> 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 me believe that author is not (classical) Mathematician but rather control systems 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. Augustine You can always check out the whois entry: Registrant: MathPages (MATHPAGES-DOM) 1605 W JAMES LN # I-7 KENT, WA 98032-4351 US Domain Name: 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 people ExCUSE me, but I prefer the term knowbies for the yang to the newbies' yin. dave ==== In sci.math, Dave Rusin : > >>I recently posted a message to newbies > >>Now I wish to exhort oldies not to work homework problems for people > > ExCUSE me, but I prefer the term knowbies > for the yang to the newbies' yin. > > dave > So if one's just graduated from college is he a newbie knowbie, or a knowbie noveau? *dodges tomatoes* :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >> I recently posted a message to newbies > >> Now I wish to exhort oldies not to work homework problems for people > > ExCUSE me, but I prefer the term knowbies > for the yang to the newbies' yin. Certainly it has to be at least oldbies, not oldies. I agree with the exhortation, BTW. Not because I particularly care if somebody somewhere gets an unearned A; that isn't really my problem. I'm concerned, rather, that the more people go along with it, the more requests of this sort we'll get. ==== > Certainly it has to be at least oldbies, not oldies. Thereby creating a grammatical rule for changing adjectives into nouns. Maybe the rule would apply only to people. Or people and dogs, say. Hungrybies, Smartbies, Dumbbies, Stinkybies, Seedybies... hum, I could go on a long time, yukking like a Nerd all the while. Max, Mr. Maximally Maxed ==== > I wish to exhort oldies not to work homework problems for people > I agree with the exhortation, BTW. Not because I particularly > care if somebody somewhere gets an unearned A; that isn't really > my problem. I'm concerned, rather, that the more people go along > with it, the more requests of this sort we'll get. This looks like a wonderful opportunity to start a new sub-group - alt.math.homework-help. If it's getting to be that big a hassle, there's obviously a pent-up demand for it. Then let those people work out the question of whether to give hints or complete answers. ==== > I agree with the exhortation, BTW. Not because I particularly > care if somebody somewhere gets an unearned A; that isn't really > my problem. I'm concerned, rather, that the more people go along > with it, the more requests of this sort we'll get. We is not a fixed entity. I, for one, hope the homework police find it difficult to regulate what people post. ==== Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real a and b, f( a + b ) = f( a ) + f( b ). Is f necessarily linear? Can you recommend a text that discusses this problem? Note: 1) For any rational q, f( qa ) = qf( a ). 2) If f is continuous, it is linear. 3) If, in the above problem, the real field is replaced by the rationals, then f is linear. -- ==== >Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real >a and b, f( a + b ) = f( a ) + f( b ). >Is f necessarily linear? No. For example, let B be a Hamel basis of the reals over the rationals, take some b_1 in B and define f(sum_{b in B} r_b b) = r_{b_1}. On the other hand, if f is Lebesgue measurable, or if it is bounded on some set of positive Lebesgue measure, then it is linear. See e.g. the thread Difficult Problem from February 1996. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real > a and b, f( a + b ) = f( a ) + f( b ). > > Is f necessarily linear? > > > Can you recommend a text that discusses this problem? > > > Note: > > 1) For any rational q, f( qa ) = qf( a ). > > 2) If f is continuous, it is linear. > > 3) If, in the above problem, the real field is replaced by the > rationals, then f is linear. First write down a proof for (1), (2) or (3). If you find this difficult then ask for help, but don't expect that you can solve the original problem if you cannot prove these three. Then use the axiom of choice. ==== > Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real > a and b, f( a + b ) = f( a ) + f( b ). > Sidenote: f(0) = 0 is redundant. > Is f necessarily linear? Note: 1) For any rational q, f( qa ) = qf( a ). > 2) If f is continuous, it is linear. > 3) If, in the above problem, the real field is replaced by the > rationals, then f is linear. > ==== Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real >a and b, f( a + b ) = f( a ) + f( b ). Is f necessarily linear? > No. For example, let B be a Hamel basis of the reals over the rationals, > take some b_1 in B and define f(sum_{b in B} r_b b) = r_{b_1}. On the other hand, if f is Lebesgue measurable, or if it is bounded on > some set of positive Lebesgue measure, then it is linear. > See e.g. the thread Difficult Problem from February 1996. > I suppose those results have dual or parallel form for g:R -> R with g(ab) = g(a)g(b), g(1) = 1 Similarly, if g is continuous, g(x) = |x|^n for some n in R. ==== |I suppose those results have dual or parallel form for g:R -> R with | g(ab) = g(a)g(b), g(1) = 1 |Similarly, if g is continuous, g(x) = |x|^n for some n in R. Incidentally, the only possibility ruled out by g(1)=1 is g identically equal to 0. Your question is closely related to the one of the O.P., since if g is such a function, then f(x) = log g(e^x) satisfies f(x+y) = log g(e^{x+y}) = log g(e^x e^y) = log (g(e^x) g(e^y)) = log g(e^x) + log g(e^y) = f(x)+f(y), which means f is the kind of function originally considered. Thus if g satisfies your conditions, then there exists a function f satisfying f(x+y)=f(x)+f(y) for every x, y, for which g(x) = e^f(log x) for x>0. That just leaves the question of what values g has when x<=0. We know g(0)=g(0)g(x) for every x. So if g(0)<>0, then g(x)=1 for every x. So if g is not identically 1, then g(0)=0. The constant function 1 does satisfy your conditions. satisfy the original conditions if f satisfies the functional equation f(x+y)=f(x)+f(y). If g is continuous (or even measurable) then f is also continuous (measurable) and f(x)=cx for some c. Then either g(x)=1=|x|^0, or g(x)=|x|^c, or the possibility you missed, g(x)=sgn(x)*|x|^c for c>0 (since for c<=0, sgn(x)*|x|^c is discontinuous at x=0). If f is discontinuous, then it's one of these peculiar functions whose graph is dense in the plane, and the graph of g is either dense in the upper half plane y>=0, or in both the first quadrant x>=0, y>=0 and the third quadrant x<=0, y<=0. Keith Ramsay ==== > |I suppose those results have dual or parallel form for g:R -> R with > | g(ab) = g(a)g(b), g(1) = 1 > |Similarly, if g is continuous, g(x) = |x|^n for some n in R. Incidentally, the only possibility ruled out by g(1)=1 is g > identically equal to 0. > Indeed, it was stated that way in parallel contrast to OP's redundant f(0) = 0 > Your question is closely related to the one of the O.P., since if > g is such a function, then f(x) = log g(e^x) satisfies f(x+y) > = log g(e^{x+y}) = log g(e^x e^y) = log (g(e^x) g(e^y)) > = log g(e^x) + log g(e^y) = f(x)+f(y), which means f is the kind of > function originally considered. Thus if g satisfies your conditions, then there exists a function f > satisfying f(x+y)=f(x)+f(y) for every x, y, for which g(x) > = e^f(log x) for x>0. > That certainly demonstrates constructively the duality. It's taking a somewhat familiar form of other dualities. log g(x) = f(log x); e^f(x) = g(e^x) cl SA = Sint A; Scl A = int SA -lub a,b = glb -a,-b; lub -a,-b = -glb a,b -sup A = inf -A; sup -A = -inf A /{ S-X | X in C } = S - /{ X | X in C } S - /{ X | X in C } = /{ S-X | X in C } > That just leaves the question of what values g has when x<=0. We know > g(0)=g(0)g(x) for every x. So if g(0)<>0, then g(x)=1 for every x. So > if g is not identically 1, then g(0)=0. The constant function 1 does > satisfy your conditions. > Yup, noticed that. I'll take your word for the rest which points out some grit in an otherwise smooth duality. A duality expressed with continuous functions, yet applicable to discontinuous ones also. > g(x) = -g(-x) or g(x)=g(-x) for every x. It's easy to check that > both possibilities, g(x) = e^f(log x) if x > 0 > = 0 if x = 0 > = e^f(log -x) if x < 0 and > g(x) = e^f(log x) if x > 0 > = 0 if x = 0 > = e^f(log -x) if x < 0 satisfy the original conditions if f satisfies the functional > equation f(x+y)=f(x)+f(y). If g is continuous (or even measurable) then f is also continuous > (measurable) and f(x)=cx for some c. Then either g(x)=1=|x|^0, or > g(x)=|x|^c, or the possibility you missed, g(x)=sgn(x)*|x|^c for c>0 > (since for c<=0, sgn(x)*|x|^c is discontinuous at x=0). If f is discontinuous, then it's one of these peculiar functions > whose graph is dense in the plane, and the graph of g is either dense > in the upper half plane y>=0, or in both the first quadrant x>=0, y>=0 > and the third quadrant x<=0, y<=0. > ==== Dear reader, for my research I have programmed a combinatorial optimization problem in aimms 2 format. Unfortunately running this program takes a lot of time, so I want to try a more efficient solver like C-plex. However, in order to use C-plex I need to convert my aimms program to an mps format program. Can anyone tell me whether this is possible in aimms? Dion Bongaerts boundary=------------5F60FDBAB57B35E6585F0C83 ==== --------------------------------------------------------------------- x-no-archive: yes Please excuse my ignorance ! I'm the messenger ..my brother is having a building constructed and called me on his cell.. he's caught without his trigonometric function table book ! he needs to know the TANGENT of a 2o angle ! Please respond ASAP ..so I can call him back. remove (NOSPAM) of course oh thank you ..thank you ==== > x-no-archive: yes > Please excuse my ignorance ! I'm the messenger ..my brother is having a > building constructed and called me on his cell.. he's caught without his > trigonometric function table book ! > he needs to know the TANGENT of a 2o angle ! > Please respond ASAP ..so I can call him back. > remove (NOSPAM) of course > oh thank you ..thank you > This is pretty funny, actually, but anyway, here it is: 0.034920769491747730500402625773725315879174297784615 approximately boundary=------------56E0B04682C893DE51A74A90 ==== --------------------------------------------------------------------- x-no-archive: yes oh ..so funny and oh .. so appreciated ! THANK YOU SO MUCH ==== I have a simple problem, but can't come up with a satisfactorily simple answer. Maybe the question misleads intuition: A mirror reverses right and left, but why doesn't it reverse top and bottom? ==== > I have a simple problem, but can't come up > with a satisfactorily simple answer. > > Maybe the question misleads intuition: > > A mirror reverses right and left, > but why doesn't it reverse top and bottom? The mirrors I look into don't 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 can't come up > with a satisfactorily simple answer. > > Maybe the question misleads intuition: > > A mirror reverses right and left, > but why doesn't it reverse top and bottom? If you mean a flat mirror, it doesn't reverse either. It simply doesn't 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 doesn't 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 can't come up >> with a satisfactorily simple answer. >> >> Maybe the question misleads intuition: >> >> A mirror reverses right and left, >> but why doesn't it reverse top and bottom? > > The mirrors I look into don't 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. The virtual image in the mirror is front-to-back symmetric about the plane of the mirror with the object being reflected. It is a matter of psychology rather than physics that we look at the image and describe it as having been left-to-right inverted (and then rotated 180 degrees on a vertical axis (*)) rather than front-to-back inverted. Both descriptions, of course, yield identical results. A top to bottom inversion followed by a rotation on a horizontal axis (**) 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 the plane 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-bottom plane of inversion and the plane of the mirror. ==== > I have a simple problem, but can't come up with a satisfactorily simple > answer. > > Maybe the question misleads intuition: > > A mirror reverses right and left, > but why doesn't it reverse top and bottom? There is no reversion at all, it is the same as reading the back of a book. In the sense that you yourself consider 'reversion', it is an effect that happens 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 the original position. Now you have two sentences, one written horizontally and the other vertically (including the letters). look at the paper from its back, or from a mirror. It is the same. Both the horizontal and vertical sentences are flipped in the same way. The key here is that looking at a flat shape from the back is the same as looking at it from the front in a mirror. -- Christos Dimitrakakis ==== ... >> A mirror reverses right and left, >> but why doesn't it reverse top and bottom? The mirrors I look into don't 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_ waved back. How do you explain this discrepancy between your theory and the 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 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. Alois ==== > A mirror reverses right and left, > but why doesn't 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 propose several possibilities: 1. You are using one of those corner mirrors that reflect 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. 