mm-16 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?Very nice site. Some topics there (that is topics selection) make mebelieve that author is not (classical) Mathematician but rather controlsystems theorist.Goran 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. AugustineYou can always check out the whois entry:Registrant:MathPages (MATHPAGES-DOM) MATHPAGES.COM Administrative Contact: Brown, Kevin (KB11700) ksbrown@SEANET.COM MathPages 6014 S 238TH PL APT D101 KENT, WA 98032-3771 US (253) 854-2063 fax: 999 999 9999 Technical Contact: Hostmaster, Support (SH12005) hostmaster@GTE-HOSTING.NET P.O.Box 152212 Irving, TX 75015-2212 US 1-800-483-4678 fax: 123 123 1234 Record expires on 08-Aug-2012. Record created on 14-Oct-2002. Domain servers in listed order: NS05A.WEBHOSTING-VERIZON.NET 209.238.3.50 NS05B.WEBHOSTING-VERIZON.NET 209.238.3.51 I recently posted a message to newbies>Now I wish to exhort oldies not to work homework problems for peopleExCUSE me, but I prefer the term knowbies for the yang to the newbiesfi 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 newbiesfi yin.> dave> So if onefis just graduated from college is he a newbie knowbie,or a knowbie noveau?*dodges tomatoes* :-)-- #191, ewill3@earthlink.netItfis 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 newbiesfi yin.Certainly it has to be at least oldbies, not oldies.I agree with the exhortation, BTW. Not because I particularlycare if somebody somewhere gets an unearned A; that isnfit reallymy problem. Ifim concerned, rather, that the more people go alongwith it, the more requests of this sort wefill get. Certainly it has to be at least oldbies, not oldies.Thereby creating a grammatical rule for changing adjectives intonouns. Maybe the rule would apply only to people. Or people anddogs, 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 isnfit really> my problem. Ifim concerned, rather, that the more people go along> with it, the more requests of this sort wefill get.This looks like a wonderful opportunity to start a new sub-group - alt.math.homework-help. If itfis getting to be that big a hassle, therefis 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 isnfit really> my problem. Ifim concerned, rather, that the more people go along> with it, the more requests of this sort wefill get.We is not a fixed entity.I, for one, hope the homework police findit difficult to regulate what people post. =Suppose f maps the reals to the reals, f( 0 ) = 0, and for all reala 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.caDepartment 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 donfit 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) = 1Similarly, 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 gidentically equal to 0.Your question is closely related to the one of the O.P., since ifg 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 offunction originally considered.Thus if g satisfies your conditions, then there exists a function fsatisfying 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 knowg(0)=g(0)g(x) for every x. So if g(0)<>0, then g(x)=1 for every x. Soif g is not identically 1, then g(0)=0. The constant function 1 doessatisfy your conditions.satisfy the original conditions if f satisfies the functionalequation 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, org(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 itfis one of these peculiar functionswhose graph is dense in the plane, and the graph of g is either densein the upper half plane y>=0, or in both the first quadrant x>=0, y>=0and 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 OPfis 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.Itfis 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. Ifill take your word for the rest which pointsout some grit in an otherwise smooth duality. A duality expressedwith continuous functions, yet applicable to discontinuous ones also.> g(x) = -g(-x) or g(x)=g(-x) for every x. Itfis 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 itfis 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.> combinatorial optimization problem inaimms 2 format. Unfortunately running this program takes a lot of time, so Iwant to try a more efficient solver like C-plex. However, in order to useC-plex I need to convert my aimms program to an mps format program. Cananyone tell me whether this is possible in aimms?Dion Bongaerts boundary=5F60FDBAB57B35E6585F0C83 =- -- messenger ..my brother is having abuilding constructed and called me on his cell.. hefis caught without histrigonometric function table book ! he needs to know the TANGENT of a 2o angle ! Please respond ASAP ..so I can call him back. remove yes> Please excuse my ignorance ! Ifim the messenger ..my brother is having a> building constructed and called me on his cell.. hefis caught without his> trigonometric function table book !> he needs to know the TANGENT of a 2o angle !> Please course> oh thank you ..thank you>This is pretty funny, actually, but anyway, here it is:0.034920769491747730500402625773725315879174297784615 approximately boundary=56E0B04682C893DE51A74A90 =I have a simple problem, but canfit come upwith a satisfactorily simple answer.Maybe the question misleads intuition:A mirror reverses right and left,but why doesnfit it reverse top and bottom? I have a simple problem, but canfit come up> with a satisfactorily simple answer.> Maybe the question misleads intuition:> A mirror reverses right and left,> but why doesnfit it reverse top and bottom?The mirrors I look into donfit reverse left and right;they reverse back and front.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen I have a simple problem, but canfit come up> with a satisfactorily simple answer.> Maybe the question misleads intuition:> A mirror reverses right and left,> but why doesnfit it reverse top and bottom?If you mean a ?t mirror, it doesnfit reverse either. It simply doesnfit do a reversal between left and right that occurs when we look at another person. Two people facing each other percieve left to be opposite directions. The mirror doesnfit do this, causing an apparent reversal.If you mean an actual reversal, the mirror must be curved as seen from above, but composed of vertical line segments.-- Will Twentyman > I have a simple problem, but canfit come up>> with a satisfactorily simple answer. Maybe the question misleads intuition: A mirror reverses right and left,>> but why doesnfit it reverse top and bottom? The mirrors I look into donfit reverse left and right;> they reverse back and front.Yes. That is the pithiest way I know to explain things. And themost correct.Wave your hand. The hand in the mirror that is on _your right_ wavesback.The virtual image in the mirror is front-to-back symmetric aboutthe plane of the mirror with the object being re?cted. It is amatter of psychology rather than physics that we look at the imageand describe it as having been left-to-right inverted (and thenrotated 180 degrees on a vertical axis (*)) rather than front-to-backinverted. Both descriptions, of course, yield identical results.A top to bottom inversion followed by a rotation on a horizontalaxis (**) parallel to the mirror is another description that works.So in some sense, the mirror _does_ reverse top and bottom. John Briggs(*) If you want to avoid doing translation in addition to rotation,put the vertical axis of rotation at the intersection of theplane of inversion and the plane of the mirror.(**) If you want to avoid doing translation in addition to rotation,put the horizontal axis at the intersection of the top-to-bottomplane of inversion and the plane of the mirror. I have a simple problem, but canfit come up with a satisfactorily simple> answer.> Maybe the question misleads intuition:> A mirror reverses right and left,> but why doesnfit it reverse top and bottom?There is no reversion at all, it is the same as reading the back of abook.In the sense that you yourself consider Oreversionfi, it is an effect thathappens in all directions.Write a sentence on the top of a piece of paper. Rotate it 90 degrees.Then, on, the new top, write the sentence again. Rotate the paper to theoriginal position. Now you have two sentences, one written horizontallyand the other vertically (including the letters).look at the paper from its back, or from a mirror. It is the same. Boththe horizontal and vertical sentences are ?pped in the same way. The keyhere is that looking at a ?t shape from the back is the same as lookingat it from the front in a mirror.-- Christos Dimitrakakis =...>> A mirror reverses right and left,>> but why doesnfit it reverse top and bottom?>> The mirrors I look into donfit reverse left and right;> they reverse back and front.> Yes. That is the pithiest way I know to explain things. And the> most correct.> Wave your hand. The hand in the mirror that is on _your right_ waves> back....Well, when I tried this, the hand in the mirror on _my left_ wavedback. How do you explain this discrepancy between your theory andthe experimental outcome??-jiw Wave your hand. The hand in the mirror that is on _your right_ waves> back.> ...> Well, when I tried this, the hand in the mirror on _my left_ waved> back. How do you explain this discrepancy between your theory and> the experimental outcome??Really??Did _you_ see the hand on the left side?Paint your left hand blue and your right hand red and wave with thered hand. Does the blue hand wave back?You are just interpreting the right hand in the mirror as the lefthand of the mirror person.Alois A mirror reverses right and left,> but why doesnfit it reverse top and bottom?Try this:Lie on your side or just tilt your head sideways and look at the mirror. Suddenly top and bottom are reversed.Felix > Wave your hand. The hand in the mirror that is on _your right_ waves>> back.> ...> Well, when I tried this, the hand in the mirror on _my left_ waved> back. How do you explain this discrepancy between your theory and> the experimental outcome??Without knowing the details of your experimental setup, I can proposeseveral possibilities:1. You are using one of those corner mirrors that re?ct twice, thuspresenting an image that is not inverted.2. You are contorting your body, crossing your arms or facingalong the axis of the mirror rather than into the mirror.Incompetence or fraud are also possible, of course. John Briggs > Wave your hand. The hand in the mirror that is on _your right_ waves>> back.>> ... Well, when I tried this, the hand in the mirror on _my left_ waved>> back. How do you explain this discrepancy between your theory and>> the experimental outcome??> Really??> Did _you_ see the hand on the left side?> You are just interpreting the right hand in the mirror as the left> hand of the mirror person.I think there is a simpler explanation for the observed evidence.James Waldby is left-handed.(You said Wave your hand. He waved his dominant hand.)-- Dave SeamanJudge Yohnfis mistakes revealed in Mumia Abu-Jamal ruling. Wave your hand. The hand in the mirror that is on _your right_ waves> back.> ...> Well, when I tried this, the hand in the mirror on _my left_ waved> back. How do you explain this discrepancy between your theory and> the experimental outcome??>> Really??>> Did _you_ see the hand on the left side?>> You are just interpreting the right hand in the mirror as the left>> hand of the mirror person.> I think there is a simpler explanation for the observed evidence.> James Waldby is left-handed.> (You said Wave your hand. He waved his dominant hand.)Omg. So I did. John Briggs Wave your hand. The hand in the mirror that is on _your right_ waves>back.>>...>>Well, when I tried this, the hand in the mirror on _my left_ waved>back. How do you explain this discrepancy between your theory and>the experimental outcome??>Really??>>Did _you_ see the hand on the left side?You are just interpreting the right hand in the mirror as the left>>hand of the mirror person.I think there is a simpler explanation for the observed evidence.>>James Waldby is left-handed.>>(You said Wave your hand. He waved his dominant hand.)>You donfit even have to assume that. Maybe hefis right-handed and just waved his left hand.Jon Miller > Wave your hand. The hand in the mirror that is on _your right_ waves>> back.> ...>> Well, when I tried this, the hand in the mirror on _my left_ waved> back. How do you explain this discrepancy between your theory and> the experimental outcome??> Without knowing the details of your experimental setup, I can propose> several possibilities:> 1. You are using one of those corner mirrors that re?ct twice, thus> presenting an image that is not inverted.> 2. You are contorting your body, crossing your arms or facing> along the axis of the mirror rather than into the mirror.> Incompetence or fraud are also possible, of course....No, no, Ifim sure I did the experiment in a competent andhonest manner. As Dave Seaman notes, I am left-handedand waved my left hand. Got an excellent laugh from theresponses!-jiw Wave your hand. The hand in the mirror that is on _your right_ waves> back.> ...>> Well, when I tried this, the hand in the mirror on _my left_ waved> back. How do you explain this discrepancy between your theory and> the experimental outcome??> Really??> Did _you_ see the hand on the left side?Yes, I really, really did. > Paint your left hand blue and your right hand red and wave with the> red hand. Does the blue hand wave back?> You are just interpreting the right hand in the mirror as the left> hand of the mirror person.Try the experiment like that yourself if you donfit believe me. I am left-handed and can tell left from right without needing to paint my hands.-jiw I have a simple problem, but canfit come up> with a satisfactorily simple answer.> Maybe the question misleads intuition:> A mirror reverses right and left,> but why doesnfit it reverse top and bottom?A mirror reverses front and back; because we are symmetrical beings, itlooks like youfive reversed left and right.Put a glove on just one of your hands. Look in the mirror. Your head iswhere the mirror imagefis head is, your feet are where the mirror imagesfeet are; your gloved hand is where the mirror imagefis gloved hand is,and your ungloved hand is where the mirror imagefis ungloved hand is - nocontradiction involved!The weird thing would be if you head and feet matched up with the mirrorimagefis corresponding parts, but your gloved hand was matched up withthe imagefis _un_gloved hand!C Brown Systems DesignsMultimedia Environments for Museums and Theme Parks = Some of the readers of this list might be interested in the following bookPierre Baldi, Paolo Frasconi, and Padhraic Smyth, Modeling theISBN: 0-470-84906-1.It covers various models and algorithms for the Web includinggenerative models of networks, IR and machine learning algorithms fortext analysis, link analysis, focused crawling, methods for modelinguser 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. Commerceon the Web: Models and Applications - Appendix A MathematicalComplements - Appendix B List of Main Symbols and AbbreviationsThe webpage http://ibook.ics.uci.edu/ contains more details, ahyperlinked bibliography, and a sample chapter in pdf.Paolo Frasconihttp://www.dsi.unifi.it/~paolo/ =Jean-Paul Allouche and I are pleased to announce the publicationof our book (AS)^2, also known as Automatic Sequences: Theory, Applications, Generalizationscurrently 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 andtheoretical physics. The book has 571+xvi pages, 1600 citations to theliterature, 460 exercises, 85 open problems, 1 musical score, and 2jokes in the index. It will be of interest to number theorists andtheoretical computer scientists.The web page for the book, http://www.math.uwaterloo.ca/~shallit/asas.htmlhas a table of contents and other information.Jeffrey Shallit, Computer Science, University of Waterloo,Waterloo, Ontario N2L 3G1 Canada shallit@graceland.uwaterloo.caURL = 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 Complexplane.[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 waysItfis still legal to go .sigless. > 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 donfit think so:> We donfit 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 myquestion):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 xtransformed by Z1 must approach x transformed by Z2. Where might Ireasonably expect to find Zx? Well, half-way between ABx and BAx.Ifive tested a few numbers, and whatfis interesting is that Zx _doesnfit_ seemto converge to a linear combination of ABx and BAx, in general, although(Zx, ABx) does seem to equal (Zx, BAx).David Turner ...>> 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 xfis 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 xfis 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 withmy afternoon):In[27]:= WilsonRoot[2.0]; WilsonRoot[3.0]; WilsonRoot[4.0];4.154444.562154.81616Ifive checked the first one in Mathematica, and I presume(read Ohopefi) that the others are right as well.-- Dave TaylorOh yeah? Well Ifill build my own theme park! With Blackjack, andhookers ... 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 xfis 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 xfis 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-derivedgeneralized prime.