[math-fun] radical denesting puzzle
rwg writes ... 4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5 This set off a "can't be right" buzzer for me. Since it's an algebraic identity, it should also be valid for the complex 5th roots of 7. Suppose we add up all five conjugates. As we run through the 5 fifth roots of 7, the LHS radicand will have real parts between 6 and 10, and we can expect the sum of all five sqrts to be roughly 5*sqrt8, about 14. But in summing the RHS, all the 7^(k/5) terms will cancel, since the sum of the five fifth roots of 7 is 0. This leaves only the 2/5 term, times 5, = 2. Thus, an apparent contradiction, 14 = 2. So I checked the equation numerically, and it's correct in double precision, about 16 decimal places. PRIME>(sqrt (- 8 (expt 7 1/5))) 2.5542566116984893 PRIME>(/ (+ (* 2 (expt 7 4/5)) (* -1 (expt 7 3/5)) (* -2 (expt 7 2/5)) (* 6 (expt 7 1/5)) 2) 5) 2.5542566116984893 My best guess as to what's going on is that the signs of the sqrts vary as different conjugates are selected. But this creates two new puzzles: (a) There are only a few combinations of sign assignments that will still cancel the imaginary parts when summing the LHS, and it's mysterious that the real parts mostly cancel too. (b) And, what's the rule or rationale for selecting the LHS sqrt signs? Normally one expects equations with different (relatively prime) degree roots to permit separate choices of the roots. A likely related curiosity: 8^5 - 7 = 32761 = 181^2. rwg> Two days ago I found several one-parameter continua (too cumbrous to display) of solutions to sqrt(a^(1/5)+-b^(1/5)) = <five terms>, e.g., They must be huge. Rich rcs@cs.arizona.edu
--- Richard Schroeppel <rcs@CS.Arizona.EDU> wrote:
rwg writes ...
4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5
This set off a "can't be right" buzzer for me. Since it's an algebraic identity, it should also be valid for the complex 5th roots of 7. Suppose we add up all five conjugates. As we run through the 5 fifth roots of 7, the LHS radicand will have real parts between 6 and 10, and we can expect the sum of all five sqrts to be roughly 5*sqrt8, about 14. But in summing the RHS, all the 7^(k/5) terms will cancel, since the sum of the five fifth roots of 7 is 0. This leaves only the 2/5 term, times 5, = 2. Thus, an apparent contradiction, 14 = 2. So I checked the equation numerically, and it's correct in double precision, about 16 decimal places.
PRIME>(sqrt (- 8 (expt 7 1/5)))
2.5542566116984893
PRIME>(/ (+ (* 2 (expt 7 4/5)) (* -1 (expt 7 3/5)) (* -2 (expt 7 2/5)) (* 6 (expt 7 1/5)) 2) 5)
2.5542566116984893
My best guess as to what's going on is that the signs of the sqrts vary as different conjugates are selected. But this creates two new puzzles: (a) There are only a few combinations of sign assignments that will still cancel the imaginary parts when summing the LHS, and it's mysterious that the real parts mostly cancel too. (b) And, what's the rule or rationale for selecting the LHS sqrt signs? Normally one expects equations with different (relatively prime) degree roots to permit separate choices of the roots.
A likely related curiosity: 8^5 - 7 = 32761 = 181^2.
rwg> Two days ago I found several one-parameter continua (too cumbrous to display) of solutions to sqrt(a^(1/5)+-b^(1/5)) = <five terms>, e.g.,
They must be huge.
Rich rcs@cs.arizona.edu
Let w = exp(2 pi i/5). Then the denesting identity is true for the fifth root r = 7^(1/5)w^k only if Re(sqrt(8-r)) is chosen negative when k = 2 or 3, and positive in the other three cases. Gene ______________________________________________________ Click here to donate to the Hurricane Katrina relief effort. http://store.yahoo.com/redcross-donate3/
On Thursday 08 September 2005 20:56, Eugene Salamin wrote:
Let w = exp(2 pi i/5). Then the denesting identity is true for the fifth root r = 7^(1/5)w^k only if Re(sqrt(8-r)) is chosen negative when k = 2 or 3, and positive in the other three cases.