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 Seaman Judge Yohn's 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 don't even have to assume that. Maybe he's 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 reflect 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, I'm sure I did the experiment in a competent and honest manner. As Dave Seaman notes, I am left-handed and waved my left hand. Got an excellent laugh from the responses! -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 don't 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 can't come up > with a satisfactorily simple answer. > > Maybe the question misleads intuition: > > A mirror reverses right and left, > but why doesn't it reverse top and bottom? A mirror reverses front and back; because we are symmetrical beings, it looks like you've reversed left and right. Put a glove on just one of your hands. Look in the mirror. Your head is where the mirror image's head is, your feet are where the mirror images feet are; your gloved hand is where the mirror image's gloved hand is, and your ungloved hand is where the mirror image's ungloved hand is - no contradiction involved! The weird thing would be if you head and feet matched up with the mirror image's corresponding parts, but your gloved hand was matched up with the image's _un_gloved hand! --------------------------------------------------- C Brown Systems Designs Multimedia Environments for Museums and Theme Parks --------------------------------------------------- ==== Some of the readers of this list might be interested in the following book Pierre Baldi, Paolo Frasconi, and Padhraic Smyth, Modeling the ISBN: 0-470-84906-1. It covers various models and algorithms for the Web including generative models of networks, IR and machine learning algorithms for text analysis, link analysis, focused crawling, methods for modeling user behavior, and for mining Web e-commerce data. 1. Mathematical Background - 2. Basic WWW Technologies - 3. Web Graphs - 4. Text Analysis - 5. Link analysis - 6. Advanced Crawling Techniques - 7. Modeling and Understanding Human Behavior on the Web - 8. Commerce on the Web: Models and Applications - Appendix A Mathematical Complements - Appendix B List of Main Symbols and Abbreviations The webpage http://ibook.ics.uci.edu/ contains more details, a hyperlinked bibliography, and a sample chapter in pdf. Paolo Frasconi http://www.dsi.unifi.it/~paolo/ ==== Jean-Paul Allouche and I are pleased to announce the publication of our book (AS)^2, also known as Automatic Sequences: Theory, Applications, Generalizations currently selling for US $50 on amazon.com. This book is about the class of sequences generated by finite automata, their generalizations, and applications to number theory and theoretical physics. The book has 571+xvi pages, 1600 citations to the literature, 460 exercises, 85 open problems, 1 musical score, and 2 jokes in the index. It will be of interest to number theorists and theoretical computer scientists. The web page for the book, http://www.math.uwaterloo.ca/~shallit/asas.html has a table of contents and other information. Jeffrey Shallit, Computer Science, University of Waterloo, Waterloo, Ontario N2L 3G1 Canada shallit@graceland.uwaterloo.ca URL = http://www.math.uwaterloo.ca/~shallit/ ==== > The book has 571+xvi pages, fifteen imaginary pages? Cool! -- J.97n Fairbairn Jon.Fairbairn@cl.cam.ac.uk ==== >> The book has 571+xvi pages, > fifteen imaginary pages? Cool! But only two jokes, a definite downside. xanthian, would have loved to see what a joke looks like in the Complex plane. [Probably a little out of phase.] -- ==== In sci.math, Kent Paul Dolan > The book has 571+xvi pages, > >> fifteen imaginary pages? Cool! > > But only two jokes, a definite downside. > > xanthian, would have loved to see what a joke looks like in the Complex > plane. > > [Probably a little out of phase.] > Either that, or highly twisted. :-) -- #191, ewill3@earthlink.net -- we could spin this a number of ways It's still legal to go .sigless. ==== The question is: if we let U, V be skewsymmetric and let > A = exp(U), B = exp(V) and C = exp(U + V) must (Cx, ABx) = (Cx, BAx) > and must Cx, ABx and BAx be linearly dependent? I don't think so: > We don't in general have C = AB. As long as U and V are small > C will be given by the Campbell-Baker-Hausdorff formula which > starts C = (AB + BA)/2 + lots of increasingly nasty terms > involving iterated commutators. If we ignore the later terms. > we see that for small U and V, C is approximately (AB + BA)/2, so > approximately dependent on AB but BA, but exactly? .... a bit unlikely > I think. Well, let me give you a hand-waving argument (which is what motivated my question): In the limit t->0, Z1 = (A^tB^t)^(1/t) should be equal to Z2 = (B^tA^t)^(1/t). That means that as t gets small, I expect that x transformed by Z1 must approach x transformed by Z2. Where might I reasonably expect to find Zx? Well, half-way between ABx and BAx. I've tested a few numbers, and what's interesting is that Zx _doesn't_ seem to converge to a linear combination of ABx and BAx, in general, although (Zx, ABx) does seem to equal (Zx, BAx). David Turner ==== >... >> So, what can be said about the reals or complex numbers x, where >> Gamma(x)/x is an integer? >> I guess we could define the set of x's which satisfy this as >> composites >= 6, a set which contains noninteger values. >> And related to Wilsons theorem, we could define the set of >> complex/real x's where: >> (Gamma(x)+1)/x = integer >> as a set of Wilson-derived generalized primes. >... >Since for positive reals 2,3,4... we have Gamma(x)=(x-1)! >and Gamma is strictly increasing for reals >= 2 , apparently >there are no other x>=2 (other than integer values) where >(Gamma(x)+1)/x is an integer. > On the contrary, there are lots of them. (Gamma(x)+1)/x > is 1.75 for x = 4, and is 5 for x=5. There are solutions > for 2, 3, and 4 which are between 4 and 5. For reference (and because I have nothing better to do with my afternoon): In[27]:= WilsonRoot[2.0]; WilsonRoot[3.0]; WilsonRoot[4.0]; 4.15444 4.56215 4.81616 I've checked the first one in Mathematica, and I presume (read 'hope') that the others are right as well. -- Dave Taylor Oh yeah? Well I'll build my own theme park! With Blackjack, and hookers ... in fact, forget the theme park! [Futurama] ==== > If, and only if (for positive integers n), n = a composite >= 6, then: > > n divides (n-1)!. > > So, what can be said about the reals or complex numbers x, where > > Gamma(x)/x is an integer? > > I guess we could define the set of x's which satisfy this as > composites >= 6, a set which contains noninteger values. > > > And related to Wilsons theorem, we could define the set of > complex/real x's where: > > (Gamma(x)+1)/x = integer > > as a set of Wilson-derived generalized primes. > > This must be a well-known topic. Anything interesting about this kind > of continuation, or is this all useless? > (It is useless to say useless, to paraphrase a cliche...) > A few things: 1) Even considering only x = positive integers, x = 1 satisfies both equations above. So, in a way, 1 is both a composite >=6 and a Wilson-derived generalized prime. (snicker.) 2) What if we, for x = composite, allow the integers (that the equations above are equal to) to be Gaussian? 2.5) Is (Gamma(x)+1)/x, if we let x = a Gaussian (integer) prime, an (real or Gaussian)integer? 3) If we take the maximum real root, x(n), of (Gamma(x)+1)/x = n, then, what can be said about the Wilson-zeta function: WZ(y) = product{n=1 to oo} 1/(1 -1/x(n)^y) ? Does the WZ() product converge for any y? Leroy Quet ==== >For complex values of x, in my ignorance it isn't obvious >to me whether (Gamma(x)+1)/x is ever an integer. Do you >have some simple examples? For example, (Gamma(x)+1)/x = 3 for x = 5.7584660619435256965 + 3.9675461457510952995 i approximately. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > (Gamma(x)+1)/x = integer > For reference (and because I have nothing better to do with > my afternoon): > > In[27]:= WilsonRoot[2.0]; WilsonRoot[3.0]; WilsonRoot[4.0]; > 4.15444 > 4.56215 > 4.81616 > > I've checked the first one in Mathematica, and I presume > (read 'hope') that the others are right as well. All pucker: (21:50) gp > p100 realprecision = 105 significant digits (100 digits displayed) (21:51) gp > solve(x=4.1,4.2,(gamma(x)+1)/x-2) %8 = 4.154439060271062694827261583664629151822185286325427363404753300368340020397 814797012023001544277754 (21:51) gp > solve(x=4.6,4.2,(gamma(x)+1)/x-3) %9 = 4.562151433950973261499668111682367676594713372717728076006767266218451425330 181354003523067887979890 (21:51) gp > solve(x=4.6,4.9,(gamma(x)+1)/x-4) %10 = 4.816164962269824928459405177621703579279411919117658752833716798354235922037 604118825483457603002219 (21:52) gp > solve(x=5,5.2,(gamma(x)+1)/x-6) %13 = 5.143587135664545182467227708447357783039049810591881586494699613779238822779 092252706929290717662694 Phil ==== >>For complex values of x, in my ignorance it isn't obvious >>to me whether (Gamma(x)+1)/x is ever an integer. Do you >>have some simple examples? For example, (Gamma(x)+1)/x = 3 for x = 5.7584660619435256965 + 3.9675461457510952995 i But... is anything known about Leroy's question?I'm asking because some years ago I formulated the very same question and I'm very curious about it. To be fair I'm only thinking to its second part i.e. that involving generalized primes as those numbers satisfying the obvious continuous version of Wilson's (necessary and sufficient) condition for primality. Well, maybe something interesting can be said about a suitable function having those primes as zeroes or poles... Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc ==== > >>For complex values of x, in my ignorance it isn't obvious >>to me whether (Gamma(x)+1)/x is ever an integer. Do you >>have some simple examples? For example, (Gamma(x)+1)/x = 3 for x = 5.7584660619435256965 + 3.9675461457510952995 i > > But... is anything known about Leroy's question?I'm asking because > some years ago I formulated the very same question and I'm very > curious about it. > > To be fair I'm only thinking to its second part i.e. that involving > generalized primes as those numbers satisfying the obvious > continuous version of Wilson's (necessary and sufficient) condition > for primality. > > Well, maybe something interesting can be said about a suitable > function having those primes as zeroes or poles... You mean like sin(pi (Gamma(x)+1)/x)? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >> Well, maybe something interesting can be said about a suitable >> function having those primes as zeroes or poles... > >You mean like sin(pi (Gamma(x)+1)/x)? Well, this is exactly the function I had considered in my youth, along with the *real* function sin^2(pi x)+sin^2(pi(Gamma(x)+1)/x). But now I'm more sort-of-thinking of something like prod_p 1/(1-p^{-x}) with p ranging over the zeroes of the function you supplied... provided that the product does make sense. Michele -- >It's because the universe was programmed in C++. No, no, it was programmed in Forth. See Genesis 1:12: And the earth brought Forth ... - Robert Israel on sci.math, thread Why numbers? ==== Would some kind soul possibly post a complete solution to this problem, as none of us on the course can fathom it and our exam is tomorrow... *gulp* :) Darren. ==== > Would some kind soul possibly post a complete solution to this > problem, as none of us on the course can fathom it and our exam is > tomorrow... *gulp* :) Darren. My reply was essentially complete. Just follow through with a detail or two. Here it is again: ------------------------ This question occurred to be the other day; it's been a while since I took calculus, and I can't come up with an answer. Does there exist a function from R to R that is nowhere continuous, but that is defined everywhere and limited everywhere (i.e. that has a finite limit at each real number)? Foghorn Leghorn moc.enecswen @ nrohgof ==== A function f(x) is said to be continuous at x=a if 1a)limx-->a- exist, 1b)limx-->a+ exists, 2) these two limits are = to one another ,say k. and 3)f(a)=k. Now you say that your function,f(x), is limited everywhere.Let's consider x at a. So limx-->a+=lim x-->a-(=k). If f(a)=k then of course we can't say that f(x) is continuous nowhere. What type of function violates just condition 3 of the definition of a continuous function? Functions that have removable discontinuity come to mind.Now the real question is whether or not a function can have an 00 