(snicker.)2) What if we, for x = composite, allow the integers (that theequations above are equal to) to be Gaussian?2.5) Is (Gamma(x)+1)/x, if we let x = a Gaussian (integer) prime, an (real orGaussian)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 isnfit 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 iapproximately.Robert Israel israel@math.ubc.caDepartment 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> Ifive checked the first one in Mathematica, and I presume> (read Ohopefi) 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 = 4.1544390602710626948272615836646291518221852863254273634047533 00368340020397814797012023001544277754(21:51) gp > solve(x=4.6,4.2,(gamma(x)+1)/x-3)%9 = 4.5621514339509732614996681116823676765947133727177280760067672 66218451425330181354003523067887979890(21:51) gp > solve(x=4.6,4.9,(gamma(x)+1)/x-4)%10 = 4.8161649622698249284594051776217035792794119191176587528337167 98354235922037604118825483457603002219(21:52) gp > solve(x=5,5.2,(gamma(x)+1)/x-6)%13 = 5.1435871356645451824672277084473577830390498105918815864946996 13779238822779092252706929290717662694Phil >For complex values of x, in my ignorance it isnfit 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 iBut... is anything known about Leroyfis question?Ifim asking becausesome years ago I formulated the very same question and Ifim verycurious about it.To be fair Ifim only thinking to its second part i.e. that involvinggeneralized primes as those numbers satisfying the obviouscontinuous version of Wilsonfis (necessary and sufficient) conditionfor primality.Well, maybe something interesting can be said about a suitablefunction 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 isnfit 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 Leroyfis question?Ifim asking because> some years ago I formulated the very same question and Ifim very> curious about it.> To be fair Ifim only thinking to its second part i.e. that involving> generalized primes as those numbers satisfying the obvious> continuous version of Wilsonfis (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.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, 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* functionsin^2(pi x)+sin^2(pi(Gamma(x)+1)/x).But now Ifim more sort-of-thinking of something likeprod_p 1/(1-p^{-x})with p ranging over the zeroes of the function you supplied...provided that the product does make sense.Michele-- >Itfis 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? =Would some kind soul possibly post a complete solution to thisproblem, as none of us on the course can fathom it and our exam istomorrow... *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 ortwo.Here it is again:This question occurred to be the other day; itfis been a while since Itook calculus, and I canfit 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 afinite limit at each real number)?Foghorn Leghornmoc.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. and3)f(a)=k.Now you say that your function,f(x), is limited everywhere.Letfis consider xat a. So limx-->a+=lim x-->a-(=k). If f(a)=k then of course we canfit saythat f(x) is continuous nowhere. What type of function violates justcondition 3 of the definition of a continuous function? Functions that haveremovable discontinuity come to mind.Now the real question is whether or nota function can have an 00 number of removable discontinuities so thefunction is nowhere continuous. I think that you can come up with one.> This question occurred to be the other day; itfis been a while since I> took calculus, and I canfit 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; itfis been a while since I>took calculus, and I canfit 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 donfit 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, itfis definied everywhere, but I donfit know if itfiscontinuous and limited (sic). Maybe someone else does.cheerio, This question occurred to be the other day; itfis been a while since I> took calculus, and I canfit 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 functionf:R->Rwhere :(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 functionin 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 BernierRight. As for the original question, I have a feeling that no such functionexists. I wonder if you could prove this using a Baire category argument,maybe something like the proof that there is no function continuous only onthe rationals. >This question occurred to be the other day; itfis been a while since I>>took calculus, and I canfit 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 donfit 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, itfis definied everywhere, but I donfit know if itfis>continuous and limited (sic). Maybe someone else does.That function is obviously nowhere continuous. But italso obviously has a limit at no point, so it doesnfit help.>cheerio,************************David C. Ullrich This question occurred to be the other day; itfis been a while since I>took calculus, and I canfit 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 ofpoints 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 closedinterval I_1 such that V(f, I_1) < 1. Now the fact that f hasa limit at the midpoint of I_1 shows that there is a closedinterval I_2, contained in the _interior_ of I_1, such thatV(f, I_2) < 1/2. Repeat. Now if x is in the intersection ofthe I_n it foilows that f is continuous at x.(The fact that I_{n+1} is in the interior of I_n shows thatx is in the interior of I_n, and |f(x) - f(y)| < 1/n for ally 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; itfis been a while> since I took calculus, and I canfit 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, letL(f,x) be the limit of f(xfi) as xfi --> x, when this limitexists. Then the set (Ifim using /= for not equal) P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)}is countable. Thus, what youfire looking for doesnfit exist by along shot (a function f such that L(f,x) exists for each x in Rand P(f) = R). On the other hand, Ifim pretty sure that for eachcountable set Z there exists a function f such that L(f,x) existsfor each x in R and P(f) = Z (i.e. no restriction besidescountable 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. Thisimmediately 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 afunction f: R --> R and a real number x in R, let C(f,x) be the setof all extended real numbers y (i.e. y can be -oo or +oo) suchthat 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 convergingto x. Then for each f and x, C(f,x) is a nonempty closed set inthe extended real line, the minimum number in C(f,x) islim-inf(xfi--> x) of f(xfi), and the maximum number in C(f,x) islim-sup(xfi--> x) of f(xfi). Note that L(f,x) exists means thatC(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; itfis been a while since I>took calculus, and I canfit 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; itfis been a while> since I took calculus, and I canfit 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(xfi) as xfi --> x, when this limit> exists. Then the set (Ifim using /= for not equal)>> P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)}>> is countable. Thus, what youfire looking for doesnfit 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, Ifim 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 setDave, can you fill this in? I donfit see why itfis an isolated set.Sorry to be so dense :-) =An attempt at clarifying for calculus students.>This question occurred to be the other day; itfis been a while since I>>took calculus, and I canfit 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 therefis 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). Ifim at a loss as to how to explain this in a short note to someone whose background doesnfit extend beyond calculus. So Ifill beg off for today (but I hope youfill 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 youfive 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. Thatfis the definition of a dense set.Jon Miller This question occurred to be the other day; itfis been a while since I>>took calculus, and I canfit 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; itfis been a while>> since I took calculus, and I canfit 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(xfi) as xfi --> x, when this limit>exists. Then the set (Ifim 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 worddense in the result I gave could be strengthened,but I didnfit want to figure out by how much.Has countable complement is much strongerthan what I would have guessed.>Thus, what youfire looking for doesnfit 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, Ifim 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(xfi--> x) of f(xfi), and the maximum number in C(f,x) is>lim-sup(xfi--> x) of f(xfi). 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; itfis been a while>> since I took calculus, and I canfit 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(xfi) as xfi --> x, when this limit>> exists. Then the set (Ifim using /= for not equal)>> P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)}>> is countable. Thus, what youfire looking for doesnfit 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, Ifim 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 donfit see why itfis an isolated set.>Sorry to be so dense :-)Let epsilon = 1/(2n) in the definition of L(f,x) exists...************************David C. Ullrich 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 donfit see why itfis an isolated set.>Sorry to be so dense :-)>> Let epsilon = 1/(2n) in the definition of L(f,x) exists...>> ************************>> David C. Ullrich =I would be very pleased if the following statement was true :) :Given a finite group with n elements, it contains at most n maximal subgroups.Does anyone else believe it, or can someone even give me a proof?(I know it is almost trivial for finite abelian groups, but I couldnt find a proof for the non-commutative case. =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 firstorderY_1fi = F_1 [x, Y_1, Y_2, ..., Y_n]Y_2fi = F_2 [x, Y_1, Y_2, ..., Y_n]... ... ... ...... ... ... ...... ... ... ...Y_nfi = F_n [x, Y_1, Y_2, ..., Y_n](where n= max{n_1, n_2, ..., n_k) ????? =Recently on this newsgroup I have encountered quite a few people whosteadfastly hold that not even one single one-to-one mapping from N toR can exist. Surprisingly, an infinitude of such mappings exist. Indeed, the set of one-to-one mappings of N to R has the samemagnitude as R itself. To wit, aleph_null. :-)In my paper which is awaiting publication I expose the most generalsuch mapping, from which others can be derived. To expose thatmapping here would require quite a few lemmas which are presented inmy paper, and to write each of those using ASCII text would be quitetedious. Moreover, I fear that some who have more direct access tojournal publication would desire to steal the credit for my importantresults. Therefor I must ask that you be patient until such time as Irecieve a reply concerning my paper, before I can offer the full truthto you for your very respected examination.In the meantime, however, I am pleased to announce that a particular,special case mapping of N to R coincidentally happens to beexplainable in laymanfis terms without requiring any of my lemmas. Ofcourse this explanation is purely unrigorous, and loosely speaking. To put it forth more rigorously would require the work in my importantpaper, therefor we must defer that pleasure until a later date. Butin the meantime:Consider a theorem which is already known to hold true for all realnumbers. Then clearly it holds true for all integers and, morespecifically, all natural numbers. Suppose that this theorem has beenproven without the use of induction.Now, the weaker case of the theorem which applies only to naturalnumbers is important in our considerations. To illustrate this,consider the theorem, x+1 is greater than x for all real x. Theweaker natural case of this theorem would be the theorem, x+1 isgreater than x for all NATURAL x (but not necessarily all real x).process, which is reviewed in an appendix of my paper). Supposefurther that the theorem can be seen to hold for the trivial initialcase where the number in question is unity (necessary for induction infor a given real r requires a finite number of applications of thereal inductive step; applying the same number of applications of thenatural inductive step to the weaker case of the theorem for naturals,we prove the weaker case for all naturals up to a certain uniquenatural. That certain unique natural is what the given real numberwas mapped to. By the reverse process we can map a given natural toa certain unique real. And thus we have a one-to-one mapping of Nonto R. :-)I assure you that I was as shocked as you are now, when I first cameupon this. Not because of the anti-Cantorian implication, as Ialready knew that Cantor was wrong, but rather because of the sheerbeauty and simplicity of this mapping, and the fact it is easilyaccessible to even the most amateur of mathematicians who has takenonly some very introductory courses in community college! I assureyou that your astonishment now will be eclipsed by your astonishmentwhen the time comes that I may reveal the complete paper to you foryour complete analysis. :-)Your friendNathan the GreatAge 11 Recently on this newsgroup I have encountered quite a few people who> steadfastly hold that not even one single one-to-one mapping from N to> R can exist. Surprisingly, an infinitude of such mappings exist. > Indeed, the set of one-to-one mappings of N to R has the same> magnitude as R itself. To wit, aleph_null. :-)> In my paper which is awaiting publication I expose the most general> such mapping, from which others can be derived. To expose that> mapping here would require quite a few lemmas which are presented in> my paper, and to write each of those using ASCII text would be quite> tedious. Moreover, I fear that some who have more direct access to> journal publication would desire to steal the credit for my important> results. Therefor I must ask that you be patient until such time as I> recieve a reply concerning my paper, before I can offer the full truth> to you for your very respected examination.> In the meantime, however, I am pleased to announce that a particular,> special case mapping of N to R coincidentally happens to be> explainable in laymanfis terms without requiring any of my lemmas. Of> course this explanation is purely unrigorous, and loosely speaking. > To put it forth more rigorously would require the work in my important> paper, therefor we must defer that pleasure until a later date. But> in the meantime:> Consider a theorem which is already known to hold true for all real> numbers. Then clearly it holds true for all integers and, more> specifically, all natural numbers. Suppose that this theorem has been> proven without the use of induction.Ok, how about a theorem regarding the fact that there are infinitely many reals within 1/4 of any real? Does this apply to N?> Now, the weaker case of the theorem which applies only to natural> numbers is important in our considerations. To illustrate this,> consider the theorem, x+1 is greater than x for all real x. The> weaker natural case of this theorem would be the theorem, x+1 is> greater than x for all NATURAL x (but not necessarily all real x).> am really surprised that for so long noone has noticed it. Perhaps> this is because, to rigorously speak of these things, would require> the lemmas I am in the process of publishing, which are by no means> quite so shockingly simple.> To wit, consider two new proofs: a proof of the weaker case of the> known theorem for natural numbers, which is done by induction on the> naturals (the typical form of induction), and a proof of the complete> form of the known theorem, this time done by induction on the reals (a> much more complicated and less useful (and thus less well-known)> process, which is reviewed in an appendix of my paper). Suppose> further that the theorem can be seen to hold for the trivial initial> case where the number in question is unity (necessary for induction in> both cases).Nope. You only need a base case(s). It doesnfit have to be at unity.> for a given real r requires a finite number of applications of the> real inductive step; applying the same number of applications of the> natural inductive step to the weaker case of the theorem for naturals,> we prove the weaker case for all naturals up to a certain unique> natural. That certain unique natural is what the given real number> was mapped to. By the reverse process we can map a given natural to> a certain unique real. And thus we have a one-to-one mapping of N> onto R. :-)Too bad it doesnfit work.> I assure you that I was as shocked as you are now, when I first came> upon this. Not because of the anti-Cantorian implication, as I> already knew that Cantor was wrong, but rather because of the sheer> beauty and simplicity of this mapping, and the fact it is easily> accessible to even the most amateur of mathematicians who has taken> only some very introductory courses in community college! I assure> you that your astonishment now will be eclipsed by your astonishment> when the time comes that I may reveal the complete paper to you for> your complete analysis. :-)I eagerly await it. Perhaps youfill post it on JSHfis forum.> Your friend> Nathan the Great> Age 11-- Will Twentyman =A one-to-one mapping from f : N -> R is easy. f(x) = x. Remember, aone-to-one function (injection) need not cover R. ie: every point in N mapsto one in R, but there will be an uncountably infinite number of points in Rthat are not mapped-to for any x in N. If you mean a bijection byone-to-one (sometimes called Oone-to-one and ontofi), then there is no waycreate a bijective function from N -> R. There are an infinite number ofintegers, but an uncountably infinite number of reals - ie: there are waymore reals than integers. Look up the proof that there can be no bijectionbetween a set and its powerset to get an idea of the logic behind it.l8r, Mike N. Christoff> Recently on this newsgroup I have encountered quite a few people who> steadfastly hold that not even one single one-to-one mapping from N to> R can exist. Surprisingly, an infinitude of such mappings exist.