[math-fun] A transcendental puzzle
Fred Lunnon <fred.lunnon@gmail.com> wrote:
However, I did slip up again: gamma has not actually been proven transcendental! I should have employed exp(1), ...
Indeed, exp(x) is transcendental for all non-zero algebraic x. It follows from that that if exp(x) is algebraic, x must be transcendental. Hence, for instance, the natural log of 2 must be transcendental. But since the transcendental numbers are of a higher cardinality than the algebraic numbers, it must be that in "most" cases x and exp(x) are both transcendental. Puzzle: Give an example where x and exp(x) are both known to be transcendental. I'll give a solution in a week if nobody posts one sooner.
Well, I do know the answer, as it happens --- so does APG, no doubt. Anybody determined to cheat (not that anybody would, of course) can always consult https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem WFL On 8/12/15, Keith F. Lynch <kfl@keithlynch.net> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
However, I did slip up again: gamma has not actually been proven transcendental! I should have employed exp(1), ...
Indeed, exp(x) is transcendental for all non-zero algebraic x. It follows from that that if exp(x) is algebraic, x must be transcendental. Hence, for instance, the natural log of 2 must be transcendental.
But since the transcendental numbers are of a higher cardinality than the algebraic numbers, it must be that in "most" cases x and exp(x) are both transcendental.
Puzzle: Give an example where x and exp(x) are both known to be transcendental. I'll give a solution in a week if nobody posts one sooner.
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Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers: -- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic. I believe the following extension puzzle has no such elementary solution: "Find x such that x, exp(x) and exp(exp(x)) are all transcendental." Sincerely, Adam P. Goucher
Sent: Wednesday, August 12, 2015 at 5:19 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] A transcendental puzzle
Well, I do know the answer, as it happens --- so does APG, no doubt.
Anybody determined to cheat (not that anybody would, of course) can always consult https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem
WFL
On 8/12/15, Keith F. Lynch <kfl@keithlynch.net> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
However, I did slip up again: gamma has not actually been proven transcendental! I should have employed exp(1), ...
Indeed, exp(x) is transcendental for all non-zero algebraic x. It follows from that that if exp(x) is algebraic, x must be transcendental. Hence, for instance, the natural log of 2 must be transcendental.
But since the transcendental numbers are of a higher cardinality than the algebraic numbers, it must be that in "most" cases x and exp(x) are both transcendental.
Puzzle: Give an example where x and exp(x) are both known to be transcendental. I'll give a solution in a week if nobody posts one sooner.
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Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental. Rich ------ Quoting "Adam P. Goucher" <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
Sincerely,
Adam P. Goucher
Sent: Wednesday, August 12, 2015 at 5:19 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] A transcendental puzzle
Well, I do know the answer, as it happens --- so does APG, no doubt.
Anybody determined to cheat (not that anybody would, of course) can always consult https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem
WFL
On 8/12/15, Keith F. Lynch <kfl@keithlynch.net> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
However, I did slip up again: gamma has not actually been proven transcendental! I should have employed exp(1), ...
Indeed, exp(x) is transcendental for all non-zero algebraic x. It follows from that that if exp(x) is algebraic, x must be transcendental. Hence, for instance, the natural log of 2 must be transcendental.
But since the transcendental numbers are of a higher cardinality than the algebraic numbers, it must be that in "most" cases x and exp(x) are both transcendental.
Puzzle: Give an example where x and exp(x) are both known to be transcendental. I'll give a solution in a week if nobody posts one sooner.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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So now I wonder if there's an x such that the countable sequence x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental. These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012): 0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654 Would a fixed point of exp(z) be transcendental? —Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
Rich's x = (Lambert) W(1) = ProductLog[1]. log(x)=-x, exp(x) = 1/x. Dan's first x = -W(-1) = (sane) Tree(1). exp(x)=x. --rwg On 2015-08-12 09:50, Dan Asimov wrote:
So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
On Wed, Aug 12, 2015 at 12:50 PM, Dan Asimov <dasimov@earthlink.net> wrote:
So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
Doesn't a simple counting argument prove this? There are countably many x that are algebraic There are countably many x such that e^x is algebraic There are countably many x such that e^(e^x) is algebraic ... and the union of countably many countable sets is countable. So all but countably many x are the x you're looking for. I'm working only in the reals here, so that the exponential function is injective. Andy
These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
Nice existence proof, Andy. I wonder if it's possible to prove that some explicit x works. Maybe x = e ? Why did God make it so hard to prove transcendence? —Dan
On Aug 12, 2015, at 12:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Wed, Aug 12, 2015 at 12:50 PM, Dan Asimov <dasimov@earthlink.net> wrote:
So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
Doesn't a simple counting argument prove this?
There are countably many x that are algebraic There are countably many x such that e^x is algebraic There are countably many x such that e^(e^x) is algebraic ...
and the union of countably many countable sets is countable.
So all but countably many x are the x you're looking for.
I'm working only in the reals here, so that the exponential function is injective.