number of removable discontinuities so the function is nowhere continuous. I think that you can come up with one. > This question occurred to be the other day; it's been a while since I > took calculus, and I can't come up with an answer. Does there exist a function from R to R that is nowhere continuous, > but that is defined everywhere and limited everywhere (i.e. that has a > finite limit at each real number)? Foghorn Leghorn > moc.enecswen @ nrohgof ==== This question occurred to be the other day; it's been a while since I >took calculus, and I can't come up with an answer. Does there exist a function from R to R that is nowhere continuous, >but that is defined everywhere and limited everywhere (i.e. that has a >finite limit at each real number)? Foghorn Leghorn >moc.enecswen @ nrohgof > Hmm... I don't know, but what about The Dirichlet Function: f(x) = 1 [if x is rational] = 0 [if x is irrational] Plot some points... It looks like two straight, horizontal lines... Obviously, it's definied everywhere, but I don't know if it's continuous and limited (sic). Maybe someone else does. cheerio, ==== > This question occurred to be the other day; it's been a while since I > took calculus, and I can't come up with an answer. > > Does there exist a function from R to R that is nowhere continuous, > but that is defined everywhere and limited everywhere (i.e. that has a > finite limit at each real number)? Consider the function f:R->R where : (a) f(x) = 0 if x is irrational, and (b) if x is rational and we define: denominator(x):= the least integer n>=1 such that n*x is an integer, and then let f(x):= 1/denominator(x) . I think Prof. Herman Rubin recently mentioned this function in another thread. David Bernier ==== >> Does there exist a function from R to R that is nowhere continuous, >> but that is defined everywhere and limited everywhere (i.e. that has a >> finite limit at each real number)? > > > Consider the function > > f:R->R > > where : > > (a) f(x) = 0 if x is irrational, and > > (b) if x is rational and we define: > denominator(x):= the least integer n>=1 such that n*x is an integer, > and then let f(x):= 1/denominator(x) . The f above _is_ continuous at all irrational points in R, right? David Bernier ==== > Does there exist a function from R to R that is nowhere continuous, >> but that is defined everywhere and limited everywhere (i.e. that has a >> finite limit at each real number)? > Consider the function f:R->R where : (a) f(x) = 0 if x is irrational, and (b) if x is rational and we define: > denominator(x):= the least integer n>=1 such that n*x is an integer, > and then let f(x):= 1/denominator(x) . The f above _is_ continuous at all irrational points in R, right? David Bernier Right. As for the original question, I have a feeling that no such function exists. I wonder if you could prove this using a Baire category argument, maybe something like the proof that there is no function continuous only on the rationals. ==== >>This question occurred to be the other day; it's been a while since I >>took calculus, and I can't come up with an answer. >>Does there exist a function from R to R that is nowhere continuous, >>but that is defined everywhere and limited everywhere (i.e. that has a >>finite limit at each real number)? >>Foghorn Leghorn >>moc.enecswen @ nrohgof > >Hmm... I don't know, but what about The Dirichlet Function: >f(x) = 1 [if x is rational] > = 0 [if x is irrational] Plot some points... It looks like two straight, horizontal lines... >Obviously, it's definied everywhere, but I don't know if it's >continuous and limited (sic). Maybe someone else does. That function is obviously nowhere continuous. But it also obviously has a limit at no point, so it doesn't help. >cheerio, ************************ David C. Ullrich ==== >This question occurred to be the other day; it's been a while since I >took calculus, and I can't come up with an answer. Does there exist a function from R to R that is nowhere continuous, >but that is defined everywhere and limited everywhere (i.e. that has a >finite limit at each real number)? No. If f has a limit at every point then there is a dense set of points at which it is continuous: Say the variation of f on S is V(f, S) = sup {|f(x) - f(y)| : x, y in S}. The fact that f has a limit at 0 shows that there is a closed interval I_1 such that V(f, I_1) < 1. Now the fact that f has a limit at the midpoint of I_1 shows that there is a closed interval I_2, contained in the _interior_ of I_1, such that V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of the I_n it foilows that f is continuous at x. (The fact that I_{n+1} is in the interior of I_n shows that x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all y in I_n.) >Foghorn Leghorn >moc.enecswen @ nrohgof ************************ David C. Ullrich ==== Foghorn Leghorn http://mathforum.org/discuss/sci.math/m/523425/523425 > This question occurred to be the other day; it's been a while > since I took calculus, and I can't come up with an answer. > > Does there exist a function from R to R that is nowhere > continuous, but that is defined everywhere and limited > everywhere (i.e. that has a finite limit at each real number)? Given a function f: R --> R and a real number x in R, let L(f,x) be the limit of f(x') as x' --> x, when this limit exists. Then the set (I'm using /= for not equal) P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} is countable. Thus, what you're looking for doesn't exist by a long shot (a function f such that L(f,x) exists for each x in R and P(f) = R). On the other hand, I'm pretty sure that for each countable set Z there exists a function f such that L(f,x) exists for each x in R and P(f) = Z (i.e. no restriction besides countable can be proved, even when L(f,x) exists for each x in R). One way to prove this is to first prove for each n = 1, 2, 3, ... that P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} is an isolated set, and hence each P(f,n) must be countable. This immediately implies that P(f) is countable, since P(f) = UNION(n=1 to oo) P(f,n). A much stronger version of this result holds, by the way. Given a function f: R --> R and a real number x in R, let C(f,x) be the set of all extended real numbers y (i.e. y can be -oo or +oo) such that there exists a sequence {x_k} with x_k --> x and f(x_k) --> y. In other words, C(f,x) is the set of all numbers (including -oo and +oo) that can be obtained as the limit of some sequence converging to x. Then for each f and x, C(f,x) is a nonempty closed set in the extended real line, the minimum number in C(f,x) is lim-inf(x'--> x) of f(x'), and the maximum number in C(f,x) is lim-sup(x'--> x) of f(x'). Note that L(f,x) exists means that C(f,x) = {y} for some y in R. Then the set Q(f) = {x in R: f(x) does not belong to C(f,x)} is countable. This can be proved in the same way as the result above. First, a bit of notation. If y belongs to R and E is a subset of R, let DIST(y,E) be the extended real number inf{|y-e|: e is in E}. Now define Q(f,n) for each n = 1, 2, 3, ... by Q(f,n) = {x in R: DIST[f(x), C(f,x)] > 1/n}. Then each Q(f,n) winds up being an isolated set, and hence Q(f) is countable since Q(f) = UNION(n=1 to oo) Q(f,n). Dave L. Renfro ==== This question occurred to be the other day; it's been a while since I >took calculus, and I can't come up with an answer. Does there exist a function from R to R that is nowhere continuous, >but that is defined everywhere and limited everywhere (i.e. that has a >finite limit at each real number)? No. If f has a limit at every point then there is a dense set of > points at which it is continuous: Say the variation of f on S is V(f, S) = sup {|f(x) - f(y)| : x, y in S}. The fact that f has a limit at 0 shows that there is a closed > interval I_1 such that V(f, I_1) < 1. Now the fact that f has > a limit at the midpoint of I_1 shows that there is a closed > interval I_2, contained in the _interior_ of I_1, such that > V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of > the I_n it foilows that f is continuous at x. (The fact that I_{n+1} is in the interior of I_n shows that > x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all > y in I_n.) Very nice argument. Compact and to the point :-) ==== > Foghorn Leghorn This question occurred to be the other day; it's been a while > since I took calculus, and I can't come up with an answer. Does there exist a function from R to R that is nowhere > continuous, but that is defined everywhere and limited > everywhere (i.e. that has a finite limit at each real number)? Given a function f: R --> R and a real number x in R, let > L(f,x) be the limit of f(x') as x' --> x, when this limit > exists. Then the set (I'm using /= for not equal) P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} is countable. Thus, what you're looking for doesn't exist by a > long shot (a function f such that L(f,x) exists for each x in R > and P(f) = R). On the other hand, I'm pretty sure that for each > countable set Z there exists a function f such that L(f,x) exists > for each x in R and P(f) = Z (i.e. no restriction besides > countable can be proved, even when L(f,x) exists for each x in R). One way to prove this is to first prove for each n = 1, 2, 3, ... > that P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} is an isolated set Dave, can you fill this in? I don't see why it's an isolated set. Sorry to be so dense :-) ==== An attempt at clarifying for calculus students. >This question occurred to be the other day; it's been a while since I >>took calculus, and I can't come up with an answer. >>Does there exist a function from R to R that is nowhere continuous, >>but that is defined everywhere and limited everywhere (i.e. that has a >>finite limit at each real number)? >> >No. If f has a limit at every point then there is a dense set of >points at which it is continuous: Say the variation of f on S is V(f, S) = sup {|f(x) - f(y)| : x, y in S}. The fact that f has a limit at 0 shows that there is a closed >interval I_1 such that V(f, I_1) < 1. Now the fact that f has >a limit at the midpoint of I_1 shows that there is a closed >interval I_2, contained in the _interior_ of I_1, such that >V(f, I_2) < 1/2. Repeat. > Which you can do, because there's a point in I_2 (possibly different from the point 0 above) where f has a limit. > Now if x is in the intersection of the I_n it foilows that f is continuous at x. > And there is at least one x in the intersection because the sets are closed (and bounded). I'm at a loss as to how to explain this in a short note to someone whose background doesn't extend beyond calculus. So I'll beg off for today (but I hope you'll be interested enough to make me [or someone else] do it Monday. >(The fact that I_{n+1} is in the interior of I_n shows that x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all y in I_n.) > > So now you've got a point x such that f is continuous at x. Now, pick any point z, and any tolerance level epsilon. By simply picking the first I_1 to be within the interval (z-epsilon, z+epsilon), we can find a point of continuity of f that is close enough to z. Since we can do this for any z, the set of points of continuity of f are arbitrarily close to any point of R. That's the definition of a dense set. Jon Miller ==== >This question occurred to be the other day; it's been a while since I >>took calculus, and I can't come up with an answer. >>Does there exist a function from R to R that is nowhere continuous, >>but that is defined everywhere and limited everywhere (i.e. that has a >>finite limit at each real number)? >> No. If f has a limit at every point then there is a dense set of >> points at which it is continuous: >> Say the variation of f on S is >> V(f, S) = sup {|f(x) - f(y)| : x, y in S}. >> The fact that f has a limit at 0 shows that there is a closed >> interval I_1 such that V(f, I_1) < 1. Now the fact that f has >> a limit at the midpoint of I_1 shows that there is a closed >> interval I_2, contained in the _interior_ of I_1, such that >> V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of >> the I_n it foilows that f is continuous at x. >> (The fact that I_{n+1} is in the interior of I_n shows that >> x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all >> y in I_n.) Very nice argument. Compact and to the point :-) heh-heh. ************************ David C. Ullrich ==== >Foghorn Leghorn > This question occurred to be the other day; it's been a while >> since I took calculus, and I can't come up with an answer. >> >> Does there exist a function from R to R that is nowhere >> continuous, but that is defined everywhere and limited >> everywhere (i.e. that has a finite limit at each real number)? Given a function f: R --> R and a real number x in R, let >L(f,x) be the limit of f(x') as x' --> x, when this limit >exists. Then the set (I'm using /= for not equal) P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} is countable. Well duh. I considered conjecturing that the word dense in the result I gave could be strengthened, but I didn't want to figure out by how much. Has countable complement is much stronger than what I would have guessed. >Thus, what you're looking for doesn't exist by a >long shot (a function f such that L(f,x) exists for each x in R >and P(f) = R). On the other hand, I'm pretty sure that for each >countable set Z there exists a function f such that L(f,x) exists >for each x in R and P(f) = Z (i.e. no restriction besides >countable can be proved, even when L(f,x) exists for each x in R). This is clear - if Z = {x_1, ...