mm-2029 === Subject: Bisecting a line with Cabri or GSP Obviously one of the fundamental compasses and straightedge constructions is the perpendicular bisector of a line segment. Does anyone here know how to do this with either Cabri or Geometer's Sketch Pad in a way that simulates the pencil and paper construction? The problem I have is that afaics you can't keep your virtual compasses open at a constant width. cheers dd === Subject: Re: Bisecting a line with Cabri or GSP > Obviously one of the fundamental compasses and straightedge > constructions is the perpendicular bisector of a line segment. Does > anyone here know how to do this with either Cabri or Geometer's Sketch > Pad in a way that simulates the pencil and paper construction? The > problem I have is that afaics you can't keep your virtual compasses > open at a constant width. > cheers > dd See www.icehouse.net/nungester/Bisect.htm. This uses sliders from the ToolsSliders.gsp file provided with GSP. Once the basic horizontal slider exists, create a separate line segment (to be bisected). Click on an endpoint and the slider segment and Construct, Circle from point and radius. Repeat for the other end. Now the two circles are the same radius, set by the slider. Make the circles big enough to intersect each other, highlight both circles, and Construct, Intersections. Make a segment between the two circle intersection points. Construct the intersection of that segment and the original segement. Then (if you are fortunate enough to have the complete non-student version of GSP with Java Sketchpad), Save As HTML and post it on your website. Cool, huh? I'm a big GSP fan and recently gave a lunchtime talk on it to about 100 engineers/techs at work. Rick http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: Bisecting a line with Cabri or GSP Originator: mtx014@linux.services.coventry.ac.uk (Robert Low) >Obviously one of the fundamental compasses and straightedge >constructions is the perpendicular bisector of a line segment. Does >anyone here know how to do this with either Cabri or Geometer's Sketch >Pad in a way that simulates the pencil and paper construction? The >problem I have is that afaics you can't keep your virtual compasses >open at a constant width. If you have a finite line segment, then you have a way of drawing one circle centred on each end so that both circles have the same radius. (I think that this is standard for classical ruler and compasses constructions; you aren't supposed to be able to use the compasses to transport distances.) -- Rob. http://www.mis.coventry.ac.uk/~mtx014/ === Subject: Re: Bisecting a line with Cabri or GSP >Obviously one of the fundamental compasses and straightedge >constructions is the perpendicular bisector of a line segment. Does >anyone here know how to do this with either Cabri or Geometer's Sketch >Pad in a way that simulates the pencil and paper construction? The >problem I have is that afaics you can't keep your virtual compasses >open at a constant width. > If you have a finite line segment, Is there such a thing as an infinite line segment? > then you have a way > of drawing one circle centred on each end so that both > circles have the same radius. > (I think that this is standard for classical ruler and > compasses constructions; you aren't supposed to be > able to use the compasses to transport distances.) If this is the case then I would have to restate the question. Basically what I want to do is use GSP/Cabri with pupils who can just about cope with the idea of basic constructions: line and angle bisectors, perpendiculars from a point to and a point on a line. The way I do this on paper, and the way the textbooks cover it, involves transporting distances. This is something I can't figure out how to do easily with GSP or Cabri. At first glance using the line segment as a radius, which, aiui, is what you are suggesting, would not readily fit in the little angels' books or on their GCSE papers. cheers dd === Subject: Re: Bisecting a line with Cabri or GSP Originator: mtx014@linux.services.coventry.ac.uk (Robert Low) >>constructions is the perpendicular bisector of a line segment. Does >>anyone here know how to do this with either Cabri or Geometer's Sketch >>Pad in a way that simulates the pencil and paper construction? The >>problem I have is that afaics you can't keep your virtual compasses >>open at a constant width. >> If you have a finite line segment, >Is there such a thing as an infinite line segment? Actually, I'm not sure if you'd call a ray a line segment. Still, it's nice if the premis is guaranteed to be true. >> (I think that this is standard for classical ruler and >> compasses constructions; you aren't supposed to be >> able to