The following is a suggestion which Becki -- Instead of using biblical names for the next 387,420,345 constants (there are not enough names), why not use AT&T, Kingston, Verizon, Surely there are at least 144,000 unique names for the new constants there. Some of them might even be physicists' names. Then if you need more names, you can always resort to their yellow pages. I am positive you will find companies favorable to having a physical property, and its corresponding constant, named after them. White Pages -- > As I recall, there are theories that go up to 8 or 10 dimensions. > > The question is only up to four (4) dimensions. > How many unique combinations would be possible? > > > Ok, you got 9 fundamental units, any of which can appear > in the numerator or the denominator up to a power of 4. > A unit cannot appear in _both_ numerator and denominator > because the smaller power would cancel. > > Labeling the constants a..i and looking at just the a > constant, there are 9 ways (out of 25 possible) that a > appears in the fraction (na is the power of a in the > numerator and da is the power of a in the denominator): > > na da > 0 0 > 0 1 > 0 2 > 0 3 > 0 4 > 1 0 > 2 0 > 3 0 > 4 0 > > And since you have 9 constants, there are > > 9^9 or 387,420,489 combinations. > > A typical example is combination #24,376: > > na nb nc nd ne nf ng nh ni da db dc dd de df dg dh di > 0 4 0 2 0 0 0 0 0 3 0 3 0 3 0 0 0 0 > > which is > > a^0 b^4 c^0 d^2 e^0 f^0 g^0 h^0 i^0 > ----------------------------------- > a^3 b^0 c^3 d^0 e^3 f^0 g^0 h^0 i^0 > > > or > > s^4 kmol^2 > ------------- > cd^3 m^3 kg^3 > > > 1) luminous intensity = 1.9720204(06) x 10^-45 cd > 2) Einstein time = 1.3511888(38) x 10^-43 s > 3) wavelength of gravity = 4.0507622(30) x 10^-35 m > 4) unified substance = 1.6605402(10) x 10^-27 kmol > 5) gravitational mass = 5.4563086(06) x 10^-8 kg > 6) electric current = 1.1857538(29) x 10^24 A > 7) Celsius temperature = 3.5518470(84) x 10^32 K > 8) relative permeability = 6.8517994(75) x 10^1 sr > 9) inverse fine-structure = 1.3703598(95) x 10^2 rad > > > Good luck coming up with names for the other 387,420,345 constants. ==== >Message-id: <4da5bf5c.0307170826.35a9aa19@posting.google.com> > > >The following is a suggestion which I hope you name one after me. The Mensanator Constant. I like the sound of that. > > >Becki > >-- > > >Instead of using biblical names for the next 387,420,345 constants >(there are not enough names), why not use AT&T, Kingston, Verizon, >Surely there are at least 144,000 unique names for the new constants >there. Some of them might even be physicists' names. Then if you >need more names, you can always resort to their yellow pages. I am >positive you will find companies favorable to having a physical >property, and its corresponding constant, named after them. > >White Pages > >-- > >First 144 Primary Fundamental Physical Constants > >001) radiant volume = 1.3554076(23) x 10^-113 m-s^2/kg >002) volume of gravity = 6.6467639(49) x 10^-104 m^3 >003) gravitational volume = 1.2181796(21) x 10^-96 m^3/kg >004) luminous efficacy = 3.7229891(12) x 10^-96 cd-sr-s^3/kg-m^2 >005) current volume = 1.3838179(77) x 10^-93 m^2/A >006) luminous energy = 1.8257112(76) x 10^-86 cd-sr-s >007) charge volume = 4.1485819(27) x 10^-85 m^3/A-s >008) moment of inertia = 8.9530792(67) x 10^-77 kg-m^2 >009) gravitational fluidity = 1.0031222(77) x 10^-70 m-s/kg >010) area of gravity = 1.6408674(64) x 10^-69 m^2 >011) Joshua constant= 1.3234483(74) x 10^-63 A-s^3/kg-m >012) Ruth constant = 4.9191969(04) x 10^-61 m^3/s >013) Samuel constant = 3.4161915(66) x 10^-59 m/A >014) electric moment = 6.4900394(48) x 10^-54 A-s-m >015) Ezra constant = 1.6752612(69) x 10^-49 kg-m^3/A-s^2 >016) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 >017) Nehemiah constant = 1.1634040(12) x 10^-47 kg-m/A^2-s >018) magnetic moment = 1.9456648(79) x 10^-45 A-m^2 >019) luminous intensity = 1.9720204(06) x 10^-45 cd >020) Einstein time = 1.3511888(38) x 10^-43 s >021) luminous flux = 1.3511888(38) x 10^-43 cd-sr >022) gravitational moment = 2.2102208(82) x 10^-42 kg-m >023) self-mutual inductance = 3.4877974(86) x 10^-39 kg-m^2/A^2-s^2 >024) absorption-emission = 2.4763790(61) x 10^-36 s/kg >025) wavelength of gravity = 4.0507622(30) x 10^-35 m >026) Esther constant = 1.2379901(47) x 10^-34 s^4/m^4 >027) Planck constant = 6.6260755(09) x 10^-34 kg-m^2/s >028) relative expansion = 2.8154365(22) x 10^-33 /K >029) electric resistivity = 1.0456153(81) x 10^-30 kg-m^3/A^2-s^3 >030) Job constant = 3.2671588(69) x 10^-29 A-s^3/kg-m^2 >031) unified substance = 1.6605402(10) x 10^-27 kmol >032) kinematic viscosity = 1.2143879(66) x 10^-26 m^2/s >033) Solomon constant = 3.7114010(91) x 10^-26 s^3/m^3 >034) Isaiah constant = 1.9864474(64) x 10^-25 kg-m^3/s^2 >035) inverse electric current = 8.4334536(86) x 10^-25 /A >036) Jeremiah constant = 1.3795107(62) x 10^-23 kg-m/A-s >037) heat capacity constant = 1.3806578(67) x 10^-23 kg-m^2/s^2-K >038) thermal resistance = 9.7865580(66) x 10^-21 s^3-K/kg-m^2 >039) Ezekiel constant = 9.7946958(79) x 10^-21 A-s^2/kg-m >040) gravitational molality = 3.0433399(76) x 10^-20 kmol/kg >041) elementary charge = 1.6021773(38) x 10^-19 A-s >042) Daniel constant = 1.1126500(56) x 10^-17 s^2/m^2 >043) first radiation = 5.9552196(79) x 10^-17 kg-m^4/s^3 >044) Hosea constant = 2.5282858(10) x 10^-16 m/A-s >045) specific heat = 2.5303881(55) x 10^-16 m^2/s^2-K >046) magnetic flux = 4.1356692(24) x 10^-15 kg-m^2/A-s^2 >047) electric permittivity = 1.2922426(95) x 10^-13 A^2-s^4/kg-m^3 >048) magnetic exposure = 2.9363759(53) x 10^-12 A-s/kg >049) permittivity of vacuum = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 >050) magnetic pole strength = 4.8032068(25) x 10^-11 A-m >051) Joel constant = 4.8682699(55) x 10^-11 cd/m >052) Newton constant = 6.6723563(41) x 10^-11 m^3/kg-s^2 >053) S-B primary constant = 1.3897405(80) x 10^-10 kg/s^3-K^4 >054) density of states = 2.0391992(76) x 10^-10 s^2/kg-m^2 >055) radiant distribution constant = 3.335640952... x 10^-9 s/m >056) gravitational mass constant = 5.4563086(06) x 10^-8 kg >057) permeability of vacuum = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr >058) electric conductance = 3.8740461(38) x 10^-5 A^2-s^3/kg-m^2 >059) magnetic permeability = 8.6102251(57) x 10^-5 kg-m/A^2-s^2 >060) Amos constant = 1.9689919(12) x 10^-3 A-kmol >061) fine-structure constant = 7.2973530(80) x 10^-3 /rad >062) second radiation constant = 1.4387688(01) x 10^-2 m-K >063) dielectric constant = 1.4594706(16) x 10^-2 /sr >064) Obadiah constant = 1.4594706(16) x 10^-2 cd/s >065) spin half constant = 5.000000000.89¥Ï x 10^-1 sr/rad >066) length fraction = 1.000000000.89¥Ï m/m >067) mass fraction = 1.000000000.89¥Ï kg/kg >068) time fraction = 1.000000000.89¥Ï s/s >069) current fraction = 1.000000000.89¥Ï amp/amp >070) temperature fraction = 1.000000000.89¥Ï K/K >071) spin two constant = 2.000000000.89¥Ï rad/sr >072) gravitational momentum = 1.6357601(69) x 10^1 kg-m/s >073) Jonah constant = 6.8517994(74) x 10^1 s/cd >074) relative permeability = 6.8517994(75) x 10^1 sr >075) inverse fine-structure = 1.3703598(95) x 10^2 rad >076) molar heat capacity = 8.3145102(91) x 10^3 kg-m^2/s^2-kmol-K >077) spin angle constant = 9.3894312(09) x 10^3 sr-rad >078) Micah constant = 1.1614098(14) x 10^4 A^2-s^2/kg-m >079) electric resistance = 2.5812805(64) x 10^4 kg-m^2/A^2-s^3 >080) Nahum constant = 1.8997879(14) x 10^5 A^2-s >081) Habakkuk constant = 5.8979849(03) x 10^5 kmol-K >082) inverse gravitational mass = 1.8327409(10) x 10^7 /kg >083) Faraday constant = 9.6485308(14) x 10^7 A-s/kmol >084) speed of light in vacuum = 2.99792458 x 10^8 m/s >085) gravitational energy = 4.9038856(17) x 10^9 kg-m^2/s^2 >086) Zephaniah constant = 1.4987209(15) x 10^10 kg-s^2/m^3 >087) Haggai constant = 2.0541178(06) x 10^10 m/cd >088) Zechariah constant = 5.6954208(84) x 10^13 A^2-s >089) Josephson primary = 2.4179883(50) x 10^14 A-s^2/kg-m^2 >090) electric displacement = 3.9552490(31) x 10^15 A-s/m >091) absorbed dose = 8.9875517(87) x 10^16 m^2/s^2 >092) luminous density = 2.7467671(33) x 10^17 cd-sr-s/m^3 >093) Malachi constant = 1.4701479(23) x 10^18 kg-m^3/s^3 >094) gravity displacement = 4.4930522(70) x 10^18 kg-s/m^2 >095) molar mass constant = 3.2858635(84) x 10^19 kg/kmol >096) magnetic potential = 1.0209607(45) x 10^20 kg-m/A-s^2 >097) thermal conductance = 1.0218097(04) x 10^20 kg-m^2/s^3-K >098) Matthew constant = 7.2489467(08) x 10^22 A-s/kg-m >099) electric current constant = 1.1857538(29) x 10^24 A >100) luminance constant = 1.2018157(77) x 10^24 cd/m^2 >101) Mark constant = 2.6944002(42) x 10^25 m^3/s^3 >102) luminous flux density = 8.2346007(05) x 10^25 cd-sr/m^2 >103) Luke constant = 4.4073925(95) x 10^26 kg-m^4/s^4 >104) Avogadro constant = 6.0221366(15) x 10^26 /kmol >105) gravitational field = 1.3469831(84) x 10^27 kg/m >106) electric potential = 3.0607633(12) x 10^28 kg-m^2/A-s^3 >107) electric conductivity = 9.5637460(76) x 10^29 A^2-s^3/kg-m^3 >108) Celcius temperature = 3.5518470(84) x 10^32 K >109) John constant = 8.0776087(15) x 10^33 m^4/s^4 >110) gravity wave number = 2.4686711(86) x 10^34 /m >111) mass flow rate constant = 4.0381539(96) x 10^35 kg/s >112) molar energy = 2.9531869(13) x 10^36 kg-m^2/s^2-kmol >113) Timothy constant = 6.3723329(52) x 10^38 kg-m/A^2-s^3 >114) surface concentration = 1.0119892(35) x 10^42 kmol/m^2 >115) frequency of gravity = 7.4008900(30) x 10^42 /s >116) gravitational force = 1.2106081(12) x 10^44 kg-m/s^2 >117) inverse luminous intensity = 5.0709414(41) x 10^44 /cd >118) angular velocity = 1.0141882(88) x 10^45 rad/s >119) Titus constant = 8.5954663(19) x 10^46 A^2-s/kg-m >120) Philemon constant = 1.9103773(59) x 10^47 kg-m^2/A^2-s^4 >121) James constant = 5.9692181(67) x 10^48 A-s^2/kg-m^3 >122) electric flux density = 9.7642093(17) x 10^49 A-s/m^2 >123) radiant intensity = 5.2968739(54) x 10^50 kg-m^2/s^3-sr >124) gravity field strength = 2.2187310(14) x 10^51 m/s^2 >125) gravitational power = 3.6293118(17) x 10^52 kg-m^2/s^3 >126) magnetic flux density = 2.5204163(73) x 10^54 kg/A-s^2 >127) thermal conductivity = 2.5225121(74) x 10^54 kg-m/s^3-K >128) Peter constant = 1.7895265(87) x 10^57 A-s/kg-m^2 >129) magnetic field strength = 2.9272363(12) x 10^58 A/m >130) absorbed dose rate = 6.6515882(43) x 10^59 m^2/s^3 >131) Jude constant = 1.0880403(11) x 10^61 kg-m^3/s^4 >132) surface density constant = 3.3252585(75) x 10^61 kg/m^2 >133) electric field strength = 7.5560181(98) x 10^62 kg-m/A-s^3 >134) dynamic viscosity = 9.9688744(17) x 10^69 kg/m-s >135) molar concentration = 2.4982686(65) x 10^76 kmol/m^3 >136) surface tension constant = 2.9885933(65) x 10^78 kg/s^2 >137) electric charge density = 2.4104622(20) x 10^84 A-s/m^3 >138) angular acceleration = 7.5058959(93) x 10^87 rad/s^2 >139) thermal transfer constant = 6.2272531(21) x 10^88 kg/s^3-K >140) current density constant = 7.2263839(39) x 10^92 A/m^2 >141) gravitational density = 8.2089700(31) x 10^95 kg/m^3 >142) energy density = 7.3778543(28) x 10^112 kg/m-s^2 >143) radiance constant = 3.2280937(18) x 10^119 kg/s^3-sr >144) irradiance constant = 2.2118250(84) x 10^121 kg/s^3 > >> As I recall, there are theories that go up to 8 or 10 dimensions. >> >> The question is only up to four (4) dimensions. >> How many unique combinations would be possible? >> >> >> Ok, you got 9 fundamental units, any of which can appear >> in the numerator or the denominator up to a power of 4. >> A unit cannot appear in _both_ numerator and denominator >> because the smaller power would cancel. >> >> Labeling the constants a..i and looking at just the a >> constant, there are 9 ways (out of 25 possible) that a >> appears in the fraction (na is the power of a in the >> numerator and da is the power of a in the denominator): >> >> na da >> 0 0 >> 0 1 >> 0 2 >> 0 3 >> 0 4 >> 1 0 >> 2 0 >> 3 0 >> 4 0 >> >> And since you have 9 constants, there are >> >> 9^9 or 387,420,489 combinations. >> >> A typical example is combination #24,376: >> >> na nb nc nd ne nf ng nh ni da db dc dd de df dg dh di >> 0 4 0 2 0 0 0 0 0 3 0 3 0 3 0 0 0 0 >> >> which is >> >> a^0 b^4 c^0 d^2 e^0 f^0 g^0 h^0 i^0 >> ----------------------------------- >> a^3 b^0 c^3 d^0 e^3 f^0 g^0 h^0 i^0 >> >> >> or >> >> s^4 kmol^2 >> ------------- >> cd^3 m^3 kg^3 >> >> >> 1) luminous intensity = 1.9720204(06) x 10^-45 cd >> 2) Einstein time = 1.3511888(38) x 10^-43 s >> 3) wavelength of gravity = 4.0507622(30) x 10^-35 m >> 4) unified substance = 1.6605402(10) x 10^-27 kmol >> 5) gravitational mass = 5.4563086(06) x 10^-8 kg >> 6) electric current = 1.1857538(29) x 10^24 A >> 7) Celsius temperature = 3.5518470(84) x 10^32 K >> 8) relative permeability = 6.8517994(75) x 10^1 sr >> 9) inverse fine-structure = 1.3703598(95) x 10^2 rad >> >> >> Good luck coming up with names for the other 387,420,345 constants. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm ==== While surfing the Web, I stumbled upon the following site: http://www.mathpages.com/ The *.com made me wonder who might be affiliated to this site. and so on, I thought it might be worth mentioning it in this NG. Would anyone care to share their views/reviews of this site? David Bernier ___________________________________________________________ Then assuredly the world was made, not in time, but simultaneously with time. --St. Augustine ==== > While surfing the Web, I stumbled upon the > following site: > http://www.mathpages.com/ Oooh, thanks for that! > The *.com made me wonder who might > be affiliated to this site. Unless it is hidden in the HTML source, you are apparently doomed to keep wondering; I couldn't find a trace of ownership claim. > and so on, I thought it might be worth mentioning > it in this NG. > Would anyone care to share their views/reviews > of this site? I am most struck by how _sane_ and _readable_ the site is. The parts I found weren't overwhelmingly technical, either, a relief for someone with my modest math skills. The reader will gain a lot of historical perspective fairly painlessly. xanthian. -- ==== David Bernier > While surfing the Web, I stumbled upon the > following site: > > http://www.mathpages.com/ > > The *.com made me wonder who might > be affiliated to this site. Indeed. He has a lot to say but is not trying to fatten his CV or sell us any book on which he makes a commission. Most unusual :) > > Would anyone care to share their views/reviews > of this site? I've seen some but not all of these interesting bits and pieces before. IMO it's a good site for any teacher who is looking for enrichment material. Larry ==== >While surfing the Web, I stumbled upon the >following site: > > http://www.mathpages.com/ > >The *.com made me wonder who might >be affiliated to this site. http://www.coolwhois.com/?d=mathpages.com is a way to answer that question. (Who oh why do so many authors put no contact information on their sites?) -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== > While surfing the Web, I stumbled upon the > following site: > > http://www.mathpages.com/ > > The *.com made me wonder who might > be affiliated to this site. This was set up by Kevin S. Brown. It was started as a conributor in the early to mid nineties. He I believe holds the dubious honor of being the first person on sci.math to attempt to explain math to JSH. > > and so on, I thought it might be worth mentioning > it in this NG. > > Would anyone care to share their views/reviews > of this site? > > David Bernier > ___________________________________________________________ > > Then assuredly the world was made, not in time, but > simultaneously with time. > > --St. Augustine > ==== (Frequently Exhorted Exhortations :-) I recently posted a message to newbies exhorting them to be more thoughtful in their choice of subject headings. What with its ephemeral nature, I didn't expect the message to do any good, and it didn't, as witness recently: math question. Now I wish to exhort oldies not to work homework problems for people, but merely to provide hints. In general, sci.math has a remarkable gentlepersons'-agreement on this. In fact, it's one of the most impressive aspects of sci.math. But often it seems that 5 repliers give hints, and the 6th provides the complete solution. I know this is all WAY OLD HAT, but perhaps it bears repeating. If not, you can send some heat my way. I will fry eggs on my monitor. Lest I be hoisted by my own petard, let me hasten to throw in a few mea culpas. ==== >I recently posted a message to newbies >Now I wish to exhort oldies not to work homework problems for people ExCUSE me, but I prefer the term knowbies for the yang to the newbies' yin. dave ==== In sci.math, Dave Rusin : > >>I recently posted a message to newbies > >>Now I wish to exhort oldies not to work homework problems for people > > ExCUSE me, but I prefer the term knowbies > for the yang to the newbies' yin. > > dave > So if one's just graduated from college is he a newbie knowbie, or a knowbie noveau? *dodges tomatoes* :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >> I recently posted a message to newbies > >> Now I wish to exhort oldies not to work homework problems for people > > ExCUSE me, but I prefer the term knowbies > for the yang to the newbies' yin. Certainly it has to be at least oldbies, not oldies. I agree with the exhortation, BTW. Not because I particularly care if somebody somewhere gets an unearned A; that isn't really my problem. I'm concerned, rather, that the more people go along with it, the more requests of this sort we'll get. ==== > Certainly it has to be at least oldbies, not oldies. Thereby creating a grammatical rule for changing adjectives into nouns. Maybe the rule would apply only to people. Or people and dogs, say. Hungrybies, Smartbies, Dumbbies, Stinkybies, Seedybies... hum, I could go on a long time, yukking like a Nerd all the while. Max, Mr. Maximally Maxed ==== > I wish to exhort oldies not to work homework problems for people > I agree with the exhortation, BTW. Not because I particularly > care if somebody somewhere gets an unearned A; that isn't really > my problem. I'm concerned, rather, that the more people go along > with it, the more requests of this sort we'll get. This looks like a wonderful opportunity to start a new sub-group - alt.math.homework-help. If it's getting to be that big a hassle, there's obviously a pent-up demand for it. Then let those people work out the question of whether to give hints or complete answers. ==== > I agree with the exhortation, BTW. Not because I particularly > care if somebody somewhere gets an unearned A; that isn't really > my problem. I'm concerned, rather, that the more people go along > with it, the more requests of this sort we'll get. We is not a fixed entity. I, for one, hope the homework police find it difficult to regulate what people post. ==== possibly of interest, http://www.geocities.com/jongiff2000/aa_index_math.html I haven't quite figured out yet the meaning of the angle between two vectors in n-space. If projections still hold, then the cosine must still hold, but exactly what does that angle measure? ==== > possibly of interest, > > http://www.geocities.com/jongiff2000/aa_index_math.html > > I haven't quite figured out yet the meaning of the > angle between two vectors in n-space. If projections > still hold, then the cosine must still hold, but exactly > what does that angle measure? > > Two non-parallel vectors in n-space determine a two dimensional subspace. The angle is measured and meaningful in that two dimensional subspace, i. e., in a plane. ==== > > possibly of interest, > > http://www.geocities.com/jongiff2000/aa_index_math.html > > I haven't quite figured out yet the meaning of the > angle between two vectors in n-space. If projections > still hold, then the cosine must still hold, but exactly > what does that angle measure? Do you suspect that Equivalence of Zero and Infinity casts some doubt on the balance of the content? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! ==== In sci.physics, Jon : > possibly of interest, > > http://www.geocities.com/jongiff2000/aa_index_math.html > > I haven't quite figured out yet the meaning of the > angle between two vectors in n-space. If projections > still hold, then the cosine must still hold, but exactly > what does that angle measure? > What does any angle measure? In N-dimensional Euclidean space (N > 2), two lines intersecting will describe a plane. That plane is 2-dimensional regardless of N. Rotate that plane onto your dissertation paper. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== Meissel's formula is given in the CRC Concise Encyclopedia of Mathematics and on the webpage http://mathworld.wolfram.com/MeisselsFormula.html I am not at all convinced that formula (4) is correct. I think in the second sum, the indices should be 1 <= i < j <= c and not 1 <= i <= j <= c, and in the third sum the indices should be c < i <= b and not c <= i <= b. I would appreciate it if anyone who has a different source or who is willing to derive the formula could check this. The formula given there for pi (x) = number of primes <= x: pi (x) = floor (x) - sum (floor (x / p(i)), 1 <= i <= c) + sum (floor (x / p(i) / p(j)), 1 <= i <= j <= c) - ... + ... + (1/2) (b + c - 2) (b - c + 1) - sum (pi (x / p (i)), c <= i <= b) where b = pi (x ^ (1/2)) and c = pi (x ^ (1/3)), and p (i) is the i-th prime number, that is p(1) = 2, p(2) = 3, p(3) = 5 etc. ==== > Meissel's formula is given in the CRC Concise Encyclopedia of > Mathematics and on the webpage > > http://mathworld.wolfram.com/MeisselsFormula.html > > I am not at all convinced that formula (4) is correct. I think in the > second sum, the indices should be 1 <= i < j <= c and not 1 <= i <= j <= > c, and in the third sum the indices should be c < i <= b and not c <= i > <= b. > > I would appreciate it if anyone who has a different source or who is > willing to derive the formula could check this. For reference, Riesel gives a permitaion of the top formula identically, but it deviates from Mathworld later on, exactly as you indicate. If you didn't crib the above from Riesel, then I'd put my money on you and Hans being right. Phil ==== Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real a and b, f( a + b ) = f( a ) + f( b ). Is f necessarily linear? Can you recommend a text that discusses this problem? Note: 1) For any rational q, f( qa ) = qf( a ). 2) If f is continuous, it is linear. 3) If, in the above problem, the real field is replaced by the rationals, then f is linear. -- ==== >Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real >a and b, f( a + b ) = f( a ) + f( b ). >Is f necessarily linear? No. For example, let B be a Hamel basis of the reals over the rationals, take some b_1 in B and define f(sum_{b in B} r_b b) = r_{b_1}. On the other hand, if f is Lebesgue measurable, or if it is bounded on some set of positive Lebesgue measure, then it is linear. See e.g. the thread Difficult Problem from February 1996. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real > a and b, f( a + b ) = f( a ) + f( b ). > > Is f necessarily linear? > > > Can you recommend a text that discusses this problem? > > > Note: > > 1) For any rational q, f( qa ) = qf( a ). > > 2) If f is continuous, it is linear. > > 3) If, in the above problem, the real field is replaced by the > rationals, then f is linear. First write down a proof for (1), (2) or (3). If you find this difficult then ask for help, but don't expect that you can solve the original problem if you cannot prove these three. Then use the axiom of choice. ==== > Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real > a and b, f( a + b ) = f( a ) + f( b ). > Sidenote: f(0) = 0 is redundant. > Is f necessarily linear? > > Note: > > 1) For any rational q, f( qa ) = qf( a ). > 2) If f is continuous, it is linear. > 3) If, in the above problem, the real field is replaced by the > rationals, then f is linear. > ==== > >Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real >a and b, f( a + b ) = f( a ) + f( b ). > >Is f necessarily linear? > No. > > For example, let B be a Hamel basis of the reals over the rationals, > take some b_1 in B and define f(sum_{b in B} r_b b) = r_{b_1}. > > On the other hand, if f is Lebesgue measurable, or if it is bounded on > some set of positive Lebesgue measure, then it is linear. > See e.g. the thread Difficult Problem from February 1996. > I suppose those results have dual or parallel form for g:R -> R with g(ab) = g(a)g(b), g(1) = 1 Similarly, if g is continuous, g(x) = |x|^n for some n in R. ==== |I suppose those results have dual or parallel form for g:R -> R with | g(ab) = g(a)g(b), g(1) = 1 |Similarly, if g is continuous, g(x) = |x|^n for some n in R. Incidentally, the only possibility ruled out by g(1)=1 is g identically equal to 0. Your question is closely related to the one of the O.P., since if g is such a function, then f(x) = log g(e^x) satisfies f(x+y) = log g(e^{x+y}) = log g(e^x e^y) = log (g(e^x) g(e^y)) = log g(e^x) + log g(e^y) = f(x)+f(y), which means f is the kind of function originally considered. Thus if g satisfies your conditions, then there exists a function f satisfying f(x+y)=f(x)+f(y) for every x, y, for which g(x) = e^f(log x) for x>0. That just leaves the question of what values g has when x<=0. We know g(0)=g(0)g(x) for every x. So if g(0)<>0, then g(x)=1 for every x. So if g is not identically 1, then g(0)=0. The constant function 1 does satisfy your conditions. satisfy the original conditions if f satisfies the functional equation f(x+y)=f(x)+f(y). If g is continuous (or even measurable) then f is also continuous (measurable) and f(x)=cx for some c. Then either g(x)=1=|x|^0, or g(x)=|x|^c, or the possibility you missed, g(x)=sgn(x)*|x|^c for c>0 (since for c<=0, sgn(x)*|x|^c is discontinuous at x=0). If f is discontinuous, then it's one of these peculiar functions whose graph is dense in the plane, and the graph of g is either dense in the upper half plane y>=0, or in both the first quadrant x>=0, y>=0 and the third quadrant x<=0, y<=0. Keith Ramsay ==== > |I suppose those results have dual or parallel form for g:R -> R with > | g(ab) = g(a)g(b), g(1) = 1 > |Similarly, if g is continuous, g(x) = |x|^n for some n in R. > > Incidentally, the only possibility ruled out by g(1)=1 is g > identically equal to 0. > Indeed, it was stated that way in parallel contrast to OP's redundant f(0) = 0 > Your question is closely related to the one of the O.P., since if > g is such a function, then f(x) = log g(e^x) satisfies f(x+y) > = log g(e^{x+y}) = log g(e^x e^y) = log (g(e^x) g(e^y)) > = log g(e^x) + log g(e^y) = f(x)+f(y), which means f is the kind of > function originally considered. > > Thus if g satisfies your conditions, then there exists a function f > satisfying f(x+y)=f(x)+f(y) for every x, y, for which g(x) > = e^f(log x) for x>0. > That certainly demonstrates constructively the duality. It's taking a somewhat familiar form of other dualities. log g(x) = f(log x); e^f(x) = g(e^x) cl SA = Sint A; Scl A = int SA -lub a,b = glb -a,-b; lub -a,-b = -glb a,b -sup A = inf -A; sup -A = -inf A /{ S-X | X in C } = S - /{ X | X in C } S - /{ X | X in C } = /{ S-X | X in C } > That just leaves the question of what values g has when x<=0. We know > g(0)=g(0)g(x) for every x. So if g(0)<>0, then g(x)=1 for every x. So > if g is not identically 1, then g(0)=0. The constant function 1 does > satisfy your conditions. > Yup, noticed that. I'll take your word for the rest which points out some grit in an otherwise smooth duality. A duality expressed with continuous functions, yet applicable to discontinuous ones also. > g(x) = -g(-x) or g(x)=g(-x) for every x. It's easy to check that > both possibilities, > > g(x) = e^f(log x) if x > 0 > = 0 if x = 0 > = e^f(log -x) if x < 0 > > and > g(x) = e^f(log x) if x > 0 > = 0 if x = 0 > = e^f(log -x) if x < 0 > > satisfy the original conditions if f satisfies the functional > equation f(x+y)=f(x)+f(y). > > If g is continuous (or even measurable) then f is also continuous > (measurable) and f(x)=cx for some c. Then either g(x)=1=|x|^0, or > g(x)=|x|^c, or the possibility you missed, g(x)=sgn(x)*|x|^c for c>0 > (since for c<=0, sgn(x)*|x|^c is discontinuous at x=0). > > If f is discontinuous, then it's one of these peculiar functions > whose graph is dense in the plane, and the graph of g is either dense > in the upper half plane y>=0, or in both the first quadrant x>=0, y>=0 > and the third quadrant x<=0, y<=0. > ==== Robert, I've been puzzling over what you mean by the reals over the rationals. I guess you mean to somehow decompose the reals into algebraically closed subsets. Do you mean some kind of quotient space? I know what is meant by the quotient of a vector space with a vector subspace, but the rationals aren't a vector subspace of the reals (in any sense I know), Can you expand on this please? > > For example, let B be a Hamel basis of the reals over the rationals, > take some b_1 in B and define f(sum_{b in B} r_b b) = r_{b_1}. > -- ==== > I guess you mean to somehow decompose the reals into algebraically closed > subsets. Do you mean some kind of quotient space? I know what is meant > by the quotient of a vector space with a vector subspace, but the > rationals aren't a vector subspace of the reals (in any sense I know), Sure they are. Any field K with a subfield k is a vector space over k. All the vector space axioms are satisfied if the elements in K are vectors, the elements in k are scalars, and addition and multiplication are as defined in K. Many of the basic results in field theory rely on this fact. ==== >Robert, > >I've been puzzling over what you mean by the reals over the rationals. > >I guess you mean to somehow decompose the reals into algebraically closed >subsets. Do you mean some kind of quotient space? I know what is meant >by the quotient of a vector space with a vector subspace, but the >rationals aren't a vector subspace of the reals (in any sense I know), > >Can you expand on this please? > > >> >> For example, let B be a Hamel basis of the reals over the rationals, >> take some b_1 in B and define f(sum_{b in B} r_b b) = r_{b_1}. >> I think you have missed something essential. The reals are a vector space over the rationals. Check the definition of a vectors space over a field. Larry (this space unintentially left blank ..... ==== Let S be the set of natural numbers n such that mu(n) = 1 . What is the density of S ? ==== Let S be the set of natural numbers n such that mu(n) = 1 . What is the density of S ? ==== > Let S be the set of natural numbers n such that mu(n) = 1 . > What is the density of S ? The set of natural numbers n such that |mu(n)| = 1 is 6/(pi^2). The sum of mu(n), n from 1 to N, is little-oh of N. Does it not follow that the number you're after is 3/(pi^2)? -- ==== Dear reader, for my research I have programmed a combinatorial optimization problem in aimms 2 format. Unfortunately running this program takes a lot of time, so I want to try a more efficient solver like C-plex. However, in order to use C-plex I need to convert my aimms program to an mps format program. Can anyone tell me whether this is possible in aimms? Dion Bongaerts ==== > to this question, which has been driving me a bit nuts. Some quick > background: as a role-player, I use a lot of dice. At times, the > rules call for one to roll multiple dice (say, five six-sided dice, or > 5d6), then to drop the lowest two and total the other three. The > basic question is: is there a formula for determining the probability > of rolling a certain result, given these conditions? > > Determining the probability of a particular outcome when just rolling > multiple dice is relatively straightforward (there's a brief > discussion here: http://mathforum.org/library/drmath/view/52207.html). > I can find a pattern to the summation needed when dropping a single > die from a set; but once I try to remove two dice from the set, the > pattern disappears and I find myself lost again. (The numbers can be > determined by brute force, of course, but that's neither practical nor > interesting.) > > So, I guess the base question is: Is there a formula for calculating > the probability of achieving a result R on n dice with d sides, > dropping the k lowest dice? show us your pattern for dropping one dice and we might be able to work it out. Herc ==== Define S:= {(x,y) in R : 0 Define S:= {(x,y) in R : 0 Define T:= {x in R : 0 > Problem: Use the fact that every real number has a decimal expansion > to produce an injective function that maps S into T. Think about how you can take two infinite decimal expansions and turn them that might give you an idea.) -- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman ==== > > Define S:= {(x,y) in R : 0 Define T:= {x in R : 0 > Problem: Use the fact that every real number has a decimal expansion > to produce an injective function that maps S into T. > > thanks. For the reals which can have terminating decimal representations, require them to be so represented, then for x with decimal representation .abcdefg... and y with decimal representation .ABCDEFG... map(x,y) onto z with decimal representation .aAbBcCdDeEfFgG... ==== >Define S:= {(x,y) in R : 0Define T:= {x in R : 0Problem: Use the fact that every real number has a decimal expansion >to produce an injective function that maps S into T. The usual solution is a bit of a trick, hard to hint at without simply giving it away. Given (x,y), take the decimal expansions of x and y and do a riffle shuffle. Brian ==== Can anyone please help me with this difficult question on probability? The probability that there is 'x' amount of items in a box (where x is in 1000's): P(x) = -0.03x^2 + 0.12x + 0.15 What is the probability that the box contains at least 4000 items? ==== > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? I don't think that anybody can help. For large x, your probability will become negative (???), and the sum of P over all possible x should equal one, but isn't even finite in this case. Weird question, who asked you? ==== Since there are 4 thousands in 4000 simply calculate P(4) > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? > > > ==== >Can anyone please help me with this difficult question on probability? > >The probability that there is 'x' amount of items in a box (where x is in >1000's): > >P(x) = -0.03x^2 + 0.12x + 0.15 > >What is the probability that the box contains at least 4000 items? > > P(x) = -0.03 * (x^2 -4x -5) = -0.03*(x+1)(x-5) because probabilities have to be non-negative and numbers of items in a box also, P(x) is defined for 0<=x<=5. Because for x in [0,5] 0<=P(x)<=1 and int[0,5]P(x)dx=1 is P(x) indeed a density function and Prob(x>=4) = int [4,5]P(x)dx. I leave that to you. ==== Sorry, I misread the problem. P(at least 4000) = P(x > 4) = 1- P( x <= 4). Now you can solve either of the probabilities listed in the line above by integration. But wait a minute: integral (-.03x^2 + .12x + .15) from 0 to oo is NEGATIVE. This problem makes no sense > Since there are 4 thousands in 4000 simply calculate P(4) > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? > > > > > ==== > Can anyone please help me with this difficult question on probability? > P(x) = -0.03x^2 + 0.12x + 0.15 > What is the probability that the box contains at least 4000 items? Sure. Show us the work you did on the problem first. Also, should we call you Marc Rice, Mary Watts, David Jones, or Frank Dewhurst? Paul Guertin pg@sff.net ==== >Can anyone please help me with this difficult question on probability? > >The probability that there is 'x' amount of items in a box (where x is in >1000's): > >P(x) = -0.03x^2 + 0.12x + 0.15 > >What is the probability that the box contains at least 4000 items? > > Well, here's one interpretation. Find the zeros of P(x) (-1 and 5). Figure that x>=0, so integrate P from 0 to 5 to see if the total probability is 1. Then integrate P from 4 to 5 to get the probability that x>=4. This is not too far removed from what would happen with real data -- you'd have a model to fit it into (not a polynomial, simply because we don't do that in probabilistic models, but something) and some data, and you'd see how well the data fit the model, and adjust it (either the model or the data) to get something useful (the probability that total losses exceed a billion dollars, for example, or that the lifetime of an appliance will exceed 5 years). Jon Miller ==== > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? > > > Probabilities must be values between 0 and 1, and the sum of all probabilities must add up to 1. You alleged probability function, as defined, does not have either of these properties. ==== > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? Who knows if this is the real intention or not, it's just a guess (the OP should clarify). Probability, in general, must be between 0 and 1 inclusive. Thus, the largest possible range of P is [0,1]. This doesn't necessarily mean that's the actual range. It just means the range of P will be a subset of [0,1] ie the actual range is taken from a codomain of [0,1]. (who knows where this model came from, apparently more is known to them about P than simply it must be between 0 and 1 in this particular case). Also, x is clearly >=0, possibly even just greater (noninclusive). Under this very reasonable assumption, we can reasonably assume the domain of P, with just the given information, to be implied [0,5] (or possibly (0,5]). The range of P is therefore [0,.27] accordingly. We have: P(x) = -0.03x^2 + 0.12x + 0.15 domain: [0,5] range: [0,.27] IOW, the graph of P is just that part of the parabola in Q1. The question asks, I think, for: integral(x=4 to x=5) P(x) dx. Hope that helps. -- Darrell ==== Note: Apparently I'm too far off base, since integral(0 to 5) P(x)dx = 1. Makes perfect sense, right? -- Darrell > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? > > > Who knows if this is the real intention or not, it's just a guess (the OP > should clarify). Probability, in general, must be between 0 and 1 > inclusive. Thus, the largest possible range of P is [0,1]. This doesn't > necessarily mean that's the actual range. It just means the range of P will > be a subset of [0,1] ie the actual range is taken from a codomain of > [0,1]. (who knows where this model came from, apparently more is known to > them about P than simply it must be between 0 and 1 in this particular > case). > > Also, x is clearly >=0, possibly even just greater (noninclusive). Under > this very reasonable assumption, we can reasonably assume the domain of P, > with just the given information, to be implied [0,5] (or possibly (0,5]). > The range of P is therefore [0,.27] accordingly. > > We have: > P(x) = -0.03x^2 + 0.12x + 0.15 > domain: [0,5] > range: [0,.27] > > IOW, the graph of P is just that part of the parabola in Q1. > > The question asks, I think, for: > integral(x=4 to x=5) P(x) dx. Hope that helps. > > -- > Darrell > > > > > > ==== Oops.... should be _not_ too far off base. > Note: Apparently I'm too far off base, since integral(0 to 5) P(x)dx = 1. > Makes perfect sense, right? > > -- > Darrell > > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is > in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? > > > Who knows if this is the real intention or not, it's just a guess (the OP > should clarify). Probability, in general, must be between 0 and 1 > inclusive. Thus, the largest possible range of P is [0,1]. This doesn't > necessarily mean that's the actual range. It just means the range of P > will > be a subset of [0,1] ie the actual range is taken from a codomain of > [0,1]. (who knows where this model came from, apparently more is known to > them about P than simply it must be between 0 and 1 in this particular > case). > > Also, x is clearly >=0, possibly even just greater (noninclusive). Under > this very reasonable assumption, we can reasonably assume the domain of P, > with just the given information, to be implied [0,5] (or possibly (0,5]). > The range of P is therefore [0,.27] accordingly. > > We have: > P(x) = -0.03x^2 + 0.12x + 0.15 > domain: [0,5] > range: [0,.27] > > IOW, the graph of P is just that part of the parabola in Q1. > > The question asks, I think, for: > integral(x=4 to x=5) P(x) dx. Hope that helps. > > -- > Darrell > > > > > > > > > ==== > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? Two answers have been given (integrate this function, or add up some values of P(x) for some values of x), but neither is correct, because you haven't said what the possible values of x are. In fact, if all the possible values of x are less than 4, the correct answer is zero. If the set of possible values is finite (like {1, 2, 3}), you will need to add up P(x) over all _possible_ values of x which are at least 4. If the set of possible values of x is an interval (like [0, 5]), then you have to have a probability _density_ function, and you integrate it from 4 to whatever the upper limit is. Without further information, people can only guess at the answer. -- Christopher Heckman ==== > Can anyone please help me with this difficult question on probability? > > The probability that there is 'x' amount of items in a box (where x is in > 1000's): > > P(x) = -0.03x^2 + 0.12x + 0.15 > > What is the probability that the box contains at least 4000 items? > > Two answers have been given (integrate this function, or add up some values > of P(x) for some values of x), but neither is correct, because you haven't > said what the possible values of x are. In fact, if all the possible values > of x are less than 4, the correct answer is zero. I count no less than 5 answers (maybe your news feed is slow). > If the set of possible values is finite (like {1, 2, 3}), you will need > to add up P(x) over all _possible_ values of x which are at least 4. > If the set of possible values of x is an interval (like [0, 5]), then > you have to have a probability _density_ function, and you integrate it from > 4 to whatever the upper limit is. I don't pretend to know much about probability, much less much about a probability density function, but if it's what I think it is (?), then that's what we have with P. > Without further information, people can only guess at the answer. ...But we can make an educated guess. I, and at least one other, guessed that integral(0,5)P(x)dx=1 is no coincidence. Yeah, practically speaking since these are items then x can in reality take on only increments of 1/1000 (x was in thousands of items) but I see the continuous parabolic curve as, well, just a model. It fits the data well, so we pretend x takes on all values in a certain interval. Close enough. Under this interpretation (which is the only one that seems to make any sense out of the problem), the question asks for integral(4,5)P(x)dx. The question could be better worded. -- Darrell ==== [...] > I don't pretend to know much about probability, much less much about a > probability density function, but if it's what I think it is (?), then > that's what we have with P. > If the distributed variable X is discrete, it is meaningful to have an equation for P(X=x). The summation of all the probabilities will be one. If the distributed variable X is continuous a probability distribution function (pdf) is useful. In this case P(X=x) is always zero. The integral over the domain of the pdf is one. For a uniform distribution we could have a pdf such that: f(x) = (1/2 0 If the distributed variable X is discrete, it is meaningful to have an > equation for > P(X=x). The summation of all the probabilities will be one. I stated earlier. The variable X is indeed discrete (is given in thousands of units; makes no sense to speak of a fractional units, so it can vary _at least_ in increments of 1/1000). At least that's a reasonable assumption. The notation P(X=x), which I may very well be misinterpreting, suggests to me that we have fit the data to some continuous function of x. and are using P(x) as a model for the actual P(X) where X is discrete. The summation of all (discrete) probabilities is 1, as in P(1)+P(2)+P(3)+...+P(n) for some n. Am I interpreting correctly? If so, then I think I can see where the OP's problem, as stated, doesn't make much sense, since the discrete functional values of the given equation don't seem to add up to 1 under different reasonable assumptions on how much X can really vary by (discretely). > If the > distributed variable X is continuous a probability distribution function > (pdf) is useful. In this case P(X=x) is always zero. > > The integral over the > domain of the pdf is one. For a uniform distribution we could have a pdf > such that: > > f(x) = (1/2 0 ( 0 elsewhere Are you saying that the given equation, with assumed domain [0,5], is or is not a probability distribution function? It's integral is 1, but there is only one point where P(x)=0. If I had to guess, it appears to be saying by way of your example that for all reals except [0,5] the function should indeed be defined and integrable, but with integral=0. IOW, I interpret you to imply that the (nonuniform) probability distribution function for this problem may look something like: f(x) = -.03x^2 + .12x + .15 on [0,5] = 0 on (-oo,0)U(5,oo) ..since it's 0 elsewhere yet the integral over the entire domain is 1. Please straighten this dummy out ;-). -- Darrell ==== I am a hardware logic designer and I would greatly appreciate some help on defining a boolean function from the given conditions below. I am trying to formulate a boolean function g which maps a 8-bit variable a[7:0] to another 8 bit variable b[7:0], i.e. g(a[7:0]) = b[7:0] ... (i) and, (b[0] + b[1].x + b[2].(x^2) + b[3].(x^3)) + (b[4] + b[5].x + b[6].(x^2) + b[7].(x^3)).y = 1/((a[0] + a[1].x + a[2].(x^2) + a[3].(x^3)) + (a[4] + a[5].x + a[6].(x^2) + a[7].(x^3)).y) ... (ii) y is an element in GF(2^8) that satisfies y^8 + y^6 + y^5 + y^3 + 1 = 0 ... (iii) and x = y^238 = y^6 + y^5 + y^3 + y^2 ... (iv) is an element in GF(2^4) that satisfies x^4 + x + 1 = 0 ... (v) Apparently, you can also assume 1/0 = 0, though not sure why (or how) that is true. - Swapnajit. boundary=------------5F60FDBAB57B35E6585F0C83 ==== --------------------------------------------------------------------- x-no-archive: yes Please excuse my ignorance ! I'm the messenger ..my brother is having a building constructed and called me on his cell.. he's caught without his trigonometric function table book ! he needs to know the TANGENT of a 2o angle ! Please respond ASAP ..so I can call him back. remove (NOSPAM) of course oh thank you ..thank you ==== > x-no-archive: yes > Please excuse my ignorance ! I'm the messenger ..my brother is having a > building constructed and called me on his cell.. he's caught without his > trigonometric function table book ! > he needs to know the TANGENT of a 2o angle ! > Please respond ASAP ..so I can call him back. > remove (NOSPAM) of course > oh thank you ..thank you > This is pretty funny, actually, but anyway, here it is: 0.034920769491747730500402625773725315879174297784615 approximately boundary=------------56E0B04682C893DE51A74A90 ==== --------------------------------------------------------------------- x-no-archive: yes oh ..so funny and oh .. so appreciated ! THANK YOU SO MUCH ==== I'm sure the proof is just under my nose and I just can't seem to see it. The problem is to prove that for p prime, no nonzero integers a,b exist such that a^2 = pb^2. This is supposed to be done without assuming that the square roots of the odd primes are irrational (you can still assume sqrt(2) is irrational). So for p=2 it's trivial. For the other p, we know p is odd, which means that pb^2 is odd that if any such a and b exist, they have equal parity. I've managed to prove that no even a and b exist which satisfy the equation: for, assume they do (for a given odd prime p), and construct the set M defined as {a+b | a^2 = pb^2 and a,b even and positive} ie, M is the set of the sums of the positive even a,b which solve the equation. (Note that there's no loss of generality in restricting it to the positive solutions due to the squaring). Clearly M is a set of positive integers, thus by the well ordering principle it has a lowest member m, which by the definition of M can be written m=i+j for some i,j such that i^2=pj^2 (note that i and j are not necessarily unique, but it turns out any choice thereof which satisfies the equation will suffice for the proof). But since i and j are even, we have i=2x, j=2y, for some positive integers x and y, and thus 4x^2=4py^2, dividing by four we get x^2=py^2, but this is the original equation, and thus since x and y solve it, we must have x+y an element of M... but x+y is clearly smaller than i+j, which contradicts the fact that m=i+j was the smallest member of M. However, try as I might, I have been unable to prove it for the solve it on my own. P.S. this is not homework. ==== > > I'm sure the proof is just under my nose and I just can't seem to > see it. The problem is to prove that for p prime, no nonzero integers > a,b exist such that a^2 = pb^2. This is supposed to be done without > assuming that the square roots of the odd primes are irrational (you > can still assume sqrt(2) is irrational). So for p=2 it's trivial. Suppose to the contrary that such a and b exist. Then we can assume that they have no common prime factor, for if they did, we could just divide both sides with the squares of those common prime factors. So p divides the right-hand side. But then p divides a^2. So p divides a. So a=pc for some integer c. So (pc)^2=pb^2. So pc^2=b^2. So p divides b. So p divides both a and b. But we said a and b had no common prime factor. DANG! ==== Trishia Rose a .8ecrit dans le message de > > I'm sure the proof is just under my nose and I just can't seem to > see it. The problem is to prove that for p prime, no nonzero integers > a,b exist such that a^2 = pb^2. This is supposed to be done without > assuming that the square roots of the odd primes are if a such a and b exist, then each prime number in the decomposition of a^2 is raised to an even power, whereas p is raised to an odd power in the decomposition of pb^2. -- Julien Santini, CMI Technop.99le de Ch.89teau-Gombert, France Home page: http://www.analgebra.com ==== > > I'm sure the proof is just under my nose and I just can't seem to > see it. The problem is to prove that for p prime, no nonzero integers > a,b exist such that a^2 = pb^2. This is supposed to be done without > assuming that the square roots of the odd primes are irrational (you > can still assume sqrt(2) is irrational). So for p=2 it's trivial. > For the other p, we know p is odd, which means that pb^2 is odd > that if any such a and b exist, they have equal parity. > I've managed to prove that no even a and b exist which satisfy > the equation: for, assume they do (for a given odd prime p), and > construct the set M defined as {a+b | a^2 = pb^2 and a,b even and > positive} ie, M is the set of the sums of the positive even a,b which > solve the equation. (Note that there's no loss of generality in > restricting it to the positive solutions due to the squaring). > Clearly M is a set of positive integers, thus by the well ordering > principle it has a lowest member m, which by the definition of M can > be written m=i+j for some i,j such that i^2=pj^2 (note that i and j > are not necessarily unique, but it turns out any choice thereof which > satisfies the equation will suffice for the proof). But since i and j > are even, we have i=2x, j=2y, for some positive integers x and y, and > thus 4x^2=4py^2, dividing by four we get x^2=py^2, but this is the > original equation, and thus since x and y solve it, we must have x+y > an element of M... but x+y is clearly smaller than i+j, which > contradicts the fact that m=i+j was the smallest member of M. > However, try as I might, I have been unable to prove it for the > solve it on my own. P.S. this is not homework. Trishia, The probable intention of the problem is to encourage you to examine closely the proof of the irrationality of sqrt(2), and then to adapt this proof to the situation where 2 is replaced by a general prime. Your problem is a rewritten form of the problem of proving that sqrt(p) is irrational. Paul Epstein ==== > Dears > > Can you give me the formula to get the sum of below series with only > avalable is 'n'. > > ??(n) = (1/1 + 1/2 + 1/3 + 1/4.....