Andy
These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
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Most proofs using this sort of counting argument are actually constructive, if you unwind them, including this one. You can explicitly enumerate the algebraics, so you can explicitly enumerate the numbers of the form log^n(x), where x is algebraic, and the exponent n indicates function iteration, not powers of log(x). Let {a_i} be such an enumeration, and then define T as the number whose decimal expansion has 3 in the nth place unless a_i has a 3 in the nth place, in which case it has 6. Then all exponentiates of T are transcendental. I agree that while this is technically constructive, it would be nice to have a more "natural" explicit example. "Everybody knows" that e, or pi, would be such an example, but that doesn't mean we can prove it. Andy On Wed, Aug 12, 2015 at 3:49 PM, Dan Asimov <asimov@msri.org> wrote:
Nice existence proof, Andy.
I wonder if it's possible to prove that some explicit x works. Maybe x = e ? Why did God make it so hard to prove transcendence?
—Dan
On Aug 12, 2015, at 12:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Wed, Aug 12, 2015 at 12:50 PM, Dan Asimov <dasimov@earthlink.net> wrote:
So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
Doesn't a simple counting argument prove this?
There are countably many x that are algebraic There are countably many x such that e^x is algebraic There are countably many x such that e^(e^x) is algebraic ...
and the union of countably many countable sets is countable.
So all but countably many x are the x you're looking for.
I'm working only in the reals here, so that the exponential function is injective.
Andy
These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Andy.Latto@pobox.com
This has me wondering if there is any open set U in C such that all iterates of exp are injective on U. If not, then is there an arc A in C - R with this property? If still not, a ha'penny will do. —Dan
On Aug 12, 2015, at 12:59 PM, Andy Latto <andy.latto@pobox.com> wrote:
I'm working only in the reals here, so that the exponential function is injective.
What am I missing? Don't you want intersection rather than union? On Wed, Aug 12, 2015 at 2:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Wed, Aug 12, 2015 at 12:50 PM, Dan Asimov <dasimov@earthlink.net> wrote:
So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
Doesn't a simple counting argument prove this?
There are countably many x that are algebraic There are countably many x such that e^x is algebraic There are countably many x such that e^(e^x) is algebraic ...
and the union of countably many countable sets is countable.
So all but countably many x are the x you're looking for.
I'm working only in the reals here, so that the exponential function is injective.
Andy
These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:
rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary
solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
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If I did want intersection, it wouldn't invalidate the proof; the intersection of countably many sets is countable, too. But I do want union. I want exp^n(x) to be transcendental (again, ^n means function iteration). So I want to exclude n if n is algebraic OR e^n is algebraic OR e^(e^n) is algebraic... So I want to exclude the union of these (countably many, countable) sets, Andy On Wed, Aug 12, 2015 at 4:35 PM, James Buddenhagen <jbuddenh@gmail.com> wrote:
What am I missing? Don't you want intersection rather than union?
On Wed, Aug 12, 2015 at 2:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Wed, Aug 12, 2015 at 12:50 PM, Dan Asimov <dasimov@earthlink.net> wrote:
So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
Doesn't a simple counting argument prove this?
There are countably many x that are algebraic There are countably many x such that e^x is algebraic There are countably many x such that e^(e^x) is algebraic ...
and the union of countably many countable sets is countable.
So all but countably many x are the x you're looking for.
I'm working only in the reals here, so that the exponential function is injective.
Andy
These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:
rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary
solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
OK, got it. Thanks Andy. On Wed, Aug 12, 2015 at 3:41 PM, Andy Latto <andy.latto@pobox.com> wrote:
If I did want intersection, it wouldn't invalidate the proof; the intersection of countably many sets is countable, too.
But I do want union. I want exp^n(x) to be transcendental (again, ^n means function iteration). So I want to exclude n if n is algebraic OR e^n is algebraic OR e^(e^n) is algebraic...
So I want to exclude the union of these (countably many, countable) sets,
Andy
On Wed, Aug 12, 2015 at 4:35 PM, James Buddenhagen <jbuddenh@gmail.com> wrote:
What am I missing? Don't you want intersection rather than union?
On Wed, Aug 12, 2015 at 2:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Wed, Aug 12, 2015 at 12:50 PM, Dan Asimov <dasimov@earthlink.net> wrote:
So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
Doesn't a simple counting argument prove this?
There are countably many x that are algebraic There are countably many x such that e^x is algebraic There are countably many x such that e^(e^x) is algebraic ...
and the union of countably many countable sets is countable.
So all but countably many x are the x you're looking for.
I'm working only in the reals here, so that the exponential function is injective.
Andy
These are the first few fixed points of f(z) = exp(z) in the
complexes:
(copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto: rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (9)
-
Adam P. Goucher -
Andy Latto -
Dan Asimov -
Dan Asimov -
Fred Lunnon -
James Buddenhagen -
Keith F. Lynch -
rcs＠xmission.com -
rwg