mm-386
===
Subject:
: Re: Motion> I am interesting in knowing how
transfinite numbers solved Zeno's paradox. They don't.
Carefully defining limits and convergence does.Zeno was
unaware it was possible to add an infinite number of non zero
quantities and come up with a finite sum.===
===
Subject:
: Re:
MotionI am interesting in knowing how transfinite numbers
solved Zeno's paradox. They don't. Carefully defining limits
and convergence does.Zeno was unaware it was possible to add
an infinite number of non zero > quantities and come up with a
finite sum.0 + 0 + 0 + 0 = 1. Why? Because 0.3 = 0 and 1.2 = 1
and 0.3 + 0.3 +0.3 + 0.3 = 1.2 That's how math works.Or,
maybe, I'm just not looking at the data correctly.===
===
Subject:
:
Re: Motion <eQeXb.311222$na.464206@attbi_s04>
<4aa861fb.0402140112.54b01ad4@posting.google.com> I am
interesting in knowing how transfinite numbers solved Zeno's
paradox. They don't. Carefully defining limits and convergence
does.> Zeno was unaware it was possible to add an infinite
number of non zero > quantities and come up with a finite
sum.0 + 0 + 0 + 0 = 1. Why? Because 0.3 = 0 and 1.2 = 1 and
0.3 + 0.3 +> 0.3 + 0.3 = 1.2 That's how math works.Or, maybe,
I'm just not looking at the data correctly. Or maybe you're
being entirely irrelevant. It looks like you're trying to show
that 'zero quantities' (as opposed to nonzero) can add to a
finite sum. ===
===
Subject:
: Analtuc continuationThere is a concept
in complex variable theory called analyticcontinuation (look it
up). The essential point is that the zetafunction is analytic
(except for the pole at z=1) so that it can bedefined for all
complex numbers. The series representation is onlygood for x >
1. However, there are other representations (eulerproduct)
which allows you to compute it for other values of z. ===
===
===
Subject:
: One Of Them Sequence-Puzzles AgainHere is the
beginning of the sequence:1, 2, 4, 6, 8, 10, 12, 14, 16, 18,
20, 22, 24, 26, 28, 30, 32, 34, 36,38, 40, 42, 44, 3, 52, 54,
56, 58, 60, 62, ...It, as a whole, has these 2
characteristics:1) It is a permutation of the positive
integers.2) It relates somehow to a post I have made to
sci.math/rec.puzzles inthe last few weeks.(And, yeah, I know,
the sequence has an infinite number ofdefinitions.But this
puzzle is for FUN. And what fun is it reminding us AGAIN ofthe
fact that there is no specific answer?)Leroy Quet===
===
Subject:
:
Zeta function and uncommon numbersWell known are many facts
and conjectures regarding the zeroes of theRiemann zeta
function.But what is known about, for fixed z (*not*
necessarily =0),all x's where zeta(x) = z?For example, is
there anything of interest about the ONES of the
zetafunction??And, also I know of no study regarding many
specific values of thezeta function,such as zeta(3/2) or
zeta(pi) or zeta(sqrt(3)), zeta(i), etc...Are there any
results of interest aside from closed-forms for thesezetas,
results such as rationality, etc?Leroy Quet===
===
Subject:
:
problems with clever use of the fact that row and column rank
are equalI was thinking about the fact that the row rank and
column rank of anmXn matrix are equal, and I felt that there
must be some problemswhere the trick is to use this fact. Does
anyone know of any suchprobems?===
===
Subject:
: Re: Big Bertha Thing
blogsBig Bertha Thing Faculty(Sequel to battle)Cosmic Ray
SeriesPossible Real World System
Constructshttp://web.onetel.com/~tonylance/faculty.hAccess
page 600K ZIP fileAstrophysics net ring Access siteNewsgroup
Reviews including sci.med301 files from the second battle of
cyberspace.Students Research Faculty 1. Existence 2. Zero
interference 3. Foundation (Students, tutors or funding) 4.
Articles (See Armistice terms) 5. Publications (See overview)
6. Projects (Pastures, Moisture, Big Bertha, Strategic
Studies) 7. Staff (First Aid Tent, volunteers, moderators) 8.
Access (All user or staff only.) 9. SRF Classical Astronomy
(see OUSA Research, All user)10. SRF Net Access Policy (See
OUSA Research, All user)11. SRF Los Alamos (see Armistice
terms, OUSA Research, restricted access.)12. SRF Big Bertha
(See SRF Los Alamos, staff only.)13. SRF Strategic Studies
(See SRF Big Bertha, staff only.)14. SRF Specification (See
SRF Big Bertha, staff only.)15. SRF Mathematics (See SRF Big
Bertha, staff only.)16. SRF Prototype (See SRF Big Bertha,
staff only.)17. SRF Yesterslaggings (SRF Net Access Policy,
all user)Big Bertha Thing galiosNobody understood a thing he
said, when he was alive.After he was dead, they decided that
he had done a good job.The Galios Theory branch of mathematics
bears his name.Now they are trying to teach me, what he did in
the first place.Do you think that the penny will drop, before
it becomes posthumous?Tony
Lancejudemarie@big-bertha-thing.comBig Bertha Thing outlandish
Update V from
http://web.onetel.com/~tonylance/structur.hReviews Included
1980, 1992, 1994, 1996, 1998, 2000, 2002.Review from
http://pdg.lbl.gov/2002/contents_listings.h Review used in
1st, 3rd, last & part of 2nd columns. (Actual Error MeV detail
line number 1,3,45,436,437,803) updated(Electron Mass MeV
0.510998901 Nominal Electron Error MeV 2.0D-08 detail line
numbers 13,19,22,24,36,51,56,57,62,79,86,95,96,99,106,
119,122,123,124,130,145,147,148,152,154,158,159,161,168,172,175
,
177,180,181,183,185,187,190,191,192,194,195,196,197,200,202,204
,
205,208,210,225,226,229,231,236,238,244,252,260,263,273,276,279
,
285,292,298,305,313,315,322,328,330,332,333,337,341,344,354,357
,
363,372,373,381,384,387,388,395,398,399,400,404,409,416,419,424
,
425,427,433,435,443,445,451,458,459,469,470,472,487,496,498,499
,
503,504,509,524,530,535,540,542,548,554,560,567,569,576,580,581
,
582,592,593,595,598,599,604,605,608,611,620,625,636,638,639,640
,
641,647,651,672,683,686,688,690,692,697,699,706,713,721,725,733
, 735,738,743,744,745,767,772,776,782) updated(Electron Mass
MeV 0.510998902 Nominal Electron Error MeV 1.0D-09 detail line
numbers 6,25,38,54,55,87,171,184,193,201,219,227,
232,233,235,259,277,284,290,310,314,320,325,329,331,334,336,
338,340,342,345,370,411,420,422,430,431,434,462,465,475,492,500
,
501,507,508,510,514,515,536,544,564,587,594,600,601,627,628,637
,
645,652,653,680,687,691,696,698,700,704,710,718,720,732,737,742
, 752,753,761,763,775,777,778,783,786,788,790) updated(Offset
see detail line numbers 20,655(441,521)354,355(220,529)
86,460(91,669)93,461(245,378)94,379(247,552)672,765(383,673)
465,766(474,628)125,272(126,127)633,682(130,634)134,770(
138,570)
139,278(404,405)407,574(152,576)290,412(416,491)163,580(
164,582) 584,689(311,777)695,741(189,652)190,597)
updated(Sequential see detail line numbers
4,5,6,7(56,57)58.59(94,95)
180,181(219,220)346,347,348(395,396)425,426(518,519)532,533
(536,537,538,539)602,603(651,652)701,702(753,754))
updatedCol.5 Polygon Edge Length in Photons Type 1 Polygon III
1 000.510998902 2.1D-08 1 1 21348 1 227879226. 1.000000000
XLCType 2 Polyhedron III 4 134.97430000 .000390 0 2 173481 2
60191661684. 264.138433066 SMC 139.56782000 .000370 1 3 176085
2 62240085846. 273.127511176 XMC 493.63100000 .000600 1 4
331591 2 220133724970. 966.010499658 XMC 493.65700000 .000100
1 5 331600 2 220145662426. 966.062884670 XMC 493.67500000
.000070 1 6 331606 2 220153620910. 966.097808802 XMC
493.69600000 .000900 1 7 331613 2 220162905990. 966.138554420
XMC 493.78000000 .001000 1 8 331641 2 220200048270.
966.301545495 XMC 498.90000000 .000100 0 9 333529 2
222483854740. 976.323549300 XMC 547.30000000 .000500 0 10
349333 2 244067788444. 1071.040097547 XMC 547.45000000 .001200
0 11 349381 2 244134865084. 1071.334449258 XMC 548.20000000
.001000 0 12 349620 2 244468988040. 1072.800677496 XMC
549.00000000 .000004 0 13 349875 2 244825731000.
1074.366168858 XMC 744.00000000 .000100 0 14 407299 2
331785765400. 1455.971969117 XMC 750.00000000 .000100 0 15
408938 2 334461393564. 1467.713399922 XMC 761.10000000 .000100
0 16 411953 2 339411372324. 1489.435339420 XMC 764.10000000
.000500 0 17 412764 2 340749064920. 1495.305521706 XMC
766.00000000 .000200 1 18 413139 2 341596372146.
1499.023751055 XMC 766.80000000 .000070 1 19 413355 2
341953417986. 1500.590571542 XMC 767.00000000 .000100 1 20
413409 2 342042708606. 1500.982404627 XMC 769.50000000 .000150
0 21 414220 2 343157245240. 1505.873313963 XMC 773.00000000
.000100 0 22 415161 2 344718142164. 1512.722981445 XMC
775.00000000 .000500 0 23 415698 2 345610485804.
1516.638843613 XMC 776.09000000 .000070 0 24 415990 2
346096192180. 1518.770263771 XMC 777.40000000 .000150 0 25
416341 2 346680489244. 1521.334328404 XMC 781.00000000 .000270
0 26 417304 2 348286091440. 1528.380175558 XMC 781.96000000
.000900 0 27 417560 2 348713542320. 1530.255953739 XMC
782.60000000 .000100 0 28 417731 2 348999212184.
1531.509555785 XMC 783.20000000 .000400 0 29 417891 2
349266611544. 1532.682981572 XMC 888.00000000 .001000 1 30
444845 2 396002916966. 1737.775416904 XMC 890.40000000 .000650
1 31 445446 2 397073047950. 1742.471461396 XMC 890.70000000
.000500 1 32 445521 2 397206693150. 1743.057935215 XMC
891.00000000 .000500 1 33 445596 2 397340360850.
1743.644507771 XMC 891.70000000 .000004 1 34 445771 2
397652339650. 1745.013561043 XMC 891.90000000 .000070 1 35
445821 2 397741498950. 1745.404817857 XMC 892.20000000 .000004
1 36 445896 2 397875256650. 1745.991785359 XMC 892.60000000
.000900 1 37 445996 2 398053635250. 1746.774562285 XMC
892.80000000 .000070 1 38 446046 2 398142839550.
1747.166016572 XMC 893.50000000 .000100 1 39 446221 2
398455133350. 1748.536452156 XMC 894.90000000 .000300 0 40
446698 2 399079099804. 1751.274597554 XMC 895.00000000 .000100
0 41 446723 2 399123770904. 1751.470627270 XMC 897.00000000
.001000 0 42 447222 2 400015929012. 1755.385675270 XMC
897.10000000 .000500 0 43 447247 2 400060652512.
1755.581934932 XMC 898.40000000 .000500 0 44 447571 2
400640495224. 1758.126452580 XMC 939.56533000 .000038 0 45
457710 2 418997803620. 1838.683635076 XBC 956.10000000 .000100
0 46 461720 2 426371640240. 1871.042164414 XMC 956.30000000
.000350 0 47 461768 2 426460295184. 1871.431207968 XMC
957.80000000 .000600 0 48 462130 2 427129198060.
1874.366547392 XMC 969.80000000 .000100 0 49 465016 2
432480690544. 1897.850445323 XMC 974.00000000 .000070 0 50
466022 2 434353941012. 1906.070810562 XMC 976.00000000 .000035
1 51 466378 2 435245689750. 1909.984062128 XMC 977.00000000
.000100 0 52 466739 2 435691521720. 1911.940501852 XMC
982.00000000 .000100 0 53 467932 2 437921649112.
1921.726946326 XMC 993.10000000 .000150 1 54 470448 2
442871461530. 1943.448155867 XMC 1019.30000000 .000035 0 55
476736 2 454555380864. 1994.720575644 XMC 1019.41100000
.000004 0 56 476762 2 454604962812. 1994.938155582 XMC
1019.42000000 .000015 0 57 476764 2 454608776920.
1994.954892992 XMC 1019.48300000 .001000 0 58 476779 2
454637383240. 1995.080425804 XMC 1019.51000000 .000100 0 59
476785 2 454648826020. 1995.130640035 XMC 1019.60000000
.000300 0 60 476806 2 454688876884. 1995.306394818 XMC
1019.80000000 .000035 0 61 476853 2 454778520924.
1995.699778812 XMC 1020.10000000 .000035 0 62 476923 2
454912049704. 1996.285741746 XMC 1020.90000000 .000070 0 63
477110 2 455268858420. 1997.851521665 XMC 1115.44000000
.001000 0 64 498712 2 497428315312. 2182.859420946 XBC
1115.59000000 .000500 0 65 498746 2 497496142524.
2183.157066384 XBC 1168.00000000 .000500 0 66 510327 2
520868314512. 2285.720921801 XMC 1189.16000000 .000045 1 67
514818 2 530304055110. 2327.127682582 XBC 1189.33000000
.000500 1 68 514855 2 530380250986. 2327.462052140 XBC
1197.43000000 .000150 1 69 516606 2 533992430910.
2343.313343139 XBC 1225.00000000 .000500 0 70 522631 2
546287369584. 2397.267092631 XMC 1231.20000000 .000500 1 71
523843 2 549051904210. 2409.398670724 XBC 1232.50000000
.000800 0 72 524228 2 549631040424. 2411.940088053 XBC
1233.60000000 .000700 0 73 524462 2 550121827812.
2414.093805163 XBC 1234.90000000 .000500 1 74 524630 2
550702202286. 2416.640656336 XBC 1269.70000000 .000800 0 75
532081 2 566221445284. 2484.743586429 XMC 1274.00000000
.000004 0 76 532981 2 568138558684. 2493.156434909 XMC
1277.00000000 .000100 0 77 533608 2 569476062544.
2499.025788968 XMC 1278.00000000 .000004 0 78 533817 2
569922246612. 2500.983773800 XMC 1285.00000000 .000070 0 79
535277 2 573044004012. 2514.682948818 XMC 1299.00000000
.000004 0 80 538185 2 579287264820. 2542.080184264 XMC
1309.00000000 .000100 1 81 540147 2 583746522738.
2561.648698675 XMC 1321.67000000 .000100 1 82 542756 2
589397115810. 2586.445136557 XBC 1321.87000000 .000500 1 83
542797 2 589486131238. 2586.835762019 XBC 1330.70000000
.000700 0 84 544712 2 593423415312. 2604.113704125 XMC
1336.20000000 .001000 0 85 545837 2 595877152812.
2614.881414473 XMC 1380.00000000 .000100 1 86 554608 2
615409055770. 2700.593057877 XBC 1381.00000000 .000150 1 87
554809 2 615855041806. 2702.550173687 XBC 1382.00000000
.000100 1 88 555010 2 616301189446. 2704.507998662 XBC
1383.20000000 .001000 1 89 555251 2 616836335730.
2706.856375447 XBC 1383.50000000 .000100 1 90 555311 2
616969603290. 2707.441192073 XBC 1384.10000000 .000500 0 91
555534 2 617237161380. 2708.615314412 XBC 1384.90000000
.000070 1 92 555592 2 617593931338. 2710.180924250 XBC
1385.10000000 .000700 0 93 555735 2 617683891920.
2710.575697321 XBC 1391.00000000 .000350 0 94 556917 2
620314203612. 2722.118266331 XBC 1391.50000000 .000200 1 95
556915 2 620537627506. 2723.098715045 XBC 1392.00000000
.000100 1 96 557015 2 620760413706. 2724.076365373 XBC
1399.80000000 .000100 1 97 558574 2 624238823326.
2739.340633560 XBC 1400.00000000 .000070 0 98 558716 2
624328254744. 2739.733084507 XMC 1407.00000000 .000100 0 99
560111 2 627449784864. 2753.431262155 XMC 1409.00000000
.000100 0 100 560509 2 628341799180. 2757.345679154 XMC
1410.70000000 .000200 0 101 560847 2 629099836512.
2760.672166369 XBC 1422.00000000 .001000 0 102 563089 2
634139570020. 2782.787975680 XMC 1429.00000000 .000100 0 103
564473 2 637260664404. 2796.484241192 XMC 1436.00000000
.000004 1 104 565753 2 640381924750. 2810.181234993 XMC
1453.00000000 .001200 0 105 569193 2 647962480884.
2843.446909391 XMC 1477.00000000 .000004 0 106 573875 2
658666179000. 2890.417834753 XMC 1479.00000000 .000100 1 107
574164 2 659557625346. 2894.329759335 XBC 1485.00000000
.000100 1 108 575328 2 662233645050. 2906.072908331 XBC
1503.00000000 .000100 0 109 578904 2 670260840240.
2941.298564179 XBC 1519.00000000 3.5D-06 0 110 581977 2
677395621012. 2972.608047264 XBC 1521.00000000 .000100 0 111
582360 2 678287503920. 2976.521887607 XMC 1523.80000000
.000200 0 112 582896 2 679536659424. 2982.003543509 XMC
1532.10000000 .001000 0 113 584481 2 683237247684.
2998.242795875 XBC 1533.00000000 .000070 0 114 584653 2
683639430124. 3000.007688827 XBC 1540.00000000 .000015 1 115
585889 2 686760891646. 3013.705565447 XBC 1549.00000000
.000100 0 116 587696 2 690774352224. 3031.317792101 XBC
1610.00000000 .000100 0 117 599156 2 717977022984.
3150.690984811 XBC 1618.00000000 .000070 1 118 600548 2
721544880930. 3166.347778143 XBC 1620.00000000 .000035 0 119
601014 2 722436858420. 3170.262033539 XBC 1629.00000000
.000100 0 120 602681 2 726449980884. 3187.872776450 XMC
1667.00000000 .000100 0 121 609670 2 743396237140.
3262.237853748 XBC 1671.00000000 .000100 0 122 610401 2
745179982404. 3270.065444245 XBC 1671.90000000 .000100 1 123
610472 2 745581225738. 3271.826216129 XBC 1673.40000000
.000100 1 124 610746 2 746250453750. 3274.762982344 XBC
1682.00000000 .000200 0 125 612407 2 750085892112.
3291.593995988 XBC 1684.00000000 .000500 1 126 612678 2
750977767950. 3295.507805306 XBC 1684.00000000 .001000 0 127
612771 2 750977822424. 3295.508044353 XBC 1688.00000000
.000700 0 128 613498 2 752760819004. 3303.332349409 XBC
1695.70000000 .000300 0 129 614896 2 756195411424.
3318.404335040 XBC 1700.00000000 .000100 0 130 615675 2
758112642600. 3326.817700355 XBC 1746.00000000 .000100 0 131
623949 2 778625957100. 3416.836061660 XBC 1773.00000000
.000100 0 132 628755 2 790666957560. 3469.675456770 XMC
1778.00000000 .000100 0 133 629641 2 792896837044.
3479.460813352 XBC 1817.00000000 .001000 0 134 636509 2
810288687180. 3555.781285566 XBC 1819.00000000 .001000 0 135
636859 2 811180045480. 3559.692823777 XBC 1819.40000000
.000500 0 136 636929 2 811358375940. 3560.475389450 XBC
1821.00000000 .000500 0 137 637209 2 812071893780.
3563.606512250 XBC 1822.00000000 .001000 0 138 637384 2
812518001680. 3565.564162834 XBC 1823.00000000 .000100 0 139
637559 2 812964232080. 3567.522350984 XBC 1868.40000000
.000350 1 140 645361 2 833210810590. 3656.370197562 XMC
1894.00000000 .000100 0 141 649856 2 844626941184.
3706.467482841 XBC 1908.00000000 .001000 0 142 652253 2
850869256524. 3733.860569300 XBC 1909.00000000 .000100 0 143
652424 2 851315456400. 3735.818623502 XBC 1920.00000000
.000100 0 144 654301 2 856220905804. 3757.345155297 XBC
1935.50000000 .000300 0 145 656937 2 863133757812.
3787.680750732 XMC 1935.90000000 .001000 0 146 657005 2
863312454060. 3788.464921590 XMC 1968.30000000 .000137 1 147
662394 2 877760826486. 3851.868561665 XMC 1972.70000000
.000100 1 148 663134 2 879722609406. 3860.477432928 XMC
1978.00000000 .000850 0 149 664110 2 882085512420.
3870.846535261 XMC 2006.00000000 .000700 0 150 668794 2
894572166460. 3925.641587268 XMC 2015.00000000 .000100 0 151
670293 2 898586752284. 3943.258751827 XMC 2019.00000000
.000100 1 152 670873 2 900370385230. 3951.085849440 XBC
2020.00000000 .000100 0 153 671124 2 900816189000.
3953.042165414 XBC 2030.00000000 .000250 0 154 672783 2
905275275744. 3972.609928665 XBC 2070.00000000 .000500 0 155
679379 2 923113010040. 4050.887069627 XBC 2104.00000000
.000100 0 156 684936 2 938276018064. 4117.426737547 XBC
2107.00000000 .000100 0 157 685424 2 939613490400.
4123.295953270 XBC 2231.90000000 .000100 0 158 705447 2
995312350512. 4367.718672662 XMC 2281.70000000 .000100 1 159
713194 2 1017520668886. 4465.175201560 XBC 2286.20000000
.001000 1 160 713897 2 1019527160238. 4473.980266363 XBC
2340.00000000 .000100 0 161 722329 2 1043519813140.
4579.266971619 XBC 2422.00000000 .000100 0 162 734876 2
1080086940504. 4739.734110313 XMC 2425.00000000 .001300 1 163
735254 2 1081426238766. 4745.611338727 XMC 2453.00000000
.000500 0 164 739564 2 1093911299320. 4800.399398057 XMC
2459.00000000 .000100 0 165 740468 2 1096587198984.
4812.142020282 XMC 2465.00000000 .000100 0 166 741371 2
1099263402024. 4823.885973810 XMC 2466.50000000 .002000 1 167
741520 2 1099933183066. 4826.825166880 XBC 2473.30000000
.000200 0 168 742618 2 1102964473084. 4840.127344842 XBC
2536.60000000 .001000 1 169 751985 2 1131192263646.
4963.999059949 XMC 2578.80000000 .000100 0 170 758291 2
1150011997944. 5046.585501146 SBC 2593.60000000 .000070 1 171
760389 2 1156612262646. 5075.549373009 SBC 2593.90000000
.000070 1 172 760433 2 1156746095070. 5076.136668421 SBC
2705.90000000 .000100 0 173 776753 2 1206691999524.
5295.313753277 XBC 2719.00000000 .000100 0 174 778631 2
1212534025584. 5320.950254518 XBC 2819.00000000 .000100 0 175
792820 2 1257128690440. 5516.644551180 SBC 2880.90000000
.001000 1 176 801406 2 1284732635710. 5637.778652583 SBC
2969.00000000 .000500 0 177 813640 2 1324021726480.
5810.190554535 XMC 2980.20000000 .000100 0 178 815173 2
1329015670204. 5832.105425020 XMC 2982.00000000 .000500 0 179
815419 2 1329817921960. 5835.625937926 XMC 3096.89000000
.000100 0 180 830979 2 1381053858840. 6060.464058448 XMC
3096.95000000 .000100 0 181 830987 2 1381080450312.
6060.580749524 XMC 3417.80000000 .001000 0 182 872972 2
1524161971512. 6688.463877405 XMC 3507.40000000 .000100 0 183
884341 2 1564119777244. 6863.810294160 XMC 3526.20000000
.000350 0 184 886708 2 1572503927944. 6900.602374101 XMC
3556.40000000 .000070 0 185 890497 2 1585971595012.
6959.702395215 XMC 3557.80000000 .000500 0 186 890672 2
1586595004512. 6962.438096538 XMC 3594.00000000 .000100 0 187
895192 2 1602739224112. 7033.283604851 XMC 3685.95000000
.000150 0 188 906571 2 1643743769224. 7213.223416969 XMC
5278.20000000 .000100 0 189 1084851 2 2353805554104.
10329.180046030 XMC 5279.10000000 .000070 1 190 1084891 2
2354207012770. 10330.941762853 XMC 5280.50000000 .000100 1 191
1085035 2 2354831951746. 10333.684175959 XMC 5416.60000000
.000100 0 192 1098982 2 2415525070612. 10600.023148279 SMC
5713.00000000 .000323 0 193 1128650 2 2547703902300.
11180.062118958 XMC 9460.51000000 .001000 0 194 1452394 2
4218899567260. 18513.752399966 XMC 9460.60000000 .000100 0 195
1452401 2 4218940234404. 18513.930859165 XMC 9859.90000000
.000070 0 196 1482735 2 4397009125920. 19295.348694576 XMC
9864.00000000 .000100 0 197 1483043 2 4398836045784.
19303.365747714 XMC 9891.80000000 .002500 0 198 1485131 2
4411231144584. 19357.759028829 XMC 9893.70000000 .000200 0 199
1485274 2 4412080680700. 19361.487039192 XMC 9912.20000000
.000100 0 200 1486662 2 4420330777812. 19397.690853189 XMC
9913.20000000 .000767 0 201 1486737 2 4420776787812.
19399.648074160 XMC 9914.00000000 .000100 0 202 1486797 2
4421133612012. 19401.213922027 XMC 9914.50000000 .002000 0 203
1486835 2 4421359608120. 19402.205658360 XMC10023.60000000
.000300 0 204 1494993 2 4470011130084. 19615.702618211
XMC10268.10000000 .001000 0 205 1513116 2 4579043085144.
20094.166394720 XMCType 3 Polyhedron III 8 139.57071000
.000240 1 206 124512 3 62241329850. 273.132970225 XMC
493.90000000 .000700 1 207 234534 3 220253605986.
966.536572254 XMC 546.00000000 .000035 0 208 246722 3
243487968024. 1068.495677724 XMC 555.00000000 .000250 0 209
248747 3 247501275024. 1086.107230433 XMC 763.00000000 .000070
0 210 291658 3 340258722488. 1493.153757192 XMC 765.00000000
.000100 0 211 292040 3 341150614560. 1497.067637749 XMC
767.60000000 .000500 0 212 292536 3 342310415328.
1502.157179207 XMC 767.70000000 .000070 0 213 292555 3
342354882320. 1502.352313238 XMC 770.50000000 .000100 0 214
293088 3 343603475328. 1507.831500744 XMC 782.71000000 .000500
0 215 295401 3 349048184808. 1531.724461834 XMC 894.30000000
.000070 1 216 315667 3 398811761450. 1750.101439479 XMC
894.52000000 .000350 0 217 315796 3 398909717648.
1750.531299628 XMC 894.70000000 .000500 0 218 315828 3
398990565648. 1750.886084052 XMC 895.50000000 .000060 0 219
315969 3 399346899720. 1752.449781096 XMC 896.00000000 .000100
1 220 315967 3 399569723450. 1753.427596116 XMC 983.70000000
.000035 1 221 331078 3 438679771874. 1925.053808433 SMC
986.00000000 .000100 1 222 331465 3 439705389986.
1929.554517558 XMC 989.00000000 .000500 1 223 331969 3
441042874946. 1935.423788678 XMC 1018.90000000 .000100 0 224
337037 3 454377105624. 1993.938252292 XMC 1020.40000000
.000070 0 225 337285 3 455046034040. 1996.873703793 XMC
1107.00000000 .000035 0 226 351306 3 493665027768.
2166.345025974 XMC 1115.60000000 .000100 0 227 352668 3
497500283568. 2183.175238484 XBC 1115.86000000 .000035 0 228
352709 3 497615965560. 2183.682884547 XBC 1197.53200000
.000150 1 229 365311 3 534037847354. 2343.512643640 XBC
1213.00000000 .000500 0 230 367741 3 540935243288.
2373.780413349 XMC 1231.60000000 .000100 1 231 370473 3
549230336034. 2410.181681212 XBC 1269.00000000 .000150 0 232
376134 3 565908648360. 2483.370942992 XMC 1281.90000000
.000100 0 233 378041 3 571661502888. 2508.616133741 XMC
1318.00000000 .000700 0 234 383327 3 587759889024.
2579.260511548 XMC 1320.60000000 .000100 0 235 383705 3
588919642920. 2584.349847318 XMC 1321.20000000 .000120 1 236
383718 3 589187428194. 2585.524966607 XBC 1321.40000000
.001000 1 237 383747 3 589276454250. 2585.915638708 XBC
1321.60000000 .000100 1 238 383776 3 589365487034.
2586.306340333 XBC 1323.10000000 .000100 0 239 384068 3
590034450768. 2589.241946820 XMC 1384.00000000 .001000 1 241
392735 3 617192571066. 2708.419639208 XBC 1384.40000000
.000500 1 242 392792 3 617371671450. 2709.205583531 XBC
1384.50000000 .000150 1 243 392806 3 617415664994.
2709.398639936 XBC 1385.00000000 .000100 1 244 392877 3
617638799250. 2710.377817634 XBC 1389.00000000 .000350 1 245
393444 3 619422177546. 2718.203797770 XBC 1409.60000000
.000400 0 246 396424 3 628609536800. 2758.520589323 XBC
1420.00000000 .000100 1 247 397812 3 633247019850.
2778.871207198 XMC 1423.40000000 .001000 1 248 398288 3
634762796154. 2785.522872339 XMC 1423.80000000 .000120 1 249
398344 3 634941241946. 2786.305944123 XMC 1449.00000000
.000100 0 250 401926 3 646179645608. 2835.623312184 XMC
1475.00000000 .000500 0 251 405516 3 657774527088.
2886.505008087 XMC 1500.00000000 .000100 0 252 408938 3
668922787128. 2935.426799844 XBC 1507.00000000 .000100 0 253
409891 3 672044167088. 2949.124318546 XBC 1510.00000000
.000500 0 254 410299 3 673382718800. 2954.998270882 XBC
1519.40000000 .000350 0 255 411574 3 677574276200.
2973.392037939 XBC 1525.00000000 .001200 0 256 412332 3
680072362224. 2984.354362446 XBC 1526.80000000 .000100 0 257
412575 3 680874172800. 2987.872939326 XMC 1530.00000000
.001000 0 258 413007 3 682300780224. 2994.133305613 XBC
1532.00000000 .000070 0 259 413277 3 683193168024.
2998.049361568 XBC 1532.20000000 .000100 0 260 413304 3
683282438880. 2998.441107923 XBC 1534.00000000 .001700 0 261
413547 3 684086139024. 3001.967976774 XBC 1534.70000000
.001000 1 262 413572 3 684396730250. 3003.330940970 XBC
1535.30000000 .000035 1 263 413653 3 684664751474.
3004.507095675 XBC 1582.00000000 .000100 0 264 419967 3
705490804224. 3095.897842939 XBC 1596.00000000 .000300 0 265
421821 3 711733511448. 3123.292649098 XBC 1625.00000000
.000870 0 266 425636 3 724665720528. 3180.042925580 XBC
1633.60000000 .000100 0 267 426761 3 728501511528.
3196.875486702 XMC 1640.00000000 .001000 0 268 427596 3
731355067248. 3209.397715121 XBC 1659.00000000 .000100 0 269
430066 3 739828777688. 3246.582809124 XBC 1671.70000000
.001000 1 270 431643 3 745492323594. 3271.436087790 XBC
1676.00000000 .000700 0 271 432264 3 747410391840.
3279.853126410 XBC 1682.00000000 .000150 1 272 432971 3
750085158474. 3291.590776572 XBC 1689.00000000 .000100 0 273
433937 3 753207015624. 3305.290389322 XBC 1760.00000000
.001000 0 274 442964 3 784870193040. 3444.237576268 XBC
1774.00000000 .000100 0 275 444722 3 791112408024.
3471.630222335 XBC 1776.96000000 .000100 1 276 445029 3
792432902706. 3477.424935198 XLC 1777.00000000 .000200 0 277
445098 3 792450698808. 3477.503029644 XBC 1823.00000000
.001000 1 278 450759 3 812964386586. 3567.523029001 XBC
1826.00000000 .000100 0 279 451193 3 814302297768.
3573.394170507 XBC 1860.00000000 .000700 0 280 455374 3
829463741000. 3639.926971667 XMC 1863.30000000 .000100 0 281
455778 3 830936164248. 3646.388391051 XMC 1864.60000000
.001000 0 282 455937 3 831516015624. 3648.932946718 XMC
1869.40000000 .000350 1 283 456461 3 833656283154.
3658.325060109 XMC 1911.00000000 .000100 0 284 461575 3
852207768800. 3739.734348580 XMC 1945.00000000 .000100 0 285
465663 3 867369980928. 3806.270523878 XBC 1963.00000000
.000200 1 286 467752 3 875397484250. 3841.497531899 XMC
2008.60000000 .000700 1 287 473155 3 895732387946.