+ 1/n) > > This is a harmonic number ... it is approxaimately ln(n) ... but for an exact value you cannot do it with an elementary function. Maple uses the digamma function Psi: > Hn := sum(1/k,k=1..n); Hn := Psi(n + 1) + gamma Asymptotically, we have: > asympt(Hn,n,15); 1/2 1 1/120 1 1/240 ln(n) + gamma + --- - 1/12 ---- + ----- - 1/252 ---- + ----- n 2 4 6 8 n n n n 691 ----- 1 32760 1 1 - 1/132 --- + ----- - 1/12 --- + O(---) 10 12 14 15 n n n n > gamma is Euler's constant -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== Still trying to teach myself some math. :) I have no problem, really, with the procedure(s) for finding determinants. My actual problem is that I'm sort of number-dyslexic and not very good with arithmetic. So, if I work through any procedure that has lots of arithmetic operations, I will probably make one or more errors and have to go root it (them) out. It's a pain, but I'm used to it. My present problem with determinants is that I have no clue how to look at the matrix and figure out, ballpark, what kind of answer to expect, so I know when I might have a correct one. I can check and recheck and use different methods and see if I duplicate the same answer, but I hope there might be a better way. Is there any way to look at a matrix and figure out *approximately* what its determinant is? (I was recently off by over 100,000 so really any clue would help...) -Laurel (Yes, I know I could use a calculator, but that takes all the fun out of it.) ==== |Still trying to teach myself some math. :) | |I have no problem, really, with the procedure(s) for finding |determinants. My actual problem is that I'm sort of number-dyslexic |and not very good with arithmetic. So, if I work through any procedure |that has lots of arithmetic operations, I will probably make one or |more errors and have to go root it (them) out. It's a pain, but I'm |used to it. | |My present problem with determinants is that I have no clue how to look |at the matrix and figure out, ballpark, what kind of answer to expect, |so I know when I might have a correct one. I can check and recheck and |use different methods and see if I duplicate the same answer, but I |hope there might be a better way. | |Is there any way to look at a matrix and figure out *approximately* |what its determinant is? (I was recently off by over 100,000 so |really any clue would help...) the ability to estimate the determinant of a matrix by quickly glancing at the matrix is probably a pretty silly thing to want to have. on the other hand, the ability to estimate the determinant of a matrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) by quickly glancing at a geometric depiction of the transformation represented by the matrix is very important to have if you really want to understand determinants. is this the sort of thing that you actually wanted to learn? it's not very difficult to learn how to do it. -- ==== >|Is there any way to look at a matrix and figure out *approximately* >|what its determinant is? (I was recently off by over 100,000 so >|really any clue would help...) > > > the ability to estimate the determinant of a matrix by quickly > glancing at the matrix is probably a pretty silly thing to want to > have. Whoops. > on the other hand, the ability to estimate the determinant of a > matrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) by > quickly glancing at a geometric depiction of the transformation > represented by the matrix is very important to have if you really want > to understand determinants. is this the sort of thing that you > actually wanted to learn? it's not very difficult to learn how to do > it. What you describe sounds like an excellent thing to learn; I didn't even know the concept. In the (VERY introductory) linear algebra book I am following matrices don't represent anything in particular, so the determinant isn't anything useful, just a bunch of steps to go through to churn out some practically unrelated-seeming number. So it's hard to tell if I'm doing it right without going off and looking at the answer. (Which is, of course, lame, and I don't want to do that until I think I'm right.) Any help would be nice, but I am very uneducated, so there's a good chance I won't understand any of it. :( -Laurel ==== > Still trying to teach myself some math. :) > > I have no problem, really, with the procedure(s) for finding > determinants. My actual problem is that I'm sort of number-dyslexic > and not very good with arithmetic. So, if I work through any procedure > that has lots of arithmetic operations, I will probably make one or > more errors and have to go root it (them) out. We all do. Especially with something as boring as calculating a determinant. > It's a pain, but I'm used to it. > > My present problem with determinants is that I have no clue how to look > at the matrix and figure out, ballpark, what kind of answer to expect, > so I know when I might have a correct one. I can check and recheck and > use different methods and see if I duplicate the same answer, but I > hope there might be a better way. > > Is there any way to look at a matrix and figure out *approximately* > what its determinant is? (I was recently off by over 100,000 so > really any clue would help...) Except in very special cases, eg. triangular matrices for which you know the determinant immediately, there isn't any general way to approximate the determinant by eyeballing it. Just changing one number, slightly, can make major changes in the determinant. If you don't want to or can't use a computer to check, you can calculate the determinant two ways; Gaussian elimination and expansion by cofactors for example. -- Paul Sperry Columbia, SC (USA) ==== |> on the other hand, the ability to estimate the determinant of a |> matrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) by |> quickly glancing at a geometric depiction of the transformation |> represented by the matrix is very important to have if you really want |> to understand determinants. is this the sort of thing that you |> actually wanted to learn? it's not very difficult to learn how to do |> it. | |What you describe sounds like an excellent thing to learn; I didn't |even know the concept. In the (VERY introductory) linear algebra book |I am following matrices don't represent anything in particular, so the |determinant isn't anything useful, just a bunch of steps to go through |to churn out some practically unrelated-seeming number. So it's hard |to tell if I'm doing it right without going off and looking at the |answer. (Which is, of course, lame, and I don't want to do that until |I think I'm right.) | |Any help would be nice, but I am very uneducated, so there's a good |chance I won't understand any of it. :( the most important thing to understand about matrixes is that they represent geometric transformations of a certain kind, namely so-called linear transformations or linear operators. let me try to demonstrate an example of a linear transformation: u t g t r t o n ' p i y w i n o w s e g t n o i m n f a k l o i o h l t s o W n o n d , e f a t s v n u m i e m i o ' ) a m i h . r n c t I e r f i h e t u i h g c e h w i i roughly, distorted by a geometric transformation represented by a certain matrix. (this is an ascii graphic which probably only works if you're viewing this with a fixed font.) now, eyeballing the distorted text, i would estimate that the determinant of the matrix is, oh, say, somewhere around +5. that's an extremely rough estimate. what i'm really estimating is that text printed out in that distorted way will use up about 5 times as much paper as text printed out in the normal way. ok, now let's check how bad my estimate was, by actually calculating the matrix that represents this geometric transformation. that matrix is: 3 -1 1 3 which if i calculate correctly has a determinant of 3*3 - 1*(-1) = 10. ok, so my estimate was way off, off by a factor of 2 in fact. that is, according to this calculation, text printed out in that distorted way uses up 10 times as much paper as normal text. is the connection between the above matrix and the above geometric transformation clear? notice that moving one step to the right in the original text corresponds to 3 right, 1 up in the distorted text, while moving one step up in the original text corresponds to 1 left, 3 up in the distorted text. -- ==== Laurel Amberdine > Still trying to teach myself some math. :) > > I have no problem, really, with the procedure(s) for finding > determinants. My actual problem is that I'm sort of number-dyslexic > and not very good with arithmetic. So, if I work through any procedure > that has lots of arithmetic operations, I will probably make one or > more errors and have to go root it (them) out. It's a pain, but I'm > used to it. > > My present problem with determinants is that I have no clue how to look > at the matrix and figure out, ballpark, what kind of answer to expect, > so I know when I might have a correct one. I can check and recheck and > use different methods and see if I duplicate the same answer, but I > hope there might be a better way. > > Is there any way to look at a matrix and figure out *approximately* > what its determinant is? If each coefficient is, say, between -10 and 10, then there are n! summands all in the range (-10)^n to 10^n, which I guess is better than nothing. Better is a result called the [drum roll] Hadamard Determinant Theorem: If the column vectors in the square matrix have norms k_1, ..., k_n, then |determinant| <= k_1 k_2 ... k_n. Larry ==== > Still trying to teach myself some math. :) > > I have no problem, really, with the procedure(s) for finding > determinants. My actual problem is that I'm sort of number-dyslexic > and not very good with arithmetic. So, if I work through any procedure > that has lots of arithmetic operations, I will probably make one or > more errors and have to go root it (them) out. It's a pain, but I'm > used to it. > > My present problem with determinants is that I have no clue how to look > at the matrix and figure out, ballpark, what kind of answer to expect, > so I know when I might have a correct one. I can check and recheck and > use different methods and see if I duplicate the same answer, but I > hope there might be a better way. > > Is there any way to look at a matrix and figure out *approximately* > what its determinant is? (I was recently off by over 100,000 so > really any clue would help...) > > > -Laurel > > (Yes, I know I could use a calculator, but that takes all the fun out > of it.) If you just want a ballpark figure, the following works. Let det A equal d. choose a random vector v. Work out A.v. The entries in A.v should be about d^(1/n) times the entries of v. (n is the length of the vector, nxn is the size of the matrix). note to flamers :- He said.. ouch! he said ballpark... ouch! BALLPARK ouch!! ouch!! Stop flaming already, ok? ball - ouch! - park, ok?? Anyway, this method should detect errors of order 100000. ==== >Still trying to teach myself some math. :) >I have no problem, really, with the procedure(s) for finding >determinants. My actual problem is that I'm sort of number-dyslexic >and not very good with arithmetic. So, if I work through any procedure >that has lots of arithmetic operations, I will probably make one or >more errors and have to go root it (them) out. It's a pain, but I'm >used to it. >My present problem with determinants is that I have no clue how to look >at the matrix and figure out, ballpark, what kind of answer to expect, >so I know when I might have a correct one. I can check and recheck and >use different methods and see if I duplicate the same answer, but I >hope there might be a better way. >Is there any way to look at a matrix and figure out *approximately* >what its determinant is? (I was recently off by over 100,000 so >really any clue would help...) In general, no. I am supposed to be rather good at arithmetic, and can do more mentally than most, but unless the matrix was of a special form, no luck about the derivative. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University ==== > > >|Is there any way to look at a matrix and figure out *approximately* >|what its determinant is? (I was recently off by over 100,000 so >|really any clue would help...) > > ... > ... In the (VERY introductory) linear algebra book > I am following matrices don't represent anything in particular, so the > determinant isn't anything useful, just a bunch of steps to go through > to churn out some practically unrelated-seeming number. ... How horrifying. Find a new textbook, fast. John Mitchell ==== > Laurel Amberdine >> >> Is there any way to look at a matrix and figure out *approximately* >> what its determinant is? > > If each coefficient is, say, between -10 and 10, then there are n! summands > all in the range > (-10)^n to 10^n, which I guess is better than nothing. Doesn't narrow it down a whole lot. :) > Better is a result > called the [drum roll] Hadamard Determinant Theorem: If the column vectors > in the square matrix have norms k_1, ..., k_n, then |determinant| <= k_1 k_2 > ... k_n. I had to look up norm: the square root of the sum of the squares of the absolute values of the elements of a matrix or the components of a vector. Is that right? -Laurel ==== >> >> Is there any way to look at a matrix and figure out *approximately* >> what its determinant is? (I was recently off by over 100,000 so >> really any clue would help...) > > Except in very special cases, eg. triangular matrices for which you > know the determinant immediately, there isn't any general way to > approximate the determinant by eyeballing it. Just changing one > number, slightly, can make major changes in the determinant. > > If you don't want to or can't use a computer to check, you can > calculate the determinant two ways; Gaussian elimination and expansion > by cofactors for example. At least there *are* multiple ways, so I can just keep trying until I get consistently the same answer. Bleagh. :) -Laurel ==== >> >> >>|Is there any way to look at a matrix and figure out *approximately* >>|what its determinant is? (I was recently off by over 100,000 so >>|really any clue would help...) >> >> ... >> ... In the (VERY introductory) linear algebra book >> I am following matrices don't represent anything in particular, so the >> determinant isn't anything useful, just a bunch of steps to go through >> to churn out some practically unrelated-seeming number. ... > > How horrifying. Find a new textbook, fast. I have two textbooks that are better, but they're too sophisticated for me to start with. Since I'm teaching myself without any formal help, I have terrible trouble figuring out terminology and symbols. I get completely stuck on even the simplest things. Once I already know what a book is (generally) supposed to say, it's a lot easier to read. :) I think I need a math dictionary. -Laurel ==== > >> Is there any way to look at a matrix and figure out *approximately* >> what its determinant is? (I was recently off by over 100,000 so >> really any clue would help...) > > If you just want a ballpark figure, the following works. > > Let det A equal d. > > choose a random vector v. > > Work out A.v. > > The entries in A.v should be about d^(1/n) times the entries of v. > (n is the length of the vector, nxn is the size of the matrix). Okay, I will give that a try. Should be interesting. > note to flamers :- > > He said.. ouch! he said ballpark... ouch! BALLPARK ouch!! ouch!! > Stop flaming already, ok? ball - ouch! - park, ok?? She said, actually, but that's okay. :) > Anyway, this method should detect errors of order 100000. -Laurel ==== > >|> on the other hand, the ability to estimate the determinant of a >|> matrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) by >|> quickly glancing at a geometric depiction of the transformation >|> represented by the matrix is very important to have if you really want >|> to understand determinants. is this the sort of thing that you >|> actually wanted to learn? it's not very difficult to learn how to do >|> it. >| >|What you describe sounds like an excellent thing to learn; I didn't >|even know the concept. In the (VERY introductory) linear algebra book >|I am following matrices don't represent anything in particular, so the >|determinant isn't anything useful, just a bunch of steps to go through >|to churn out some practically unrelated-seeming number. So it's hard >|to tell if I'm doing it right without going off and looking at the >|answer. (Which is, of course, lame, and I don't want to do that until >|I think I'm right.) >| >|Any help would be nice, but I am very uneducated, so there's a good >|chance I won't understand any of it. :( > > > the most important thing to understand about matrixes is that they > represent geometric transformations of a certain kind, namely > so-called linear transformations or linear operators. let me try > to demonstrate an example of a linear transformation: > > > u t g t r t > o n ' p i > y w i n > o w s e g > t n o i m n f > a k l o i o > h l t s o > W n o n d , > e f a t s > v n u m i > e m i o ' ) > a m i h . > r n c t > I e r f i h e > t u i h g c > e h w i i > > > roughly, distorted by a geometric transformation represented by a > certain matrix. (this is an ascii graphic which probably only works > if you're viewing this with a fixed font.) No problem there. > now, eyeballing the > distorted text, i would estimate that the determinant of the matrix > is, oh, say, somewhere around +5. that's an extremely rough estimate. > what i'm really estimating is that text printed out in that distorted > way will use up about 5 times as much paper as text printed out in the > normal way. > > ok, now let's check how bad my estimate was, by actually calculating > the matrix that represents this geometric transformation. that matrix > is: > > 3 -1 > > 1 3 > > which if i calculate correctly has a determinant of 3*3 - 1*(-1) = 10. > ok, so my estimate was way off, off by a factor of 2 in fact. A lot closer than 100,000 though. :) > that > is, according to this calculation, text printed out in that distorted > way uses up 10 times as much paper as normal text. Hm. Okay. > is the connection between the above matrix and the above geometric > transformation clear? notice that moving one step to the right in the > original text corresponds to 3 right, 1 up in the distorted text, > while moving one step up in the original text corresponds to 1 left, > 3 up in the distorted text. Aha. I see... practicing on. Hope I can get to something more practical soon. -Laurel ==== Laurel Amberdine > Laurel Amberdine ... > Better is a result > called the [drum roll] Hadamard Determinant Theorem: If the column vectors > in the square matrix have norms k_1, ..., k_n, then |determinant| <= k_1 k_2 > ... k_n. > > I had to look up norm: the square root of the sum of the squares of the > absolute values of the elements of a matrix or the components of a vector. > Is that right? Yes. Norm just means length in this setting. The determinant can be thought of as the volume of a certain n-dimensional parallelogram, and HDT just says that the volume is <= the product of the lengths of the edges that define it. HDT is easy to prove along those lines, too. Larry ==== > He said.. ouch! he said ballpark... ouch! BALLPARK ouch!! ouch!! > Stop flaming already, ok? ball - ouch! - park, ok?? > > She said, actually, but that's okay. :) > Oops! Sorry abt that... Blame centuries of culturally-ingrained gender bias still as manifest in western civilisation as elsewhere.... :-) ==== Some of the readers of this list might be interested in the following book Pierre Baldi, Paolo Frasconi, and Padhraic Smyth, Modeling the ISBN: 0-470-84906-1. It covers various models and algorithms for the Web including generative models of networks, IR and machine learning algorithms for text analysis, link analysis, focused crawling, methods for modeling user behavior, and for mining Web e-commerce data. 1. Mathematical Background - 2. Basic WWW Technologies - 3. Web Graphs - 4. Text Analysis - 5. Link analysis - 6. Advanced Crawling Techniques - 7. Modeling and Understanding Human Behavior on the Web - 8. Commerce on the Web: Models and Applications - Appendix A Mathematical Complements - Appendix B List of Main Symbols and Abbreviations The webpage http://ibook.ics.uci.edu/ contains more details, a hyperlinked bibliography, and a sample chapter in pdf. Paolo Frasconi http://www.dsi.unifi.it/~paolo/ ==== Jean-Paul Allouche and I are pleased to announce the publication of our book (AS)^2, also known as Automatic Sequences: Theory, Applications, Generalizations currently selling for US $50 on amazon.com. This book is about the class of sequences generated by finite automata, their generalizations, and applications to number theory and theoretical physics. The book has 571+xvi pages, 1600 citations to the literature, 460 exercises, 85 open problems, 1 musical score, and 2 jokes in the index. It will be of interest to number theorists and theoretical computer scientists. The web page for the book, http://www.math.uwaterloo.ca/~shallit/asas.html has a table of contents and other information. Jeffrey Shallit, Computer Science, University of Waterloo, Waterloo, Ontario N2L 3G1 Canada shallit@graceland.uwaterloo.ca URL = http://www.math.uwaterloo.ca/~shallit/ ==== > The book has 571+xvi pages, fifteen imaginary pages? Cool! -- J.97n Fairbairn Jon.Fairbairn@cl.cam.ac.uk ==== >> The book has 571+xvi pages, > fifteen imaginary pages? Cool! But only two jokes, a definite downside. xanthian, would have loved to see what a joke looks like in the Complex plane. [Probably a little out of phase.] -- ==== In sci.math, Kent Paul Dolan > > > The book has 571+xvi pages, > >> fifteen imaginary pages? Cool! > > But only two jokes, a definite downside. > > xanthian, would have loved to see what a joke looks like in the Complex > plane. > > [Probably a little out of phase.] > Either that, or highly twisted. :-) -- #191, ewill3@earthlink.net -- we could spin this a number of ways It's still legal to go .sigless. ==== [trim self-serving bollocks] Wrong newsgroup sonny. I trimmed out the one ng this 'should' have gone to. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) ==== Could anyone point me to a reference for the noncentral t-distribution, that is its probability density as well as its cumulative distribution function? I did find the probability density p(t) -- no original author given -- as follows: ... (p + t^2) ^ (-(p+1)/2) series ... (2 t^2 / (p + t^2)) ^ (i/2), where p is the degree of freedom and i the index of the series. However, this function is even, although the probability density is not. A missing sign? Also, I did find a cumulative distribution function P(-t0 <= t <= t0) -- likewise no original author given --, which basically is a series in the incomplete beta function. That's fine, but is there a way to compute the left tail P(-oo < t < t0) and right tail P (t0 < t < +oo) other than by numerical integration? After all the distribution function is not even around its maximum. Bjoern ==== > > The question is: if we let U, V be skewsymmetric and let > A = exp(U), B = exp(V) and C = exp(U + V) must (Cx, ABx) = (Cx, BAx) > and must Cx, ABx and BAx be linearly dependent? I don't think so: > We don't in general have C = AB. As long as U and V are small > C will be given by the Campbell-Baker-Hausdorff formula which > starts C = (AB + BA)/2 + lots of increasingly nasty terms > involving iterated commutators. If we ignore the later terms. > we see that for small U and V, C is approximately (AB + BA)/2, so > approximately dependent on AB but BA, but exactly? .... a bit unlikely > I think. Well, let me give you a hand-waving argument (which is what motivated my question): In the limit t->0, Z1 = (A^tB^t)^(1/t) should be equal to Z2 = (B^tA^t)^(1/t). That means that as t gets small, I expect that x transformed by Z1 must approach x transformed by Z2. Where might I reasonably expect to find Zx? Well, half-way between ABx and BAx. I've tested a few numbers, and what's interesting is that Zx _doesn't_ seem to converge to a linear combination of ABx and BAx, in general, although (Zx, ABx) does seem to equal (Zx, BAx). David Turner ==== If, and only if (for positive integers n), n = a composite >= 6, then: n divides (n-1)!. So, what can be said about the reals or complex numbers x, where Gamma(x)/x is an integer? I guess we could define the set of x's which satisfy this as composites >= 6, a set which contains noninteger values. And related to Wilsons theorem, we could define the set of complex/real x's where: (Gamma(x)+1)/x = integer as a set of Wilson-derived generalized primes. This must be a well-known topic. Anything interesting about this kind of continuation, or is this all useless? (It is useless to say useless, to paraphrase a cliche...) Leroy Quet ==== ... > So, what can be said about the reals or complex numbers x, where > Gamma(x)/x is an integer? > > I guess we could define the set of x's which satisfy this as > composites >= 6, a set which contains noninteger values. > > And related to Wilsons theorem, we could define the set of > complex/real x's where: > (Gamma(x)+1)/x = integer > as a set of Wilson-derived generalized primes. ... Since for positive reals 2,3,4... we have Gamma(x)=(x-1)! and Gamma is strictly increasing for reals >= 2 , apparently there are no other x>=2 (other than integer values) where (Gamma(x)+1)/x is an integer. For complex values of x, in my ignorance it isn't obvious to me whether (Gamma(x)+1)/x is ever an integer. Do you have some simple examples? -jiw ==== >Since for positive reals 2,3,4... we have Gamma(x)=(x-1)! >and Gamma is strictly increasing for reals >= 2 , apparently >there are no other x>=2 (other than integer values) where >(Gamma(x)+1)/x is an integer. ??? What do you mean? (Gamma(x)+1)/x = 1, 5, 103, 329891, ..., for x = 3, 5, 7, 11, .... By the Intermediate Value Theorem, there must be non-integer reals where (Gamma(x)+1)/x takes the other positive integer values 2, 3, 4, 6, 7, .... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >... >> So, what can be said about the reals or complex numbers x, where >> Gamma(x)/x is an integer? >> I guess we could define the set of x's which satisfy this as >> composites >= 6, a set which contains noninteger values. >> And related to Wilsons theorem, we could define the set of >> complex/real x's where: >> (Gamma(x)+1)/x = integer >> as a set of Wilson-derived generalized primes. >... >Since for positive reals 2,3,4... we have Gamma(x)=(x-1)! >and Gamma is strictly increasing for reals >= 2 , apparently >there are no other x>=2 (other than integer values) where >(Gamma(x)+1)/x is an integer. On the contrary, there are lots of them. (Gamma(x)+1)/x is 1.75 for x = 4, and is 5 for x=5. There are solutions for 2, 3, and 4 which are between 4 and 5. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University ==== >Since for positive reals 2,3,4... we have Gamma(x)=(x-1)! >and Gamma is strictly increasing for reals >= 2 , apparently >there are no other x>=2 (other than integer values) where >(Gamma(x)+1)/x is an integer. > > ??? What do you mean? (Gamma(x)+1)/x = 1, 5, 103, 329891, ..., > for x = 3, 5, 7, 11, .... By the Intermediate Value Theorem, there > must be non-integer reals where (Gamma(x)+1)/x takes the other positive > integer values 2, 3, 4, 6, 7, .... Yes, sorry, missed the obvious. -jiw ==== >... >> So, what can be said about the reals or complex numbers x, where >> Gamma(x)/x is an integer? >> I guess we could define the set of x's which satisfy this as >> composites >= 6, a set which contains noninteger values. >> And related to Wilsons theorem, we could define the set of >> complex/real x's where: >> (Gamma(x)+1)/x = integer >> as a set of Wilson-derived generalized primes. >... >Since for positive reals 2,3,4... we have Gamma(x)=(x-1)! >and Gamma is strictly increasing for reals >= 2 , apparently >there are no other x>=2 (other than integer values) where >(Gamma(x)+1)/x is an integer. > On the contrary, there are lots of them. (Gamma(x)+1)/x > is 1.75 for x = 4, and is 5 for x=5. There are solutions > for 2, 3, and 4 which are between 4 and 5. For reference (and because I have nothing better to do with my afternoon): In[27]:= WilsonRoot[2.0]; WilsonRoot[3.0]; WilsonRoot[4.0]; 4.15444 4.56215 4.81616 I've checked the first one in Mathematica, and I presume (read 'hope') that the others are right as well. -- Dave Taylor Oh yeah? Well I'll build my own theme park! With Blackjack, and hookers ... in fact, forget the theme park! [Futurama] ==== > If, and only if (for positive integers n), n = a composite >= 6, then: > > n divides (n-1)!. > > So, what can be said about the reals or complex numbers x, where > > Gamma(x)/x is an integer? > > I guess we could define the set of x's which satisfy this as > composites >= 6, a set which contains noninteger values. > > > And related to Wilsons theorem, we could define the set of > complex/real x's where: > > (Gamma(x)+1)/x = integer > > as a set of Wilson-derived generalized primes. > > This must be a well-known topic. Anything interesting about this kind > of continuation, or is this all useless? > (It is useless to say useless, to paraphrase a cliche...) > A few things: 1) Even considering only x = positive integers, x = 1 satisfies both equations above. So, in a way, 1 is both a composite >=6 and a Wilson-derived generalized prime. (snicker.) 2) What if we, for x = composite, allow the integers (that the equations above are equal to) to be Gaussian? 2.5) Is (Gamma(x)+1)/x, if we let x = a Gaussian (integer) prime, an (real or Gaussian)integer? 3) If we take the maximum real root, x(n), of (Gamma(x)+1)/x = n, then, what can be said about the Wilson-zeta function: WZ(y) = product{n=1 to oo} 1/(1 -1/x(n)^y) ? Does the WZ() product converge for any y? Leroy Quet ==== >For complex values of x, in my ignorance it isn't obvious >to me whether (Gamma(x)+1)/x is ever an integer. Do you >have some simple examples? For example, (Gamma(x)+1)/x = 3 for x = 5.7584660619435256965 + 3.9675461457510952995 i approximately. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >>> (Gamma(x)+1)/x = integer > For reference (and because I have nothing better to do with > my afternoon): > > In[27]:= WilsonRoot[2.0]; WilsonRoot[3.0]; WilsonRoot[4.0]; > 4.15444 > 4.56215 > 4.81616 > > I've checked the first one in Mathematica, and I presume > (read 'hope') that the others are right as well. All pucker: (21:50) gp > p100 realprecision = 105 significant digits (100 digits displayed) (21:51) gp > solve(x=4.1,4.2,(gamma(x)+1)/x-2) %8 = 4.154439060271062694827261583664629151822185286325427363404753300368340020397 814797012023001544277754 (21:51) gp > solve(x=4.6,4.2,(gamma(x)+1)/x-3) %9 = 4.562151433950973261499668111682367676594713372717728076006767266218451425330 181354003523067887979890 (21:51) gp > solve(x=4.6,4.9,(gamma(x)+1)/x-4) %10 = 4.816164962269824928459405177621703579279411919117658752833716798354235922037 604118825483457603002219 (21:52) gp > solve(x=5,5.2,(gamma(x)+1)/x-6) %13 = 5.143587135664545182467227708447357783039049810591881586494699613779238822779 092252706929290717662694 Phil ==== >>For complex values of x, in my ignorance it isn't obvious >>to me whether (Gamma(x)+1)/x is ever an integer. Do you >>have some simple examples? > >For example, (Gamma(x)+1)/x = 3 for > > x = 5.7584660619435256965 + 3.9675461457510952995 i But... is anything known about Leroy's question?I'm asking because some years ago I formulated the very same question and I'm very curious about it. To be fair I'm only thinking to its second part i.e. that involving generalized primes as those numbers satisfying the obvious continuous version of Wilson's (necessary and sufficient) condition for primality. Well, maybe something interesting can be said about a suitable function having those primes as zeroes or poles... Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc ==== > >>For complex values of x, in my ignorance it isn't obvious >>to me whether (Gamma(x)+1)/x is ever an integer. Do you >>have some simple examples? > >For example, (Gamma(x)+1)/x = 3 for > > x = 5.7584660619435256965 + 3.9675461457510952995 i > > But... is anything known about Leroy's question?I'm asking because > some years ago I formulated the very same question and I'm very > curious about it. > > To be fair I'm only thinking to its second part i.e. that involving > generalized primes as those numbers satisfying the obvious > continuous version of Wilson's (necessary and sufficient) condition > for primality. > > Well, maybe something interesting can be said about a suitable > function having those primes as zeroes or poles... You mean like sin(pi (Gamma(x)+1)/x)? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >> Well, maybe something interesting can be said about a suitable >> function having those primes as zeroes or poles... > >You mean like sin(pi (Gamma(x)+1)/x)? Well, this is exactly the function I had considered in my youth, along with the *real* function sin^2(pi x)+sin^2(pi(Gamma(x)+1)/x). But now I'm more sort-of-thinking of something like prod_p 1/(1-p^{-x}) with p ranging over the zeroes of the function you supplied... provided that the product does make sense. Michele -- >It's because the universe was programmed in C++. No, no, it was programmed in Forth. See Genesis 1:12: And the earth brought Forth ... - Robert Israel on sci.math, thread Why numbers? ==== I have to solve the following problem: Assume that A and B are square matrixes with ||A||=m and ||B||=d and that (I-A) and (I+B) are non-singular, show that: ||((I-A)^-1)B((I+B)^-1)<=d/(1-d-m(1+d)) where m<(1-d)/(1+d) X=((I-A)^-1)B((I+B)^-1) Premultiply then both sides first by (I-A) and then postmultiply both sides by (I+B). Whatever I try in my result I have a wrong sign in the denominator and I have a wrong relation symbol. I would be very grateful for some hints how can I solve this task. Michael p.s. I put the task in the WWW in a more convenient manner: http://www.rdg.ac.uk/~sip02mm2/matrix.html ==== > > > > I have to solve the following problem: > > > > Assume that A and B are square matrixes with > > ||A||=m > > and > > ||B||=d > > and that > > (I-A) > > and > > (I+B) > > are non-singular, show that: > > > > ||((I-A)^-1)B((I+B)^-1)<=d/(1-d-m(1+d)) > > > > where > > > > m<(1-d)/(1+d) > > > > > > > X=((I-A)^-1)B((I+B)^-1) > > > > Premultiply then both sides first by (I-A) and then postmultiply both sides > by (I+B). > > > > Whatever I try in my result I have a wrong sign in the denominator and I > have a wrong relation symbol. > > > > I would be very grateful for some hints how can I solve this task. > > Michael > > > > p.s. I put the task in the WWW in a more convenient manner: > http://www.rdg.ac.uk/~sip02mm2/matrix.html > ==== Would some kind soul possibly post a complete solution to this problem, as none of us on the course can fathom it and our exam is tomorrow... *gulp* :) Darren. ==== > Would some kind soul possibly post a complete solution to this > problem, as none of us on the course can fathom it and our exam is > tomorrow... *gulp* :) > > Darren. My reply was essentially complete. Just follow through with a detail or two. Here it is again: ------------------------ This question occurred to be the other day; it's been a while since I took calculus, and I can't come up with an answer. Does there exist a function from R to R that is nowhere continuous, but that is defined everywhere and limited everywhere (i.e. that has a finite limit at each real number)? Foghorn Leghorn moc.enecswen @ nrohgof ==== A function f(x) is said to be continuous at x=a if 1a)limx-->a- exist, 1b)limx-->a+ exists, 2) these two limits are = to one another ,say k. and 3)f(a)=k. Now you say that your function,f(x), is limited everywhere.Let's consider x at a. So limx-->a+=lim x-->a-(=k). If f(a)=k then of course we can't say that f(x) is continuous nowhere. What type of function violates just condition 3 of the definition of a continuous function? Functions that have removable discontinuity come to mind.Now the real question is whether or not a function can have an 00 number of removable discontinuities so the function is nowhere continuous. I think that you can come up with one. > This question occurred to be the other day; it's been a while since I > took calculus, and I can't come up with an answer. > > Does there exist a function from R to R that is nowhere continuous, > but that is defined everywhere and limited everywhere (i.e. that has a > finite limit at each real number)? > > Foghorn Leghorn > moc.enecswen @ nrohgof ==== > >This question occurred to be the other day; it's been a while since I >took calculus, and I can't come up with an answer. > >Does there exist a function from R to R that is nowhere continuous, >but that is defined everywhere and limited everywhere (i.e. that has a >finite limit at each real number)? > >Foghorn Leghorn >moc.enecswen @ nrohgof > Hmm... I don't know, but what about The Dirichlet Function: f(x) = 1 [if x is rational] = 0 [if x is irrational] Plot some points... It looks like two straight, horizontal lines... Obviously, it's definied everywhere, but I don't know if it's continuous and limited (sic). Maybe someone else does. cheerio, ==== > This question occurred to be the other day; it's been a while since I > took calculus, and I can't come up with an answer. > > Does there exist a function from R to R that is nowhere continuous, > but that is defined everywhere and limited everywhere (i.e. that has a > finite limit at each real number)? Consider the function f:R->R where : (a) f(x) = 0 if x is irrational, and (b) if x is rational and we define: denominator(x):= the least integer n>=1 such that n*x is an integer, and then let f(x):= 1/denominator(x) . I think Prof. Herman Rubin recently mentioned this function in another thread. David Bernier ==== >> Does there exist a function from R to R that is nowhere continuous, >> but that is defined everywhere and limited everywhere (i.e. that has a >> finite limit at each real number)? > > > Consider the function > > f:R->R > > where : > > (a) f(x) = 0 if x is irrational, and > > (b) if x is rational and we define: > denominator(x):= the least integer n>=1 such that n*x is an integer, > and then let f(x):= 1/denominator(x) . The f above _is_ continuous at all irrational points in R, right? David Bernier ==== > > > >> Does there exist a function from R to R that is nowhere continuous, >> but that is defined everywhere and limited everywhere (i.e. that has a >> finite limit at each real number)? > > > Consider the function > > f:R->R > > where : > > (a) f(x) = 0 if x is irrational, and > > (b) if x is rational and we define: > denominator(x):= the least integer n>=1 such that n*x is an integer, > and then let f(x):= 1/denominator(x) . > > The f above _is_ continuous at all irrational points in R, right? > > David Bernier Right. As for the original question, I have a feeling that no such function exists. I wonder if you could prove this using a Baire category argument, maybe something like the proof that there is no function continuous only on the rationals. ==== >> >>This question occurred to be the other day; it's been a while since I >>took calculus, and I can't come up with an answer. >> >>Does there exist a function from R to R that is nowhere continuous, >>but that is defined everywhere and limited everywhere (i.e. that has a >>finite limit at each real number)? >> >>Foghorn Leghorn >>moc.enecswen @ nrohgof >> > >Hmm... I don't know, but what about The Dirichlet Function: > > >f(x) = 1 [if x is rational] > = 0 [if x is irrational] > >Plot some points... It looks like two straight, horizontal lines... >Obviously, it's definied everywhere, but I don't know if it's >continuous and limited (sic). Maybe someone else does. That function is obviously nowhere continuous. But it also obviously has a limit at no point, so it doesn't help. >cheerio, ************************ David C. Ullrich ==== >This question occurred to be the other day; it's been a while since I >took calculus, and I can't come up with an answer. > >Does there exist a function from R to R that is nowhere continuous, >but that is defined everywhere and limited everywhere (i.e. that has a >finite limit at each real number)? No. If f has a limit at every point then there is a dense set of points at which it is continuous: Say the variation of f on S is V(f, S) = sup {|f(x) - f(y)| : x, y in S}. The fact that f has a limit at 0 shows that there is a closed interval I_1 such that V(f, I_1) < 1. Now the fact that f has a limit at the midpoint of I_1 shows that there is a closed interval I_2, contained in the _interior_ of I_1, such that V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of the I_n it foilows that f is continuous at x. (The fact that I_{n+1} is in the interior of I_n shows that x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all y in I_n.) >Foghorn Leghorn >moc.enecswen @ nrohgof ************************ David C. Ullrich ==== Foghorn Leghorn http://mathforum.org/discuss/sci.math/m/523425/523425 > This question occurred to be the other day; it's been a while > since I took calculus, and I can't come up with an answer. > > Does there exist a function from R to R that is nowhere > continuous, but that is defined everywhere and limited > everywhere (i.e. that has a finite limit at each real number)? Given a function f: R --> R and a real number x in R, let L(f,x) be the limit of f(x') as x' --> x, when this limit exists. Then the set (I'm using /= for not equal) P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} is countable. Thus, what you're looking for doesn't exist by a long shot (a function f such that L(f,x) exists for each x in R and P(f) = R). On the other hand, I'm pretty sure that for each countable set Z there exists a function f such that L(f,x) exists for each x in R and P(f) = Z (i.e. no restriction besides countable can be proved, even when L(f,x) exists for each x in R). One way to prove this is to first prove for each n = 1, 2, 3, ... that P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} is an isolated set, and hence each P(f,n) must be countable. This immediately implies that P(f) is countable, since P(f) = UNION(n=1 to oo) P(f,n). A much stronger version of this result holds, by the way. Given a function f: R --> R and a real number x in R, let C(f,x) be the set of all extended real numbers y (i.e. y can be -oo or +oo) such that there exists a sequence {x_k} with x_k --> x and f(x_k) --> y. In other words, C(f,x) is the set of all numbers (including -oo and +oo) that can be obtained as the limit of some sequence converging to x. Then for each f and x, C(f,x) is a nonempty closed set in the extended real line, the minimum number in C(f,x) is lim-inf(x'--> x) of f(x'), and the maximum number in C(f,x) is lim-sup(x'--> x) of f(x'). Note that L(f,x) exists means that C(f,x) = {y} for some y in R. Then the set Q(f) = {x in R: f(x) does not belong to C(f,x)} is countable. This can be proved in the same way as the result above. First, a bit of notation. If y belongs to R and E is a subset of R, let DIST(y,E) be the extended real number inf{|y-e|: e is in E}. Now define Q(f,n) for each n = 1, 2, 3, ... by Q(f,n) = {x in R: DIST[f(x), C(f,x)] > 1/n}. Then each Q(f,n) winds up being an isolated set, and hence Q(f) is countable since Q(f) = UNION(n=1 to oo) Q(f,n). Dave L. Renfro ==== > >This question occurred to be the other day; it's been a while since I >took calculus, and I can't come up with an answer. > >Does there exist a function from R to R that is nowhere continuous, >but that is defined everywhere and limited everywhere (i.e. that has a >finite limit at each real number)? > > No. If f has a limit at every point then there is a dense set of > points at which it is continuous: > > Say the variation of f on S is > > V(f, S) = sup {|f(x) - f(y)| : x, y in S}. > > The fact that f has a limit at 0 shows that there is a closed > interval I_1 such that V(f, I_1) < 1. Now the fact that f has > a limit at the midpoint of I_1 shows that there is a closed > interval I_2, contained in the _interior_ of I_1, such that > V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of > the I_n it foilows that f is continuous at x. > > (The fact that I_{n+1} is in the interior of I_n shows that > x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all > y in I_n.) Very nice argument. Compact and to the point :-) ==== > Foghorn Leghorn > http://mathforum.org/discuss/sci.math/m/523425/523425 > > > This question occurred to be the other day; it's been a while > since I took calculus, and I can't come up with an answer. > > Does there exist a function from R to R that is nowhere > continuous, but that is defined everywhere and limited > everywhere (i.e. that has a finite limit at each real number)? > > Given a function f: R --> R and a real number x in R, let > L(f,x) be the limit of f(x') as x' --> x, when this limit > exists. Then the set (I'm using /= for not equal) > > P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} > > is countable. Thus, what you're looking for doesn't exist by a > long shot (a function f such that L(f,x) exists for each x in R > and P(f) = R). On the other hand, I'm pretty sure that for each > countable set Z there exists a function f such that L(f,x) exists > for each x in R and P(f) = Z (i.e. no restriction besides > countable can be proved, even when L(f,x) exists for each x in R). > > One way to prove this is to first prove for each n = 1, 2, 3, ... > that > > P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} > > is an isolated set Dave, can you fill this in? I don't see why it's an isolated set. Sorry to be so dense :-) ==== An attempt at clarifying for calculus students. > >>This question occurred to be the other day; it's been a while since I >>took calculus, and I can't come up with an answer. >> >>Does there exist a function from R to R that is nowhere continuous, >>but that is defined everywhere and limited everywhere (i.e. that has a >>finite limit at each real number)? >> >> >No. If f has a limit at every point then there is a dense set of >points at which it is continuous: > >Say the variation of f on S is > > V(f, S) = sup {|f(x) - f(y)| : x, y in S}. > >The fact that f has a limit at 0 shows that there is a closed >interval I_1 such that V(f, I_1) < 1. Now the fact that f has >a limit at the midpoint of I_1 shows that there is a closed >interval I_2, contained in the _interior_ of I_1, such that >V(f, I_2) < 1/2. Repeat. > Which you can do, because there's a point in I_2 (possibly different from the point 0 above) where f has a limit. > Now if x is in the intersection of the I_n it foilows that f is continuous at x. > And there is at least one x in the intersection because the sets are closed (and bounded). I'm at a loss as to how to explain this in a short note to someone whose background doesn't extend beyond calculus. So I'll beg off for today (but I hope you'll be interested enough to make me [or someone else] do it Monday. >(The fact that I_{n+1} is in the interior of I_n shows that x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all y in I_n.) > > So now you've got a point x such that f is continuous at x. Now, pick any point z, and any tolerance level epsilon. By simply picking the first I_1 to be within the interval (z-epsilon, z+epsilon), we can find a point of continuity of f that is close enough to z. Since we can do this for any z, the set of points of continuity of f are arbitrarily close to any point of R. That's the definition of a dense set. Jon Miller ==== > >> >>This question occurred to be the other day; it's been a while since I >>took calculus, and I can't come up with an answer. >> >>Does there exist a function from R to R that is nowhere continuous, >>but that is defined everywhere and limited everywhere (i.e. that has a >>finite limit at each real number)? >> >> No. If f has a limit at every point then there is a dense set of >> points at which it is continuous: >> >> Say the variation of f on S is >> >> V(f, S) = sup {|f(x) - f(y)| : x, y in S}. >> >> The fact that f has a limit at 0 shows that there is a closed >> interval I_1 such that V(f, I_1) < 1. Now the fact that f has >> a limit at the midpoint of I_1 shows that there is a closed >> interval I_2, contained in the _interior_ of I_1, such that >> V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of >> the I_n it foilows that f is continuous at x. >> >> (The fact that I_{n+1} is in the interior of I_n shows that >> x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all >> y in I_n.) > >Very nice argument. Compact and to the point :-) heh-heh. ************************ David C. Ullrich ==== >Foghorn Leghorn >http://mathforum.org/discuss/sci.math/m/523425/523425 > > >> This question occurred to be the other day; it's been a while >> since I took calculus, and I can't come up with an answer. >> >> Does there exist a function from R to R that is nowhere >> continuous, but that is defined everywhere and limited >> everywhere (i.e. that has a finite limit at each real number)? > >Given a function f: R --> R and a real number x in R, let >L(f,x) be the limit of f(x') as x' --> x, when this limit >exists. Then the set (I'm using /= for not equal) > > P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} > >is countable. Well duh. I considered conjecturing that the word dense in the result I gave could be strengthened, but I didn't want to figure out by how much. Has countable complement is much stronger than what I would have guessed. >Thus, what you're looking for doesn't exist by a >long shot (a function f such that L(f,x) exists for each x in R >and P(f) = R). On the other hand, I'm pretty sure that for each >countable set Z there exists a function f such that L(f,x) exists >for each x in R and P(f) = Z (i.e. no restriction besides >countable can be proved, even when L(f,x) exists for each x in R). This is clear - if Z = {x_1, ...} let f(x_n) = 1/n and f(x) = 0 for other x. >One way to prove this is to first prove for each n = 1, 2, 3, ... >that > > P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} > >is an isolated set, and hence each P(f,n) must be countable. This >immediately implies that P(f) is countable, since > > P(f) = UNION(n=1 to oo) P(f,n). > >A much stronger version of this result holds, by the way. Given a >function f: R --> R and a real number x in R, let C(f,x) be the set >of all extended real numbers y (i.e. y can be -oo or +oo) such >that there exists a sequence {x_k} with x_k --> x and f(x_k) --> y. >In other words, C(f,x) is the set of all numbers (including -oo and >+oo) that can be obtained as the limit of some sequence converging >to x. Then for each f and x, C(f,x) is a nonempty closed set in >the extended real line, the minimum number in C(f,x) is >lim-inf(x'--> x) of f(x'), and the maximum number in C(f,x) is >lim-sup(x'--> x) of f(x'). Note that L(f,x) exists means that >C(f,x) = {y} for some y in R. > >Then the set > > Q(f) = {x in R: f(x) does not belong to C(f,x)} > >is countable. This can be proved in the same way as the result above. >First, a bit of notation. If y belongs to R and E is a subset of R, >let DIST(y,E) be the extended real number inf{|y-e|: e is in E}. > >Now define Q(f,n) for each n = 1, 2, 3, ... by > > Q(f,n) = {x in R: DIST[f(x), C(f,x)] > 1/n}. > >Then each Q(f,n) winds up being an isolated set, and hence Q(f) >is countable since > > Q(f) = UNION(n=1 to oo) Q(f,n). > > >Dave L. Renfro ************************ David C. Ullrich ==== > >> Foghorn Leghorn >> http://mathforum.org/discuss/sci.math/m/523425/523425 >> >> >> This question occurred to be the other day; it's been a while >> since I took calculus, and I can't come up with an answer. >> >> Does there exist a function from R to R that is nowhere >> continuous, but that is defined everywhere and limited >> everywhere (i.e. that has a finite limit at each real number)? >> >> Given a function f: R --> R and a real number x in R, let >> L(f,x) be the limit of f(x') as x' --> x, when this limit >> exists. Then the set (I'm using /= for not equal) >> >> P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} >> >> is countable. Thus, what you're looking for doesn't exist by a >> long shot (a function f such that L(f,x) exists for each x in R >> and P(f) = R). On the other hand, I'm pretty sure that for each >> countable set Z there exists a function f such that L(f,x) exists >> for each x in R and P(f) = Z (i.e. no restriction besides >> countable can be proved, even when L(f,x) exists for each x in R). >> >> One way to prove this is to first prove for each n = 1, 2, 3, ... >> that >> >> P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} >> >> is an isolated set > >Dave, can you fill this in? I don't see why it's an isolated set. >Sorry to be so dense :-) Let epsilon = 1/(2n) in the definition of L(f,x) exists... ************************ David C. Ullrich ==== [snip] >> P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} >> >> is an isolated set > >Dave, can you fill this in? I don't see why it's an isolated set. >Sorry to be so dense :-) > > Let epsilon = 1/(2n) in the definition of L(f,x) exists... > > ************************ > > David C. Ullrich ==== Simple question. Let A' denote the complement of a set A, then Oxtoby in Measure and Category says A is nowhere dense iff A' contains a dense open set,.. I understand the condition that A' contains a dense set, since A' is dense and A' contains itself. I don't understand the condition of the dense subset having to be _open_. Why is this condition of having to be open necessary? Matthew Breneman ==== >Simple question. > >Let A' denote the complement of a set A, then >Oxtoby in Measure and Category says A is nowhere dense iff A' contains a >dense open set,.. > >I understand the condition that A' contains a dense set, since A' is dense >and A' contains itself. I don't understand the condition of the dense subset >having to be _open_. Why is this condition of having to be open necessary? Clearly A' is dense is not nearly strong enough: the set of irrationals is really really dense in R, even though its complement is dense :). An equivalent definition that may be helpful is this: A is nowhere dense iff it is not dense on any open set, that is cl(A) (the closure of A) contains no open neighbourhoods. The motivation for this definition should be pretty clear. It's quite easy to see the equivalence: if cl(A) has empty interior, then its complement is dense. But the complement of cl(A) is int(A'), so A' contains a dense open set, namely its own interior. -- Erick ==== >Let A' denote the complement of a set A, then >Oxtoby in Measure and Category says A is nowhere dense iff A' contains a >dense open set,.. >I understand the condition that A' contains a dense set, since A' is dense >and A' contains itself. I don't understand the condition of the dense subset >having to be _open_. Why is this condition of having to be open necessary? Saying that A is nowhere dense means that there is no nonempty open set in which A is dense. So every nonempty open set U must have a nonempty open subset V(U) disjoint from A. The union of these V(U) for all nonempty open U is a dense open set contained in A'. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > Let A' denote the complement of a set A, then > Oxtoby in Measure and Category says A is nowhere dense iff A' contains a > dense open set,.. > Theorem: For a topological space S SA nowhere dense iff some open dense U with U subset A Proof: If SA nowhere dense: S = S - int cl SA = cl int A; int A subset A; let U = int A Conversely if some open dense U with U subset A: cl U = S int cl SA subset int cl SU = int SU = int(cl U - U) = int cl U - cl U = nulset -- Another theorem in same vein A nowhere dense iff cl A = bd cl A -- Previous thoughts upon nowhere dense Def: A somewhere dense when A not nowhere dense Theorem: A somewhere dense iff some open nonnul U with cl U = cl U/A iff some open nonnul U with U subset cl A (The second equivalence gleaned from this thread.) ---- ==== Just wondering if anybody knows an approach to the following: An n-prime set is a set of n integers such that each element can be combined in some way with 1,2,3...(n-1) other elements such that their sum is a prime. Can you form an n-prime set for any n ? It seems like the answer should be no, because the primes are such an intractable set. But for any n I've tried, I've been able to produce a set. Any thoughts would be appreciated. David ==== >Just wondering if anybody knows an approach to the following: >An n-prime set is a set of n integers such that each element can be >combined in some way with 1,2,3...(n-1) >other elements such that their sum is a prime. Can you form an n-prime >set for any n ? I'm not sure what you mean by 1,2,3...(n-1) other elements. Please state it more clearly. And perhaps it would help if you gave an example for, say, n=5. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== An example of such a set for n=5 would be 22111, for we have 1+2=3 (combines 1 (or 2) with some other element to make a prime) 1+2+2=5 (combines 1 (or 2) with two other elements to make a prime) 1+1+1+2=5 1+1+1+2+2=7 Of course, this is a very simple example because we don't need to display any more than one sum for each length. ==== Need help to complete the following number sequence: 1 4 4 20 23 15 16 15 9 14 20 23 5 14 20 25 6 9 22 5 20 15 20 8 5 14 15 18 20 8 9 14 7 9 14 1 16 16 12 5 29 1 14 4 25 15 21 1 18 5 7 15 15 4 20 15 7 15 Any help? Thanx ==== A=1 ==== Consider the following ODE system: y_1^(n_1) = f_1 [x, y_1, y_2, ..., y_k, ..., y_1^(n_1 - 1), ..., y_k^(n_1 - 1)] y_2^(n_2) = f_2 [x, y_1, y_2, ..., y_k, ..., y_1^(n_2 - 1), ..., y_k^(n_2 - 1)] ... ... ... ... ... ... ... ... ... ... ... ... y_k^(n_k) = f_2 [x, y_1, y_2, ..., y_k, ..., y_1^(n_k - 1), ..., y_k^(n_k - 1)] (y^(w) means the w-th derivative of y). Could you shom me (in detail) how to reduce it to the following of first order Y_1' = F_1 [x, Y_1, Y_2, ..., Y_n] Y_2' = F_2 [x, Y_1, Y_2, ..., Y_n] ... ... ... ... ... ... ... ... ... ... ... ... Y_n' = F_n [x, Y_1, Y_2, ..., Y_n] (where n= max{n_1, n_2, ..., n_k) ????? ==== The problem is, if we have the number 1^.5, is it 1 or -1. Going even farther, 1^.25 could be 1, -1, i, or -i. How do you fix this? Simple, convert all numbers to polar coordinates. Now, if we say (1<0)^.5, the answer is 1^0 (0*2 = 0). If we try (1<360), the answer 1<180 (180*2 = 360) which is -1. A number like (1<453)^.5 would have an answer of 1<226.5 or -.68835 + -.7253 i as opposed to getting the other possible without polar coordinates as 1<406.5 or .68835 + .7253 i. With this logic, (1<0)^.25 becomes 1 (1<0)^.5 is 1, (1<360)^.25 is i, (1<360)^.5 is -1, (1<720)^.25 is -1, (1<720)^.5 is 1, (1<1080)^.25 is -i, and (1<1080)^.5 is 1 Their are two small problems with this. The first problem is when you add two polar vectors together head to tail that are on or over 360 degree apart. Say we add the numbers 1<15 and 1<375 together. Under currently accepted mathematics, we would mod 375 by 360 then add them together to get 2<15 degrees. This works perfectly for addition and subtraction, but what if we had the problem (1<15 + 1<375)^.25. I'll tell you right now, the solution is not (2^.5)<7.5, although that is what nearly every single calculator in the world would give you. The second problem is the conversion between normal number system and polar coordinates. The number 5<390 is 5*cos(390) + 5*sin(390) i. This converts it to nice angles, which are 0, 90, 180, and 270 so it can be easily added to something like 3<0 without getting a geometry book out. Unfortunately, when the number is converted back to the normal number system, 5<30 will result, losing all additional information about the number of revolutions that have occurred. ==== > The problem is, if we have the number 1^.5, is it 1 or -1. Going even > farther, 1^.25 could be 1, -1, i, or -i. How do you fix this? CASs like Maple and Mathematica define a principal value and always compute with it. 1^(1/2) = 1, 1^(1/4) = 1, etc. But also (-1)^(1/3) is complex, not real. So some beginners are confused by it. ==== > > The problem is, if we have the number 1^.5, is it 1 or -1. Going even > farther, 1^.25 could be 1, -1, i, or -i. How do you fix this? > > CASs like Maple and Mathematica define a principal value and always > compute with it. 1^(1/2) = 1, 1^(1/4) = 1, etc. > But also (-1)^(1/3) is complex, not real. So some beginners are > confused by it. No, you are wrong. (1<0)^1/4 = 1 always. Since we are dealing with polar coordinates, this theorem no longer applies. Via. -1^.5 = 1<180 = i. You made a very common mistake and do not understand the problem. Basically, with that theorem (1 1<0 1<45 -> 1<22.5 1<90 -> 1<45 1<135 -> 1<67.5 1<180 -> 1<90 1<225 -> 1<112.5 1<270 -> 1<135 1<315 -> 1<157.5 1<360 -> ? 1<0 ? != 1<180. Uh oh, your theorem just failed. Dare you to graph it with your theorem. A = 1<360 is like saying A went around the circle one time and taking the square root would be asking where would I be one half revolution ago. B = 1<0 is like saying I have not had any revolutions yet. They may be at the same position but in theory, they are two numbers, unless you mod them by 360 like you did and said they are exactly the same. They are not. PS: Don't treat people like they are idiots and don't have an education because you think they are wrong. You just haven't thought about it long enough to realize the definition is wrong in just a few cases. It is bad judgement. ==== > How do you do induction on the reals? > > Given a real, what is the next real? > Back around 1900 or so, the Heine-Borel Theorem was viewed as a continuous form of induction. So instead of the situation with natural numbers (if P holds for n, then it holds for the next integer) we have something like: if P holds up to a value x, then it holds for some small neighborhood of x, so it holds up to x+epsilon, for some epsilon>0, possibly depending on x. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > > I have just finished a course in Representation Theory of finite > groups, ending with the classification of all representations of the > symmetric group (using the young diagrams and all that). (I am a > (pure)math honours undergraduate having completed 2 years.) > > My questioin is - > > What are the applications of representation theory of groups (in > general)? What are the reasons for studying it apart from its beauty? > > It can't be denied that this theory is beautiful enough to merit a > study solely for enjoying the same, but whichever book I go to tells > me that it finds applications as varied and far as physics and quantum > chemistry. Prefaces after prefaces say that most mathematicians would > at some point or the other come accross it. I can't readily see that. > Studying representations of a finite group does seem to magically > deduce a few properties abt the group itself (for instance, the > Burnside's p^a.q^b theorem). > > But, all said and done, I really dont know how or where these > things could get applied. And, for instance, how do they show up > their head in physics? The groups whose representations are important in modern physics are mainly, but not solely, Lie groups. A Lie group is an infinite group that is also a topological space and a differentiable manifold, with the operations of group multiplication and inverse taking both differentiable. As Greg said, representations of various finite groups are useful for classifying the symmtries of crystals. Interestingly, representations of the symmetric group, including Young diagrams, are useful for constructing representations of Lie groups on tensor spaces. The reps of the symmetric group give the symmetries of the (finite number of) tensor indices involved. As a simple example, consider the function T : R^3 -> R that gives the temperature at all points in (normal, 3-dim) space, where one point in space is chosen as the origin. Then x in R^3 represents a posiition in space with respect to the origin and 3 orthogonal axes, and T(x) is the temperature at position x (with respect to the chosen origin and axes). If we chose the same origin for space, but chose a set of orthogonal axes that are rotated with respect to the original axes, then we get a new temperature function T' : R^3 -> R. Now, the set of rotations in 3-space forms a group that is isomorphic to SO(3), the group of real 3x3 (this what the 3 means) that have determinant +1 (this is what the S, means, i.e., they are Special), and are Orthogonal (this is what the O means). It turns out that T'(x) = T( g^(-1) x). This gives a representation on the infinite-dimensional vector space of functions that map R^3 to R. Other examples of important matrix Lie groups are SU(2), SU(3), U(1), etc. SU(2) is the group of 2x2 complex Unitary matrices that determinant +1. I think you can see the pattern. A fairy introductory test, written in a style amenable to pure mathies, that gives myriad applications of reps of both finite and Lie groups is Group Theory and Physics by S. Sternberg. A pure math book that gives a readable pure math introduction to the representation theory of Lie groups and Lie algebras is Representation Theory A First Course W. Fulton and J. Harris. These are both excellent books. The second book contains no applications to physics, even though much material, with notational changes, is directly applicable to physics. I am very interested in the applications of group representation theory in physics. Here are some links to some old posts of mine on applications. These posts require moderate background in physics, or math, or both, and all have significant typos. George http://groups.google.ca/groups?hl=en&lr=&ie=UTF-8&selm=3C990919.D50FED01%40y ahoo.com&rnum=5 http://groups.google.ca/groups?q=rotation+author:George+author:Jones&hl=en&l r=&ie=UTF-8&scoring=d&selm=7c74e370.0305120951.21c52059%40posting.google.com& rnum=1 http://groups.google.ca/groups?q=quark+author:George+author:Jones&hl=en&lr=& ie=UTF-8&scoring=d&selm=3ADA0F4F.97ACB477%40uvi.edu&rnum=5 123456789012345678901234567890123456789012345678901234567890123456789012 ==== > > I have just finished a course in Representation Theory of finite > groups, ending with the classification of all representations of the > symmetric group (using the young diagrams and all that). (I am a > (pure)math honours undergraduate having completed 2 years.) > > My questioin is - > > What are the applications of representation theory of groups (in > general)? What are the reasons for studying it apart from its beauty? > > It can't be denied that this theory is beautiful enough to merit a > study solely for enjoying the same, but whichever book I go to tells > me that it finds applications as varied and far as physics and quantum > chemistry. Prefaces after prefaces say that most mathematicians would > at some point or the other come accross it. I can't readily see that. > Studying representations of a finite group does seem to magically > deduce a few properties abt the group itself (for instance, the > Burnside's p^a.q^b theorem). > > But, all said and done, I really dont know how or where these > things could get applied. And, for instance, how do they show up > their head in physics? > > > Anshul > > A S K, > your questions got me thinking: > > You have finished a course on group representations, and you are asking > very natural questions. In fact, I would hope that most students taking > the same course would want to know about the origins of representation > theory and its uses. > > So let me ask you: Did you put your questions to the professor who > taught the course? If so, what did (s)he say? > > You don't have to reveal the name of the university and the professor. I > am just curious to know what answer you get when you ask those who teach > the subject. Why should a pure math course give physical applications? Why should a pure math prof know physical applications? George ==== >> You have finished a course on group representations, and you are asking >> very natural questions. In fact, I would hope that most students taking >> the same course would want to know about the origins of representation >> theory and its uses. >> >> So let me ask you: Did you put your questions to the professor who >> taught the course? If so, what did (s)he say? >> >> You don't have to reveal the name of the university and the professor. I >> am just curious to know what answer you get when you ask those who teach >> the subject. > >Why should a pure math course give physical applications? To engage and edify the students. If you have homework problems that work with the material anyway, some effort should be made, when possible, to relate it to something useful. > >Why should a pure math prof know physical applications? Because he's a math prof teaching stuff that people use in physical applications. -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== > >> A S K, your questions got me thinking: >> >> You have finished a course on group representations, and you are >> asking very natural questions. In fact, I would hope that most >> students taking the same course would want to know about the >> origins of representation theory and its uses. >> >> So let me ask you: Did you put your questions to the professor who >> taught the course? If so, what did (s)he say? >> >> You don't have to reveal the name of the university and the >> professor. I am just curious to know what answer you get when you >> ask those who teach the subject. > > > Why should a pure math course give physical applications? > A pure math course doesn't have to give physical applications (and I didn't state it should), but pointing out connections and relationships to anything familiar helps in learning. Students learn new material better if they can relate it to something they already know. It has to do with the structure of the brain and learning mechanisms. That something familiar could be in other areas of mathematics, in physics, or in their daily life. Seeing multiple connections helps students develop intuition, which is just as important for their mathematical ability as formal derivation of results. > Why should a pure math prof know physical applications? I don't know, maybe (s)he is curious? Maybe (s)he wants to know where the subject came from, and why the researchers who started the area thought it was important? I know there are pure math profs who don't care. That's fine. It just means that students like A S K have to take their questions elsewhere. Nemo ==== > > > [...] > >their head in physics? >> >>You must know by now that transformations form a group. Something like a >>Dirac spinor isn't a vector, but you can figure out how it transforms >>under boosts or rotations by starting with the properties of the Lorentz >>group. > > > Ya, invertible vector space transformations do form a group. I don't > know what a Dirac Spinor is. I don't know what the Lorentz group is > either. (I am a pure math undergrad:|) > It's not a bad idea to acquire the habit of Googling for terms you don't understand. As for spinors, there are two trails of references to look for: function, a solution to the Schr.9adinger equation. Its standard interpretation is as a probability density function; The question of the behavior of this function to the observation that two types of symmetry may take place: even [swapping preserves the wave function] and odd [swapping reverses the sign of the wave function]. Solutions of the latter sort lead to the consideration of a phenomenon requiring a 4 pi rotation to return to its original state; that idea turns out to coincide with: 2. The so-called spinor group Spin(n), a particular double cover of the special orthogonal group (SO(n) is the Lie of proper rotations of R^n, i.e., the set of orthogonal matrices of determinant +1). See E. Cartan, The Theory of Spinors, originally published in 1966 by Hermann; currently (1981), Dover Press. The relevance of Dirac's name here is due largely to the Dirac Equation, a relativistic spin-1/2 wave equation; the solutions to the Dirac Equation are spinors in that their transformation properties are given by representations of the spinor group Spin(3), often referred to as the so-called spin representations of SO(3). The second question, as to the Lorentz group, there are generalizations of the orthogonal groups O(n) as those matrices that preserve the quadratic form x1^2 + x2^2 + ... + xn^2; if A is a non-degenerate nxn symmetric matrix, one can define O(n;A) as the set of all nxn matrices K that preserve the inner product defined by the matrix A: x'Ax = (Kx)'A(Kx) for all n-vectors x, or K'AK = A. Special relativity determines that the quantity x^2 + y^2 + z^2 - (ct)^2 is conserved under transformations between inertial coordinate systems. By suitable normalization, one can reduce this to the requirement that the matrix [ 1 0 0 0 ] [ 0 1 0 0 ] [ 0 0 1 0 ] [ 0 0 0 -1 ] The group of 4x4 matrices that preserve this form is called O(3,1). The subgroup of elements of determinant 1 is called SO(3,1), or the Lorentz group, and the subgroup that preserves the orientation of the 4th coordinate is the orthochronous Lorentz group. > >>The standard model appears by applying certain symmetries to a generic >>Lagrangian. Decide you want it invariant under U(1), figure out what >>extra terms you need to make it so, and there's electromagnetism. >>Theorists find that more satisfying than just putting a bunch of terms in >>way before they were discovered. > > I vaguely remember from my mechanics course that a lagrangian used to > be the kinitec energy less the potential energy. What is U(1)? I > would be grateful if you could explain that keeping my background in > mind. > U(n) is the group of unitary nxn matrices: those (complex) nxn matrices A for which A^* A = A A^* = Id(nxn), the nxn identity matrix, where (...)^* is the complex-conjugate transpose, also called the Hermitian adjoint. SU(n) is the subgroup of U(n) consisting of matrices of determinant 1. If n = 1, then you've got a 1x1 complex matrix, or a complex number z. Unitarity requires z^* z = 1, so the square magnitude of z is 1. Thus, U(1) is the set of complex numbers of modulus 1, or the circle group. > > Anshul Dale. ==== >It's not a bad idea to acquire the habit of Googling for terms you don't >understand. However (as Dale surely knows; this is addressed to the same newbie he was addressing, whose name I seem already to have deleted) using this habit wisely involves also developing a good nose for cranks, frauds, and those who are simply deluded (or ignorant), and thus post misinformation which--on the face of it--looks like it might be right, but isn't. You also have to watch out for physicists (who aren't *necessarily* subsumed under any or all of the previous categories), whose use of mathematical terms doesn't always agree with the mathematicians' use of the same terms. (The asymmetry in this statement isn't purely generated by my prejudices; the newbie was a self-described maths undergraduate.) Lee Rudolph ==== > > >>It's not a bad idea to acquire the habit of Googling for terms you don't >>understand. > > > However (as Dale surely knows; this is addressed to the same newbie > he was addressing, whose name I seem already to have deleted) using > this habit wisely involves also developing a good nose for cranks, > frauds, and those who are simply deluded (or ignorant), and thus post > misinformation which--on the face of it--looks like it might be right, > but isn't. > > You also have to watch out for physicists (who aren't *necessarily* > subsumed under any or all of the previous categories), whose use > of mathematical terms doesn't always agree with the mathematicians' > use of the same terms. (The asymmetry in this statement isn't > purely generated by my prejudices; the newbie was a self-described > maths undergraduate.) > > Lee Rudolph > blind search process. Dale. ==== Hey sci/alt.math, I'm a physics grad student, but I have been doing some summer reading in order to get a more rigorous background in mathematics... long story short, I'm started reading Calculus on Manifold by Spivak, and I'm wondering if people can critique my answer to one of the problems I'm working on. The problem is very similar to one found in the Spivak book, though a bit less general. Here it is: (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the first component of the n-tuple of numbers, x) Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and 0 < x2} is Open. So, basically, this is the part of a circle of radius 'r' located in the first quadrent, and our job is to prove that it is open. Now, I understand that all I have to do is find some open rectangle, A, that is a subset of S such that ( y within S ) --> ( y within A ) So, what I came up with is this. Given any old y within S choose the open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a ) where 'a' satisifys 0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) } where min{ } means, choose the smallest number in the set. and 0 < b = (r^2) - (y1^2) - (y2^2) Obviously, the y1 and y2 are in there to keep the rectangles lower left corner inside S, and the third term is in there to keep the rectangles upper right corner inside S. It is true that y1 and y2 are greater than zero, and so is the third term because b is greater than zero. Okay, So I have found that S is open. Because for any y within S there is an open rectangle, A, for which y is within A and A is a subset of S. So, am I done? Is this a good and rigorous proof? Comments and Critisism (constructive, hopefully) would be very much appreciated. adrock ==== I didn't go into the computations, but you have the right. You can also use open circles as well. For example, if (x_0,y_0) is any point in the region, then let d_1 be the distance from (x_0,y_0) to the edge of the quarter-cirlce, let d_2 be the distance from (x_0,y_0) to the x-axis, and let d_3 be the distance from (x_0,y_0) to the y-axis (you can come up with formulas for each of these quite easily for you I think). In any event, let d = min (d_1, d_2, d_3). Then, ( x - x_0)^2 + (y - y_0)^2 = (d/2)^2 gives the equation of a circle, centered at (x_0,y_0) that is completely inside of the region at hand. As I stated before, I didn't go into your computations, but your idea is right on (get some open neighborhood around any point and you've shown that the set is open). I just thought I would throw in the open circle to give you another example to use either now or in the future. I hope this helps, Brian > Hey sci/alt.math, > > I'm a physics grad student, but I have been doing some summer reading > in order to get a more rigorous background in mathematics... long story > short, I'm started reading Calculus on Manifold by Spivak, and I'm > wondering if people can critique my answer to one of the problems > I'm working on. > The problem is very similar to one found in the Spivak book, though > a bit less general. Here it is: > > (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the > first component of the n-tuple of numbers, x) > > Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and 0 < x2} > is Open. > > So, basically, this is the part of a circle of radius 'r' located in the > first quadrent, and our job is to prove that it is open. > > Now, I understand that all I have to do is find some open rectangle, A, > that is a subset of S such that ( y within S ) --> ( y within A ) > > So, what I came up with is this. Given any old y within S choose > the