largest" function, but proving the same thing without bringing in hyperoperators is about as obvious.) – user2357112 Jun 30 '18 at 19:27
@user2357112 other users had already discussed theoretical functions that are "larger" than anything computable and that there is no "largest" function. But, the OP was specifically asking for an example of what such a function might look like. Enter hyperoperators. They can be used to produce functions that grow much faster than $n!$, which is specifically what the OP sought. – SlipEternal Jun 30 '18 at 22:55
This doesn't answer the question at all. – Don Hatch Jul 02 '18 at 06:47
@DonHatch clearly, you and the OP have a different understanding of what "the question" is. Perhaps if you would like to provide your own answer, you could elaborate on how you think "the question" should be answered. – SlipEternal Jul 02 '18 at 12:36
@InterstellarProbe, the question I see is "is there a largest function". Your example of an infinite sequence of functions, each increasing faster than the previous, does not answer that question, any more than an example of an increasing sequence of numbers answers the question "is there a largest number". Yes, I see the green checkmark; congratulations. Apparently either the OP didn't notice that you didn't answer the question, or the two of you actually have a different question in mind, in which case it would help if you would state the question you're answering instead. – Don Hatch Jul 02 '18 at 18:01
@DonHatch Others have already answered that there is no "largest function". But, part of the OP's question was specifically about finding a function that grows faster than $n!$. I indicated that I answered that part of the question when I stated quite clearly that the sequence of functions I offered increases much faster than $n!$. Only one answer may be marked as an answer. I do not need to replicate answers that have already been given. That is not needed on this site. I made it obvious what part of the question I was answering, and I answered it. – SlipEternal Jul 02 '18 at 19:12

score 31 · Answer 3 · edited Jul 22 '18 at 05:36

31

There is none. Given any function from $\mathbb R$ to $\mathbb R$, it is not hard to construct another function that grows faster than it.

edited Jul 22 '18 at 05:36

Parcly Taxel

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answered Jun 29 '18 at 17:30

Robert Israel

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My question is moreso about parent functions. For example, is there a parent function that grows faster than $n!$? – Prime Jun 29 '18 at 17:35
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Indeed, given a countably infinite set of functions from $\mathbb R$ to $\mathbb R$ one can define another function that grows faster than any function in the set, where "grows faster" can be interpreted as eventually greater than, or as differences approach infinity, or as ratio to any in the set approaches infinity, or as ratio of logarithm to the logarithm of any in the set approaches infinity --- pretty much any manner of pointwise comparison of your choosing. – Dave L. Renfro Jun 29 '18 at 17:37
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You need to define what a "parent function" is better. – fleablood Jun 29 '18 at 17:47
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@SincerelyPrime In regard to Dave's comment there a number of cardinal invariants of the continuum which deal with "the largest function" in some way. $\mathfrak {d}$ and $\mathfrak{b}$ come to mind. See https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum for more information. – DRF Jun 29 '18 at 18:39
@Prime, I think fleablood's comment was for you. – grovkin Jul 01 '18 at 23:24
Haha I'm well aware @grovkin . I am seeking a definition from analysis...and will probably post an edit or something like that here as soon as I can...defining a parent function is proving to be more difficult than I thought, but I am looking into it. – Prime Jul 01 '18 at 23:25

score 11 · Answer 4 · edited Jun 30 '18 at 15:29

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There is no largest function, any more than there is a largest number. Take the largest function you can think of, call its slope $f'(x)$. Now double that function - you have doubled your slope to $2f'(x)$.

edited Jun 30 '18 at 15:29

Kenta S

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answered Jun 29 '18 at 17:32

Nuclear Hoagie

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The new slope is actually $2f(x)f'(x)$: it doesn't just double. – Randall Jun 29 '18 at 17:37
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Yeah, if you want to double the slope, you need to take $f(2x)$. – probably_someone Jun 29 '18 at 21:35
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@probably_someone Even with $g(x) = f(2x)$ we have $g'(x) = 2 f'(2x) \not = 2 f'(x)$ = doubling the slope. – Ovi Jun 30 '18 at 09:30
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To double the slope, all you need to do is to double the function. – Eric Duminil Jun 30 '18 at 13:30

wlad · Answer 5 · 2018-07-02T07:58:35.657

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Given a sequence of functions $f_i:\mathbb N \to \mathbb N$, let $f_\infty:\mathbb N \to \mathbb N$ be defined by $$f_\infty(n) = 1+\max\{f_1(n),f_2(n),\dotsc,f_n(n)\}$$ $f_\infty$ is eventually greater than any function in the sequence $f_i$.

Alternatively we can define $f_\infty$ by $$f_\infty(n)=1+\sum_{i=1}^nf_i(n)$$ This type of argument is called a diagonal argument.

edited Jul 02 '18 at 07:58

answered Jun 30 '18 at 22:28

wlad

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score 4 · Answer 6 · answered Jul 01 '18 at 14:54

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Take a look at Hardy's translation and edition of Du Bois-Reymond's Orders of Infinity, page 10:

answered Jul 01 '18 at 14:54

Red Banana

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for all functions $\phi_1, \phi_2$ such that $\phi_1 \prec \phi_2$, which are there more of? – Mitch Jul 01 '18 at 23:42

score 1 · Answer 7 · answered Jul 07 '18 at 01:47

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I remember that this one gets mighty big mighty fast: https://en.wikipedia.org/wiki/Ackermann_function

However whatever function you claim is the biggest, I can beat it by squaring it.

It is like trying to name the smallest positive number. Whatever number you claim is smallest, I can cut in half.

answered Jul 07 '18 at 01:47

richard1941

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Well, referring to a smallest positive number, what you said is only true for $\mathbb{R}$ ... this is obvious. And squaring a function is simply the square of that function...I am speaking about largest parent functions which I will define a little bit later with a largely requested edit. – Prime Jul 07 '18 at 02:27

score 0 · Answer 8 · answered Aug 28 '19 at 23:59

There is no 'largest' function, as one can always add, multiply, or exponentiate again to produce a larger one, there are some that get very large, very quickly. For example, the Ackermann function $A(x, y)$, which is defined as follows (for $x, y \geq 0$): $$ A(x,y) = \left\{ \begin{array}{ll1} x + 1 & \text{if }y=0 \\ A(x-1,1)&\text{if }x>0, y=0\\ A(x-1,A(x,y-1))&\text{if }x,y>0 \end{array} \right. $$ Powered by a recursive definition, this function grows astonishingly fast. For example, while $A(1, 1) = 3$ and $A(2, 2) = 7$, $A(3, 3) = 61$, $A(4, 4)$ is: $$2^{2^{2^{65536}}} - 3$$ and will overload most computers (to give an idea of scale, a computer with a 128 bit processor can handle numbers with about $40$ digits, and this is many, many times larger than that). And while this is an astoundingly large function, there are infinitely larger ones.

score -2 · Answer 9 · edited May 13 '20 at 12:56

I'm just saying, the function that I think is the LARGEST ar aka the one that increases the value of the number is, what I invented, is Pxn("number").That kinda makes in infifnite because how it works for example if I took $1$, it would give me $\pi$, but if I take $3$, that means take every single number of $\pi$ and multiply it by $3$ so $``3.14"$ would be $``6.312"$. So there you go, bye (I wrote this and I'm 9yr old)

Is there a "largest function"?

9 Answers9