Re: [math-fun] Messages in pi
On 2018-12-05 16:44, Keith F. Lynch wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
Pi appears in the value of the Riemann zeta function at every even positive integer argument, and the Riemann zeta function is interesting because it involves primes. Also, pi appears in the natural log of -1. So pi would certainly be considered an important mathematical constant even in a world without geometry (e.g. a world that consists only of texts and their interactions).
I thought of another example: Pi appears in the probability that two randomly selected integers have a factor in common. Again, nothing to do with circles or geometry.
But the proof needs ζ(2) = π²/6. OtOH, Stirling's formula: In[105]:= Limit[n!/√(2 n)/(n/E)^n, n -> ∞] Out[105]= √π = (-½)!
Of course pi is also the sum of various infinite series, but that alone doesn't make it interesting. For probably at least half of the hundreds of thousands of sequences in OEIS, the sum of their reciprocals converges to some constant. If not, the sum of the reciprocals of their squares or factorials probably does. Most of these constants are of no interest to anyone. If a message is hidden in one of them nobody will ever notice.
(Homework problem: Contrive an infinite integer sequence such that the sum of its reciprocals will converge to a number that contains an interesting message when expressed in binary. Submit it to OEIS, and get it accepted.)
James Propp <jamespropp@gmail.com> wrote:
I like Keith?s idea of a God whose messages are the strings of digits that DON?T appear in pi. That way the messages would only be read by really smart beings and not by dorks with big computers.
It's true that no finite mindless computer search could ever determine that a string of digits doesn't appear in pi. But I don't see how a smart being could tell either, unless they found a proof that the string couldn't appear in pi. And if they did, the proof would seem to explain away the message.
On second thought, saying that a proof explains away a message makes no more sense than saying that a calculation that finds a message explains away the message. It's hard to reason about such things.
Mike Speciner <ms@alum.mit.edu> wrote:
If you remove all occurrences of some finite sequence of digits from a transcendental number it may or may not still be transcendental. But can you turn an algebraic number into a transcendental by removing all occurrences of some finite sequence of digits?
It would almost certainly be transcendental. But it would also almost certainly be impossible to prove it in any specific case. Similarly with most other ways of mangling numbers whose digits neither terminate nor repeat, e.g. reading the binary representation of the cube root of three as if it were ternary.
The state of math is quite scandalous. For most numbers, we have no clue how to determine whether they're rational, algebraic, or transcendental. And, last I heard, nobody was able to prove *any* specific number to be normal,
In[108]:= N[ChampernowneNumber[], 42] Out[108]= 0.123456789101112131415161718192021222324253 Mathematical constants treated as numeric by NumericQ and as constants by D. ChampernowneNumber[b] is a *normal* transcendental real number whose base-b representation is obtained by concatenating base-b digits of consecutive integers. ChampernowneNumber can be evaluated to arbitrary numerical precision. ChampernowneNumber automatically threads over lists. —rwg
though it's long been known that the vast majority of numbers are normal. And for most OEIS sequences, we have no idea whether the sum of the reciprocals of the terms converges, much less which sort of number it converges to, if any.
As an aside, it's well known that the harmonic series, the sum of the reciprocals of the positive integers, diverges. Less well known is that if you discard the integers which contain some specific integer string, e.g. your ten-digit phone number, the resulting sequence will converge. Which sort of number will it converge to?
One way in which an algebraic irrational number wouldn't turn into a transcendental by removing all occurrences of some finite sequence of digits is to choose an algebraic number which doesn't contain that sequence of digits. But I doubt there is any such algebraic irrational number. I suspect that all algebraic irrationals are normal, though I have no clue how to even start thinking about the possibility of considering an approach to finding a proof. (Of course a *rational* number without a specific string can always be found, and all rational numbers are algebraic.)
(But it's still weird to think that every algebraic number, expressed in binary, contains infinitely many copies of this very email.)
Allan Wechsler <acwacw@gmail.com> wrote:
In general, I'd expect the results of global digit-editing on an irrational number to be transcendental,
As would I.
with a very few special-case exceptions. For instance, express phi in binary and then retain only the even-indexed bits; change every 1 to a 6 in the base-5 expansion of Conway's "audioactive" constant; aso. asf.