open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a ) > where 'a' satisifys > > 0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) } > > where min{ } means, choose the smallest number in the set. > > and 0 < b = (r^2) - (y1^2) - (y2^2) > > Obviously, the y1 and y2 are in there to keep the rectangles lower > left corner inside S, > and the third term is in there to keep the rectangles upper > right corner inside S. > It is true that y1 and y2 are greater than zero, and so is > the third term because b is greater than zero. > > Okay, So I have found that S is open. Because for any y within S > there is an open rectangle, A, for which y is within A and A is a > subset of S. > > So, am I done? Is this a good and rigorous proof? Comments and > Critisism (constructive, hopefully) would be very much > appreciated. > > > adrock > ==== > (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the > first component of the n-tuple of numbers, x) > > Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and > 0 < x2} is Open. > Here's an approach you might like. B = { x : ||x|| < r } is an open ball. A = { x : 0 < x } is open line segement in R. Thus AxA is open in RxR S = B / AxA is the intersection of two open sets, hence open. ==== > >> (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the >> first component of the n-tuple of numbers, x) >> >> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and >> 0 < x2} is Open. >> >Here's an approach you might like. >B = { x : ||x|| < r } is an open ball. >A = { x : 0 < x } is open line segement in R. >Thus AxA is open in RxR >S = B / AxA is the intersection of two open sets, hence open. > Ah, that is just great... to bad I have not proven that the open ball is open. Could you enclude a proof that the open ball is open? ==== > Hey sci/alt.math, > > I'm a physics grad student, but I have been doing some summer reading > in order to get a more rigorous background in mathematics... long story > short, I'm started reading Calculus on Manifold by Spivak, An excellent choice for summer reading. and I'm > wondering if people can critique my answer to one of the problems > I'm working on. > The problem is very similar to one found in the Spivak book, though > a bit less general. Here it is: > > (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the > first component of the n-tuple of numbers, x) > Note that Spivak's use of superscripts in this context is a little nonstandard. Most books prefer subscripts to denote the components of an element of n-space. The standard ASCII convention is x_1 for x-subscript-1, so that an n-vector is denoted (x_1, x_2, . . ., x_n). > Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and 0 < > x2} > is Open. > > So, basically, this is the part of a circle of radius 'r' located in the > first quadrent, and our job is to prove that it is open. > Note that most books define an open set in terms of balls, not rectangles. In other words if x is a point (in n-space) then we define the *open ball* centered at x of radius d, B(x,d) = {b element of R^n such that |x-b| < d}. Then we say that a set U is open if each point of U is contained in an open ball that is a subset of U. With this definition it's very easy to show that the interior of a circle is an open set. Here Spivak is having a little fun with us. By defining open sets in terms of open rectangles and then asking us to show that the interior of a (quarter) circle is open, he's forcing us to mess around with square roots and figure out the distance from a point inside a circle, to the circle. You might find it instructive to convince yourself that a set is open using the rectangle definition if and only if it is open using the ball definition. Actually if you do that first, this problem gets a lot easier, since (using the ball definition) if x is an element of the set S = {b such that |b| < r}, then B(x, r/2} is an open ball that contains x and is a subset of S. Ok, back to the original problem. > Now, I understand that all I have to do is find some open rectangle, A, > that is a subset of S such that ( y within S ) --> ( y within A ) > statement ( y within S ) --> ( y within A ) is the DEFINITION of 'S is a subset of A', which is not what you mean to say. Spivak's definition of an open set U is that if x is any arbitrary element of U, there exists an open rectangle A with (x element of A) and (A subset of U). choose some open rectangle A subset of S that DOES NOT contain y. Then ( y within S ) --> ( y within A ) is VACUOUSLY TRUE since (y within S) is false. Therefore you have just proven that ANY SET that contains an open rectangle is an open set, which is false. So again, just to be clear, here is Spivak's definition of an open set. A set U is open if for any x in U, there exists an open rectangle A with x in A and A subset of U. And as a reminder, an open rectangle is the cartesian product of open intervals. > So, what I came up with is this. Given any old y within S choose > the open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a ) > where 'a' satisifys > > 0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) } > > where min{ } means, choose the smallest number in the set. > > and 0 < b = (r^2) - (y1^2) - (y2^2) > > Obviously, the y1 and y2 are in there to keep the rectangles lower > left corner inside S, > and the third term is in there to keep the rectangles upper > right corner inside S. > It is true that y1 and y2 are greater than zero, and so is > the third term because b is greater than zero. > I have to admit I didn't work all the way through your solution. It seems too complicated. That doesn't mean it's wrong, just that I found it easier to work this out for myself than work through your reasoning. The idea is that for a point inside the circle {|x| < r} we want to find the horizontal and vertical distance to the circle, and use those distances to construct the open intervals. The other thing I did was to use the standard notational convention for R^2, which is to label points (x,y) instead of (x_1, x_2). This lets us use the familiar concepts of the x-axis, the y-axis, and functions denoted as y = f(x). Why make life complicated? As you're following along with this it's incredibly helpful to have a diagram in front of you, a big circle centered at the origin, with a random point (a,b) in the first quadrant inside the circle. We need to figure the horizontal and vertical distances from (a,b) to the circle. If (a,b) is a point inside the circle, what is the horizontal distance to the circle? The horizontal line through (a,b) has equation y=b, and the circle has equation x^2 + y^2 = r^2 => x^2 + b^2 = r^2 => x = sqrt(r^2 - b^2). In that last equation we know r^2 - b^2 is positive since a^2 + b^2 < r^2 is given, therefore the sqrt exists in the reals. Then the distance between (a,b) and (sqrt(r^2 - b^2), b) is sqrt(r^2 - b^2) - a. We know the sign is positive from the geometry: sqrt(r^2 - b^2) is the x-coordinate of a point on the circle horizontally to the right of (a,b). On the other hand, the distance between (a,b) and the y-axis is just a. So we can choose our x-interval around a to be (a/2, a + (sqrt(r^2 - b^2) - a) / 2). Likewise, going through the same reasoning in the vertical direction, a workable y-interval is (b/2, b + (sqrt(r^2 - a^2) - b)/2). The open rectangle we need is the cartesian product of the x- and y-intervals. If this is the same answer you had, that's good. On the other hand yours doesn't look quite right to me, but I haven't taken the time to figure out why. > Okay, So I have found that S is open. Because for any y within S > there is an open rectangle, A, for which y is within A and A is a > subset of S. > Yes, that's the correct statement that you mis-stated above. > So, am I done? Is this a good and rigorous proof? Comments and > Critisism (constructive, hopefully) would be very much > appreciated. > Another way to go would be to prove that the interior of a circle is open, the upper open half plane {y>0} is open, and the right open half plane {x>0} is open, and a finite intersection of open sets is open (a Spivak exercise, at least in my ancient edition). Therefore the original set is open. I hope my long-winded post helped. ==== > >> (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the >> first component of the n-tuple of numbers, x) >> >> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and >> 0 < x2} is Open. >> >Here's an approach you might like. >B = { x : ||x|| < r } is an open ball. >A = { x : 0 < x } is open line segement in R. >Thus AxA is open in RxR >S = B / AxA is the intersection of two open sets, hence open. > > > Ah, that is just great... to bad I have not proven that the open ball > is open. Could you enclude a proof that the open ball is open? > Ah, but that's the whole point of the exercise, as I pointed out in my detailed response in the original thread. If you define open sets in terms of open balls, it's trivial to show that the interior of a circle is open. If you define open sets in terms of open rectangles, you have to do a little work, and that's what Spivak wants the reader to do. ==== > Hey sci/alt.math, > > I'm a physics grad student, but I have been doing some summer reading > in order to get a more rigorous background in mathematics... long story > short, I'm started reading Calculus on Manifold by Spivak, and I'm > wondering if people can critique my answer to one of the problems > I'm working on. > The problem is very similar to one found in the Spivak book, though > a bit less general. Here it is: > > (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the > first component of the n-tuple of numbers, x) > > Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and 0 < x2} > is Open. > > So, basically, this is the part of a circle of radius 'r' located in the > first quadrent, and our job is to prove that it is open. > > Now, I understand that all I have to do is find some open rectangle, A, > that is a subset of S such that ( y within S ) --> ( y within A ) I don't think that last phrase such that ... is what you want to say. I suspect that this is your own phraseolgy (which is not standard). In brief, you need to get the quantifiers correct. That is, for each y within S, there is some open rectangle A that is a subset of S, such that y is in A. In particular, A depends on the y. > > So, what I came up with is this. Given any old y within S choose > the open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a ) Right. The y is given first and then you choose the A. > where 'a' satisifys > > 0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) } > > where min{ } means, choose the smallest number in the set. > > and 0 < b = (r^2) - (y1^2) - (y2^2) > > Obviously, the y1 and y2 are in there to keep the rectangles lower > left corner inside S, > and the third term is in there to keep the rectangles upper > right corner inside S. > It is true that y1 and y2 are greater than zero, and so is > the third term because b is greater than zero. The above paragraph gives a geometric reasoning for the proof. I didn't check it out. But, you may want to give a more analytic reasoning for the proof. That is, you want to show that A is a subset of S. To do this, proceed the standard way. That is, something like the following: Now let (u1, u2) be in A. I claim that (u1, u2) is in S. For first, 0 < u1 because ... Seconding, 0 < u2 because ... Finally, u1^2 + u2^2 < r because ... Hence, A is a subset of S. As an aside, the last sentence of the above paragraph can be improved by stating explicitly the motivation for it. That is, it justifies that a can indeed be found. Thus, you can say something like: I can find an a since the mim is greater than zero because y1 and y2 are greater than zero, and so is the third term because b is greater than zero. > Okay, So I have found that S is open. Because for any y within S > there is an open rectangle, A, for which y is within A and A is a > subset of S. This is the correct statement of what you are trying to prove. > So, am I done? Is this a good and rigorous proof? Comments and > Critisism (constructive, hopefully) would be very much > appreciated. > > > adrock ==== > You might find it instructive to convince yourself that a set is open > using the rectangle definition if and only if it is open using the ball > definition. Actually if you do that first, this problem gets a lot > easier, since (using the ball definition) if x is an element of the set > S = {b such that |b| < r}, then B(x, r/2} is an open ball that contains > x and is a subset of S. > Correction. Of course what I meant to write was that B(x, 1/2 * min(|x|, r - |x|) is the required open ball. ==== I think that the neatest suggestion was to show that the open ball is open under Spivak's definition (which is basically Spivak problem 1-15), and show that the 1st quadrent is open, and then that the interesction of two (or a finite number) of open sets is open...Then my problem is solved. get stuck again...ciao, adrock ==== > I think that the neatest suggestion was to show that > the open ball is open under Spivak's definition (which is > basically Spivak problem 1-15), and show that the 1st quadrent > is open, and then that the interesction of two > (or a finite number) of open sets is open...Then my problem > is solved. Indeed you really learned something from that problem. Fishfry pointed out what was important. Another uniformally equivalent metric to circles and rectangles is diamonds or rhomboids D((x,y) - (a,b)) = |x-a| + |y-b| Also there's the box metric which is about the same except instead of open rectangles it uses open squares. > get stuck again...ciao, > You're welcome. You don't have to wait to get stuck before returning. The approach I suggested that was to your liking was a topological approach the topology of metric spaces which is easier handled just topologically. Of course, there is some basic transitions to understand, the rectangle, circle topologies being equivalent is one, another being the equivalence of the rectangle topology to the product topology which seemed instantly clear to you. ==== I have to minimize something like tr(X'AX) where A in R^{n*n} is a definite positive matrix and X is a n*k binary matrix, such that x_ij={0,1} and the sum of each row is 1 (i.e. sum_i x_ij=1). The problem is well known to be NP-complete. Do you know any reference to a continuous approximation? Or an efficient way to solve the problem when X is a big matrix? ==== > William Elliot ha scritto nel messaggio > > Consider the subposet ({a,b}, <=) and the function F such that > * * ------------> * F(a)=F(b) > a b F > (a and b are not comparable). Is F order preserving? I think so. > > a,b not comparable, written a || b, > when not a <= b & not b <= a > What's order preserving, an order isomorphism? > > F is order preserving iff x<=y ==> F(x)<=F(y); > F is an order embedding iff x<=y <==> F(x)<=F(y); > F is an order isomorphism iff x<=y <==> F(x)<=F(y) and F is surjective. > > 1 for a,b, (a || b ==> f(a) = f(b)) > imply > 2 f is order preserving. > > Now I know that if F:(S,<=) --> (S',<=') is a function such that F(x)=K for > all x in S, F is an order preserving function. > > Consider x < y, x < z, y || z; f(x) = a, f(y) = f(z) = b; b < a > f has 1 but lacks 2, does it not? > Also consider same f with a < b, b || a or a = b instead. > > You mean (the first case) > > *y *z *f(x) > / > / ----------> > / f > / > *x *f(y)=f(z) > > so f is not order preserving. > Exactly, nor in the case a || b > Now consider the following: > a* > | > | -----------> * G(a)=G(b) > | G > b* > > Is G order embedding? It is true that we have G(b)<=G(a) ===> b<=a, but > also > G(a)<=G(b) & a<=b. > > Does the diagram mean > for all a,b, (a <= b ==> g(a) = g(b)) ? > Now g(a) = g(b) ==> g(a) <= g(b). Thus > for all a,b, (a <= b ==> g(a) <= g(b)) > Is that last property order embedding? > > I think this last is not an order embedding because: > 1) b<=a ==> G(b)<=G(a) is true > 2) G(a)<=G(b) ==> a<=b is false; > > so the equivalence 1)<==>2) does not hold. > Sometimes it would, but in general it wouldn't. What's needed to cinch that is a counter example. As you're understanding this stuff, have you a counter example to present? Easy theorem to prove, if you haven't already order embedding maps are injective. Exercise: when is an order preserving injection, order embedding? Include proofs and/or counterexamples of claims. ==== As I understand it, probably the most important open question in finite geometry is: which numbers occur as the order of a finite (affine or projective) plane? In particular, a proof is sought for what might be called the Order Conjecture (OC). OC: If n is the order of a finite affine/projective plane, then n is a prime power. In the case that an affine plane A^2 does have an order q that is a prime power, then A^2 can easily be coordinatized by a finite field of the same order (i.e., A^2 = GF(q)^2, where GF(q) is the Galois field of order q). Conversely, if it could be proved that _every_ finite affine plane can be coordinatized by a finite field, then OC would follow, since the only orders for GF(q) are prime powers. Has this approach been taken? I know Bruck and Ryser's result on numbers of the form 4k+1 and 4k+2, and I've heard of the work at Concordia University (Montreal) to eliminate 10 as a possibility; neither of these seems to use this line of reasoning. Can someone point me in the right direction? -- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman ==== > As I understand it, probably the most important open question in finite > geometry is: which numbers occur as the order of a finite (affine or > projective) plane? In particular, a proof is sought for what might be > called the Order Conjecture (OC). > > OC: If n is the order of a finite affine/projective plane, then n is a > prime power. The n = 10 case was difficult enough .... > In the case that an affine plane A^2 does have an order q that is a prime > power, then A^2 can easily be coordinatized by a finite field of the same > order (i.e., A^2 = GF(q)^2, where GF(q) is the Galois field of order q). > Conversely, if it could be proved that _every_ finite affine plane can be > coordinatized by a finite field, then OC would follow, since the only > orders for GF(q) are prime powers. But there are finite affine/projective planes which are *not* isomorphic to the ones constructed in the classical way from finite fields. The smallest order for which such exist is 9. See http://www.math.uni-kiel.de/geometrie/klein/math/geometry/smallproj.html -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > But there are finite affine/projective planes which are *not* isomorphic > to the ones constructed in the classical way from finite fields. > The smallest order for which such exist is 9. See > http://www.math.uni-kiel.de/geometrie/klein/math/geometry/smallproj.html Aha. I was unaware of the different isomorphism types -- hadn't seen them number of new (to me) ideas to explore. So I have a couple more questions. Is there a reference that describes _how_ the order-10 case was eliminated (one that says more than The proof took thousands of hours of computer verification.?) Secondly, in the above linked site's description of quasifields and semifields, it uses two distinct existence symbols: the usual backwards E, and backwards E^1. Do these have different meanings? -- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman ==== > >> But there are finite affine/projective planes which are *not* isomorphic >> to the ones constructed in the classical way from finite fields. >> The smallest order for which such exist is 9. See >> http://www.math.uni-kiel.de/geometrie/klein/math/geometry/smallproj.html > > Aha. I was unaware of the different isomorphism types -- hadn't seen them > number of new (to me) ideas to explore. So I have a couple more questions. > Is there a reference that describes _how_ the order-10 case was eliminated > (one that says more than The proof took thousands of hours of computer > verification.?) Pehaps you should consult this account by one of the main protagonists. 92b:51013 Lam, C. W. H.(3-CONC-C) The search for a finite projective plane of order $10$. Amer. Math. Monthly 98 (1991), no. 4, 305--318. > Secondly, in the above linked site's description of > quasifields and semifields, it uses two distinct existence symbols: the > usual backwards E, and backwards E^1. Do these have different > meanings? Dunno, but I would guess that the second means there exists exactly one. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== I'm trying to calculate the expected value of the magnitude of a Gaussian random vector with given mean (let's assume zero for simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma). Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear function. (I just wanted to address a common answer I've received from people I've asked already.) Note also that if sigma is the identity matrix, the problem is easily solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is relatively simple to deal with because x'*x appears as a group everywhere. Contrast this with what I'd like, which is of the form: integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that helps anyone when making substitutions, but it's something I noticed. Afsheen ==== >I'm trying to calculate the expected value of the magnitude of a >Gaussian random vector with given mean (let's assume zero for >simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where >x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma). >Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't >imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because >E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear >function. (I just wanted to address a common answer I've received from >people I've asked already.) >Note also that if sigma is the identity matrix, the problem is easily >solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is >relatively simple to deal with because x'*x appears as a group >everywhere. Contrast this with what I'd like, which is of the form: >integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that >helps anyone when making substitutions, but it's something I noticed. About the only way to get accurate numerical answers is to numerically integrate the derivative of the Laplace transform of the distribution of x'x. If L is the Laplace transform, the -int sqrt(1/t) L'(t) dt/sqrt(pi) is the expected value of sqrt(x'x). -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University ==== Is there no way to solve this analytically? Since we're given that x is a Gaussian random vector of known mean and covariance, I would think that there's a nice expression for E(sqrt(x'x)). Afsheen > >> > >>I'm trying to calculate the expected value of the magnitude of a >>Gaussian random vector with given mean (let's assume zero for >>simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where >>x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma). >> > >>Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't >>imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because >>E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear >>function. (I just wanted to address a common answer I've received from >>people I've asked already.) >> > >>Note also that if sigma is the identity matrix, the problem is easily >>solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is >>relatively simple to deal with because x'*x appears as a group >>everywhere. Contrast this with what I'd like, which is of the form: >>integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that >>helps anyone when making substitutions, but it's something I noticed. >> > > About the only way to get accurate numerical answers > is to numerically integrate the derivative of the > Laplace transform of the distribution of x'x. If > L is the Laplace transform, the > > -int sqrt(1/t) L'(t) dt/sqrt(pi) > > is the expected value of sqrt(x'x). > ==== >Is there no way to solve this analytically? Since we're given that x is >a Gaussian random vector of known mean and covariance, I would think >that there's a nice expression for E(sqrt(x'x)). >Afsheen If the characteristic roots of the covariance matrix, assuming the mean vector is 0, all occur with even multiplicities, the distribution is a linear combination, some coefficients being negative, of Gamma distributions with integer exponents. In this case, the integration can be done in closed form. I am not sure if the integral representation given can be done in elementary terms if there is one root of odd multiplicity. If there are two such roots, we are already at least at the stage of elliptic integrals. >I'm trying to calculate the expected value of the magnitude of a >Gaussian random vector with given mean (let's assume zero for >simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where >x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma). >Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't >imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because >E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear >function. (I just wanted to address a common answer I've received from >people I've asked already.) >Note also that if sigma is the identity matrix, the problem is easily >solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is >relatively simple to deal with because x'*x appears as a group >everywhere. Contrast this with what I'd like, which is of the form: >integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that >helps anyone when making substitutions, but it's something I noticed. >> About the only way to get accurate numerical answers >> is to numerically integrate the derivative of the >> Laplace transform of the distribution of x'x. If >> L is the Laplace transform, the >> -int sqrt(1/t) L'(t) dt/sqrt(pi) >> is the expected value of sqrt(x'x). -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University ==== > Wow! 