> Indeed, the set of one-to-one mappings of N to R has the same> magnitude as R itself. To wit, aleph_null. :-)> In my paper which is awaiting publication I expose the most general> such mapping, from which others can be derived. To expose that> mapping here would require quite a few lemmas which are presented in> my paper, and to write each of those using ASCII text would be quite> tedious. Moreover, I fear that some who have more direct access to> journal publication would desire to steal the credit for my important> results. Therefor I must ask that you be patient until such time as I> recieve a reply concerning my paper, before I can offer the full truth> to you for your very respected examination.> In the meantime, however, I am pleased to announce that a particular,> special case mapping of N to R coincidentally happens to be> explainable in laymanfis terms without requiring any of my lemmas. Of> course this explanation is purely unrigorous, and loosely speaking.> To put it forth more rigorously would require the work in my important> paper, therefor we must defer that pleasure until a later date. But> in the meantime:>> Consider a theorem which is already known to hold true for all real> numbers. Then clearly it holds true for all integers and, more> specifically, all natural numbers. Suppose that this theorem has been> proven without the use of induction.> Now, the weaker case of the theorem which applies only to natural> numbers is important in our considerations. To illustrate this,> consider the theorem, x+1 is greater than x for all real x. The> weaker natural case of this theorem would be the theorem, x+1 is> greater than x for all NATURAL x (but not necessarily all real x).> am really surprised that for so long noone has noticed it. Perhaps> this is because, to rigorously speak of these things, would require> the lemmas I am in the process of publishing, which are by no means> quite so shockingly simple.> To wit, consider two new proofs: a proof of the weaker case of the> known theorem for natural numbers, which is done by induction on the> naturals (the typical form of induction), and a proof of the complete> form of the known theorem, this time done by induction on the reals (a> much more complicated and less useful (and thus less well-known)> process, which is reviewed in an appendix of my paper). Suppose> further that the theorem can be seen to hold for the trivial initial> case where the number in question is unity (necessary for induction in> for a given real r requires a finite number of applications of the> real inductive step; applying the same number of applications of the> natural inductive step to the weaker case of the theorem for naturals,> we prove the weaker case for all naturals up to a certain unique> natural. That certain unique natural is what the given real number> was mapped to. By the reverse process we can map a given natural to> a certain unique real. And thus we have a one-to-one mapping of N> onto R. :-)>> I assure you that I was as shocked as you are now, when I first came> upon this. Not because of the anti-Cantorian implication, as I> already knew that Cantor was wrong, but rather because of the sheer> beauty and simplicity of this mapping, and the fact it is easily> accessible to even the most amateur of mathematicians who has taken> only some very introductory courses in community college! I assure> you that your astonishment now will be eclipsed by your astonishment> when the time comes that I may reveal the complete paper to you for> your complete analysis. :-)>> Your friend> Nathan the Great> Age 11 Your friend> Nathan the Great> Age 11Werenfit you age 11 a couple of years ago? Your friend> Nathan the Great> Age 11> Werenfit you age 11 a couple of years ago?> Hefis referring to his _mental_ age.F. = I think that the neatest suggestion was to show that the open ball is open under Spivakfis definition (which isbasically Spivak problem 1-15), and show that the 1st quadrentis open, and then that the interesction of two (or a finite number) of open sets is open...Then my problemis solved. get stuck again...ciao,adrock I think that the neatest suggestion was to show that> the open ball is open under Spivakfis 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 isdiamonds or rhomboids D((x,y) - (a,b)) = |x-a| + |y-b|Also therefis the box metric which is about the same except instead ofopen rectangles it uses open squares.> get stuck again...ciao,>Youfire welcome. You donfit have to wait to get stuck before returning. Theapproach I suggested that was to your liking was a topological approachthe 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 equivalenceof the rectangle topology to the product topology which seemed instantlyclear to you. = I have to minimize something like tr(XfiAX) where A in R^{n*n} is adefinite positive matrix and X is a n*k binary matrix, such that x_ij={0,1} and thesum of each row is 1 (i.e. sum_i x_ij=1). The problem is well known to be NP-complete. Do youknow any reference to a continuous approximation? Or an efficient way to solve the problem when Xis a big matrix? 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.htmlAha. I was unaware of the different isomorphism types -- hadnfit 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 sitefis 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 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 -- hadnfit 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:51013Lam, 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 sitefis 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 have also posted this separately to moderated groupssci.math.research and sci.physics.research .-I know it was once believed that variational problems just havea minimal solution but for a long while now it has been knownthat it is a stationary (extremal, or maybe extremal andin?ction point) solution and that while the nature of thesolution can be determined mathematically that can be laboriousand that usually whether it is a minimum of maximum isdetermined from the physical context.So what I was wondering is, I know that there are manyphysical and other applied mathematical examples of minimal solutions but I would like to know of a few maximalsolutions and perhaps in?ction point ones if such exist.My math background in the area includes stuff by Weinstock, Arnold and I guess a section in Arfken a good while ago,and it may have applications to my Ph.D. thesis inseismic ray theory for which I am currently beginningto write a thesis proposal and preparing for my Ph.D.comprehensive exam and proposal defense (so I am not farfrom the end of my first year of my Ph.D.).Some books that I glanced at today included The Parsimonious Universe: Shape and Form in the Natural WorldSpringer-Verlag, NY, NY, 1996,ISBN 0-387-97991-3which is not very mathematical but is a very good general bookbut has very little on maxima.Arnold, V.I., 1989. Mathematical Methods of Classical Mechanics, Springer.(a fairly advanced text which I studied Chs. 3, 4, 7, 8, and a bit ofthe Appendix, as part of a recent graduate reading course on differential geometry) certainly specifies extremal on e.g. p. 57. Another,Weinstock, R., 1952/1974. Calculus of Variations with applications to physics and engineering, Dover, NY.on p. 23 talks about maximum, minimum and stationary so I thinkhe by stationary means in?ction point whereas usuallystationary means maximum, minimum or in?ction point.But I just remembered a statement about a quantity that is thoughtto be minimized but is sometimes lavishly expended and forgotthe exact wording or who expended foundIreland, one of the foremost Irish scientists of all time.but will read it eventually. An extended version of my misrememberedquote above by Hamilton isBut although the law of least action has thus attained a rank among the highest theorems of physics, yet its pretensions to a cosmological necessity, on the ground of economy in the universe, are now generally rejected. And the rejection appears just, for this, among other reasons, that the quantity pretended to be economised is in fact often lavishly expended. In optics, for example, though the sum of the incident and re?cted portions of the path of light, in a single ordinary re?xion at a plane, is always the shortest of any, yet in re?xion at a curved mirror this economy is often violated. If an eye be placed in the interior but not at the centre of a re?cting hollow sphere, it may see itself re?cted in two opposite points, of which one indeed is the nearest to it, but the other on the contrary is the furthest; so that of the two different paths of light, corresponding to these two opposite points, the one indeed is the shortest, but the other is the longest of any.So anyway Ifim interested in maxima examples and, if any, in?ctionpoint (or analogy to in?ction point) examples. I will alsoask local colleagues more advanced in these areas than I amwhen they get back from conferences/etc.This is indeed related to my Ph.D. thesis but also spun offof philosophical discussions on talk.atheism entitledbipolar religious figures? and Parsimonious nature? (was Re: bipolar religious figures?)andOT: love balancing/supplantingwhich sort of update the shamanic component of my web pagehttp://www.n?.com/~dalton a bit, for any who might beinterested in that sort of stuff, but that is not the mainfocus of this post.Davidhttp://www.n?.com/~dalton So anyway Ifim interested in maxima examples and, if any, in?ction> point (or analogy to in?ction point) examples. I will also> ask local colleagues more advanced in these areas than I am> when they get back from conferences/etc.> Certainly, if you can formulate a thermodynamic problem in variational form,solutions will maximize production of entropy over a path.Bill =I have this problem: S- items (sorted) G- groups, and I want to dividethe items into G groups. for this Ifim using the permutation: (S-1choose G-1).My question is what is the is advance,Michal I have this problem: S- items (sorted) G- groups, and I want to divide>the items into G groups. for this Ifim using the permutation: (S-1>choose G-1).>My question is what is the complexity of this problem is NP-complete?>or polynomial.I think youfill have to explain what exactly your problem is. Perhapstherefis some reason why one division might be better than another?Or are you just trying to count the number of ways to do it?Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =Suppose a mass m is on a vertical spring (at rest) and the spring iscompressed a length x by the mass. What is the spring constant?I have mg = kx k = mg / xHowever, I want to show this using the conservation of mechanical energy.Setting my GPE = 0 reference base to be where the mass lies on the spring, IhaveKE0 + GPE0 + EPE0 = KE + GPE + EPE0 + mgh + 0 = 0 + 0 + 1/2*k*x^2mgh = 1/2*k*x^2Since h = x, k = 2mg / xHere is my problem, 2mg / x != mg / xI have looked and cannot figure out the ?w in my thoughts, I think Please help. Suppose a mass m is on a vertical spring (at rest) and the spring is>compressed a length x by the mass. What is the spring constant?>I have mg = kx k = mg / x>However, I want to show this using the conservation of mechanical energy.>Setting my GPE = 0 reference base to be where the mass lies on the spring, I>have>KE0 + GPE0 + EPE0 = KE + GPE + EPE>0 + mgh + 0 = 0 + 0 + 1/2*k*x^2You canfit assume this since the mass stationary where it has compressedthe spring sufficiently to support it is not in the same motion as the mass stationary on an uncompressed spring. If you have a stationary mass resting on the spring at the place to which it has compressed the spring (i.e.sufficiently to support it), then it remains stationary. If you have the mass stationary on an uncompressed spring, then it will start to compress the spring, and it will end up oscillating. Since the two situations never occur within the same motion of the mass, then there is no reason to assume that the mechanical energies of the two cases are equal. And in fact, if you mark the point where mass has compressed the spring just enough to support it, and then you switch to the problem where you start the mass at rest on the uncompressed spring, then the mass will be moving as it passes the marked point. This means that the nett mechanical energy is higher for the mass at rest on the uncompressed spring than it is for the mass at rest when it has compressed the spring sufficiently to support it (i.e. the increase in the springfis energydue to its compression is less than the decrease in the gravitational potential energy of the mass).>mgh = 1/2*k*x^2>Since h = x, k = 2mg / x>Here is my problem, 2mg / x != mg / x>I have looked and cannot figure out somewhere but I am missing something. Please help.What you missed was that you canfit equate mgh with 1/2*k*x^2.David McAnally-- can not solve this, as what I have learnt in school seems thatI 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.111112122123111222223331332333Many 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 isnfit 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> Twentyman www.YeOldeCoffeeShoppe.com has a great many visitors and sections> for talk, tv, reviews and romance.but Ifive already got 1000s dropping in, all with nothing to read,just jump the buck, Ifill have to pay a dozen people just to postsomething there for a few months otherwise, give the site a KICK,its a GREAT traffic getter a few minutes of telling people drawsreal crowds, but without the first couple of people standing round thebusker nooone will gather. And if you like usenet youfill 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.--Seanhttp://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.comHerc 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 OStanding Outside Imaginationfis Doorfi would be cool.Even better would be ORinging the Bell of Life and Runningfi.>ps need some writing talent at www.Yeoldecoffeeshoppe.comIfill let you know as soon as I have the motivation and time toactually look at the site and see what the hell it is.--Seanhttp://www.livejournal.com/users/spclsd223/ Why are certain polyhedra that are self intersecting considered REAL> polyhedra? > See Imre Lakatosfi _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> mathematiciansfi views on such issues are of the whatever genre?> How can they not care? Isnfit it their job to, you know, care?> cdj> Alright.> Clearly polyhedrons which are self intersecting shouldnfit> 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? Doesnfit that kinda> violate the rules, about being able to be constructed as if from> boundaries from algebraic expressions, and such?> This is what I donfit 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 Lakatosfi _Proofs and Refutations_ for a variety of detailed> views about this exact question.> And where will I find that?Any decent library.-- =Suppose you have a finite abelian cyclic group . Is there acanonical way (or any way, really) to express this alternatively interms 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 Ifid appreciate it ifanyone could post with a way to do this (if it indeed can be done). Isthere any kind of _Mathematica_ or Maple command that does thisautomatically, perhaps? (say once you enter the generator of thecyclic group and its order)Or perhaps if it can be done in certain special cases, maybe someonecould 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, withgenerator (1,1) as prototype of theorem. That should work unless order ofgroup 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.> ?> Itfis not clear to me exactly what youfire 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 toZ_n; then itfis not clear that a non-trivial presentation of the givenform exists.By non-trivial here, I mean a presentation where neither = G nor = G; clearly, isalways going to give a representation of Z_a, but thatfis probably notwhat 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 yourgiven form is ; if we take g = 3, then .But these are both equivalent to . There is ahomomorphism from this group onto D_3 (dihedral group) withpresentation , and onto Z_6 withpresentation ; so it would seemthere is no non-trivial representation of Z_6 of your form.> If so, this would be very useful to me, and Ifid 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.>> Itfis not clear to me exactly what youfire given in this problem.See my other post where I try to work this out in general, startingfrom the hints I gave in my first post. In working thru this, thereason I disallowed the order of the group to be a power of a primeis 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 itfis 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 thatfis 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. Donfit 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 isnfit a power of a prime.Thus some cofinite j,k > 1 with n = rsLet (1,0) and (0,1) be generators of Z/r & Z/s.r(1,0) = (0,0) = s(0,1)If Ifive 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. 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.>> 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. Donfit take h = 2, take h = 4,> as in accordance to my work in the other post.> Thus Z_6 = = <-1>, does it not?> Yes, it works in the sense that does indeed generate Z_6.But, as I read it, the poster wants a presentation of the group of theform:G = such that G is isomorphic to Z_n for a given n. In the case n = 6, nosuch non-trivial (in the sense of my post) presentation is possible,unless we _require_ that the group be Abelian (i.e., unless there is anadditional, _implict_ relation g+h = h+g).C Brown Systems DesignsMultimedia Environments for Museums and Theme Parks <3F1C7264.303BBB4D@cbrownsystems.com> 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.