use the compasses to transport distances.) >If this is the case then I would have to restate the question. Basically >what I want to do is use GSP/Cabri with pupils who can just about cope with >the idea of basic constructions: line and angle bisectors, perpendiculars >from a point to and a point on a line. The way I do this on paper, and the >way the textbooks cover it, involves transporting distances. This is Yes, they tend to assume 'stiff' compasses rather than 'collapsing' ones; I seem to recall that with such compasses it isn't that hard to trisect the angle, but it's been a long time since I knew anything at all about this. >something I can't figure out how to do easily with GSP or Cabri. At first >glance using the line segment as a radius, which, aiui, is what you are >suggesting, would not readily fit in the little angels' books or on their >GCSE papers. So, first I need to apologise: I was in 'give a hint and don't do the homework for you' mode, inappropriately. Sorry. Second: I don't understand why it's a problem. You put the point of the compasses on one end and draw an arc passing through the other; then you repeat at the other end. Join the intersections. Why won't that fit into their books/papers readily? (OK, so it takes up slightly more room than doing it with a substantially smaller circle, but is paper in that short supply?) -- Rob. http://www.mis.coventry.ac.uk/~mtx014/ === Subject: Re: Bisecting a line with Cabri or GSP > Second: I don't understand why it's a problem. You > put the point of the compasses on one end and draw an > arc passing through the other; then you repeat at > the other end. Join the intersections. Why won't that > fit into their books/papers readily? (OK, so it takes > up slightly more room than doing it with a substantially > smaller circle, but is paper in that short supply?) You might actually be right. I've been playing around with this and it seems a viable practical approach. The only concern is do the exam boards allow enough room on their GCSE papers? I'll dig out some past papers and check it out. We are talking about borderline understanding where kids could be put off by, say, arcs going through the question. cheers dd === Subject: Re: Bisecting a line with Cabri or GSP Originator: mtx014@linux.services.coventry.ac.uk (Robert Low) >> Second: I don't understand why it's a problem. You >> put the point of the compasses on one end and draw an >> arc passing through the other; then you repeat at >> the other end. Join the intersections. Why won't that >> fit into their books/papers readily? (OK, so it takes >You might actually be right. I've been playing around with this and it >seems a viable practical approach. The only concern is do the exam >boards allow enough room on their GCSE papers? I'll dig out some past Ouch. It's a bit grim if the paper doesn't allow you to fit in the answer they ask for. I'm sufficiently old that I don't think in terms of fitting things on to the exam paper; all my exams were booklets with questions on them, and we had separate blank booklets for writing the answers in. We kept the exam paper, they got the script. Seemed like a fair exchange to me. >papers and check it out. We are talking about borderline understanding >where kids could be put off by, say, arcs going through the question. Sure: not your fault or theirs if the exam paper layout is bad. If it did turn out to be the situation that the layout didn't allow the construction to fit in, I'd be tempted to write something aggrieved to the exam board... -- Rob. http://www.mis.coventry.ac.uk/~mtx014/ === Subject: Re: Bisecting a line with Cabri or GSP In my experience the big problem is kids not owning a compass or bringing it to lessons to practice and to the exams to use it. I've tried all sorts with some success. Here are some of the things I've tried: Buying 99p stationery sets (compass, protractor, set squares and ruler - plus a couple of pencils or a crappy calculator if you're lucky) and passing them on at cost. No credit! I know we're not surposed to sell things to pupils but a couple of mine asked me to buy compasses for them as they were unsure exactly what to buy. Lending pupils equipment (not my favorite approach). Giving notice of the need for compasses, saying that if they are not brought it's no problem as I will give them some string. Come the lesson I will give out the string and show how to use it. Of course there will be some with compasses who do the job so much easier that it convinces a few more to buy compasses (from me or a shop, I care not). The success of this approach surprised me. If they do not bring a compass to the exams I warn them that marks are for the construction marks and the result. If they