} let f(x_n) = 1/n and f(x) = 0 for other x. >One way to prove this is to first prove for each n = 1, 2, 3, ... >that P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} is an isolated set, and hence each P(f,n) must be countable. This >immediately implies that P(f) is countable, since P(f) = UNION(n=1 to oo) P(f,n). A much stronger version of this result holds, by the way. Given a >function f: R --> R and a real number x in R, let C(f,x) be the set >of all extended real numbers y (i.e. y can be -oo or +oo) such >that there exists a sequence {x_k} with x_k --> x and f(x_k) --> y. >In other words, C(f,x) is the set of all numbers (including -oo and >+oo) that can be obtained as the limit of some sequence converging >to x. Then for each f and x, C(f,x) is a nonempty closed set in >the extended real line, the minimum number in C(f,x) is >lim-inf(x'--> x) of f(x'), and the maximum number in C(f,x) is >lim-sup(x'--> x) of f(x'). Note that L(f,x) exists means that >C(f,x) = {y} for some y in R. Then the set Q(f) = {x in R: f(x) does not belong to C(f,x)} is countable. This can be proved in the same way as the result above. >First, a bit of notation. If y belongs to R and E is a subset of R, >let DIST(y,E) be the extended real number inf{|y-e|: e is in E}. Now define Q(f,n) for each n = 1, 2, 3, ... by Q(f,n) = {x in R: DIST[f(x), C(f,x)] > 1/n}. Then each Q(f,n) winds up being an isolated set, and hence Q(f) >is countable since Q(f) = UNION(n=1 to oo) Q(f,n). >Dave L. Renfro ************************ David C. Ullrich ==== > Foghorn Leghorn http://mathforum.org/discuss/sci.math/m/523425/523425 >> This question occurred to be the other day; it's been a while >> since I took calculus, and I can't come up with an answer. >> Does there exist a function from R to R that is nowhere >> continuous, but that is defined everywhere and limited >> everywhere (i.e. that has a finite limit at each real number)? >> Given a function f: R --> R and a real number x in R, let >> L(f,x) be the limit of f(x') as x' --> x, when this limit >> exists. Then the set (I'm using /= for not equal) >> P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)} >> is countable. Thus, what you're looking for doesn't exist by a >> long shot (a function f such that L(f,x) exists for each x in R >> and P(f) = R). On the other hand, I'm pretty sure that for each >> countable set Z there exists a function f such that L(f,x) exists >> for each x in R and P(f) = Z (i.e. no restriction besides >> countable can be proved, even when L(f,x) exists for each x in R). >> One way to prove this is to first prove for each n = 1, 2, 3, ... >> that >> P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} >> is an isolated set Dave, can you fill this in? I don't see why it's an isolated set. >Sorry to be so dense :-) Let epsilon = 1/(2n) in the definition of L(f,x) exists... ************************ David C. Ullrich ==== [snip] >> P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n} >> is an isolated set Dave, can you fill this in? I don't see why it's an isolated set. >Sorry to be so dense :-) Let epsilon = 1/(2n) in the definition of L(f,x) exists... ************************ David C. Ullrich ==== 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 first order Y_1' = F_1 [x, Y_1, Y_2, ..., Y_n] Y_2' = F_2 [x, Y_1, Y_2, ..., Y_n] ... ... ... ... ... ... ... ... ... ... ... ... Y_n' = F_n [x, Y_1, Y_2, ..., Y_n] (where n= max{n_1, n_2, ..., n_k) ????? ==== 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 layman's 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). 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 ==== > 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 layman's 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 doesn't 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 doesn't 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 you'll post it on JSH's 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, a one-to-one function (injection) need not cover R. ie: every point in N maps to one in R, but there will be an uncountably infinite number of points in R that are not mapped-to for any x in N. If you mean a bijection by one-to-one (sometimes called 'one-to-one and onto'), then there is no way create a bijective function from N -> R. There are an infinite number of integers, but an uncountably infinite number of reals - ie: there are way more reals than integers. Look up the proof that there can be no bijection between 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 layman's 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 11 Weren't you age 11 a couple of years ago? ==== > > Your friend > Nathan the Great > Age 11 > > Weren't you age 11 a couple of years ago? > He's referring to his _mental_ age. F. ==== I think that the neatest suggestion was to show that the open ball is open under Spivak's definition (which is basically Spivak problem 1-15), and show that the 1st quadrent is open, and then that the interesction of two (or a finite number) of open sets is open...Then my problem is solved. get stuck again...ciao, adrock ==== > I think that the neatest suggestion was to show that > the open ball is open under Spivak's definition (which is > basically Spivak problem 1-15), and show that the 1st quadrent > is open, and then that the interesction of two > (or a finite number) of open sets is open...Then my problem > is solved. Indeed you really learned something from that problem. Fishfry pointed out what was important. Another uniformally equivalent metric to circles and rectangles is diamonds or rhomboids D((x,y) - (a,b)) = |x-a| + |y-b| Also there's the box metric which is about the same except instead of open rectangles it uses open squares. > get stuck again...ciao, > You're welcome. You don't have to wait to get stuck before returning. The approach I suggested that was to your liking was a topological approach the topology of metric spaces which is easier handled just topologically. Of course, there is some basic transitions to understand, the rectangle, circle topologies being equivalent is one, another being the equivalence of the rectangle topology to the product topology which seemed instantly clear to you. ==== I have to minimize something like tr(X'AX) where A in R^{n*n} is a definite positive matrix and X is a n*k binary matrix, such that x_ij={0,1} and the sum of each row is 1 (i.e. sum_i x_ij=1). The problem is well known to be NP-complete. Do you know any reference to a continuous approximation? Or an efficient way to solve the problem when X is a big matrix? ==== > But there are finite affine/projective planes which are *not* isomorphic > to the ones constructed in the classical way from finite fields. > The smallest order for which such exist is 9. See > http://www.math.uni-kiel.de/geometrie/klein/math/geometry/smallproj.html Aha. I was unaware of the different isomorphism types -- hadn't seen them number of new (to me) ideas to explore. So I have a couple more questions. Is there a reference that describes _how_ the order-10 case was eliminated (one that says more than The proof took thousands of hours of computer verification.?) Secondly, in the above linked site's description of quasifields and semifields, it uses two distinct existence symbols: the usual backwards E, and backwards E^1. Do these have different meanings? -- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman ==== > >> But there are finite affine/projective planes which are *not* isomorphic >> to the ones constructed in the classical way from finite fields. >> The smallest order for which such exist is 9. See >> http://www.math.uni-kiel.de/geometrie/klein/math/geometry/smallproj.html > > Aha. I was unaware of the different isomorphism types -- hadn't seen them > number of new (to me) ideas to explore. So I have a couple more questions. > Is there a reference that describes _how_ the order-10 case was eliminated > (one that says more than The proof took thousands of hours of computer > verification.?) Pehaps you should consult this account by one of the main protagonists. 92b:51013 Lam, C. W. H.(3-CONC-C) The search for a finite projective plane of order $10$. Amer. Math. Monthly 98 (1991), no. 4, 305--318. > Secondly, in the above linked site's description of > quasifields and semifields, it uses two distinct existence symbols: the > usual backwards E, and backwards E^1. Do these have different > meanings? Dunno, but I would guess that the second means there exists exactly one. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== I have also posted this separately to moderated groups sci.math.research and sci.physics.research . ------------------------- I know it was once believed that variational problems just have a minimal solution but for a long while now it has been known that it is a stationary (extremal, or maybe extremal and inflection point) solution and that while the nature of the solution can be determined mathematically that can be laborious and that usually whether it is a minimum of maximum is determined from the physical context. So what I was wondering is, I know that there are many physical and other applied mathematical examples of minimal solutions but I would like to know of a few maximal solutions and perhaps inflection 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 in seismic ray theory for which I am currently beginning to write a thesis proposal and preparing for my Ph.D. comprehensive exam and proposal defense (so I am not far from 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 World Springer-Verlag, NY, NY, 1996, ISBN 0-387-97991-3 which is not very mathematical but is a very good general book but 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 of the 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 think he by stationary means inflection point whereas usually stationary means maximum, minimum or inflection point. But I just remembered a statement about a quantity that is thought to be minimized but is sometimes lavishly expended and forgot the exact wording or who it was by and did a www.google.ca search for calculus lavishly expended found Ireland, one of the foremost Irish scientists of all time. but will read it eventually. An extended version of my misremembered quote above by Hamilton is But 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 reflected portions of the path of light, in a single ordinary reflexion at a plane, is always the shortest of any, yet in reflexion at a curved mirror this economy is often violated. If an eye be placed in the interior but not at the centre of a reflecting hollow sphere, it may see itself reflected 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 I'm interested in maxima examples and, if any, inflection point (or analogy to inflection point) examples. I will also ask local colleagues more advanced in these areas than I am when they get back from conferences/etc. This is indeed related to my Ph.D. thesis but also spun off of philosophical discussions on talk.atheism entitled bipolar religious figures? and Parsimonious nature? (was Re: bipolar religious figures?) and OT: love balancing/supplanting which sort of update the shamanic component of my web page http://www.nfld.com/~dalton a bit, for any who might be interested in that sort of stuff, but that is not the main focus of this post. David http://www.nfld.com/~dalton ==== > > So anyway I'm interested in maxima examples and, if any, inflection > point (or analogy to inflection 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 divide the items into G groups. for this I'm using the permutation: (S-1 choose G-1). My question is what is the complexity of this problem is NP-complete? or polynomial. thanks is advance, Michal ==== >I have this problem: S- items (sorted) G- groups, and I want to divide >the items into G groups. for this I'm 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 you'll have to explain what exactly your problem is. Perhaps there's 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.ca Department 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 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^2 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 the flaw in my thoughts, I think the two needed to cancel somewhere but I am missing something. 