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June 24[edit]

A big number[edit]

Hi, I have defined a big number N like this:

Define

f_{1}(x)=x^{x}\,

Define

f_{i}(x)=f_{i-1}^{x}(x),\ i>1\,

Define

N=f_{f_{10}(10)}(10)\,

The notation $f^{n}(x)\,$ means the n-fold iteration of $f\,$ . For example, $f^{3}(x)=f(f(f(x)))\,$ .

My number N seems beyond incomprehensible, but how much bigger could we get with a definition of comparable length? Obviously we could tinker a bit, like changing "10" to "99", or writing $f_{1}(x)=x^{x^{x}}\,$ , or whatever, but can we make a quantum leap to a whole new domain of incomparable vastness? 86.176.208.236 (talk) 02:46, 24 June 2011 (UTC)[reply]

Well, this kind of question is a bit boring to those of us who have been around for awhile, I'm afraid. You could make it more interesting by asking, for a given formal language such as the language of Principia Mathematica, what is the largest number that can be specified by an expression of length N? Looie496 (talk) 02:56, 24 June 2011 (UTC)[reply]

Very sorry to have bored you. You may find learning the difference between "a while" and "awhile" more interesting. 86.176.211.230 (talk) 13:25, 24 June 2011 (UTC)[reply]

Well, I wouldn't call it a quantum leap. It's quite well know. There are different kinds of infinity. For example, there are infinitly many integers and there are infinitely many real numbers. But some how the real numbers are more infinite than the integers; for example we think of integers as dots on a line while the real numbers are the whole line itself. A mathematician called Georg Cantor did a lot of work on this in the late 1800's and early 1900's. You should look at our articles on cardinals and ordinals. These guys are really crazy, but they're really cool too. I can't remember which one, I think it's an ordinal, but it's a fixed point of exponentiation; so if you take it to it's own power then you get it back again, i.e. x^x = x (and it's not one!) Take a look and see what you make of it. Feel free to pop back and ask some questions if you get stuck. — Fly by Night (talk) 03:29, 24 June 2011 (UTC)[reply]

What you've defined is basically the Ackermann function.--Antendren (talk) 04:08, 24 June 2011 (UTC)[reply]

"Incomprehensible" is a subjective and time-dependent concept. There is no such thing as a big number - every number is small because most numbers are bigger. Bo Jacoby (talk) 12:37, 24 June 2011 (UTC).[reply]

Take a look at Robert Munafo's large number pages for examples of definitions of large numbers. —Bkell (talk) 13:35, 24 June 2011 (UTC)[reply]

I like this number. Count Iblis (talk) 15:06, 24 June 2011 (UTC)[reply]

You may also be interested in the busy beaver function, which grows "faster asymptotically than does any computable function." SemanticMantis (talk) 13:42, 28 June 2011 (UTC)[reply]

Short form of the axiom of dependent choice[edit]

Resolved

I recently got the idea to convert the axiom of dependent choice to primitive symbols. I used Metamath to help me do this. I used the formulation of DC given on Wikipedia's page. When I was done, my formula had 559 quantifiers. Now, note this page. There is a formulation of full AC with only 5 quantifiers! Even if I converted my formula to prenex normal form, and was able to squash all quantifiers quantifying identical variables together (which the prenex normal form article makes it clear it is not), I still would have 18 quantifiers, and a formula approximately 100 times longer than the one given at the external link. Is there a known "short form" of dependent choice, like the one given at the external link for the full axiom of choice? If not, do you have any tips for creating a shorter formula (because I really can't believe mine is the shortest possible; I have the rest of ZF I can create equivalences over, and most of the equivalences I used in creating my formula in primitives were purely logical)? Thank you! JamesMazur22 (talk) 21:04, 24 June 2011 (UTC)[reply]

