April 2012 - mathfun@mailman.xmission.com

[math-fun] pi^4 + pi^5 ~ e^6
by Robert Munafo 29 Apr '12

29 Apr '12

On 2012-04-27, David Wilson <davidwwilson(a)comcast.net> wrote: > Yes, and then there was > > pi^4 + pi^5 = e^6 > > which was found by me, however, I found it long ago in an email > signature on a USENET group. > > I was impressed, and later mentioned it to Eric Weisstein, and it ended > up in MathWorld (Almost Integer) crediting me. > > From there it went to Wikipedia (Mathematical Coincidence). > > Does it have a prior history in the literature? My entry on the number has a footnote giving this page: http://zhurnaly.com/cgi-bin/wiki/CoincidentalTaxonomy which (after the article but before the comment section) says "TopicScience - Datetag20011019", suggesting a date of Oct 19, 2001. -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

4 8

Re: [math-fun] Is e^e^e^79 an integer?
by Michael Reid 29 Apr '12

29 Apr '12

> Start with owning the numbers 0 and 1. > Each step, with two numbers you already own (perhaps identical), > compute A+B, A-B, A*B, A/B, or 2^A. > CONJECTURE: the greatest finite real number reachable in N>0 steps is > 2^2^2^...^2 with N-1 uses of "^" and evaluating right to left. OK, in the first step, I compute 1 + 1 = 2 . In the next step, compute 2^2 = 4 (which may as well be 2 + 2 or 2 * 2 ). But in the third step you want me to do 2^4 , whereas I'd rather do 4^4 , since I already "pwn" 4 . Did I misunderstand something?

2 1

[math-fun] The biggest reachable number problem...
by Warren Smith 29 Apr '12

29 Apr '12

> Michael Reid: >> WDS: Start with owning the numbers 0 and 1. >> Each step, with two numbers you already own (perhaps identical), >> compute A+B, A-B, A*B, A/B, or 2^A. >> CONJECTURE: the greatest finite real number reachable in N>0 steps is >> 2^2^2^...^2 with N-1 uses of "^" and evaluating right to left. > > OK, in the first step, I compute 1 + 1 = 2 . In the next step, > compute 2^2 = 4 (which may as well be 2 + 2 or 2 * 2 ). > > But in the third step you want me to do 2^4 , whereas I'd > rather do 4^4 , since I already "own" 4 . > Did I misunderstand something? --the operation that is allowed is 2^x, not x^y. >Henry Brady: > Brute force, do the first few cases exhaustive search, then "obvious" > must continue --actually, I did do some brute force on this to prove my conjecture for, I think it was, N<=6 or something. But the "obvious" part, I was never able to do. You might be able to generate two very close real numbers then 1/(A-B) is very large. I doubt it, but... >Schanuel's conjecture... I would think proving e^e^e^79 is noninteger and proving Schanuel's conjecture would be approximately equally difficult. If you could do one, it would probably be easy to do the other.

1 0

[math-fun] Conjecture on Skewes' number
by Allan Wechsler 29 Apr '12

29 Apr '12

Skewes' number, e^e^e^79, is a notoriously flamboyant upper bound in analytic number theory. As an upper bound, it has since been replaced by much more moderate results. A <a href=" http://garden.irmacs.sfu.ca/?q=op/is_skewes_number_e_e_e_79_an_integer">conjecture</a> on the Open Problem Garden is that Skewes' number is not integral. I am boggled. Of course we know that e^79 is nonintegral because e is not algebraic. (Is there an easier proof?) But do we even know that e^e^79 is nonintegral? Do we know that e^e^10 is nonintegral?

4 4

Re: [math-fun] Is e^e^e^79 an integer?
by Henry Baker 29 Apr '12

29 Apr '12

This problem sounds like it would succumb to a little brute force, followed by induction. The brute force solves all cases <k, for some relatively small k, by which time it is obvious that no other approach will do as well, and then simply show that your approach continues doing better & better as N gets bigger. The brute force part could be an exhaustive search by a computer. At 01:49 PM 4/28/2012, Warren Smith wrote: >A related conjecture I made a long time ago, which nobody >has ever resolved, is this. > >Start with owning the numbers 0 and 1. >Each step, with two numbers you already own (perhaps identical), >compute A+B, A-B, A*B, A/B, or 2^A. >CONJECTURE: the greatest finite real number reachable in N>0 steps is >2^2^2^...^2 with N-1 uses of "^" and evaluating right to left.

