Re: [math-fun] record computation of Pi
Bill Gosper <billgosper@gmail.com> wrote:
Digits. Feh. What a waste of chips and neurons. The only excuse: Now that they've done it, converting to the continued fraction might be slightly easier than extracting the cf directly.
Why decimal digits? Why pi? Why not 3^3^3 balanced ternary digits of the cube root of 3? Some people like continued fractions. More people like decimal digits. Some people like e, phi, or Euler's constant. More people like pi. Pi was proven irrational in the 18th century, and transcendental in the 19th. The 20th belonged to digit hunters. I hope the 21st will find something new to do with the number, such as a proof (or disproof) of normality. (Tests of normality are a common excuse for digit hunters, as if they needed one.) If trends continue, a mole of decimal digits of pi will be known before the end of the century. Perhaps we could all agree to stop there. So we're a third of the way to the next term of this series: 1400 1706 1949 1958 1961 1973 1983 1987 1989 1997 1999 2002 2011... (The nth term is when pi was first known to 10^n digits.) Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary, something that previously appeared completely opaque in every base. At the very least, it's much easier to store, or even memorize, the quadrillionth binary digit of pi than to store, email, or memorize trillions of decimal digits. There is, however, a shortcut: Here are all the decimal digits of pi: 0123456789. (Some assembly (and repetition) required.) Instead of digit hunting a constant, I've been searching for an order-five additive-multiplicative magic square. This has the advantage of a definite end point. (Unless, of course, there isn't one.) So far, my search has resembled digit hunting. I've found more than a thousand semi-magic squares (rows and columns work, but diagonals do not). Once I have enough such squares, I plan to do statistical tests on near-misses to see where fully magic squares might be lurking, or at least where I can find lots more semi-magic squares. For instance is there any pattern to the common sums or the common products? What about their parity? For the products, there's a clear pattern there: All but one of the more than a thousand squares I've found has an even product. For the sums, odd is more likely than even by a statisticly significant amount, but not enough that you'd likely notice by eyeballing a list of the sums.
Normality to a given base is surely a desirable thing to know about a number. But my definition of a "random" number is as follows: Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables) such that for each k, the set of real numbers not satisfying P_k has measure 0. There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where S_k = {x in R | P_k(x) is true} Since each one of these has full measure, the same is true of their countable intersection S. Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S. Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further. —Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
That's such a beautiful definition! It depends on your language of propositions in an interesting way: if your propositions are first-order predicate logic over the language of rings, then 'random' = algebraic. Probably you want second-order arithmetic? Reals are, a la descriptive set theory, sets of naturals. -- APG.
Sent: Wednesday, December 14, 2016 at 8:23 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] record computation of Pi
Normality to a given base is surely a desirable thing to know about a number.
But my definition of a "random" number is as follows:
Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables)
such that
for each k, the set of real numbers not satisfying P_k has measure 0.
There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where
S_k = {x in R | P_k(x) is true}
Since each one of these has full measure, the same is true of their countable intersection S.
Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S.
Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further.
—Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
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Of course pi fails Asimov's strict "normality" test. One of the P_k's is a rigorized equivalent to "x is the ratio of the circumference of a circle to its diameter". Almost all numbers fail this test; but pi and only pi passes it. So pi fails to be "strictly normal". In fact, every number that can be rigorously described or defined fails. While the vast preponderance of real numbers is "strictly normal", we will never be able to say anything interesting about any one of them. They are ineffable and unknowable. On Wed, Dec 14, 2016 at 8:09 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
That's such a beautiful definition!
It depends on your language of propositions in an interesting way: if your propositions are first-order predicate logic over the language of rings, then 'random' = algebraic.
Probably you want second-order arithmetic? Reals are, a la descriptive set theory, sets of naturals.
-- APG.
Sent: Wednesday, December 14, 2016 at 8:23 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] record computation of Pi
Normality to a given base is surely a desirable thing to know about a number.
But my definition of a "random" number is as follows:
Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables)
such that
for each k, the set of real numbers not satisfying P_k has measure 0.
There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where
S_k = {x in R | P_k(x) is true}
Since each one of these has full measure, the same is true of their countable intersection S.
Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S.
Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further.
—Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
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https://mathwithbaddrawings.com/2016/11/09/pick-a-truly-random-number/ On Wed, Dec 14, 2016 at 9:50 AM, Allan Wechsler <acwacw@gmail.com> wrote:
Of course pi fails Asimov's strict "normality" test. One of the P_k's is a rigorized equivalent to "x is the ratio of the circumference of a circle to its diameter". Almost all numbers fail this test; but pi and only pi passes it. So pi fails to be "strictly normal". In fact, every number that can be rigorously described or defined fails. While the vast preponderance of real numbers is "strictly normal", we will never be able to say anything interesting about any one of them. They are ineffable and unknowable.
On Wed, Dec 14, 2016 at 8:09 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
That's such a beautiful definition!
It depends on your language of propositions in an interesting way: if your propositions are first-order predicate logic over the language of rings, then 'random' = algebraic.
Probably you want second-order arithmetic? Reals are, a la descriptive set theory, sets of naturals.
-- APG.
Sent: Wednesday, December 14, 2016 at 8:23 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] record computation of Pi
Normality to a given base is surely a desirable thing to know about a number.
But my definition of a "random" number is as follows:
Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables)
such that
for each k, the set of real numbers not satisfying P_k has measure 0.
There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where
S_k = {x in R | P_k(x) is true}
Since each one of these has full measure, the same is true of their countable intersection S.
Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S.
Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further.
—Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
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-- Forewarned is worth an octopus in the bush.