>> 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. Donfit take h = 2, take h = 4,> as in accordance to my work in the other post.> Thus Z_6 = = <-1>, does it not?> Yes, it works in the sense that does indeed generate Z_6.>> But, as I read it, the poster wants a presentation of the group of the> form:>> G = such that G is isomorphic to Z_n for a given n. In the case n = 6, no> such non-trivial (in the sense of my post) presentation is possible,> unless we _require_ that the group be Abelian (i.e., unless there is an> additional, _implict_ relation g+h = h+g).>Indeed, either a technical oversight or a catagorical presumption ofAbelian groups, by the OP. James Harrisfi PrimeCountH program was used> to compute pi(10^15). The result is correct.>> Also, I get pi(1)=0 using PrimeCountH from> AmateurMath, his forum...>> To tell Java to use more memory than the default,> there is an option -Xmx ; with -Xmx500m,> the maximum allowed total memory> for the program is set to 500 Mbytes.> command = java -Xmx500m PrimeCountH 1000000000000000> Sieve Time: 22493 /*** 22.5 seconds ***/> m_max=31622777> pi(1000000000000000)=29844570422669 /*** correct ***/>> Total Time: 17830829 /*** about 5 hours ***/ I used the j2sdk1.4.2 on a PC with an Athlon 1800+ processor.The pi(x) project seems to be stuck on pi(10^23).http://numbers.computation.free.fr/Constants/Primes/ Pix/results.htmlhttp://numbers.computation.free.fr/Constants/ Primes/Pix/pixproject.htmlSo... there is an opportunity!-- Clive Toothhttp://www.clivetooth.dk >James Harrisfi PrimeCountH program was used>>to compute pi(10^15). The result is correct.>>Also, I get pi(1)=0 using PrimeCountH from>>AmateurMath, his forum...>>To tell Java to use more memory than the default,>>there is an option -Xmx ; with -Xmx500m,>>the maximum allowed total memory>>for the program is set to 500 Mbytes.>>command = java -Xmx500m PrimeCountH 1000000000000000>>Sieve Time: 22493 /*** 22.5 seconds ***/>>m_max=31622777>>pi(1000000000000000)=29844570422669 /*** correct ***/>>Total Time: 17830829 /*** about 5 hours ***/I used the j2sdk1.4.2 on a PC with an Athlon 1800+ processor.> The pi(x) project seems to be stuck on pi(10^23).> http://numbers.computation.free.fr/Constants/Primes/Pix/ results.html> http://numbers.computation.free.fr/Constants/Primes/Pix/ pixproject.html> So... there is an opportunity!Indeed, James Harris has some top brass buddys. He could very likely gethis PrimeCountH program to run on half of NSAfis computers.David Bernier > http://www.msnusers.com/AmateurMath/Documents/CountViewer.html >> Very stupid to put it at a MSN site... like most reasonable people, Idonfit> have a Microsoft Passport account so I canfit go to your site. And Ifimreally> not going to create a Passport account...>> =- Brian Dickens, the NetherlandsI have also tried try the applet,and I also do not intend to create a Passport account.It seems to me,that if Harris is interested in folks seeing his works,that he should make them more accessible.--Tom Potter http://tompotter.us I donfit think it works for p = 3 either.<2> = {2, 1} Z modulo 3.GREG>> Lurch>> Conjecture:> For every prime p, the multiplicative group Z(modulo p) contains at> least> one prime q

=Consider a number p less than n!If p is not divisible by the integers 2 .. n, then say that x is n-prime(note that 1 will be in this set). x is not necessarily prime but ispotentially prime in that it cannot be divided by 2 through n.Let N be the set of all p that are n-prime.All prime numbers less than n! will be in the set of N. This makes sensebecause all primes less than N will also be n-prime -- that is notdivisible by 2 .. n.There is a simple symmetry about n! Construct a new set M by adding n! toeach member of N. The set of primes between n! and 2n! will be in M.Again, not all the numbers in M will be prime but M will contain all theprimes. In fact the number of primes in M will be less than the number ofprimes in N. Even so, in many cases a prime less than n! will have a mirrorprime between n! and 2n!.For example, consider 4! = 24. The following numbers are n-prime:1, 3, 5, 7, 11,13,17,19,23and form the set N.The set M will be these numbers plus 24 or25, 27, 29, 31, 37, 41, 43, 47All of these are the prime except for 25 and all the primes between 24 and48 fall in the set M.N and M can be fairly easily constructed for 5!.In short, if you know an n-prime less than n! then you can find a candidateprime greater than n! by simply adding n! to the n-prime. >Consider a number p less than n!>>If p is not divisible by the integers 2 .. n, then say that x is n-prime>(note that 1 will be in this set). x is not necessarily prime but is>potentially prime in that it cannot be divided by 2 through n.>>Let N be the set of all p that are n-prime.>>All prime numbers less than n! will be in the set of N. This makes sense>because all primes less than N will also be n-prime -- that is not[SNIP]God damn, I hate it when people start out a post with: Consider... and then proceed to out some random math problem, with no introduce yourself... and ONLY THEN give us your god-damned homework problem. ing-A, this happens all the time. I donfit give a about somerandom math problem... but if you would like some help, then thatfisa different story... but you have to ASK FOR IT...I donfit know... sorry about the rant, but I just find it so annoying.AS God damn, I hate it when people start out a post with: Consider... > and then proceed to out some random math problem, with no > introduce yourself... and ONLY THEN give us your god-damned > homework problem.Then again, this is a maths newgroup. :-]--J K Hauglandhttp://www.neutreeko.com God damn, I hate it when people start out a post with: Consider... >> and then proceed to out some random math problem, with no introduce yourself... and ONLY THEN give us your god-damned >> homework problem.>>Then again, this is a maths newgroup. :-]Maybe you are being a bit unfair... It wasnfit a homework problem, itwas a result that he found, and overall not an unisnteresting one forat least some people here. Before complaining, why not actually reasthe post. God damn, I hate it when people start out a post with: Consider... > and then proceed to out some random math problem, with no> introduce yourself... and ONLY THEN give us your god-damned> homework problem.>>Then again, this is a maths newgroup. :-]> Maybe you are being a bit unfair... It wasnfit a homework problem, it> was a result that he found, and overall not an unisnteresting one for> at least some people here. Before complaining, why not actually reas> the post.I complaining, why not actually read his post?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen > God damn, I hate it when people start out a post with: Consider... >> and then proceed to out some random math problem, with no>> introduce yourself... and ONLY THEN give us your god-damned>> homework problem.>>Then again, this is a maths newgroup. :-]>Good Point... I guess thatfis why I encluded the tags.AS >Consider a number p less than n!>>If p is not divisible by the integers 2 .. n, then say that x is n-prime>>(note that 1 will be in this set). x is not necessarily prime but is>>potentially prime in that it cannot be divided by 2 through n.>>Let N be the set of all p that are n-prime.>>All prime numbers less than n! will be in the set of N. This makes sense>>because all primes less than N will also be n-prime -- that is not>>[SNIP]>>>God damn, I hate it when people start out a post with: Consider... >and then proceed to out some random math problem, with no >introduce yourself... and ONLY THEN give us your god-damned >homework problem.> ing-A, this happens all the time. I donfit give a about some>random math problem... but if you would like some help, then thatfis>a different story... but you have to ASK FOR IT.>..I donfit know... sorry about the rant, but I just find it so annoying.>>This looks like a rant with introduce yourself... and ONLY THEN give us your god-damned homework problem. ing-A, this happens all the time.But fortunately not that often on sci.math.Jon Miller all.>Before complaining, why not actually read his post?Nah, Gartogg just replied to the wrong post. He was JKs post.Jon Miller Consider a number p less than n!> If p is not divisible by the integers 2 .. n, then say that x is n-prime> (note that 1 will be in this set). x is not necessarily prime but is> potentially prime in that it cannot be divided by 2 through n.> Let N be the set of all p that are n-prime.> All prime numbers less than n! will be in the set of N. This makes sense> because all primes less than N will also be n-prime -- that is not> divisible by 2 .. n.> There is a simple symmetry about n! Construct a new set M by adding n! to> each member of N. The set of primes between n! and 2n! will be in M.> Again, not all the numbers in M will be prime but M will contain all the> primes. In fact the number of primes in M will be less than the number of> primes in N. Even so, in many cases a prime less than n! will have a mirror> prime between n! and 2n!.> For example, consider 4! = 24. The following numbers are n-prime:> 1, 3, 5, 7, 11,13,17,19,23> and form the set N.> The set M will be these numbers plus 24 or> 25, 27, 29, 31, 37, 41, 43, 47> All of these are the prime except for 25 and all the primes between 24 and> 48 fall in the set M.> N and M can be fairly easily constructed for 5!.> In short, if you know an n-prime less than n! then you can find a candidate> prime greater than n! by simply adding n! to the n-prime.This is known as reinventing the wheel.You cross Oem, Ifill knock Oem in.Phil all.>>Before complaining, why not actually read his post?> Nah, Gartogg just replied to the wrong post. He was JKs post.He snipped asfis name too.Symptomatic though of someone who doesnfit read carefullybefore sounding off.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen =Conjecture 1:If p is any odd prime & p-1 contains at least 1 Quadratic non-residue, thenat least one prime q which divides p-1 is a primitive root.Conjecture 2:If p is any prime bigger than 3, then the multiplicative group Z(modulo p-1)contains at least one primitive root of p.Conjecture 3:Suppose p is any odd prime and p-1 is of the form 2q^x, where q is an oddprime. If g is a quadratic non residue of both p and p-1, then g is aprimitive root.--So, Ifim looking for counterexamples, or reasons why these wouldnfit be true.So far, Ifim thinking that both 1 & 3 are false, and 2 is true - although Ihavenfit found any counterexamples, or been able to prove them either way.Any help would be great.GREG =I have a function proportional to a probability distribution of interest that is giving me fits.y = x * I(1-(x^2); y, 1/2)where OIfi is the regularized beta function. What I need is the form of this distribution as y->+oo and x>0. For large y, it looks awfully like a gamma or beta distribution, and Ifid really like to know if it *is* one of those (or something similar). Can anyone help with this?Zeus =Suppose X, Y, Z are positive random variable with the pdf f_X(t), f_Y(t),f_Z(t), respectively. And F_X(t), F_Y(t), F_Z(t) are respective cdffunction. The quantity a is a positive real number. I need to evaluate thefollowing probabiltiy.P( X f_Z(t), respectively. And F_X(t), F_Y(t), F_Z(t) are respective cdf> function. The quantity a is a positive real number. I need to evaluate the> following probabiltiy.> P( X I develop the following expresion.> P( X =int_{0}^{a} f_X(t) F_Y(t) [1-F_Z(t) ] dt> No. As you have stated it, you have definite values for Y and Z, such that 0 < Y < X < min(a,Z) Why the last inequality? Because if a < Z then X < a guarantees x < Z, and vice-versa. Of course the probability must be 0 if min(a,Z) < Y . Thus P( Y < X < min(a,Z) ) = int _Y ^{min(a,Z)} {dt f_X (t) ) = [ F_X ( min(a,Z) ) - F_X (Y) ] theta ( min(a,Z) - Y ) You can then multiply by the pdffis f_Y (u) and f_Z (v), and integrate over u and v to get the probability of finding an X that satisfies the restrictions.-- Julian V. NobleProfessor Emeritus of ^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. =Hey,Im curious, what would you guys/gals say the probability of someoneentering a Ph.D. program in Math or Stats and not finishing it. i.e.dropping out. Im curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.Of not leaving with a Ph.D.? 75% would be my guess, based on the eightyears Ifive been at Colorado.Doug Hey,>>Im curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.Wouldnfit know about stat, but sad to say a large majority of thepeople who enter the PhD program in math here at OSU endup without a PhD, one way or another.************************David C. Ullrich >Im curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.>> Of not leaving with a Ph.D.? 75% would be my guess, based on the eight> years Ifive been at Colorado.>> DougDepends a lot on the school, of course. If you want the highestprobability, I would guess in the states it might be University of Montanaor University of Idaho or Idaho State. Not disparaging, thatfis just howthey are. Hey,> Im curious, what would you guys/gals say the probability of someone> entering a Ph.D. program in Math or Stats and not finishing it. i.e.> dropping out.I just read that it was about 50-50. Long ago, I heard that it isanother 50-50 that one who finishes will do nothing after theirthesis. This suggests that a lot of theses are written by theadvisor. Hey,>> Im curious, what would you guys/gals say the probability of someone> entering a Ph.D. program in Math or Stats and not finishing it. i.e.> dropping out.>> I just read that it was about 50-50. Long ago, I heard that it is> another 50-50 that one who finishes will do nothing after their> thesis. This suggests that a lot of theses are written by the> advisor.Not in the least. In graduate school you are surrounded by excellentmathematicians and the spirit of mathematics. Mathematics is everywhere; itis the whole world. Everybody around you thinks that itfis the only thingworth learning.Then you get a job at Podunk, and discover that your newfound colleaguesthink that knowing mathematics is knowing the difference between additionand subtraction. Discussions in the faculty lounge are about football.You teach 12 to 15 credits a week, same old stuff year after year. You getnumb and tired and disillusioned (Pirsig mentions this in Zen and the Art ofMotorcycle Maintenance). You have no real contact with the living world ofmathematics and mathematicians; all youfive got is your Calculus I textbookand your colleagues. With great effort you can scare up money to go to theoccasional convention.Some people overcome these obstacles, bless them. =...>> I just read that it was about 50-50. Long ago, I heard that it is>> another 50-50 that one who finishes will do nothing after their>> thesis. This suggests that a lot of theses are written by the>> advisor.>>Not in the least. In graduate school you are surrounded by excellent>mathematicians and the spirit of mathematics. Mathematics is everywhere; it>is the whole world. Everybody around you thinks that itfis the only thing>worth learning.>>Then you get a job at Podunk, and discover that your newfound colleagues>think that knowing mathematics is knowing the difference between addition>and subtraction. Discussions in the faculty lounge are about football.>>You teach 12 to 15 credits a week, same old stuff year after year. You get>numb and tired and disillusioned (Pirsig mentions this in Zen and the Art of>Motorcycle Maintenance). You have no real contact with the living world of>mathematics and mathematicians; all youfive got is your Calculus I textbook>and your colleagues. With great effort you can scare up money to go to the>occasional convention.>>Some people overcome these obstacles, bless them.I like to believe that the advent of Usenet, later the web, arxiv.org, etc., are helping more people overcome those obstaclesmore effectively.Lee Rudolph > Hey, Im curious, what would you guys/gals say the probability of someone>> entering a Ph.D. program in Math or Stats and not finishing it. i.e.>> dropping out.>>I just read that it was about 50-50.50-50 for finishing or not finishing?;-)-- Rouben Rostamian I just read that it was about 50-50. Long ago, I heard that it is> another 50-50 that one who finishes will do nothing after their> thesis. This suggests that a lot of theses are written by the> advisor.Most of those PhDs go to positions where research is not encouraged,rather teaching and service are encouraged. 4-year colleges, communitycolleges, even high schools. That could also be a reason for notwriting more papers. The idea that writing no research papers equalsdoing nothing shows a warped view of the world.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ >Not in the least. In graduate school you are surrounded by excellent>mathematicians and the spirit of mathematics. Mathematics is everywhere; it>is the whole world. Everybody around you thinks that itfis the only thing>worth learning.>>Then you get a job at Podunk, and discover that your newfound colleagues>think that knowing mathematics is knowing the difference between addition>and subtraction. Discussions in the faculty lounge are about football.