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^2 You can't assume this since the mass stationary where it has compressed the 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 spring's energy due 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 the flaw in my thoughts, I think the two >needed to cancel somewhere but I am missing something. Please help. What you missed was that you can't equate mgh with 1/2*k*x^2. David McAnally -------------- ==== Gotcha, thanks! ==== Dear All, Sorry I can not solve this, as what I have learnt in school seems that I have given them back to my teacher :..) It is about my project: 3 possible workloads, running on 4 processors, how many combinations (not permutation, as it is 3^4) are there? What is the formula for this? e.g. 111 112 122 123 111 222 223 331 332 333 Many thanks!! Jialin ==== > Dear All, > > Sorry I can not solve this, as what I have learnt in school seems that > I have given them back to my teacher :..) > > It is about my project: > > 3 possible workloads, running on 4 processors, > > how many combinations (not permutation, as it is 3^4) are there? > What is the formula for this? If you mean you have 4 processors, each of which can be running one of 3 tasks, then you will use the multiplication principle. 3*3*3*3 = 3^4. If you wish to use combinations in this problem, then there is some information missing, but it isn't clearly stated (in my mind). Note: your example below does not have an obvious connection with the problem stated above. > e.g. > 111 > 112 > 122 > 123 > 111 > 222 > 223 > 331 > 332 > 333 > > > Many thanks!! > Jialin > -- Will Twentyman ==== > www.YeOldeCoffeeShoppe.com has a great many visitors and sections > for talk, tv, reviews and romance. but I've already got 1000s dropping in, all with nothing to read, just jump the buck, I'll have to pay a dozen people just to post something there for a few months otherwise, give the site a KICK, its a GREAT traffic getter a few minutes of telling people draws real crowds, but without the first couple of people standing round the busker nooone will gather. And if you like usenet you'll love forums. Herc ==== >just jump the buck, Cool band name. >without the first couple of people standing round the >busker nooone will gather. Cool album name. --Sean http://www.livejournal.com/users/spclsd223/ ==== just jump the buck, Cool band name. without the first couple of people standing round the >busker nooone will gather. Cool album name. --Sean > http://www.livejournal.com/users/spclsd223/ Dude. I love it when two worlds within my field of interest collide and celebrate with each other. Do you like Standing Outside Imaginations Door as a book title? ps need some writing talent at www.Yeoldecoffeeshoppe.com Herc ==== >Dude. >I love it when two worlds within my field of interest collide and celebrate with each other. Quoting my journal?! My plan is working perfectly! >Do you like Standing Outside Imaginations Door as a book title? No, but 'Standing Outside Imagination's Door' would be cool. Even better would be 'Ringing the Bell of Life and Running'. >ps need some writing talent at www.Yeoldecoffeeshoppe.com I'll let you know as soon as I have the motivation and time to actually look at the site and see what the hell it is. --Sean http://www.livejournal.com/users/spclsd223/ ==== > Why are certain polyhedra that are self intersecting considered REAL > polyhedra? > > See Imre Lakatos' _Proofs and Refutations_ for a variety of detailed > views about this exact question. And where will I find that? > > Question to all - am I much mistaken in thinking that nowadays, most > mathematicians' views on such issues are of the whatever genre? > How can they not care? Isn't it their job to, you know, care? > > cdj > Alright. > > > > Clearly polyhedrons which are self intersecting shouldn't > be considered REAL polyhedrons. Then we could do all sorts of weird > stuff to them. Take a look at > http://mathworld.wolfram.com/GreatDodecahedron.html and > http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for > example. > > Why do we consider figures which intersect themselves to be > polyhedrons or polyhedra in the first place? Doesn't that kinda > violate the rules, about being able to be constructed as if from > boundaries from algebraic expressions, and such? > > This is what I don't get about this weird geometry mumbo jumbo about > things which are self intersecting being considered actual shapes. > What is wrong with these theoretical geometricians? > > (...Starblade Riven Darksquall...) (...Starblade Riven Darksquall...) ==== > Why are certain polyhedra that are self intersecting considered REAL > polyhedra? > > See Imre Lakatos' _Proofs and Refutations_ for a variety of detailed > views about this exact question. > > And where will I find that? Any decent library. -- ==== Suppose you have a finite abelian cyclic group . Is there a canonical way (or any way, really) to express this alternatively in terms of two generators and two relations as: Generators = {g,h} Relations = {ag=0, bh=cg} where a, b, and c are integers. ? If so, this would be very useful to me, and I'd appreciate it if anyone could post with a way to do this (if it indeed can be done). Is there any kind of _Mathematica_ or Maple command that does this automatically, perhaps? (say once you enter the generator of the cyclic group and its order) Or perhaps if it can be done in certain special cases, maybe someone could point to those? ==== > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, with generator (1,1) as prototype of theorem. That should work unless order of group is power of prime. ==== > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > > ? > It's not clear to me exactly what you're given in this problem. If by this you mean we are given n, and told only that the group G = , where g and h are unspecified elements of G, is isomorphic to Z_n; then it's not clear that a non-trivial presentation of the given form exists. By non-trivial here, I mean a presentation where neither = G nor = G; clearly, is always going to give a representation of Z_a, but that's probably not what you want. Consider the cyclic group Z_6. We must select g and h from {2,3,4}, since otherwise one of g or h will generate Z_6 on its own; so either {g,h} = {2,3} or {g,h} = {3,4}. If we take g = 2 or 4, then the only possible presentation of your given form is ; if we take g = 3, then . But these are both equivalent to . There is a homomorphism from this group onto D_3 (dihedral group) with presentation , and onto Z_6 with presentation ; so it would seem there is no non-trivial representation of Z_6 of your form. > If so, this would be very useful to me, and I'd appreciate it if > anyone could post with a way to do this (if it indeed can be done). Is > there any kind of _Mathematica_ or Maple command that does this > automatically, perhaps? (say once you enter the generator of the > cyclic group and its order) > > Or perhaps if it can be done in certain special cases, maybe someone > could point to those? ==== > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. It's not clear to me exactly what you're given in this problem. See my other post where I try to work this out in general, starting from the hints I gave in my first post. In working thru this, the reason I disallowed the order of the group to be a power of a prime is elucidated. More comments below. > If by this you mean we are given n, and told only that the group G = > , where g and h are unspecified elements of G, is isomorphic to > Z_n; then it's not clear that a non-trivial presentation of the given > form exists. By non-trivial here, I mean a presentation = G nor = G; clearly, is > always going to give a representation of Z_a, but that's probably not > what you want. Consider the cyclic group Z_6. We must select g and h from {2,3,4}, > since otherwise one of g or h will generate Z_6 on its own; so either > {g,h} = {2,3} or {g,h} = {3,4}. If we take g = 2 or 4, then the only possible presentation of your > given form is ; if we take g = 3, then 2g = 0, 2g = 3h>. > And h = 2 or h = 4. Don't take h = 2, take h = 4, as in accordance to my work in the other post. Thus Z_6 = = <-1>, does it not? > But these are both equivalent to . There is a > homomorphism from this group onto D_3 (dihedral group) with > presentation , and onto Z_6 with > presentation ; so it would seem > there is no non-trivial representation of Z_6 of your form. > ==== Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, with > generator (1,1) as prototype of theorem. That should work unless order of > group is power of prime. > Let G be a finite cyclic group with order n. Assume n > 1 isn't a power of a prime. Thus some cofinite j,k > 1 with n = rs Let (1,0) and (0,1) be generators of Z/r & Z/s. r(1,0) = (0,0) = s(0,1) If I've done this right, G = <(1,1)> = <(1,0) + (0,1)> is a cyclic group of order n with generator (1,1) Oh, if you want it in the - form then use (1,1) = (1,0) - (0,s-1) Whence (0,s-1) is a generator of Z/s and r(1,0) = (0,0) = s(0,s-1) When n the order of G is a power of a prime, then be contented with trivial solution, g = 0, h = 1, G = <0-1> = <-1>, 1g = 0 = nh. ==== > > 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. > since otherwise one of g or h will generate Z_6 on its own; so either > {g,h} = {2,3} or {g,h} = {3,4}. If we take g = 2 or 4, then the only possible presentation of your > given form is ; if we take g = 3, then 2g = 0, 2g = 3h>. And h = 2 or h = 4. Don't take h = 2, take h = 4, > as in accordance to my work in the other post. > Thus Z_6 = = <-1>, does it not? > 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). --------------------------------------------------- C Brown Systems Designs Multimedia 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. Don't take h = 2, take h = 4, > as in accordance to my work in the other post. > Thus Z_6 = = <-1>, does it not? > 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 of Abelian groups, by the OP. ==== > James Harris' 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! -- Clive Tooth http://www.clivetooth.dk ==== > > >>James Harris' 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 get his PrimeCountH program to run on half of NSA's computers. David Bernier ==== http://www.msnusers.com/AmateurMath/Documents/CountViewer.html Very stupid to put it at a MSN site... like most reasonable people, I don't > have a Microsoft Passport account so I can't go to your site. And I'm really > not going to create a Passport account... =- Brian Dickens, the Netherlands I 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 don't 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

It fails for p=2 :-) > ==== 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. ==== 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 shit out some random math problem, with no introduce yourself... and ONLY THEN give us your god-damned homework problem. Fucking-A, this happens all the time. I don't give a shit about some random math problem... but if you would like some help, then that's a different story... but you have to ASK FOR IT. ..I don't 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 shit 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 Haugland http://www.neutreeko.com ==== > God damn, I hate it when people start out a post with: Consider... >> and then proceed to shit 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 wasn't 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. ==== > > God damn, I hate it when people start out a post with: Consider... > and then proceed to shit 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 wasn't 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 don't see Jan Kristian being unfair at all. Before 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 shit 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 that's 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] and then proceed to shit out some random math problem, with no > >introduce yourself... and ONLY THEN give us your god-damned >homework problem. > > Fucking-A, this happens all the time. I don't give a shit about some >random math problem... but if you would like some help, then that's >a different story... but you have to ASK FOR IT. >..I don't 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. Fucking-A, this happens all the time. But fortunately not that often on sci.math. Jon Miller ==== >I don't see Jan Kristian being unfair at all. >Before complaining, why not actually read his post? > > Nah, Gartogg just replied to the wrong post. He was responding to as, not Jan Kristian, even though he replied to 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 'em, I'll knock 'em in. Phil ==== > >>I don't see Jan Kristian being unfair at all. >>Before complaining, why not actually read his post? >> > Nah, Gartogg just replied to the wrong post. He was responding to as, > not Jan Kristian, even though he replied to JKs post. He snipped as's name too. Symptomatic though of someone who doesn't read carefully before 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, then at 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 odd prime. If g is a quadratic non residue of both p and p-1, then g is a primitive root. -- So, I'm looking for counterexamples, or reasons why these wouldn't be true. So far, I'm thinking that both 1 & 3 are false, and 2 is true - although I haven't 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 'I' 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 I'd 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 cdf function. The quantity a is a positive real number. I need to evaluate the following probabiltiy. P( X > 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 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 pdf's 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. Noble Professor Emeritus of Physics jvn@spamfree.virginia.edu ^^^^^^^^ 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 someone entering 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 eight years I've 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. Wouldn't know about stat, but sad to say a large majority of the people who enter the PhD program in math here at OSU end up 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 I've been at Colorado. Doug Depends a lot on the school, of course. If you want the highest probability, I would guess in the states it might be University of Montana or University of Idaho or Idaho State. Not disparaging, that's just how they 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 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. ==== > 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 it's 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 you've 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 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 it's 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 you've 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. 