Just an unrelated question: since you seem to know a lot of converting everyday math into primitives, do you know if there's a primitive formula for the generalized associative law? The one that says the order of operation (e.g. in a group) doesn't matter if it's associative. Money is tight (talk) 03:06, 25 June 2011 (UTC)[reply]

\forall x\forall y\forall z(x+y)+z=x+(y+z)

,

\forall x\forall y\forall z(x*y)*z=x*(y*z)

. Are those what you're looking for? Or do you mean in ZFC primitives? JamesMazur22 (talk) 13:18, 25 June 2011 (UTC)[reply]

I think Money is tight is asking about the generalization of that law to arbitrary-length expressions (dunno whether he's asking for ZFC primitives). So it seems you would first need to have an encoding of arbitrary-length expressions, and then state that any such expression with the same sequence of terms has a value independent of the parentheses in the expression. --COVIZAPIBETEFOKY (talk) 15:08, 25 June 2011 (UTC)[reply]

Yes COVIZAPIBETEFOKY is right I'm asking for the generalized associative law in ZFC primitives, and the functions symbols + or * are obviously not allowed. Can it even be expressed? I know that in the language of group theory the generalized associative law is really a metatheorem, I wonder if it can be converted into one formula in ZFC. Money is tight (talk) 02:35, 26 June 2011 (UTC)[reply]

That is a very long and difficult problem, albeit it is a possible one (create a sequence of sets, each set containing one set for each possible arrangement of parentheses for that many terms being added or multiplied, etc.). I could help you with it, but it might take me a while. In the meantime, is there anyone who can help me with my question? It is likely already published somewhere in the literature! JamesMazur22 (talk) 21:05, 26 June 2011 (UTC)[reply]

I didn't find a short form, but it turns out I really don't need it. And if I do, I will derive it myself. As for Money is tight's question, I think that answer is best continued on their talk page. I am capable of answering it myself, given some time. JamesMazur22 (talk) 15:21, 2 July 2011 (UTC)[reply]

Sum[edit]

Is there a closed form expression for $\sum _{k=1}^{n}{\frac {a^{k}}{k}}$ or for $\sum _{k=1}^{\infty }{\frac {a^{k}}{k}}$ ? Widener (talk) 22:48, 24 June 2011 (UTC)[reply]

From

\log(1+x)=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {x^{k}}{k}}

It follows that:

-\log(1-x)=\sum _{k=1}^{\infty }{\frac {x^{k}}{k}}

Count Iblis (talk) 23:04, 24 June 2011 (UTC)[reply]

First we have:

{\begin{aligned}{\frac {d}{da}}\sum _{k=1}^{n}{\frac {a^{k}}{k}}&=\sum _{k=1}^{n}{\frac {d}{da}}\,{\frac {a^{k}}{k}}=\sum _{k=1}^{n}a^{k-1}={\begin{cases}{\dfrac {1}{1-a}}&{\text{if }}n=\infty ,\\[15pt]{\dfrac {1-a^{n}}{1-a}}&{\text{if }}n<\infty .\end{cases}}\end{aligned}}

Then look for antiderivatives. It's easier in the infinite case; you get a logarithm. In the infinite case it converges of −1 < a < 1, and uniform convergence of power series is needed to justify interchanging the d/da with the sum. Michael Hardy (talk) 00:29, 25 June 2011 (UTC)[reply]

I think it converges when a = −1 as well. Widener (talk) 04:06, 25 June 2011 (UTC)[reply]

Mathematica gives a closed form in terms of the Lerch transcendent,

\sum _{k=1}^{n}{\frac {a^{k}}{k}}=-a^{n+1}\Phi (a,1,n+1)-\log(1-a)

Also in terms of a hypergeometric function:

\sum _{k=1}^{n}{\frac {a^{k}}{k}}=-{\frac {a^{n+1}{}_{2}F_{1}(1,n+1;n+2;a)}{n+1}}-\log(1-a)

-- Meni Rosenfeld (talk) 14:31, 26 June 2011 (UTC)[reply]