put the line in the right place, then fake freehand compass marks, they will get some of the marks. A protractor can often be used in the place of a compass. ____ http://mathz.com === Subject: arc length I'm having troubles integrating the expression sqrt(t^4-2t^2+2t+1)dt, in the interval [0,2]. This is what I arrived at when attempting to calculate the arc length of the curve g(t) = [(t^3)/3-t, t^2] in the interval [0,2]. This is the first of a series of problems like it, and obviously I haven't gotten off to a good start. I will be extremely grateful if anyone can provide me with hints on how to solve this. So far I've tried some substitutions and integration by parts. Joe === Subject: Re: arc length > I'm having troubles integrating the expression sqrt(t^4-2t^2+2t+1)dt, in the > interval [0,2]. This is what I arrived at when attempting to calculate the > arc length of the curve g(t) = [(t^3)/3-t, t^2] in the interval [0,2]. This > is the first of a series of problems like it, and obviously I haven't gotten > off to a good start. I will be extremely grateful if anyone can provide me > with hints on how to solve this. So far I've tried some substitutions and > integration by parts. Back up a moment, where are you getting sqrt(t^4-2t^2+2t+1)? Regardless of whether x=t^3/(3-t) or x=(t^3/3)-t, I'm not getting that expression for the integrand. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: arc length > I'm having troubles integrating the expression sqrt(t^4-2t^2+2t+1)dt, in the > interval [0,2]. This is what I arrived at when attempting to calculate the > arc length of the curve g(t) = [(t^3)/3-t, t^2] in the interval [0,2]. It should be t^4 - 2t^2 + 1 + (2t)^2, which is t^4 + 2t^2 + 1. That makes it easier. It's a hint that math teachers and textbook writers use to make it easy for you: if the integral is too hard to compute, then you made a mistake setting it up. Unfortunately, if the class allows calculators, then this hint disappears. Who cares if it's hard to calculate it -- just have the machine do it! > This > is the first of a series of problems like it, and obviously I haven't gotten > off to a good start. I will be extremely grateful if anyone can provide me > with hints on how to solve this. So far I've tried some substitutions and > integration by parts. The time it takes to check for mistakes in your work is much less than the time it takes to try to do something you don't know how to do, both on exams and in homework. The fact that you don't know how to do it doesn't prove that it's not doable, but it does give you a hint that you ought to check your work. Jon Miller === Subject: Re: arc length I did check over every attempt I made at it. In all honesty, Ive spent more than two hours reworking the same problem. I've looked at texts, went to get help from the assisstance centre at school, but not really getting any help in the end. Ill give the assisstance centre another shot tomorrow, but I'll be glad to get help from anywhere at this point. > I'm having troubles integrating the expression sqrt(t^4-2t^2+2t+1)dt, in > the > interval [0,2]. This is what I arrived at when attempting to calculate > the > arc length of the curve g(t) = [(t^3)/3-t, t^2] in the interval [0,2]. > It should be t^4 - 2t^2 + 1 + (2t)^2, which is t^4 + 2t^2 + 1. That makes > it easier. > It's a hint that math teachers and textbook writers use to make it easy for > you: if the integral is too hard to compute, then you made a mistake > setting it up. Unfortunately, if the class allows calculators, then this > hint disappears. Who cares if it's hard to calculate it -- just have the > machine do it! > This > is the first of a series of problems like it, and obviously I haven't > gotten > off to a good start. I will be extremely grateful if anyone can provide > me > with hints on how to solve this. So far I've tried some substitutions and > integration by parts. > The time it takes to check for mistakes in your work is much less than the > time it takes to try to do something you don't know how to do, both on exams > and in homework. The fact that you don't know how to do it doesn't prove > that it's not doable, but it does give you a hint that you ought to check > your work. > Jon Miller === Subject: Re: arc length Cc: linjoe@telus.net > It should be t^4 - 2t^2 + 1 + (2t)^2, which is t^4 + 2t^2 + 1. That makes > it easier. > It's a hint that math teachers and textbook writers use to make it easy > for > you: if the integral is too hard to compute, then you made a mistake > setting it up. Unfortunately, if the class allows calculators, then this > I did check over every attempt I made at it. In all honesty, Ive spent > more than two hours reworking the same problem. I've looked at texts, went > to get help from the assisstance centre at school, but not really getting > any help in the end. Ill give the assisstance centre another shot > tomorrow, but I'll be glad to get help from anywhere at this point. Look closely at t^4 + 2t^2 + 1. That's the same as (t^2 + 1)^2 So your integral is: 2 Int sqrt[(t^2 + 1)^2] dt 0 Now you should be able to move ahead. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: arc length (t^2-1)^2+2t, or t^4-2t^2+2t+1, which isn't t^4-2t^2+1+(2t)^2, as you > It should be t^4 - 2t^2 + 1 + (2t)^2, which is t^4 + 2t^2 + 1. That makes > it easier. > It's a hint that math teachers and textbook writers use to make it easy > for > you: if the integral is too hard to compute, then you made a mistake > setting it up. Unfortunately, if the class allows calculators, then this > I did check over every attempt I made at it. In all honesty, Ive spent > more than two hours reworking the same problem. I've looked at texts, went > to get help from the assisstance centre at school, but not really getting > any help in the end. Ill give the assisstance centre another shot > tomorrow, but I'll be glad to get help from anywhere at this point. > Look closely at t^4 + 2t^2 + 1. That's the same as (t^2 + 1)^2 > So your integral is: > 2 > Int sqrt[(t^2 + 1)^2] dt > 0 > Now you should be able to move ahead. > -- > Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: arc length > (t^2-1)^2+2t, or t^4-2t^2+2t+1, which isn't t^4-2t^2+1+(2t)^2, as you I think what everyone has been trying to tell you is that given a curve (x(t), y(t) for 0 <= t <= 2 the arclength is int(sqrt((x'(t))^2 + (y'(t))^2,t=0..2). In your case (x'(t))^2 + (y'(t))^2 = (t^2 - 1)^2 + (2t)^2 = [NOTE: (2t)^2<---] t^4 - 2t^2 + 1 +4t^2 = t^4 +2t^2 + 1 = (t^2 + 1)^2 -- Paul Sperry Columbia, SC (USA) === Subject: Re: arc length > I did check over every attempt I made at it. In all honesty, Ive spent > more than two hours reworking the same problem. I've looked at texts, went > to get help from the assisstance centre at school, but not really getting > any help in the end. Ill give the assisstance centre another shot > tomorrow, but I'll be glad to get help from anywhere at this point. You seem to have missed the point. This is an extremely easy problem to do. Your error is a simple algebraic one. Read Jonathan's first line again: It's t^4 + 2t^2 + 1 under the square root. === Subject: integral ambiguity Hi. I'm returning to mathematical studies after a long respite, and have a question about integrals. I've heard at least 2 definitions for them so far, plus I need to clear up the meaning of improper ones. Basically, I've always thought of an integral as an antiderivative, where you take each term of a polynomial ax^b and make it: (ax^(b+1))/b+1. So this says that it's an operator which takes a polynomial and changes its form, generating another polynomial equation with variables and the like. Yet more commonly I read it as meaning the area of the curve along the certain interval, and returning a single, constant scalar, being the area or volume. So which is it? Are the definite and indefinite integrals really two completely separate things? Further, is an indefinite integral the same as an improper integral which ranges from -infinity to infinity? How to calculate such an area when it is written with a variable being assigned to the values of the range, as in: Int(x=-inf, +inf) f(x)e^(2(pi)ikx/p)dx ? -- Blitzen === Subject: Re: integral ambiguity >Basically, I've always thought of an integral as an antiderivative, >where you take each term of a polynomial ax^b and make it: >(ax^(b+1))/b+1. So this says that it's an operator which takes >a polynomial and changes its form, generating another polynomial >equation with variables and the like. Actually, a _family_ of polynomials. The antiderivatives (note plural) of 3x^2 are x^3, x^3+1, x^3-6.210, etc: that's why that +c we tack on every antiderivative is important. If you graph several antiderivatives of a given function, you have a set of parallel- looking curves. >Yet more commonly I read it as meaning the area of the curve along >the certain interval, and returning a single, constant scalar, being >the area or volume. >So which is it? Are the definite and indefinite integrals really >two completely separate things? Conceptually, yes. An indefinite integral is a family of functions. A definite integral is a number. Two very different animals. I don't think this is emphasized enough in our calculus courses. We just call them integrals off the bat and students think of them as the same sort of thing. I did, when I studied calculus for the first time. The Fundamental Theorem of Calculus is what ties them together. It's