>>You teach 12 to 15 credits a week, same old stuff year after year. You get>numb and tired and disillusioned (Pirsig mentions this in Zen and the Art of>Motorcycle Maintenance). You have no real contact with the living world of>mathematics and mathematicians; all youfive got is your Calculus I textbook>and your colleagues. At this point I would suggest: Guyfis UPINT, GP/PARI, some decent coffee, asupply of good Scotch Whiskey, and a great wife. Perhaps time on a troutstream just outside Podunk two evenings a week may be of some benefit. Asummer working the wheat harvest might help also. Clearly Podunk ainfit MSRI. Southwest will, however, get you to Oakland for$99. I donfit know what AC Transit costs these days, but it canfit be much. Even the numb and tired and disillusioned have choices, I would think.Rich Im curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.> Of not leaving with a Ph.D.? 75% would be my guess, based on the eight> years Ifive been at Colorado.> DougWow, i never would have thought it be that high. I would have guessedmaybe 30% drop out rate. I figured after someone got their Masters inMathematics, got straight Afis in their graduate courses that it wouldbe sufficient to prepare them for the doctorate program in mathematics/or statistics. I just read that [chance of completing Ph.D.] was about 50-50. Long> ago, I heard that it is another 50-50 that one who finishes will do> nothing after their thesis.You mean 50% of mathematics Ph.D.s are unemployed, spending theirentire lives sitting in their room staring at the walls? I thinknot. Perhaps there is a much more narrow (-minded) interpretation ofthe word nothing?Ifim going to speculate that nothing is interpreted along the linesnot inconsistent with the simple observation that something on theorder of 50% of math or science Ph.D. graduates enter careers otherthan academics.> This suggests that a lot of theses are written by the advisor.I would suggest that one for whom this is suggested by the 50-50figure should consider investing some time in the study of logic orstatistics.Kevin. Im curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.> Of not leaving with a Ph.D.? 75% would be my guess, based on the eight> years Ifive been at Colorado.> Doug> Wow, i never would have thought it be that high. I would have guessed> maybe 30% drop out rate. I figured after someone got their Masters in> Mathematics, got straight Afis in their graduate courses that it would> be sufficient to prepare them for the doctorate program in mathematics> /or statistics.One thing is that most people enter without a Masters degree in thefirst place and switch to a Masters rather than finish the PhD. Thataccounts for a big portion of the discrepancy you think is present. Ion the other hand was one of the few people that had passed most ofthe hurdles of a math PhD program without actually getting such adegree. I believe there was one other out of maybe two or three dozenPhD graduates during the five year period I was trying for a PhD.Karl Hallowell >Then you get a job at Podunk, and discover that your newfound colleagues>think that knowing mathematics is knowing the difference between addition>and subtraction. Discussions in the faculty lounge are about football.>>You teach 12 to 15 credits a week, same old stuff year after year. You get>numb and tired and disillusioned (Pirsig mentions this in Zen and the Art of>Motorcycle Maintenance). You have no real contact with the living world of>mathematics and mathematicians; all youfive got is your Calculus I textbook>and your colleagues. With great effort you can scare up money to go to the>occasional convention.>>Some people overcome these obstacles, bless them.> I like to believe that the advent of Usenet, later the web, > arxiv.org, etc., are helping more people overcome those obstacles> more effectively.I believe that is quite accurate. I currently (to be cured in a fewmonths) have no access to a nearby college library, community, etc.The nearest college is more than fourty miles away and there I haveonly a few informal contacts in the aerospace engineering community.My real connections (as such) are online.Any serious math or physics concept is available models, the inverse Galoisproblem, or the Eight Vertex model and quickly find relevant researchand expository material. The USENET might not be able to answer myquestions, but they never have failed to come up with some insight.Ifim still trying to figure out how to use arXiv.org (even after yearsof playing with it), but itfis proving to be an amazing research tooleven with my limited experience.Karl Hallowell >>Im curious, what would you guys/gals say the probability of someone>>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>>dropping out. Of not leaving with a Ph.D.? 75% would be my guess, based on the eight>> years Ifive been at Colorado. Doug>>Wow, i never would have thought it be that high. I would have guessed>maybe 30% drop out rate. I figured after someone got their Masters in>Mathematics, got straight Afis in their graduate courses that it would>be sufficient to prepare them for the doctorate program in mathematics>/or statistics.Again, I have no idea how things are in stat, but no thatfis not howit is in math at all. A masterfis degree requires that you learn acertain amount of mathematics - how much and how well youfirerequired to learn it varies from place to place. A PhD requires_much_ more. First, it requires that you learn much moremathematics - much deeper mathematics, and youfire requiredto understand it much better than a masterfis student (tooversimplify that last point, a masterfis student gets creditfor knowing facts, while a PhD student only gets credit forknowing how to _prove_ those facts).And then therefis the much more significant difference: APhD requires a thesis, which is supposed to be significant original research. Of course some theses aremore significant and original than others, but regardless,itfis a totally different sort of requirement from anythingthatfis required in a typical masterfis degree - at leasttheoretically, when you finish your PhD therefis supposedto be _something_ that you understand better thananyone else on the planet.************************David C. Ullrich >...> I just read that it was about 50-50. Long ago, I heard that it is> another 50-50 that one who finishes will do nothing after their> thesis. This suggests that a lot of theses are written by the> advisor.>>Not in the least. In graduate school you are surrounded by excellent>>mathematicians and the spirit of mathematics. Mathematics is everywhere; it>>is the whole world. Everybody around you thinks that itfis the only thing>>worth learning.>>Then you get a job at Podunk, and discover that your newfound colleagues>>think that knowing mathematics is knowing the difference between addition>>and subtraction. Discussions in the faculty lounge are about football.>>You teach 12 to 15 credits a week, same old stuff year after year. You get>>numb and tired and disillusioned (Pirsig mentions this in Zen and the Art of>>Motorcycle Maintenance). You have no real contact with the living world of>>mathematics and mathematicians; all youfive got is your Calculus I textbook>>and your colleagues. With great effort you can scare up money to go to the>>occasional convention.>>Some people overcome these obstacles, bless them.>>I like to believe that the advent of Usenet, later the web, >arxiv.org, etc., are helping more people overcome those obstacles>more effectively.It can certainly help people stay in touch, or at least that seemsplausible. Hard to see how it can help with the huge teachingloads at Podunk, though.>Lee Rudolph************************David C. Ullrich >I like to believe that the advent of Usenet, later the web, >>arxiv.org, etc., are helping more people overcome those obstacles>>more effectively.>>It can certainly help people stay in touch, or at least that seems>plausible. Hard to see how it can help with the huge teaching>loads at Podunk, though.Why, by providing the students^Wclients^Wenrollees at Podunkwith sci.math to do their homework for them, of course.And if youfid read Hyman Bassfis report to the Carnegie Foundationin the latest _Notices_, youfid realize that teaching load isa doubleplusungood phrase. Time to talk about research burdeninstead!Lee Rudolph Most of those PhDs go to positions where research is not encouraged,> rather teaching and service are encouraged. 4-year colleges, community> colleges, even high schools. That could also be a reason for not> writing more papers. The idea that writing no research papers equals> doing nothing shows a warped view of the world.Adding to this ... A PhD program in mathematics that ONLY prepares theparticipant for writing research papers is a seriously incompleteprogram at best. Data shows that only about 20% of math PhDs in the USwill end up at PhD-granting universities. >> Most of those PhDs go to positions where research is not encouraged,>> rather teaching and service are encouraged. 4-year colleges, community>> colleges, even high schools. That could also be a reason for not>> writing more papers. The idea that writing no research papers equals>> doing nothing shows a warped view of the world.>>Adding to this ... A PhD program in mathematics that ONLY prepares the>participant for writing research papers is a seriously incomplete>program at best. Data shows that only about 20% of math PhDs in the US>will end up at PhD-granting universities.There is neither a logical nor a pragmatic connection between yourlast two sentences. Many universities and colleges which do notgrant PhDs (in mathematics) nonetheless have (however unreasonablyand/or unrealistically) a requirement that their (mathematics)faculty members write and publish research papers, at leastif they expect to get tenure and/or merit raises. Lee Rudolph Im curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.> Of not leaving with a Ph.D.? 75% would be my guess, based on the eight> years Ifive been at Colorado. DougWell, at my University you need an A average in your graduate coursesto be aloud entrance into the PH.D. program. Would it still be a 75%failure rate you think ?? > Hey,>> Im curious, what would you guys/gals say the probability of someone>> entering a Ph.D. program in Math or Stats and not finishing it. i.e.>> dropping out.>I just read that it was about 50-50. Long ago, I heard that it is>another 50-50 that one who finishes will do nothing after their>thesis. This suggests that a lot of theses are written by the>advisor.How in the hell does it suggest that? If I go into industry after finishing,maybe Ifim not writing papers, but that doesnfit mean that my thesis waswritten for me.Doug Clearly Podunk ainfit MSRI. Southwest will, however, get you to Oakland for> $99. I donfit know what AC Transit costs these days, but it canfit be much. > Even the numb and tired and disillusioned have choices, I would think.if youfive got the patience to ride the bus from Oakland, kudos to you. Itfis only $2.25, but it takes over an hour and a half just to get todowntown Berkeley. if you take the BART, itfis 4.25 total, and worththe $2.Ben > Most of those PhDs go to positions where research is not encouraged,>> rather teaching and service are encouraged. 4-year colleges, community>> colleges, even high schools. That could also be a reason for not>> writing more papers. The idea that writing no research papers equals>> doing nothing shows a warped view of the world.>Adding to this ... A PhD program in mathematics that ONLY prepares the>participant for writing research papers is a seriously incomplete>program at best. Data shows that only about 20% of math PhDs in the US>will end up at PhD-granting universities.Such a program will only prepare the participant for doingresearch in a narrow area. Unfortunately, these seem to bemost of what is being done now, especially in statistics.The emphasis on interdisciplinary programs mainly producesthose who do not know the basics of anything, but theseprograms have high rates of finishing.Students are not getting the basics of set theory, algebra,analysis, and topology these days. Learning how to computeand how to solve certain types of problems fails if basicmaterial not covered in that is needed. Abstract conceptsare needed for understanding, even if the details of themare not used. -- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University Hey,>Im curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.No one seems to have mentioned it, but I think it may depend on thespecific university and what its criteria are for being admitted intothe program. Averaging over all U.S. schools would probably not providea meaningful statistic. My guess is that there is a significant variation.Hard data could be obtained by asking the Director of Graduate Studiesfor various programs. At my school Ifill ask him the next time wefire onthe tennis courts.--John E. PrussingUniversity of Illinois at Urbana-ChampaignDepartment of Aerospace Engineeringhttp://www.uiuc.edu/~prussing =available online in PDF format at http://www.ams.org/employment/asst.pdfEveryone thinking about graduate school in mathematics should look atthis booklet. For example, the first institution that I looked at in the booklet had72 full time graduate students, 15 part time graduate students, and 15full time first year graduate students. The department had graduated18 MS students in the past year, and an average of 3 PhDfis per yeargets an MS, but only about one in five entering graduate students goeson to get a PhD. Of course, you should go back to previous years booklets to see whether there have been any significant changes inenrollment patterns. Looking at about a dozen schools in this booklet, the ratio of full time first year graduate students to PhDfis per year (notethat PhDfis for the last four years are given in the book, so this hasto be divided by four) runs from about 5-to-1 down to 2-to-1. Of course, some students enter the graduate program intending to getan MS degree. Unfortunately, I canfit think of any way to distinguishthose students from students who were given an MS as a consolationprize. In many cases, the total number of MS and PhD degrees per yearis very similar to the number of full time first year students, indicatingthat most students get at least an MS. In other cases, far fewer degreesare awarded than there are entering students. For the big picture, itfis worth pointing out that there areapproximately 15,000 graduate students in PhD granting departments ofmath and statistics in the US, and that these departments producesomething like 1,000-1,200 PhD graduates per year. These numbershavenfit changed dramatically for the numbers.)-- Brian Borchers borchers@nmt.eduDepartment of Mathematics http://www.nmt.edu/~borchers/Socorro, NM 87801 FAX: 505-835-5366-- Brian Borchers borchers@nmt.eduDepartment of Mathematics http://www.nmt.edu/~borchers/Socorro, NM 87801 FAX: 505-835-5366 In curious, what would you guys/gals say the probability of someone>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>dropping out.>> No one seems to have mentioned it, but I think it may depend on the> specific university and what its criteria are for being admitted into> the program. Averaging over all U.S. schools would probably not provide> a meaningful statistic. My guess is that there is a significant variation.>> Hard data could be obtained by asking the Director of Graduate Studies> for various programs. At my school Ifill ask him the next time wefire on> the tennis courts.>> --> John E. Prussing> University of Illinois at Urbana-Champaign> Department of Aerospace Engineering> http://www.uiuc.edu/~prussingI did mention it, when this thread started. I also suggested a few schoolswhere I guessed (I have no data) that the probability of finishing might behighest. Wow, i never would have thought it be that high. I would have guessed>maybe 30% drop out rate. I figured after someone got their Masters in>Mathematics, got straight Afis in their graduate courses that it would>be sufficient to prepare them for the doctorate program in mathematics>/or statistics.But that ability to do well in classes is not particularly well correlatedwith the ability to generate new mathematical ideas. You donfit get aPhD for taking a lot of classes, you know. (In some places, you donfittake _any_ classes to get a PhD.)I would also object to a phrase like drop out. In secondary schooland below, there is a clear expectation that degree completion is thenecessary goal for everyone of that age. At the graduate level, and evenat the undergraduate level, leaving a program is not necessarily anindication of some kind of failure. Studentsfi eyes are opened in schoolto the reality of the career choices for which they are preparing, andthey may well decide they donfit like that image -- even if theyfire doingwell and can continue to do well. Even if your mathematical skills aresuperb, if what you want to do is make a lot of money, or to have timeto raise a family, or to work with some of the worldfis needy people,then you would be making a mistake to complete a PhD in mathematics.dave > This suggests that a lot of theses are written by the advisor.Right. After all, why *wouldnfit* a professor want to forego his orher own research activities for a couple years to write an enormouspaper under a studentfis name?Maybe you meant to say something less hilarious.-- Kevin Hey,>> Im curious, what would you guys/gals say the probability of someone> entering a Ph.D. program in Math or Stats and not finishing it. i.e.> dropping out.>> I just read that it was about 50-50. Long ago, I heard that it is> another 50-50 that one who finishes will do nothing after their> thesis. This suggests that a lot of theses are written by the> advisor.> Not in the least. In graduate school you are surrounded by excellent> mathematicians and the spirit of mathematics. Mathematics is everywhere; it> is the whole world. Everybody around you thinks that itfis the only thing> worth learning.