1 0

[math-fun] Is e^e^e^79 an integer?
by Warren Smith 28 Apr '12

28 Apr '12

A perhaps useful attack on a problem of this ilk is: Read http://en.wikipedia.org/wiki/E-function the function F(x)=exp(exp(x)-1) perhaps is an E-function, and indeed iterating the function G(X)=exp(x)-1 any fixed finite number of times, perhaps yields an E-function. If so, then by the Siegel–Shidlovsky theorem it would follow that if x is a nonzero integer then G(G(G(...x))) must be non-algebraic. To investigate whether this line of attack could work, I ran the following MAPLE9 script: foo := series( exp(exp(x)-1), x, 301 ); for k from 1 to 300 do print( k, evalf(log(coeff(foo,x,k)*k!),4) ); od; with the conclusion that k! times the series coefficient of x^k [coefficients indeed all are of the form integer/k!] seems to be growing too fast as a function of k to qualify for E-function status. ... Sorry. ---- A related conjecture I made a long time ago, which nobody has ever resolved, is this. Start with owning the numbers 0 and 1. Each step, with two numbers you already own (perhaps identical), compute A+B, A-B, A*B, A/B, or 2^A. CONJECTURE: the greatest finite real number reachable in N>0 steps is 2^2^2^...^2 with N-1 uses of "^" and evaluating right to left.

1 0

Re: [math-fun] xkcd
by Dan Asimov 28 Apr '12

28 Apr '12

What's enigmatic about this kind of very simple expression for a near-integer is whether it is a coincidence, or whether it begs for an explanation. To explain or not to explain: that is the question. For example, it's fairly well known that exp(pi*sqrt(163)) is an integer up to < 10^(-12), and this can be explained by the q-expansion of the j-invariant. This explanation also holds for the near-integers exp(pi*sqrt(67), up to < 10^(-5), exp(pi*sqrt(43), up to < 10^(-3), exp(pi*sqrt(19), up to < 1/4. Wikipedia also mentions this amazing discovery of D. Bailey & J. Borwein: Integral_{x=0 to oo} (cos(2x) * Prod_{n=1 to oo} cos(x/n)) dx = pi/8, up to < 10^(-42). Wikipedia claims this is just a coincidence, but I'd call that chutzpah. --Dan Simon wrote: << exp(Pi)-Pi = 19,9990999... . . . found in the 80's by me, Neil and a certain John H. C. >> ________________________________________________________________________________________ It goes without saying that .

2 1

Re: [math-fun] xkcd
by Dan Asimov 27 Apr '12

27 Apr '12

xkcd is wonderful (though once in a while I don't get it at all, when it involves too much insider computer knowledge). And speaking of approximations, an MIT alum told me around 1993 that pi seconds ~ 1 nanocentury. --Dan Rich wrote: << Today's XKCD cartoon has a nice collection of approximations. http://xkcd.com/1047/ There's the ancient expression 6 pi^5 for the mass ratio of the proton to the electron, and the more accurate (e^8 - 10)/phi. Robert Munafo & REIS are at least partially responsible. >> ________________________________________________________________________________________ It goes without saying that .

5 5

[math-fun] divisor question
by David Wilson 27 Apr '12

27 Apr '12

For 1 <= j <= k, does there always exist n with exactly j divisors <= k?

5 5

Re: [math-fun] xkcd
by Phil Carmody 27 Apr '12

27 Apr '12

> Today's XKCD cartoon has a nice collection of approximations. > > http://xkcd.com/1047/ Nice! My own favourite light year expression used to be Pi^(29+Pi) metres, 1 part in 132 too large, until I discovered (Pi^8)*10^9 kilometres, 1 Part in 337 too large. I don't know how those rate using RIES's metrics. Phil

2 1