really pretty remarkable that you should be able to find the area under a nonnegative function between x=a and x=b by evaluating any antiderivative of that function at x=a and the same antiderivative at x=b, and subtracting. >Further, is an indefinite integral the same as an improper integral >which ranges from -infinity to infinity? No. An integral with one or both limits of integration at infinity either is a definite number (like any other definite integral), or does not exist. The only way it can exist is if the integrand gets closer and closer to 0 as x approaches infinity. Example: the definite integral from -inf to +inf of exp(-0.5x^2) is a number, specifically sqrt(2pi). The definite integral from -inf to +inf of x^2 does not exist. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com reduces the number of useful answers you get. http://www.cs.tut.fi/~jkorpela/usenet/laws.html === Subject: Re: integral ambiguity alt.math.undergrad: [...] > So which is it? Are the definite and indefinite integrals > really two completely separate things? Yes, as far as their definitions are concerned. The indefinite integral is defined to be the antiderivative, which is a function (or more accurately, a family of functions, since f(x) and f(x) + c always have the same derivative). The definite integral of a function over an interval is defined to be the limit (in a specific sense) of Riemann sums and is therefore a number (when it exists). The two definitions are completely different, and there is no reason a priori to expect one concept to be related to the other. It turns out, however, that they are related, by the Fundamental Theorem, which, ignoring technical details, says that if g(x) = f'(x), then the integral of g from a to b is f(b) - f(a). > Further, is an indefinite integral the same as an improper > integral which ranges from -infinity to infinity? No. An improper integral is a limit of definite integrals, which are numbers, and is therefore a number itself (if it exists); an indefinite integral is an antiderivative and therefore a function. [...] Brian === Subject: Re: integral ambiguity > Basically, I've always thought of an integral as an antiderivative, > where you take each term of a polynomial ax^b and make it: > (ax^(b+1))/b+1. So this says that it's an operator which takes > a polynomial and changes its form, generating another polynomial > equation with variables and the like. No, you have error. integral ax^b = (ax^(b+1)/(b+1) or to be exact integral ax^b = (ax^(b+1)/(b+1) + arbitary constant > Yet more commonly I read it as meaning the area of the curve along > the certain interval, and returning a single, constant scalar, being > the area or volume. This is a definite integral. Care must be taken when thinking this area stuff. For example integral(-1,1) 2x = x^2 evalutated from x = -1 to x = 1 = 1^2 - (-1)^2 = 0 integral(-2,1) 2x = x^2 evalutated from x = -2 to x = 1 = 1^2 - (-2)^2 = -3 gives some weird notions for area. > So which is it? Are the definite and indefinite integrals really > two completely separate things? > Further, is an indefinite integral the same as an improper integral > which ranges from -infinity to infinity? How to calculate such an > area when it is written with a variable being assigned to the values > of the range, as in: No, improper integrals are integrals don't evaluate. integral(-oo,oo) 0 dx = 0 is quite proper. > Int(x=-inf, +inf) f(x)e^(2(pi)ikx/p)dx A complex integrand can't be used for a definite integral. For complex integrand, you have to use complex integration over a region for example or around a closed curve. === Subject: Re: integral ambiguity >> Basically, I've always thought of an integral as an antiderivative, >> where you take each term of a polynomial ax^b and make it: >> (ax^(b+1))/b+1. So this says that it's an operator which takes >> a polynomial and changes its form, generating another polynomial >> equation with variables and the like. >No, you have error. > integral ax^b = (ax^(b+1)/(b+1) >or to be exact > integral ax^b = (ax^(b+1)/(b+1) + arbitary constant Your answer is closer, but your parentheses are not balanced! :-) -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com reduces the number of useful answers you get. http://www.cs.tut.fi/~jkorpela/usenet/laws.html === Subject: Re: integral ambiguity >> Basically, I've always thought of an integral as an antiderivative, >> where you take each term of a polynomial ax^b and make it: >> (ax^(b+1))/b+1. So this says that it's an operator which takes >>No, you have error. > integral ax^b = (ax^(b+1)/(b+1) >or to be exact > integral ax^b = (ax^(b+1)/(b+1) + arbitary constant > Your answer is closer, but your parentheses are not