3930.732974958 XMC 2022.00000000 .000100 1 288 474731 3
901707867594. 3956.955109168 XBC 2036.00000000 .000100 0 289
476432 3 907951708224. 3984.354889041 XBC 2044.00000000
.000100 0 290 477367 3 911518920224. 4000.008847774 XBC
2082.00000000 .000100 0 291 481784 3 928465217760.
4074.374106221 XBC 2100.00000000 .000300 0 292 483862 3
936491675624. 4109.596526469 XBC 2110.00000000 .000200 1 293
484954 3 940951347506. 4129.166857474 SMC 2137.00000000
.001000 0 294 488106 3 952991821368. 4182.003941720 XBC
2204.50000000 .000070 0 295 495755 3 983094063120.
4314.101291181 XBC 2281.00000000 .000700 1 296 504227 3
1017209366250. 4463.809115492 XBC 2283.00000000 .000100 1 297
504448 3 1018101035834. 4467.722019707 XBC 2283.10000000
.000070 1 298 504459 3 1018145427786. 4467.916824441 XBC
2285.80000000 .001000 1 299 504757 3 1019348414450.
4473.195878110 XBC 2290.00000000 .001000 1 300 505221 3
1021222935474. 4481.421818916 XBC 2307.00000000 .001000 0 301
507149 3 1028802461400. 4514.682972462 XMC 2356.00000000
.001000 1 302 512451 3 1050654038634. 4610.574017984 XBC
2359.00000000 .000100 0 303 512833 3 1051992794888.
4616.448867910 XMC 2414.00000000 .000500 0 304 518777 3
1076520378024. 4724.083001862 XMC 2428.00000000 .000100 0 305
520279 3 1082763032480. 4751.477576460 XMC 2451.30000000
.000100 1 306 522715 3 1093153854986. 4797.075513088 SBC
2452.10000000 .001500 0 307 522855 3 1093511495520.
4798.644943265 SBC 2455.00000000 .000400 0 308 523164 3
1094804376240. 4804.318478070 XMC 2462.10000000 .000800 0 309
523920 3 1097970761280. 4818.213492089 XBC 2465.80000000
.000070 1 310 524259 3 1099619972586. 4825.450708640 XBC
2500.00000000 .000300 0 311 527937 3 1114872015624.
4892.381087972 XBC 2534.20000000 .000100 1 312 531482 3
1130122470450. 4959.304497769 XMC 2534.80000000 .000100 1 313
531545 3 1130390353506. 4960.480046154 XMC 2535.00000000
.000100 1 314 531566 3 1130479654914. 4960.871926579 XMC
2535.20000000 .000100 1 315 531587 3 1130568959850.
4961.263822486 XMC 2574.50000000 .001000 1 316 535692 3
1148093697450. 5038.167443354 XMC 2644.50000000 .000100 0 317
542980 3 1179311293520. 5175.159290387 SBC 2694.60000000
.000600 0 318 548099 3 1201652247600. 5273.197863152 XBC
2699.90000000 .000035 0 319 548638 3 1204016814728.
5283.574268099 XBC 2976.00000000 .000500 0 320 576008 3
1327143168288. 5823.888344644 XMC 3097.00000000 .000350 0 321
587601 3 1381102091208. 6060.675716039 XMC 3170.00000000
.000100 0 322 594486 3 1413656794728. 6203.535177568 XBC
3416.50000000 .000100 0 323 617167 3 1523582892224.
6685.922709883 XMC 3510.10000000 .000100 0 324 625564 3
1565323774640. 6869.093783213 XMC 3511.30000000 .000350 0 325
625671 3 1565859303648. 6871.443839501 XMC 3512.30000000
.000100 0 326 625760 3 1566304813440. 6873.398865415 XMC
3525.40000000 .000700 0 327 626926 3 1572147345608.
6899.037587604 XMC 3555.90000000 .000100 0 328 629632 3
1585748340224. 6958.722688588 XMC 3770.00000000 .000015 0 329
648310 3 1681226017640. 7377.706371708 XMC 3772.00000000
.000100 0 330 648482 3 1682118211224. 7381.621575387 XMC
4040.00000000 .000500 0 331 671124 3 1801632378000.
7906.084330829 XMC 5279.20000000 .000100 0 332 767178 3
2354251403448. 10331.136561996 XMC 9460.59000000 .000100 0 333
1027002 3 4218936540024. 18513.914647156 XMC 9858.10000000
.000150 0 334 1048356 3 4396205404368. 19291.821731780 XMC
9860.00000000 .000500 0 335 1048457 3 4397052517224.
19295.539108177 XMC 9890.70000000 .000100 0 336 1050088 3
4410743431328. 19355.618801900 XMC 9892.00000000 .000100 0 337
1050157 3 4411323099224. 19358.162552404 XMC 9893.60000000
.001000 0 338 1050242 3 4412037235224. 19361.296387868 XMC
9911.90000000 .000070 0 339 1051213 3 4420199290328.
19397.113848052 XMC 9912.30000000 .000200 0 340 1051234 3
4420375895960. 19397.888844681 XMC 9913.30000000 .000500 0 341
1051287 3 4420821630624. 19399.844857398 XMC10023.10000000
.000630 0 342 1057093 3 4469786670968. 19614.717626643
XMC10254.70000000 .000150 0 343 1069236 3 4573066771728.
20067.940601694 XMC10355.20000000 .000100 0 344 1074463 3
4617887251328. 20264.625838812 XMC10580.00000000 .000100 0 345
1086063 3 4718135704128. 20704.545065148 XMCType 4 Polyhedron
IV 6 493.63800000 .000070 1 346 178161 4 220137085512.
966.025246689 XMC 493.70900000 .000570 1 347 178174 4
220169167446. 966.166031501 XMC 493.75300000 .000300 1 348
178182 4 220188664620. 966.251590744 XMC 498.10000000 .000100
0 349 179057 4 222127108626. 974.758044097 XMC 547.12000000
.000230 0 350 187661 4 243987174396. 1070.686339772 XMC
757.50000000 .000070 0 351 220813 4 337806028500.
1482.390626077 XMC 762.60000000 .000350 0 352 221555 4
340080397086. 1492.371213715 XMC 764.00000000 .000070 0 353
221758 4 340704674880. 1495.110725363 XMC 771.00000000 .000004
0 354 222772 4 343826424186. 1508.809864862 XMC 771.00000000
.000100 1 355 222698 4 343826515794. 1508.810266865 XMC
775.80000000 .000100 0 356 223464 4 345967084554.
1518.203702140 XMC 777.00000000 .000100 0 357 223637 4
346501945830. 1520.550828227 XMC 782.08000000 .000600 0 358
224367 4 348767852124. 1530.494280878 XMC 893.70000000 .001000
0 359 239844 4 398543626770. 1748.924786896 XMC 894.00000000
.000500 0 360 239884 4 398677741194. 1749.513319806 XMC
984.00000000 .000100 1 361 251604 4 438813539496.
1925.640819475 XMC 1019.70000000 3.5D-07 0 362 256194 4
454733727150. 1995.503210767 XMC 1115.67800000 .000100 0 363
267980 4 497535027372. 2183.327704352 XBC 1189.48000000
.000350 1 364 276642 4 530446724088. 2327.753755351 XBC
1192.65000000 .000350 0 365 277070 4 531860395644.
2333.957355305 XBC 1233.80000000 .000630 0 366 281809 4
550211369658. 2414.486740700 XBC 1268.00000000 .000500 0 367
285688 4 565462411500. 2481.412726494 XMC 1280.00000000
.000035 0 368 287037 4 570813987582. 2504.896991277 XMC
1282.00000000 .000100 0 369 287261 4 571705847100.
2508.810728978 XMC 1286.00000000 .000120 0 370 287709 4
573489928470. 2516.639794406 XMC 1314.82000000 .000350 0 371
290915 4 586342247556. 2573.039490471 XBC 1317.90000000
.000035 1 372 291199 4 587715611406. 2579.066208545 XBC
1320.69000000 .000040 1 373 291507 4 588959653746.
2584.525426403 XBC 1321.46000000 .000700 1 374 291592 4
589302777078. 2586.031150896 XBC 1382.60000000 .000500 1 375
298264 4 616568523534. 2705.681138017 XBC 1383.00000000
.000150 1 376 298307 4 616746871974. 2706.463782592 XBC
1388.00000000 .000100 0 377 298901 4 618976904826.
2716.249812197 XMC 1389.00000000 .000500 0 378 299009 4
619423018518. 2718.207488198 XBC 1391.00000000 .000100 1 379
299169 4 620314429176. 2722.119256171 XBC 1405.00000000
.000100 0 380 300726 4 626557970544. 2749.517722796 XMC
1413.00000000 .000004 0 381 301581 4 630125312508.
2765.172251849 XMC 1421.60000000 .001680 0 382 302497 4
633959459850. 2781.997600123 XMC 1428.00000000 .000200 1 383
303123 4 636814228986. 2794.525153363 XMC 1433.40000000
.000004 0 384 303750 4 639222597780. 2805.093772699 XMC
1501.00000000 .000700 0 385 310830 4 669368219958.
2937.381488026 XBC 1512.00000000 .000100 0 386 311967 4
674274410172. 2958.911270710 XMC 1528.70000000 .000035 0 387
313685 4 681721740684. 2991.592312517 XBC 1535.70000000
.000100 1 388 314350 4 684843368286. 3005.290917944 XBC
1536.20000000 .000340 1 389 314401 4 685065737946.
3006.266740374 XBC 1572.00000000 .000070 1 390 318045 4
701031301908. 3076.328256039 XBC 1606.00000000 .000100 0 391
321518 4 716193734988. 3142.865400938 XBC 1655.00000000
.000700 0 392 326386 4 738045086814. 3238.755457305 XBC
1668.90000000 .001000 0 393 327754 4 744243638100.
3265.956494428 XBC 1670.00000000 .000800 0 394 327862 4
744734774358. 3268.111742481 XBC 1673.00000000 .000004 1 395
328106 4 746072052138. 3273.980104435 XBC 1673.10000000
.000100 1 396 328116 4 746116439604. 3274.174889483 XBC
1686.00000000 .000500 0 397 329429 4 751869819264.
3299.422384663 XBC 1723.00000000 .000300 0 398 333024 4
768369136422. 3371.826163838 XBC 1731.00000000 .000035 0 399
333796 4 771936808482. 3387.482141448 XMC 1772.00000000
.000100 0 400 337726 4 790220932116. 3467.718168026 XBC
1786.70000000 .000100 0 401 339124 4 796776089226.
3496.484094720 XBC 1813.00000000 .001000 1 402 341563 4
808505477976. 3547.956047455 XBC 1831.00000000 .001200 0 403
343302 4 816530945502. 3583.174121813 XBC 1838.00000000
.000100 0 404 343958 4 819653474160. 3596.876681335 XBC
1840.00000000 .000100 0 406 344146 4 820549557108.
3600.808952669 XBC 1863.00000000 .000100 1 407 346242 4
830802376674. 3645.801292453 XMC 1897.00000000 .000150 0 408
349435 4 845964151644. 3712.335549376 XMC 1905.00000000
.000100 0 409 350171 4 849531841866. 3727.991606686 XBC*
1937.00000000 .000500 0 410 353100 4 863802843228.
3790.616891195 XBC 2024.00000000 .000250 1 411 360897 4
902599773762. 3960.869051583 XBC 2044.00000000 .001050 1 412
362676 4 911519706672. 4000.012298936 XBC 2091.00000000
.000100 0 413 366868 4 932478864510. 4091.987149851 XBC
2203.20000000 .000100 0 414 376582 4 982513937274.
4311.555531060 XBC 2358.00000000 .000100 0 415 389587 4
1051546947264. 4614.492359492 XBC 2373.00000000 .000100 1 416
390782 4 1058235964830. 4643.845704610 XBC 2421.00000000
.000100 0 417 394757 4 1079641658214. 4737.780082744 XMC
2461.00000000 .000070 0 418 398005 4 1097479896552.
4816.059435589 XMC 2505.00000000 .000004 1 419 401506 4
1117101349656. 4902.164050952 XBC 2573.20000000 .000200 1 420
406936 4 1147514970186. 5035.627820616 XMC 2984.00000000
.000100 0 421 438260 4 1330710985866. 5839.544960829 XMC
2988.30000000 .000500 0 422 438576 4 1332628322622.
5847.958789460 XMC 3096.93000000 .000700 0 423 446476 4
1381071948738. 6060.543442157 XMC 3098.40000000 .000004 0 424
446582 4 1381726733652. 6063.416827877 XMC 3510.40000000
.000100 0 425 475347 4 1565457677760. 6869.681388860 XMC
3510.53000000 .000070 0 426 475356 4 1565515838772.
6869.936616214 XMC 3563.00000000 .000100 0 427 478895 4
1588914483660. 6972.616642379 XMC 3684.00000000 .000100 0 428
486959 4 1642874291544. 7209.407897252 XMC 3686.00000000
.000100 0 429 487091 4 1643766775092. 7213.324373377 XMC
5275.80000000 .000100 1 430 582714 4 2352735512886.
10324.484395458 XMC 5278.80000000 .000700 1 431 582880 4
2354073661356. 10330.356578252 XMC 5281.30000000 .000100 0 432
583046 4 2355187809342. 10335.245782088 XMC 9915.70000000
.000100 0 433 798903 4 4421891783598. 19404.540998388
XMC10023.50000000 .000200 0 434 803234 4 4469965511094.
19615.502428879 XMC11019.00000000 .000100 0 435 842177 4
4913907861258. 21563.649954024 XMCType 5 Polyhedron III 20
105.65835680 5.2D-06 1 436 68476 5 47118189746. 206.768254277
XMC 139.57022000 .000140 1 437 78748 5 62241141746.
273.132144770 XMC 497.44000000 .000070 0 438 148940 5
221832725400. 973.466205296 XMC 497.66100000 .000800 0 439
148973 5 221931037020. 973.897625139 XMC 548.00000000 .000100
0 440 156326 5 244379746020. 1072.409057682 XMC 768.00000000
.000105 0 441 185064 5 342488691600. 1502.939507088 XMC
774.50000000 .000750 1 442 185784 5 345386683626.
1515.656734879 XMC 782.20000000 .000004 0 443 186767 5
348820990560. 1530.727467716 XMC 783.70000000 .000070 0 444
186946 5 349489938620. 1533.663005420 XMC 886.00000000 .000004
1 445 198716 5 395110352946. 1733.858587645 XMC 894.20000000
.000100 1 446 199634 5 398767215126. 1749.905957316 XMC
894.63000000 .000500 0 447 199739 5 398958678600.
1750.746154456 XMC 898.00000000 .000335 0 448 200115 5
400462133400. 1757.343749272 XMC 957.46000000 .000200 0 449
206634 5 426978165900. 1873.703774560 XMC 959.00000000 .000200
0 450 206800 5 427664468000. 1876.715466815 XMC 991.00000000
.000100 0 451 210222 5 441934995060. 1939.338669950 XMC
1019.40000000 .000100 0 452 213213 5 454599965820.
1994.916227335 XMC 1255.00000000 .000100 0 453 236572 5
559665477560. 2455.974102527 XMC 1272.00000000 .001000 0 454
238169 5 567247107300. 2489.244488219 XMC 1280.10000000
.000100 0 455 238926 5 570858724020. 2505.093307716 XMC
1281.00000000 .000200 0 456 239010 5 571260191100.
2506.855061461 XMC 1282.20000000 .000500 0 457 239122 5
571795700060. 2509.205029773 XMC 1321.12000000 .000100 1 458
242677 5 589151569286. 2585.367607340 XBC 1323.80000000
.000100 0 459 242970 5 590346638700. 2590.611917824 XMC
1380.00000000 .000700 0 460 248074 5 615409575500.
2700.595338603 XBC 1385.10000000 .000100 1 461 248486 5
617683286046. 2710.573038571 XBC 1388.30000000 .000100 1 462
248773 5 619110422246. 2716.835725280 XBC 1390.70000000
.000070 1 463 248988 5 620180610546. 2721.532021291 XBC
1425.00000000 .001000 0 464 252086 5 635476034820.
2788.652770042 XMC 1430.00000000 .000035 0 465 252528 5
637706433120. 2798.440403339 XMC 1431.20000000 .000150 0 466
252634 5 638241905900. 2800.790212882 XMC 1434.00000000
.000100 0 467 252881 5 639490530420. 2806.269538672 XMC
1515.00000000 .000650 0 468 259925 5 675612655500.
2964.783878545 XBC 1517.80000000 .000070 0 469 260165 5
676860873900. 2970.261422162 XBC 1531.40000000 .000150 0 470
261328 5 682925849120. 2996.876288846 XBC 1535.00000000
.000035 0 471 261635 5 684531348600. 3003.921685253 XBC
1608.00000000 .000035 0 472 267784 5 717085384400.
3146.778216633 XBC 1624.00000000 .000100 0 473 269113 5
724220758820. 3178.090304818 XBC 1665.00000000 3.5D-06 1 474
272447 5 742504281786. 3258.323695491 XBC 1665.10000000
.000200 0 475 272497 5 742548875060. 3258.519383684 XBC
1669.00000000 .000100 0 476 272816 5 744288426720.
3266.153039856 XMC 1670.60000000 .000100 1 477 272905 5
745001998526. 3269.284399474 XBC 1670.80000000 .001000 0 478
272963 5 745090723320. 3269.673749550 XBC 1671.80000000
.000100 1 479 273003 5 745536989346. 3271.632094037 XBC
1679.00000000 .000350 0 480 273632 5 748747450560.
3285.720527066 XBC 1735.00000000 .001000 0 481 278158 5
773721511220. 3395.313933619 XBC 1775.00000000 .000100 0 482
281346 5 791558530620. 3473.587937410 XMC 1778.20000000
3.5D-06 1 483 281559 5 792985399626. 3479.849451595 XLC
1825.00000000 .001000 0 484 285281 5 813855342420.
3571.432801075 XBC 1841.00000000 .000004 0 485 286529 5
820991543700. 3602.748517761 XBC 1868.30000000 .001000 1 486
288606 5 833164997646. 3656.169157104 XMC 1882.00000000
.000070 0 487 289702 5 839275385060. 3682.983305639 XBC
1900.00000000 .000500 0 488 291084 5 847301861400.
3718.205806965 XBC 2189.00000000 .000070 1 489 312402 5
976181099286. 4283.765205021 XBC 2250.00000000 .000300 0 490
316762 5 1003384814060. 4403.142979167 XBC 2373.00000000
.001000 0 491 325305 5 1058236683300. 4643.848857465 XBC
2457.00000000 .000100 1 492 330978 5 1095695553846.
4808.229223343 XBC 2463.00000000 .000100 1 493 331382 5
1098371492286. 4819.972015729 XMC 2467.00000000 .000100 1 494
331651 5 1100155053746. 4827.798799641 XBC 2470.00000000
.001000 0 495 331887 5 1101493126560. 4833.670650435 XBC
2472.10000000 .000100 0 496 332028 5 1102429248120.
4837.778622787 XBC 2517.50000000 .000100 0 497 335063 5
1122675490320. 4926.624993539 SBC 2574.10000000 .000100 1 498
338775 5 1147916273226. 5037.388854507 SBC 2626.60000000
.000004 1 499 342213 5 1171328675046. 5140.129250070 XBC
2643.80000000 .000350 0 500 343365 5 1178998665900.
5173.787389905 SBC 2975.80000000 .000100 0 501 364287 5
1327053826560. 5823.496287283 XMC 2982.60000000 .000100 0 502
364703 5 1330086429120. 5836.804225059 XMC 3416.00000000
.000100 0 503 390302 5 1523360415060. 6684.946415695 XMC
3417.40000000 .000100 0 504 390382 5 1523984963060.
6687.687113085 XMC 3556.15000000 .000035 0 505 398228 5
1585859382120. 6959.209972567 XMC 3556.90000000 .000700 0 506
398270 5 1586193911700. 6960.677985189 XMC 3559.90000000
.000100 0 507 398438 5 1587532382820. 6966.551583864 XMC
4414.00000000 .000800 0 508 443668 5 1968417378920.
8637.985188347 XMC 5278.30000000 .000100 1 509 485141 5
2353850629446. 10329.377849677 XMC 5325.00000000 .000300 0 510
487306 5 2374676249420. 10420.766697795 SMC 9890.80000000
.000070 0 511 664137 5 4410786189060. 19355.806435204 XMC
9894.40000000 .000100 0 512 664258 5 4412393548220.
19362.859992424 XMC 9912.10000000 .000070 0 513 664852 5
4420288467560. 19397.505183557 XMC 10255.30000000 .000700 0
514 676264 5 4573336739600. 20069.125298855 XMC10577.40000000
.000700 0 515 686802 5 4716976740060. 20699.459195372 XMCType
6 Polyhedron III 8 IV 6 134.97660000 .000250 0 516 74216 6
60192629712. 264.142681053 SMC 139.56995000 .000150 1 517
75330 6 62240931540. 273.131222326 XMC 493.65800000 .000500 1
518 141859 6 220146288590. 966.065632459 XMC 493.74200000
.000100 1 519 141871 6 220183618688. 966.229447734 XMC
761.00000000 .000100 1 520 176163 6 339366763782.
1489.239584226 XMC 768.00000000 .000100 1 521 176972 6
342488636574. 1502.939265618 XMC 769.00000000 .000500 0 522
177146 6 342934201998. 1504.894535661 XMC 769.20000000 .000004
0 523 177169 6 343023728270. 1505.287402854 XMC 774.00000000
.000100 0 524 177721 6 345164134370. 1514.680124330 XMC
775.10000000 .000700 0 525 177847 6 345654276464.
1516.831009703 XMC 781.78000000 .000300 0 526 178612 6
348633930194. 1529.906592688 XMC 782.70000000 .000500 0 527
178717 6 349043690060. 1531.704737579 XMC 889.00000000 .000400
1 528 190412 6 396447933942. 1739.728280199 XMC 896.00000000
.000004 0 529 191215 6 399569850212. 1753.428152385 XMC
900.70000000 .000004 0 530 191716 6 401665828070.
1762.625909876 XMC 957.00000000 .000100 0 531 197617 6
426772988636. 1872.803397340 XMC 957.90000000 .000350 0 532
197710 6 427174309058. 1874.564507508 XMC 958.00000000 .001200
0 533 197720 6 427218059802. 1874.756498436 XMC 969.00000000
.000500 0 534 198852 6 432123796950. 1896.284292935 XMC
994.00000000 .000070 0 535 201401 6 443272833282.
1945.209491286 XMC 1019.36000000 .000070 0 536 203954 6
454582166388. 1994.838118276 XMC 1019.50000000 .000035 0 537
203968 6 454644373376. 1995.111100544 XMC 1019.52000000
.001200 0 538 203970 6 454654134588. 1995.153935568 XMC
1019.67000000 .000500 0 539 203985 6 454720615530.
1995.445673183 XMC 1115.52000000 .000070 0 540 213357 6
497464575684. 2183.018541953 XBC 1116.00000000 .000500 0 541
213403 6 497678957888. 2183.959313114 XBC 1197.41700000
.000070 1 542 221003 6 533986487154. 2343.287260217 XBC
1222.00000000 .000700 1 543 223261 6 544948670594.
2391.392493996 XMC 1231.80000000 .000100 1 544 224155 6
549319524514. 2410.573066077 XBC 1321.10000000 .000500 1 545
232141 6 589142327996. 2585.327053884 XBC 1381.90000000
.000070 1 546 237425 6 616256609664. 2704.312369676 XBC
1384.30000000 .000500 1 547 237631 6 617326247330.
2709.006249345 XBC 1385.30000000 .000100 1 548 237717 6
617772557262. 2710.964786505 XBC 1390.00000000 .000800 1 549
238120 6 619869109478. 2720.165064445 XBC 1403.00000000
.000100 0 550 239275 6 625665969230. 2745.603362853 XBC
1404.00000000 .000100 0 551 239360 6 626111792916.
2747.559766225 XMC 1420.00000000 .000800 0 552 240720 6
633246345486. 2778.868247894 XMC 1424.00000000 .000500 0 553
241059 6 635030531010. 2786.697770379 XMC 1426.00000000
.000100 0 554 241228 6 635922488138. 2790.611936421 XMC
1430.80000000 .000500 0 555 241634 6 638063702574.
2800.008205109 XMC 1446.00000000 .000100 0 556 242914 6
644841577892. 2829.751483762 XMC 1454.00000000 .000100 0 557
243585 6 648409272612. 2845.407560810 XMC 1518.00000000
.000100 0 558 248888 6 676950106806. 2970.653001981 XBC
1520.00000000 .000700 0 559 249052 6 677841973244.
2974.566770049 XBC 1526.00000000 .000100 0 560 249543 6
680517370056. 2986.307185614 XBC 1531.30000000 .000700 0 561
249976 6 682881071102. 2996.679789943 XBC 1577.80000000
.000030 0 562 253743 6 703617802806. 3087.678570604 XMC
1646.00000000 .000100 0 563 259169 6 734031020802.
3221.140573832 XBC 1652.00000000 .000100 0 564 259641 6
736707070392. 3232.883853976 XMC 1672.00000000 .001000 1 565
261168 6 745625518644. 3272.020586221 XBC 1699.00000000
.000100 1 566 263269 6 757666623302. 3324.860438582 XBC
1775.10000000 .000100 1 567 269102 6 791603185038.
3473.783893921 XLC 1778.40000000 .000100 0 568 269391 6
793074856386. 3480.242013750 XBC 1814.00000000 .000100 0 569
272074 6 808950696086. 3549.909793383 XBC 1822.00000000
.000100 1 570 272635 6 812518380146. 3565.565823653 XBC
1830.00000000 .000700 1 571 273233 6 816086462604.
3581.223602208 XBC 1856.60000000 .000100 0 572 275250 6
827948294010. 3633.276751651 XMC 1861.00000000 .000100 0 573
275576 6 829909948182. 3641.885057930 XMC 1863.00000000
.000100 0 574 275724 6 830801778498. 3645.798667484 XMC
2018.00000000 .000300 0 575 286965 6 899924689506.
3949.130007603 XMC 2019.00000000 .000100 0 576 287036 6
900370086048. 3951.084536543 XBC 2097.00000000 .001000 0 577
292528 6 935154837278. 4103.730092878 XBC 2123.00000000
.000100 0 578 294336 6 946748973228. 4154.608517180 XBC
2193.00000000 .000070 0 579 299149 6 977965506860.
4291.595701927 XMC 2425.00000000 .000100 0 580 314575 6
1081425341636. 4745.607401861 XMC 2452.20000000 .000100 0 581
316334 6 1093554786834. 4798.834918081 SBC 2453.00000000
.000375 1 582 316353 6 1093912251654. 4800.403577174 XMC
2466.00000000 .000100 0 583 317223 6 1099708896306.
4825.840931661 XMC 2469.00000000 .000100 1 584 317383 6
1101046863704. 4831.712319858 XMC 2535.30000000 .001000 1 585
321617 6 1130613391782. 4961.458802664 XMC 2535.90000000
.000100 1 586 321655 6 1130880758654. 4962.632085884 XMC
2976.30000000 .000070 0 587 348503 6 1327276650288.
5824.474102295 XMC 2980.40000000 .000100 0 588 348743 6
1329105096486. 5832.497853429 XMC 3096.87000000 .000500 0 589
355492 6 1381045373438. 6060.426822048 XMC 3097.50000000
.000100 0 590 355528 6 1381326018368. 6061.658373230 XMC
3505.00000000 .002000 0 591 378192 6 1563051098226.
6859.120621315 XMC 3509.40000000 .000100 0 592 378429 6
1565011496892. 6867.723418071 XMC 3553.40000000 .000300 0 593
380794 6 1584634070906. 6953.832952311 XMC 3557.00000000
.000500 0 594 380987 6 1586238816702. 6960.875041334 XMC
3764.00000000 .000100 0 595 391916 6 1678550571576.
7365.965740010 XMC 4417.00000000 .000100 0 596 424553 6
1969755477216. 8643.857150963 XMC 5279.10000000 .000100 0 597
464139 6 2354207012814. 10330.941763046 XMC 5280.60000000
.000035 0 598 464205 6 2354876005782. 10333.877497820 XMC
5368.60000000 .000500 0 599 468057 6 2394120230862.
10506.092516138 XMC 9872.80000000 .000200 0 600 634729 6
4402761234374. 19320.590611335 XMC10355.30000000 .001000 0 601
650054 6 4617930659940. 20264.816328365 XMCType 7 Polyhedron
III 4 VI 4 493.63600000 .000035 1 602 125330 7 220136158450.
966.021178473 XMC 493.69100000 .000100 1 603 125337 7
220160723914. 966.128978839 XMC 762.30000000 .000035 0 604
155826 7 339946573432. 1491.783956788 XMC 763.70000000 .000130
0 605 155969 7 340570789024. 1494.523195475 XMC 770.00000000
.000100 0 606 156611 7 343380267052. 1506.851998225 XMC
782.40000000 .000200 0 607 157867 7 348910065788.
1531.118355598 XMC 895.90000000 .000100 0 608 168930 7
399525193624. 1753.232186351 XMC 958.20000000 .000700 0 609
174705 7 427308164224. 1875.151902719 XMC 971.10000000 .000100
0 610 175877 7 433060530088. 1900.394949068 XMC 973.00000000
.000035 0 611 176049 7 433907970304. 1904.113762015 XMC
979.00000000 .000100 0 612 176591 7 436583810212.
1915.856122014 XMC 997.00000000 .000100 0 613 178207 7
444610782788. 1951.080800968 XMC 1019.38000000 .000100 0 614
180196 7 454590900572. 1994.876446403 XMC 1115.65000000
.000100 0 615 188513 7 497522755552. 2183.273852054 XBC
1234.35000000 .000500 0 616 198288 7 550456609252.
2415.562923020 XBC 1282.80000000 .001000 0 617 202142 7
572062264288. 2510.374790759 XMC 1315.20000000 .001000 0 619
204679 7 586511768084. 2573.783395613 XBC 1317.00000000
.000100 0 620 204819 7 587314386124. 2577.305515879 XMC
1387.00000000 .001000 1 621 210153 7 618530789098.
2714.292127260 XBC 1420.10000000 .000100 0 622 212685 7
633291706744. 2779.067306223 XMC 1519.70000000 .000150 0 623
220017 7 677707804288. 2973.977997836 XBC 1533.60000000
.000500 0 624 221021 7 683907048472. 3001.182075596 XBC
1534.50000000 .000100 1 625 221049 7 684308219530.
3002.942530312 XBC 1553.00000000 .000500 1 626 222378 7
692558640898. 3039.147767239 XBC 1600.00000000 .000100 0 627
225755 7 713517640924. 3131.121925629 XBC 1665.00000000
.000200 0 628 230295 7 742504242484. 3258.323523023 XBC
1673.30000000 .000500 1 629 230833 7 746205345338.
3274.565033576 XBC 1680.00000000 .000200 0 630 231330 7
749193203224. 3287.676618772 XBC 1685.00000000 .000100 0 631
231674 7 751423035304. 3297.461767331 XBC 1691.10000000
.000070 1 632 232058 7 754143943138. 3309.401898434 XBC
1694.00000000 .000200 0 633 232292 7 755437277788.
3315.077425215 XMC 1700.00000000 .000100 1 634 232668 7
758112711718. 3326.818003665 XBC 1713.00000000 .000100 0 635
233591 7 763909844212. 3352.257498944 XMC 1767.00000000
.000200 1 636 237210 7 787991377570. 3457.934237367 XMC
1863.80000000 .000100 0 637 243656 7 831158859892.
3647.365643992 XMC 1913.00000000 .000300 0 638 246851 7
853099282732. 3743.646569749 XBC 1940.00000000 .000070 0 639
248587 7 865140432188. 3796.486618697 XMC 1956.00000000
.000100 0 640 249610 7 872275623944. 3827.797905299 XBC
1973.60000000 .000100 1 641 250698 7 880124209858.
3862.239771948 XMC 1975.00000000 .001000 1 642 250787 7
880749061414. 3864.981801430 XMC 2008.00000000 .000100 1 643
252874 7 895465057730. 3929.559852595 XMC 2112.40000000
.000035 1 644 259366 7 942021617738. 4133.863513026 SMC
2115.00000000 .000200 0 645 259557 7 943181341288.
4138.952715628 XBC 2147.00000000 .000100 1 646 261482 7
957451248514. 4201.573198752 XMC 2299.00000000 .000100 0 647
270612 7 1025235752188. 4499.031220108 XBC 3513.00000000
.000200 0 648 334516 7 1566618042812. 6874.773406559 XMC
3526.14000000 .001000 0 649 335141 7 1572477550312.
6900.486621418 SMC 3553.00000000 .000004 0 650 336415 7
1584455440964. 6953.049072424 XMC 5278.00000000 .000700 0 651
410027 7 2353715710588. 10328.785786678 XMC 5278.20000000
.000100 1 652 410015 7 2353805822590. 10329.181224224
XMC10232.30000000 .000500 0 653 570906 7 4563079244392.
20024.112441000 XMCType 8 Polyhedron III 32 IV 6 549.30000000
.000500 0 654 103362 8 244959312984. 1074.952365267 XMC
767.00000000 .000300 0 655 122139 8 342042273786.