For instance of what? Are those special-case exceptions, or examples of the general case, i.e. presumed transcendental?
Nitpick: If you remove enough short strings, you'll get a rational number. For instance pi in binary with all instances of the string "0" removed equals four.
On Thu, Dec 6, 2018 at 6:48 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2018-12-05 16:44, Keith F. Lynch wrote:
The state of math is quite scandalous. For most numbers, we have no clue how to determine whether they're rational, algebraic, or transcendental. And, last I heard, nobody was able to prove *any* specific number to be normal,
In[108]:= N[ChampernowneNumber[], 42]
Out[108]= 0.123456789101112131415161718192021222324253
Mathematical constants treated as numeric by NumericQ and as constants by D.
ChampernowneNumber[b] is a *normal* transcendental real number whose base-b representation is obtained by concatenating base-b digits of consecutive integers.
It's normal to base 10. I think the claim is that nobody knows a specific number to be normal to every base. That said, the claim has to be restricted to computable numbers, since an algorithmically random real like the halting probability of a prefix-free universal Turing machine has to be normal to every base; if not, you could predict infinitely many (not necessarily contiguous) digits of it, which contradicts the definition of algorithmically random. -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
On Thu, Dec 6, 2018 at 10:02 AM Mike Stay <metaweta@gmail.com> wrote:
It's normal to base 10. I think the claim is that nobody knows a specific number to be normal to every base. That said, the claim has to be restricted to computable numbers, since an algorithmically random real like the halting probability of a prefix-free universal Turing machine has to be normal to every base; if not, you could predict infinitely many (not necessarily contiguous) digits of it, which contradicts the definition of algorithmically random.
I don't follow this. If I tell you that X has no 3s in it's base-10 expansion, which means it is not normal, how does this let you predict infinitely many digits (or even one digit!) of X? Andy Latto
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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It's not about unpredictability, it's about the stronger notion of incompressibility. You can show that Chaitin's constant is algorithmically incompressible*, which implies normal. * proof: if not, you could create a program of N bits which knows how many programs (other than itself) of length <= N bits terminate, and then (by dovetailing) algorithmically determine *which* programs terminate. Then, it can decide to terminate if and only if it doesn't terminate. Best wishes, Adam P. Goucher
Sent: Thursday, December 06, 2018 at 8:33 PM From: "Andy Latto" <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Messages in pi
On Thu, Dec 6, 2018 at 10:02 AM Mike Stay <metaweta@gmail.com> wrote:
It's normal to base 10. I think the claim is that nobody knows a specific number to be normal to every base. That said, the claim has to be restricted to computable numbers, since an algorithmically random real like the halting probability of a prefix-free universal Turing machine has to be normal to every base; if not, you could predict infinitely many (not necessarily contiguous) digits of it, which contradicts the definition of algorithmically random.
I don't follow this. If I tell you that X has no 3s in it's base-10 expansion, which means it is not normal, how does this let you predict infinitely many digits (or even one digit!) of X?
Andy Latto
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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e is 2.718281828459... Maybe Gauss put 1828 into this decimal expansion (twice in succession for emphasis!) to celebrate the year in which his differential geometry masterpiece “Disquisitiones generales circa superficies curva” was published. Jim Propp On Thursday, December 6, 2018, Adam P. Goucher <apgoucher@gmx.com> wrote:
It's not about unpredictability, it's about the stronger notion of incompressibility. You can show that Chaitin's constant is algorithmically incompressible*, which implies normal.
* proof: if not, you could create a program of N bits which knows how many programs (other than itself) of length <= N bits terminate, and then (by dovetailing) algorithmically determine *which* programs terminate. Then, it can decide to terminate if and only if it doesn't terminate.
Best wishes,
Adam P. Goucher
Sent: Thursday, December 06, 2018 at 8:33 PM From: "Andy Latto" <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Messages in pi
On Thu, Dec 6, 2018 at 10:02 AM Mike Stay <metaweta@gmail.com> wrote:
It's normal to base 10. I think the claim is that nobody knows a specific number to be normal to every base. That said, the claim has to be restricted to computable numbers, since an algorithmically random real like the halting probability of a prefix-free universal Turing machine has to be normal to every base; if not, you could predict infinitely many (not necessarily contiguous) digits of it, which contradicts the definition of algorithmically random.
