[math-fun] FWL more confusion re arctrig
FWL, you are still confused. Of course nonrigid cases are forbidden.
What exactly is meant by the term "non-rigid" here? The belt length is well-defined, and the analysis still applies. So why exactly does it fail to deliver the expected result? [ However, I did slip up again: gamma has not actually been proven transcendental! I should have employed exp(1), or zeta(3), or indeed a continuum of other possibilities. ] WFL On 8/7/15, Warren D Smith <warren.wds@gmail.com> wrote:
FWL, you are still confused. Of course nonrigid cases are forbidden.
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Ah, is not this vigorous repartée redolent of some long-ago halcyon school mathematics class, where the pedagogue had only to pose a question for the entire room to lapse into terrified silence, speculating feverishly over which unfortunate would on this occasion catch the basilisk eye and be summoned to blurt forth his inevitably dismal apology for a response --- apart, that is, from one unruly fellow in the corner engaged in engraving " Teacher don't kno' nuffink " on his desk with a penknife. Well then, we might try breaking the analysis down into smaller segments; for example, focus for one moment on the transparently innocuous final step: ** Given real number z and polynomial f with rational coefficients such that f(z) = 0 , does it necessarily follow that z is algebraic? ** (spoiler below). And while the class is pondering that conundrum, permit me to digress concerning an entertaining neologism, encountered recently in a related post, though strictly provisional pending further clarification from its source. Given a set S and proposition P , for which the author has previously established to his personal satisfaction that P(x) is true for all x in S , then y in S is (so I infer) defined to be "non-rigid" when P(y) inconveniently turns out to be false and a mole could perceive this fact with one eye shut. However, while an ingeniously improbable coinage, and more succinct than "self-evident counterexample", I somehow doubt that it will catch on. Such phenomena are of course not infrequently encountered in statistics, in which discipline they are known as "outliers" --- a word that for many years I fancied must be French, and mispronounced "oot-lee-ehs" accordingly. SPOILER: The deduction fails when f = 0 identically, so that f(z) = 0 for all z . There now --- that wasn't so difficult, was it? This is the situation for a Somsky pair of planets: the belt length is a constant integer, so its fractional part vanishes for any sunset (offset), and the proof that sunset is algebraic falls over. It is fairly easy to see that belt length is a continuous, weakly monotonic function of sunset; however, to exclude constant length might involve evaluating it at two distinct points, which for a general class of trains looks a tougher prospect. Incidentally, even implementing belt length as a continuous function --- rather than merely its fractional part --- proves an unexpectedly challenging algorithmic exercise. In fact, writing a train-searching program constitutes an excellent project for a student to whom one has taken a dislike! Fred Lunnon On 8/7/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What exactly is meant by the term "non-rigid" here?
The belt length is well-defined, and the analysis still applies.
So why exactly does it fail to deliver the expected result?
On 8/7/15, Warren D Smith <warren.wds@gmail.com> wrote:
FWL, you are still confused. Of course nonrigid cases are forbidden.
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<< It is fairly easy to see that belt length is a continuous, weakly monotonic function of sunset; however, to exclude constant length might involve evaluating it at two distinct points, which for a general class of trains looks a tougher prospect. >> Or one could always try that business invented a while back by some chap name of Newton, called "differentiation" ... Class dismissed. WFL On 8/10/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Ah, is not this vigorous repartée redolent of some long-ago halcyon school mathematics class, where the pedagogue had only to pose a question for the entire room to lapse into terrified silence, speculating feverishly over which unfortunate would on this occasion catch the basilisk eye and be summoned to blurt forth his inevitably dismal apology for a response --- apart, that is, from one unruly fellow in the corner engaged in engraving " Teacher don't kno' nuffink " on his desk with a penknife.
Well then, we might try breaking the analysis down into smaller segments; for example, focus for one moment on the transparently innocuous final step:
** Given real number z and polynomial f with rational coefficients such that f(z) = 0 , does it necessarily follow that z is algebraic? **
(spoiler below).
And while the class is pondering that conundrum, permit me to digress concerning an entertaining neologism, encountered recently in a related post, though strictly provisional pending further clarification from its source.
Given a set S and proposition P , for which the author has previously established to his personal satisfaction that P(x) is true for all x in S , then y in S is (so I infer) defined to be "non-rigid" when P(y) inconveniently turns out to be false and a mole could perceive this fact with one eye shut.
However, while an ingeniously improbable coinage, and more succinct than "self-evident counterexample", I somehow doubt that it will catch on. Such phenomena are of course not infrequently encountered in statistics, in which discipline they are known as "outliers" --- a word that for many years I fancied must be French, and mispronounced "oot-lee-ehs" accordingly.
SPOILER: The deduction fails when f = 0 identically, so that f(z) = 0 for all z . There now --- that wasn't so difficult, was it?
This is the situation for a Somsky pair of planets: the belt length is a constant integer, so its fractional part vanishes for any sunset (offset), and the proof that sunset is algebraic falls over.
It is fairly easy to see that belt length is a continuous, weakly monotonic function of sunset; however, to exclude constant length might involve evaluating it at two distinct points, which for a general class of trains looks a tougher prospect.
Incidentally, even implementing belt length as a continuous function --- rather than merely its fractional part --- proves an unexpectedly challenging algorithmic exercise. In fact, writing a train-searching program constitutes an excellent project for a student to whom one has taken a dislike!
Fred Lunnon
On 8/7/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What exactly is meant by the term "non-rigid" here?
The belt length is well-defined, and the analysis still applies.
So why exactly does it fail to deliver the expected result?
On 8/7/15, Warren D Smith <warren.wds@gmail.com> wrote:
FWL, you are still confused. Of course nonrigid cases are forbidden.
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* Fred Lunnon <fred.lunnon@gmail.com> [Aug 10. 2015 20:17]:
[...]
** Given real number z and polynomial f with rational coefficients such that f(z) = 0 , does it necessarily follow that z is algebraic? **
[...]
SPOILER: The deduction fails when f = 0 identically, so that f(z) = 0 for all z . There now --- that wasn't so difficult, was it?
Aaargh! Of course I guessed wrong. So, "non-rigid algebraic" it is.
[...]
participants (3)
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Fred Lunnon -
Joerg Arndt -
Warren D Smith