> Then you get a job at Podunk, and discover that your newfound colleagues> think that knowing mathematics is knowing the difference between addition> and subtraction. Discussions in the faculty lounge are about football.> You teach 12 to 15 credits a week, same old stuff year after year. You get> numb and tired and disillusioned (Pirsig mentions this in Zen and the Art of> Motorcycle Maintenance). You have no real contact with the living world of> mathematics and mathematicians; all youfive got is your Calculus I textbook> and your colleagues. With great effort you can scare up money to go to the> occasional convention.> Some people overcome these obstacles, bless them.Ok, let me put it this way. I KNOW that a lot of PhD theses arewritten by the advisors. Let me see, I have had 8 PhD students. Ofhad only the most minimal help; four had a lot of help and explanationdescribed a PhD thesis as a work by the advisor under adversecircumstances and I know for a fact that that was true in his case. Wherever I have been there is always one supervisor who is known towrite all or nearly all of his studentsfi theses. One once complainedthat he didnfit mind writing them, it was having to explain them thathe objected to.But yes, there are other explanations for why people donfit go on to dotheir own work, but as I look at my students there is a strongcorrelation between what they did in grad school and what they didafterwards. > This suggests that a lot of theses are written by the advisor.>>Right. After all, why *wouldnfit* a professor want to forego his or>her own research activities for a couple years to write an enormous>paper under a studentfis name?>>Maybe you meant to say something less hilarious.A lot of people have pointed out that this does not necessarilysuggest that. Barr just posted a reply, saying let me put itthis way and then asserting that in _fact_ a lot of PhD thesesare written by advisors. Thatfis not really putting it anotherway, itfis a separate assertion.And whether you believe it or not, itfis a _fact_ that a lot ofPhD theses are essentially written by the advisor. Barrsays hefis seen a lot of this - so have I. Have you spent alot of time on the faculty in a PhD-granting math department,or is your disbelief just motivated by your wonderment asto why a professor would do such a thing?(Regarding why a professor would do such a thing: First,it doesnfit mean hefis putting his own research on hold forthose years. Anyway, there are all sorts of reasons: youhave a student who possibly should have been kickedout years ago but wasnfit - after the guyfis been here forfive or six years, passed his exams and courses andall, you really hate to kick him out just because hecanfit do the thesis. Or in more cynical vein: If noneof the students get degrees then sooner or later thebean counters will remove the PhD program from thedepartment, and then the professor will have to teachtrigonometry instead of advanced course. All sorts ofreasons it happens.Not that _I_five ever done such a thing of course...)************************David C. Ullrich > Hey,>> Im curious, what would you guys/gals say the probability of someone>> entering a Ph.D. program in Math or Stats and not finishing it. i.e.>> dropping out.>>I just read that it was about 50-50.> 50-50 for finishing or not finishing?> ;-)Yes.;-) <877k6cyh1x.fsf@saurus.asaurus.invalid> <6vrnhvs?fhciu879crdm26mc61pqabf6@4ax.com> This suggests that a lot of theses are written by the advisor.>>Right. After all, why *wouldnfit* a professor want to forego his or>>her own research activities for a couple years to write an enormous>>paper under a studentfis name?>>Maybe you meant to say something less hilarious.>> A lot of people have pointed out that this does not necessarily> suggest that. Barr just posted a reply, saying let me put it> this way and then asserting that in _fact_ a lot of PhD theses> are written by advisors. Thatfis not really putting it another> way, itfis a separate assertion.>> And whether you believe it or not, itfis a _fact_ that a lot of> PhD theses are essentially written by the advisor. Barr> says hefis seen a lot of this - so have I. Have you spent a> lot of time on the faculty in a PhD-granting math department,> or is your disbelief just motivated by your wonderment as> to why a professor would do such a thing?What does it mean when you and he say that a thesis has been(essentially) written by an advisor? You surely donfit mean that theadvisor has contributed some non-negligible portion of the LaTeXsource file, do you? Do you mean that the advisor has contributedalmost every original idea in the thesis? And also most (orsubstantial parts) of the presentation decisions?I suspect that Ifim not alone in being unclear on what you and MichaelBarr mean.-- [R]eality has a fascinating ability to check us when we get a little toobig for our britches... Make no mistake. There isnfit a mathematician alivetoday that I canfit now touch, and not a mathematical career on the planetthat I canfit now affect. --James Harris, render of worlds Not that _I_five ever done such a thing of course...)> I believe Hans Zassenhaus was once quoted as saying he didnfit mindwriting the thesis for the student, but he refused to then TEACH it tothe student so the student would be able to defend it.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > This suggests that a lot of theses are written by the advisor.> Right. After all, why *wouldnfit* a professor want to forego his or> her own research activities for a couple years to write an enormous> paper under a studentfis name?> Maybe you meant to say something less hilarious.At some universities itfis a written rule, and at othersitfis an unwritten rule, that a criterion for tenure/promotion,one must have produced Ph.D. students. Besides that, itfis personally embarrassing if onefis studentsdonfit make it, and itfis ego-boosting if one produces manystudents. Further, one hopes that if one feeds the student one ideahefill get going. So one primes the pump. Then primes itagain. Then again. Then again. Then the student has enough to call it dissertation.Itfis easy to see how it happens. Especially if youfire ina department that produces few Ph.D.s and youfid like to increase that number. How can you attract good studentsto your program if youfire record for number of Ph.Dfis graduated per year is only 3 or 4? The pressure could betremendous to get the graduates out there, get the numbersup, so that better students can be recruited.Or maybe Ifim the department chair working under a dean whomisunderstands what affirmative action is about and Ifim underthe gun to produce quotas from under-represented groups. Soby golly, this one advisee of mine is going to graduate if Ihave to type the damn thing myself and walk across the stagefor him.There are lots of motives, none very honest of course, forsidelining onefis own research a bit for pushing a slow studentthrough. Exasperation being the chief of these.Bart >>Im curious, what would you guys/gals say the probability of someone>>entering a Ph.D. program in Math or Stats and not finishing it. i.e.>>dropping out. Of not leaving with a Ph.D.? 75% would be my guess, based on the>> eight years Ifive been at Colorado. Doug> Well, at my University you need an A average in your graduate courses> to be aloud entrance into the PH.D. program. Would it still be a 75%> failure rate you think ??> I think this points up a misunderstanding. By Entrance to thePh.D. program here, I think you originally meant someone whohas graduate courses, I suppose an MS, has passes qualifyingexams, and a certain amount of Ph.D. coursework done(?)Every school has a difference set of hurdles for admission tothe program. Some have orals for qualifying, some have oralsas defense of the dissertaion. But I think you meant someonepost-masters degree.In which case, 50-50 might be the right answer.But some people are taking entrance as right out of undergrad,in which case the chances of finishing are much lower. Bart But some people are taking entrance as right out of undergrad,>in which case the chances of finishing are much lower. I would say that the majority of students entering the Ph.D. program hereat Colorado have nothing more than a bachelorfis degree. Not the vastmajority, but a majority nonetheless.Doug >But some people are taking entrance as right out of undergrad,>>in which case the chances of finishing are much lower. > I would say that the majority of students entering the Ph.D. program here> at Colorado have nothing more than a bachelorfis degree. Not the vast> majority, but a majority nonetheless.My point was that does one mean by entering the Ph.D programeither A. starting grad school with the intent of getting aPh.D. or B. Being finally accepted into the program, havingpassed qualifiers or the equivalent. The answer to the OPfis originial question depends on what one means. Bart This suggests that a lot of theses are written by the advisor.>>Right. After all, why *wouldnfit* a professor want to forego his or>her own research activities for a couple years to write an enormous>paper under a studentfis name?>>Maybe you meant to say something less hilarious.>> A lot of people have pointed out that this does not necessarily>> suggest that. Barr just posted a reply, saying let me put it>> this way and then asserting that in _fact_ a lot of PhD theses>> are written by advisors. Thatfis not really putting it another>> way, itfis a separate assertion.>> And whether you believe it or not, itfis a _fact_ that a lot of>> PhD theses are essentially written by the advisor. Barr>> says hefis seen a lot of this - so have I. Have you spent a>> lot of time on the faculty in a PhD-granting math department,>> or is your disbelief just motivated by your wonderment as>> to why a professor would do such a thing?>>What does it mean when you and he say that a thesis has been>(essentially) written by an advisor? You surely donfit mean that the>advisor has contributed some non-negligible portion of the LaTeX>source file, do you? No.>Do you mean that the advisor has contributed>almost every original idea in the thesis? Yes. Sometimes it appears to be somewhat morethan almost every. Honest, Ifive seen it happen(much too close to naming names already, althoughI donfit think the people I have in mind were herewhen you were here anyway... but I have two specificstudents in mind, two different advisors.)>And also most (or>substantial parts) of the presentation decisions?>>I suspect that Ifim not alone in being unclear on what you and Michael>Barr mean.************************David C. Ullrich suggest that. Barr just posted a reply, saying let me put it> this way and then asserting that in _fact_ a lot of PhD theses> are written by advisors. Thatfis not really putting it another> way, itfis a separate assertion.>> And whether you believe it or not, itfis a _fact_ that a lot of> PhD theses are essentially written by the advisor. Barr> says hefis seen a lot of this - so have I. Have you spent a> lot of time on the faculty in a PhD-granting math department,> or is your disbelief just motivated by your wonderment as> to why a professor would do such a thing?>>What does it mean when you and he say that a thesis has been>>(essentially) written by an advisor? You surely donfit mean that the>>advisor has contributed some non-negligible portion of the LaTeX>>source file, do you? > No.>>Do you mean that the advisor has contributed>>almost every original idea in the thesis? > Yes. Sometimes it appears to be somewhat more> than almost every. Honest, Ifive seen it happen> (much too close to naming names already, although> I donfit think the people I have in mind were here> when you were here anyway... but I have two specific> students in mind, two different advisors.)> Well, how many original ideas are in an average thesis anyway? I wouldbet no more than one, or the germ of one. And I wouldnfit be too surprisedto learn if a given random thesisfi one original idea was heavily hinted ator suggested outright by the advisor. My impressions are that Ph.D.theses are *usually* an indicator of hard work and persistence rather thanoriginality. But all this raises an interesting point. Some people, on their own,write terrific theses, and even soon after their theses are touted to the community at large (or perhaps a smaller, more specialized community). But some good mathematicians do not fall into this category. Theirtheses problems are suggested by their advisors, who also occasionallygive hints of various kinds. The boundary between an advisorfiscontributions and a studentfis can get quite blurred. In fact, I wouldargue that a good advisor is defined by how blurred this boundary is. Itfis not so easy to strike the right balance, and I think thatfis why a lotof people end up suggesting too much to their students. The other extremeof not suggesting anything is a luxury that only a minority of professorscan get away with. ...>Do you mean that the advisor has contributed>almost every original idea in the thesis? Yes. Sometimes it appears to be somewhat more>> than almost every. ...>Well, how many original ideas are in an average thesis anyway? I would>bet no more than one, or the germ of one. And I wouldnfit be too surprised>to learn if a given random thesisfi one original idea was heavily hinted at>or suggested outright by the advisor. My impressions are that Ph.D.>theses are *usually* an indicator of hard work and persistence rather than>originality. But, but...it says right there on my diploma that my Ph.D. isin recognition of scientific attainments and the ability to carry on original research as demonstrated by a thesis. And itfis *signed*, and everything. Are you calling Jerry Wiesner a liar, sir? (For that matter, are you calling the me of 30 yearsago either hard working or persistent?)>But all this raises an interesting point. Some people, on their own,>write terrific theses, and even soon after their theses are touted to >the community at large (or perhaps a smaller, more specialized community).> But some good mathematicians do not fall into this category. Their>theses problems are suggested by their advisors, who also occasionally>give hints of various kinds. The boundary between an advisorfis>contributions and a studentfis can get quite blurred. In fact, I would>argue that a good advisor is defined by how blurred this boundary is. >Itfis not so easy to strike the right balance, and I think thatfis why a lot>of people end up suggesting too much to their students. The other extreme>of not suggesting anything is a luxury that only a minority of professors>can get away with. Not having been in a position to be an advisor myself, I have had to rely on other people (some of whom I donfit even know...) to give their students hints of the form see if you can go any further withthis half-baked idea of Rudolphfis. Ifill tell you, *that* is a luxury. And it saves me the trouble of using my (non-existent) pull to get first jobs for my (non-existent) advisees, too.Lee Rudolph (what, me bitter?) > suggest that. Barr just posted a reply, saying let me put it> this way and then asserting that in _fact_ a lot of PhD theses> are written by advisors. Thatfis not really putting it another> way, itfis a separate assertion.>> And whether you believe it or not, itfis a _fact_ that a lot of> PhD theses are essentially written by the advisor. Barr> says hefis seen a lot of this - so have I. Have you spent a> lot of time on the faculty in a PhD-granting math department,> or is your disbelief just motivated by your wonderment as> to why a professor would do such a thing?>>What does it mean when you and he say that a thesis has been>(essentially) written by an advisor? You surely donfit mean that the>advisor has contributed some non-negligible portion of the LaTeX>source file, do you? No.>>Do you mean that the advisor has contributed>almost every original idea in the thesis? Yes. Sometimes it appears to be somewhat more>> than almost every. Honest, Ifive seen it happen>> (much too close to naming names already, although>> I donfit think the people I have in mind were here>> when you were here anyway... but I have two specific>> students in mind, two different advisors.)Well, how many original ideas are in an average thesis anyway? I would>bet no more than one, or the germ of one. And I wouldnfit be too surprised>to learn if a given random thesisfi one original idea was heavily hinted at>or suggested outright by the advisor. My impressions are that Ph.D.>theses are *usually* an indicator of hard work and persistence rather than>originality. Of course. But after the advisor hints at or suggests the result thestudent is supposed to have at least _something_ to do withactually coming up with the proof. In the cases I have in mindthat was not so - instead there was an endless series of:Ok, why not try this:Ok, Ifill look at that.[week passes]So did that work?Donfit know, couldnfit figure anything out either way.Hmm, letfis see...[delay of minutes or days]No, that doesnfit work [or does work]. Why not try this:Ok...(Or so the advisor claimed during endless bitch&moan sessionsduring those years, and I believe him, because itfis _exactly_ whathappened earlier when I was helping the same student, goingthrough all the exercises in some book one summer - he simplynever made any progress on any of them, with maybe twoexceptions, all the solutions were due to me the week afterthe exercise was assigned.)>But all this raises an interesting point. Some people, on their own,>write terrific theses, and even soon after their theses are touted to >the community at large (or perhaps a smaller, more specialized community).