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That might even be funny. Euclidean space : Non-Euclidean space :: Conventional math : Non-conventional math [ 2^(7/12) = 3/2 ] (cross-posted to sci.math) ObPuzzle: Why did I choose this particular equation? ==== > > consider the following statement, where * is some arbitrary operation: > if a and b are in the empty set, then a*b is in the empty set > > Why not > Because * isn't defined at a,b, that's why. > if a and b are in the empty set, then c is in the empty set > > ? > As you've bestowed existence upon c, yes, c in nulset. > Immediately a philosophical/logical/set theoretical nightmare arises > as one attempts to decide whether the statement is true or false or > whether it depends on one's choice of *. > > logic's equation FALSE->FALSE = TRUE unequivically says the > statement is true, but all common sense screams otherwise. > > Mine doesn't? So why does yours? > Philosophical conundrum: are non existence things in the nulset? nulset = { x | x /= x } if x doesn't exist, does x /= x ? ==== > if a and b are in the empty set, then a*b is in the empty set > > logic's equation FALSE->FALSE = TRUE unequivically says the > statement is true, but all common sense screams otherwise. > > Mine doesn't? So why does yours? So you would hold, then, that if a and b are both prime numbers divisible by 4, then a+b is also a prime number divisible by 4? if a finite set M of integers has two distinct largest elements, a and b, then a^b is also the largest element of M? if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well? (Ironically if we replace 5 here with aleph_0, it *almost* becomes true) OK, but more seriously, if indeed the empty set possesses the closure property, can it also include an identity with regard to *? this seems to depend on how one phrases the definition of a group... if one says, the set in question must possess some 'identity' i... then clearly the empty set does not possess an identity. But if one says, for any a in the set in question, there exists some 'identity' i, which is the same for all other elements in the set... then right away it seems the statement is true, since the false statement the empty set contains an identity i is cancelled out by the false for any a in the empty set... See, what I'm really getting at here is, is the empty set a group? ==== > Philosophical conundrum: are non existence things in the nulset? Yes! No element of the emptyset exists. So... all nonexistent things are in the emptyset. [Phrased more technically: The emptyset is a subset of the emptyset.] ==== > consider the following statement, where * is some arbitrary operation: > > if a and b are in the empty set, then a*b is in the empty set > > Immediately a philosophical/logical/set theoretical nightmare arises > as one attempts to decide whether the statement is true or false or > whether it depends on one's choice of *. > > logic's equation FALSE->FALSE = TRUE unequivically says the > statement is true, but all common sense screams otherwise. I prefer to say that the sentence a*b is in the empty set is not meaningful, rather than that it is true (or false). I believe the logicians have a name for what I am thinking of (rathen than just meaningful). In more detail, first I require that you tell me the type of a and b. For example, you could say that a and b are in the set Q of rational numbers. Then, your statement is: If a in Q and b in Q are in the empty set, then a*b is in the empty set Second, I require that you tell me the type of operation * is. I need to know the domain and range of the operation *. I think that you are willing to say that the operation is a binary operation from the same set to the same set: that is, * : AxA -> A where A is some set. For example, you could say that A is a cyclic group of order 2 and the operation * is the group product. Then, your statement is: Let A be the cyclic group of order 2 with multiplication *. If a in Q and b in Q are in the empty set, then a*b is in the empty set I prefer to say that the statement a*b is in the empty set is not meaningful, since the rational numbers a and b are not in the domain of * (that is, (a,b) is not in AxA). Since the conclusion is not a meaningful statement, I take the whole implication to not be a meaningful statement. But, I suspect that some (many?) may wish to say that the statement a*b is in the empty set is just false. As an aside, I would take the following to be a meaningful statement: Let A be the cyclic group of order 2 with multiplication *. If a in Q and b in Q are in the empty set, then if a is in A and b is in A, then a*b is in the empty set I take it to be meaningful since the hypothesis a is in A and b in is A allows be to do the group multiplication a*b to get an element of A. -- Bill Hale ==== > if a and b are in the empty set, then a*b is in the empty set > > logic's equation FALSE->FALSE = TRUE unequivically says the > statement is true, but all common sense screams otherwise. > > Mine doesn't? So why does yours? > > So you would hold, then, that if a and b are both prime numbers > divisible by 4, then a+b is also a prime number divisible by 4? Yes. > if a finite set M of integers has two distinct largest elements, a and > b, then a^b is also the largest element of M? Yes. > if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well? Yes. The above three statements are not that strange. In fact, they occur all the times. Consider the standard epsilon-delta definition of continuity. The condition statement |x - x0| < delta is false for a great many x's, but the implication statement if |x-x0| < delta, then |f(x) - f(x0)| < epsilon is still true for those x's. Indeed, we need to prove the implication even for those x's because that what the definition says. > (Ironically if we replace 5 here with aleph_0, it *almost* becomes > true) > > OK, but more seriously, if indeed the empty set possesses the closure > property, can it also include an identity with regard to *? this > seems to depend on how one phrases the definition of a group... if one > says, the set in question must possess some 'identity' i... then > clearly the empty set does not possess an identity. And the definition of group *does* require such existence of the unit element. > But if one says, > for any a in the set in question, there exists some 'identity' i, > which is the same for all other elements in the set... then right > away it seems the statement is true, since the false statement the > empty set contains an identity i is cancelled out by the false for > any a in the empty set... > See, what I'm really getting at here is, is the empty set a group? But, that is why the definition of group is not phrase in your second way. -- Bill Hale ==== > OK, but more seriously, if indeed the empty set possesses the closure > property, can it also include an identity with regard to *? this > seems to depend on how one phrases the definition of a group... if one > says, the set in question must possess some 'identity' i... then > clearly the empty set does not possess an identity. But if one says, > for any a in the set in question, there exists some 'identity' i, > which is the same for all other elements in the set... then right > away it seems the statement is true, since the false statement the > empty set contains an identity i is cancelled out by the false for > any a in the empty set... But who would be perverse enough to use the second formulation? This is simply not a real problem. > See, what I'm really getting at here is, is the empty set a group? Of course it is not a group, because no one would ever (even by accident) define a group using the clause for any a in the set in question, there exists some 'identity' i, which is the same for all other elements in the set... -- Jesse Hughes How come there's still apes running around loose and there are humans? Why did some of them decide to evolve and some did not? Did they choose to stay as a monkey or what? -Kans. Board of Ed member ==== > > > consider the following statement, where * is some arbitrary operation: > > if a and b are in the empty set, then a*b is in the empty set > > Why not > > if a and b are in the empty set, then c is in the empty set > > ? Indeed, why not if a and b are members of the empty set, then a*b is the King of France ? --------------------------------------------------- C Brown Systems Designs Multimedia Environments for Museums and Theme Parks --------------------------------------------------- ==== >> if a and b are in the empty set, then a*b is in the empty set >> >> logic's equation FALSE->FALSE = TRUE unequivically says the >> statement is true, but all common sense screams otherwise. >> >> Mine doesn't? So why does yours? > > So you would hold, then, that if a and b are both prime numbers > divisible by 4, then a+b is also a prime number divisible by 4? Yes. > if a finite set M of integers has two distinct largest elements, a and > b, then a^b is also the largest element of M? Yes. > if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well? > (Ironically if we replace 5 here with aleph_0, it *almost* becomes > true) Yes. > OK, but more seriously, if indeed the empty set possesses the closure > property, can it also include an identity with regard to *? this > seems to depend on how one phrases the definition of a group... if one > says, the set in question must possess some 'identity' i... then > clearly the empty set does not possess an identity. But if one says, > for any a in the set in question, there exists some 'identity' i, > which is the same for all other elements in the set... then right > away it seems the statement is true, since the false statement the > empty set contains an identity i is cancelled out by the false for > any a in the empty set... > See, what I'm really getting at here is, is the empty set a group? No. A group has an identity element. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== >> >> >> consider the following statement, where * is some arbitrary operation: >> >> if a and b are in the empty set, then a*b is in the empty set >> >> Why not >> >> if a and b are in the empty set, then c is in the empty set >> >> ? > > Indeed, why not > > if a and b are members of the empty set, then a*b is the King of > France > > ? That empty set better not accidentally get any members, because if it does then /all/ kinds of weird shit is going to happen! Phil ==== > That empty set better not accidentally get any members, because > if it does then /all/ kinds of weird shit is going to happen! lol This seems to be related to not playing with a full deck, but I don't know which is the cause and which the effec'. ==== > > So you would hold, then, that if a and b are both prime numbers > divisible by 4, then a+b is also a prime number divisible by 4? > Right. > > if a finite set M of integers has two distinct largest elements, a and > b, then a^b is also the largest element of M? > Right. > > if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well? > (Ironically if we replace 5 here with aleph_0, it *almost* becomes > true) > Right. If 0 = 1 then I'm gonna eat my heat! (No, I'm really not eager to eat my heat!) > > OK, but more seriously, if indeed the empty set possesses the closure > property, can it also include an identity with regard to *? > No. Note: its certainly true that (if our domain is R and * is the usual multiplication defined for real numbers) (x)(y)(x e 0 & y e 0 -> x*y e 0) BUT this does in no way imply the EXISTENCE of some element in 0. > > this seems to depend on how one phrases the definition of a group... > if one says, the set in question must possess some 'identity' i... > then clearly the empty set does not possess an identity. > Exactly. > > But if one says, for any a in the set in question, there exists some > 'identity' i [in the set], which is the same for all other elements > in the set... then right away it seems the statement is true [...] > Right. BUT a group just ISN'T defined this way. (Know why? ;-) > > See, what I'm really getting at here is, is the empty set a group? > The answer is simply NO, just because... (see explanation above). F. ==== > > consider the following statement, where * is some arbitrary operation: > if a and b are in the empty set, then a*b is in the empty set > > Why not > > Because * isn't defined at a,b, that's why. > Well... of course it should (better) read: For any to real numbers a,b: if a and b are in the empty set, then a*b is in the empty set , where * is the usual multiplication defined for real numbers. But I guess this was meant. (For example, a,b could be qualified variables always meant to denote real numbers, etc.) > > Philosophical conundrum: are non existence things in the nulset? > No. In Free Logic we actually can deal with such things. If there exists such a thing as t, we write E!t And we might adopt the following Rule of Atomic Denotation: A(t) where A(t) is atomic ---- E!t Now we might actually do set theory in a framework of free logic. Then we would have t e u |- E!t. With other words, if something is an element of a set, it exists. The other way round: ~E!t |- t !e u for any set u. Hence, if u is the empty set, 0, we have: ~E!t |- t !e 0. or |- ~E!t -> t !e 0. If t doesn't exist, it's not an element of the empty set. > > 0 = {x | x =/= x} > Right. > > if x doesn't exist, does x =/= x ? > Yes. But it's still not an element of 0. (You surely know that native comprehension usually won't work in a properly defined set theory. ;-) F. ==== Let d be a positive integer. What are the sufficient and necessary condition that the Pell's equation x^2-dy^2=-1 has integer solutions? It's easy to see that there are no integer solutions when d is perfect square or when d contain a prime divisor of the form 4k+3. But what positive integer value of d give rises to integers solutions? ==== sinisa escribi.97 en el mensaje|n22401dcf.0307130403.6994fe3d@posting.google.com: > Let d be a positive integer. What are the sufficient and necessary > condition that the Pell's equation x^2-dy^2=-1 has integer solutions? > It's easy to see that there are no integer solutions when d is perfect > square or when d contain a prime divisor of the form 4k+3. But what > positive integer value of d give rises to integers solutions? It must be a complex question. At least it seems yet from the title of the Lagarias, J.C. On the Computational Complexity of Determining the Solvability or Unsolvability of the Equation X^2-DY^2=-1. Trans. Amer. Math Soc. 260, 485-508, 1980. Althought some things can be easily saw. By example, If d is prime and equal to 1 (mod 4) the equation has solutions. If d has some factor equal to 3 (mod 4) there aren't solutions. If d = 0 (mod 4) there aren't solutions. Then, in order to exist solutions, d must be equal to 1, 2 or 5 (mod 8) and hasn't factors equals to 3 (mod 4). Furthermore, studuing the length of the period of the develop in continued fraction of Sqrt(d), is easy to see that: If d=m^2+1, there are solutions. If d=m^2 + k (2m/k integer and k > 1), there aren't solutions. Ifi d=(m+1)^2 - k, m>1 and 2(m+1)/k integer, there aren't solutions. Also it can got conditions progresivaly more complex for d equal to (m^2 +a*2m) and ((m+1)^2-a*(m+1)), where a is a fraction as 2/3, 3/4, 2/5, 3/5, 4/5, .... In other way, if d can be expressed as d = ((4h^2 + 1)k + h)^2 + (4hk + 1)) the period length is 3 and there are solutions. But for a any large d, the more easy way to decide if there are solutions or not must be develop Sqrt(d) in continued fraction and see if the length of the period is odd or even. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== > Let d be a positive integer. What are the sufficient and > necessary conditions that the Pell's equation x^2-dy^2=-1 > has integer solutions? Many sufficient conditions are known. See: A Numerical Investigation of the Diophantine Equation x^2 - dy^2 = -1 R. D. Beach and H. C. Williams Department of Computer Science, University of Manitoba which is in Congressus Numerantium VI Proceedings of the Third Southeastern Conference of Combinatorics, Graph Theory, and Computing Florida Atlantic University Boca Raton February 28 - March 2, 1972 Edited by F. Hoffman, R. B. Levow, and R. S. D. Thomas Utilitas Mathematica Publishing Inc. P. O. Box 7, University Centre University of Manitoba Winnipeg, Manitoba, Canada R3T 2N2 ISBN 0 919628 06 0 Gives tables of values of d for which x^2 - dy^2 = -1 has a solution, other than those that can be predicted by a set of criteria that are listed. One criterion listed is when d = p and p == 1 (mod 4). In fact, if d is a prime, then x^2 - dy^2 = -1 has a solution if and only if p is 2 or p==1 (mod 4). Another criterion is d=pq, where p==1 mod 4, q==1 mod 4, and (p/q)=-1. Here, x^2 - dy^2 = -1 has a solution when the criterion is met, but there are solutions to (1) for some pq that do not meet this criterion. Yet another criterion is d=pq, where p==1 mod 4, q==1 mod 4, and (p/q)=1, (p/q)_4=(q/p)_4=-1. (No other criteria are listed for pq.) Some d of the form pq for which the equation has a solution, but not covered by the above criteria are 2305, 2605, 2705, 5305, 5713, 6757, 6953, 8005, 8653, 9305, 9953, 10001, 10405, 10817, 10961, 11029, 11629, 12181, 12961, 13105, 14305, 15529, 16601, 17305, 17693, 20005. A criterion given for d=2p is p==5 (mod 8). Another criterion for d=2p is p==9 (mod 16) and (2/p)_4 = -1. A few d's of the form 2p for which the equation has a solution, but not covered by the above criteria are 226, 2402, 3202, 3554, 4226, 6178, 6242, 6274, 6626, 7522. [(*/*) is the Legendre symbol, and (*/*)_4 is the biquadr symbol) Other pq: 5305 - 5, 1061 5713 - 29, 197 6757 - 29, 233 6953 - 17, 409 8005 - 5, 1601 2p: 1. p==5 (mod 8) 2. p==9 (mod 16), (2/p)_4=-1 John Robertson ==== Of course the main criterion is whether or not the length of the period of the continued fraction expansion of d is odd or even. John Robertson ==== Dear All, Sorry I can not solve this, as what I have learnt in school seems that I have given them back to my teacher :..) It is about my project: 3 possible workloads, running on 4 processors, how many combinations (not permutation, as it is 3^4) are there? What is the formula for this? e.g. 111 112 122 123 111 222 223 331 332 333 Many thanks!! Jialin ==== > Dear All, > > Sorry I can not solve this, as what I have learnt in school seems that > I have given them back to my teacher :..) > > It is about my project: > > 3 possible workloads, running on 4 processors, > > how many combinations (not permutation, as it is 3^4) are there? > What is the formula for this? If you mean you have 4 processors, each of which can be running one of 3 tasks, then you will use the multiplication principle. 3*3*3*3 = 3^4. If you wish to use combinations in this problem, then there is some information missing, but it isn't clearly stated (in my mind). Note: your example below does not have an obvious connection with the problem stated above. > e.g. > 111 > 112 > 122 > 123 > 111 > 222 > 223 > 331 > 332 > 333 > > > Many thanks!! > Jialin > -- Will Twentyman ==== www.YeOldeCoffeeShoppe.com has a great many visitors and sections for talk, tv, reviews and romance. Now add free, open to anyone to post but signing up first is free and you can use an avatar. Dont let my avatar scare you off, I trained in the gym 3 days running before that photo! Herc -- www.winternet.com/~mikelr/flame76.html __ / / / / / / / / / / / /__/__ /________ ___________/ ==== > www.YeOldeCoffeeShoppe.com has a great many visitors and sections > for talk, tv, reviews and romance. but I've already got 1000s dropping in, all with nothing to read, just jump the buck, I'll have to pay a dozen people just to post something there for a few months otherwise, give the site a KICK, its a GREAT traffic getter a few minutes of telling people draws real crowds, but without the first couple of people standing round the busker nooone will gather. And if you like usenet you'll love forums. Herc ==== >just jump the buck, Cool band name. >without the first couple of people standing round the >busker nooone will gather. Cool album name. --Sean http://www.livejournal.com/users/spclsd223/ ==== > >just jump the buck, > > Cool band name. > >without the first couple of people standing round the >busker nooone will gather. > > Cool album name. > > --Sean > http://www.livejournal.com/users/spclsd223/ Dude. I love it when two worlds within my field of interest collide and celebrate with each other. Do you like Standing Outside Imaginations Door as a book title? ps need some writing talent at www.Yeoldecoffeeshoppe.com Herc ==== >Dude. >I love it when two worlds within my field of interest collide and celebrate with each other. Quoting my journal?! My plan is working perfectly! >Do you like Standing Outside Imaginations Door as a book title? No, but 'Standing Outside Imagination's Door' would be cool. Even better would be 'Ringing the Bell of Life and Running'. >ps need some writing talent at www.Yeoldecoffeeshoppe.com I'll let you know as soon as I have the motivation and time to actually look at the site and see what the hell it is. --Sean http://www.livejournal.com/users/spclsd223/ ==== Task: Put n different points so that there are as many lines with 3 points on it as possible. (example for 7 points - the familiar configuration * * * * * * * with 6 lines) Does it help allowing higher dimensions (Leech lattice and all that funny stuff)? Can one even overcome the Sylvester theorem (there will always be a line with only 2 points) - the proof in the 2D case was a metric one after all. -- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the address For spamming, remove anything else ==== For a given point (x,y) lying on a cartesian plane what would the vertecies of the corners of a regular ploygon be? (In the simple case I'm assuming that the 'base' of the ploygon is parralel to the x-axis.) If the baseline is then rotated what adjustment would need to be made to the fixed points identifed above? (I concede that if then drawing the lines on a computer screen Bresnham or simmilar would be used for actual drawing.) ii) There are 4 towns a,b,c,d located on a flat plane. A highway has been proposed between the two major towns a and b. and with only these towns it would be a straight line. Towns c and d are perpendicular to AB and it is proposed that the highway routing should also serve thse communities (which are comparitivly smaller than A or B) c and and d are not of equal size, but are the same distance from the highway. reflects the relative'size/traffic/distance from highway #' of c and d. Ideally the engineers want an ideal route. (I'm assuimng they would have access to likley traffic usage.) What model could be used to determine the routing of the new highway? iii) What would be a good model to use when attempting to plan a timetiablke for an out and back metro line? (Assuming an unequal traffic distribution over time.) FMNT80 ==== > >For a given point (x,y) lying on a cartesian plane what >would the vertecies of the corners of a regular ploygon be? Um, that depends on which polygon you want. How many sides, is (x,y) the center or one of the vertices, how long are the sides (or how far is it from the center to the vertices...) I mean really, if you're going to post homework problems you could at least include the statement of the problem... >(In the simple case I'm assuming that the 'base' of the ploygon >is parralel to the x-axis.) > >If the baseline is then rotated what adjustment would need to be >made to the fixed points identifed above? > >(I concede that if then drawing the lines on a computer screen >Bresnham or simmilar would be used for actual drawing.) > > >ii) > >There are 4 towns a,b,c,d located on a flat plane. > >A highway has been proposed between the two major towns a and b. >and with only these towns it would be a straight line. > >Towns c and d are perpendicular to AB and it is proposed >that the highway routing should also serve thse communities >(which are comparitivly smaller than A or B) >c and and d are not of equal size, but are the same distance >from the highway. > >reflects the relative'size/traffic/distance from highway >#' of c and d. > >Ideally the engineers want an ideal route. >(I'm assuimng they would have access to likley traffic usage.) >What model could be used to determine the routing of the new highway? > >iii) >What would be a good model to use when attempting to plan a >timetiablke >for an out and back metro line? (Assuming an unequal traffic >distribution over time.) > > >FMNT80 ************************ David C. Ullrich ==== Why are certain polyhedra that are self intersecting considered REAL polyhedra? Clearly polyhedrons which are self intersecting shouldn't be considered REAL polyhedrons. Then we could do all sorts of weird stuff to them. Take a look at http://mathworld.wolfram.com/GreatDodecahedron.html and http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for example. Why do we consider figures which intersect themselves to be polyhedrons or polyhedra in the first place? Doesn't that kinda violate the rules, about being able to be constructed as if from boundaries from algebraic expressions, and such? This is what I don't get about this weird geometry mumbo jumbo about things which are self intersecting being considered actual shapes. What is wrong with these theoretical geometricians? (...Starblade Riven Darksquall...) ==== > Why are certain polyhedra that are self intersecting considered REAL > polyhedra? Clearly polyhedrons which are self intersecting shouldn't > be considered REAL polyhedrons. Then we could do all sorts of weird > stuff to them. Take a look at > http://mathworld.wolfram.com/GreatDodecahedron.html and > http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for > example. > > Why do we consider figures which intersect themselves to be > polyhedrons or polyhedra in the first place? Doesn't that kinda > violate the rules, about being able to be constructed as if from > boundaries from algebraic expressions, and such? > > This is what I don't get about this weird geometry mumbo jumbo about > things which are self intersecting being considered actual shapes. > What is wrong with these theoretical geometricians? > > (...Starblade Riven Darksquall...) Some work I've done with polygons started out with a restriction that they be convex polygons. The results I ended up with were easier to work with if I dropped that restriction and allowed them to be convex or self intersecting. All I was doing was working with the vertices, and dropping the restrictions allowed me to do less work to prove my result. I suspect there are similar reasons for working with self-intersecting polyhedra. -- Will Twentyman ==== > Why are certain polyhedra that are self intersecting considered REAL > polyhedra? See