[math-fun] Algebraic number question
By happenstance, I noticed that v8 + v5 = 5.064495+ is just over 5, so I computed 1 / (v8 + v5 - 5) = 1/4 * (5v10 + 7v5 + 11v2 + 15) = 15.505053+ I noticed that the terms 5v10 = v250 = 15.811388+ 7v5 = v245 = 15.652475+ 11v2 = v242 = 15.556349+ 15 = v225 = 15.000000 are all in the same ball park. Is there some reason this should be the case? Can other similar examples be formulated?
[Wilson is using 'v' to mean square-root. I typically use q or r. --rcs] This sort of thing happens in some (all?) extension fields based on pure radicals. In your example, imagine running through the sign changes on +-v8 and +-v5. If you flip v5, your equation changes into 1 / (v8 - v5 - 5) = 1/4 * (-5v10 - 7v5 + 11v2 + 15) = small The small term is small because the denominator v8-v5-5 is no longer close to 0. Similar things happen when you flip v8 to -v8, and when you flip both of them. Now add up signed combinations of these results: A straight Add cancels all terms except the +15, which isn't affected by the sign flips: The sum is 1/4 * (15+15+15+15) = 15 = 15.505053+ + small + small + small. True! Next, instead of Adding, use Subtract with the equations where v5 is flipped: Various things cancel out, except the v5 flipped terms: 1/4 * (7v5 + 7v5 + 7v5 + 7v5) = 7v5 = 15.505053 - small + small - small so 7v5 = roughly 15.5. The other terms 11v2 and 5v10 behave similarly, with other combinations of sign flips. I think this generalizes to fields with more square roots. For cube roots, such as cbrt2, instead of sign flips +1 -> -1, substitute +1 -> w and w2, with w & w2 being the two complex cube roots of 1. Rich --------- Quoting David Wilson <davidwwilson@comcast.net>:
By happenstance, I noticed that v8 + v5 = 5.064495+ is just over 5, so I computed
1 / (v8 + v5 - 5) = 1/4 * (5v10 + 7v5 + 11v2 + 15) = 15.505053+
I noticed that the terms
5v10 = v250 = 15.811388+ 7v5 = v245 = 15.652475+ 11v2 = v242 = 15.556349+ 15 = v225 = 15.000000
are all in the same ball park.
Is there some reason this should be the case? Can other similar examples be formulated?
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Actually, I was using the Unicode square root symbol, which got ASCIIcked to a v.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rcs@xmission.com Sent: Thursday, July 06, 2017 7:10 PM To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: Re: [math-fun] Algebraic number question
[Wilson is using 'v' to mean square-root. I typically use q or r. --rcs]
This sort of thing happens in some (all?) extension fields based on pure radicals. In your example, imagine running through the sign changes on +-v8 and +-v5. If you flip v5, your equation changes into
1 / (v8 - v5 - 5) = 1/4 * (-5v10 - 7v5 + 11v2 + 15) = small
I guess I see. My intuition was that if you changed signs in terms on the left, you would end up with entirely different terms on the right, but as you tacitly imply below, term sign changes on the left result only in term sign changes on the right. In this case, there are 4 linearly independent sign changes on the left, and 4 (I assume linearly independent) term changes on the right. One of the sign changes on the left results in a small right-hand value, the others in large right hand values. Solving the right side for term magnitudes leads to terms of about the same size. I assume this would work for any similar example (where the left side has one large-valued sign change). I also presume the fact that there are four sign changes on the left and 4 terms on the right is not coincidence.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rcs@xmission.com Sent: Thursday, July 06, 2017 7:10 PM To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: Re: [math-fun] Algebraic number question
[Wilson is using 'v' to mean square-root. I typically use q or r. --rcs]
This sort of thing happens in some (all?) extension fields based on pure radicals. In your example, imagine running through the sign changes on +-v8 and +-v5. If you flip v5, your equation changes into
1 / (v8 - v5 - 5) = 1/4 * (-5v10 - 7v5 + 11v2 + 15) = small
The small term is small because the denominator v8-v5-5 is no longer close to 0. Similar things happen when you flip v8 to -v8, and when you flip both of them. Now add up signed combinations of these results: A straight Add cancels all terms except the +15, which isn't affected by the sign flips: The sum is 1/4 * (15+15+15+15) = 15 = 15.505053+ + small + small + small. True! Next, instead of Adding, use Subtract with the equations where v5 is flipped: Various things cancel out, except the v5 flipped terms: 1/4 * (7v5 + 7v5 + 7v5 + 7v5) = 7v5 = 15.505053 - small + small - small so 7v5 = roughly 15.5. The other terms 11v2 and 5v10 behave similarly, with other combinations of sign flips.