1500.980496511 XMC 782.50000000 .000100 0 656 123366 8
348949186002. 1531.290026420 XMC 783.30000000 .000100 0 657
123430 8 349311409628. 1532.879568531 XMC 784.10000000 .000350
0 658 123493 8 349668272810. 1534.445587462 XMC 886.60000000
.000900 1 659 131279 8 395377477026. 1735.030805423 XMC
897.60000000 .000100 0 660 132129 8 400283555022.
1756.560095662 XMC 1021.00000000 .000100 0 661 140919 8
455313592092. 1998.047825965 XMC 1115.74000000 .000100 0 662
147312 8 497562563268. 2183.448539833 XBC 1189.38000000
.000200 1 663 152063 8 530401996866. 2327.557479355 XBC
1197.24000000 .000035 1 664 152565 8 533907654108.
2342.941317994 XBC 1233.10000000 .000070 0 665 154866 8
549899204994. 2413.116871803 XBC 1273.30000000 .000350 0 666
157370 8 567825980478. 2491.784751270 XMC 1279.00000000
.000300 0 667 157722 8 570368575908. 2502.942395934 XMC
1378.00000000 .000100 1 668 163682 8 614517019392.
2696.678544064 XBC 1384.10000000 .000350 1 669 164044 8
617237149790. 2708.615263552 XBC 1419.00000000 .000700 0 670
166131 8 632808684348. 2776.947664146 XMC 1421.00000000
.000150 0 671 166247 8 633692756334. 2780.827227902 XMC
1423.00000000 .000004 0 672 166364 8 634584919860.
2784.742299678 XMC 1428.00000000 .000100 0 673 166656 8
636814330596. 2794.525599258 XMC 1517.30000000 .000700 0 674
171788 8 676638322890. 2969.284803916 XBC 1529.00000000
.000100 0 675 172449 8 681855160866. 2992.177798892 XMC
1532.30000000 .000500 0 676 172635 8 683327056950.
2998.636904928 XBC 1547.00000000 .000700 0 677 173461 8
689881896800. 3027.401439392 XBC 1638.00000000 .000500 0 678
178490 8 730463995014. 3205.487432251 XMC 1671.43000000
.000100 1 679 180273 8 745358659296. 3270.849530163 XBC
1674.20000000 .000100 1 680 180424 8 746607061664.
3276.327881086 XBC 1690.00000000 .000100 0 681 181301 8
753653124876. 3307.248045840 XBC 1694.00000000 .000100 1 682
181488 8 755436568038. 3315.074310626 XBC 1771.00000000
.000100 0 683 185595 8 789775123422. 3465.761830444 XMC
1783.00000000 .000100 1 684 186196 8 795126260018.
3489.244166636 XLC 2079.00000000 .000500 0 685 201087 8
927126601704. 4068.499871524 XMC 2268.00000000 .000035 1 686
210005 8 1011411306990. 4438.365553295 XBC 2452.40000000
.000150 0 687 218400 8 1093644232242. 4799.227430420 SBC
2464.40000000 .000100 1 688 218911 8 1098995719826.
4822.711306848 XBC 2469.00000000 .000300 0 689 219138 8
1101047603262. 4831.715565253 XBC 2472.00000000 .000100 0 690
219271 8 1102385070008. 4837.584756445 XBC 2515.90000000
.000100 1 691 221187 8 1121961942798. 4923.493740487 SBC
2568.60000000 .000100 1 692 223492 8 1145463584756.
5026.625747605 XMC 2647.40000000 .000100 1 693 226895 8
1180604313114. 5180.833434611 SBC 2710.00000000 .000100 0 694
229584 8 1208520197340. 5303.336414439 SBC 2790.00000000
.000100 0 695 232948 8 1244196358160. 5459.893734061 SBC
2814.90000000 .000200 1 696 233964 8 1255300742784.
5508.622987793 SBX 2974.40000000 .000100 0 697 240523 8
1326429634652. 5820.757152528 XMC 3414.80000000 .000350 0 698
257715 8 1522825187364. 6682.597681651 XMC 3686.02000000
.000100 0 699 267754 8 1643775686498. 7213.363479205 XMC
5279.60000000 .000700 0 700 320448 8 2354429684820.
10331.918912257 XMCType 9 Polyhedron III 8 IV 18 493.64000000
.000100 1 701 94196 9 220138012266. 966.029313554 XMC
493.67000000 .000500 1 702 94199 9 220151233218. 966.087330918
XMC 759.00000000 .000500 0 703 116862 9 338474610204.
1485.324556105 XMC 773.50000000 .000035 0 704 117973 9
344941208432. 1513.701860792 XMC 775.90000000 .000004 0 705
118156 9 346011528734. 1518.398736066 XMC 776.40000000 .000035
0 706 118194 9 346234384194. 1519.376690326 XMC 781.80000000
.001300 0 707 118604 9 348641867256. 1529.941422813 XMC
783.50000000 .000070 0 708 118733 9 349400776782.
1533.271737469 XMC 897.90000000 .000400 0 709 127106 9
400416897708. 1757.145241963 XMC 1019.50400000 .000080 0 710
135440 9 454646293272. 1995.119525603 XMC 1020.30000000
.000430 0 711 135495 9 455016816936. 1996.745490684 XMC
1189.61000000 .000070 1 712 146272 9 530505070550.
2328.009796514 XBC 1192.55000000 .000100 0 713 146484 9
531815963904. 2333.762375970 SBC 1310.00000000 .000500 1 714
153498 9 584192233254. 2563.604605424 XMC 1321.30000000
.000070 1 715 154159 9 589231742966. 2585.719432652 XBC
1387.60000000 .001100 1 716 157981 9 618798947174.
2715.468882512 XBC 1398.00000000 .000100 0 717 158601 9
623436044232. 2735.817806543 XMC 1412.00000000 .000100 0 718
159393 9 629679433692. 2763.215606551 XMC 1459.00000000
.001000 0 719 162024 9 650638417056. 2855.189691824 XMC
1524.00000000 .000150 0 720 165594 9 679625530842.
2982.393537013 XBC 1527.00000000 .000004 0 721 165757 9
680963505416. 2988.264956701 XBC 1542.00000000 .000100 0 722
166569 9 687652733850. 3017.619227169 XBC 1544.70000000
.000500 0 723 166715 9 688856978574. 3022.903801569 XMC
1583.00000000 3.5D-06 0 724 168769 9 705936358070.
3097.853062174 XBC 1672.10000000 .000200 1 725 173427 9
745670819562. 3272.219379760 XBC 1675.00000000 .000100 0 726
173604 9 746964126660. 3277.894785635 XBC 1712.80000000
.000100 0 727 175552 9 763820363276. 3351.864830698 XBC
1864.70000000 .000150 0 728 183171 9 831560366640.
3649.127571813 XMC 1869.00000000 .000700 0 729 183382 9
833478453350. 3657.544691459 XMC 1872.00000000 .000100 0 730
183529 9 834816087962. 3663.414619295 XBC 1874.00000000
.000100 1 731 183602 9 835707239394. 3667.325249709 XMC
1967.00000000 .000250 1 732 188104 9 877180692104.
3849.322764086 XMC 1969.30000000 .000100 1 733 188214 9
878206590456. 3853.824702985 XMC 1972.40000000 .001000 1 734
188362 9 879588267866. 3859.887903367 XMC 2110.30000000
.000035 1 735 194837 9 941085154596. 4129.754041713 SMC
2121.00000000 .000100 0 736 195354 9 945857106414.
4150.694747462 XBC 2127.00000000 .000200 0 737 195630 9
948532715034. 4162.436092503 XBC 2284.70000000 .000100 1 738
202730 9 1018858875798. 4471.047640815 XBC 2625.60000000
.001470 1 739 217332 9 1170881396010. 5138.166460202 SBC
2765.00000000 .000100 1 740 223028 9 1233047623824.
5410.969860956 SBC 2790.00000000 .000300 1 741 224034 9
1244195843262. 5459.891474539 SBC 2980.00000000 .000300 0 742
231558 9 1328926553988. 5831.714357271 XMC 5279.50000000
.000100 0 743 308211 9 2354385489936. 10331.724972315 XMC
9459.97000000 .000100 0 744 412569 9 4218659786898.
18512.700174337 XMC10868.00000000 .000100 0 745 442208 9
4846569436392. 21268.149455589 XMCType 10 Polyhedron IV 6 VI 8
497.74200000 .000900 0 746 84716 10 221966978492. 974.055346721
XMC 552.00000000 .000200 0 747 89214 10 246163420466.
1080.236337410 XMC 763.00000000 .000100 1 748 104853 10
340258874720. 1493.154425230 XMC 783.40000000 .000200 0 749
106281 10 349355906942. 1533.074835624 XMC 890.00000000
.000100 1 750 113249 10 396894340322. 1741.687240600 XMC
984.45000000 .000500 1 751 119110 10 439014413426.
1926.522312420 XMC 1019.90000000 .000140 0 752 121267 10
454822862528. 1995.894362604 XMC 1115.39000000 .000100 0 753
126817 10 497406524420. 2182.763796205 XBC 1115.69000000
.000150 0 754 126834 10 497540170826. 2183.350275316 XBC
1239.00000000 .000300 1 755 133632 10 552530610434.
2424.664240496 XMC 1261.00000000 .000500 0 756 134841 10
562341518036. 2467.717342677 XMC 1283.00000000 .000300 0 757
136012 10 572151688412. 2510.767209697 XMC 1284.00000000
.000440 0 758 136065 10 572597514086. 2512.723621793 XMC
1287.00000000 .000100 0 759 136224 10 573936032888.
2518.597429710 XMC 1306.00000000 .000300 1 760 137199 10
582409134230. 2555.779850814 XMC 1311.30000000 .000100 0 761
137504 10 584772409766. 2566.150587882 XMC 1313.40000000
.000500 0 762 137614 10 585708376910. 2570.257882612 XBC
1375.00000000 .000070 1 763 140778 10 613179308462.
2690.808281322 XBC 1416.00000000 .000700 0 764 142888 10
631462728776. 2771.041221528 XMC 1423.00000000 .000130 1 765
143215 10 634584476984. 2784.740356210 XMC 1430.00000000
.000100 1 766 143567 10 637706209202. 2798.439420722 XMC
1496.00000000 .000200 0 767 146869 10 667139293010.
2927.600311447 XMC 1692.00000000 .001000 0 768 156194 10
754544870408. 3311.161283337 XBC 1787.00000000 .000100 0 769
160519 10 796909824092. 3497.070962019 XBC 1817.00000000
.000500 1 770 161838 10 810288007052. 3555.778300967 XBC
1870.00000000 3.5D-06 1 771 164182 10 833923476290.
3659.497580925 XMC 2106.60000000 .000004 1 772 174262 10
939435327158. 4122.514121397 XMC 2214.00000000 .000035 1 773
178650 10 987330323276. 4332.691226869 XBC 2452.50000000
.000150 0 774 188048 10 1093688397074. 4799.421238485 SBC
2465.10000000 .000100 1 775 188511 10 1099307810804.
4824.080852390 XBC 2471.00000000 .000100 0 776 188756 10
1101938753564. 4835.626190709 XBC 2500.00000000 .000100 1 777
189841 10 1114871439152. 4892.378558246 XBC 2626.40000000
.000200 1 778 194582 10 1171239193568. 5139.736579446 SBC
3414.10000000 .001000 0 779 221872 10 1522511963810.
6681.223166038 XMC 3683.00000000 .000100 0 780 230444 10
1642428435404. 7207.451351463 XMC 5278.60000000 .000100 1 781
275869 10 2353984267160. 10329.964290646 XMC 5369.90000000
.000500 0 782 278258 10 2394698514416. 10508.630191749 XMC
5383.30000000 .000800 0 783 278605 10 2400675355166.
10534.858298869 XMCType 77 Two Structure Types Within Ratio
Range 893.00000000 .000100 1 784 446096 72 398232053850.
1747.557514742 XMC 893.00000000 .000100 1 785 190840 76
398232053420. 1747.557512856 XMC 957.40000000 .000100 0 786
174632 77 426951140788. 1873.585180546 XMC 957.40000000
.000100 0 787 326707 73 426951162224. 1873.585274614
XMC10231.20000000 .001000 0 788 1510395 72 4562589132840.
20021.961689654 XMC10231.20000000 .001000 0 789 646147 76
4562589050198. 20021.961326997 XMC10268.50000000 .000500 0 790
812991 74 4579222648626. 20094.954371251 XMC10268.50000000
.000500 0 791 571915 77 4579222747964. 20094.954807175 XMCType
88 Many Structure Types Within Ratio Range 1445.00000000
.000100 0 792 161245 89 644395584878. 2827.794337330 XMC
1445.00000000 .000100 0 793 242830 86 644395672838.
2827.794723324 XMC 1445.00000000 .000100 0 794 253849 85
644395686500. 2827.794783277 XMC 2040.00000000 .000100 0 795
288525 86 909735059976. 3992.180752694 XBC 2040.00000000
.000100 0 796 199192 88 909735068942. 3992.180792039 XBC
2040.00000000 .000100 0 797 171506 80 909735146270.
3992.181131377 XBC 2976.60000000 .000100 0 798 240612 88
1327410335442. 5825.060751444 XMC 2976.60000000 .000100 0 799
576066 83 1327410449688. 5825.061252788 XMC 2976.60000000
.000100 0 800 307920 87 1327410480484. 5825.061387930 XMC
2976.60000000 .000100 0 801 364336 85 1327410852320.
5825.063019654 XMCType 99 Zero Structure Types Within Ratio
Range 938.27199800 .000038 1 802 LO 99 418421016631.
1836.152526827 XBC 938.27199800 .000038 1 803 HI 99
418421084914. 1836.152826472 XBC Type Total 01 01 02 204 03
140 04 90 05 80 06 86 07 52 08 47 09 45 10 38 783 No. Matched
77 04 88 03 99 01 791 Grand Total===
===
Subject:
: THATS IT!!Like
so: http://osj.cjb.net ? That's the usual notation for a
germanright angle.Oswald===
===
Subject:
: Theory DevelopmentNo
theoretical system can survive without being aware to
itslimitations.It means that anyx outputcan be only
amodel([b:4857bd9020][i:4857bd9020]x[/i:4857bd9020][/b:
4857bd9020])input.Shortly
speaking,[b:4857bd9020][i:4857bd9020]x[/i:4857bd9020][/b:
4857bd9020]=model([b:4857bd9020][i:4857bd9020]x[/i:4857bd9020]
[/b:4857bd9020]).Math is first of all a form of theory,
therefore any concept that canbe used by it is only a
model([b:4857bd9020]concept[/b:4857bd9020]).For example, let
us take infinity concept.If
[b:4857bd9020][i:4857bd9020]INF[/i:4857bd9020][/b:4857bd9020]
isinfinity itself (= actual infinity) ,
then[b:4857bd9020][i:4857bd9020]inf[/i:4857bd9020][/b:
4857bd9020]=model([b:4857bd9020][i:4857bd9020]INF[/i:
4857bd9020][/b:4857bd9020])=potentialinfinity.Please look at
this model for better
understanding:http://www.geocities.com/complementarytheory/
RiemannsLimits.pdfIn this way we first of all aware to our
input limitations, whichare:No input
=model([b:4857bd9020][i:4857bd9020]EMPTINESS[/i:4857bd9020][/b
:4857bd9020])= lowest limit.No input
=model([b:4857bd9020][i:4857bd9020]FULLNESS[/i:4857bd9020][/b:
4857bd9020])= highest limit.If we translate this to set's
representation then:{}
=model([b:4857bd9020][i:4857bd9020]EMPTINESS[/i:4857bd9020][/b
:4857bd9020]) = lowest limit.{__}
=model([b:4857bd9020][i:4857bd9020]FULLNESS[/i:4857bd9020][/b:
4857bd9020])= highest limit.Between these limits we can
find[b:4857bd9020][i:4857bd9020]inf[/i:4857bd9020][/b:
4857bd9020]=model([b:4857bd9020][i:4857bd9020]INF[/i:
4857bd9020][/b:4857bd9020])=potentialinfinity,
where[b:4857bd9020][i:4857bd9020]inf[/i:4857bd9020][/b:
4857bd9020] has twoinput forms:{.} = singleton, which is a
localized element.{.__.} = non-singleton, which is a
non-localized element (connect atleast two different
singletons).{.} and {._.} can appear in two basic
collections:Collection{[b:4857bd9020][i:4857bd9020]a[/i:
4857bd9020][/b:4857bd9020],[b:4857bd9020][i:4857bd9020]b[/i:
4857bd9020][/b:4857bd9020],[b:4857bd9020][i:4857bd9020]c[/i:
4857bd9020][/b:4857bd9020]} isfinitely many
elements.Collection{[b:4857bd9020][i:4857bd9020]a[/i:
4857bd9020][/b:4857bd9020],[b:4857bd9020][i:4857bd9020]b[/i:
4857bd9020][/b:4857bd9020],[b:4857bd9020][i:4857bd9020]c[/i:
4857bd9020][/b:4857bd9020], ...} isinfinitely many
elements.Any non-empty collection which is not a singleton, is
an associationbetween {.} and {._.}, for example: b b {a , a} .
. | | |___|_ | {a , b} . . | | |___| | For more details please
look
at:http://www.geocities.com/complementarytheory/CATpage.hI'll
be glad to get your remarks and insights.Thank you.Doron----==
Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet
News==----http://www.newsfeed.com The #1 Newsgroup Service in
the World! >100,000 Newsgroups---= 19 East/West-Coast
Specialized Servers - Total Privacy via Encryption
=---===
===
Subject:
: Re: Transcendental or algebraic?> If x^x=2
then x is transcendental or algebraic?<snip>x is algebraic and
irrationalLook up Hilbert's Seventh Problem. x^x would end up
transcendentalrather than 2.Negative. Gelfond-Schneider says
that is ais RATIONAL and b is algebraic irrational thena^b is
transcendental. It says nothing aboutwhen a is irrational.You
can lead a horse's ass to knowledge, but you can't make him
think.===
===
Subject:
: Re: Transcendental or algebraic?> Negative.
Gelfond-Schneider says that is a> is RATIONAL and b is
algebraic irrational then> a^b is transcendental. It says
nothing about> when a is
irrational.http://mathworld.wolfram.com/GelfondsTheorem.hhttp:
//www.math.sdu.edu.cn/mathency/math/g/g104.htmhttp://
en.wikipedia.org/wiki/Hilbert%27s_seventh_problemhttp://
planetmath.org/encyclopedia/3952.hWhere, may I ask, did you
find it limited to rational a?-- iel W.
sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039
53 36 N / 086 11 55 W===
===
Subject:
: Re: Volume of Astroid
Help!!>The problem is to get volume of solid formed by
rotation of x^2/3 +>y^2/3=a^2/3 around the x-axis and y-axis
with a>0.>1) I have never heard of an astroid or hypocycloid,
so had to do some>digging.>2) I found that x^2/3 + y^2/3=a^2/3
is the formula for an astroid (a four>sided diamond shaped
thing with rounded sides.The points of the diamond>fall on the
axes. It is formed from rotating a circle within an circle.>3)
I also found some other formulas : x=a cos ^3 theta and y= a
sin ^3>theta.>4) That's all I got. I was flying through disks
and washers and shells, but>when it came to this I hit a wall
because 1) I had never heard of an astroid>or the shape of the
graph and 2) this formula has three variables (x,y,a)>vice the
normal x and y functions.>5) Any help on how to work this
appreciated!The formula for disks seems easiest to apply: | a
2 | pi y dx | -aWhere y^2 = ( a^(2/3) - x^(2/3) )^3. Using the
binomial theorem toexpand y^2 in terms of x leaves a pretty
simple integral.Rob son <rob@trash.whim.org>take out the trash
before replying===
===
Subject:
: Re: Volume of Astroid Help!!> The
problem is to get volume of solid formed by rotation of x^2/3
+> y^2/3=a^2/3 around the x-axis and y-axis with a>0.1) I have
never heard of an astroid or hypocycloid, so had to do some>
digging.2) I found that x^2/3 + y^2/3=a^2/3 is the formula for
an astroid (a four> sided diamond shaped thing with rounded
sides.The points of the diamond> fall on the axes. It is
formed from rotating a circle within an circle.3) I also found
some other formulas : x=a cos ^3 theta and y= a sin ^3>
theta.4) That's all I got. I was flying through disks and
washers and shells, but> when it came to this I hit a wall
because 1) I had never heard of an astroid> or the shape of
the graph and 2) this formula has three variables (x,y,a)>
vice the normal x and y functions.Surely you could handle the
circle x^2 + y^2 = a^2; it also has 3 variables. The idea is
to fix a and then consider the curve you get, just like in the
case of the circle. Having done that, just treat it like any
other volume of rotation problem. Solve for y as a function of
x (which you can do for y >= 0), then for the rotation about
the x-axis calculate the integral of Pi*y(x)^2 dx over the
appropriate interval. (Hint for rotating about the y-axis: By
symmetry, there's nothing left to do.)===
===
Subject:
: Re: Volume
of Astroid Help!!> 2) this formula has three variables (x,y,a)
vice the normal x and yfunctions.In this case, the a is a
constant, and your astroid has just
twovariables.Doug===
===
Subject:
: Re: Volume of Astroid Help!!1)
You are correct and I am mistaken about my notation.It should
be (as you suggested) x raised to the 2/3 power plus y raised
tothe 2/3 power equals a raised to the 2/3 power2) I was also
not clear about the axis issue (English is my 2nd language,and
I have only been in America for 5 years, so I do get the syntax
wrongand get corrected often.) The problem asks for you to a)
rotate the equationaround the x -axis and then part 2 of the
question is rotate around they-axis.Hopefully, this
clarification will make it easier for help to be rendered.I
apologize for causing all the confusion and take all your
comments aboutimproving my question to heart.> The problem is
to get volume of solid formed by rotation of x^2/3 +>
y^2/3=a^2/3 around the x-axis and y-axis with a>0.> Do you
mean (x^2)/3 + (y^2)/3=(a^2)/3, which is what your notation>
suggests, or (x^2)^(1/3) + (y^2)^(1/3)=(a^2)^(1/3), which is
what your> description suggests?> And what do you mean by
rotating it about _both_ axes? Do you mean> around one axis OR
the other, or something like the solid bounded by> the 3D
surface |x|^(2/3) + |y|^(2/3) + |z|^(2/3) = |a|^(2/3) ?> 1) I
have never heard of an astroid or hypocycloid, so had to do
some> digging.2) I found that x^2/3 + y^2/3=a^2/3 is the
formula for an astroid (afour> sided diamond shaped thing with
rounded sides.The points of thediamond> fall on the axes. It is
formed from rotating a circle within an circle.3) I also found
some other formulas : x=a cos ^3 theta and y= a sin ^3>
theta.4) That's all I got. I was flying through disks and
washers and shells,but> when it came to this I hit a wall
because 1) I had never heard of anastroid> or the shape of the
graph and 2) this formula has three variables(x,y,a)> vice the
normal x and y functions.>5) Any help on how to work this
appreciated!>>===
===
Subject:
: A series for the inverse sine
cardinal functionThe sine cardinal function, ( 1 if x = 0
sinc(x) = ( ( sin(x)/x otherwise,arises in several
applications, and so its inverse should also be ofinterest.
This note gives a series expansion for the inverse of
sinc(x),0 <= x <= x0, where x0 (approx. 4.4934) denotes the
positive value of xat which sinc(x) reaches its absolute
minimum.[Besides posting this to sci.math, I have also posted
it tosci.math.num-analysis and comp.dsp in the hope that some
people might beable to give references to previous
mathematical treatments of the inversesinc function. In
particular, I'd be interested to know if the series givenhere
is already known.]With f(x) = 2*x + 3*x^3/10 + 321*x^5/2800 +
3197*x^7/56000 + 445617*x^9/13798400 +
1766784699*x^11/89689600000 + 317184685563*x^13/25113088000000
+ 14328608561991*x^15/1707689984000000 +
6670995251837391*x^17/1165411287040000000 +
910588298588385889*x^19/228420612259840000000 + ...,it can be
shown that the desired inverse, abbreviated as Asinc here,
isgiven by Asinc(x) = Sqrt(3/2) * f(Sqrt(1 -
x))-----------------------------------------------------------
------------A simple example:Find the positive root of the
equation sin(x) = x/2.Rewriting the equation as sinc(x) =
1/2,the solution is x = Asinc(1/2) = Sqrt(3/2) *
f(Sqrt(1/2)).Using just those terms shown above in the series
for f, we getx = 1.8954905... (For comparison, solving by a
different method, a muchmore accurate approximation is x =
1.8954942668788...)-------------------------------------------
----------------------------The radius of convergence of the
above series for f is slightly larger than1.1 . As expected,
when x is near that value, convergence is very slow.The series
for f was obtained by reversion of series, etc. To be
morespecific, for anyone interested, in Mathematica the series
may be obtainedusingSimplify[Normal[InverseSeries[1 -
Series[Sin[x]/x, {x, 0, n}]]/Sqrt[3/2]/. x -> x^2], x >
0]where n (even) should be chosen to be 1 more than the
highest power of xdesired in the result.Unfortunately, I do
not know a nice formula for the coefficients in theseries for
f.Thoughtful comments will be appreciated.l===
===
Subject:
: Re: A
series for the inverse sine cardinal functionhere it is in
Maple...> z := u*sqrt(3/2); z := 1/2*6^(1/2)*u> AA :=
sin(z)/z=1-y^2; AA := 1/3*sin(1/2*6^(1/2)*u)*6^(1/2)/u =
1-y^2> AA1 := solve(AA,u); AA1 :=
-1/3*sin(RootOf(-_Z+_Z*y^2+sin(_Z)))*6^(1/2)/(-1+y^2)>
series(AA1,y,10);
2*y+3/10*y^3+321/2800*y^5+3197/56000*y^7+445617/13798400*y^9+O
(y^11)> series(AA1,y,50);
2*y+3/10*y^3+321/2800*y^5+3197/56000*y^7+445617/13798400*y^9+
1766784699/89689600000*y^11+317184685563/25113088000000*y^13+
14328608561991/1707689984000000*y^15+6670995251837391/
1165411287040000000*y^17+910588298588385889/
228420612259840000000*y^19+1889106915501879285127263/
669318078043783168000000000*y^21+122684251268939994619239/
60571771768668160000000000*y^23+86578199631805319180104483967/
58899990867852918784000000000000*y^25+
36790913563978761395277930686421/
34161994703354692894720000000000000*y^27+
295479400033606079171291233070109663/
371437974410411888261201920000000000000*y^29+
5429197579977012936689051219262425180781/
9174517967937173640051687424000000000000000*y^31+
111912180845235829957717919886518013347855667/
252655431286439247725093999083520000000000000000*y^33+
21623582556767547163123245489615552651038372109/
64865414807824606864932296091238400000000000000000*y^35+
1686950689722579328034933293949678511289600201851/
6691379632807169971329857912569856000000000000000000*y^37+
362964877894310955248925785551248746981943835416147/
1895485357802467420969439750506151936000000000000000000*y^39+
241119375769652087142546687133690376324455849576103305306113/
165132849541924608926394092646417466709245952000000000000000000
0*y^41+
62733744440681157939079200023293588467527441653821853933262223
/
561451688442543670349739914997819386811436236800000000000000000
000*y^43+
905410732614926003422654073745394071490598621469546605760977273
43/
105552917427198210025751104019590044720550012518400000000000000
0000000*y^45+
126749063504538759034999649107808731423757688715140874157349767
781/
191914395322178563683183825490163717673727295488000000000000000
0000000*y^47+
249693053510060031661928074632553530869877798468407353253082594
529911/
489709179353451023617580383462989683786482353963008000000000000
0000000000*y^49+O(y^50)-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: Sine
without approximating> Is it possible to solve:> sin x = x/2>
Without graphing, and without approximating??> If you mean a
formula for the positive solution,> x = ...formula in terms of
known constants...> Then... no.In other words, there is no
solution in closed form in terms of familiarfunctions.> See
Kepler's Equation...>
http://mathworld.wolfram.com/KeplersEquation.hYes, it is
related to Kepler's equation, E - e sin(E) = M. But in
itsclassic form, 0 <= e <= 1, as is appropriate for the
eccentricity of anelliptical orbit. The equation sin(E) = E/2
is instead equivalent tothe hyperbolic form of Kepler's
equation, with e = 2.Since we happen to have the special case
with M = 0, let me suggestanother way of solving the original
equation. See my recently postedsolution in terms of an
infinite series.l===
===
Subject:
: Re: Help to verify an open
mappingf: R^n{0} -> S^(n-1)> x |-> x/||x||> where ||.|| is the
euclidean norm.> How to prove, analytically, that f is an open
mapping?Any response would be appreciated.----== Posted via
Newsfeed.Com - Unlimited-Uncensored-Secure Usenet
News==----http://www.newsfeed.com The #1 Newsgroup Service in
the World! >100,000 Newsgroups---= 19 East/West-Coast
Specialized Servers - Total Privacy via Encryption
=---===
===
Subject:
: beginner wavelet question -- how to restore
the input?My sloppy self-studying wavelets hit the following
problem. Here is inputvector 1.0 2.0 3.0 4.0By applying Haar
wavelet matrix I got5.00 -2.0 -0.7 -0.7To doublecheck myself,
I applied Pyramid scheme and got0.70 0.70 2.0 5.00which is
somewhat consistent with previous result. Now, I want to
restore mysignal back, so I thought that I just have to apply
Haar or Pyramidetransformation twice. It doesn't work however,
as the result is different.Therefore, is my assumption that
Haar matrix is square root of unit matrixwrong? What is
inverse of Haar matrix, then?===
===
Subject:
: Korea by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1E3qoa26347;Hi Mr. Gilden,The army
has sent me to Korea. So far they have nothing for me to do so
I am just sight seeing. I am having a lot of fun and have
visited a traditional Korean village and a 600 year old fort
so far. The fort is not that old by Korean standards as they
trace their history back 5000 years. Korea is a great place.
The people are the friendliest I have met anywhere and there
are beutiful parks throughout the city. The only bad part is
that there are no emmissions controlls so the air is very
poluted.I am trying to learn the language , but the only word
I have truely mastered is the word for hello. Because of this
I say hello a lot. Hopefully I will be able to learn new words
soon.Dewey===
===
Subject:
: Re: Korea> Hi Mr. Gilden,> The army has
sent me to Korea. So far they have nothing for me to do so I
am > just sight seeing. I am having a lot of fun and have
visited a traditional > Korean village and a 600 year old fort
so far. The fort is not that old by > Korean standards as they
trace their history back 5000 years. > Korea is a great place.
The people are the friendliest I have met anywhere > and there
are beutiful parks throughout the city. The only bad part is
that > there are no emmissions controlls so the air is very
poluted.> I am trying to learn the language , but the only
word I have truely mastered > is the word for hello. Because
of this I say hello a lot. Hopefully I will be > able to learn
new words soon.embryos. Don't take any wooden won.===
===
Subject:
:
discontinuous linear functional existance? by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1E3qn526327;I've seen a
construction of a discont. linear fucntional f:R->R but
Iwondered do there exist discont. linear functions over say a
realinfinite dimensional real vector space?===
===
Subject:
: Re:
discontinuous linear functional
existance?Content-transfer-encoding: 8bit> I've seen a
construction of a discont. linear fucntional f:R->R but I>
wondered do there exist discont. linear functions over say a
real> infinite dimensional real vector space?> There are no
discontinuous linear functions from R to R, as vectorspaces
over R. This is because f(c) = f(c * 1) = c f(1) is
acontinuous function of c.There are discontinuous linear
functions from R to R, regarding R as avector space over Q,
the rationals. And for a similar reason thereexist
discontinuous linear functionals from an
infinite-dimensionalBanach space X to R: pick a Hamel basis
{u_a : a in A} of X over R;since this is infinite, it contains
particular basis elements u_1, u_2,... (countable in number).
WLOG we may assume ||u_i|| = 1.Now take f(u_i) = i, f(u_a) = 0
for all other u_a than u_1, ... This fextends uniquely to a
linear functional on X, but it isn't continuous,because
|f(u_i)| = i||u_i|| <= ||f|| ||u_i||.--Ron Bruck===
===
Subject:
:
Re: need help!!> For your example of -1 to 1, the integral of
arcsin(2/(3+cos(x))) is > about 1.097132985 .> Jhow you find
this answer[1.097132985]for [-1,1]? do you plot it? > if you
dont plot it please say me that ,how do you find this>
answer,and if you plot it ,please show me There are computer
programs that use numerical methods to evaluate > areas. I
used Maple, some people use Mathematica, there are probably
many > more. Perhaps there are even java applets available on
the web to do > these.> Or, if you only have that one function
to work on, you can write your > own program to implement one
of the many integration approximation > routines given in a
standard calculus book.> Or, you could write out a taylor
series for your function, and then > integrate them
term-by-term, and keep as many terms as you want (more > terms
-> more accuracy.) I did not plot the graph... but I
could.Jdear jim thank you for your help and i only have one
qestion, can you show methis plot? and can maple plot this
function?or mathematica? thank you hupo===
===
Subject:
: Re: need
help!!