balanced! :-) (ax^(b+1))/(b+1) + c === Subject: Fourier Analysis questions unfortunately, the author does not provide any answers in the back of the book, or a solutions manual that I know of. Let N be a positive integer and let f() be defined on Z (integers) by: f[n] := {1 if n = 0, +/-1, +/-2, +/-N {0 otherwise. (a) Show that f() has the p-periodic Fourier transform { 2N + 1 if s = 0 F(s) = 1/p{ sin{(2N + 1)(PI)s/p} { -------------------- if 0 < s < p { sin((PI)s/p) (Hint: Set z = e^(-i(PI)s/p) in formula: sum(n=-N,N)z^(2n) = z^(-2N) sum(n=0,2N)z^(2n) = z^(2N+1) - z^(-2N-1) -------------------- , z =/= +/- 1. z - z^-1 (b) Verify that f has the Fourier representation...: f[n] = Int(0,p)(sin{2N + 1)(PI)s/p}) (-------------------) e^(2(PI)ins/p)ds, ( p sin((PI)s/p ) n=+/-1,+/-2,... of the Fourier Synthesis/Analysis equations for Z (integers). (let me know if you want me to type these). I've never been good at proving or showing; here he seems to give both the question and the answer, and wants you to show how to get from A to be. Sure would be nice if he provided the answers elsewhere so you'd know if you're on the right track or not.. (or to help you along when stuck). I have more-or-less understood what he's said in the chapters; I just always have a tough time applying them to practice in the exercise (as is typical of me in any math class ;o) ). TIA for any words, urls or refereces to other publications. -- Blitzen === Subject: Re: Linear Algebra question I received a not-so-nice letter regarding my question - I am new to Linear Algebra and not a math genius like some of you. I only posted this since I was having a difficult time and I did not wish to waste anyones effort by being lazy. This newsgroup is one of the few resources I have and I am most greatful. For those of you who think I ask stupid questions, please set me to ignore. I have a question I hope someone can help me with. How do you show that > the > set of 2 x 2 matrices with real coefficients form a linear space over R > with > dimension 4? Tom Risager's response is certainly correct and probably what you were > intended to do. > However, there is a shortcut if you have enough background. > There is a natural one-to-one onto function, f, from R^4 to the 2 x 2's > such that f(u + v) = f(u) + f(v) and f(a*v) = a*f(v). It follows > immediately that, since R^4 is a v.s. of dim 4, so are the 2 x 2's. > -- > Paul Sperry > Columbia, SC (USA) > Neat. Is it always true that if a linear transform is onto and one-to-one > then it maps between spaces of the same dimension? > Tom > Yes. The image of a basis is a basis. > But note that you can't have a linear transformation until you have > both the domain and codomain vector spaces. The real thrust of my > suggestion was that you can avoid laboriously (and trivially) verifying > all those axioms for the 2 x 2's by using f and the fact that R^4 > satisfies them. > -- > Paul Sperry > Columbia, SC (USA) === Subject: Re: Linear Algebra question > I received a not-so-nice letter regarding my question - I am new to Linear > Algebra and not a math genius like some of you. I only posted this since I > was having a difficult time and I did not wish to waste anyones effort by > being lazy. This newsgroup is one of the few resources I have and I am most > greatful. For those of you who think I ask stupid questions, please set me > to ignore. Don't let not-so-nice e-mails bother you. Unfortunately, it goes with the USENET territory. -- Paul Sperry Columbia, SC (USA) === Subject: Re: David Peter Robbins - August 12th 1942 > Could you tell me what this is all about? First two words: mental illness. Dar's been over the hill for quite a while now, and recently he's followed his hero Haris into cross-posting to all of the *.math.* hierachy. Also, some advise to you, if you don't mind. It'll probably save you some stomach acid if you don't top post (posting above the message that you wish to reply to) and that you snip all of the post that you don't specifically want to quote. In Dar's case, that would be all of it, as illustrated below. HTH. Cheerts, Drieux {snip} of the usual bigotry and insanity. === Subject: seatings around a table I'm in a discrete math class studying right now about seatings in a line and at tables, etc, and I'm working the following practice problem: (1) How many ways can you arrange 7 distinguishable men and 5 distinguishible women in a row such that no two women are adjacent? (2) ... at a round table ... They say two round-table arrangements are the same iff each person has the same neighbor to the right in each seating. The back of the text says the answer to (1) is 7!5!(8c5) and the answer two (2) is 6!5!