I don't follow this. If I tell you that X has no 3s in it's base-10 expansion, which means it is not normal, how does this let you predict infinitely many digits (or even one digit!) of X?
Andy Latto
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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-- Andy.Latto@pobox.com
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For π Day. 2015, WRI stirred up publicity with https://blog.stephenwolfram.com/2015/03/pi-or-pie-celebrating-pi-day-of-the-... Off-list I grumbled that no self-respecting Deity would bother sending clues to worshipers dumb enough to use decimal instead of continued fractions. However, in 2018 we're barely able to afford a full CF version of Wolfram's birthday games. Assuming https://en.wikipedia.org/wiki/Gauss%E2%80%93Kuzmin_distribution, define In[1486]:= cfprob[L_List] := Abs@Log[2, (1 + 1/FromContinuedFraction@L)/(1 + 1/FromContinuedFraction@MapAt[# + 1 &, L, -1])] In[1487]:= cfprob[r : (Integer_ | Rational_)] := cfprob@ContinuedFraction@r Then In[1477]:= cfprob /@ Range@3 Out[1477]= {Log[4/3]/Log[2], Log[9/8]/Log[2], Log[16/15]/Log[2]} In[1478]:= N@% Out[1478]= {0.415037499278844, 0.169925001442312, 0.0931094043914815} I.e, 41.5% of terms should be 1, 17% should be 2, etc. But cfprob also gives us the probabilities of term *sequences:* In[1505]:= cfprob /@ {{1, 2, 3}, 1 + 1/(2 + 1/3), {3, 2, 1}, {2, 1, 3}} Out[1505]= {Log[221/220]/Log[2], Log[221/220]/Log[2], Log[221/220]/Log[2], Log[210/209]/Log[2]} In[1506]:= N@% Out[1506]= {0.00654284586677209, 0.00654284586677209, 0.00654284586677209, 0.00688638558523976} (Invariant under reversal but not shuffling.) This says to expect about six 1,2,3's in every burst of 1000 terms: In[1499]:= SequencePosition[ContinuedFraction[\[Pi], 10^3], {1, 2, 3}] // tim During evaluation of In[1499]:= 0.000353,6 Out[1499]= {{47, 49}, {293, 295}, {512, 514}, {542, 544}, {841, 843}, {987, 989}} In[1500]:= SequencePosition[ContinuedFraction[\[Pi], 10^3], {3, 2, 1}] // tim During evaluation of In[1500]:= 0.000403,6 Out[1500]= {{190, 192}, {585, 587}, {621, 623}, {855, 857}, {903, 905}, {922, 924}} Try a million: In[1498]:= SequencePosition[ContinuedFraction[\[Pi],10^6],{1,2,3}]//tim During evaluation of In[1498]:= 0.805228, 6561 Out[1498]= {{47,49},{293,295},{512,514},{542,544},{841,843},{987,989},{1026,1028},{1199,1201},{1237,1239}, \[CenterEllipsis]6543\[CenterEllipsis] , {999141,999143},{999146,999148},{999179,999181},{999580,999582},{999592,999594},{999689,999691}, {999708,999710},{999749,999751},{999810,999812}} (.8 seconds for a million terms. Have I actually lived to see this?) In[1501]:= 3^8 Out[1501]= 6561 In[1503]:= SequencePosition[ContinuedFraction[\[Pi],10^6],{3,2,1}]//tim;0; During evaluation of In[1503]:= 0.378853, 6477 Continued fractions accommodate fancier date formats: In[1507]:= cfprob@{3, 14, 15} Out[1507]= Log[593569/593568]/Log[2] In[1508]:= N@% Out[1508]= 2.4305451534818*10^-6 In[1509]:= SequencePosition[ContinuedFraction[\[Pi], 10^6], {3, 14, 15}] // tim During evaluation of In[1509]:= 0.39352, 2 Out[1509]= {{415314, 415316}, {607114, 607116}} Sure enough, there were two of them. But for really fancy dates, In[1511]:= cfprob@{14, 3, 2015} // N Out[1511]= 1.79490870081438*10^-10 we'll need Eric Weisstein's record π CF calculation. —rwg
participants (5)
-
Adam P. Goucher -
Andy Latto -
Bill Gosper -
James Propp -
Mike Stay