<Pine.LNX.4.44.0402120052390.7468-100000@eva117.cs.ualberta.ca
> <54798887.0402120736.5ee1a279@posting.google.com>
<3ced55cd.0402121359.87e94c@posting.google.com>
<54798887.0402122110.326a16a4@posting.google.com>
<Pine.LNX.4.44.0402122355350.17517-100000@eva117.
cs.ualberta.ca>
<54798887.0402130232.1776a935@posting.google.com>
<Pine.LNX.4.44.0402131123580.22178-100000@eva117.
cs.ualberta.ca>
<54798887.0402132018.4ad321f7@posting.google.comdear jim >
thank you for your help and i only have one qestion, can you
show me> this plot? and can maple plot this function?or
mathematica? The graph of f(x) = arcsin(2/(3+cos(x))) can be
found at http://www.cs.ualberta.ca/~nastos/plots There are two
images, one is of f(x) for x from -1 to 1 and the other is on
the domain -8 to 8.J===
===
Subject:
: Re: need help!!dear jim >
thank you for your help and i only have one qestion, can you
show me> this plot? and can maple plot this function?or
mathematica? The graph of f(x) = arcsin(2/(3+cos(x))) can be
found at > http://www.cs.ualberta.ca/~nastos/plots There are
two images, one is of f(x) for x from -1 to 1 and the other is
> on the domain -8 to 8.Jdear jim thank you for your help but
you dont answer my second question,do youplot it whit
MAPLEthank you again hupo===
===
Subject:
: Re: need help!!
<3ced55cd.0402121359.87e94c@posting.google.com>
<54798887.0402122110.326a16a4@posting.google.com>
<Pine.LNX.4.44.0402122355350.17517-100000@eva117.
cs.ualberta.ca>
<54798887.0402130232.1776a935@posting.google.com>
<Pine.LNX.4.44.0402131123580.22178-100000@eva117.
cs.ualberta.ca>
<54798887.0402132018.4ad321f7@posting.google.com>
<Pine.LNX.4.44.0402140017020.27830-100000@eva117.
cs.ualberta.ca>
<54798887.0402140608.4f67bf94@posting.google.comthank you for
your help but you dont answer my second question,do you> plot
it whit MAPLE> thank you again Yes, I plotted it with Maple.
But I don't think that is important, as there are many
programs that will plot graphs and you might be able to find
java applets on the web to do the same thing.J===
===
Subject:
: Re:
please help> I need to prove that (1+X)^n >= (1+nX), n is a
natural #, X is a real > number>= -1.> Here is what I have
(and don't have). I know that (1+X)^n and (1+nX)> intersect at
(0,1). Since [ (1+X)^n ] ' > n for all X >0 and since (1 + >
nX)> ' = n for all X we know that (1+X)^n > (1+nX) ( but we
already knew this> from the binomial theorem).> The real
question is whether or not (1 + X) ^n and (1 + nX ) intersect
> below> X =0 ?> For n = 0 or 1, (1+x)^n = (1+n*x).> For
integers n > 1, let f(x) = (1+x)^n - (1+n*x), then show> f(0)
= 0, and > f'(x) = 0 at x = 0 (and only at x = 0), and >
f''(x) > 0 at x = 0.hi dear virgil > i think that i could > if
f(x)=(1+x)^n - (1+nx) > f'(x)=n(1+X)^[n-1] -n > f'(0)=n-n=0>
f(x)=n(n-1)(1+x)^[n-2] >f(0)=n(n-1) we know that n>1 at result
n(n-1)>0> do you thonk that it is true ?> thank you > hupoIt is
when n > 1, which I mentioned.hi then if n<1 at result f(0)<0
it is normale.do you have aproblem whit this? thank you
hupo===
===
Subject:
: Re: please help> I need to prove that (1+X)^n
>= (1+nX), n is a natural #, X is a real > number>= -1.> Here
is what I have (and don't have). I know that (1+X)^n and
(1+nX)> intersect at (0,1). Since [ (1+X)^n ] ' > n for all X
>0 and since (1 > + > nX)> ' = n for all X we know that
(1+X)^n > (1+nX) ( but we already knew > this> from the
binomial theorem).> The real question is whether or not (1 +
X) ^n and (1 + nX ) > intersect > below> X =0 ?> For n = 0
or 1, (1+x)^n = (1+n*x).> For integers n > 1, let f(x) =
(1+x)^n - (1+n*x), then show> f(0) = 0, and > f'(x) = 0 at x =
0 (and only at x = 0), and > f''(x) > 0 at x = 0.> hi dear
virgil > i think that i could > if f(x)=(1+x)^n - (1+nx) >
f'(x)=n(1+X)^[n-1] -n > f'(0)=n-n=0> f(x)=n(n-1)(1+x)^[n-2]
>f(0)=n(n-1) we know that n>1 at result n(n-1)>0> do you thonk
that it is true ?> thank you > hupoIt is when n > 1, which I
mentioned.hi > then if n<1 at result f(0)<0 it is normale.> do
you have aproblem whit this? > thank you > hupoIf n = 0 or n =
1 then f(x) = 0 for all x.===
===
Subject:
: Re: Problem of the Week
Trig Question>> We are told that the observer is at latitude
42 deg. N. and that the>> sun's rays make an angle of 12
degrees with the plane of the equator.>>> Are we supposed to
assume that the problem statement means the sun is 12>> degrees
*south* of the equator? It doesn't say that, and it surely
makes>> a difference.> Good question :) > Well, I initially
thought the suns rays would be beaming down parallel> (as in
North of the equator). But, I'm not certain, so I'm going to>
assume they are from the North and will try and work the
problem out> that way, still open for debate I suppose.The sun
is south of the equator this time of year, and I estimate that
12degrees south is probably about right. That's why I wondered
whether we weresupposed to assume that.I solved it both ways
using spherical trig, and it makes a difference in theanswer
if I have done it correctly. In particular, the time of
daylight isless than 12 hours if the sun is at 12 degrees
south, and is greater than 12hours if the sun is at 12 degrees
north.-- SeamanJudge Yohn's mistakes revealed in Mumia
Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>===
===
Subject:
: Re: help in complex analysis> this
is a question from my functions of a complex variable class:>
verify that the function > g(z)= ln(r) + i*t (r>0, 0<t<2PI)>
is analytic in the indicated domain with derivative g'(z)=1/z.
then> show that G(z)= g(z^2 + 1) is analytic in the first
quadrant (x,y>0)> with derivative:> G'(z)= (2z)/(z^2 + 1)There
is a hint that Im(z^2 + 1)>0 when x,y>0I really don't
understand this question so any help is greatly> appreciated.I
can certainly help you understand the question. You have
toundertand the meaning of the definition of g(z), namely that
g(z) =ln(r) + i*t. Remember that any complex number z can be
written as z =r*(e^(i*t)), where, when written as z = x + i*y,
we would have r =sqrt(x^2 + y^2) and t = tan(y/x), so we can
have r>=0 and t can berestricted to lie in any half-open
interval of length 2*pi. Now youhave to prove that the limit
as h goes to 0 in any way in the complexplane, that (g(z+h) -
g(z))/h goes to 1/z and you will have provedthat g(z) is
analytic with derivative 1/z. This is surely moreannoying than
difficult, even if you take the least imaginativeapproach and
represent h as, say u + i*v where u and v are both verylittle
and try to see what happens to the difference quotient with
1/zsubtracted from it. I expect that the second part of the
problem is apiece of cake after taking care of the first
part.You are being up front that this is a homework problem,
but if youcontinue to have difficulties after all the hints
that I expect youwill get, then ask
again.HTH,Achava===
===
Subject:
: Re: Oh, how I
wonder...===>
===
Subject:
: Re: Oh, how I wonder...>Message-id:
<9661bda2.0402130903.f8f7b9b@posting.google.com> It's a joke.
Imaginary means not real, where this use of real>> refers to
existence. The mathematical use of real doesn't refer to>>
existence just as the mathematical use of primary doesn't
refer to>> order. -1 happens to be a real number
mathematically. But the>> square root of -1 is not a real
number in the mathematical sense.>> So it's called imaginary
even though it does exist. Try explaining >> that to a bunch
of drunks at a party, it's a barrel of laughs.>Better yet,
explain to an AC Circuits student that the charge on
a>capacitor is imaginary, then watch him touch the two
leads...I was about to say nobody's that dumb but then
remembered that I graduated from DeVry, so I know better than
to say that.--MensanatorAce of Clubs===
===
Subject:
: Re: Zeta
function questionContent-transfer-encoding: 8bit> I posted
this earlier but it doesn't seem to have got there, so i will
post> it again.I am trying to understand the Zeta function.I
understand that it is defined as Zeta(s) = sum(1/n^s),
n=1..infinity.> Therefore, Zeta(2) = Pi^2/6, Zeta(4) =
Pi^4/90, and so on.> Also Zeta(1) = infinity. These make
sense.However, the books also tell me that when s = 0, Zeta(0)
= -1/2> Also when s = -2k for k = 1,2,3,...,then Zeta(-2k) =
0When i substitute s = 0 or s = -2k into the sum function
above, i get> divergence to infinity, not the specific values
above.Is the sum function for Zeta(s) only valid for s =
positive numbers, s> greater than or equal to 1?> If so, how
is the function for Zeta(s) defined when s is less than 1?>
[What I am going to say depends on considering Zeta as a
function of acomplex variable s. It won't make any sense
unless you are familiarwith some of the theory of analytic
functions of a complex variable.]That definition of Zeta is
fine for values of s with Re(s) > 1. Thereare other formulas
which agree with this one in that domain, and makesense in
larger domains. The upshot of it all is that Zeta is definedon
the .whole. complex plane, except for a pole at s =1. You speak
of the books. there are a bunch of books about Zeta,including
several that are aimed at the popular audience. I could,if I
chose, rant and rave until I frothed at the mouth about
somepopularizations. I will try instead to be nice, and say
that _PrimeObsession_ by Derbyshire seems to be to be decent.
He assumesthat the reader does not know college math, and
guides him or her to anunderstanding of what Zeta is, and why
it is interesting.There are also many books aimed at students
of mathematics and dealingwith Zeta. if you have taken a
university course in complexvariables then you can read Harold
Edwards's book on the Riemann Zeta Function. Available from
Dover Press.-- Chris HenrichThe wonderful thing about not
planning, is that failure comes as a completesurprise, and is
not preceded by a period of worry or depression. --
Kiltannen===
===
Subject:
: Re: combinations and permutations> I know
my logic is wrong, but can someone please explain why? This
example> is explained in the Grimaldi book, but I find his
brief explanation to be> of limited help.How many arrangements
of the letters of the word TALLAHASSEE have no> adjacent
A's?The step I don't understand is how to determine the number
ways the > adjacent A's can be arranged. Here's how I think it
should be done:There are 9 places to put the A's:
1-T-2-L-3-L-4-H-5-S-6-S-7-E-8-E-9.> There are 6 ways to
arrange the A's so 2 are together.???Are you counting
AATLLHSSEEA as different from AATLLHSSEEA?I get 72. There 9
different spots to put the pair of A's, and 8different spots
to put the other A.> There are 6 ways to > arrange the A's so
all 3 are together.Again, ??I get 9, period.> Doesn't this
mean there are (6 + 6) > x 9 = 108 ways of arranging the A's
so at least 2 are adjacent?72 + 9 = 81.Of course, for this
step, you need arrangements that have NO adjacentA's, rather
than SOME adjacent A's. The total number of ways to mix A'sin
once the other 8 letters are placed is C(11,3) = 165.
Subtractingthe 81 ways that you don't want leaves 84.> The
book says there are C(9,3) = 84 ways. I just don't understand
why you> use the combination formula for this.One
justification: Since the goal is no adjacent A's, each digit
isreplaced with either a single A or nothing. Looking just at
the digits,that seems like a clear case of C(9,3).-- iel W.
sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039
53 36 N / 086 11 55 W===
===
Subject:
: Re: combinations and
permutationsContent-transfer-encoding: 8bit> I know my logic
is wrong, but can someone please explain why? This example >
is explained in the Grimaldi book, but I find his brief
explanation to be > of limited help.How many arrangements of
the letters of the word TALLAHASSEE have no > adjacent A's?The
step I don't understand is how to determine the number ways the
> adjacent A's can be arranged. Here's how I think it should be
done:There are 9 places to put the A's:
1-T-2-L-3-L-4-H-5-S-6-S-7-E-8-E-9. > There are 6 ways to
arrange the A's so 2 are together. There are 6 ways to >
arrange the A's so all 3 are together. Doesn't this mean there
are (6 + 6) > x 9 = 108 ways of arranging the A's so at least 2
are adjacent?The book says there are C(9,3) = 84 ways. I just
don't understand why you > use the combination formula for
this.> Well, I'll go out on a limb and say your book is
wrong.There are 8!/(2!*2!*2!) distinct arrangements of
TLLHSSEE. That'salready 5040.For each of those arrangements,
how many distinct ways can we insertthe A's so that no two are
together? We'll count the number of lettersN_1 before the first
A, the number of letters N_2 between the first andsecond A's,
the number of letters N_3 between the second and third A'sand
the number of letters N_4 after the last A.We have N_1 + N_2 +
N_3 + N_4 = 8 with N_1 and N_4 >=0 , N_2 and N_3 >= 1 and N_1,
N_2, N_3 and N_4 all integers. There are C(9, 3) possible
solutions.So, there are C(9, 3)*8!/(2!*2!*2!) possibilities
altogether.-- ===
===
Subject:
: Re: combinations and permutations>
Well, I'll go out on a limb and say your book is wrong.The
original poster refers to a particular step.> There are
8!/(2!*2!*2!) distinct arrangements of TLLHSSEE. That's>
already 5040.> So, there are C(9, 3)*8!/(2!*2!*2!)
possibilities altogether.Specifically, the step where you went
from 8!/(2!*2!*2!) toC(9,3)*8!/(2!*2!*2!). -- iel W.
sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039
53 36 N / 086 11 55 W===
===
Subject:
: Re: combinations and
permutations>How many arrangements of the letters of the word
TALLAHASSEE have no >adjacent A's?>The step I don't understand
is how to determine the number ways the >adjacent A's can be
arranged. Here's how I think it should be done:>There are 9
places to put the A's: 1-T-2-L-3-L-4-H-5-S-6-S-7-E-8-E-9.
>There are 6 ways to arrange the A's so 2 are together. There
are 6 ways to >arrange the A's so all 3 are together. Doesn't
this mean there are (6 + 6) >x 9 = 108 ways of arranging the
A's so at least 2 are adjacent?>The book says there are C(9,3)
= 84 ways. I just don't understand why you >use the combination
formula for this.Think of grouping things this way:
xxx(Ax)x(Ax)xx(A)That is, you have 3 spots among 9. In the
first two spots you will putan Ax and in the third you will
place an A. This enumerates all thepositions where no two As
are next to each other. Thus, you have C(9,3)ways of placing
the As so that no two of them are next to each other.Rob son
<rob@trash.whim.org>take out the trash before
replying===
===
Subject:
: Re: combinations and
permutationsX-AuthenticatedUsername:
news.13101@buckeye-express.com>>How many arrangements of the
letters of the word TALLAHASSEE have no>>adjacent A's?>>The
step I don't understand is how to determine the number ways
the >>adjacent A's can be arranged. Here's how I think it
should be done:>>There are 9 places to put the A's:
1-T-2-L-3-L-4-H-5-S-6-S-7-E-8-E-9. >>There are 6 ways to
arrange the A's so 2 are together. There are 6>>ways to
arrange the A's so all 3 are together. Doesn't this
mean>>there are (6 + 6) x 9 = 108 ways of arranging the A's so
at least 2>>are adjacent? >>The book says there are C(9,3) = 84
ways. I just don't understand why>>you use the combination
formula for this.Think of grouping things this way:
xxx(Ax)x(Ax)xx(A)That is, you have 3 spots among 9. In the
first two spots you will> put an Ax and in the third you will
place an A. This enumerates all> the positions where no two As
are next to each other. Thus, you have> C(9,3) ways of placing
the As so that no two of them are next to each> other. Rob son
<rob@trash.whim.orgtake out the trash before replying> The rest
of the problem involves determining the number of arrangements
of the letters without the A's. This is 8!/(2!x2!x2!) = 5040.
I understand this. But then you take 5040 x C(9,3) to get the
final answer.The problem is asking about arrangements. How
many arrangements of the letters have no adjacent A's? C(9,3)
does not distinguish A1xA2xA3xxxxxx from A2xA1xA3xxxxxx, so
how does 5040 x C(9,3) distinguish these two
permutations?===
===
Subject:
: Re: combinations and permutations>
The problem is asking about arrangements. How many
arrangements of the> letters have no adjacent A's? C(9,3) does
not distinguish > A1xA2xA3xxxxxx from A2xA1xA3xxxxxx, so how
does 5040 x C(9,3) distinguish> these two permutations?Why
should it distinguish between AxAxAxxxxxx and AxAxAxxxxxx?--
iel W.
sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039
53 36 N / 086 11 55 W===
===
Subject:
: Re: combinations and
permutationsX-AuthenticatedUsername:
news.13101@buckeye-express.com>> The problem is asking about
arrangements. How many arrangements of>> the letters have no
adjacent A's? C(9,3) does not distinguish >> A1xA2xA3xxxxxx
from A2xA1xA3xxxxxx, so how does 5040 x C(9,3)>> distinguish
these two permutations?Why should it distinguish between
AxAxAxxxxxx and AxAxAxxxxxx?Ok, I feel stupid now. Even with
arrangements you DON'T distinguish patience.===
===
Subject:
: Re:
combinations and permutations> Ok, I feel stupid now. Even
with arrangements you DON'T distinguish > patience.There were
a few entertaining lapses in the first printing of theGrimaldi
book itself. I remember one excercise with
theinclusion-exclusion formula; if you worked out the
intermediate resultsrather than simply plugging the numbers
into the formula, you got a fewnegative numbers. (Take a look
at the acknowledgements page.)-- iel W.
sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039
53 36 N / 086 11 55 W===
===
Subject:
: Re: need help in
understanding Torkel's ZFC comment> How about if you point
me to the best example or two of proofs using>> ZFC that is
given - preferably not just proving part of ZFC with>> another
part?> The square root of 2 is irrational:>>
http://au.metamath.org/mpegif/sqr2irr.h> Q: What does that
have to do with ZFC? There are 42 lines and>> offhand it
doesn't look like any of them are from ZFC.> A: This theorem
was proved from axioms: ax-1 ax-2 ax-3 ax-mp ax-4>> ax-5 ax-6
ax-7 ax-gen ax-8 ax-9 ax-10 ax-11 ax-12 ax-13 ax-14>> ax-15
ax-16 ax-17 ax-ext ax-rep ax-un ax-pow ax-reg ax-inf.>
(http://au.metamath.org/mpegif/sqr2irr.h)>As I said elsewhere,
19 of these are axioms outside of ZFC,Ax-1 to ax-16, ax-mp and
ax-gen are the rules of propositional calculus> and predicate
calculus. They represent the rules of logic under which ALL>
first-order theories work. . . . There is no need to list>
these axioms as axioms for ZFC since the axioms are
automatically axioms> of *every* first-order theory.> The
axioms used by the above proof depends only on the above
proof.> Talking about other theories is a red herring.You seem
to be having difficulty in comprehending the concept
thatcertain rules are applicable universally to ALL theories,
and sothere is no need to state these rules for individual
theories.In the case of Metamath, ax-1 to ax-17, ax-gen and
ax-mp, and ALLthe consequences of these alone are such rules
(which go all theway to 2eu5 (number 1001 in the mpegif
version of the rules)).ALL formal theories that Metamath can
handle have all these rulesas default. The statement of the
axioms of ZFC are, as always,the statements of those rules
peculiar to ZFC (in that thestatement of axioms for ANY theory
is the statement of thoseaxioms peculiar to the theory). The
theorems of ZFC are theconsequences of the axioms of ZFC,
freely using the rules thathave been derived from ax-1 to
ax-17, ax-gen and ax-mp, under theappropriate restrictions.>
Have you ever heard about C-B Systems? They include C-B1
which> consists of the rules of propositional calculus and
predicate> calculus. There are also C-B2 to C-B99, each with
its own rules. ZFC> is the set of axioms under which all C-B
systems work. There is no> need to list the ZFC axioms as
axioms for C-B1 since they are> automatically axioms of every
C-B system. So C-B1 alone> (propositional calculus plus
predicate calculus) proves the above> theorem.No, the rules of
propositional calculus and predicate calculus aloneare NOT
enough to prove that the square root of 2 is irrational,
sincethe square root of 2 is not defined within the rules of
propositionalcalculus and predicate calculus, nor is the
concept of a rationalnumber or an irrational number. You need
MORE than just propositionalcalculus and predicate calculus
before the statement The square rootof 2 is irrational even
has a meaning. You can only prove thestatement The square root
of 2 is irrational within the context ofa structure in which
the statement actually has a meaning. Such astructure is
external to propositional calculus and predicate
calculus.Propositional calculus and predicate calculus are not
sufficient.>and I>don't think you really need ZFC to prove that
sqrt(2) is irrational>anyway.> These sets identified with R, C,
i, 1, 0, complex addition, complex> multiplication, and < on R,
all have their existence guaranteed by ZFC.> ZFC is a way to
define which sets we are going to allow to exist.Like the sets
R, C, i, 1, 0, complex addition, complex multiplication,and
less than on R, that I listed above. Metamath goes to a lot
oftrouble to identify these specific sets, using the axioms of
ZFC.> This is carefully done to avoid the paradoxes. For
example, if we> have the Separation axiom and ~(x e x) is a
formula, then there can be> no set of all sets.I know that.>
But the set of complex numbers (and its subsets)> is not
involved in the paradoxes.But there is no GUARANTEE *within*
the theory of complex numbers that aset satisfying the
appropriate axioms even exists. Similarly, *within*the theory
of real numbers, there is no guarantee that a set R
satisfyingthe appropriate conditions even exists. *Within* the
theories, thereis no proof that we are not working in a
vacuum.> It is not prohibited by ZFC or any> of its
variations. So there is no question as to whether or not
there> is a set of complex numbers.It is not prohibited by
ZFC. But that is not the point. The point isthat if ZFC is
consistent, then it can be used to guarantee the existenceof
sets R and C which are models for the theories of real numbers
andcomplex numbers. This means that ZFC is enough to assure us
that when weare working in the theories of real numbers or
complex numbers, then weare not working in a vacuum because we
KNOW that there are models forthese theories. We also know that
the consistency of ZFC implies theconsistency of the theories
of real and complex numbers.You really should try to
understand what people are trying to say and whatthe point is
for what people do, rather then criticise people for
claimingstuff that they never have claimed.The point is not
that ZFC is NECESSARY to do these things (although youcontinue
to falsely claim that that is what we are saying, in spite
ofthe many times people have pointed out to you that your
claim is false).The point is that ZFC is SUFFICIENT to do
these things. Until youcan make the distinction between the
necessity of ZFC (which NOBODYclaims) and the sufficiency of
ZFC clear in your mind, then you willnot understand what
people are trying to do.> Mathematicians used complex numbers
long before the advent of ZFC> without the contradictions that
prompted ZFC. We don't need ZFC to be> assured of the existence
of the complex numbers, nor is it the purpose> of ZFC.See
comment above about the guarantee that a MODEL for C exists.
Youhave completely missed the point.> ZFC relies on the
existence of the mathematical concept (object)> called a set,
and defines particular ones and their properties.> Likewise,
mathematics relies on the existence of the mathematical>
concept (object) called a complex number, and defines
particular ones> and their properties.But without ZFC, how are
you going to guarantee that there even existsa set of complex
numbers in which to work???> The fact that there is no
rational number whose square is 2 is due> only to the nature
of the natural numbers and multiplication, and> principles
concerning factorization of the natural numbers.And requires a
structure in which the statement actually has
meaning.Propositional calculus and predicate calculus are not
sufficient tosupply such a structure. You seem to be missing
this important point.> Charlie VolkstorfDavid McAnally Despite
anything you may have heard to the contrary, the rain in Spain
stays almost invariably in the hills.===
===
Subject:
: Re: JSH: Open
letter to Jim Ferry> We've all been having a good laugh> at
your expense this whole time. You've been twisting and turning
to the tune called by Harris thiswhole time. I suspect that
James Harris will be hailed in the futureas a genius of the
internet.===
===
Subject:
: Re: universal set with 3 valued logic>
yes, in ZF you can prove the universal set does not exist and
you don'tneed a computer program to do that (i would be
interested in what happens ifyour program is equipped with
three logical values and the o logicalconnective which is an
extension of the iff biconditional). but this is anextension
of ZF and an enlargement of perspective. read the paper
beforeyou disagree on this point, especially the first couple
of pages.> why does one NEED to do this? well, the short
answer is that one doesn't.one doesn't need to not do it,
either. i find it satisfying on aphilosophical level that the
universal set no longer need to be of a typedifferent than a
set so that's why i personally need to do it.It just doesn't
make sense to introduce a strange new system of logic justto
salvage the Universal Set. Is there anything really wrong with
the usual2-value logic (T or F)-- other than prohibiting the
existence of theUniversal Set, I mean?===
===
Subject:
: order type
of a countable subset of RI'm stuck on the following
problem:Let A be a countable subset of R (real numbers) with
the followingproperty:for every x, y in A with x<y, there are
u, v, w in A such thatu<x<v<y<w. Show that there is a
bijection f of A onto the set Q ofrational numbers, such that
x<y implies f(x)<f(y).===
===
Subject:
: Re: order type of a
countable subset of R> I'm stuck on the following problem:Let
A be a countable subset of R (real numbers) with the
following> property:> for every x, y in A with x<y, there are
u, v, w in A such that> u<x<v<y<w. Show that there is a
bijection f of A onto the set Q of> rational numbers, such
that x<y implies f(x)<f(y).> Since every open interval of
rationals is order isomorphic to any other such interval,
including Q itself, you could do it by mapping A onto any open
interval of reals.But a direct construction onto Q is also
possible:Start with any x and y, members, of A with x < y, and
assign to them the values 0 and 1, respectively.Then,
recursively, (1) pick a member of A less that any picked so
far and assign to it the largest negative integer not yet
assigned;(2) Pick a member of A larger than any yet picked and
assign to it the smallest positive integer not yet assigned;(3)
for each pair of consecutive members of A already picked, which
have been assigned rational values, in standard form, of r/s
and u/v, pick a member of A between them and assign to it the
rational (r+u)/(s+v).The proof that this method works can be
made but is not trivial.The method of picking new fractions in
step (3) may be referenced under the rubric Farey
Fractions.===
===
Subject:
: Re: e is transcendental (was: classes of
transcendental numbers ?<snip>>My question :>What is the
implication of this second value of e^[ipi]=0 ?>>No
implication. It is not true.> e^[iPi]=mod 1, arg
180>>e^[iPi]=mod 1, arg 180 degrees> This is correct in
polar form.>>But it has absolutely nothing to do with your
earlier extraordinary>>claim that exp(i pi) = 0. That
particular claim of yours is still>>unjustified, and it is
still false.>It has to do.Not from what I have seen.>cos pi +
i sin pi=-1+i*0 =-1 is correct>However the phasor e^[ipi] =cos
pi + i sin pi ,> is properly defined as MOD 1 , ARG [THETA]No.
You are just comparing two different ways of specifying a
nonzero complex number. Both characterizations are
equallyvalid.> If I state that a phasor A=-1 IS THIS
CORRECT?>No it must be accompanied by its length and its
direction.>A=-1 shows a length of unity directed towards the
[0,0] point >[of an X-Y coordinates system crossing]>from any
angular direction around a unit circle,from >the positive
X-axis (taken as zero with the usual notation.You phasor
points to [-1,0] from [0,0].> e^[ipi] cannot loose its value
as 1 (unity),Whatever THAT is supposed to mean.>and with the
usual>notation it is positive ,leaving away the [0,0] point>
e^[ipi] = -1 it is driven towards the [0,0] pointNo. On the
complex plane, you are thinking of an arrow pointing from
[0,0] to [-1,0].> since> e^[ipi] =-1+i*{0) is a
technicalityNo, it is not.> and SELECTIVELY one may choose
either of the two components>according to its interest, to
give the real ,or the imaginary>solution.This looks like the
claim of an electrical engineer who works withcomplex numbers,
and then takes the real or imaginary part at theend. It has
nothing to do with the fact that exp(i pi) = -1.> Simply
taking care of the angle ,it is fair for one to state>tha
unity (mod 1 of e^[ipi]) at 180 deg ,anticlockwise from
the>positive X-axis ,has two components ,one of which the
real(by >convention)has the value of one, and the imaginary
(by convention)>Zero.So? The real part of exp(i pi) is
cos(pi), and its imaginary partis sin(pi), so all you are
saying is that cos(pi) = -1 and sin(pi) = 0, and we were
already aware of these facts. There is NO reason to conclude
that exp(i pi) = 0. If you take the real part or imaginary
part of a complex linear combination of exp(i theta_k), k =
1,...,n, then you get a real linear combination of
cos(theta_k) and sin(theta_k), k = 1,...,n.Once you go to real
and imaginary parts, expressions like exp(i theta_k) are no
longer appropriate, and you have to usecos(theta_k) and
sin(theta_k) instead.Alternatively, if you want the imaginary
part of exp(i pi), theappropriate expression is [exp(i pi) -
exp(-i pi)]/(2i), andnot just exp(i pi).David
McAnally--------------===
===
Subject:
: Metamath Axiom of
ChoiceWhile looking at the Metamath site, and specifically at
their statement ofthe Axiom of Choice, I note that their
statement of the Axiom of Choice iseffectively the statement
that for any set x, there exists a set y suchthat for any
nonempty element w of x, there is at least one element t of
ysuch that w is an element of t, and that w has exactly one
element incommon with the union of all elements of y which
have w as an element,i.e. there is a set y such that w is an
element of some element t of y,and if s = {t in y : w in t},
then w n (Us) has exactly one element, whereA n B denotes the
intersection of A and B, and where Us is the union ofthe
elements of s.That this implies the Axiom of Choice is obvious
(if w is a nonempty element of x, let f(w) be the unique
element of w n (Us), where s = {t in y : w in t}). The
question is how can one prove the Metamathstatement of the
Axiom of Choice in ZFC?David McAnally -------===
===
Subject:
: Re:
Metamath Axiom of Choice>While looking at the Metamath site,
and specifically at their statement of>the Axiom of Choice, I
note that their statement of the Axiom of Choice
is>effectively the statement that for any set x, there exists
a set y such>that for any nonempty element w of x, there is at
least one element t of y>such that w is an element of t, and
that w has exactly one element in>common with the union of all
elements of y which have w as an element,>i.e. there is a set y
such that w is an element of some element t of y,>and if s = {t
in y : w in t}, then w n (Us) has exactly one element, where>A
n B denotes the intersection of A and B, and where Us is the
union of>the elements of s.>That this implies the Axiom of
Choice is obvious (if w is a nonempty >element of x, let f(w)
be the unique element of w n (Us), where >s = {t in y : w in
t}). The question is how can one prove the Metamath>statement
of the Axiom of Choice in ZFC?How does this sound for a proof.
Let f : x-{empty set} -> Ux be a choice function (so that f(v)
is an element of v for all nonempty elements v of x). Let y =
{{v,f(v)} : v in x, v nonempty}. The first thing to note is
that if w is a nonempty element of x, then there is an element
of y which has w as an element, specifically, {w,f(w)}. Let w
be a nonempty element of x, then s = {t in y : w in t} =
{{w,f(w)}} u {{v,w} : v in x, v nonempty, f(v) = w}, where A u
B denotes the union of A and B. It follows that Us = {w,f(w)} u
{v in x : v nonempty, f(v) = w}, and so the elements of Us are
w, f(w), and nonempty elements v of x such that f(v) = w.