And is there some obvious (to those sufficiently initiated) reason why? I'm guessing the Right Way to describe this is in terms of the Legendre symbol (k/5), but just how that works out is not at all clear to me. -- g
[Dropped eavesdroppers undropped to correctify egregium.] 4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5 rcs>This set off a "can't be right" buzzer for me. Since it's an algebraic identity, it should also be valid for the complex 5th roots of 7. Suppose we add up all five conjugates. As we run through the 5 fifth roots of 7, the LHS radicand will have real parts between 6 and 10, and we can expect the sum of all five sqrts to be roughly 5*sqrt8, about 14. But in summing the RHS, all the 7^(k/5) terms will cancel, since the sum of the five fifth roots of 7 is 0. This leaves only the 2/5 term, times 5, = 2. Thus, an apparent contradiction, 14 = 2. So I checked the equation numerically, and it's correct in double precision, about 16 decimal places. [...] My best guess as to what's going on is that the signs of the sqrts vary as different conjugates are selected. Bingo. Formally substituting for 7, 2 i k pi 8 i k pi 6 i k pi -------- -------- -------- 1/5 5 4/5 5 3/5 5 sqrt(8 - 7 e ) = (2 7 e - 7 e 4 i k pi 2 i k pi -------- -------- 2/5 5 1/5 5 - 2 7 e + 6 7 e + 2)/5 [k=0..4] (c42) expand(dfloat(%)) (d42) [2.55425661169849d0 = 2.55425661169849d0, 2.75838494663133 - 0.25441403376677 %i = 2.75838494663133 - 0.25441403376677 %i, 3.03551325248057 - 0.14288155378756 %i = 0.14288155378756 %i - 3.03551325248057, 0.14288155378756 %i + 3.03551325248057 = - 0.14288155378756 %i - 3.03551325248057, 0.25441403376677 %i + 2.75838494663133 = 0.25441403376677 %i + 2.75838494663133] (c43) apply("+",%) (d43) 14.1420530099223d0 = 2.0d0 Just like you said. rcs>But this creates two new puzzles: (a) There are only a few combinations of sign assignments that will still cancel the imaginary parts when summing the LHS, and it's mysterious that the real parts mostly cancel too. (b) And, what's the rule or rationale for selecting the LHS sqrt signs? Normally one expects equations with different (relatively prime) degree roots to permit separate choices of the roots. No clue. In this case, the empirical truth is 4 %i %pi 2 %i %pi -------- -------- 1/5 5 1/5 5 - sqrt(8 - 7 %e ) + sqrt(8 - 7 %e ) 2 %i %pi 4 %i %pi - -------- - -------- 1/5 5 1/5 5 + sqrt(8 - 7 %e ) - sqrt(8 - 7 %e ) 1/5 + sqrt(8 - 7 ) = 2 Just like Gene said. This might be a big clue for finding denestings, if we knew how to interpret it. rcs>A likely related curiosity: 8^5 - 7 = 32761 = 181^2. YOW! And from the "+" example, (c44) sqrt(41^5*29+33^5*2^2) (d44) 59299 rwg> Two days ago I found several one-parameter continua (too cumbrous to display) of solutions to sqrt(a^(1/5)+-b^(1/5)) = <five terms>, e.g., rcs>They must be huge. Huge FRAUDS! If b is the parameter, they contain both b and |b|. The good news: They remain true with |b| <- +or- b. The very bad news: When you do this, b comp[letely scales out of the expressions, leaving me with only a few dozen miscellaneous identities instead of a continuum. (But I really do have a continuum of four-termers.) --rwg PS: My pessimism about finding six-termers extends to (a^(1/n+-b^(1/n))^(1/k), k>2, not just sqrts (k=2). Of course k=-2, -1, and 2/3 constitute cheating. EAVESDROPS <-> OVERPASSED PPS: For "eavesdropper", my MW has Etymology: Middle English evesdropper, from evesdrop, n. + -er and for "eavesdrop" Etymology: probably back-formation from eavesdropper More probably than from evesdrop?
On Friday 09 September 2005 02:08, Bill Gosper wrote:
(c43) apply("+",%)
(d43) 14.1420530099223d0 = 2.0d0 Just like you said.
Yay for computer algebra!
rcs>A likely related curiosity: 8^5 - 7 = 32761 = 181^2.
YOW!
Closely related to a puzzle I saw on one of Noam Elkies's web pages: "use 2, 5, 7 and basic arithmetic to write 181". Unfortunately it's rather too easy with the hint above, though it does require one further idea. -- g
participants (4)
-
Eugene Salamin -
Gareth McCaughan -
R. William Gosper -
Richard Schroeppel