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, 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. -- 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 it's 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 you've got is your Calculus I textbook >and your colleagues. At this point I would suggest: Guy's UPINT, GP/PARI, some decent coffee, a supply of good Scotch Whiskey, and a great wife. Perhaps time on a trout stream just outside Podunk two evenings a week may be of some benefit. A summer working the wheat harvest might help also. Clearly Podunk ain't MSRI. Southwest will, however, get you to Oakland for $99. I don't know what AC Transit costs these days, but it can't 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 I've 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 A's in their graduate courses that it would be 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 their entire lives sitting in their room staring at the walls? I think not. Perhaps there is a much more narrow (-minded) interpretation of the word nothing? I'm going to speculate that nothing is interpreted along the lines not inconsistent with the simple observation that something on the order of 50% of math or science Ph.D. graduates enter careers other than 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 50-50 figure should consider investing some time in the study of logic or statistics. 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 I've 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 A's 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 the first place and switch to a Masters rather than finish the PhD. That accounts for a big portion of the discrepancy you think is present. I on the other hand was one of the few people that had passed most of the hurdles of a math PhD program without actually getting such a degree. I believe there was one other out of maybe two or three dozen PhD graduates during the five year period I was trying for a PhD. Karl Hallowell ==== > 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 you've 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 few months) have no access to a nearby college library, community, etc. The nearest college is more than fourty miles away and there I have only a few informal contacts in the aerospace engineering community. My real connections (as such) are online. Any serious math or physics concept is available online. Ie, I can google for Borel subgroups, Kaluza Klein models, the inverse Galois problem, or the Eight Vertex model and quickly find relevant research and expository material. The USENET might not be able to answer my questions, but they never have failed to come up with some insight. I'm still trying to figure out how to use arXiv.org (even after years of playing with it), but it's proving to be an amazing research tool even 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 I've 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 A's 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 that's not how it is in math at all. A master's degree requires that you learn a certain amount of mathematics - how much and how well you're required to learn it varies from place to place. A PhD requires _much_ more. First, it requires that you learn much more mathematics - much deeper mathematics, and you're required to understand it much better than a master's student (to oversimplify that last point, a master's student gets credit for knowing facts, while a PhD student only gets credit for knowing how to _prove_ those facts). And then there's the much more significant difference: A PhD requires a thesis, which is supposed to be significant original research. Of course some theses are more significant and original than others, but regardless, it's a totally different sort of requirement from anything that's required in a typical master's degree - at least theoretically, when you finish your PhD there's supposed to be _something_ that you understand better than anyone 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 it's 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 you've 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 seems plausible. Hard to see how it can help with the huge teaching loads 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 Podunk with sci.math to do their homework for them, of course. And if you'd read Hyman Bass's report to the Carnegie Foundation in the latest _Notices_, you'd realize that teaching load is a doubleplusungood phrase. Time to talk about research burden instead! 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 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. ==== >> >> 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 your last two sentences. Many universities and colleges which do not grant PhDs (in mathematics) nonetheless have (however unreasonably and/or unrealistically) a requirement that their (mathematics) faculty members write and publish research papers, at least if 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 I've 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 ?? ==== >> 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 I'm not writing papers, but that doesn't mean that my thesis was written for me. Doug ==== > Clearly Podunk ain't MSRI. Southwest will, however, get you to Oakland for > $99. I don't know what AC Transit costs these days, but it can't be much. > Even the numb and tired and disillusioned have choices, I would think. if you've got the patience to ride the bus from Oakland, kudos to you. It's only $2.25, but it takes over an hour and a half just to get to downtown Berkeley. if you take the BART, it's 4.25 total, and worth the $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 doing research in a narrow area. Unfortunately, these seem to be most of what is being done now, especially in statistics. The emphasis on interdisciplinary programs mainly produces those who do not know the basics of anything, but these programs have high rates of finishing. Students are not getting the basics of set theory, algebra, analysis, and topology these days. Learning how to compute and how to solve certain types of problems fails if basic material not covered in that is needed. Abstract concepts are needed for understanding, even if the details of them are not used. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University ==== >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 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 I'll ask him the next time we're on the tennis courts. -- John E. Prussing University of Illinois at Urbana-Champaign Department of Aerospace Engineering http://www.uiuc.edu/~prussing ==== available online in PDF format at http://www.ams.org/employment/asst.pdf Everyone thinking about graduate school in mathematics should look at this booklet. For example, the first institution that I looked at in the booklet had 72 full time graduate students, 15 part time graduate students, and 15 full time first year graduate students. The department had graduated 18 MS students in the past year, and an average of 3 PhD's per year gets an MS, but only about one in five entering graduate students goes on to get a PhD. Of course, you should go back to previous years booklets to see whether there have been any significant changes in enrollment patterns. Looking at about a dozen schools in this booklet, the ratio of full time first year graduate students to PhD's per year (note that PhD's for the last four years are given in the book, so this has to 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 get an MS degree. Unfortunately, I can't think of any way to distinguish those students from students who were given an MS as a consolation prize. In many cases, the total number of MS and PhD degrees per year is very similar to the number of full time first year students, indicating that most students get at least an MS. In other cases, far fewer degrees are awarded than there are entering students. For the big picture, it's worth pointing out that there are approximately 15,000 graduate students in PhD granting departments of math and statistics in the US, and that these departments produce something like 1,000-1,200 PhD graduates per year. These numbers haven't changed dramatically in the last 10 years. (See the latest AMS annual survey report for the numbers.) -- Brian Borchers borchers@nmt.edu Department of Mathematics http://www.nmt.edu/~borchers/ Socorro, NM 87801 FAX: 505-835-5366 -- Brian Borchers borchers@nmt.edu Department of Mathematics http://www.nmt.edu/~borchers/ Socorro, NM 87801 FAX: 505-835-5366 ==== > In <1eb18a7f.0307180852.4f4c5c6b@posting.google.com >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 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 I'll ask him the next time we're on > the tennis courts. -- > John E. Prussing > University of Illinois at Urbana-Champaign > Department of Aerospace Engineering > http://www.uiuc.edu/~prussing I did mention it, when this thread started. I also suggested a few schools where I guessed (I have no data) that the probability of finishing might be highest. ==== >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 A's 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 correlated with the ability to generate new mathematical ideas. You don't get a PhD for taking a lot of classes, you know. (In some places, you don't take _any_ classes to get a PhD.) I would also object to a phrase like drop out. In secondary school and below, there is a clear expectation that degree completion is the necessary goal for everyone of that age. At the graduate level, and even at the undergraduate level, leaving a program is not necessarily an indication of some kind of failure. Students' eyes are opened in school to the reality of the career choices for which they are preparing, and they may well decide they don't like that image -- even if they're doing well and can continue to do well. Even if your mathematical skills are superb, if what you want to do is make a lot of money, or to have time to raise a family, or to work with some of the world's 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 *wouldn't* a professor want to forego his or her own research activities for a couple years to write an enormous paper under a student's 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 it's 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 you've 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 are written by the advisors. Let me see, I have had 8 PhD students. Of had only the most minimal help; four had a lot of help and explanation described a PhD thesis as a work by the advisor under adverse circumstances 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 to write all or nearly all of his students' theses. One once complained that he didn't mind writing them, it was having to explain them that he objected to. But yes, there are other explanations for why people don't go on to do their own work, but as I look at my students there is a strong correlation between what they did in grad school and what they did afterwards. ==== >> This suggests that a lot of theses are written by the advisor. Right. After all, why *wouldn't* a professor want to forego his or >her own research activities for a couple years to write an enormous >paper under a student's 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. That's not really putting it another way, it's a separate assertion. And whether you believe it or not, it's a _fact_ that a lot of