I think this generalizes to fields with more square roots. For cube roots, such as cbrt2, instead of sign flips +1 -> -1, substitute +1 -> w and w2, with w & w2 being the two complex cube roots of 1.
Rich
--------- Quoting David Wilson <davidwwilson@comcast.net>:
By happenstance, I noticed that v8 + v5 = 5.064495+ is just over 5, so I computed
1 / (v8 + v5 - 5) = 1/4 * (5v10 + 7v5 + 11v2 + 15) = 15.505053+
I noticed that the terms
5v10 = v250 = 15.811388+ 7v5 = v245 = 15.652475+ 11v2 = v242 = 15.556349+ 15 = v225 = 15.000000
are all in the same ball park.
Is there some reason this should be the case? Can other similar examples be formulated?
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Your expression can be shoehorned into the shape a + b v2 + c v5 + d v2 v5 with a,b,c,d all rational numbers, and v10 replaced by v2*v5. If you consider the equation (a+bv2+cv5+dv2v5) * (e+fv2+gv5+hv2v5) = k + lv2 + mv5 + nv2v5, you can work out the expressions for k,l,m,n by multiplying out the LHS and sorting the terms by their radicals. For example, k = ae + 2bf + 5cg + 10dh. We can change v2 into -v2 by negating b,d,f,h and l,n. Notice that this also negates the v10 = v2v5 terms. Similar changes get us v5 -> -v5, also negating the v10 terms. Negating both v2 and v5 leaves v10 unchanged (or doubly negated, same thing). Now, suppose a+bv2+... and e+fv2+... are reciprocals. Then their product is 1, so k=1, while l,m,n are 0. Then negating v2 will negate b,d,f,h and l,n. But l and n are both 0, so the RHS is unaffected by the negations. So negating v2 in both the a+bv2+... expression and its reciprocal e+fv2+... produces two new expressions which are also reciprocals. I'm implicitly assuming reciprocals are unique here. But we've already assumed that by speaking of "the" reciprocal. Rich --------- Quoting David Wilson <davidwwilson@comcast.net>:
I guess I see.
My intuition was that if you changed signs in terms on the left, you would end up with entirely different terms on the right, but as you tacitly imply below, term sign changes on the left result only in term sign changes on the right.
In this case, there are 4 linearly independent sign changes on the left, and 4 (I assume linearly independent) term changes on the right. One of the sign changes on the left results in a small right-hand value, the others in large right hand values. Solving the right side for term magnitudes leads to terms of about the same size.
I assume this would work for any similar example (where the left side has one large-valued sign change). I also presume the fact that there are four sign changes on the left and 4 terms on the right is not coincidence.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rcs@xmission.com Sent: Thursday, July 06, 2017 7:10 PM To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: Re: [math-fun] Algebraic number question
[Wilson is using 'v' to mean square-root. I typically use q or r. --rcs]
This sort of thing happens in some (all?) extension fields based on pure radicals. In your example, imagine running through the sign changes on +-v8 and +-v5. If you flip v5, your equation changes into
1 / (v8 - v5 - 5) = 1/4 * (-5v10 - 7v5 + 11v2 + 15) = small
The small term is small because the denominator v8-v5-5 is no longer close to 0. Similar things happen when you flip v8 to -v8, and when you flip both of them. Now add up signed combinations of these results: A straight Add cancels all terms except the +15, which isn't affected by the sign flips: The sum is 1/4 * (15+15+15+15) = 15 = 15.505053+ + small + small + small. True! Next, instead of Adding, use Subtract with the equations where v5 is flipped: Various things cancel out, except the v5 flipped terms: 1/4 * (7v5 + 7v5 + 7v5 + 7v5) = 7v5 = 15.505053 - small + small - small so 7v5 = roughly 15.5. The other terms 11v2 and 5v10 behave similarly, with other combinations of sign flips.
I think this generalizes to fields with more square roots. For cube roots, such as cbrt2, instead of sign flips +1 -> -1, substitute +1 -> w and w2, with w & w2 being the two complex cube roots of 1.
Rich
--------- Quoting David Wilson <davidwwilson@comcast.net>:
By happenstance, I noticed that v8 + v5 = 5.064495+ is just over 5, so I computed
1 / (v8 + v5 - 5) = 1/4 * (5v10 + 7v5 + 11v2 + 15) = 15.505053+
I noticed that the terms
5v10 = v250 = 15.811388+ 7v5 = v245 = 15.652475+ 11v2 = v242 = 15.556349+ 15 = v225 = 15.000000
are all in the same ball park.
Is there some reason this should be the case? Can other similar examples be formulated?
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