Since w is an element of v for all nonempty elements v of x
such that f(v) = w, then the elements of Us are w, f(w) and
some sets which have w as an element. By the axiom of
regularity, the only common element of w and Us is f(w),
completing the proof of the Metamath statement of the Axiom of
Choice.David McAnally -------===
===
Subject:
: Re: Metamath Axiom of
Choice>While looking at the Metamath site, and specifically
at their statement of>the Axiom of Choice, I note that their
statement of the Axiom of Choice is>effectively the statement
that for any set x, there exists a set y such>that for any
nonempty element w of x, there is at least one element t of
y>such that w is an element of t, and that w has exactly one
element in>common with the union of all elements of y which
have w as an element,>i.e. there is a set y such that w is an
element of some element t of y,>and if s = {t in y : w in t},
then w n (Us) has exactly one element, where>A n B denotes the
intersection of A and B, and where Us is the union of>the
elements of s.>That this implies the Axiom of Choice is
obvious (if w is a nonempty >element of x, let f(w) be the
unique element of w n (Us), where >s = {t in y : w in t}). The
question is how can one prove the Metamath>statement of the
Axiom of Choice in ZFC?How does this sound for a proof. Let f
: x-{empty set} -> Ux be a choice > function (so that f(v) is
an element of v for all nonempty elements v of x). > Let y =
{{v,f(v)} : v in x, v nonempty}. The first thing to note is
that > if w is a nonempty element of x, then there is an
element of y which has w > as an element, specifically,
{w,f(w)}. Let w be a nonempty element of x, > then s = {t in y
: w in t} = {{w,f(w)}} u {{v,w} : v in x, v nonempty, > f(v) =
w}, where A u B denotes the union of A and B. It follows that
> Us = {w,f(w)} u {v in x : v nonempty, f(v) = w}, and so the
elements of > Us are w, f(w), and nonempty elements v of x
such that f(v) = w. Since > w is an element of v for all
nonempty elements v of x such that f(v) = w, > then the
elements of Us are w, f(w) and some sets which have w as an >
element. By the axiom of regularity, the only common element
of w and Us > is f(w), completing the proof of the Metamath
statement of the Axiom of > Choice.David McAnally At the
moment, they (the Time Lords) are far from being
all-powerful.> That's why it's been left up to me and me and
me.> quote by: Patrick Troughton in The Three
Doctors-------Uhm... the Axiom of Choice is an _axiom_. For
specificity, it is awff of the FOL of set theory which is
(roughly speaking) stipulatedtrue, and can thus be used in
deductions. If you decide that youdon't like the axiom of
choice, you can choose to not work with FOlanguages which
include it. But even then, it's hard to deny it's usein axiom
systems which _do_ use it. (Since, by construction,
you'resetting up a conditional... If the axiom of choice is
true, and...blah..., then ...blah'...)In any case, there is
nothing to prove.'cid 'ooh===
===
Subject:
: Re: Metamath Axiom of
Choice>[...]>Uhm... the Axiom of Choice is an _axiom_. For
specificity, it is a>wff of the FOL of set theory which is
(roughly speaking) stipulated>true, and can thus be used in
deductions. If you decide that you>don't like the axiom of
choice, you can choose to not work with FO>languages which
include it. But even then, it's hard to deny it's use>in axiom
systems which _do_ use it. (Since, by construction,
you're>setting up a conditional... If the axiom of choice is
true, and>...blah..., then ...blah'...)>In any case, there is
nothing to prove.Uhm, he wants to know how to prove the
Metamath version of AC_from_ what he's taking as the standard
version.>'cid 'ooh===
===
Subject:
: Re: Metamath Axiom of Choice
<d7ba1f79.0402140358.4c3180c3@posting.google.com> Discussion,
linux)>>>While looking at the Metamath site, and
specifically at their statement of>>the Axiom of Choice, I
note that their statement of the Axiom of Choice
is>>effectively the statement that for any set x, there exists
a set y such>>that for any nonempty element w of x, there is at
least one element t of y>>such that w is an element of t, and
that w has exactly one element in>>common with the union of
all elements of y which have w as an element,>>i.e. there is a
set y such that w is an element of some element t of y,>>and if
s = {t in y : w in t}, then w n (Us) has exactly one element,
where>>A n B denotes the intersection of A and B, and where Us
is the union of>>the elements of s.>>>That this implies the
Axiom of Choice is obvious (if w is a nonempty >>element of x,
let f(w) be the unique element of w n (Us), where >>s = {t in y
: w in t}). The question is how can one prove the
Metamath>>statement of the Axiom of Choice in ZFC?>>> How
does this sound for a proof. Let f : x-{empty set} -> Ux be a
choice >> function (so that f(v) is an element of v for all
nonempty elements v of x). >> Let y = {{v,f(v)} : v in x, v
nonempty}. The first thing to note is that >> if w is a
nonempty element of x, then there is an element of y which has
w >> as an element, specifically, {w,f(w)}. Let w be a nonempty
element of x, >> then s = {t in y : w in t} = {{w,f(w)}} u
{{v,w} : v in x, v nonempty, >> f(v) = w}, where A u B denotes
the union of A and B. It follows that >> Us = {w,f(w)} u {v in
x : v nonempty, f(v) = w}, and so the elements of >> Us are w,
f(w), and nonempty elements v of x such that f(v) = w. Since >>
w is an element of v for all nonempty elements v of x such that
f(v) = w, >> then the elements of Us are w, f(w) and some sets
which have w as an >> element. By the axiom of regularity, the
only common element of w and Us >> is f(w), completing the
proof of the Metamath statement of the Axiom of >> Choice.>> David McAnally>>> At the moment, they (the Time Lords)
are far from being all-powerful.>> That's why it's been left
up to me and me and me.>> quote by: Patrick Troughton in The
Three Doctors>>> -------> Uhm... the Axiom of Choice is an
_axiom_. For specificity, it is a> wff of the FOL of set
theory which is (roughly speaking) stipulated> true, and can
thus be used in deductions. If you decide that you> don't like
the axiom of choice, you can choose to not work with FO>
languages which include it. But even then, it's hard to deny
it's use> in axiom systems which _do_ use it. (Since, by
construction, you're> setting up a conditional... If the axiom
of choice is true, and> ...blah..., then ...blah'...)> In any
case, there is nothing to prove.You should read more
carefully.(1) The author did not argue that the axiom should
be avoided.(2) Instead, David saw that a particular formal
theory stated adifferent formulation of Choice than he was
used to. He wondered howto prove that the two formulations are
equivalent. Hence, there *is*something to prove here.Seems
pretty clear to me that David McAnally knows what an axiom
isand that he's absolutely correct to wonder about a proof
here.-- But he himself was not to blame for his vices. They
grew out of a personaldefect in his mother. She did her best
in the way of flogging him while aninfant... but, poor woman!
she had the misfortune to be left-handed, and achild flogged
left-handedly had better be left unflogged. -- E.A.
Poe===
===
Subject:
: Even number as primesCan someone give me a HINT
on how to prove:For every even number n >=6 it can be
represented n = xy + z, where x, yand z are primes.===
===
Subject:
:
Re: Even number as primes> Can someone give me a HINT on how to
prove:> For every even number n >=6 it can be represented n =
xy + z, where x, y> and z are primes.Chen can.Phil-- Unpatched
IE vulnerability: DNSError folder disclosureDescription:
Gaining access to local security zonesReference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.h===

===
Subject:
: Re: Even number as primes> Can someone give me a HINT
on how to prove: For every even number n >=6> it can be
represented n = xy + z, where x, y and z are primes.That's
Chen's Theorem. Googling doesn't turn up any obvious hints on
how toprove it--just references to journals and books where
proofs have beengiven, except there was a post to sci.math a
few years ago that said this,which might be a bit of a hint
(post from Chris Thompson): There's also a proof in Chapter 10
(and last) of Nathanson Additive Number Theory - The Classical
Bases Springer 1996 [GTM 164], but be warned that although the
proofs of the Jurkat-Richert bounds for the linear sieve are
included, the Bombieri-Vinogradov theorem is used but not
proved here.One could take this is a hint that you should use
a linear sieve, use theJurkat-Richert bounds, and the
Bombieri-Vinogradov theorem, assuming any ofthat makes sense
(it makes no sense to me :-)).-- --Tim Smith===
===
Subject:
: Re:
Even number as primes> Can someone give me a HINT on how to
prove:> For every even number n >=6 it can be represented n =
xy + z, where x, y> and z are primes.I don't immediately see
how to represent 10 in this way: the only products of primes
that are < 10 are 2*2, 2*3 and 3*3, BUT 6, 4, and 1 are not
prime===
===
Subject:
: Re: Even number as primes> Can someone give
me a HINT on how to prove:> For every even number n >=6 it can
be represented n = xy + z, where x,y> and z are primes.> I
don't immediately see how to represent 10 in this way:> the
only products of primes that are < 10 are> 2*2, 2*3 and 3*3,
BUT 6, 4, and 1 are not primeOh, sorry, i forgot to mention n
<> 10, sorry.===
===
Subject:
: Re: Even number as primes> Can
someone give me a HINT on how to prove:> For every even number
n >=6 it can be represented n = xy + z, where x,> y> and z are
primes.I don't immediately see how to represent 10 in this
way:the only products of primes that are < 10 are> 2*2, 2*3
and 3*3, BUT 6, 4, and 1 are not prime> Oh, sorry, i forgot to
mention n <> 10, sorry. Or perhaps the statement should be that
any even number n can be written as p + q, where p is a prime
and q is a 2-prime (i.e. either a prime itself or a product of
exactly two prime factors.) J===
===
Subject:
: Re: Help find equation
for pianist's possibilities of sound?Mime-version:
1.0Content-type: text/plain;
charset=US-ASCIIContent-transfer-encoding: 7bit>>> >
Given these measurements, V and T are continuous variables.
The number of> values they can take are uncountably large.>> But the number of discernable differences to the human ear
is not that>> large. MIDI for instance uses 127 possible
volumes. For my purposes I just>> need a hard number to
illustrate my real point which is that there are>> almost an
incomprehendable number of possibilities for the piano's
sound.>I think Greg's point is that there is no hard number,
though: for> all intents and purposes you might as well pick
one out of the air.But suppose we start with a simple scenario
and see what numbers> follow from a few assumptions. Consider
first the number of single> notes, assuming 100 each of
distinguishable volumes and durations: in> this case there
will be 88*100^2 = 880,000 possibilities. That's> probably not
incomprehensible enough for you (even if we double or> triple
it to account for the pedals), so let's consider the two-note>
chords: there are 88*87*100^3 (assuming both notes have the
same> duration) = 7,656,000,000 of them. Still comprehending
that fine?> Well, how about three-note chords? Trying to keep
this realistic,> most pianists' hands only span a portion of
the keyboard (an interval> something like an tenth, I would
guess), so a fair number of the keys> are bound to be out of
reach once the first two are chosen: let's say> there are
still fifty available, giving 88*87*50*100^4 =>
38,280,000,000,000 'triad sounds'. By the same token the
four-note> chords include 88*87*50*49*100^5 =
187,572,000,000,000,000> possibilities, in words approaching
two hundred quadrillion. In> seconds of time that's about the
age of the solar system.Got a head for large numbers? If both
hands are already stretched any> additional notes will have to
be chosen from within those spanned, so> the number of
available keys will shrink some more, yielding 'only'> about
88*87*50*49*24*23*100^7 ~= 10^24 six-note chord sounds. (Note>
that I'm assuming the pressure of each finger, hence the volume
of> each component note, is independently controllable over its
full> range -- which may not be the case in practice.) But
that's about the> number of molecules in a small glass of
water.Using sequences of notes or chords you can inflate the
figures much> further. Assuming complete independence the
numbers multiply> together: throwing in 100 possible rests
between each pair a run of> three six-note chords would yield
something like 10^80 possibilities,> which I believe is
roughly on the order of the number of subatomicThank you
Odysseus! I'm practically dying of laughter. That explanation
isThat the already created universe is the playing field for
the pianist ormusician or artistic discoverer and explorer.
And thanks for the belly ache!Dennis Hammhamm0123456789@i.am
(remove the integers to reply)===
===
Subject:
: Re: Got a speeding
ticket and need to fight backThe Lord of Chaos (Suresh
Devanathan)> If the cop really wants to get the speeder, he
should speed clock him,> with his own car. That's the only way
out. Instead of hiding on some> obscure corner and pulling a
radar gun, isnt just going to cut it.> -sure does not
matter how they caught you. You were speeding and
engeringothers and yourself. You should be ashamed of yourself
for braking the law.Lying about it makes it worse. Trying to
avoid the fine is attempted theftfrom the state or local
community. You don't make the rules. The legislatureor local
authority do that and they will enforce them.===
===
Subject:
: Re:
Got a speeding ticket and need to fight backThe last i
checked, I dictate the laws of physics. I am God, to
manypeople. Your court is beneath me. Your constitution is
beneath me. In fact,you cannot even prosecute me, because the
constitution depends on me. I tellwhat the laws are. And
that's final.-suresh> The Lord of Chaos (Suresh
Devanathan)message> If the cop really wants to get the
speeder, he should speed clock him,> with his own car. That's
the only way out. Instead of hiding on some> obscure corner
and pulling a radar gun, isnt just going to cut it.-suresh>It
does not matter how they caught you. You were speeding and
engering> others and yourself. You should be ashamed of
yourself for braking thelaw.> Lying about it makes it worse.
Trying to avoid the fine is attempted theft> from the state or
local community. You don't make the rules. Thelegislature> or
local authority do that and they will enforce them.===
===
Subject:
:
Re: Got a speeding ticket and need to fight backis your other
name ape ?-suresh> The Lord of Chaos (Suresh
Devanathan)message> If the cop really wants to get the
speeder, he should speed clock him,> with his own car. That's
the only way out. Instead of hiding on some> obscure corner
and pulling a radar gun, isnt just going to cut it.-suresh>It
does not matter how they caught you. You were speeding and
engering> others and yourself. You should be ashamed of
yourself for braking thelaw.> Lying about it makes it worse.
Trying to avoid the fine is attempted theft> from the state or
local community. You don't make the rules. Thelegislature> or
local authority do that and they will enforce them.===
===
Subject:
:
Re: Got a speeding ticket and need to fight backhow old are
you?-suresh> The Lord of Chaos (Suresh Devanathan)message>
Everybody has to learn to play by my rules. Nobody has other
options.> Otherwise, you will goto .Here's a secret> 1) I make
the rules> 2) I make the ruleswhat is do u think the 3rd rule
is?> I make the rulesThat's all everyone, has to know. I
always have it my way. Otherwise, iam> going to torture you
alive, just to have fun and make a buffon out ofyou.>-suresh>
I was willing to help you. I had an idea and asked you for
your courtdate.> Instead you go off on a temper tantrum. No
help for you. So sorry.There> is a confident ignorance that
comes with youth. The sooner you recognise> that, the better.>
Rob===
===
Subject:
: Re: Got a speeding ticket and need to fight
backyou know what increases with age? fratulence. For example:
take you.-sureshThe Lord of Chaos (Suresh Devanathan)> how old
are you?> -sureshThe Lord of Chaos (Suresh Devanathan)>
message> Everybody has to learn to play by my rules. Nobody
has other options.> Otherwise, you will goto .> Here's a
secret> 1) I make the rules> 2) I make the rules> what is do u
think the 3rd rule is?> I make the rules> That's all everyone,
has to know. I always have it my way. Otherwise,i> am> going
to torture you alive, just to have fun and make a buffon out
of> you.> -sureshI was willing to help you. I had an idea and
asked you for your court> date.> Instead you go off on a
temper tantrum. No help for you. So sorry.> There> is a
confident ignorance that comes with youth. The sooner
yourecognise> that, the better.>Rob>===
===
Subject:
: Re: Got a
speeding ticket and need to fight backThe Lord of Chaos
(Suresh Devanathan)> Thank God that people like you and Rob
Duncan are not engineers.You> should never build a computer.
Dont ever send your children to> college for engineering. They
are most likely to drive the companies,> where they work,
bankrupt. If 15 out of every 100 instructions, a> computer
executes is in error, you know something, the computer is> no
more than random number generator.That's the restatement of
your idiocy. You do not understand> statistics. Go away. You
do not understand pecentages. You know> nothing about them.
You do not even know how to use them. If you ever> got a Ph.D
in psychology, remember that all you did was bull your> way
thru college. How about if i told u, that there is 15% chance
you are going to die> tomorrow? Let's see: 85% chance that you
wont die. Let's see the power> of the 15% chance. You know 3/20
people hate you, and think you> are ugly. Out of 20 times you
drive, 3 times, the chances are you are> going to hit
something. chance are that our of 3 of 20 people, in> public,
are going to mug you.-sureshI can't quit seem to figure this
out, since your posts don't follow the correct threads, but I
think you are talking to me here, seeing as I was the one
calling your 15% argument bull.If this is the case, I think I
know a little but about statistics. I have been teaching stats
for 3 years now, and at the college level, not to some
snot-nosed high school brat trying to get out of a speeding
ticket. Not only that, but I've taught stats at a top 25
university in the nation.Secondly, before you start telling
people they don't know stats, you need to go back to class.
See, you said early on that the SD of the radar gun was +/-
5mph. That is impossible, standard deviation MUST be positive
since it involves taking the POSTIVE square root. Maybe you
should have paid attention in stats class.Oh, and I'm not
getting a PhD in psychology, I'm working on a PhD in math.If
3/20 people hate me (although, what three-twentieths of
aperson is, I will never know), then I'm doing better than I
thought. I've sure failed more than 15% of my students, some
for spweing out random thougths that they called statistics.
And, rest assured, you surely would have failed my class. Oh,
and I've driven hundreds of thousands of times (8 years of
driving) and never been in an accident or hit anything.
Although, I have gotten 2 speeding tickets, and paid them. And
I've been out in public way more often than that and never been
mugged.Maybe your stats class should have taught you about
things like margin of error. Small expirements don't mean a
whole lot. Since I'm guessing your basing your stats on your
personal experiences, your life must suck, getting mugged once
a week or so (I guess you actually go out 3 times a day).Oh,
and about building computers, yeah, I can do that do. In fact,
I've spent a fair amount time working on a project to reduce
the size of circuits. I can build an ALU from NAND gates if I
have to, and have.As for driving a company to bankruptcy, not
likely from me. Look up operations research and math modeling.
In fact, I plan to work with companie to keep them from going
bankrupt. Yeah, that might involve laying off people, and you
would sure make the top of the list.Now, I would suggest going
out and buying a nice stats book, try An Introduction to
Statistical Inference, William ston. I like that book, but
that's only probably because I helped write it.Now, if you
would like to actually learn stats and why your arguement is
flawed, we can talk some more. ===Timothy M. BrauchGraduate
StudentDepartment of MathematicsWake Forest
University===
===
Subject:
: Re: Got a speeding ticket and need to
fight backidiot, you smell like : the same way you talk. I
think it's because youhave fungus growing in your
mouth-suresh> The Lord of Chaos (Suresh Devanathan)> Everybody
has to learn to play by my rules. Nobody has other options.>
Otherwise, you will goto .Here's a secret> 1) I make the
rules> 2) I make the ruleswhat is do u think the 3rd rule is?>
I make the rules> And yet, you still have to ask for help to
get out of a speeding ticket.> What's the matter? Can't tell
the judge your rules?> Doug===
===
Subject:
: Re: Got a speeding
ticket and need to fight backThe Lord of Chaos (Suresh
Devanathan)> idiot, you smell like : the same way you talk. I
think it's becauseyou> have fungus growing in your
mouthTrollin', trollin', trollin'...keep them posters
trollin'...Doug===
===
Subject:
: Re: Got a speeding ticket and need
to fight back> Trollin', trollin', trollin'...God my ass is
swollen.Rawhide!Slainte,Fletch===
===
Subject:
: Re: Got a speeding
ticket and need to fight backApproximatelyf lamda = c f lamda
= 186 000 mpsf -> frequency in hertzh = 60min> = 60 * 60 secf
lamda = 186 000> = 186 000 mpsf -> 1000,000,000 s^-1
(gigahertz)lamda = .000186 milesThat's the radar's wavelength
approximately. Take the radar's time window,> it internally
uses to compute the speed: 1/10 second ( the human eye can>
see about 30 frames / sec ). Assuming Reaction time of a
driver = 1/10 sec> approx> 0.000186/ .1s = .00186 mps = 6.6
mphThis has nothing to do with how these radar units actually
work. Googleon 'Doppler'.And then understand that most traffic
court judges have probably heardthem all. And thrown them all
out of court. If you can't identify someprocedural error on
the part of the officer, pay the fine.-- Hovnanian
mailto:@Hovnanian.comnote to spammers: a Washington State
resident------------------------------------------------------
------------Bureaucrat, n.: A person who cuts red tape
sideways. -- J. McCabe===
===
Subject:
: Re: Got a speeding ticket
and need to fight backX-NewsReader: GRn 3.2n February 9,
1999Approximatelyf lamda = c f lamda = 186 000 mpsf ->
frequency in hertzh = 60min> = 60 * 60 secf lamda = 186 000> =
186 000 mpsf -> 1000,000,000 s^-1 (gigahertz)lamda = .000186
milesThat's the radar's wavelength approximately. Take the
radar's time window,> it internally uses to compute the speed:
1/10 second ( the human eye can> see about 30 frames / sec ).
Assuming Reaction time of a driver = 1/10 sec> approx0.000186/
.1s = .00186 mps = 6.6 mphThis has nothing to do with how these
radar units actually work. Google> on 'Doppler'.And then
understand that most traffic court judges have probably heard>
them all. And thrown them all out of court. If you can't
identify some> procedural error on the part of the officer,
pay the fine.Amid all the noise associated with this posting,
I amwondering why nobody has bothered to question
whatcomputational time and human reaction time have to dowith
a Doppler velocity measurement.If your case is weak, try to
baffle them with BS.===
===
Subject:
: Re: Got a speeding ticket and
need to fight backI will answer your question why reaction
time is importantA) even for DSP analysis, you know there are
going to be processingdelays. Your MP3 player for example,
actually is always a few milli secondsahead, processing all
the crappy songs you listen to.B) it gives a good accurate
measure of spontaneous velocity. You point theradar at some
big stupid object. Internally, it(the object) is composed
ofions and ions of molecules. Even air, for example, could
causesinterference to measurement. The way, you get rid of the
error, is byaveraging, a lot of data, over a period of time.C)
Syncrozning with the Speedometer reading. Speedometer is
usally analog.If you are not that blind, you will notice that
needle that does thereading, is actually bumping up and down,
around a certain value. That isbecause the car, is actually
vibrating at a very high angular velocity.However, by
averaging out the measurement and by slowing the
speedometerneedle down to your reaction time, you have a
steady-state reading.If things were not measured around your
reaction time, you know the speed ofthe car can be anywhere
from a million 1,000,000,000 mph to 0 mphLearn some science
first, before you spew your crap. To talk smart, you needto
know how smart people talk. Too bad, catch-22 for u.-suresh>
Approximately> f lamda = c> f lamda = 186 000 mps> f ->
frequency in hertz> h = 60min> = 60 * 60 sec> f lamda = 186
000> = 186 000 mps> f -> 1000,000,000 s^-1 (gigahertz)> lamda
= .000186 miles> That's the radar's wavelength approximately.
Take the radar's timewindow,> it internally uses to compute
the speed: 1/10 second ( the human eyecan> see about 30 frames
/ sec ). Assuming Reaction time of a driver = 1/10sec> approx>
0.000186/ .1s = .00186 mps = 6.6 mphThis has nothing to do
with how these radar units actually work. Google> on
'Doppler'.And then understand that most traffic court judges
have probably heard> them all. And thrown them all out of
court. If you can't identify some> procedural error on the
part of the officer, pay the fine.> Amid all the noise
associated with this posting, I am> wondering why nobody has
bothered to question what> computational time and human
reaction time have to do> with a Doppler velocity
measurement.> If your case is weak, try to baffle them with
BS.===
===
Subject:
: Re: Got a speeding ticket and need to fight
backI am sorry that I moved into Idiot Zone. Speed Up! I am
sorry, you can notkeep up with you. Go away. Dictionary.com
has made it easy for you . Lookup Go away.-suresh>
Approximately> f lamda = c> f lamda = 186 000 mps> f ->
frequency in hertz> h = 60min> = 60 * 60 sec> f lamda = 186
000> = 186 000 mps> f -> 1000,000,000 s^-1 (gigahertz)> lamda
= .000186 miles> That's the radar's wavelength approximately.
Take the radar's timewindow,> it internally uses to compute
the speed: 1/10 second ( the human eyecan> see about 30 frames
/ sec ). Assuming Reaction time of a driver = 1/10sec> approx>
0.000186/ .1s = .00186 mps = 6.6 mphThis has nothing to do
with how these radar units actually work. Google> on
'Doppler'.And then understand that most traffic court judges
have probably heard> them all. And thrown them all out of
court. If you can't identify some> procedural error on the
part of the officer, pay the fine.> Amid all the noise
associated with this posting, I am> wondering why nobody has
bothered to question what> computational time and human
reaction time have to do> with a Doppler velocity
measurement.> If your case is weak, try to baffle them with
BS.===
===
Subject:
: Re: Got a speeding ticket and need to fight
backDid you pass HSPT? or did you parents pay the school
board?-sureshThe Lord of Chaos (Suresh Devanathan)> I am sorry
that I moved into Idiot Zone. Speed Up! I am sorry, you cannot>
keep up with you. Go away. Dictionary.com has made it easy for
you .Look> up Go away.> -suresh> Approximately> f lamda = c> f
lamda = 186 000 mps> f -> frequency in hertz> h = 60min> = 60 *
60 sec> f lamda = 186 000> = 186 000 mps> f -> 1000,000,000
s^-1 (gigahertz)> lamda = .000186 miles> That's the radar's
wavelength approximately. Take the radar's time> window,> it
internally uses to compute the speed: 1/10 second ( the
humaneye> can> see about 30 frames / sec ). Assuming Reaction
time of a driver =1/10> sec> approx> 0.000186/ .1s = .00186
mps = 6.6 mph> This has nothing to do with how these radar
units actually work.Google> on 'Doppler'.> And then understand
that most traffic court judges have probably heard> them all.
And thrown them all out of court. If you can't identify some>
procedural error on the part of the officer, pay the fine.Amid
all the noise associated with this posting, I am> wondering why
nobody has bothered to question what> computational time and
human reaction time have to do> with a Doppler velocity
measurement.If your case is weak, try to baffle them with
BS.===
===
Subject:
: Re: Got a speeding ticket and need to fight
backThe Lord of Chaos (Suresh Devanathan)> Did you pass HSPT?
or did you parents pay the school board?> -sureshThe Lord of
Chaos (Suresh Devanathan)>> I am sorry that I moved into Idiot
Zone. Speed Up! I am sorry, you>> can > not>> keep up with you.
Go away. Dictionary.com has made it easy for you>> . > Look>>
up Go away.>> -sureshWho the hell are you talking to? You
reply to yourself and make no sense? Are you asking yourself
if you passed the HPST? Shouldn't you already know that
answer? Or, are we dealing with a feud between two of your
personalities? Please, if you expect any of this to make
sense, reply to actual message you want to comment on.Or,
perhaps you are setting up the case for an insanity plea....
===Timothy M. BrauchGraduate StudentDepartment of
MathematicsWake Forest University===
===
Subject:
: Re: Got a
speeding ticket and need to fight backThis is clearly a troll.
Insults come with every post. Killfile.Slainte,Fletch> The Lord
of Chaos (Suresh Devanathan)> Did you pass HSPT? or did you
parents pay the school board?> -sureshThe Lord of Chaos
(Suresh Devanathan)>> I am sorry that I moved into Idiot Zone.
Speed Up! I am sorry, you>> can> not>> keep up with you. Go
away. Dictionary.com has made it easy for you>> .> Look>> up
Go away.>> -suresh>Who the hell are you talking to? You reply
to yourself and make no> sense? Are you asking yourself if you
passed the HPST? Shouldn't you> already know that answer? Or,
are we dealing with a feud between two of> your personalities?
Please, if you expect any of this to make sense,> reply to
actual message you want to comment on.> Or, perhaps you are
setting up the case for an insanity plea....> ===> Timothy
M. Brauch> Graduate Student> Department of Mathematics> Wake
Forest University===
===
Subject:
: Re: Got a speeding ticket and
need to fight backOk now, Did you pass HSPT? or did you
parents pay the school board?-suresh> The Lord of Chaos
(Suresh Devanathan)> -sureshThe Lord of Chaos (Suresh
Devanathan)>> I am sorry that I moved into Idiot Zone. Speed
Up! I am sorry, you>> can> not>> keep up with you. Go away.
Dictionary.com has made it easy for you>> .> Look>> up Go
away.>> -suresh>Who the hell are you talking to? You reply to
yourself and make no> sense? Are you asking yourself if you
passed the HPST? Shouldn't you> already know that answer? Or,
are we dealing with a feud between two of> your personalities?
Please, if you expect any of this to make sense,> reply to
actual message you want to comment on.> Or, perhaps you are
setting up the case for an insanity plea....> ===> Timothy
M. Brauch> Graduate Student> Department of Mathematics> Wake
Forest University===
===
Subject:
: Re: Got a speeding ticket and
need to fight backBtw, a wonderful sign of your ignorance. Do
you know how the droppler shiftis analysed?Approximatelyf
lamda = c f lamda = 186 000 mpsf -> frequency in hertzh =
60min> = 60 * 60 secf lamda = 186 000> = 186 000 mpsf ->
1000,000,000 s^-1 (gigahertz)lamda = .000186 milesThat's the
radar's wavelength approximately. Take the radar's
timewindow,> it internally uses to compute the speed: 1/10
second ( the human eyecan> see about 30 frames / sec ).
Assuming Reaction time of a driver = 1/10sec> approx0.000186/
.1s = .00186 mps = 6.6 mph> This has nothing to do with how
these radar units actually work. Google> on 'Doppler'.> And
then understand that most traffic court judges have probably
heard> them all. And thrown them all out of court. If you
can't identify some> procedural error on the part of the
officer, pay the fine.> --> Hovnanian mailto:@Hovnanian.com>
note to spammers: a Washington State resident>
--------------------------------------------------------------
----> Bureaucrat, n.: A person who cuts red tape sideways. --
J. McCabe===
===
Subject:
: Re: Got a speeding ticket and need to
fight back>Thank God that people like you and Rob Duncan are
not engineers.You should>never build a computer. Dont ever
send your children to college for>engineering. They are most
likely to drive the companies, where they work,>bankrupt. If
15 out of every 100 instructions, a computer executes is
in>error, you know something, the computer is no more than
random number>generator.>That's the restatement of your
idiocy. You do not understand statistics.>Go away. You do not
understand pecentages. You know nothing about them.>You do not
even know how to use them. If you ever got a Ph.D in
psychology,>remember that all you did was bull your way thru
college.>How about if i told u, that there is 15% chance you
are going to die>tomorrow? Let's see: 85% chance that you wont
die. Let's see the power of>the 15% chance. You know 3/20
people hate you, and think you are ugly.>Out of 20 times you
drive, 3 times, the chances are you are going to
hit>something. chance are that our of 3 of 20 people, in
public, are going to>mug you.If your meds say that they have
to be taken every day then taking 7 pills oncea week is *not*
an acceptable substitute.-- ===
===
Subject:
: Re: 3 Squares Covering
1 Circle> Yet another coverings-problem...> (I myself might
have already asked this, but I do not believe that I,> at
least, have asked this specific question before.)If we have a
unit circle (radius = 1),and we have 3 unit squares (sides
=1),then what is the maximum area of the circle we can ever
cover with the> squaresa) if overlapping of the squares is not
allowed?b) if overlap is allowed?> I may be way off,> but I
conjecture that the (a) case issquares arranged like: ____> !
!> ! !> ---------> ! ! !> ! ! !> ---------But I bet tilting
the squares works better.As for the (b) case,if we start with
the (a) solution I give, then raise the bottom> squares (or
lower the top square by the same amount), then tilt the>
squares somewhat somehow,PERHAPS we might have an arrangement
which maximally covers a unit> circle placed properly on top
of the squares.Any better ideas or arrangements? (or more
precise descriptions of> possible/actual ideal
arrangement(s)?...)> Leroy QuetSorry, accidentally posted
message prematurely. Area of circle is [pi], area of 3 squares
is is 3. Maximum possible area is 3/[pi].===
===
Subject:
: Re: 3
Squares Covering 1 Circle|| Yet another coverings-problem...||
(I myself might have already asked this, but I do not believe
that I,|| at least, have asked this specific question
before.)|||| If we have a unit circle (radius = 1),|||| and we
have 3 unit squares (sides =1),|||| then what is the maximum
area of the circle we can ever cover with|| the squares|||| a)
if overlapping of the squares is not allowed?|||| b) if overlap
is allowed?|||||| I may be way off,|| but I conjecture that the
(a) case is|||| squares arranged like:|||| ____|| ! !|| ! !||
---------|| ! ! !|| ! ! !|| ---------|||| But I bet tilting
the squares works better.|||| As for the (b) case,|||| if we
start with the (a) solution I give, then raise the bottom||
squares (or lower the top square by the same amount), then
tilt the|| squares somewhat somehow,|||| PERHAPS we might have
an arrangement which maximally covers a unit|| circle placed
properly on top of the squares.|||| Any better ideas or
arrangements? (or more precise descriptions of||
possible/actual ideal arrangement(s)?...)|||||| Leroy Quet||
Sorry, accidentally posted message prematurely.|| Area of
circle is [pi], area of 3 squares is is 3. Maximum possible|
area is 3/[pi].Well yes the theoretical maximum is 3/pi
assuming we can get all of thearea of the squares within the
circle. But we clearly can't, I think theproblem means what is
the maximum area of the circle we *can* cover.-- Rob===
===
Subject:
:
Re: 3 Squares Covering 1 Circle> Yet another
coverings-problem...> (I myself might have already asked this,
but I do not believe that I,> at least, have asked this
specific question before.)If we have a unit circle (radius =
1),and we have 3 unit squares (sides =1),then what is the
maximum area of the circle we can ever cover with the>
squaresa) if overlapping of the squares is not allowed?b) if
overlap is allowed?> I may be way off,> but I conjecture that
the (a) case issquares arranged like: ____> ! !> ! !>
---------> ! ! !> ! ! !> ---------But I bet tilting the
squares works better.As for the (b) case,if we start with the
(a) solution I give, then raise the bottom> squares (or lower
the top square by the same amount), then tilt the> squares
somewhat somehow,PERHAPS we might have an arrangement which
maximally covers a unit> circle placed properly on top of the
squares.Any better ideas or arrangements? (or more precise
descriptions of> possible/actual ideal arrangement(s)?...)>
Leroy QuetThe area of the circle is PiI. the area===
===
Subject:
:
Re: A question about Kripke semantics and physics (Was: Re:
Study groups in science)> By the way, Franz. The vocabulary
word for the day is impudence.> Quite. It was impudent of you
to waste such a long space in this thread as> that which I
have just snipped without reading.Glad you recognized the
word.As for the rest of your reply, it is to be expected of
someone who spent moretime at faculty teas than at the
library.:-)===
===
Subject:
: Re: Columbia University uses False
Arrest, Document Destruction!But the whole situation is
competely unrelated to Mathematics. This> must not be a
subject of discussion in Sci.math section. Will the> editor
respond?Who?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.hLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Study groups
in science> if you are interested in scientific online
workshops, please visit> site>
http://de.geocities.com/scienceworkshops/> Its goal is to
organize study groups on scientific topics like quantum> field
theory,> probabilistic inference or neural nets, to name a few
topics I am> personally interested in> (of course, arbitray
topics may be suggested).What positive precautions are you
taking to prevent the idiots morons and> kooks from taking it
over, as has happened in sci.physics?Probably merely
formalizing it as a study group to study a particulartext is
sufficient kook-repellent. It doesn't seem like their
venue.===
===
Subject:
: Re: real continuous function nowhere
differentiable on R^1.iel C Bastos <iel@slashlog.org> a
.8ecrit s le message de> When 0 <= n <= m, the equation
|phi(s) - phi(t)| <= |s - t| implies> that |gamma_n| <= 4^n.