> But some good mathematicians do not fall into this category. Their>theses problems are suggested by their advisors, who also occasionally>give hints of various kinds. The boundary between an advisorfis>contributions and a studentfis can get quite blurred.Yes, no doubt itfis quite blurry in the typical case. Why you thinkit woud be blurry in the cases Ifim referring to, given mycharacterizations of them, is beyond me.> In fact, I would>argue that a good advisor is defined by how blurred this boundary is. >Itfis not so easy to strike the right balance, and I think thatfis why a lot>of people end up suggesting too much to their students. The other extreme>of not suggesting anything is a luxury that only a minority of professors>can get away with. >>************************David C. Ullrich =You guys make getting a math Ph.d sound as though it isnfit even worth thetime! As a senior math student intending to go to grad school, I donfit findanything in everyonesfi comments very encouraging.It seems as though getting a math Ph.d boils down to the following:1) Forego homeownership for student loans.2) Work incredibly hard to understand something that nobody else cares tounderstand because they are only interested in making lots of money.3)Assiduously toil during your undergraduate years in order to attain a nearperfect gpa that will get you into a respectable grad program.4)Do the same as (3), but insert MA/MS and Ph.d program.5)Enter a Ph.d program and have an advisor write a thesis for you.6)Go to Podunk U. and attempt to do some original research.7)When (6) fails, develop a drinking problem.8)Retire9)Die, but still in student loan debt and living in an off-campus apartmentunfulfilled and anonymous.> Hey,>> Im curious, what would you guys/gals say the probability of someone> entering a Ph.D. program in Math or Stats and not finishing it. i.e.> dropping out. You guys make getting a math Ph.d sound as though it isnfit even worth the>time! As a senior math student intending to go to grad school, I donfit find>anything in everyonesfi comments very encouraging.He asked a question - people tried to answer as accurately as theycould.I donfit think anyonefis said itfis not worth the time. People have saiditfis not easy. Itfis not. When he asks what proportion of PhD studentsget PhDfis do you think wefid really be doing him or anyone else afavor by _lying_, saying itfis no problem for most students?I mean really, by all means go to grad school in math! We did,and we all think it would be great if you did too.>It seems as though getting a math Ph.d boils down to the following:>>1) Forego homeownership for student loans.>2) Work incredibly hard to understand something that nobody else cares to>understand because they are only interested in making lots of money.>3)Assiduously toil during your undergraduate years in order to attain a near>perfect gpa that will get you into a respectable grad program.>4)Do the same as (3), but insert MA/MS and Ph.d program.>5)Enter a Ph.d program and have an advisor write a thesis for you.>6)Go to Podunk U. and attempt to do some original research.>7)When (6) fails, develop a drinking problem.>8)Retire>9)Die, but still in student loan debt and living in an off-campus apartment>unfulfilled and anonymous.Thatfis more or less the procedure, yes. Step 4 is optional - in a lotof PhD programs they pay no attention to whether you have aMasterfis dergree, many of the students are straight out ofundergrad. (At Wisconsin the Masterfis was more or less aconsolation prize for students whofid done the courseworkbut didnfit finish the PhD.)>> Hey,>> Im curious, what would you guys/gals say the probability of someone>> entering a Ph.D. program in Math or Stats and not finishing it. i.e.>> dropping out.>************************David C. Ullrich You guys make getting a math Ph.d sound as though it isnfit even worth> the time! As a senior math student intending to go to grad school, I> donfit find anything in everyonesfi comments very encouraging.> It seems as though getting a math Ph.d boils down to the following:> 1) Forego homeownership for student loans.> 2) Work incredibly hard to understand something that nobody else cares> to understand because they are only interested in making lots of> money. > 3)Assiduously toil during your undergraduate years in order to> attain a near perfect gpa that will get you into a respectable grad> program. > 4)Do the same as (3), but insert MA/MS and Ph.d program.> 5)Enter a Ph.d program and have an advisor write a thesis for you.> 6)Go to Podunk U. and attempt to do some original research.> 7)When (6) fails, develop a drinking problem.> 8)Retire> 9)Die, but still in student loan debt and living in an off-campus> apartment unfulfilled and anonymous.You have Step 7 put off waaaaaaay to long. Youfill want to get thatdrinking problem going pretty much at the beginning of Step 3.(Itfis a lot easier to get tenure if your colleagues percieve youas a nonthreatening, harmless sot. And youfill make a better Deanthat way.)Bart >But all this raises an interesting point. Some people, on their own,>>write terrific theses, and even soon after their theses are touted to >>the community at large (or perhaps a smaller, more specialized community).>> But some good mathematicians do not fall into this category. Their>>theses problems are suggested by their advisors, who also occasionally>>give hints of various kinds. The boundary between an advisorfis>>contributions and a studentfis can get quite blurred.> Yes, no doubt itfis quite blurry in the typical case. Why you think> it woud be blurry in the cases Ifim referring to, given my> characterizations of them, is beyond me.> I didnfit mean to imply that your cases were of that kind, but I only meantto provide some kind of explanation of why that might happen, as I think Idid with the snippet below. >> In fact, I would>>argue that a good advisor is defined by how blurred this boundary is. >>Itfis not so easy to strike the right balance, and I think thatfis why a lot>>of people end up suggesting too much to their students. The other extreme>>of not suggesting anything is a luxury that only a minority of professors>>can get away with. > ************************> David C. Ullrich Thatfis more or less the procedure, yes. Step 4 is optional - in a lot>of PhD programs they pay no attention to whether you have a>Masterfis dergree, many of the students are straight out of>undergrad. (At Wisconsin the Masterfis was more or less a>consolation prize for students whofid done the coursework>but didnfit finish the PhD.)The same at Colorado - when I entered after completing my bachelorfisdegree, I chose the Masterfis degree program (assuming that it had to bedone before the doctorate), and was convinced otherwise by the chair ofthe department.Doug It seems as though getting a math Ph.d boils down to the following:>>1) Forego homeownership for student loans.>2) Work incredibly hard to understand something that nobody else cares to>understand because they are only interested in making lots of money.>3)Assiduously toil during your undergraduate years in order to attain a near>perfect gpa that will get you into a respectable grad program.>4)Do the same as (3), but insert MA/MS and Ph.d program.>5)Enter a Ph.d program and have an advisor write a thesis for you.>6)Go to Podunk U. and attempt to do some original research.>7)When (6) fails, develop a drinking problem.>8)Retire>9)Die, but still in student loan debt and living in an off-campus apartment>unfulfilled and anonymous.Oh, itfis not that bad. OCuz if you post to sci.math then you wonfit be anonymous when you die.Anon. Oh, itfis not that bad. >fiCuz if you post to sci.math then you wonfit be anonymous when you die.>>Anon.On the Internet, no one knows youfire dead.Lee Rudolph You guys make getting a math Ph.d sound as though it isnfit even worth the> time! As a senior math student intending to go to grad school, I donfit find> anything in everyonesfi comments very encouraging.> It seems as though getting a math Ph.d boils down to the following:> 1) Forego homeownership for student loans.> 2) Work incredibly hard to understand something that nobody else cares to> understand because they are only interested in making lots of money.> 3)Assiduously toil during your undergraduate years in order to attain a near> perfect gpa that will get you into a respectable grad program.> 4)Do the same as (3), but insert MA/MS and Ph.d program.> 5)Enter a Ph.d program and have an advisor write a thesis for you.> 6)Go to Podunk U. and attempt to do some original research.> 7)When (6) fails, develop a drinking problem.> 8)Retire> 9)Die, but still in student loan debt and living in an off-campus apartment> unfulfilled and anonymous.#7 should occur during graduate school. Immediately after myundergraduateI went to graduate school drinking about the same amount as I did asanundergraduate (which seemed like a lot at the time). After 2 years, Ihadto quit graduate school because the material is so much harder than itisas an undergraduate. I donfit think people have stressed that enoughon thisthread. Graduate level mathematics is much harder than undergraduatelevelmathematics. If you are the top student in your class, you will beput inyour proper place quickly in graduate school. Herein lies theimportanceof developing a drinking problem. It is why I had to quit after 2years.Shortly before coming back to graduate school, I was talking tosomeonewho gave me this piece of wisdom: The only thing that got me throughgrad school was alchohol. As it stands now, I drink far more than I ever did as an undergraduate. In my department, it seems to be ageneralrule that the ones who make it (or who are going to make it) havedrinking problems, and those who donfit, donfit. Is alchoholism worththe opportunityto pursue mathematical knowledge on a daily basis? No doubt its ahardquestion, but its one you have to ask.On the more serious note of advisorfis writing their studentfis thesis:I have only attended one PhD defence and in that talk the studentfisadvisornever once had to steer the student in the right direction during thequestion/answer session. I do have to admit, though, that they didasksome fairly trivial questions - for example one could be done simplybyusing Fatoufis lemma (which the candidate confidently answered). Hugh Oh, itfis not that bad.>fiCuz if you post to sci.math then you wonfit be anonymous when you die.>>Anon.>> On the Internet, no one knows youfire dead.>> Lee RudolphHeard in an English pub, to the tune of Irish Washerwoman:Oh, McTavish is dead and his brother donfit know it,Theyfire both of them dead and theyfire in the same bed,And neither one knows that the other is dead. =I have a Russian friend who was a Doctor in quantum physics.I asked him about his education, and I was blown away aboutwhat he told me about getting a PhD in Russia. (During thecold war). He said to get a PhD in Russia, you have to had 4-5 publishedpapers in a respectable journal of your field; that a PhDis looked upon like a Masters in the States. Therefisanother level of education above PhD called Doctor. And,of course, you need to write even more papers. And when youbecome a Doctor, then you earn the title Doctor.He said after Doctor, therefis another step called Academicbut he said that such a title is mostly political.I could be wrong about what he said, because I was just so shockedto learn how hard it is there.So I think the probability is determined by the country.(Now for my very long aside. Sorry about it, but Ifim justso fond of my Russian friend.... My friend is a Olowlyfi programmerin the States, and back in Russia he was a respected quantumphysicist.He is so eager to solve challenging math problems. I only knowa few Olympiad problems to give him. He pretty much solves themin his head. So I went out one day, and bought an olympiad bookjust to rattle off a problem when he would stop in my cube witha happy smile asking for a math problem.It was so entertaining/jaw dropping to see him usually solvemost of these olympiad problem in his head. I donfit know,perhaps these problems are easy for professional mathematicians.I always thought it impressive.I told him, Man, what are you doing as a corporate programmer?You are too talented to be a programmer? He wasnfit so sure himself.But, it was along the lines of making ends meet. He was moreconcerned about other issues like spending time with friends andfamily, and how to live a good happy life.This seems typical of most Russian emmigrants Ifive met so far.Taxicab drivers, sofware testers, guitar bums all having anextremely good education, and all happy to make some money,and all more concerned about questions of living a meaningfullife. I contrast this with alot of my American friends whoseem to want a Masters, or an MBA just as an Oedgefi in a ratrace.Just the kinda thing that makes you go Hmmmm...Well, those Russians, you gotta love them!)> I donfit think anyonefis said itfis not worth the time. People have said> itfis not easy. Itfis not. >I have found an elegant proof that the derivative of sin(x) is cos(x).>I have studied two a-levels in maths and read lots of math books but>have not come across this particular proof before.>>Obviously, this is not a groundbreaking proof, it is simply a>different way of proving a fundamental result. (without using limits>or infintesimals). However, for personal interest I would like to know>if it is original.>>Is there anywhere I can find a catalogue of existing proofs for the>derivative of sin(x)?>>If it turned out this proof was original should I consider getting it>published or is it not worth it, since it is such a tiny proof.>>Among these things, I am also working on some integration techniques.>Again, I have a similiar problem: I do not know whether this stuff I>am finding is original. I have done 6 modules of pure maths at school>so I am not a complete novice, but on the other hand I am aware that>there is many things that I do not know of pure maths since I have yet>to start my maths degree. Can anyone suggest a website that provides>information on advanced integration techniques, and for that matter>information on higher level maths?>>Any responses to the above would be gratefully received.>>Flame.> 1. Therefis no money in math. So if someone steals your idea, you > havenfit lost any money.I made no reference to money. To quote myself: ...for personalinterest I would like to know if it is original> 2. If youfire doing original work, youfill continue to do original work. > If someone steals an idea, just stay away from that person in the > future and keep doing your original work.If someone steals an idea as you put it, then i would lose priorityas the discoverer. I think that as unfair, strange though it maysound.> 3. The only fame you can expect from doing math is among other > mathematicians.I do not expect fame. I do not understand where you derived this ideafrom my post.> 4. There is money in applying mathematical ideas to other disciplines. Please see my response to 1.> Not deep mathematical ideas, but a little math and some logic and > organization to business problems (and translating your results to > English for your colleagues) will keep you steadily employed at quite > reasonable rates. Thinking original thoughts is using time that could > be spent thinking profitable thoughts. Please see my response to 1.(Of course, if you get paid well > enough, you can afford to spend some time thinking original thoughts. > Itfis much easier to take the lower salary and be a university professor.)> Jon MillerI am currently making enquiries with other professional mathematiciansregarding my work so far. If any developments occur, I will post themhere along with my work.Flame =Can someone give me a hint (not solution) on the following problem. It is number 2.29 in Rotmanfis Introduction to Homological Algebra.Given: g A>B | |f| Prove the following diagram is a pushout: | V C g A-->B | | | | f| ffi| | | V gfi V C-->DWhere D=(C /osum B)/W, W={(fa,-ga):a /in A}, ffi:b-->(0,b)+W and gfi:c-->(c,0)+W.What do I need to prove to prove the diagram is a pushout and what is the significance of the set W? Can someone give me a hint (not solution) on the following >problem. It is number 2.29 in Rotmanfis Introduction to >Homological Algebra.>>Given:> g> A>B> |> |>f| Prove the following diagram is a pushout:> |> V> C>> g> A-->B> | |> | | >f| ffi|> | |> V gfi V> C-->D>>Where D=(C /osum B)/W, W={(fa,-ga):a /in A}, ffi:b-->(0,b)+W and >gfi:c-->(c,0)+W.>What do I need to prove to prove the diagram is a pushoutThe diagram is a pushout if it satisfies the following two conditions:1. COMMUTATIVITY: ffig = gfif (I am applying functions on the left, so they should be read right-to-left; ffig means g first, then ffi).2. UNIVERSAL PROPERTY: Given any K and maps b:B->K, c:C->K such that bg=cf, there exists a unique map d:D->K such that b=dffi and c=dgfi.> and >what is the significance of the set W?You can think of a pushout as a co-equalizer; you are finding thelargest object on which you can make f and g Oequalfi. W is a measureof how far they are on being Oequalfi (not exactly, since we aredealing with the dual notion, but maybe that makes some sense toyou?). In order for ffig(a) to be equal to gfif(a) for all a, you needto make sure that (f(a),0) is the same as (0,g(a)); for them to bethe same, you need to mod out by (f(a),-g(a)); so W is the closure ofall those identities in Cosum B; moding out by W is the same asimposing those identities on Cosum B. = selective about what I accept as reality. Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu pushout if it satisfies the following two conditions:> 1. COMMUTATIVITY: ffig = gfif (I am applying functions on the left, so> they should be read right-to-left; ffig means g first, then ffi).