(7c5) ( (XcB) = X choose B ). Could someone please explain to me how they get these answers? I can almost see (1), but it would still be helpful if, given the answer to (1), someone could reason how they get (2). === Subject: Re: seatings around a table > I'm in a discrete math class studying right now about seatings in a > line and at tables, etc, and I'm working the following practice > problem: > (1) How many ways can you arrange 7 distinguishable men and 5 > distinguishible women in a row such that no two women are adjacent? > (2) ... at a round table ... > They say two round-table arrangements are the same iff each person has > the same neighbor to the right in each seating. > The back of the text says the answer to (1) is 7!5!(8c5) and the > answer two (2) is 6!5!(7c5) ( (XcB) = X choose B ). Could someone > please explain to me how they get these answers? I can almost see (1), > but it would still be helpful if, given the answer to (1), someone > could reason how they get (2). a) You can sit the men in 7! ways. You must choose five places among 8 where to sit the women: before the first men, before the second, ..., after the last ---> 8c5 ways. Then you can permute the five women ---> 5! b)You can sit the men in 6! ways, it is irrelevant who is the first. You must choose five sites among 8 where sit the women: at right that the first men, at right that the second, ..., at right that the seventh ---> 7c5 ways. Then you can permute the five women ---> 5! -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Linear Dependence/Independence Newbie Question Alert: I am trying to conceptualize linear dependence and independence in a matrix. be zero for this to be true, the linear equation is linearly independent. This also means that in a linear dependent case all the equations are related to each other (I am taking it that way) since they sum up to zero. In a linearly independent matrix the equations will not sum to zero no matter what an is. This gets me confused... It seems, intuitively at least, that I should be able to find some an term that would make them add up to zero, the books are heavy on theory but short on examples...Can you judge whether or not a matrix is linearly independent/dependent by inspection? What is a good test for this? === Subject: Re: Linear Dependence/Independence > Newbie Question Alert: > I am trying to conceptualize linear dependence and independence in a matrix. Your terminology is perhaps misleading you. One can speak of the rows of marix, or the columns, as being linearly independent but, to me anyway, linear dependence and independence in a matrix has no meaning. But never mind; let's see what we can do. > be zero for this to be true, the linear equation is linearly independent. For vectors x_1, x_2,...,x_n, if the only way a_1*x_1 + a_2*x_2 +...+a_n*x_n = 0-vector is for scalars a_1, a_2, ..., a_n is for a_1 = a_2 = ... = a_n = 0-scalar then the vectors are linearly independent. > This also means that in a linear dependent case all the equations are > related to each other (I am taking it that way) since they sum up to zero. > In a linearly independent matrix the equations will not sum to zero no > matter what an is. This gets me confused... It seems, intuitively at > least, that I should be able to find some an term that would make them add > up to zero, the books are heavy on theory but short on examples...Can you > judge whether or not a matrix is linearly independent/dependent by > inspection? What is a good test for this? 1*(1, 2, 3) + (-2)*(4, 5, 6) + 1*(7, 8 , 9) = (0, 0, 0) So (1, 2, 3), (4, 5, 6), (7, 8, 9) are linearly dependent vectors. I guess one could say the equations x + 2y + 3z = 1, 4x + 5y + 6z = 1, 7x + 8y + 9z = 1 are linearly dependent (but I don't think I've ever seen it done). In deciding whether (1, 2, 3), (4, 5, 6) and (7, 8 , 9) are linearly independent, I asked myself Self, what about solutions to x(1, 2, 3) + y(4, 5, 6) + z(7, 8, 9) = (0, 0, 0)? Well that gives x + 4y + 7z = 0; 2x + 5y + 8z = 0; 3x + 6y + 9z = 0. A little wheeling and dealing gets x - z = 0; y + 2z = 0 ( and 0 + 0 + 0 = 0 ). So, for example z = 1, x = 1, y = -2 works. One way to work with the three equations is to put the coefficients in a matrix: 1 4 7 2 5 8 3 6 9 and work with the matrix. The progression linear independence -> equations -> matrix may be what is confusing you. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Linear Dependence/Independence > Newbie Question Alert: > Easiest thing is to look at two examples in 2 dimensions. The vectors (1,1) and (2,2) are linearly dependent. The vectors (1,0) and (0,1) are linearly independent. Check the formal definition and you can see that's true. What LI means is that you can't express one in terms of the other(s).