Wait — each P_k is a proposition satisfied by all reals except for a set of measure 0. I don't see what this has to do with pi. Also, "normal" has a strict definition. This is what I'd call a "truly random" number, not "normal" in some sense. —Dan
On Dec 14, 2016, at 6:50 AM, Allan Wechsler <acwacw@gmail.com> wrote:
Of course pi fails Asimov's strict "normality" test. One of the P_k's is a rigorized equivalent to "x is the ratio of the circumference of a circle to its diameter". Almost all numbers fail this test; but pi and only pi passes it. So pi fails to be "strictly normal". In fact, every number that can be rigorously described or defined fails. While the vast preponderance of real numbers is "strictly normal", we will never be able to say anything interesting about any one of them. They are ineffable and unknowable.
On Wed, Dec 14, 2016 at 8:09 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
That's such a beautiful definition!
It depends on your language of propositions in an interesting way: if your propositions are first-order predicate logic over the language of rings, then 'random' = algebraic.
Probably you want second-order arithmetic? Reals are, a la descriptive set theory, sets of naturals.
-- APG.
Sent: Wednesday, December 14, 2016 at 8:23 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] record computation of Pi
Normality to a given base is surely a desirable thing to know about a number.
But my definition of a "random" number is as follows:
Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables)
such that
for each k, the set of real numbers not satisfying P_k has measure 0.
There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where
S_k = {x in R | P_k(x) is true}
Since each one of these has full measure, the same is true of their countable intersection S.
Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S.
Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further.
—Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
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I got the definition of P_k inverted, but it doesn't matter. P_107 (or something) reads, "X is *not* the limit of 4(1 - 1/3 + 1/5 - 1/7 + ...)" (rigorized, with appropriate epsilon-delta language). This proposition is false for all numbers except pi, and the singleton set containing pi is definitely of measure 0. Therefore, pi is not in the intersection of the truth domains of all the P_k, and hence, pi is not truly random. *Nor is any number you can unambiguously name or describe*. On Wed, Dec 14, 2016 at 12:22 PM, Dan Asimov <asimov@msri.org> wrote:
Wait — each P_k is a proposition satisfied by all reals except for a set of measure 0. I don't see what this has to do with pi.
Also, "normal" has a strict definition. This is what I'd call a "truly random" number, not "normal" in some sense.
—Dan
On Dec 14, 2016, at 6:50 AM, Allan Wechsler <acwacw@gmail.com> wrote:
Of course pi fails Asimov's strict "normality" test. One of the P_k's is a rigorized equivalent to "x is the ratio of the circumference of a circle to its diameter". Almost all numbers fail this test; but pi and only pi passes it. So pi fails to be "strictly normal". In fact, every number that can be rigorously described or defined fails. While the vast preponderance of real numbers is "strictly normal", we will never be able to say anything interesting about any one of them. They are ineffable and unknowable.
On Wed, Dec 14, 2016 at 8:09 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
That's such a beautiful definition!
It depends on your language of propositions in an interesting way: if your propositions are first-order predicate logic over the language of rings, then 'random' = algebraic.
Probably you want second-order arithmetic? Reals are, a la descriptive set theory, sets of naturals.
-- APG.
Sent: Wednesday, December 14, 2016 at 8:23 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] record computation of Pi
Normality to a given base is surely a desirable thing to know about a number.
But my definition of a "random" number is as follows:
Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables)
such that
for each k, the set of real numbers not satisfying P_k has measure 0.
There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where
S_k = {x in R | P_k(x) is true}
Since each one of these has full measure, the same is true of their countable intersection S.
Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S.
Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further.
—Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
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Yes — that is certainly true. —Dan
On Dec 14, 2016, at 10:00 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I got the definition of P_k inverted, but it doesn't matter. P_107 (or something) reads, "X is *not* the limit of 4(1 - 1/3 + 1/5 - 1/7 + ...)" (rigorized, with appropriate epsilon-delta language). This proposition is false for all numbers except pi, and the singleton set containing pi is definitely of measure 0. Therefore, pi is not in the intersection of the truth domains of all the P_k, and hence, pi is not truly random. *Nor is any number you can unambiguously name or describe*.
On Wed, Dec 14, 2016 at 12:22 PM, Dan Asimov <asimov@msri.org> wrote:
Wait — each P_k is a proposition satisfied by all reals except for a set of measure 0. I don't see what this has to do with pi.
Also, "normal" has a strict definition. This is what I'd call a "truly random" number, not "normal" in some sense.
—Dan
On Dec 14, 2016, at 6:50 AM, Allan Wechsler <acwacw@gmail.com> wrote:
Of course pi fails Asimov's strict "normality" test. One of the P_k's is a rigorized equivalent to "x is the ratio of the circumference of a circle to its diameter". Almost all numbers fail this test; but pi and only pi passes it. So pi fails to be "strictly normal". In fact, every number that can be rigorously described or defined fails. While the vast preponderance of real numbers is "strictly normal", we will never be able to say anything interesting about any one of them. They are ineffable and unknowable.
On Wed, Dec 14, 2016 at 8:09 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
That's such a beautiful definition!
It depends on your language of propositions in an interesting way: if your propositions are first-order predicate logic over the language of rings, then 'random' = algebraic.
Probably you want second-order arithmetic? Reals are, a la descriptive set theory, sets of naturals.
-- APG.
Sent: Wednesday, December 14, 2016 at 8:23 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] record computation of Pi
Normality to a given base is surely a desirable thing to know about a number.
But my definition of a "random" number is as follows:
Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables)
such that
for each k, the set of real numbers not satisfying P_k has measure 0.
There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where
S_k = {x in R | P_k(x) is true}
Since each one of these has full measure, the same is true of their countable intersection S.
Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S.
Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further.
—Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
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participants (5)
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Adam P. Goucher -
Allan Wechsler -
Dan Asimov -
Keith F. Lynch -
Michael Kleber