> 2. UNIVERSAL PROPERTY: Given any K and maps b:B->K, c:C->K such that> bg=cf, there exists a unique map d:D->K such commutativity. That part was really obvious. Theuniversal property is the part giving me the most trouble. I amassuming that one must use a previously proved universal property toobtain the one in question. I am not seeing how to construct or provethe existence of such a map d:D->K, for any K. I donfit mind if youspoil the problem now, unless you think you can give me a suitablehint.Chris it satisfies the following two conditions: 1. COMMUTATIVITY: ffig = gfif (I am applying functions on the left, so>> they should be read right-to-left; ffig means g first, then ffi). 2. UNIVERSAL PROPERTY: Given any K and maps b:B->K, c:C->K such that>> bg=cf, there exists a unique map d:D->K the commutativity. That part was really obvious. The>universal property is the part giving me the most trouble. I am>assuming that one must use a previously proved universal property to>obtain the one in question. >> I am not seeing how to construct or prove>the existence of such a map d:D->K, for any K. I donfit mind if you>spoil the problem now, unless you think you can give me a suitable>hint.I donfit follow what you mean. Assume you have an object K, and mapsb:B->K and c:C->K such that bg = cf. f A > C | | g | | gfi | | V V B -> D ffiWe know that D is defined as (Boplus C)/W; so to define a map from Dto K, we can define a map from Boplus C to K whose kernel contains W,and factor it through the quotient. ffi and gfi are the obviousinclusions, and W is the subgroup generated by allpairs (g(a),-f(a)) for a in A.So letfis consider what the d HAS to be. First, we want b=dffi andc=dgfi. So given any x in B, we know what b(x) is (we are GIVEN themaps b and c); and we know that ffi(x) = (x,0) in Boplus C. So we map(x,0) to b(x).Likewise, we will need to map (0,y) to c(y) for all y in C.That means that we need to map an element (x,y) in Boplus C tob(x)+c(y).That defines a map, call it e: B oplus C -> K.Now we need to verify that e factors through the quotient D, that is,that W is contained in the kernel of e.So letfis take an element of W, which is of the form (g(a),-f(a)) for ain A. According to the definition of e, we mape(g(a),-f(a)) = b(g(a))+c(-f(a)) = b(g(a)) - c(f(a)) = bg(a) - cf(a).But we are assuming futher that b and c are such that bg=cf; sobg(a)-cf(a)=0 for all a in A. Therefore, e takes W to 0, and so W iscontained in the kernel of e.Therefore, e factors through the quotient p:Boplus C -> (Boplus C)/W = D.So define d to be the unique map from (Boplus C)/W to K such that commutativity condition, b=dffi and c=dgfi.Moreover, since the definition of e was forced by the commutativity ofthe diagram, the choice of d is also forced, so that d is the onlyfunction that will fit in that diagram. Thus, d is unique.In general, when you have a universal construction, IF you have an->explicit<- construction of the object, then the universal propertyis easy to verify, because you will have no choice about how to definethe map in question. it should be obvious what the map has to be inan object. Itfis not denial. Ifim just very selective about what I accept as reality. Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu =to this question, which has been driving me a bit nuts. Some quickbackground: as a role-player, I use a lot of dice. At times, therules call for one to roll multiple dice (say, five six-sided dice, or5d6), then to drop the lowest two and total the other three. Thebasic question is: is there a formula for determining the probabilityof rolling a certain result, given these conditions?Determining the probability of a particular outcome when just rollingmultiple dice is relatively straightforward (therefis a briefdiscussion here: http://mathforum.org/library/drmath/view/52207.html). I can find a pattern to the summation needed when dropping a singledie from a set; but once I try to remove two dice from the set, thepattern disappears and I find myself lost again. (The numbers can bedetermined by brute force, of course, but thatfis neither practical norinteresting.)So, I guess the base question is: Is there a formula for calculatingthe probability of achieving a result R on n dice with d sides,dropping the k lowest dice? =How can the following Wiener-Hopf operator W be bounded?Define W = P M_f P, where M_f is multiplication by a function f in C0(S^1),P is the projection of L2(S^1) onto the subspace spanned by z^k, k >= 0,when f is only assumed to be continuous (e.g what if f is not integrable?)(This is an exercise in Booss: Topology and Analysis)Andreas How can the following Wiener-Hopf operator W be bounded?Bounded on what space? (Or: Bounded in what norm?)>Define W = P M_f P, where M_f is multiplication by a function >f in C0(S^1),>>P is the projection of L2(S^1) onto the subspace spanned by z^k, k >= 0,If youfire asking about boundedness in L2 this is obvious, because Pand M_f are both bounded. (M_f is bounded if and only if f isbounded...)>when f is only assumed to be continuous (e.g what if f is not integrable?)??? A continuous function on S^1 is not integrable???I must be missing what you mean by S^1 - thatfis not the unit circle?In any case, if S^1 is locally compact then f in C0(S^1) implies thatf is bounded.>(This is an exercise in Booss: Topology and Analysis)>>Andreas>************************David C. Ullrich Bounded on what space? (Or: Bounded in what norm?)With respect to the L2 norm.>Define W = P M_f P, where M_f is multiplication by a function>f in C0(S^1),>>P is the projection of L2(S^1) onto the subspace spanned by z^k, k >= 0,>> If youfire asking about boundedness in L2 this is obvious, because P> and M_f are both bounded. (M_f is bounded if and only if f is> bounded...)>>when f is only assumed to be continuous (e.g what if f is not integrable?)>> ??? A continuous function on S^1 is not integrable???It dawned on me later that f and M_f must be bounded because f is continuousand defined on a circle... it was my lack of experience in such things.> I must be missing what you mean by S^1 - thatfis not the unit circle?> In any case, if S^1 is locally compact then f in C0(S^1) implies that> f is bounded.(9.4, page 97 of the German edition): If you know the book... I canfit quite seethe significance of the condition sup_S^1 |fg - 1| <1does it mean that T_g is an inverse of P_n T_f , modulo a compact operatorbecause T_(fg -1) is compact?)Not to worry - Ifill figure it out or skip this (minor) point in the book.Andreas Bounded on what space? (Or: Bounded in what norm?)>>With respect to the L2 norm.>Define W = P M_f P, where M_f is multiplication by a function>>f in C0(S^1),>>P is the projection of L2(S^1) onto the subspace spanned by z^k, k >= 0,>> If youfire asking about boundedness in L2 this is obvious, because P>> and M_f are both bounded. (M_f is bounded if and only if f is>> bounded...)>>when f is only assumed to be continuous (e.g what if f is not integrable?)>> ??? A continuous function on S^1 is not integrable???>>It dawned on me later that f and M_f must be bounded because f is continuous>and defined on a circle... it was my lack of experience in such things.In fact, for future reference, if X is just locally compact and f is in C0(X) then f is bounded - thatfis a large part of the differencebetween C0 and C...>> I must be missing what you mean by S^1 - thatfis not the unit circle?>> In any case, if S^1 is locally compact then f in C0(S^1) implies that>> f is bounded.>>(9.4, page 97 of the German edition): If you know the book... Nope, sorry.>I canfit quite see>the significance of the condition sup_S^1 |fg - 1| <1>does it mean that T_g is an inverse of P_n T_f , modulo a compact operator>because T_(fg -1) is compact?)>>Not to worry - Ifill figure it out or skip this (minor) point in the book.>>Andreas************************David C. Ullrich =I place a quarter (coin) on the table. Exactly how many quarters can I putaround this centered quarter? I place a quarter (coin) on the table. Exactly how many quarters can I put> around this centered quarter?Six. Think honeycomb.-- Ioannishttp://users.forthnet.gr/ath/jgal/_ __Eventually, _everything_ is understandable. I place a quarter (coin) on the table. Exactly how many quarters can I put> around this centered quarter?Define put. I place a quarter (coin) on the table. Exactly how many quarters can I put>around this centered quarter?That depends, of course, on the size of the table. :-)-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You find yourself amusing, Blackadder.I try not to ? in the face of public opinion. >I place a quarter (coin) on the table. Exactly how many quarters can I put>>around this centered quarter?> That depends, of course, on the size of the table. :-)It also depends on the definition of around. If we consider it ina three-dimensional sense and place nolimits on distance, then everyquarter on Earth is around it.And since quarter isnfit defined, we could take it to mean one-fourthof anything, which raises the total a bit higher... :-)-- Wayne Brown | When your tailfis in a crack, you improvisefwbrown@bellsouth.net | if youfire good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock =hi all,I am actually trying to understand this mathematical notion that is soweird to me (I am far from being a god at maths...).could someone drop the light onto the following for me ?We will compute a rotation about the unit vector, u by an angle . Thequaternion that computes this rotation is q = (s,v) s = cos(teta/2) v = u * sin(teta/2)We will represent a point p in space by the quaternion P=(0,p) Wecompute the desired rotation of that point by this formula: P = (0,p) Protated = qPq^-1The first thing I donfit understand at all here is where the s and vvalues come from ?!? It might sound stupid but I donfit understandthis.Any help ?thanxSam =hi all,I am actually trying to understand this mathematical notion that is soweird to me (I am far from being a god at maths...).could someone drop the light onto the following for me ?We will compute a rotation about the unit vector, u by an angle . Thequaternion that computes this rotation is q = (s,v) s = cos(teta/2) v = u * sin(teta/2)We will represent a point p in space by the quaternion P=(0,p) Wecompute the desired rotation of that point by this formula: P = (0,p) Protated = qPq^-1The first thing I donfit understand at all here is where the s and vvalues come from ?!? It might sound stupid but I donfit understandthis.Any help ?thanxSam hi all,> I am actually trying to understand this mathematical notion that is so> weird to me (I am far from being a god at maths...).> could someone drop the light onto the following for me ?> We will compute a rotation about the unit vector, u by an angle . The> quaternion that computes this rotation is> q = (s,v)> s = cos(teta/2)> v = u * sin(teta/2)> We will represent a point p in space by the quaternion P=(0,p) We> compute the desired rotation of that point by this formula:> P = (0,p)> Protated = qPq^-1> The first thing I donfit understand at all here is where the s and v> values come from ?!? It might sound stupid but I donfit understand> this.> Any help ?> thanx> Samexpanding q = (s,v) gives a unit quaternion, which rotates R^3 in theform you gave. rotation in R^3 requires a axis of rotation, which thiscase is u, and an angle of rotation, theta. in general, the quaternionwhich gives the rotation is q = cos (theta/2) - sin (theta/2) a(i*j*k). itfis much nicer to consider rotations in the clifford algebraframework, where the unit ball of the even subalgebra rotates theunderlying quadratic space. M.T. =ok thanx, but I still donfit understand why we use cos(theta/2) and u *sin(theta/2) as values for s and v...besides does anyone could enlight me on this :To rotate a vector v an angle of θ around about an arbitrary unitaxis w, you can use the formula:vfi = w(v.w) + (v - w(v.w))cos(θ) + (v^w)sin(θ)how can we end up to this formula ?> hi all,> I am actually trying to understand this mathematical notion that is so> weird to me (I am far from being a god at maths...).> could someone drop the light onto the following for me ?> We will compute a rotation about the unit vector, u by an angle . The> quaternion that computes this rotation is> q = (s,v)> s = cos(teta/2)> v = u * sin(teta/2)> We will represent a point p in space by the quaternion P=(0,p) We> compute the desired rotation of that point by this formula:> P = (0,p)> Protated = qPq^-1> The first thing I donfit understand at all here is where the s and v> values come from ?!? It might sound stupid but I donfit understand> this.> Any help ?> thanx> Sam> expanding q = (s,v) gives a unit quaternion, which rotates R^3 in the> form you gave. rotation in R^3 requires a axis of rotation, which this> case is u, and an angle of rotation, theta. in general, the quaternion> which gives the rotation is q = cos (theta/2) - sin (theta/2) a> (i*j*k). itfis much nicer to consider rotations in the clifford algebra> framework, where the unit ball of the even subalgebra rotates the> underlying quadratic space.> M.T. ok thanx, but I still donfit understand why we use cos(theta/2) and u *>sin(theta/2) as values for s and v...The product of two quaternions s+v and t+w, where s and t are scalarsand v and w are vectors, is (s+v)(t+w) = (st-v.w)+(sw+tv+v^w). Itfollows that (s+v)(s-v) = s**2+v.v, where s**2 denotes the square of s.So if w is a unit vector, then (cos(theta/2)+sin(theta/2)w)^{-1}= cos(theta/2)-sin(theta/2)w, and so (cos(theta/2)+sin(theta/2)w) v (cos(theta/2)+sin(theta/2)w)^{-1}= (cos(theta/2)v-sin(theta/2)w.v+sin(theta/2)w^v) (cos(theta/2)-sin(theta/2)w)= cos(theta/2)**2 v + sin(theta/2) cos(theta/2) v.w + sin(theta/2) cos(theta/2) w^v - sin(theta/2) cos(theta/2) v.w + sin(theta/2)**2 (w.v)w + sin(theta/2) cos(theta/2) w^v - sin(theta/2)**2 (w^v)^w= cos(theta/2)**2 v + 2 sin(theta/2) cos(theta/2) w^v + sin(theta/2)**2 (w.v)w - sin(theta/2)**2 (w.w)v + sin(theta/2)**2 (w.v)w= [cos(theta/2)**2 - sin(theta/2)**2] v + 2 sin(theta/2) cos(theta/2) w^v + 2 sin(theta/2)**2 (w.v)w = cos(theta) v + sin(theta) w^v + (1 - cos(theta)) (w.v)w= (w.v)w + (v - (w.v)w) cos(theta) + w^v sin(theta),which is the result when v is rotated about w by an angle of theta in the right handed sense.>besides does anyone could enlight me on this :>To rotate a vector v an angle of θ around about an arbitrary unit>axis w, you can use the formula:>vfi = w(v.w) + (v - w(v.w))cos(θ) + (v^w)sin(θ)The first thing to note is that this rotation looks like a left handed rotation. For right handed rotations, which I will be dealing with below,a right handed rotation of an angle theta about a unit vector w transforms v to vfi = (v.w)w + (v - (v.w)w) cos(theta) + w^v sin(theta). Note that the only difference between this formula and the one you supplied above is that the sign of the coefficient of v^w is reversed (recall thatw^v = -v^w).When you rotate about the unit vector w, then w and all its multiplesremain fixed, so in particular, for any vector v, (v.w)w remains fixed under the rotation. The plane orthogonal to w turns about the origin,and if a unit vector r is orthogonal to w, then the plane has basis r and w^r, and a right handed rotation about w rotates the plane so thatr moves towards w^r. It follows that r is transformed by a right handed rotation about u through an angle of theta to rfi = r cos(theta) + w^r sin(theta).If v is a multiple of w, then (v.w)w = v and w^v = 0, so that(v.w)w + (v - (v.w)w) cos(theta) + w^v sin(theta) = v,which is the result of rotating v about w by any angle as a consequence of the fact that v is a multiple of w.If v is not a multiple of w, then v - (v.w)w is orthogonal to w, and w^(v - (v.w)w) = v^w, with the result that under right handed rotation about w through an angle of theta, v - (v.w)w transforms to (v-(v.w)w) cos(theta) + w^v sin(theta). Since (v.w)w transforms toitself, then v = (v.w)w + (v-(v.w)w) transforms to(v.w)w + (v - (v.w)w) cos(theta) + w^v sin(theta).>how can we end up to this formula ?David McAnally>> hi all,>> I am actually trying to understand this mathematical notion that is so>> weird to me (I am far from being a god at maths...).>> could someone drop the light onto the following for me ? We will compute a rotation about the unit vector, u by an angle . The>> quaternion that computes this rotation is>> q = (s,v)>> s = cos(teta/2)>> v = u * sin(teta/2) We will represent a point p in space by the quaternion P=(0,p) We>> compute the desired rotation of that point by this formula:>> P = (0,p)>> Protated = qPq^-1>> The first thing I donfit understand at all here is where the s and v>> values come from ?!? It might sound stupid but I donfit understand>> this.>> Any help ?>> thanx>> Sam expanding q = (s,v) gives a unit quaternion, which rotates R^3 in the>> form you gave. rotation in R^3 requires a axis of rotation, which this>> case is u, and an angle of rotation, theta. in general, the quaternion>> which gives the rotation is q = cos (theta/2) - sin (theta/2) a>> (i*j*k). itfis much nicer to consider rotations in the clifford algebra>> framework, where the unit ball of the even subalgebra rotates the>> underlying quadratic space.>> M.T. ok thanx, but I still donfit understand why we use cos(theta/2) and u *> sin(theta/2) as values for s and v...> besides does anyone could enlight me on this :> To rotate a vector v an angle of θ around about an arbitrary unit> axis w, you can use the formula:> vfi = w(v.w) + (v - w(v.w))cos(θ) + (v^w)sin(θ)> how can we end up to this formula ? work out the geometry. in R^n, let nfi be the greatest even number <=n, the a rotation in R^n is a product of nfi re?ctions. so work out aformula for re?ctions and the result for rotation follows directly.in doing so, one may want to note that successive re?ctions aboutvectors a and b is a rotation about a X b by the angle between a andb. again, it is more convenient to consider this in the cliffordalgebra framework. M.T.