Since |gamma_m| = 4^m, we conclude that> [i dont this either,
not to mention the equation below]> | f(x + delta_m) - f(x) |
| oo |> | --------------------- | = | Sum( (3/4)^n * gamma_n)
|> | delta_m | | n=0 |> (m - 1)> 3^m - Sum ( 3^n )> n=0> = 1/2
(3^m + 1).> As m -> oo, delta_m -> 0. It follows that f is not
differentiable at> x.Why f is not differentiable at x.Only |
f(x + delta_m) - f(x) | | --------------------- | <= 1/2 (3^m
+ 1). | delta_m |===
===
Subject:
: Re: real continuous function
nowhere differentiable on R^1.>iel C Bastos <iel@slashlog.org>
a .8ecrit s le message de>> When 0 <= n <= m, the equation
|phi(s) - phi(t)| <= |s - t| implies>> that |gamma_n| <= 4^n.
Since |gamma_m| = 4^m, we conclude that>> [i dont this either,
not to mention the equation below]>> | f(x + delta_m) - f(x) |
| oo |>> | --------------------- | = | Sum( (3/4)^n * gamma_n)
|>> | delta_m | | n=0 |>> (m - 1)>> 3^m - Sum ( 3^n )>> n=0>> =
1/2 (3^m + 1).>> As m -> oo, delta_m -> 0. It follows that f is
not differentiable at>> x.>Why f is not differentiable at
x.>Only> | f(x + delta_m) - f(x) |> | --------------------- |
<= 1/2 (3^m + 1).> | delta_m |There was a >= missing in the
original, and you incorrectlyguessed it was supposed to be <=.
We actually have | f(x + delta_m) - f(x) | |
--------------------- | >= 1/2 (3^m + 1), | delta_m |which
shows that f is not differentiable at x, right?(We're using
the reverse triangle inequality: |a + b| >= |a| - |b|,which
implies for example |a + b + c| >= |a| - (|b| + |c|)by the
triangle inequality for |b + c|.)===
===
Subject:
: Generating
Functions and Counting ProblemsI'm going mad trying to
understand a problem in a textbook I have. It isbasically a
counting problem where you end up with (1 + x + x^2 + x^3)^5
andyou need to find the coefficient for x^12. The only
strategy that I havehad for working these types of problems is
to try and manipulate thegenerating function into a formula
that can be expressed as a summation, andthen using the
summation to find the coefficient of the term I need. But
Ican't for the life of me manipulate the above into anything
that I can use.Does anyone have any insight?-- peace
Kazuya===
===
Subject:
: Re: Generating Functions and Counting
Problems> I'm going mad trying to understand a problem in a
textbook I have. It> is basically a counting problem where you
end up with (1 + x + x^2 +> x^3)^5 and you need to find the
coefficient for x^12. The only> strategy that I have had for
working these types of problems is to try> and manipulate the
generating function into a formula that can be> expressed as a
summation, and then using the summation to find the>
coefficient of the term I need. But I can't for the life of
me> manipulate the above into anything that I can use. Does
anyone have> any insight? (1 + x + x^2 + x^3)^5 =
(1+x)^5(1+x^2)^5 =
[1+5x+10x^2+10x^3+5x^4+x^5].[1+5x^2+10x^4+10x^6+5x^8+x^10]10 +
25 = 35 ie 35x^12also [(x^4-1)/(x-1)]^5 = (1 + x + x^2 +
x^3)^5Carl===
===
Subject:
: Re: Generating Functions and Counting
ProblemsEn el
mensaje:xrmXb.3609$WW3.893@newsread2.news.pas.earthlink.net,DJ
Kazuya <djkazuya@earthlink.net> escribi.97:> I'm going mad
trying to understand a problem in a textbook I have.> It is
basically a counting problem where you end up with (1 + x +
x^2> + x^3)^5 and you need to find the coefficient for x^12.
The only> strategy that I have had for working these types of
problems is to> try and manipulate the generating function
into a formula that can be> expressed as a summation, and then
using the summation to find the> coefficient of the term I
need. But I can't for the life of me> manipulate the above
into anything that I can use. Does anyone have> any insight?(1
+ x + x^2 + x^3)^5 = (1 - x^4)^5/(1 - x)^51/(1 - x)^5 =
(1/24)d^4(1/(1-x))/dx^4As 1/(1 - x) = 1 + x + x^2 + x^3 + x^4
+ ... + x^n + ...d(1/(1-x))/dx = 1 + 2x + 3x^2 + 4x^3 + ... +
nx^(n-1) + ...d^2(1/(1-x))/dx^2 = 2 + 6x + 12x^2 + ... +
n(n-1)x^(n-2) + ...d^3(1/(1-x))/dx^3 = 6 + 24x + ... +
n(n-1)(n-2)x^(n-3) + ...d^4(1/(1-x))/dx^4 = 24 + ... +
n(n-1)(n-2)(n-3)x^(n-4) + ...(1/24)d^4(1/(1-x))/dx^4 =
Sum(C(n+4, 4)x^n, n, 0, inf)In general,1/(1 - x)^(k+1) =
(1/k!)d^k(1/(1-x))/dx^k = Sum(C(n+k, k)x^n, n, 0, inf)Then,(1
+ x + x^2 + x^3)^5 = (1 - x^4)^5/(1 - x)^5 = (1 -
x^4)^5*Sum(C(n+4, 4)x^n, n, 0, inf) = (1 - 5x^4 + 10x^8 -
10x^12 + ...)Sum(C(n+4, 4)x^n, n, 0, inf)Then the coefficient
of x^12 isC(16, 4) - 5C(12, 4) + 10C(8, 4) - 10C(4, 4) = 1820
- 5*495 + 10*70 - 10*1 = 35As it can be saw developping (1 + x
+ x^2 + x^3)^5. Perhaps more quickly, inthat particular case
...-- ===
===
Subject:
: Re: Generating Functions and Counting
Problems> I'm going mad trying to understand a problem in a
textbook I have. It> is basically a counting problem where you
end up with (1 + x + x^2 +> x^3)^5 and you need to find the
coefficient for x^12. The only strategy> that I have had for
working these types of problems is to try and> manipulate the
generating function into a formula that can be expressed> as a
summation, and then using the summation to find the coefficient
of> the term I need. But I can't for the life of me manipulate
the above> into anything that I can use. Does anyone have any
insight?1 + x + x^2 + x^3 = (1+x)(1+x^2)(1 + x + x^2 + x^3)^5
= (1+x)^5 (1+x^2)^5Let [x^n]f(x) mean coefficient of x^n in
f(x)And N-ch-i to mean 'N choose iThen [x^n](left side) = SUM
( [x^i](1+x)^5 * [x^j](1+x^2)^5 ) over all pairs i,j with
i+j=12.[x^i](1+x)^5 = 5-ch-i[x^j](1+x^2)^5 = 5-ch-(j/2) if j
is even, and 0 otherwisei=0 j=12 + i=1 j=11 + ... + i=11 j=1 +
i=12 j=0(all odd j terms drop to 0)(5-ch-0)(5-ch-6) +
(5-ch-2)(5-ch-5) + (5-ch-4)(5-ch-4) + (5-ch-6)(5-ch-3)= 0 + 10
+ 25 + 0 + ...= 35( since N-ch-k is 0 when k > N )Answer of 35
is verified by expansion.You should have a formula
representing such a sum of binomial coefficients for
extracting roots in your textbook. You might also be
interested in the multinomial theorem.J===
===
Subject:
: Continuum
hypotheses and Lebesgue measureHi all,Consider this statement:
If a Lebesgue-measurable subset M of the set R of real numbers
has cardinal less than the cardinal of R then the Lebesgue
measure of M is zero.This is trivial if we assume the
continuum hypothsis. And what if wedon't? It's easy to see
that if there were subsets of R whose cardinalwas in between
the cardinals of N (the naturals) and R, then, amongthose
sets, there would be Lebesgue-measurable sets with
Lebesguemeasure zero. But would there be necessarily, among
those sets,Lebesgue-measurable sets with Lebesgue measure
greater than zero?Best ,Jose Carlos Santos===
===
Subject:
: Re:
Continuum hypotheses and Lebesgue measure> Hi all,Consider
this statement: If a Lebesgue-measurable subset M of the set R
of real numbers has> cardinal less than the cardinal of R then
the Lebesgue measure of> M is zero.This is trivial if we
assume the continuum hypothsis. And what if we> don't? It's
easy to see that if there were subsets of R whose cardinal>
was in between the cardinals of N (the naturals) and R, then,
among> those sets, there would be Lebesgue-measurable sets
with Lebesgue> measure zero. But would there be necessarily,
among those sets,> Lebesgue-measurable sets with Lebesgue
measure greater than zero?> (*) Every subset of R with
cardinal < c is Lebesgue measurable[and therefore, as Ullrich
noted, of measure zero](*) is independent of ZFC. (*) is a
consequence of Martin's Axiom.But there are also models of ZFC
+ notCH where (*) fails:were there are subsets of cardinal
strictly between aleph_0 and c thathave positive Lebesgue
outer measure but (of course) zero innermeasure. As I recall
(It has been a long time...) the random realstype models do
this.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re:
Continuum hypotheses and Lebesgue measure>Hi all,>Consider
this statement:> If a Lebesgue-measurable subset M of the set
R of real numbers has> cardinal less than the cardinal of R
then the Lebesgue measure of> M is zero.>This is trivial if we
assume the continuum hypothsis. And what if we>don't? It's easy
to see that if there were subsets of R whose cardinal>was in
between the cardinals of N (the naturals) and R, then,
among>those sets, there would be Lebesgue-measurable sets with
Lebesgue>measure zero. But would there be necessarily, among
those sets,>Lebesgue-measurable sets with Lebesgue measure
greater than zero?Hmm... [five mnutes pass]. Huh, my first
guess was wrong.In fact regardless of CH any set of positive
measure has cardinalityc.Pf: If m(A) > 0 then A has a compact
subset K with m(K) > 0. HenceK is uncountable, and it follows
that K has a subset homeomorphicto the Cantor set. (Hint:
There exist a < b < c < d such that [a, b] intersect K and [c,
d] intersect K are both uncountable - repeatuntil done.)
QED.>Best ,>Jose Carlos Santos===
===
Subject:
: Re: Continuum
hypotheses and Lebesgue measure>>Consider this statement:>> If
a Lebesgue-measurable subset M of the set R of real numbers
has>> cardinal less than the cardinal of R then the Lebesgue
measure of>> M is zero.>>This is trivial if we assume the
continuum hypothsis. And what if we>>don't? It's easy to see
that if there were subsets of R whose cardinal>>was in between
the cardinals of N (the naturals) and R, then, among>>those
sets, there would be Lebesgue-measurable sets with
Lebesgue>>measure zero. But would there be necessarily, among
those sets,>>Lebesgue-measurable sets with Lebesgue measure
greater than zero?> Hmm... [five mnutes pass]. Huh, my first
guess was wrong.In fact regardless of CH any set of positive
measure has cardinality> c.Pf: If m(A) > 0 then A has a
compact subset K with m(K) > 0. Hence> K is uncountable, and
it follows that K has a subset homeomorphic> to the Cantor
set. (Hint: There exist a < b < c < d such that [a, b] >
intersect K and [c, d] intersect K are both uncountable -
repeat> until done.) QED.OK, I got it. I thought about a
similar approach (after posting myquestion): since K could not
have the cardinality of the reals, it didnot contain any
interval; therefore K, and any closed subset of K, wouldbe
totally disconnected. If I could prove (but I was unable to do
it)that K contained some perfect subset K* with more than one
point, thenK*, being compact, perfect, and totally
disconected, would behomeomorphic to the Cantor set.Jose
Carlos Santos===
===
Subject:
: Re: Kerry and lowest/highest common
denominator> Kerry likes to say ``I don't think Presidents
aren't supposed to>go for the lowest common denominator, I
think they're supposed to look>for the highest common
denominator.''>eg.
http://home.att.net/~rhhardin6/imuscut.kerrydenominator.ra
(32kb) Internet Explorer warns that *.ra files can be gerous.A
search for denominator at www.kerry.com finds no match.>Is this
a ``potatoe'' thing?>-- >Ron Hardin>rhhardin@mindspring.com>On
the internet, nobody knows you're a jerk. Is KERRY a HORNY
JERK?-- Adams served two terms as Vice President and one as
President, but lostreelection. Later his son became President
despite losing the popular vote.That son lost his reelection
attempt badly. Now history is repeating itself.pmontgom@cwi.nl
Microsoft Research and CWI Home: San Rafael,
California===
===
Subject:
: Re: Kerry and lowest/highest common
denominatorI'm pretty sure that ``lowest common denominator''
is used in puttingfractions in common terms, ie with the same
denominator, so the LCDof 9 and 12 is 36.And that ``highest
common denominator'' is the ``highest common divisor'', thus
the HCD of 9 and 12 is 3.They have different uses, and a
history of phrase corruption, so thatHCD <= LCD today.To
contrast them, since they really come from different
situations,is what makes it a ``potatoe'' incident. The
rhetoric has a nicesweep but it makes zero sense.-- Ron
Hardinrhhardin@mindspring.comOn the internet, nobody knows
you're a jerk.===
===
Subject:
: Re: Kerry and lowest/highest common
denominator>Another well-known example:>I COULD care
less!...>when what is actually meant is>I could *not* care
less!...Not necessarily. If caring is rated on a 0-100 scale,
I would saycould not care lessif my concern was exactly 0,
whereas I might sayI could care lessif my concern was merely
low, such as 10 or 12.--MensanatorAce of Clubs===
===
Subject:
: Re:
Kerry and lowest/highest common denominator> if my concern was
merely low, such as 10 or 12. If your concern is 'low' but not
aboslute-zero low, then why bother saying that you could care
less and just go ahead and start caring less. When one loves
someone, it is sometimes said I can *not* love you more ...
never does anyone tell their significant other I could love
you more The I could care less expression is just carelessness
passed from person to person.J===
===
Subject:
: Re: Kerry and
lowest/highest common denominator===>
===
Subject:
: Re: Kerry and
lowest/highest common denominator>Message-id:
<Pine.LNX.4.44.0402132359340.27770-100000@eva117.
cs.ualberta.ca> if my concern was merely low, such as 10 or
12.> If your concern is 'low' but not aboslute-zero low, then
why bother >saying that you could care less and just go ahead
and start caring less.> When one loves someone, it is
sometimes said I can *not* love you >more ... never does
anyone tell their significant other I could love you >moreBut
isn't that a commonly used phrase in snake-oil marketing?I
could love you more...if you had a bigger penis.> The I could
care less expression is just carelessness passed from >person
to person.That may be, but the world isn't black and white and
I wouldn't wantto be denied a certain subtlety of
expression.>J--MensanatorAce of Clubs===
===
Subject:
: Re: Kerry and
lowest/highest common denominatorthat's Least Common and
Greatest Common Denominator. the whole potatoe thing was just
an archiac English usagethat Quayle got from his tutor.> Kerry
likes to say ``I don't think Presidents aren't supposed to> go
for the lowest common denominator, I think they're supposed to
look> for the highest common denominator.'' > Is this a
``potatoe'' thing? thus quoth:A COORDINATED WALL ST.
ASSAULTWho Killed U.S. Nuclear Power? by Marsha
Freeman--zero-dimensional holograms.utterly compressed
information.DemConK2 cloning malfunction:(Citizen -mick)
Kane/Bore/Gush/Nadir;LaRouche et al versus Fowler et al, '96
--Supreme Decision, March 27, 2000. but, hey;the Voting Rights
Act is upfor republication in '07! of course, a lot of this is
moot, if we allow Cheenyto grind us into Tony's McCrusade
(Usama's MacJihad). see my sig.Give the World a Trickier Dick
Cheeny -- out of office after GIGA
years.http://www.benlinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanac===
===
Subject:
: Re:
Integrationism and pareto-optimal policiesIlya Shambat>
...Over 40 kilobytes! And look at the crossposting. Sci.math
certainly isacquiring a dubious reputation. I won't be
surprised if some psychologyLH===
===
Subject:
: (p-1)'th cyclotomic
polynomialHow come, the sum of the roots of the (p-1)th
cyclotomic polynomial(mod p)where p is a prime, are always
equal to 1, -1 or 0?===
===
Subject:
: Re: (p-1)'th cyclotomic
polynomial> How come, the sum of the roots of the (p-1)th
cyclotomic polynomial(mod p)> where p is a prime, are always
equal to 1, -1 or 0?The sum of the roots (zeroes) of the n-th
cyclotomic polynomial(the primitive n-th roots of unity) is
mu(n) where mu is theMobius function.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.hLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Math Too
Advanced For Mainstream EconomistsMy URL suggests that
explaining wages and employment by the supplyand demand for
labor is problematic:
<http://www.dreamscape.com/rvien/Sraffa4/Sraffa4.h>Poor Mark
Witte seems to believe that the supply and demand modelis
sometimes reasonable. But he cannot and will not present
anyargument whatsoever that addresses my argument (which
merelyechoes well-established results in the
literature).Instead he tries to shift the burden of proof:>
Question: Does the originator of this thread believe that a>
well-behaved supply and demand model is never appropriate for>
explaining how labor markets function?This raise an empirical
question. Does poor Mark Witte realizehe has no point?-- Try
http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/
Bukharin.h To solve Linear Programs: .../LPSolver.hr c A game:
.../Keynes.h v s a Whether strength of body or of mind, or
wisdom, or i m p virtue, are found in proportion to the power
or wealth e a e of a man is a question fit perhaps to be
discussed by n e . slaves in the hearing of their masters, but
highly @ r c m unbecoming to reasonable and free men in search
of d o the truth. -- Rousseau===
===
Subject:
: Help to verify an
open mappingf: R^n{0} -> S^(n-1) x |-> x/||x|| where ||.|| is
the euclidean norm. How to prove, analytically, that f is an
open mapping?----== Posted via Newsfeed.Com -
Unlimited-Uncensored-Secure Usenet
News==----http://www.newsfeed.com The #1 Newsgroup Service in
the World! >100,000 Newsgroups---= 19 East/West-Coast
Specialized Servers - Total Privacy via Encryption
=---===
===
Subject:
: Image ComparisonX-User-Info: 202.72.138.67
202.72.138.67 oliver1@westnet.com.au Wondering if anyone can
advise me here please.Im working on a computer vision project
and need to compareto images each image is an ouline one pixel
wide and the coordinatesfor every pixel in each image is
available.My question is this, if view both images as
matricies is theresome way of comparing the two
images/matricies to see whetherthey represent the same object
regardless of their orientation/rotationor scaling??forum for
this question.===
===
Subject:
: God is the Talk Show Host on Cosmic
RadioIn simple terms: Einstein's gravity is FM cosmic radio
and dark That's my theory in a nutshell! :-)Jack, you
say,about my WMAP ratio paper that was recently removed by
arXiv andwhich is on my web site
athttp://www.innerx.net/personal/tsmith/WMAPpaper.pdfthatyou
... cannot understand it ... and you ... do not understandthe
correspondence you make ....The correspondence
is:-----------------------------------------------------------
-----DE (dark energy, cosmological constant) - the 10
Rotations, Boosts, and Special Conformal generatorsDM (dark
matter) - the 4 TranslationsOM (ordinary matter) - the 1
Dilatation----------------------------------------------------
-----------Yes, I do not understand why you do that. It seems
completely arbitrary to me without compelling reason.What you
do here is not consistent with my theory, which is much
simpler.What I would have here isEinstein gravity spin 1
tetrad field from locally gauging the 4 Translations. Shipov
torsion field, which is maybe spin 3/2 (supersymmetry), from
locally gauging the 6 Lorentz Rotations not done by Einstein
in the original 1915 geometrodynamics.Spin 0 scalar gravity
from locally gauging the 1 Dilatation.A second spin 1 tetrad
field from locally gauging the 4 special conformal
transformations to uniformly accelerated hyperbolic motion of
the relativistic rocket in special relativity.I am not sure of
any of the above as yet since I have not worked through the
algebra.TSThe correspondence is clear to me in terms of
Segal's intuitiveconformal model, and I have a link to my web
site for more detailsat the word correspondence in my WMAP
ratio paper (on the lineimmediately above the statement of the
correspondence).It's not clear to me and it seems wrong in the
light of what I have done.Einstein's gravity (spin 2 metric
field) emerges from the phase modulation of the macro-quantum
superfluid ODLRO vacuum coherence.Both dark energy and dark
matter are w = -1 residual micro-quantum zero point energy
normal fluid artifacts emergent from the amplitude modulation
of the same macro-quantum superfluid ODLRO vacuum coherence.
Dark energy anti-gravitates and dark matter gravitates from
battle-tested standard physics of the equivalence principle,
covariance of the local laws of physics and the Heisenberg
uncertainty principle. They do so strongly at short range and
are not limited by G(Newton) that applies only in the
long-range.I do not understand Segal. I have not read him. His
work may be important but not for dark energy/matter in my
opinion.In simple terms: Einstein's gravity is FM cosmic radio
and dark That's my theory in a nutshell! :-)TA: The link is
tohttp://www.innerx.net/personal/tsmith/
coscongraviton.hHowever,it is fair for you to ask me for an
explanation of the correspondenceterms of your model, so here
is my attempt at such an
explanation:--------------------------Dark Energy DE:DE is the
NORMAL state of stuff in our universe (it is now,according to
WMAP, about 73% of it).I do not understand you.It looks more
like deSitter spacetime than Minkowski spacetime.I thought
that would be anti-deSitter hyperbolic since the anti-gravity
makes the universe accelerate in its expansion not
decelerate?Also I do not think there is anything fundamental
in the 73%. It is probably contingent unless iel can somehow
get it from an error-correcting code sphere packing in
hyperspace information flow model, but that model would have
to be compelling in terms battle-tested physics like general
relativity and quantum theory.TS:In Segal's model and as
Aldrovandi and Peireira show in somemathematical detail in
their paper athttp://xxx.lanl.gov/abs/gr-qc/9809061the DE
spacetime structure comes from ... the group Q,formed by a
semi-direct product between Lorentzand special conformal
transformation groups ....Those arethe 10 Rotations, Boosts
and Special Conformal generatorsthatcorrespond to DE in my
WMAP ratio paper.You mean Poincare group not Lorentz group. In
any case what you did here is to replace one mystery by another
mystery.I do not understand your explanation. I have not read
the papers you cite. In any case I have a simple explanation
of Einstein's gravity with both dark energy and dark matter
from the condensed matter physics point of view of More is
different so I do not need the stuff you mention since it is
excess theoretical baggage in my view. I do more with less
here I think. Of course, I may be completely wrong but I
understand my picture and I have no clue about the Segal stuff
you cite until you can explain it clearly and compellingly in
your own words.--------------------------TS: Dark Matter DM:DM
is a lesser part (it is now, according to WMAP, about 23% of
it)of our universe, and differs from the dominant DE by
beingbased on the 4 Translations.Again I do not understand
you.TS: In your terms,... Locally gauge the 4 parameter
translation subgroup T4, to get alocal compensating field that
is the spin 1 Einstein tetrad field ....Those4
Translationsthereforecorrespond to DM in my WMAP paper.formula
based on the two-fluid model well known in superfluid
theory./zpf = (Area Quantum)^-1[(Area Quantum)^3/2|Vacuum
Coherence|^2 - 1]DM is attractive /zpf < 0DE is repulsive /zpf
> 0from Einstein's exotic vacuum equationGuv + /zpfguv = 0End
of story. Very simple. We don't need excess theoretical
baggage from Segal to, solve that simple observational problem
in my opinion. I could be wrong but I see no compelling reason
to think I am at this moment.--------------------------TS:
Ordinary Matter OM:OM (the stuff of which we and Earth are
made) is sort of weird andexceptional (it is now, according to
WMAP, only about 4% of it).For us to call it ordinary is quite
provincial,because it is only ordinary in the context of our
physical bodiesand the planet on which we live.What
characterizes all OM is that its mass comes from the
Higgsmechanism.You say ... Finally what about the dilaton? Who
ordered that? :-)I guess its compensating gauge field is spin 0
gravity? ....Your guess is close,but the Dilatation actually
gives the spin 0 Higgs field,Well that should give my/zpf =
(Area Quantum)^-1[(Area Quantum)^3/2|Vacuum Coherence|^2 -
1]Because the Higgs bosons come from linear noise in the
|Vacuum Coherence| background in terms of the renormalizable
Mexican Hat potential for the Vacuum Coherence (Inflation)
local field. It's the same as I did in my 1966 paper with
Stoneham on the Goldstone Theorem and the Jahn-Teller Effect
in Proceedings of the Physical Society of London. It's same
simple game with some variations that all the Pundits have
been playing for a long time. I got it from Goldstone and
Higgs and mainly Kibble from going down to Imperial College
from Harwell in the summer of 1966.TS: and therefore all the
mass of OM,sothe 1 Dilatationthereforecorresponds to OM in my
WMAP paper.Your associations here are too vague for my mind to
grasp. The problem seems different to me, but I could be wrong
of course.------------------------------TS:If you want to see
our universe in your terms,you might say thatDE and DM are ...
TWO types ... of spacetimeandthatOM is something like a little
frothy foam on/in the DE/DM system.No, I do not say anything
like that.The DE and DM are simply the high and low superfluid
density non-equilibrium exotic phases of the physical vacuum
relative to the non-gravitating /zpf = 0 equilibrium vacuum.
DE is like a high density peak and DM is like a low density
peak in a longitudinal spin 0 dilaton standing sound wave in
the ether (physical vacuum). shell so frequencies and
wavelengths are not tied together as they are in far field on
mass shell propagation.TS: As my WMAP ratio paper says,if you
use the above correspondence you get a present-day ratio DE :
DM : OM = 75.3% : 20.2% : 4.5%which isclose to the WMAP
observation of 73% : 23% : 4%.I do not understand how you
compute those numbers. In any case, I think those numbers are
contingent in sense of WAP in Max Tegmark's infinity of
infinity of parallel universes. My Father's House has many
mansions.===
===
Subject:
: Automatic Problem Solver 3.0 New
Release,The third edition of APS has been released and a free
demo of APS 3.0 is nowavailable for
download:http://www.gepsoft.com/gepsoft/General features of
version 3.0:o Translates the evolved models into 8 different
languages (C, C#, C++,Java, JavaScript, Visual Basic, VB.NET,
and Fortran).o Draws the parse trees of the evolved models.o
Translates the evolved models virtually into any programming
languagethrough User Defined Grammars.o A total of 70
different built-in mathematical functions and comparisonrules
plus Dynamic UDFs and Static UDFs for modeling.o A total of 11
built-in fitness functions for Function Finding.o A total of 10
built-in fitness functions for Classification.o A total of 11
built-in fitness functions for Times Series Prediction.o User
Defined Fitness Functions for all problem categories.o
Implements a new algorithm for handling random numerical
constants.o Data screening engine for preprocessing.o Time
series transformation engine.o Evolution from seed models.o
Change seed utilities.o Saves all the best-of-generation
models of a run.o Plots the evolutionary dynamics of the run.o
Complexity increase engine.o Supports Databases and Text Files
both for loading input data andscoring.o Recursive testing and
prediction for Time Series.o Implements an extensive package of
statistical indexes for modelevaluation.o and much more.The
demo allows you to use your own datasets and all the features
areoperational except scoring/prediction and visualization of
the models.However, for the included datasets and sample runs
all APS features arefully operational in the demo.Enjoy!All
the best,Candida
Ferreira------------------------------------------------------
-----Candida Ferreira, Ph.D.Chief Scientist, Gepsoft73 Elmtree
DriveBristol BS13 8NA, UKph: +44 (0) 117 330
9272http://www.gepsoft.com/gepsofthttp://
www.gene-expression-programming.com/author.asp----------------
-------------------------------------------===
===
Subject:
: short
exact sequence of vector spaces by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i1EDOdo00481;I'm trying to show every short exact sequence of
vector spaces splits.Any suggestions?===
===
Subject:
: Re: short
exact sequence of vector spaces> I'm trying to show every
short exact sequence of vector spaces splits.> Any
suggestions?> Bases. (For the infinite-dimensional case, Hamel
bases.)-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: e is
transcendental (was: classes of transcendental numbers ? by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1EDOg000563;><snip>My question
:>>What is the implication of this second value of
e^[ipi]=0 ?>>No implication. It is not true.>>
e^[iPi]=mod 1, arg 180>e^[iPi]=mod 1, arg 180 degrees>>
This is correct in polar form.>But it has absolutely nothing
to do with your earlier extraordinary>claim that exp(i pi) =
0. That particular claim of yours is still>unjustified, and
it is still false.>>It has to do.>Not from what I have
seen.>>cos pi + i sin pi=-1+i*0 =-1 is correct>>However the
phasor e^[ipi] =cos pi + i sin pi ,>> is properly defined as
MOD 1 , ARG [THETA]>No. You are just comparing two different
ways of specifying >a nonzero complex number. Both
characterizations are equally>valid.>> If I state that a
phasor A=-1 IS THIS CORRECT?>>No it must be accompanied by its
length and its direction.>>A=-1 shows a length of unity
directed towards the [0,0] point >>[of an X-Y coordinates
system crossing]>>from any angular direction around a unit
circle,from >>the positive X-axis (taken as zero with the
usual notation.>You phasor points to [-1,0] from [0,0].>>
e^[ipi] cannot loose its value as 1 (unity),>Whatever THAT is
supposed to mean.>>and with the usual>>notation it is positive
,leaving away the [0,0] point>> e^[ipi] = -1 it is driven
towards the [0,0] point>No. On the complex plane, you are
thinking of an arrow >pointing from [0,0] to [-1,0].>> since>>
e^[ipi] =-1+i*{0) is a technicality>No, it is not.>> and
SELECTIVELY one may choose either of the two
components>>according to its interest, to give the real ,or
the imaginary>>solution.>This looks like the claim of an
electrical engineer who works with>complex numbers, and then
takes the real or imaginary part at the>end. It has nothing to
do with the fact that exp(i pi) = -1.>> Simply taking care of
the angle ,it is fair for one to state>>tha unity (mod 1 of
e^[ipi]) at 180 deg ,anticlockwise from the>>positive X-axis
,has two components ,one of which the real(by >>convention)has
the value of one, and the imaginary (by convention)>>Zero.>So?
The real part of exp(i pi) is cos(pi), and its imaginary
part>is sin(pi), so all you are saying is that cos(pi) = -1
and >sin(pi) = 0, and we were already aware of these facts.
There is >NO reason to conclude that exp(i pi) = 0. I,have a
reason , with my due respect. Panagiotis Stefanides If you
take the real >part or imaginary part of a complex linear
combination of >exp(i theta_k), k = 1,...,n, then you get a
real linear >combination of cos(theta_k) and sin(theta_k), k =
1,...,n.>Once you go to real and imaginary parts, expressions
like >exp(i theta_k) are no longer appropriate, and you have
to use>cos(theta_k) and sin(theta_k) instead.>Alternatively,
if you want the imaginary part of exp(i pi), the>appropriate
expression is [exp(i pi) - exp(-i pi)]/(2i), and>not just
exp(i pi).>David McAnally>--------------===
===
Subject:
: Re: e is
transcendental (was: classes of transcendental numbers ?
charset=iso-8859-7 Eur Ing Panagiotis Stefanides
<panamars@otenet.gr>   > I,have a reason , with my due
respect.> Panagiotis
Stefanideshttp://www.crank.net/maths.h--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: hyperbolas> I would like to know
how to find the gradient of the tangent at the> turning point
of a hyperbola that doesn't have asymptotes going> straight up
and straight across. Finding the turning point of a nice>
simple hyperbola is easy since the gradient is 1 or -1,
that's> obvious, but I'm not quite sure how to find them for
otherThe turning point (apex) are on one bissector of the
asymptotes.The tangent in these points are parallel to the
other bissector.Calculation is rather messy in the most
general case when hyperbolais defined by a general quadratic
f(x,y)=0.But everything depends on how is defined your
hyperbola...First find the asymptotes, then the
bisectors.Taking these bisectors as coordinate axis,
hyperbola'sequation is X^2/A^2-Y^2/B^2=1. apex are X=+/-A,
Y=0,and tangent is parallel to Y axis.-- philippe(chephip at
free dot fr)===
===
Subject:
: re:hyperbolasYeah I know it will be
but I don't know how to find it!! I'm no mathswhiz, I'm only
17, please can someone tell me how to do it? :(----== Posted
via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet
News==----http://www.newsfeed.com The #1 Newsgroup Service in
the World! >100,000 Newsgroups---= 19 East/West-Coast
Specialized Servers - Total Privacy via Encryption
=---===
===
Subject:
: Re: easy...analysis problem....>
hello.......genius teacher........> let function f :E -> R is
differentiable on open set E in R> 1. let any a in E , f''(a)
exist.> show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) /
(h^2)} , h->0> 2. let any a in E , n-derivative f^(n) (a)
exist.> express f^(n) (a) by limit of f(x)>
-------------------------> um.....i solved first problem.> lim
{f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0> = lim [{f(a+h)-f(a)} -
{f(a-h+h)-f(a-h)} / (h^2)]> = lim {f'(a) - f'(a-h)} / h , let
h=(-t)> = lim {f'(a) - f'(a+t)} / (-t)> t->0> = lim {f'(a+t) -
f'(a)} / t> = f''(a)> it's right??> but i can't solve second
problem....> You haven't solved the FIRST problem, either.