Imre Lakatos' _Proofs and Refutations_ for a variety of detailed views about this exact question. Question to all - am I much mistaken in thinking that nowadays, most mathematicians' views on such issues are of the whatever genre? cdj Clearly polyhedrons which are self intersecting shouldn't > be considered REAL polyhedrons. Then we could do all sorts of weird > stuff to them. Take a look at > http://mathworld.wolfram.com/GreatDodecahedron.html and > http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for > example. > > Why do we consider figures which intersect themselves to be > polyhedrons or polyhedra in the first place? Doesn't that kinda > violate the rules, about being able to be constructed as if from > boundaries from algebraic expressions, and such? > > This is what I don't get about this weird geometry mumbo jumbo about > things which are self intersecting being considered actual shapes. > What is wrong with these theoretical geometricians? > > (...Starblade Riven Darksquall...) ==== there's a much-simpler example, but i can't recall the particulars. last that I saw it was in Stillman's book on 3-manifolds, which is introductory. was it a tetrahedron? > Why are certain polyhedra that are self intersecting considered REAL > polyhedra? Clearly polyhedrons which are self intersecting shouldn't > be considered REAL polyhedrons. Then we could do all sorts of weird > stuff to them. Take a look at > http://mathworld.wolfram.com/GreatDodecahedron.html and > http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for > example. --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac ==== > Why are certain polyhedra that are self intersecting considered REAL > polyhedra? > > See Imre Lakatos' _Proofs and Refutations_ for a variety of detailed > views about this exact question. And where will I find that? > > Question to all - am I much mistaken in thinking that nowadays, most > mathematicians' views on such issues are of the whatever genre? > How can they not care? Isn't it their job to, you know, care? > > cdj > Alright. > > > > Clearly polyhedrons which are self intersecting shouldn't > be considered REAL polyhedrons. Then we could do all sorts of weird > stuff to them. Take a look at > http://mathworld.wolfram.com/GreatDodecahedron.html and > http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for > example. > > Why do we consider figures which intersect themselves to be > polyhedrons or polyhedra in the first place? Doesn't that kinda > violate the rules, about being able to be constructed as if from > boundaries from algebraic expressions, and such? > > This is what I don't get about this weird geometry mumbo jumbo about > things which are self intersecting being considered actual shapes. > What is wrong with these theoretical geometricians? > > (...Starblade Riven Darksquall...) (...Starblade Riven Darksquall...) ==== > Why are certain polyhedra that are self intersecting considered REAL > polyhedra? > > See Imre Lakatos' _Proofs and Refutations_ for a variety of detailed > views about this exact question. > > And where will I find that? Any decent library. -- ==== I am trying to derive a very simple equation but for some reason I am having trouble with it. I want to derive the equation S = So + vo*(t-to) + 1/2*a*(t-to)^2 where o = not I can derive the equation if I dont use 't' and 'to' but just use 't' to represent delta t. Any help? ==== Differentiate S with respect to t, dS/dt = vo + a*(t-to) Michael Leung ???????:d93c7061.0307252129.29ebe8b0@posting.google.com... > I am trying to derive a very simple equation but for some reason I am > having trouble with it. > > I want to derive the equation > > S = So + vo*(t-to) + 1/2*a*(t-to)^2 > where o = not > > I can derive the equation if I dont use 't' and 'to' but just use 't' > to represent delta t. > > Any help? ==== This argument works. We know that v = ds/dt = vo + at = (vo + ato) + (at - ato) = (vo + ato) + a(t - to)-- just rename vo + ato by v'o IE ds/dt = v'o + a(t - to). Integrating gives S= So +v'ot + (1/2)a(t - to)^2 which is your desired result. > I am trying to derive a very simple equation but for some reason I am > having trouble with it. > > I want to derive the equation > > S = So + vo*(t-to) + 1/2*a*(t-to)^2 > where o = not > > I can derive the equation if I dont use 't' and 'to' but just use 't' > to represent delta t. > > Any help? ==== Oops, it is not your desired result. > This argument works. > We know that v = ds/dt = vo + at = (vo + ato) + (at - ato) = (vo + ato) + > a(t - to)-- just rename vo + ato by v'o > IE ds/dt = v'o + a(t - to). Integrating gives S= So +v'ot + (1/2)a(t - to)^2 > which is your desired result. > I am trying to derive a very simple equation but for some reason I am > having trouble with it. > > I want to derive the equation > > S = So + vo*(t-to) + 1/2*a*(t-to)^2 > where o = not > > I can derive the equation if I dont use 't' and 'to' but just use 't' > to represent delta t. > > Any help? > > ==== If a is constant: a = (v- vo)/( t - to)=>a( t- to) = v - vo => v = vo + a( t - to) => S = So + vot + (1/2)a( t - to )^2 = (So + voto) + vo( t - to) + (1/2)a( t - to)^2. Now let S'o = So + voto. > I am trying to derive a very simple equation but for some reason I am > having trouble with it. > > I want to derive the equation > > S = So + vo*(t-to) + 1/2*a*(t-to)^2 > where o = not > > I can derive the equation if I dont use 't' and 'to' but just use 't' > to represent delta t. > > Any help? ==== Does anybody know if the following problem has been solved, or if any progress has been made towards a solution? Let k be a positive integer. Let G(k) be the least n such that all sufficiently large positive integers are the sum of <= n kth powers. Let H(k) be the least n such that (the set of positive integers which can not be expressed as the sum of <= n kth powers) has Schnirelmann density 0. Does G(k) = H(k) ? Paul Epstein ==== Suppose you have a finite abelian cyclic group . Is there a canonical way (or any way, really) to express this alternatively in terms of two generators and two relations as: Generators = {g,h} Relations = {ag=0, bh=cg} where a, b, and c are integers. ? If so, this would be very useful to me, and I'd appreciate it if anyone could post with a way to do this (if it indeed can be done). Is there any kind of _Mathematica_ or Maple command that does this automatically, perhaps? (say once you enter the generator of the cyclic group and its order) Or perhaps if it can be done in certain special cases, maybe someone could point to those? ==== > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, with generator (1,1) as prototype of theorem. That should work unless order of group is power of prime. ==== > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > > ? > It's not clear to me exactly what you're given in this problem. If by this you mean we are given n, and told only that the group G = , where g and h are unspecified elements of G, is isomorphic to Z_n; then it's not clear that a non-trivial presentation of the given form exists. By non-trivial here, I mean a presentation where neither = G nor = G; clearly, is always going to give a representation of Z_a, but that's probably not what you want. Consider the cyclic group Z_6. We must select g and h from {2,3,4}, since otherwise one of g or h will generate Z_6 on its own; so either {g,h} = {2,3} or {g,h} = {3,4}. If we take g = 2 or 4, then the only possible presentation of your given form is ; if we take g = 3, then . But these are both equivalent to . There is a homomorphism from this group onto D_3 (dihedral group) with presentation , and onto Z_6 with presentation ; so it would seem there is no non-trivial representation of Z_6 of your form. > If so, this would be very useful to me, and I'd appreciate it if > anyone could post with a way to do this (if it indeed can be done). Is > there any kind of _Mathematica_ or Maple command that does this > automatically, perhaps? (say once you enter the generator of the > cyclic group and its order) > > Or perhaps if it can be done in certain special cases, maybe someone > could point to those? ==== > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > > It's not clear to me exactly what you're given in this problem. See my other post where I try to work this out in general, starting from the hints I gave in my first post. In working thru this, the reason I disallowed the order of the group to be a power of a prime is elucidated. More comments below. > If by this you mean we are given n, and told only that the group G = > , where g and h are unspecified elements of G, is isomorphic to > Z_n; then it's not clear that a non-trivial presentation of the given > form exists. > > By non-trivial here, I mean a presentation > where neither = G nor = G; clearly, is > always going to give a representation of Z_a, but that's probably not > what you want. > > Consider the cyclic group Z_6. We must select g and h from {2,3,4}, > since otherwise one of g or h will generate Z_6 on its own; so either > {g,h} = {2,3} or {g,h} = {3,4}. > > If we take g = 2 or 4, then the only possible presentation of your > given form is ; if we take g = 3, then 2g = 0, 2g = 3h>. > And h = 2 or h = 4. Don't take h = 2, take h = 4, as in accordance to my work in the other post. Thus Z_6 = = <-1>, does it not? > But these are both equivalent to . There is a > homomorphism from this group onto D_3 (dihedral group) with > presentation , and onto Z_6 with > presentation ; so it would seem > there is no non-trivial representation of Z_6 of your form. > ==== > > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > > I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, with > generator (1,1) as prototype of theorem. That should work unless order of > group is power of prime. > Let G be a finite cyclic group with order n. Assume n > 1 isn't a power of a prime. Thus some cofinite j,k > 1 with n = rs Let (1,0) and (0,1) be generators of Z/r & Z/s. r(1,0) = (0,0) = s(0,1) If I've done this right, G = <(1,1)> = <(1,0) + (0,1)> is a cyclic group of order n with generator (1,1) Oh, if you want it in the - form then use (1,1) = (1,0) - (0,s-1) Whence (0,s-1) is a generator of Z/s and r(1,0) = (0,0) = s(0,s-1) When n the order of G is a power of a prime, then be contented with trivial solution, g = 0, h = 1, G = <0-1> = <-1>, 1g = 0 = nh. ==== I am looking for software to carry out primality testing of positive integers. I have the following requirements: 1) It must be an exact test. Probabilistic tests commonly used in cryptographic applications are not an option. 2) It must be able to tackle integers up to about one thousand decimal digits long. 3) It should return an answer within a reasonable time when running ona modern PC - say, at worst no more than a few hours for any arbitrary integer as above. Any pointers would be much appreciated. ==== >I am looking for software to carry out primality testing of positive >integers. I have the following requirements: > > 1) It must be an exact test. Probabilistic tests commonly used in >cryptographic applications are not an option. > > 2) It must be able to tackle integers up to about one thousand decimal >digits long. > > 3) It should return an answer within a reasonable time when running ona >modern PC - say, at worst no more than a few hours for any arbitrary integer as >above. If you perform an exact primality test on a computer there is always a positive probability that the answer will be wrong because of a random hardware error, cosmic rays, etc. A probabilistic test that says n is prime with probability p can hence be just as reliable as an exact test, depending on the value of p... > Any pointers would be much appreciated. ************************ David C. Ullrich ==== > >I am looking for software to carry out primality testing of positive >integers. I have the following requirements: > > 1) It must be an exact test. Probabilistic tests commonly used in >cryptographic applications are not an option. > > 2) It must be able to tackle integers up to about one thousand decimal >digits long. > > 3) It should return an answer within a reasonable time when running ona >modern PC - say, at worst no more than a few hours for any arbitrary integer as >above. > > If you perform an exact primality test on a computer there is always a > positive probability that the answer will be wrong because of a > random hardware error, cosmic rays, etc. A probabilistic test > that says n is prime with probability p can hence be just as > reliable as an exact test, depending on the value of p... > David's point is a good one. Apart from exhaustively testing an implementation with all possible inputs of length less than 1000 digits, I see no way to prove an implementation attains your goal of establishing primality in the given time limit of no more than a few hours. Bear in mind that assuming a generalized Riemann hypothesis in 1976, Miller showed that testing strong pseudoprimality for O((log N)^2) bases is deterministic. For this reason the tests commonly used in cryptographic applications might be said to be probabilistic in name only. A good summary of primality testing algorithms is provided by the Primes in P FAQ found here: http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_FAQ.html Here's a page with links to various primality testing programs: http://directory.google.com/Top/Science/Math/Number_Theory/Prime_Numbers/Pri mality_Tests/Primality_Proving/Software/ I've used the PRIMO program by Marcel Martin, based on the ECPP algorithm. This approach is attractive for many purposes because (for prime inputs) the algorithm outputs a primality certificate whose validity is much easier to check in practice than running the ECPP implementation itself. In my experience the software handles inputs of a couple of hundred digits very quickly, but I have no experience in the range of inputs with several hundred digits contemplated by you. (replace bellsouth by mpinet and .net by .com to reply) ==== > > I am looking for software to carry out primality testing of positive > integers. I have the following requirements: > > 1) It must be an exact test. Probabilistic tests commonly used in > cryptographic applications are not an option. > > 2) It must be able to tackle integers up to about one thousand decimal > digits long. > > 3) It should return an answer within a reasonable time when running ona > modern PC - say, at worst no more than a few hours for any arbitrary integer as > above. > > Any pointers would be much appreciated. Look at PRIMO - Primality proving using elliptic curves. http://www.ellipsa.net/ Hugo Pfoertner ==== > > >I am looking for software to carry out primality testing of positive >integers. I have the following requirements: > > 1) It must be an exact test. Probabilistic tests commonly used in >cryptographic applications are not an option. > > 2) It must be able to tackle integers up to about one thousand decimal >digits long. > > 3) It should return an answer within a reasonable time when running ona >modern PC - say, at worst no more than a few hours for any arbitrary > integer as >above. > > If you perform an exact primality test on a computer there is always a > positive probability that the answer will be wrong because of a > random hardware error, cosmic rays, etc. A probabilistic test > that says n is prime with probability p can hence be just as > reliable as an exact test, depending on the value of p... > > > David's point is a good one. Apart from exhaustively testing > an implementation with all possible inputs of length less than > 1000 digits, I see no way to prove an implementation attains > your goal of establishing primality in the given time limit of no > more than a few hours. > > Bear in mind that assuming a generalized Riemann hypothesis > in 1976, Miller showed that testing strong pseudoprimality for > O((log N)^2) bases is deterministic. > > For this reason the tests commonly used in cryptographic > applications might be said to be probabilistic in name only. > > A good summary of primality testing algorithms is provided by > the Primes in P FAQ found here: > > http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_FAQ.html > > Here's a page with links to various primality testing programs: > > http://directory.google.com/Top/Science/Math/Number_Theory/Prime_Numbers/Prim ality_Tests/Primality_Proving/Software/ > > I've used the PRIMO program by Marcel Martin, based on the > ECPP algorithm. This approach is attractive for many purposes > because (for prime inputs) the algorithm outputs a primality > certificate whose validity is much easier to check in practice > than running the ECPP implementation itself. In my experience > the software handles inputs of a couple of hundred digits very > quickly, but I have no experience in the range of inputs with > several hundred digits contemplated by you. > > (replace bellsouth by mpinet and .net by .com to reply) To give an example, PRIMO takes 53min CPU time on an 800 MHz PC to prove (5^1301-2^1301)/3 (908 decimal digits) prime. A strong probable prime test as used in PFGW http://www.primeform.net/openpfgw/ takes 0.49s on the same PC: Primality testing (5^1301-2^1301)/3 [N-1, Brillhart-Lehmer-Selfridge] (5^1301-2^1301)/3 is PRP! (0.490000 seconds) The probability that a random 1000 digit PRP is composite is 1.2*10^(-123) (from http://www.utm.edu/research/primes/notes/prp_prob.html ) So David Ullrich's comment on hardware reliablility is really justified. Hugo Pfoertner ==== > >> >>I am looking for software to carry out primality testing of positive >>integers. I have the following requirements: >> >> 1) It must be an exact test. Probabilistic tests commonly used in >>cryptographic applications are not an option. >> >> 2) It must be able to tackle integers up to about one thousand decimal >>digits long. >> >> 3) It should return an answer within a reasonable time when running ona >>modern PC - say, at worst no more than a few hours for any arbitrary >integer as >>above. >> >> If you perform an exact primality test on a computer there is always a >> positive probability that the answer will be wrong because of a >> random hardware error, cosmic rays, etc. A probabilistic test >> that says n is prime with probability p can hence be just as >> reliable as an exact test, depending on the value of p... >> > >David's point is a good one. Apart from exhaustively testing >an implementation with all possible inputs of length less than >1000 digits, I see no way to prove an implementation attains >your goal of establishing primality in the given time limit of no >more than a few hours. > >Bear in mind that assuming a generalized Riemann hypothesis >in 1976, Miller showed that testing strong pseudoprimality for >O((log N)^2) bases is deterministic. Couldn't figure out what that sentence meant at first - the next sentence clarified it. Is the result that checking just any old set of that many bases is conclusive, or that there exists such a set of bases? >For this reason the tests commonly used in cryptographic >applications might be said to be probabilistic in name only. > >A good summary of primality testing algorithms is provided by >the Primes in P FAQ found here: > >http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_FAQ.html > >Here's a page with links to various primality testing programs: > >http://directory.google.com/Top/Science/Math/Number_Theory/Prime_Numbers/Pr imality_Tests/Primality_Proving/Software/ > >I've used the PRIMO program by Marcel Martin, based on the >ECPP algorithm. This approach is attractive for many purposes >because (for prime inputs) the algorithm outputs a primality >certificate whose validity is much easier to check in practice >than running the ECPP implementation itself. In my experience >the software handles inputs of a couple of hundred digits very >quickly, but I have no experience in the range of inputs with >several hundred digits contemplated by you. > >(replace bellsouth by mpinet and .net by .com to reply) > ************************ David C. Ullrich ==== David C. Ullrich escribi.97 en el mensaje|n7fl2hvcbbdl0c0pem1fdg8jvmsm0ap4clo@4ax.com: > >> > >> I am looking for software to carry out primality testing of >> positive >> integers. I have the following requirements: >> >> 1) It must be an exact test. Probabilistic tests commonly used in >> cryptographic applications are not an option. >> >> 2) It must be able to tackle integers up to about one thousand >> decimal >> digits long. >> >> 3) It should return an answer within a reasonable time when >> running ona >> modern PC - say, at worst no more than a few hours for any >> arbitrary >> integer as >> above. > > If you perform an exact primality test on a computer there is > always a > positive probability that the answer will be wrong because of a > random hardware error, cosmic rays, etc. A probabilistic test > that says n is prime with probability p can hence be just as > reliable as an exact test, depending on the value of p... > >> >> David's point is a good one. Apart from exhaustively testing >> an implementation with all possible inputs of length less than >> 1000 digits, I see no way to prove an implementation attains >> your goal of establishing primality in the given time limit of no >> more than a few hours. >> >> Bear in mind that assuming a generalized Riemann hypothesis >> in 1976, Miller showed that testing strong pseudoprimality for >> O((log N)^2) bases is deterministic. > > Couldn't figure out what that sentence meant at first - the > next sentence clarified it. Is the result that checking just > any old set of that many bases is conclusive, or that there > exists such a set of bases? > ==== >David C. Ullrich escribi.97 en el >mensaje|n7fl2hvcbbdl0c0pem1fdg8jvmsm0ap4clo@4ax.com: >> > > [...] > > Bear in mind that assuming a generalized Riemann hypothesis > in 1976, Miller showed that testing strong pseudoprimality for > O((log N)^2) bases is deterministic. >> >> Couldn't figure out what that sentence meant at first - the >> next sentence clarified it. Is the result that checking just >> any old set of that many bases is conclusive, or that there >> exists such a set of bases? >> > > >Millers Test: If the extended Riemann hypothesis is true, then if n is an >a-SPRP for all integers a with 1 < a < 2(log n)^2, then n is prime. Keen. ************************ David C. Ullrich ==== > >David C. Ullrich escribi.97 en el >mensaje|n7fl2hvcbbdl0c0pem1fdg8jvmsm0ap4clo@4ax.com: >> > > [...] > > Bear in mind that assuming a generalized Riemann hypothesis > in 1976, Miller showed that testing strong pseudoprimality for > O((log N)^2) bases is deterministic. >> >> Couldn't figure out what that sentence meant at first - the >> next sentence clarified it. Is the result that checking just >> any old set of that many bases is conclusive, or that there >> exists such a set of bases? >> > > >Millers Test: If the extended Riemann hypothesis is true, then if n is an >a-SPRP for all integers a with 1 < a < 2(log n)^2, then n is prime. > > Keen. > Yes. ==== > David's point is a good one. Apart from exhaustively testing > an implementation with all possible inputs of length less than > 1000 digits, I see no way to prove an implementation attains > your goal of establishing primality in the given time limit of no > more than a few hours. > Are you testing an implementation of an algorithm or are you testing a number for primality ? Algorithms are not proven with all possible inputs but proven primes are tested will all necessary imputs...with the algorithm itself not usually in question. > Bear in mind that assuming a generalized Riemann hypothesis > in 1976, Miller showed that testing strong pseudoprimality for > O((log N)^2) bases is deterministic. > For this reason the tests commonly used in cryptographic > applications might be said to be probabilistic in name only. ==== > David's point is a good one. Apart from exhaustively testing > an implementation with all possible inputs of length less than > 1000 digits, I see no way to prove an implementation attains > your goal of establishing primality in the given time limit of no > more than a few hours. > > > Are you testing an implementation of an algorithm or are you testing a > number for primality ? Algorithms are not proven with all possible inputs > but proven primes are tested will all necessary imputs...with the algorithm > itself not usually in question. > > > Bear in mind that assuming a generalized Riemann hypothesis > in 1976, Miller showed that testing strong pseudoprimality for > O((log N)^2) bases is deterministic. > For this reason the tests commonly used in cryptographic > applications might be said to be probabilistic in name only. > If you're asking me, I was responding to the original poster's quest for primality testing software. See the OP's requirements, which were left in my response, specifying that inputs up to a thousand digits should be handled in no more than a few hours. In terse fashion I was making a case for Miller's test, involving at most 94 strong pseudoprime computations to prove primality in the given range, as being more deterministic with respect to running time than ECPP. However the sufficiency of Miller's test rests on the extended Riemann hypothesis, and ECPP has no such theoretical dependence. My brevity was no doubt confusing. It seemed likely to me that the OP knew about strong pseudoprime tests and was explicitly rejecting them, so in that case one of the ECPP implementations would be my best suggestion. -- chip ==== > David's point is a good one. Apart from exhaustively testing > an implementation with all possible inputs of length less than > 1000 digits, I see no way to prove an implementation attains > your goal of establishing primality in the given time limit of no > more than a few hours. > > Are you testing an implementation of an algorithm or are you testing a > number for primality ? Algorithms are not proven with all possible inputs > but proven primes are tested will all necessary imputs...with the > algorithm > itself not usually in question. > > If you're asking me, I was responding to the original poster's quest > for primality testing software. See the OP's requirements, which > were left in my response, specifying that inputs up to a thousand > digits should be handled in no more than a few hours. > Yeah, I was responding to your wording of a situation. I guess it can be ====