PhD theses are essentially written by the advisor. Barr says he's 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? (Regarding why a professor would do such a thing: First, it doesn't mean he's putting his own research on hold for those years. Anyway, there are all sorts of reasons: you have a student who possibly should have been kicked out years ago but wasn't - after the guy's been here for five or six years, passed his exams and courses and all, you really hate to kick him out just because he can't do the thesis. Or in more cynical vein: If none of the students get degrees then sooner or later the bean counters will remove the PhD program from the department, and then the professor will have to teach trigonometry instead of advanced course. All sorts of reasons it happens. Not that _I_'ve 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> <6vrnhvsflefhciu879crdm26mc61pqabf6@4ax.com> ==== > This suggests that a lot of theses are written by the advisor. >>Right. After all, why *wouldn't* a professor want to forego his or >>her own research activities for a couple years to write an enormous >>paper under a student's 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. That's not really putting it another > way, it's a separate assertion. And whether you believe it or not, it's a _fact_ that a lot of > PhD theses are essentially written by the advisor. Barr > says he's 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 don't mean that the advisor has contributed some non-negligible portion of the LaTeX source file, do you? Do you mean that the advisor has contributed almost every original idea in the thesis? And also most (or substantial parts) of the presentation decisions? I suspect that I'm not alone in being unclear on what you and Michael Barr mean. -- [R]eality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isn't a mathematician alive today that I can't now touch, and not a mathematical career on the planet that I can't now affect. --James Harris, render of worlds ==== > Not that _I_'ve ever done such a thing of course...) > I believe Hans Zassenhaus was once quoted as saying he didn't mind writing the thesis for the student, but he refused to then TEACH it to the 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 *wouldn't* a professor want to forego his or > her own research activities for a couple years to write an enormous > paper under a student's name? > > Maybe you meant to say something less hilarious. At some universities it's a written rule, and at others it's an unwritten rule, that a criterion for tenure/promotion, one must have produced Ph.D. students. Besides that, it's personally embarrassing if one's students don't make it, and it's ego-boosting if one produces many students. Further, one hopes that if one feeds the student one idea he'll get going. So one primes the pump. Then primes it again. Then again. Then again. Then the student has enough to call it dissertation. It's easy to see how it happens. Especially if you're in a department that produces few Ph.D.s and you'd like to increase that number. How can you attract good students to your program if you're record for number of Ph.D's graduated per year is only 3 or 4? The pressure could be tremendous to get the graduates out there, get the numbers up, so that better students can be recruited. Or maybe I'm the department chair working under a dean who misunderstands what affirmative action is about and I'm under the gun to produce quotas from under-represented groups. So by golly, this one advisee of mine is going to graduate if I have to type the damn thing myself and walk across the stage for him. There are lots of motives, none very honest of course, for sidelining one's own research a bit for pushing a slow student through. 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 I've 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 the Ph.D. program here, I think you originally meant someone who has graduate courses, I suppose an MS, has passes qualifying exams, and a certain amount of Ph.D. coursework done(?) Every school has a difference set of hurdles for admission to the program. Some have orals for qualifying, some have orals as defense of the dissertaion. But I think you meant someone post-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 here at Colorado have nothing more than a bachelor's degree. Not the vast majority, 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 bachelor's degree. Not the vast > majority, but a majority nonetheless. My point was that does one mean by entering the Ph.D program either A. starting grad school with the intent of getting a Ph.D. or B. Being finally accepted into the program, having passed qualifiers or the equivalent. The answer to the OP's originial question depends on what one means. Bart ==== > This suggests that a lot of theses are written by the advisor. Right. After all, why *wouldn't* a professor want to forego his or >her own research activities for a couple years to write an enormous >paper under a student's 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. That's not really putting it another >> way, it's a separate assertion. >> And whether you believe it or not, it's a _fact_ that a lot of >> PhD theses are essentially written by the advisor. Barr >> says he's 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 don't 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, I've seen it happen (much too close to naming names already, although I don't 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.) >And also most (or >substantial parts) of the presentation decisions? I suspect that I'm 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. That's not really putting it another > way, it's a separate assertion. And whether you believe it or not, it's a _fact_ that a lot of > PhD theses are essentially written by the advisor. Barr > says he's 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 don't 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, I've seen it happen > (much too close to naming names already, although > I don't 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 wouldn't be too surprised to learn if a given random thesis' 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 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 advisor's contributions and a student's can get quite blurred. In fact, I would argue that a good advisor is defined by how blurred this boundary is. It's not so easy to strike the right balance, and I think that's 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. ==== > ... >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 wouldn't be too surprised >to learn if a given random thesis' 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. is in recognition of scientific attainments and the ability to carry on original research as demonstrated by a thesis. And it's *signed*, and everything. Are you calling Jerry Wiesner a liar, sir? (For that matter, are you calling the me of 30 years ago 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 advisor's >contributions and a student's can get quite blurred. In fact, I would >argue that a good advisor is defined by how blurred this boundary is. >It's not so easy to strike the right balance, and I think that's 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 don't even know...) to give their students hints of the form see if you can go any further with this half-baked idea of Rudolph's. I'll 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. That's not really putting it another > way, it's a separate assertion. And whether you believe it or not, it's a _fact_ that a lot of > PhD theses are essentially written by the advisor. Barr > says he's 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 don't 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, I've seen it happen >> (much too close to naming names already, although >> I don't 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 wouldn't be too surprised >to learn if a given random thesis' 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 the student is supposed to have at least _something_ to do with actually coming up with the proof. In the cases I have in mind that was not so - instead there was an endless series of: Ok, why not try this: Ok, I'll look at that. [week passes] So did that work? Don't know, couldn't figure anything out either way. Hmm, let's see... [delay of minutes or days] No, that doesn't work [or does work]. Why not try this: Ok... (Or so the advisor claimed during endless bitch&moan sessions during those years, and I believe him, because it's _exactly_ what happened earlier when I was helping the same student, going through all the exercises in some book one summer - he simply never made any progress on any of them, with maybe two exceptions, all the solutions were due to me the week after the 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 advisor's >contributions and a student's can get quite blurred. Yes, no doubt it's quite blurry in the typical case. Why you think it woud be blurry in the cases I'm referring to, given my characterizations of them, is beyond me. > In fact, I would >argue that a good advisor is defined by how blurred this boundary is. >It's not so easy to strike the right balance, and I think that's 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 isn't even worth the time! As a senior math student intending to go to grad school, I don't find anything in everyones' 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. > 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 isn't even worth the >time! As a senior math student intending to go to grad school, I don't find >anything in everyones' comments very encouraging. He asked a question - people tried to answer as accurately as they could. I don't think anyone's said it's not worth the time. People have said it's not easy. It's not. When he asks what proportion of PhD students get PhD's do you think we'd really be doing him or anyone else a favor by _lying_, saying it's 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. That's 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 Master's dergree, many of the students are straight out of undergrad. (At Wisconsin the Master's was more or less a consolation prize for students who'd done the coursework but didn't 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 isn't even worth > the time! As a senior math student intending to go to grad school, I > don't find anything in everyones' 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. You'll want to get that drinking problem going pretty much at the beginning of Step 3. (It's a lot easier to get tenure if your colleagues percieve you as a nonthreatening, harmless sot. And you'll make a better Dean that 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 advisor's >>contributions and a student's can get quite blurred. > > Yes, no doubt it's quite blurry in the typical case. Why you think > it woud be blurry in the cases I'm referring to, given my > characterizations of them, is beyond me. > I didn't mean to imply that your cases were of that kind, but I only meant to provide some kind of explanation of why that might happen, as I think I did with the snippet below. >> In fact, I would >>argue that a good advisor is defined by how blurred this boundary is. >>It's not so easy to strike the right balance, and I think that's 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 ==== >That's 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 >Master's dergree, many of the students are straight out of >undergrad. (At Wisconsin the Master's was more or less a >consolation prize for students who'd done the coursework >but didn't finish the PhD.) The same at Colorado - when I entered after completing my bachelor's degree, I chose the Master's degree program (assuming that it had to be done before the doctorate), and was convinced otherwise by the chair of the 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, it's not that bad. 