Your passage> = lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} /
(h^2)]> = lim {f'(a) - f'(a-h)} / h , let h=(-t)> is flawed.>
Hint: apply L'Hopital's Rule to the original quotient, once.
Then use> the definition of the second derivative.> Another
hint for the first problem: since f''(a) exists, f'(a) exists>
and f(a+h) = f(a) + h f'(a) + (h2/2)f''(a) + o(h^2).Is that
true? Taylor's theorem only guarantees that f(a+h) = f(a) + h
f'(a) + h^2/2 f''(z)for some z between a and a+h; but we're
NOT told that f'' is> continuous, only that it exists
throughout the open set.That was why I suggested L'H.--Ron
BruckUh, the hypotheses state that f is differentiable on E.
So f'(a)> exists for any a in E. (You're technically right,
Ron, but not> particularly helpful. :] )I think he's
technically wrong.... All that`s required for f(a+h) =f(a) + h
f'(a) + (h2/2)f''(a) + o(h^2) to be true is existence
off''(a).Artur===
===
Subject:
: Re: easy...analysis problem....>
hello.......genius teacher........> let function f :E -> R is
differentiable on open set E in R> 1. let any a in E , f''(a)
exist.> show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) /
(h^2)} , h->0> 2. let any a in E , n-derivative f^(n) (a)
exist.> express f^(n) (a) by limit of f(x)>
-------------------------> um.....i solved first problem.> lim
{f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0> = lim [{f(a+h)-f(a)} -
{f(a-h+h)-f(a-h)} / (h^2)]> = lim {f'(a) - f'(a-h)} / h , let
h=(-t)> = lim {f'(a) - f'(a+t)} / (-t)> t->0> = lim {f'(a+t) -
f'(a)} / t> = f''(a)> it's right??> but i can't solve second
problem....> You haven't solved the FIRST problem, either.
Your passage> = lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} /
(h^2)]> = lim {f'(a) - f'(a-h)} / h , let h=(-t)> is flawed.>
Hint: apply L'Hopital's Rule to the original quotient, once.
Then use> the definition of the second derivative.> Another
hint for the first problem: since f''(a) exists, f'(a) exists>
and f(a+h) = f(a) + h f'(a) + (h2/2)f''(a) + o(h^2).Is that
true? Taylor's theorem only guarantees that f(a+h) = f(a) + h
f'(a) + h^2/2 f''(z)for some z between a and a+h; but we're
NOT told that f'' is> continuous, only that it exists
throughout the open set.But this is not Taylor's theorem. I'm
not assuming f'' is continuous.All I assume is f''(a) exists.
I'm not even assuming f'' exists, letalone is continuous, at
any point of the open set other than a itself.where o(h^2)
goes to zero faster than h^2 goes. All that's requiredfor this
equation to be true is the existence of f''(a), nothinh
more.Artur===
===
Subject:
: Re: easy...analysis problem....>
hello.......genius teacher........> 2. let any a in E ,
n-derivative f^(n) (a) exist.> express f^(n) (a) by limit of
f(x)> let me advice, please.....sir~==========Denote
C(n,k)=n!/(k!(n-k)!) and D(n,h;f)= SUM_{k=0 to
k=n}(-1)^{n-k}C(n,k)*f(a+k*h)when a+k*h , k in {0,1,...,n}
belong to E. When f:E-->R is suitable restricted (e.g. f in
C^{m}[p,q], p<a< q in E , m >= n), try to verify if equality
D(n,h;f) f^{(n)}(a)= lim_{h-->0} -------- h^n is verified.
Perhaps help, Alex.======================
===
Subject:
: Re:
easy...analysis problem....> hello.......genius
teacher........> 2. let any a in E , n-derivative f^(n) (a)
exist.> express f^(n) (a) by limit of f(x)> let me advice,
please.....sir~> ==========> Denote C(n,k)=n!/(k!(n-k)!) and D(n,h;f)= SUM_{k=0 to k=n}(-1)^{n-k}C(n,k)*f(a+k*h)> when
a+k*h , k in {0,1,...,n} belong to E. > When f:E-->R is
suitable restricted> (e.g. f in C^{m}[p,q], p<a< q in E , m >=
n), try to> verify if equality> D(n,h;f)> f^{(n)}(a)=
lim_{h-->0} --------> h^n > is verified. Perhaps help, Alex.>
=================== For instance, books written by Steffensen
(,,Interpolation), Norlund , Ch.Jor (,,Calculus of Finite
Differences?, Chelsea). By the way, there is the following
nice ,,asymptotic formula : In the following [0,1] is a subset
of E , a in (0,1), f:E-->R is such that f(a) exists.Denote
(B_nf)(a)=SUM_{k=0 to k=n}C(n,k)a^k(1-a)^{n-k}f(k/n)
(Bernstein polynomial of f) W(n,f;a) = n*( f(a) - (B_nf)(a)
).Then W(n,f;a)f(a)=2*lim_{n-->infty} -----------
(E.Voronovskaja-1932) . a(1-a)The above result was generalized
by S.N.Bernstein -1932 for derivatives of even orders [see
G.G.Lorentz ,,Bernstein polynomials, Toronto Univ.Press-1953]
.===
===
Subject:
: Re: easy...analysis
problem....Content-transfer-encoding: 8bit> hello.......genius
teacher........> let function f :E -> R is differentiable on
open set E in R> 1. let any a in E , f''(a) exist.> show that
f''(a) = lim {f(a+h) - 2f(a) + f(a-h) / (h^2)} , h->0> 2. let
any a in E , n-derivative f^(n) (a) exist.> express f^(n) (a)
by limit of f(x)> -------------------------> um.....i solved
first problem.> lim {f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0> =
lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]> = lim {f'(a) -
f'(a-h)} / h , let h=(-t)> = lim {f'(a) - f'(a+t)} / (-t)>
t->0> = lim {f'(a+t) - f'(a)} / t> = f''(a)> it's right??>
but i can't solve second problem....> You haven't solved the
FIRST problem, either. Your passage> = lim [{f(a+h)-f(a)} -
{f(a-h+h)-f(a-h)} / (h^2)]> = lim {f'(a) - f'(a-h)} / h , let
h=(-t)> is flawed.> Hint: apply L'Hopital's Rule to the
original quotient, once. Then use> the definition of the
second derivative.> Another hint for the first problem: since
f''(a) exists, f'(a) exists> and f(a+h) = f(a) + h f'(a) +
(h2/2)f''(a) + o(h^2).Is that true? Taylor's theorem only
guarantees that f(a+h) = f(a) + h f'(a) + h^2/2 f''(z)for some
z between a and a+h; but we're NOT told that f'' is>
continuous, only that it exists throughout the open set.That
was why I suggested L'H.--Ron BruckUh, the hypotheses state
that f is differentiable on E. So f'(a)> exists for any a in
E. (You're technically right, Ron, but not> particularly
helpful. :] )?? Don't understand the remark, since my L'H
suggestion still works. We have lim_{h to 0} [f(a+h) + f(a-h)
- 2f(a)]/h^2 = lim_{h-->0} [f'(a+h) - f'(a-h)]/(2h)IF this
limit exists; and we have lim_{h -> 0} [f'(a+h) -
f'(a-h)]/(2h) = (1/2) lim_{h -> 0} [f'(a+h)-f'(a)]/h + 1/2
lim_{h -> 0}[f'(a-h)-f'(a)]/(-h) = f''(a)by the definition of
f''.My objection to the suggestion that f(a+h) = f(a) + h
f'(a) + h^2/2 f''(a) + o(h^2)holds with even greater force if
f'' is only defined at a. The OP isobviously not a native
English speaker; I took her remark let any a inE, f''(a) exist
to be a garbled way of saying f''(a) exists for all ain E. In
context, your interpretation (a is fixed, f''(a) exists)
ismore natural.So here's the question: if f is differentiable
in a neighborhood of a,and twice differentiable at a, does it
follow that f(a+h) = f(a) + h f'(a) + h^2/2 f''(a) +
o(h^2)?--Ron Bruck===
===
Subject:
: Re: easy...analysis problem....>
hello.......genius teacher........> let function f :E -> R is
differentiable on open set E in R> 1. let any a in E , f''(a)
exist.> show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) /
(h^2)} , h->0> 2. let any a in E , n-derivative f^(n) (a)
exist.> express f^(n) (a) by limit of f(x)>
-------------------------> um.....i solved first problem.> lim
{f(a+h) - 2f(a) + f(a-h) / (h^2)}> h->0> = lim [{f(a+h)-f(a)} -
{f(a-h+h)-f(a-h)} / (h^2)]> = lim {f'(a) - f'(a-h)} / h , let
h=(-t)> = lim {f'(a) - f'(a+t)} / (-t)> t->0> = lim {f'(a+t) -
f'(a)} / t> = f''(a)> it's right??> but i can't solve second
problem....> You haven't solved the FIRST problem, either.
Your passage> = lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} /
(h^2)]> = lim {f'(a) - f'(a-h)} / h , let h=(-t)> is flawed.>
Hint: apply L'Hopital's Rule to the original quotient, once.
Then use> the definition of the second derivative.> Another
hint for the first problem: since f''(a) exists, f'(a) exists>
and f(a+h) = f(a) + h f'(a) + (h2/2)f''(a) + o(h^2).> Is that
true? Taylor's theorem only guarantees that> f(a+h) = f(a) + h
f'(a) + h^2/2 f''(z)> for some z between a and a+h; but we're
NOT told that f'' is> continuous, only that it exists
throughout the open set.> That was why I suggested L'H.> --Ron
BruckUh, the hypotheses state that f is differentiable on E. So
f'(a)> exists for any a in E. (You're technically right, Ron,
but not> particularly helpful. :] )?? Don't understand the
remark, since my L'H suggestion still works. > We have lim_{h
to 0} [f(a+h) + f(a-h) - 2f(a)]/h^2 = lim_{h-->0} [f'(a+h) -
f'(a-h)]/(2h)IF this limit exists; and we have lim_{h -> 0}
[f'(a+h) - f'(a-h)]/(2h) = (1/2) lim_{h -> 0}
[f'(a+h)-f'(a)]/h + 1/2 lim_{h -> 0}[f'(a-h)-f'(a)]/(-h) =
f''(a)by the definition of f''.My objection to the suggestion
that f(a+h) = f(a) + h f'(a) + h^2/2 f''(a) + o(h^2)holds with
even greater force if f'' is only defined at a. The OP is>
obviously not a native English speaker; I took her remark let
any a in> E, f''(a) exist to be a garbled way of saying f''(a)
exists for all a> in E. In context, your interpretation (a is
fixed, f''(a) exists) is> more natural.So here's the question:
if f is differentiable in a neighborhood of a,> and twice
differentiable at a, does it follow that f(a+h) = f(a) + h
f'(a) + h^2/2 f''(a) + o(h^2)?> Yes, sir!If f is n-1 times
differentiable at a and its n_th derivative justexists
(nothing more assumed) at a (nothing more assumed), then,
forevery h such that a+h remains in a neighborhood of a, we
have f(a+h) =f(a) + hf'(a)....+ h^n/(n!) f_n(a) + o(h^n),
where f_n means the n_thderivative of f. The prrof I know is
by induction on n.Artur===
===
Subject:
: Re: easy...analysis
problem....Content-transfer-encoding: 8bit> hello.......genius
teacher........> let function f :E -> R is differentiable on
open set E in R> 1. let any a in E , f''(a) exist.> show
that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) / (h^2)} , h->0>
2. let any a in E , n-derivative f^(n) (a) exist.> express
f^(n) (a) by limit of f(x)> ------------------------->
um.....i solved first problem.> lim {f(a+h) - 2f(a) + f(a-h)
/ (h^2)}> h->0> = lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} /
(h^2)]> = lim {f'(a) - f'(a-h)} / h , let h=(-t)> = lim
{f'(a) - f'(a+t)} / (-t)> t->0> = lim {f'(a+t) - f'(a)} / t = f''(a)> it's right??> but i can't solve second
problem....> You haven't solved the FIRST problem, either.
Your passage> = lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} /
(h^2)]> = lim {f'(a) - f'(a-h)} / h , let h=(-t)> is
flawed.> Hint: apply L'Hopital's Rule to the original
quotient, once. Then> use> the definition of the second
derivative.> Another hint for the first problem: since f''(a)
exists, f'(a) exists> and f(a+h) = f(a) + h f'(a) +
(h2/2)f''(a) + o(h^2).> Is that true? Taylor's theorem only
guarantees that> f(a+h) = f(a) + h f'(a) + h^2/2 f''(z)> for
some z between a and a+h; but we're NOT told that f'' is>
continuous, only that it exists throughout the open set.> That
was why I suggested L'H.> --Ron Bruck> Uh, the hypotheses state
that f is differentiable on E. So f'(a)> exists for any a in E.
(You're technically right, Ron, but not> particularly helpful.
:] )?? Don't understand the remark, since my L'H suggestion
still works. > We have lim_{h to 0} [f(a+h) + f(a-h) -
2f(a)]/h^2 = lim_{h-->0} [f'(a+h) - f'(a-h)]/(2h)IF this limit
exists; and we have lim_{h -> 0} [f'(a+h) - f'(a-h)]/(2h) =
(1/2) lim_{h -> 0} [f'(a+h)-f'(a)]/h + 1/2 lim_{h ->
0}[f'(a-h)-f'(a)]/(-h) = f''(a)by the definition of f''.My
objection to the suggestion that f(a+h) = f(a) + h f'(a) +
h^2/2 f''(a) + o(h^2)holds with even greater force if f'' is
only defined at a. The OP is> obviously not a native English
speaker; I took her remark let any a in> E, f''(a) exist to be
a garbled way of saying f''(a) exists for all a> in E. In
context, your interpretation (a is fixed, f''(a) exists) is>
more natural.So here's the question: if f is differentiable in
a neighborhood of a,> and twice differentiable at a, does it
follow that f(a+h) = f(a) + h f'(a) + h^2/2 f''(a) +
o(h^2)?Yes, sir!> If f is n-1 times differentiable at a and
its n_th derivative just> exists (nothing more assumed) at a
(nothing more assumed), then, for> every h such that a+h
remains in a neighborhood of a, we have f(a+h) => f(a) +
hf'(a)....+ h^n/(n!) f_n(a) + o(h^n), where f_n means the
n_th> derivative of f. The prrof I know is by induction on
n.OK. It follows from L'Hopital's rule, which in turn follows
from themean value theorem. For n = 2, for example, it says
thatlim_{h->0} [f(a+h)-f(a)-hf'(a)-h^2/2 f''(a)]/h^2= lim_{h
-> 0} [f'(a+h)-f'(a)-h f''(a)]/(2h)if the latter limit exists;
which of course it does, since f''(a)exists; and the limit is
0.I was trying to see this from the alternative approach,
f'(a+h) = f'(a) + h f''(a) + h b(h), where b(h) --> 0 as h -->
0.If we knew f' were integrable, we'd have f(a+h) = f(a) + h
f'(a) + h^2/2 f''(a) + int_0^h t b(t) dt,where the last term
is o(h^2); but f' doesn't have to be integrable, soI doubted
the truth of the conclusion. But the conclusion is OK, evenif
this reason isn't.--Ron Bruck===
===
Subject:
: Lattice points,
balls, number theory.Consider R^n with the usual Euclidean
metric. Call a point a latticepoint if it has integer
coordinates, so the lattice points in R^n aresimply Z^n.It is
not very hard to show that given any non-negative integer
mthere exist a closed ball B in R^n (usual metric) such that B
containsexactly m lattice points in its interior. I remember
having seen orheard somewhere, that the same result is true
when the interior isreplaced with the boundary. My question is
the following :What is known about the coupled problem where
one looks for pairs ofnon-negative integers (k,m) such that
there exist a closed ball in R^nhaving exactly k lattice
points on its boundary and exactly m in itsinterior
??-Anders===
===
Subject:
: Re: Lattice points, balls, number
theory.> Consider R^n with the usual Euclidean metric. Call a
point a lattice> point if it has integer coordinates, so the
lattice points in R^n are> simply Z^n.> It is not very hard to
show that given any non-negative integer m> there exist a
closed ball B in R^n (usual metric) such that B contains>
exactly m lattice points in its interior. I remember having
seen or> heard somewhere, that the same result is true when
the interior is> replaced with the boundary. My question is
the following :Not so when n = 1 as boundary is only two
points.> What is known about the coupled problem where one
looks for pairs of> non-negative integers (k,m) such that
there exist a closed ball in R^n> having exactly k lattice
points on its boundary and exactly m in its> interior
??===
===
Subject:
: Re: Lattice points, balls, number
theory.Consider R^n with the usual Euclidean metric. Call a
point a lattice> point if it has integer coordinates, so the
lattice points in R^n are> simply Z^n.It is not very hard to
show that given any non-negative integer m> there exist a
closed ball B in R^n (usual metric) such that B contains>
exactly m lattice points in its interior. I remember having
seen or> heard somewhere, that the same result is true when
the interior is> replaced with the boundary. My question is
the following :Not so when n = 1 as boundary is only two
points.What is known about the coupled problem where one looks
for pairs of> non-negative integers (k,m) such that there exist
a closed ball in R^n> having exactly k lattice points on its
boundary and exactly m in its> interior ??>Allright, allright
:) n>1.===
===
Subject:
: Re: JSH: how you can prove to the
mathematicians you are right>> No. In my experience, if
one's trying to formalize a proof, but the>> result keeps
getting spit out by the checker, one always figures that>> his
formalization is off, not that his basic argument is wrong.>>
No??? I mean, yes, most people would conclude they had
screwed> up the translation and the formalism, but JSH is not
most people. > Or do you disagree on that also?Oh, no, no, no,
I do *not* disagree that JSH isn't most people. I> don't quote
most people, but I quote him regularly.But nothing about JSH
indicates that he's more likely to decide that> his argument
is wrong rather than his attempt to formalize it. On the>
contrary, he's already shown a certain stubbornness regarding
his> argument (maybe you've even noticed this). Why would he
decide that> his formalization is correct and his argument
wrong? That's not what I'm saying. He would decide that his
argument iscorrect AND his formalism is correct, but that the
proof-checker,and the whole theory of proof-checkers, had a
fundamental flaw. >> This is one reason that suggestions James
formalizes his argument>> aren't so helpful. When you're in the
thick of things, it's hard to>> tell the difference between a
bad argument and problems turning a good>> argument into a
formal proof.>> Look: he is not going to put in the kind of
effort it would take> to do this. Why? (1) Because he resists
learning from anyone else;> (2) I think he knows he is wrong.
I agree on (1) and (2) is not implausible.> It's hard to be
sure. He lives in a half-lit dream world where there is a gray
zone between truth and lies. But don't we all?> Nobody can come
as close as he has to understanding our arguments> without
actually understanding them. Hence his refusal to EVER even>
quote Decker's main point. If he really believed he is right,
the> motivation to have his proof validated by machine would
be enormous.> It is quite possible that he would become famous
[Letterman, Leno,> Larry King, etc.], even rich.And that's just
the letter L! When he gets to O, there's Oprah, and> at R he
gets Rose (or does he get Charlie Rose at C?). > Barbara
Walters ... maybe his *own* talk show ... books ... The
Sayings of Chairman James ... SNL host ... endorsements (own
your own Object Ring!)... rub shoulders with the mighty ...
trips on Air Force One (General,I'd like you to meet our most
brilliant national security asset) ...beautiful women *with
cleavage* and very short skirts ... and then, the elder
statesman ... only person to win both Fields Medal and Nobel
Prize (in Physics, Economics, Literature, Peace, and
Prime-Counting) ... new breed of dog named after him (the
Harris Lap-dog) ... seven ex-wives, all still adoring him {the
Best We Ever Had, by Britney S., Jennifer L. and Barbara W.)
... capital of State of Pennsylvania renamed in his honor (no,
wait, that wouldn't work) ...>> Well, you're more impressed by
his efforts than I am. Just look at a typical JSH non-rant
posting: lots of algebra [often > carelessly done to be sure],
some ideas [usually unsound], some evidence> of thinking and
calculating and head-scratching: evidence of > effort, I would
say, but not evidence of a worthwhile result, and> never
evidence of an effort to understand much beyond high-school >
algebra. He agonizes over things, thinks about them, works at>
them. But when it comes to learning any of the existing body
of> mathematical theory, he is extremely lazy. No, I am not
impressed.It's the theory that would keep him from learning to
formalize his> argument. Sure, he diddles about with
computations and twisted> arguments, but he rarely learns much
or puts in any real effort to> further his understanding. ...
face added to Mt. Rushmore ... picture on Wheaties box ...
JamesHarris Secret De-Coder Ring ... Jimmy Harris Country
Sausage ... Nora B.===
===
Subject:
: Re: JSH: how you can prove to
the mathematicians you are right [.snip.]>my approach to
proving things in the integer>case was essentially to prove
Dedekind's Prauge Theorem >for the case F(x) and G(x) in Z[x].
(I >note that if p|ab then p|a or p|b then do>a lot of messy
bookkeeping). Well, ->that's<- certainly too hard. Just use
the version of Gauss's> Lemma that says that the content of
the product of two polynomials is> the product of their
contents. If F(x)*G(x) has a content which is a> multiple of
7, then 7 must divide either the content of F(x) OR the>
content of G(x), which means one of the two polynomials has>
coefficients all divisible by 7.The Prague Theorem can be
interpreted as a generalization of that> result on the
contents.> Incredulous Number Theorist: What? You know what an
algebraic integer is but you do not know Gauss's Lemma?OODA: It
comes from learning most of your number theory from reading JSH
threads. Doesn't Gauss's lemma have something to do with
quadratic residues.INT: Wrong Gauss's lemma. What about the
discussion of Gauss's Lemma (product of primitive polynomials
is primitive) a year or so back?OODA: <mostly inaudible,
somthing about not reading all posts carefully mumble mumble
time preasure mumble> [1]I guess the upshot is that OODA's
should not attemptnumber theory. As Pope puts it A little
number theory is a gerous thing. Drink deep or put too much
effort into your proofs [2] -William Hughes[1] On sci.math
nobody can see your face turn red. This is a good thing.[2]
Something like this anyway.P.S. Slighly more seriously.
Clearly the result in the integer case (fg=pP implies p|f or
p|g) follows from Gauss's lemma. However, the contrapositive
of Gauss's lemma in the form product of two primitive
polynomials is primitive, follows immediately from this
result. Thus I was trying to prove Gauss's lemma for integer
polynomials and believed I had succeeded. This means either I
am incredibly smart or incredibly stupid. I am not sure
which.===
===
Subject:
: Re: puzzle: GCDs of Infinite Set of Integer
Pairs> Let us say we have 2 spherical dice which somehow each
represent all> positive integers, each integer represented
exactly once per die, on> their countably-infinite number of
sides. And each integer is equally> likely to be rolled on
both dice.So, the dice-pair is rolled an infinite number of
times.And how, pray tell, do you intend to stop the dice's
rolling? :-)-- Ted Schuerzinger <fedya at bestweb dot net>The
way I see it, you raised three children who could knock out
and hog-tie a perfect stranger, you must be doing *something*
right.Marge Simpson,
<http://www.snpp.com/episodes/7G01.h>===
===
Subject:
: Re: puzzle:
GCDs of Infinite Set of Integer Pairs> (I am posting this for
fun, for it is probably easily solved by those> with some
background in number theory.)> Let us say we have 2 spherical
dice which somehow each represent all> positive integers, each
integer represented exactly once per die, on> their
countably-infinite number of sides. And each integer is
equally> likely to be rolled on both dice.So, the dice-pair is
rolled an infinite number of times.What is the closed-form
representation of the probability thatfor *at least one* roll
of the dice-pair> the m_th rolling of the pair produces
integers j and k where> GCD(j,k) = m?(ie. there is at least
one m, where m = GCD(j,k) on the m_th rolling> of the
dice-pair.)> Hopefully the solution to this puzzle is not too
well known...Leroy QuetMy gut says zero.Chuck===
===
Subject:
: Re:
puzzle: GCDs of Infinite Set of Integer PairsI'm inclined to
go with Chuck. The whole infinite thing is kind ofhinky.
Especially the countably-infinite part. Also, somehowmakes me
doublecheck my wallet. But, knowing nothing about
numbertheory...and upon rereading the problem, I say 100% is
theprobability. But I don't know what a closed-form
representation is.> (I am posting this for fun, for it is
probably easily solved by those> with some background in
number theory.)> Let us say we have 2 spherical dice which
somehow each represent all> positive integers, each integer
represented exactly once per die, on> their countably-infinite
number of sides. And each integer is equally> likely to be
rolled on both dice.So, the dice-pair is rolled an infinite
number of times.What is the closed-form representation of the
probability thatfor *at least one* roll of the dice-pair> the
m_th rolling of the pair produces integers j and k where>
GCD(j,k) = m?(ie. there is at least one m, where m = GCD(j,k)
on the m_th rolling> of the dice-pair.)> Hopefully the
solution to this puzzle is not too well known...Leroy QuetMy
gut says zero.Chuck===
===
Subject:
: Re: puzzle: GCDs of Infinite
Set of Integer Pairsath: meganewsservers.com>I'm inclined to
go with Chuck. The whole infinite thing is kind of>hinky.
Especially the countably-infinite part. Also, somehow>makes me
doublecheck my wallet. But, knowing nothing about
number>theory...and upon rereading the problem, I say 100% is
the>probability. But I don't know what a closed-form
representation is.I don't think the puzzle as stated has a
well-defined answer. I thinkto turn it in to a wel-defined
puzzle, you reinterpret it as follows:imagine you have two n
sided dice, which you will rolln times. What is the
probability that on the mth roll the GCD ofthe number of the
two dice is m? Find the limit as n approachesinfinity.I expect
the answer is probably zero, but I can't proe
it.George===
===
Subject:
: Hausdorff dimension of graph of
Weierstrass function.The lower bound of the Hausdorff
dimension of the graph
off(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 1<s<2 and
t>1.is s. Is it equal to s?-- G.C.===
===
Subject:
: Re: Hausdorff
dimension of graph of Weierstrass function.> The lower bound
of the Hausdorff dimension of the graph
off(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 1<s<2 and
t>1.is s. Is it equal to s?I believe the Hausdorff dimension
is an open problem.There are known estimates, and values for
other dimensions.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re:
Hausdorff dimension of graph of Weierstrass function.>> The
lower bound of the Hausdorff dimension of the graph of>>>
f(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 1<s<2 and t>1.>> is s. Is it equal to s?>I believe the Hausdorff dimension is
an open problem.If so then he must have meant upper bound,
because itonly took me a few minutes to check that s _is_ an
upper bound.(Just looked again, I think my scribbles are
right.)>There are known estimates, and values for other
dimensions.===
===
Subject:
: Re: Hausdorff dimension of graph of
Weierstrass function.>The lower bound of the Hausdorff
dimension of the graph
of>f(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 1<s<2 and
t>1.>is s. Is it equal to s?If what you say is so and my
scribblings are correct then yes.But I wonder if you really
meant to say that s is a lower bound;getting an upper bound is
much easier, it's hard for me toimagine how a person could know
a lower bound but beunable to find an upper bound (it's easier
to cover thegraph by a bunch of little squares than to show it
can'tbe covered by an even smaller bunch of squares):If I did
the calculations right then f is in Lip_alpha for alpha = 2 -
s, and for any f in Lip_alpha the (2-alpha)-dimensionalmeasure
is sigma-finite (the measure of the part of thegraph above a
compact interval is finite.) Hence thes-dimensional measure of
the graph of this functionis sigma-finite, which shows that s
is an _upper_ boundfor the dimension.Do you really know that s
is a lower bound? How?(Years ago I read a paper on the
dimension of thesegraphs; the argument was not nearly as
trivial as thecalculations I did to show that s was an upper
bound...)===
===
Subject:
: Re: Hausdorff dimension of graph of
Weierstrass function.>The lower bound of the Hausdorff
dimension of the graph
off(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 1<s<2 and
t>1.is s. Is it equal to s?If what you say is so and my
scribblings are correct then yes.> But I wonder if you really
meant to say that s is a lower bound;> getting an upper bound
is much easier, it's hard for me to> imagine how a person
could know a lower bound but be> unable to find an upper bound
(it's easier to cover the> graph by a bunch of little squares
than to show it can't> be covered by an even smaller bunch of
squares):If I did the calculations right then f is in
Lip_alpha for alpha => 2 - s, and for any f in Lip_alpha the
(2-alpha)-dimensional> measure is sigma-finite (the measure of
the part of the> graph above a compact interval is finite.)
Hence the> s-dimensional measure of the graph of this
function> is sigma-finite, which shows that s is an _upper_
bound> for the dimension.Do you really know that s is a lower
bound? How?> (Years ago I read a paper on the dimension of
these> graphs; the argument was not nearly as trivial as the>
calculations I did to show that s was an upper bound...)Let
N_x(S) be the smallest number of sets of diameter at most x
which can cover S and defineBox(S)=lim_{x -> 0} inf (-log
N_x(S)/log x)Let S be the graph of f , then Box(S) >= s , and
the Hausdorffdimension of f is >= s . But I can't prove
more.-- G.C.===
===
Subject:
: Re: Hausdorff dimension of graph of
Weierstrass function.Hausdorff dimension <= Box dimension in
general.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re:
universal set with 3 valued logicI have seen this (or
something very like this) before. Long ago.A 3-valued system
of logic, with some Polish name associated with it.But there
are some three-step self-referential set definitionsthat are
still contradictory. So you can introduce more truth-valuesfor
them.This would have been published in the 1960's, perhaps. Or
earlier.===
===
Subject:
: Re: universal set with 3 valued logic by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1EDOeR00493;>It just doesn't make
sense to introduce a strange new system of logic just>to
salvage the Universal Set. Is there anything really wrong with
the usual>2-value logic (T or F)-- other than prohibiting the
existence of the>Universal Set, I mean?>i never said there was
anything 'wrong' with either binary logic orZFC. i don't even
know what 'right' and 'wrong' mean in math otherthan
consistency. to me, the prohibition of a universal set
is'wrong', if anything is, but maybe that's just where you and
i differ.to me, this is better than making the object {x|x=x} a
proper class,whatever that is but maybe that's just
me.cheers===
===
Subject:
: e*pi is irrationale*pi is
irrational.Reference [
http://mathworld.wolfram.com/GelfondsTheorem.h ]Comments?-- G.
A. Edgar http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re:
e*pi is irrational> e*pi is irrational.> Reference [
http://mathworld.wolfram.com/GelfondsTheorem.h ]> Comments?See
also:http://mathworld.wolfram.com/IrrationalNumber.hIn Hardy
and Wright (Eric's Source?), An Introduction to theTheory of
Numbers, Fifth Edition, they say, on page 39, Itwas not until
1929 that theorems were discovered which gobeyond those of
[sections] 11.13-14 in any very importantway. It has been
shown recently that further classes ofnumbers in which e^pi,
2^sqrt(2), e*pi are included areirrational...Sections 11.13
and 11.14 are about the transcendence ofe and pi.===
===
Subject:
:
Re: e*pi is irrational>e*pi is irrational.>Reference [
http://mathworld.wolfram.com/GelfondsTheorem.h
]>Comments?Well, yes - given that you're the kind of guy
people might tendto believe, you might at least include a
question mark or something...(Today must be your first visit
to mathworld - it's really full oferrors. I admit this is a
good one.)===
===
Subject:
: Re: e*pi is irrationale*pi is
irrational.Reference [
http://mathworld.wolfram.com/GelfondsTheorem.h ]Comments?Well,
yes - given that you're the kind of guy people might tend> to
believe, you might at least include a question mark or>
something...e*pi _is_ irrational, so why the need for a
question mark? But (how)does that follow from the
Gelfond-Schneider theorem?-- G.C.===
===
Subject:
: Re: e*pi is
irrational> e*pi _is_ irrational, so why the need for a
question mark?G. A. Edgar cited MathWorld. So he did *not* put
a questionmark. Instead, he gave the reference and asked for
comments.He could as well have written to the editors, hinting
atthis non-sequitur and asking them to remove it.> But (how)
does that follow from the Gelfond-Schneider theorem?That's the
point :-)When called e*pi is irrational an error,then indeed he
didn't mean it (as I jokingly supposed), buthe called the
implication an error.The fact that at least one of e+pi and
e*pi must be irrationalis so funny that I think I can get it
from my memory:(x-e)(x-pi) = 0 has no rational roots, OK?The
left side is x^2 - (e+pi)*x + e*pi. Assuming e+pi and e*piboth
rational you have x^2 - Ax + B = 0 with rationalA=e+pi and
B=e*pi. But then x is rational. Contradiction.Best ,Rainer
Rosenthalr.rosenthal@web.de===
===
Subject:
: Re: e*pi is irrational>
The left side is x^2 - (e+pi)*x + e*pi. Assuming e+pi and e*pi>
both rational you have x^2 - Ax + B = 0 with rational> A=e+pi
and B=e*pi. But then x is rational. Contradiction.Actually,
that would just let you conclude that x is algebraic.