'Cuz if you post to sci.math then you won't be anonymous when you die. Anon. ==== >Oh, it's not that bad. >'Cuz if you post to sci.math then you won't be anonymous when you die. Anon. On the Internet, no one knows you're dead. Lee Rudolph ==== > You guys make getting a math Ph.d sound as though it isn't even worth the > time! As a senior math student intending to go to grad school, I don't find > anything in everyones' 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 my undergraduate I went to graduate school drinking about the same amount as I did as an undergraduate (which seemed like a lot at the time). After 2 years, I had to quit graduate school because the material is so much harder than it is as an undergraduate. I don't think people have stressed that enough on this thread. Graduate level mathematics is much harder than undergraduate level mathematics. If you are the top student in your class, you will be put in your proper place quickly in graduate school. Herein lies the importance of developing a drinking problem. It is why I had to quit after 2 years. Shortly before coming back to graduate school, I was talking to someone who gave me this piece of wisdom: The only thing that got me through grad 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 a general rule that the ones who make it (or who are going to make it) have drinking problems, and those who don't, don't. Is alchoholism worth the opportunity to pursue mathematical knowledge on a daily basis? No doubt its a hard question, but its one you have to ask. On the more serious note of advisor's writing their student's thesis: I have only attended one PhD defence and in that talk the student's advisor never once had to steer the student in the right direction during the question/answer session. I do have to admit, though, that they did ask some fairly trivial questions - for example one could be done simply by using Fatou's lemma (which the candidate confidently answered). Hugh ==== >Oh, it's not that bad. >'Cuz if you post to sci.math then you won't be anonymous when you die. Anon. On the Internet, no one knows you're dead. Lee Rudolph Heard in an English pub, to the tune of Irish Washerwoman: Oh, McTavish is dead and his brother don't know it, They're both of them dead and they're 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 about what he told me about getting a PhD in Russia. (During the cold war). He said to get a PhD in Russia, you have to had 4-5 published papers in a respectable journal of your field; that a PhD is looked upon like a Masters in the States. There's another level of education above PhD called Doctor. And, of course, you need to write even more papers. And when you become a Doctor, then you earn the title Doctor. He said after Doctor, there's another step called Academic but he said that such a title is mostly political. I could be wrong about what he said, because I was just so shocked to 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 I'm just so fond of my Russian friend.... My friend is a 'lowly' programmer in the States, and back in Russia he was a respected quantum physicist. He is so eager to solve challenging math problems. I only know a few Olympiad problems to give him. He pretty much solves them in his head. So I went out one day, and bought an olympiad book just to rattle off a problem when he would stop in my cube with a happy smile asking for a math problem. It was so entertaining/jaw dropping to see him usually solve most of these olympiad problem in his head. I don't 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 wasn't so sure himself. But, it was along the lines of making ends meet. He was more concerned about other issues like spending time with friends and family, and how to live a good happy life. This seems typical of most Russian emmigrants I've met so far. Taxicab drivers, sofware testers, guitar bums all having an extremely good education, and all happy to make some money, and all more concerned about questions of living a meaningful life. I contrast this with alot of my American friends who seem to want a Masters, or an MBA just as an 'edge' in a rat race. Just the kinda thing that makes you go Hmmmm... Well, those Russians, you gotta love them!) > I don't think anyone's said it's not worth the time. People have said > it's not easy. It's 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. There's no money in math. So if someone steals your idea, you > haven't lost any money. I made no reference to money. To quote myself: ...for personal interest I would like to know if it is original > 2. If you're doing original work, you'll 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 priority as the discoverer. I think that as unfair, strange though it may sound. > 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 idea from 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. > It's much easier to take the lower salary and be a university professor.) > > Jon Miller I am currently making enquiries with other professional mathematicians regarding my work so far. If any developments occur, I will post them here along with my work. Flame ==== Can someone give me a hint (not solution) on the following problem. It is number 2.29 in Rotman's Introduction to Homological Algebra. Given: g A------>B | | f| Prove the following diagram is a pushout: | V C g A-------->B | | | | f| f'| | | V g' V C-------->D Where D=(C /osum B)/W, W={(fa,-ga):a /in A}, f':b-->(0,b)+W and g':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 Rotman's Introduction to >Homological Algebra. Given: > g > A------>B > | > | >f| Prove the following diagram is a pushout: > | > V > C g > A-------->B > | | > | | >f| f'| > | | > V g' V > C-------->D Where D=(C /osum B)/W, W={(fa,-ga):a /in A}, f':b-->(0,b)+W and >g':c-->(c,0)+W. >What do I need to prove to prove the diagram is a pushout The diagram is a pushout if it satisfies the following two conditions: 1. COMMUTATIVITY: f'g = g'f (I am applying functions on the left, so they should be read right-to-left; f'g means g first, then f'). 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=df' and c=dg'. > and >what is the significance of the set W? You can think of a pushout as a co-equalizer; you are finding the largest object on which you can make f and g 'equal'. W is a measure of how far they are on being 'equal' (not exactly, since we are dealing with the dual notion, but maybe that makes some sense to you?). In order for f'g(a) to be equal to g'f(a) for all a, you need to make sure that (f(a),0) is the same as (0,g(a)); for them to be the same, you need to mod out by (f(a),-g(a)); so W is the closure of all those identities in Cosum B; moding out by W is the same as imposing those identities on Cosum B. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== [snip] > The diagram is a pushout if it satisfies the following two conditions: > > 1. COMMUTATIVITY: f'g = g'f (I am applying functions on the left, so > they should be read right-to-left; f'g means g first, then f'). > > 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=df' and c=dg'. > [snip] Ok, I got 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 don't mind if you spoil the problem now, unless you think you can give me a suitable hint. Chris ==== >[snip] >> The diagram is a pushout if it satisfies the following two conditions: >> >> 1. COMMUTATIVITY: f'g = g'f (I am applying functions on the left, so >> they should be read right-to-left; f'g means g first, then f'). >> >> 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=df' and c=dg'. >> >[snip] Ok, I got 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 don't mind if you >spoil the problem now, unless you think you can give me a suitable >hint. I don't follow what you mean. Assume you have an object K, and maps b:B->K and c:C->K such that bg = cf. f A ---> C | | g | | g' | | V V B ----> D f' We know that D is defined as (Boplus C)/W; so to define a map from D to K, we can define a map from Boplus C to K whose kernel contains W, and factor it through the quotient. f' and g' are the obvious inclusions, and W is the subgroup generated by all pairs (g(a),-f(a)) for a in A. So let's consider what the d HAS to be. First, we want b=df' and c=dg'. So given any x in B, we know what b(x) is (we are GIVEN the maps b and c); and we know that f'(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 to b(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 let's take an element of W, which is of the form (g(a),-f(a)) for a in A. According to the definition of e, we map e(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; so bg(a)-cf(a)=0 for all a in A. Therefore, e takes W to 0, and so W is contained 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=df' and c=dg'. Moreover, since the definition of e was forced by the commutativity of the diagram, the choice of d is also forced, so that d is the only function 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 property is easy to verify, because you will have no choice about how to define the map in question. it should be obvious what the map has to be in an object. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== ==== to this question, which has been driving me a bit nuts. Some quick background: as a role-player, I use a lot of dice. At times, the rules call for one to roll multiple dice (say, five six-sided dice, or 5d6), then to drop the lowest two and total the other three. The basic question is: is there a formula for determining the probability of rolling a certain result, given these conditions? Determining the probability of a particular outcome when just rolling multiple dice is relatively straightforward (there's a brief discussion here: http://mathforum.org/library/drmath/view/52207.html). I can find a pattern to the summation needed when dropping a single die from a set; but once I try to remove two dice from the set, the pattern disappears and I find myself lost again. (The numbers can be determined by brute force, of course, but that's neither practical nor interesting.) So, I guess the base question is: Is there a formula for calculating the probability of achieving a result R on n dice with d sides, dropping the k lowest dice? ==== 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 you're 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??? I must be missing what you mean by S^1 - that's not the unit circle? In any case, if S^1 is locally compact then f in C0(S^1) implies that f 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 you're 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. > I must be missing what you mean by S^1 - that's 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 can't 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 - I'll 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 you're 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 - that's a large part of the difference between C0 and C... >> I must be missing what you mean by S^1 - that's 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 can't 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 - I'll 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 put around 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. -- Ioannis http://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 fly 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 in a three-dimensional sense and place noÊlimits on distance, then every quarter on Earth is around it. And since quarter isn't defined, we could take it to mean one-fourth of anything, which raises the total a bit higher... :-) -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're 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 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 don't understand at all here is where the s and v values come from ?!? It might sound stupid but I don't understand this. Any help ? thanx Sam ==== 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 don't understand at all here is where the s and v values come from ?!? It might sound stupid but I don't understand this. Any help ? thanx Sam ==== > 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 don't understand at all here is where the s and v > values come from ?!? It might sound stupid but I don't 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). it's 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 don't 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: v' = 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 don't understand at all here is where the s and v > values come from ?!? It might sound stupid but I don't 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). it's 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 don't 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 scalars and v and w are vectors, is (s+v)(t+w) = (st-v.w)+(sw+tv+v^w). It follows 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: >v' = 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 v' = (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 that w^v = -v^w). When you rotate about the unit vector w, then w and all its multiples remain 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 that r moves towards w^r. It follows that r is transformed by a right handed rotation about u through an angle of theta to r' = 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 to itself, 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 don't understand at all here is where the s and v >> values come from ?!? It might sound stupid but I don't 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). it's 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 don't 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: > v' = 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 n' be the greatest even number <= n, the a rotation in R^n is a product of n' reflections. so work out a formula for reflections and the result for rotation follows directly. in doing so, one may want to note that successive reflections about vectors a and b is a rotation about a X b by the angle between a and b. again, it is more convenient to consider this in the clifford algebra framework. M.T.