Butthat's still enough to contradict what is known about pi
and e.This proof actually supports the slightly stronger
statement that e+piand e*pi are not both algebraic.-- iel W.
sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039
53 36 N / 086 11 55 W===
===
Subject:
: Re: e*pi is irrational> This
proof actually supports the slightly stronger> statement that
e+pi and e*pi are not both algebraic.Well, thanks. Jokes and
proofs must be rememberedcorrectly - or they lack someting
:-)Rainer Rosenthalr.rosenthal@web.de===
===
Subject:
: Re: e*pi is
irrational>>>>e*pi is irrational.>>Reference [
http://mathworld.wolfram.com/GelfondsTheorem.h ]>>Comments?>> Well, yes - given that you're the kind of guy people might
tend>> to believe, you might at least include a question mark
or>> something...>e*pi _is_ irrational, so why the need for a
question mark? I may very well be wrong, this is just my
memory of things I'veread. But I thought it was unknown
whether e*pi was irrational.>But (how)>does that follow from
the Gelfond-Schneider theorem?What _does_ it follow
from?===
===
Subject:
: Re: e*pi is irrational>>e*pi is
irrational.>>Reference [
http://mathworld.wolfram.com/GelfondsTheorem.h ]>>Comments?>>
Well, yes - given that you're the kind of guy people might
tend>> to believe, you might at least include a question mark
or>> something...e*pi _is_ irrational, so why the need for a
question mark?I may very well be wrong, this is just my memory
of things I've> read. But I thought it was unknown whether e*pi
was irrational.>But (how)>does that follow from the
Gelfond-Schneider theorem?What _does_ it follow from?Um... I
thought it was stated (but not proved) in Hardy & Wright,
butnow I can't find it. Sorry if I've misremembered.--
G.C.===
===
Subject:
: Re: e*pi is irrational...Um... I thought it
was stated (but not proved) in Hardy & Wright, but> now I
can't find it. Sorry if I've misremembered.It seems that I did
remember correctly, see Nat Silver's contribution. H&W give no
reference (in the 4th ed) but they mention 1929 which wasthe
date of Gelfond's paper.The e + pi or e*pi result refers to
transcendentality(transcendence?).-- G.C.===
===
Subject:
: Re: e*pi
is irrationale*pi is irrational.(Today must be your first
visit to mathworld - it's> really full of errors. I admit this
is a good one.)So you think e*pi is irrational is an
error?Proof ;-?Rainer Rosenthalr.rosenthal@web.de===
===
Subject:
:
Re: e*pi is irrational>>e*pi is irrational.>> (Today must be
your first visit to mathworld - it's>> really full of errors.
I admit this is a good one.)>So you think e*pi is irrational
is an error?>Proof ;-?I don't have a proof. I may well be
wrong, but my recollectionis that it's known that one of e +
pi or e*pi must be irrationalbut it's not known which.>Rainer
Rosenthal>r.rosenthal@web.de===
===
Subject:
: mathworld peculiarityI
suppose these are too numerous to deserve mention.But here: [
http://mathworld.wolfram.com/Lindemann-WeierstrassTheorem.h
]we see the symbol Q_p, which is used for the p-adic
numbers.Presumably it should be Q, the rationals.-- G. A.
Edgar http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re:
Ring of algebraic integers, comparison checkCORRECTION POST:It
turns out that there's an incredibly simple way to check the
ringof algebraic integers and see that there's a problem with
how it'scurrently taught.You can take two quadratics:x^2 - x +
42 = 0andy^2 - by - 7 = 0where by how algebraic integers are
currently taught, you'd think thatthere exists an algebraic
integer b, such that a root of the secondquadratic is a factor
of a root of the first. Intriguingly, theredoes not, and it's
easy to show.Now x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on
of its roots, andy^2 - by - 7 = 0, has as one of its roots (b +
sqrt(b^2 + 28))/2.So I can simply introduce z, where(1 +
sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2which isz = (1 +
sqrt(-167))/(b + sqrt(b^2 + 28))where now the question is, can
an algebraic b exist such that z is analgebraic integer?The
square roots don't tell me much, but working to eliminate
themadds solutions, as a first step consider:28z^2 +
2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0where I have *two*
solutions, where they arez_1 = (1 + sqrt(-167))/(b + sqrt(b^2
+ 28))and z_2 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))and
working still further I get196 z^4 + 28b z^3 + (168b^2 + 2324)
z^2 - 168bz + 7056 = 0and I can divide both sides by 28 to
finally get7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0where I
have four solutions, which arez_1 = (1 + sqrt(-167))/(b +
sqrt(b^2 + 28))z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))z_3
= (1 + sqrt(-167))/(b - sqrt(b^2 + 28))z_4 = (1 -
sqrt(-167))/(b - sqrt(b^2 + 28)).(Those not sure can simply
multiply out(Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to
verify.)Now then our question was, could what is now z_1 be an
algebraicinteger for an algebraic integer b?And now an
important theorem comes into play, as an algebraic
integercannot be the root of a non-monic primitive with
integer coefficientsthat is irreducible over Q.So integer b's
are eliminated as a possibility immediately.There are several
ways to show that none of the the z's can be afraction for an
integer b, but some of you may wonder why they'd needto be,
like why couldn't they be irrational algebraic integers?Well,
unless 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0is
reducible over Q with that integer b, the theorem I
mentionedrequires that *none* of its roots can be an algebraic
integer, as it'snon-monic and primitive.Therefore, even if you
had an irrational algebraic integer root, thepolynomial would
have to be reducible over Q, meaning that you'd needat least
one rational root.If b is irrational--remember it's an
algebraic integer--then you get apolynomial of degree equal to
the order of the monic polynomial withinteger coefficients for
which b is a root times 4.However, for any such polynomial,
you run into the same problem thatit'd need at least one root
that's rational.But you can characterize every possible root
byz_j = (1 + sqrt(-167))/W_jz_k = (1 - sqrt(-167))/W_kwhere j
and k are counting numbers, and W_j and W_k are
algebraicintegers.The problem is that (1+sqrt(-167)/U where U
is an algebraic integercannot be rational unless U has
(1+sqrt(-167) itself as a factor.Therefore, in the ring of
algebraic integers, the roots ofx^2 - x + 42 = 0andy^2 - by -
7 = 0never share non-unit factors in the ring of algebraic
integers.Notice that the key theorem here is that one that
prevents analgebraic integer from being the root of a
non-monic primitive withinteger coefficients irreducible over
Q.Take away the condition that you have an algebraic integer
factor, andyou're ok, which proves that there must be a ring
beyond algebraicintegers, where they *can* share non-unit
factors.I have called the more inclusive ring, the ring of
objects, or theObject Ring.The Object Ring is a commutative
ring that includes all numbers suchthat -1 and 1 are the only
members that are both a unit and aninteger, where no non-unit
member is a factor of any two integers thatare coprime.One of
the most surprising results that follows quickly
fromunderstanding the limitations of the ring of algebraic
integers, andhaving the fuller definition, is a proof of
Fermat's Last Theorem ofaround two to three pages.James
Harris===
===
Subject:
: Re: Ring of algebraic integers, comparison
check> CORRECTION POST:It turns out that there's an incredibly
simple way to check the ring> of algebraic integers and see
that there's a problem with how it's> currently taught.You can
take two quadratics:x^2 - x + 42 = 0andy^2 - by - 7 = 0where by
how algebraic integers are currently taught, you'd think that>
there exists an algebraic integer b, such that a root of the
second> quadratic is a factor of a root of the first.
Intriguingly, there> does not, and it's easy to show.Now x^2 -
x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, andy^2 -
by - 7 = 0, has as one of its roots (b + sqrt(b^2 + 28))/2.So
I can simply introduce z, where(1 + sqrt(-167))/2 = (b +
sqrt(b^2 + 28))z/2which isz = (1 + sqrt(-167))/(b + sqrt(b^2 +
28))where now the question is, can an algebraic b exist such
that z is an> algebraic integer?The square roots don't tell me
much, but working to eliminate them> adds solutions, as a first
step consider:28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 =
0where I have *two* solutions, where they arez_1 = (1 +
sqrt(-167))/(b + sqrt(b^2 + 28))and z_2 = (1 + sqrt(-167))/(b
- sqrt(b^2 + 28))and working still further I get196 z^4 + 28b
z^3 + (168b^2 + 2324) z^2 - 168bz + 7056 = 0and I can divide
both sides by 28 to finally get7z^4 + bz^3 + (6b^2 + 83)z^2 -
6bz + 252 = 0where I have four solutions, which arez_1 = (1 +
sqrt(-167))/(b + sqrt(b^2 + 28))z_2 = (1 - sqrt(-167))/(b +
sqrt(b^2 + 28))z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))z_4
= (1 - sqrt(-167))/(b - sqrt(b^2 + 28)).(Those not sure can
simply multiply out(Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to
verify.)Now then our question was, could what is now z_1 be an
algebraic> integer for an algebraic integer b?And now an
important theorem comes into play, as an algebraic integer>
cannot be the root of a non-monic primitive with integer
coefficients> that is irreducible over Q.So integer b's are
eliminated as a possibility immediately.> OK.> There are
several ways to show that none of the the z's can be a>
fraction for an integer b, I thought you just eliminated
integer b's.> but some of you may wonder why they'd need> to
be, like why couldn't they be irrational algebraic
integers?Well, unless 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252
= 0is reducible over Q with that integer b, the theorem I
mentioned> requires that *none* of its roots can be an
algebraic integer, as it's> non-monic and primitive.Therefore,
even if you had an irrational algebraic integer root, the>
polynomial would have to be reducible over Q, meaning that
you'd need> at least one rational root.> You must be assuming
that a polynomial that is reducible over Q must have at least
one rational root. That is false. Consider3*x^4 - 15*x^2 + 5.
Plus, if b is not an ordinary integer but it is an algebraic
integer, it must be irrational. A polynomial with irrational
coefficients cannot be reducible over Q.> If b is
irrational--remember it's an algebraic integer--then you get
a> polynomial of degree equal to the order of the monic
polynomial with> integer coefficients for which b is a root
times 4.However, for any such polynomial, you run into the
same problem that> it'd need at least one root that's
rational.> You have not shown that such a polynomial would
have to be reducible over Q. Even if it were, that does not
imply that it has a rational root.> But you can characterize
every possible root byz_j = (1 + sqrt(-167))/W_jz_k = (1 -
sqrt(-167))/W_kwhere j and k are counting numbers, and W_j and
W_k are algebraic> integers.The problem is that (1+sqrt(-167)/U
where U is an algebraic integer> cannot be rational unless U
has (1+sqrt(-167) itself as a factor.Therefore, in the ring of
algebraic integers, the roots ofx^2 - x + 42 = 0andy^2 - by - 7
= 0never share non-unit factors in the ring of algebraic
integers.Notice that the key theorem here is that one that
prevents an> algebraic integer from being the root of a
non-monic primitive with> integer coefficients irreducible
over Q.> A true theorem, but you are assuming something else:
that a polynomialreducible over Q has to have at least one
rational root: clearlya false assumption. Nora B.> Take away
the condition that you have an algebraic integer factor, and>
you're ok, which proves that there must be a ring beyond
algebraic> integers, where they *can* share non-unit factors.I
have called the more inclusive ring, the ring of objects, or
the> Object Ring.The Object Ring is a commutative ring that
includes all numbers such> that -1 and 1 are the only members
that are both a unit and an> integer, where no non-unit member
is a factor of any two integers that> are coprime.One of the
most surprising results that follows quickly from>
understanding the limitations of the ring of algebraic
integers, and> having the fuller definition, is a proof of
Fermat's Last Theorem of> around two to three pages.> James
Harris===
===
Subject:
: Re: Ring of algebraic integers, comparison
check > It turns out that there's an incredibly simple way to
check the ring > of algebraic integers and see that there's a
problem with how it's > currently taught. > You can take two
quadratics: > x^2 - x + 42 = 0 > and > y^2 - by - 7 = 0 >
where by how algebraic integers are currently taught, you'd
think that > there exists an algebraic integer b, such that a
root of the second > quadratic is a factor of a root of the
first. Intriguingly, there > does not, and it's easy to
show.No, it is not easy to show. Keith Ramsay gave the value
of the commonfactor between 7 and (1 + sqrt(-167))/2 already.
Just call it a, andset b = a - a', where a' is the complex
conjugate of a, and you are there. > Now > x^2 - x + 42 = 0
has (1 + sqrt(-167))/2 as on of its roots, and > y^2 - by - 7
= 0, has as one of its roots > (b + sqrt(b^2 + 28))/2.Yup. >
So I can simply introduce z, where > (1 + sqrt(-167))/2 = (b +
sqrt(b^2 + 28))z/2 > which is > z = (1 + sqrt(-167))/(b +
sqrt(b^2 + 28)) > where now the question is, can an algebraic
b exist such that z is an > algebraic integer?Yup, and the
answer is yes. > The square roots don't tell me much, but
working to eliminate them > adds solutions, as a first step
consider: > 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 >
where I have *two* solutions, where they are > z_1 = (1 +
sqrt(-167))/(b + sqrt(b^2 + 28)) > and > z_2 = (1 +
sqrt(-167))/(b - sqrt(b^2 + 28)) > and working still further I
get > 196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 =
0 > and I can divide both sides by 28 to finally get > 7z^4 +
5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 > where I have four
solutions, which are > z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 +
28)) > z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) > z_3 = (1
+ sqrt(-167))/(b - sqrt(b^2 + 28)) > z_4 = (1 - sqrt(-167))/(b
- sqrt(b^2 + 28)).Somebody already detected an error in the
arithmetic, but I will letthat go. > (Those not sure can
simply multiply out > (Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to
verify.)Yup, you need a premultiplication by 7. > Now then our
question was, could what is now z_1 be an algebraic > integer
for an algebraic integer b?It can be, and has been proven. >
And now an important theorem comes into play, as an algebraic
integer > cannot be the root of a non-monic primitive with
integer coefficients > that is irreducible over Q.Indeed. > So
integer b's are eliminated as a possibility immediately.Indeed.
Algebraic integer b's are still possible. > There are several
ways to show that none of the the z's can be a > fraction for
an integer b, but some of you may wonder why they'd need > to
be, like why couldn't they be irrational algebraic
integers?Integer b's are completely irrelevant, so I will skip
some stuff here.b is an algebraic integer. > If b is
irrational--remember it's an algebraic integer--then you get a
> polynomial of degree equal to the order of the monic
polynomial with > integer coefficients for which b is a root
times 4.You lost me here. Note your four z's above: > z_1 = (1
+ sqrt(-167))/(b + sqrt(b^2 + 28)) > z_2 = (1 - sqrt(-167))/(b
+ sqrt(b^2 + 28)) > z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 +
28)) > z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)).z_1 is an
algebraic integer for some algebraic integer b, and so isz_4
(because it's complex conjugate is -z_1). Neither z_2 nor
z_3are algebraic integers. The polynomial can be written as:
7(Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4)So, the subpolynomial: (Z
- z_1)(Z - z_4)is a monic polynomial with algebraic integer
coefficients (z_1 and z_4are algebraic integers), and the
subpolynomial 7(Z - z_2)(Z - z_3)is a non-monic polynomial
with algebraic integer coefficiens, z_2 andz_3 are not
algebraic integers, you need the factor 7 to make it
apolynomial with algebraic integer coefficients.So within the
ring of algebraic integers, the original polynomial splits,i.e
is the product of two polynomial with algebraic integer
cofficients.It will take me some time to show an easier
example for this, but thatmay come later.-- dik t. winter,
cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/===
===
Subject:
: Re: Ring of algebraic
integers, comparison check... > 7z^4 + 5z^3 + (6b^2 + 83)z^2
- 6bz + 252 = 0(note that 5z^3 still should be bz^3). > So
within the ring of algebraic integers, the original polynomial
splits, > i.e is the product of two polynomial with algebraic
integer cofficients. > It will take me some time to show an
easier example for this, but that > may come later.James seems
to be focussed on that 83. He appears to think that to geta
monic polynomial, or a split in two polynomials, from it all
thecoefficients should be divisible by 7. As this forces b to
be divisibleby 7, there is a problem with the value of 83. But
his thinking is notcorrect. Consider the polynomial: 3z^4 +
4z^3 + 7z^2 + 4z + 3it is not monic, not all of the
coefficients are divisible by 3,nevertheless, it has algebraic
integer roots. To wit, it can be writtenas: (x^2 + x + 1)(3x^2
+ x + 3).Now set b = 4, we have: 3z^3 + bz^3 - (b^2 - 23)z +
bz + 3.That 23 is coprime to 3 is no reason for it to have no
algebraic integerroots for any b.-- dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home:
bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/===
===
Subject:
: Re: Ring of algebraic
integers, comparison check> ...> > 7z^4 + 5z^3 + (6b^2 +
83)z^2 - 6bz + 252 = 0(note that 5z^3 still should be bz^3).  So within the ring of algebraic integers, the original
polynomial splits,> i.e is the product of two polynomial
with algebraic integer cofficients.> It will take me some
time to show an easier example for this, but that> may come
later.James seems to be focussed on that 83. He appears to
think that to get> a monic polynomial, or a split in two
polynomials, from it all the> coefficients should be divisible
by 7. As this forces b to be divisible> by 7, there is a
problem with the value of 83. But his thinking is not>
correct. Consider the polynomial:I said exactly what I'm
thinking in my post, which except for sometypos is quite
correct.Now why you think 83 is significant yourself is a
puzzle.> 3z^4 + 4z^3 + 7z^2 + 4z + 3> it is not monic, not all
of the coefficients are divisible by 3,> nevertheless, it has
algebraic integer roots. To wit, it can be written> as:> (x^2
+ x + 1)(3x^2 + x + 3).> Now set b = 4, we have:> 3z^3 + bz^3
- (b^2 - 23)z + bz + 3.> That 23 is coprime to 3 is no reason
for it to have no algebraic integer> roots for any b.Yes Dik
Winter, there are indeed non-monic polynomials with
integercoefficients that have rational roots!Any readers who
needed you to show them that probably aren'twell-versed in
mathematics.James Harris===
===
Subject:
: Re: Ring of algebraic
integers, comparison check >You can take two quadratics: x^2 - x + 42 = 0 >and >y^2 - by - 7 = 0 >where by how
algebraic integers are currently taught, you'd think that
>there exists an algebraic integer b, such that a root of the
second >quadratic is a factor of a root of the first. > Can
you explain why? It doesn't seem obvious to me.That is not
difficult. Let's have p a divisor of 7, so 7 = p.q.(And let p
and q be algebraic integers). We have: (y - p)(y + q) = y^2 -
(p - q).y - 7so set b = (p - q) and we have p a root of the
polynomial in y above.So *any* factor of 7 is the root of such
a polynomial.On the other hand, a root r of x^2 - x + 42 is not
co-prime to 7, andso there is an algebraic integer factor of x,
such that it is also afactor of 7. And by the previous
paragraph it is a root of thepolynomial in y.But we know that
common factor due to work by Keith Ramsay.Set a = (44444 -
111.sqrt(-167))^(1/11), and a' it's complex conjugate.Now b =
a - a' will do the trick.(The reason is that with z =
([(298-23.sqrt(-167))[(-3-7.sqrt(-167))/2])^(1/11)(a.z)^11 =
[(1 + sqrt(-167))/2]^11, so with the proper choice ofthe
eleven possibilities for z, a.z = (1 + sqrt(-167))/2, a rootof
the polynomial in x, and (a.a')^11 = 7^11, so a.a' = 7.)-- dik
t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/===
===
Subject:
: Re: What is a number?/What
is not a number?What is a number?/What is not a number?>
Numbers seem to be simple things (objects) or a property of
things.> Yet the deeper you study numbers the more complex
they become.As children understanding of numbers starts with
the Naturals then the> Intergers, the Rationals, the Reals,
the Complex and beyond.Mathamatics later turns this all on its
headWhat'dja mean dude? Doesn't mathematics go in the order
Naturals =>Integers => Rationals => Reals => Complex and
beyond, in the sense ofdefining each in terms of the previous
one? So if you define thenaturals `a la von Neumann and you're
happy with set theory, you'redone.-- G.C.===
===
Subject:
: Re: What
is a number?/What is not a number?<< What is a number?/What is
not a number? Numbers seem to be simple things (objects) or a
property of things. >>a number is the value of a function that
assigns to a subset of all objects aword indicating that the
people who saw some reason to create the subset alsohad some
reason to quantify it. When the usual way of counting the
entities inthe subset fails, these people hire salesmen and
women called mathematicians toinvent a term such as 'pi' or
'e' to convince the rest of us that numerosityprevails despite
no realistic quantification. It's been proven===
===
Subject:
: Re:
What is a number?/What is not a number?a number is the value
of a function that assigns to a subset of all objects aword
indicating that the people who saw some reason to create the
subset alsohad some reason to quantify it. When the usual way
of counting the entities inthe subset fails, these people hire
salesmen and women called mathematicians toinvent a term such
as 'pi' or 'e' to convince the rest of us that
numerosityprevails despite no realistic quantification. It's
been proven===
===
Subject:
: Re: What is a number?/What is not a
number?>> What is a number?/What is not a number? >> Numbers
seem to be simple things (objects) or a property of things.>>
Yet the deeper you study numbers the more complex they
become.>>> As children understanding of numbers starts with
the Naturals then>> the Intergers, the Rationals, the Reals,
the Complex and beyond.>>> Mathamatics later turns this all
on its head>>>> All Computer Data & programs that
currently Exist now can be>> considered as on very large
interger, this Interger is unknowen,>> and tecnicaly
impossable to calculate>>>> Carl Do you mean Integer with
one r?> Yes These Numbers exist but are not practical.Has a
working Computer and its data been formalised this way.Carl
===
===
Subject:
: Ttttttt....two unknowns for an Inverse
KinemaniacHello guys,For the solution of an inverse kinematics
problemI need some help. I have two equations of the forma
sin(q) + b sin(p) + c sin(q) + d cos(p)cos(q) + e cos(p)sin(q)
+ f=0g sin(q) + h sin(p) + i sin(q) + j cos(p)cos(q) + k
cos(p)sin(q) + l=0so two equations with two unknowns. Is there
a general explicit solution for these?-- Cornelis
Wessels===
===
Subject:
: Algebraic integers, simple resultWhat I
discovered was an easy demonstration by looking
at(1+sqrt(-167) since (1+sqrt(-167))/2[(1-sqrt(-167))/2] =
-7you might assume that (1+sqrt(-167)) has non-unit factors in
commonwith 7 in the ring of algebraic integers, but it does
not.The way to quickly and easily see that is to consider z =
(1+sqrt(-167))/r_1where r_1 is presumed to be an algebraic
integer factor of 7 that isalso a factor of (1+sqrt(-167)) so
that z is an algebraic integer.However, you find that if r_1
is not a unit, then z is the root of anon-monic polynomial of
order equal to 2 times the order of theminimum polynomial with
integer coefficients which has r_1 as a root.And you know that
polynomial which has z as a root, can't have arational root
because neither (1+sqrt(-167))/r_k, nor(1-sqrt(-167))/r_k,
where k is some counting number equal to orgreater than 1 that
is less than or equal to the order of thepolynomial, can be
rational.The polynomial must be non-monic because even if r_1
were a factor of(1+sqrt(-167)) it wouldn't be a factor of
(1-sqrt(-167)), and all ofthe other roots can't be a factor of
both either.I thought I'd use r_1 just to simplify things
slightly.Oh yeah, so there doesn't exist an algebraic integer
r_1 that is anon-unit factor of 7 where z can be an algebraic
integer.It's a fascinatingly simple to show result. James
Harris===
===
Subject:
: Re: Algebraic integers, simple resultJames,
can your tools settle a question that has arisen elsewhere?
Ispi*e algebraic?...-- G.C.===
===
Subject:
: combinatorical
puzzleHello.Let S := {1...n} be the set with mightiness n (| S
| = n).Let A = (A_ij), i,j=1...n be a matrix with subsets of S
as entries. A_11 A_12 .................. A_1n : : : : : : A =
A_i1 A_i2 .................. A_in : : : : : : A_n1 A_n2
.................. A_nnThe following conditions are valid:(1)
A_ij subset S for all i,j(2) | A_ij | <= log n for all i,j(3)
Let j in {1..n} be an arbitrary row number and let a_1j in
A_1j, a_2j in A_2j, .... , a_nj in A_nj be an arbitrary choice
of elements (one for each subset of row j) then |
{a_1j,a_2j,....,a_nj} | => n/2. This means: If you take one
(arbitrary) element from each subset A_ij of a fixed row j,
then the union of all this n elements has mightiness greater
then n/2.My question is:Is it possible to choose an arbitrary
subset A_ij, j=1...n for each row,in such a way, that the
union of these n subset has less then n/2 elements?This means:
Take one arbitrary subset A_1 from row one, one arbitrary
subset A_2 from row two and so on. Is it possible to choose
this subsets in that way, that: | A_1 cup A_2 cup ..... cup
A_n | <= n/2 ?Any comments or solution
stratigies?GreetingsSebastian===
===
Subject:
: JSH: What it would
takeNow it's interesting to see what it would take forz = (1 +
sqrt(-167))/(b + sqrt(b^2 + 28))to be an algebraic integer for
an algebraic integer b.First off notice that for an algebraic
integer b, (b + sqrt(b^2 + 28))is an algebraic integer, so it
fits into the format I used in myprevious post where I just
said r_1.Next, it's kind of obvious from demonstration that
what makes itimpossible is what's required to get a polynomial
with integercoefficients.You need a monic quadratic with
integer coefficients for z, but can'tget one so it's not an
algebraic integer.Now consider if you have something like z =
sqrt(2)sqrt(7)/sqrt(2),then obviously you can divide off
sqrt(2), and get that monicquadratic with integer
coefficients.What I've shown in a rather simple way is that
only such simple anddirect results will work, so if you don't
*see* the non-unit factorsbetween two irrational algebraic
integers, then you don't have any inthe ring of algebraic
integers!That's fascinating to me as for a long time I've been
trying to figureout how to show the limitations of the ring,
but kept bumping intowhat you can't physically see, but here
now I get to use it to show mypoint.James Harris===
===
Subject:
:
JSH: Splitting field, algebraic integer factorsPreviously I
posted that if you can't *see* the factors betweenirrational
algebraic integers then they're not there, but morecorrectly
the situation is that two algebraic integers have to bemembers
of the same splitting field to have a non-unit algebraicinteger
in common.(I say more correctly as there may be some
terminology issues herebecause what mathematicians currently
call a splitting field is nota true field, but something
close, like the field of rationals. Butthat's another issue
for another time.)That's a nifty and powerful result. Why am I
the one who had todiscover it?James Harris===
===
Subject:
: Re: JSH:
Splitting field, algebraic integer factors>Previously I posted
that if you can't *see* the factors between>irrational
algebraic integers then they're not there, but more>correctly
the situation is that two algebraic integers have to
be>members of the same splitting field to have a non-unit
algebraic>integer in common.Uh, right. Splitting field of
_what_, exactly?>(I say more correctly as there may be some
terminology issues here>because what mathematicians currently
call a splitting field is not>a true field, but something
close, like the field of rationals. Huh? You need to give the
definition of true field. Becausethe rationals _are_ a field,
and splitting fields _are_ fields,truly.> But>that's another
issue for another time.)>That's a nifty and powerful result.
Why am I the one who had to>discover it?>James
Harris===
===
Subject:
: Anybody want to play with a new math of
infinities?Consider the question of what happens when wedivide
1 by 0?A lot of work assumes there is an answer thatcorrelates
to infinity, which is problematic,because what happens when
you multiply thisinfinity by 0 again?A simpler solution is to
use the symbology:The result of dividing 1 by 0 is 1/0.The
result of dividing 2 by 0 is 2/0. etc.Since the / has a
pre-existing meaning, let us say,The result of dividing 1 by 0
is 1~0.The result of dividing 2 by 0 is 2~0. etc.When you
multiply 1~0 by 0, you get 1.But when you multiple 2~0 by 0,
you get 2.Does this lead to any inconsistency?What about
comparisons?It would seem 1~0 should be greater than 2~0,
whichshould be greater than 3~0. 0~0 would be greather than
1~0.Are there theorems than can be proved usingthis new
symbology?===
===
Subject:
: Stuck on two combinatorics
problemsAnyone got ideas on these two problems ?1)A is a
rectangle consisting of n * 2 numbered squares. Compute a_n,
thenumber of ways to partition A into domino bricks ( that is
2 x 1rectangles). For instance a_1 = 1, a_2 = 22) Show that if
p and q are odd primes then2^(pq+1) [equivce symbol] 2^(p+q)
(mod pq)===
===
Subject:
: Re: Stuck on two combinatorics problems>
Anyone got ideas on these two problems ?1)A is a rectangle
consisting of n * 2 numbered squares. Compute a_n, the> number
of ways to partition A into domino bricks ( that is 2 x 1>
rectangles). For instance a_1 = 1, a_2 = 2 Your initial
conditions are there; form a recurrence relation by having a_n
and extending it by 1 or 2 rows. You might need an expressions
for a_n when n is odd and when n is even... but maybe not.> 2)
Show that if p and q are odd primes then> 2^(pq+1) [equivce
symbol] 2^(p+q) (mod pq)left side 2*(2^p)^q right side 2^p
2^qCan you remove 2^p or 2^q from either side.J===
===
Subject:
: Re:
Stuck on two combinatorics problemsEn el
mensaje:q0uXb.49032$mU6.192412@newsb.telia.net,Valentino Berti
<noaddress@none.com> escribi.97:> Anyone got ideas on these two
problems ?> 1)A is a rectangle consisting of n * 2 numbered
squares. Compute> a_n, the number of ways to partition A into
domino bricks ( that is 2> x 1 rectangles). For instance a_1 =
1, a_2 = 2You can make a n*2 rectangle adding a two horixontal
dominos a one (n-2)*2rectangle, or adding a vertical domino to
a (n-1)*2 rectangle, then ...> 2) Show that if p and q are odd
primes then> 2^(pq+1) [equivce symbol] 2^(p+q) (mod pq)If p =
q odd prime it fails. Did you forgotten the condition p =/=
q?-- ===
===
Subject:
: Re: there is no such thing as infinityRichard
Herring wibbled:> In message
<4f1e888502f84d511af633bad4d7158d@news.teranews.com>, Darryl
>> You think you can reach the largest number with a FORTAN
program? How> crude! Obviously, you need to write this
program in C++.>> Anything worth doing can be done in Perl, in
a fraction of the memory>> space something like C++ would
use.And can be written in exponentially more obfuscated a
manner than even C or>assembler... :)Do you have something
against APL?You need to ask?===
===
Subject:
: Re: there is no such
thing as infinityI told you, you've got to wait for Intel to
come out with a 64googleplex chip. You can't do this job right
with just 32 crummybits!>Fortran says positive infinity =
2147483647 and negative infinity =>-2147483648. Weird thing is
2*(positive infinity) = 0. >Where should I publish my
findings?===
===
Subject:
: please help!!!hi as usuall i need help
for this integer ,and who can integrate from this function
y=1/[(e^x)(e^x-1)] if e=2.71... ofcourse i have answer but as
usuall i dont know that how i find it thank you for your help
hupo===
===
Subject:
: Re: please help!!!> as usuall i need help for
this integer ,and who can integrate from this function>
y=1/[(e^x)(e^x-1)] if e=2.71... > ofcourse i have answer but
as usuall i dont know that how i find itPerform the
substitution x = log(y) and dx = dy/y.Best ,Jose Carlos
Santos===
===
Subject:
: Re: please help!!!En el
mensaje:54798887.0402141058.141b81d3@posting.google.com,hupo
<hupo19@yahoo.com> escribi.97:> hi> as usuall i need help for
this integer ,and who can integrate from> this function
y=1/[(e^x)(e^x-1)] if e=2.71...> ofcourse i have answer but as
usuall i dont know that how i find it> thank you for your help>
hupoLet t = e^x and you get a rational integral, easily worked
by descompositionin simple fractions.-- ===
===
Subject:
: Please
help meI need to solve this equation step by step. y(x) =
y(x)/x + 3(y(x))^2Could someone help me? I think that this is
Bernoullis equation, But not sure.===
===
Subject:
: Something to
consider when asking for help here...(post titles)When you're
asking for assistance here, *please* title your post
withsomething more imaginative and informative than need help,
pleasehelp, need assistance, HELP!!!, or the like.Why?Two main
reasons - one you're more likely to get a response from people
inhere, who have to filter through hundreds of posts per day,
if you'redescriptive in your header.Secondly, these things get
archived, and in the likely event that yourpost gets answered,
it will do no good to anyone searching later if yourtitles
don't support the thread itself.Just my thoughts.Doug (reverse
address in the standard edu fashion to send e-mail)