mm-649 === Subject: Yes it does!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4I2AiG21524; I have attended Devry in Colorado for the past 2 years and have seen the the tution double in that time( $585 per credit hour). This wouldn't be a problem but 90% of the teachers don't have a clue of what's going on!! It's sad when you go to a school, pay that kind of money and you know more than the teachers do. Devry is just a money hungry school that doesn't give a Rat's Ass about it's students. I would not recommend going to Devry, especially In Colorado!!!!! === Subject: Re: Yes it does!! >I have attended Devry in Colorado for the past 2 years and have seen >the the tution double in that time( $585 per credit hour). This >wouldn't be a problem but 90% of the teachers don't have a clue of >what's going on!! It's sad when you go to a school, pay that kind of >money and you know more than the teachers do. Devry is just a money >hungry school that doesn't give a Rat's Ass about it's students. >I would not recommend going to Devry, especially In Colorado!!!!! Then why are you there??????? === Subject: Sequences and Subsequences redux OK. I'm not proud just slow evidently. Given a sequence, a_n, of length one million with a _n real and a_i <> a_j if i <> j, prove or disprove: there is a subsequence of length [CapitalThorn]ve which is increasing or a subsequence of length [CapitalThorn]ve which is decreasing. This is not homework. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Sequences and Subsequences redux alt.math.undergrad: > OK. I'm not proud just slow evidently. > Given a sequence, a_n, of length one million with a _n real and > a_i <> a_j if i <> j, prove or disprove: there is a subsequence of > length [CapitalThorn]ve which is increasing or a subsequence of length [CapitalThorn]ve which > is decreasing. > This is not homework. Let's just prove the general theorem: any real sequence of length greater than n^2 contains a monotonic subsequence of length n+1. Let s = be a real sequence of length n^2 + 1. For k = 0, ..., n de[CapitalThorn]ne u(k) and d(k) as follows: u(k) is the maximum length of all monotone non-decreasing subsequences of s with [CapitalThorn]rst term a_k, and d(k) is the maximum length of all monotone non-increasing subsequences of s with [CapitalThorn]rst term a_k. Suppose that s has no monotone subsequence of length n+1; then 1 <= u(k), d(k) <= n for all k. There are only n^2 possible pairs , so there must be distinct j and k such that j < k, u(j) = u(k), and d(j) = d(k). But this quickly leads to a contradiction: if a_j <= a_k, then clearly u(j) >= u(k) + 1, and if a_j >= a_k, then d(j) >= d(k) + 1. Brian === Subject: Re: Sequences and Subsequences redux > alt.math.undergrad: > OK. I'm not proud just slow evidently. > Given a sequence, a_n, of length one million with a _n real and > a_i <> a_j if i <> j, prove or disprove: there is a subsequence of > length [CapitalThorn]ve which is increasing or a subsequence of length [CapitalThorn]ve which > is decreasing. > This is not homework. > Let's just prove the general theorem: any real sequence of > length greater than n^2 contains a monotonic subsequence of > length n+1. Since I wasn't getting anywhere on that one, I decided to try an easier one and re[CapitalThorn]ne. > Let s = be a real > sequence of length n^2 + 1. For k = 0, ..., n de[CapitalThorn]ne u(k) > and d(k) as follows: u(k) is the maximum length of all > monotone non-decreasing subsequences of s with [CapitalThorn]rst term > a_k, and d(k) is the maximum length of all monotone > non-increasing subsequences of s with [CapitalThorn]rst term a_k. Aha! I was insisting on looking at the subsequences themselves rather than what I was really interested in - their lengths. I should have known better. > Suppose that s has no monotone subsequence of length n+1; > then 1 <= u(k), d(k) <= n for all k. There are only n^2 > possible pairs , so there must be distinct j and > k such that j < k, u(j) = u(k), and d(j) = d(k). But this > quickly leads to a contradiction: if a_j <= a_k, then > clearly u(j) >= u(k) + 1, and if a_j >= a_k, then d(j) >= > d(k) + 1. > Brian Oddly, when I [CapitalThorn]rst saw the problem, I thought Pigeon Hole Principle. Turns out it was. The pigeons are the (n + 1)^2 pairs (u(k), d(k)) k = 0...n and the pigeon holes are the n^2 possible values. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Sequences and Subsequences redux alt.math.undergrad: [...] > Oddly, when I [CapitalThorn]rst saw the problem, I thought Pigeon Hole Principle. > Turns out it was. The pigeons are the (n + 1)^2 pairs > (u(k), d(k)) k = 0...n and the pigeon holes are the n^2 possible values. Yep. The result appears as an example or a hard problem in several sophomore-level discrete math texts in the section on the pigeonhole principle, and I've also seen it used as a contest problem. The proof perhaps isn't quite an Erd.9as Ôbook proof', but I still think that it's awfully pretty. Brian === Subject: Re: Sequences and Subsequences redux >OK. I'm not proud just slow evidently. >Given a sequence, a_n, of length one million with a _n real and >a_i <> a_j if i <> j, prove or disprove: there is a subsequence of >length [CapitalThorn]ve which is increasing or a subsequence of length [CapitalThorn]ve which >is decreasing. Construct a sequence using the [CapitalThorn]rst million integers, arranged in this way: 1, 1,000,000, 2, 999,999, 3, 999,998, ..., 499,998, 500,003, 499,999, 500,002, 500,000, 500,001. I think this is a pretty good counterexample: it has no monotonic subsequence of length >= 3. >This is not homework. -- Stan Brown, Oak Road Systems Cortland County, New York, USA http://OakRoadSystems.com You need any friends you can get. The only thing standing between you and a watery grave is your wits, and that's not my idea of adequate protection. -- /Beat the Devil/ (1954) === Subject: Re: Sequences and Subsequences redux >OK. I'm not proud just slow evidently. >Given a sequence, a_n, of length one million with a _n real and >a_i <> a_j if i <> j, prove or disprove: there is a subsequence of >length [CapitalThorn]ve which is increasing or a subsequence of length [CapitalThorn]ve which >is decreasing. > Construct a sequence using the [CapitalThorn]rst million integers, arranged in > this way: > 1, 1,000,000, 2, 999,999, 3, 999,998, ..., > 499,998, 500,003, 499,999, 500,002, 500,000, 500,001. > I think this is a pretty good counterexample: it has no monotonic > subsequence of length >= 3. consider 1, 2, 3, 4, 5 to be a subsequence. Formally, in this case, is there always a montone strictly increasing n:{1, 2, 3, 4, 5} -> {1, ..., 1,000,000} such that a_n(1) < a_n(2) < ... < a_n(5) or a_n(1) > a_n(2) > .. > a_n(5)? -- Paul Sperry Columbia, SC (USA) === Subject: Re: Sequences and Subsequences redux >Formally, in this case, is there always a montone strictly increasing >n:{1, 2, 3, 4, 5} -> {1, ..., 1,000,000} such that >a_n(1) < a_n(2) < ... < a_n(5) or a_n(1) > a_n(2) > .. > a_n(5)? Gotcha: contiguous is not implied in subsequence as you use it. Sorry! I thought at the time I was writing that it seemed too easy. -- Stan Brown, Oak Road Systems Cortland County, New York, USA http://OakRoadSystems.com You need any friends you can get. The only thing standing between you and a watery grave is your wits, and that's not my idea of adequate protection. -- /Beat the Devil/ (1954) === Subject: Re: Sequences and Subsequences redux > OK. I'm not proud just slow evidently. > Given a sequence, a_n, of length one million with a _n real and > a_i <> a_j if i <> j, prove or disprove: there is a subsequence of > length [CapitalThorn]ve which is increasing or a subsequence of length [CapitalThorn]ve which > is decreasing. > This is not homework. I would start by saying: If there is an increasing subsequence of 5 or more we are done so we can star by supposing there is no such susequence and try to show that there must then be a decreasing subsequence of 5 or more. Actually, one million in your original sequence should be way more than needed. === Subject: Re: Sequences and Subsequences redux > OK. I'm not proud just slow evidently. Given a sequence, a_n, of length one million with a _n real and > a_i <> a_j if i <> j, prove or disprove: there is a subsequence of > length [CapitalThorn]ve which is increasing or a subsequence of length [CapitalThorn]ve which > is decreasing. This is not homework. > I would start by saying: > If there is an increasing subsequence of 5 or more we are done so we can > star by supposing there is no such susequence and try to show that there > must then be a decreasing subsequence of 5 or more. neither kind of subsequence and shoot for a contradiction. But, contradiction of what? > Actually, one million in your original sequence should be way more than > needed. Sure, that's why I picked it. In fact, according to Bernd's original conjecture, length 17 should be enough. -- Paul Sperry Columbia, SC (USA) === Subject: curve sketching by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4IAY8332037; can any one show me how to [CapitalThorn]nd the asymptotes for a) f(x)=sin(x)+cos(3x)sin(2x) b) 1/(ln(abs(3x^2-1)) === Subject: curve sketching by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4IAY8132057; can any one show me how to [CapitalThorn]nd the asymptotes for a) f(x)=sin(x)+cos(3x)sin(2x) b) 1/(ln(abs(3x^2-1)) === Subject: Re: curve sketching >can any one show me how to [CapitalThorn]nd the asymptotes for >a) f(x)=sin(x)+cos(3x)sin(2x) >b) 1/(ln(abs(3x^2-1)) Please don't post the same query multiple times. Do you know what asymptotes are? You have a vertical asymptote where the function value goes to in[CapitalThorn]nity (though you've probably been taught to use more accurate language). Can you imagine any x values that would make your [CapitalThorn]rst function get very very large? No, so there are no vertical asymptotes. A horizontal asymptote is a horizontal line that the function graph gets closer and closer to as x gets bigger and bigger (or smaller and smaller, for negative x). Pretty obviously that doesn't happen with your [CapitalThorn]rst function. If that's not obvious, rewrite it using cos A sin B = (1/2)sin(A+B) - (1/2)sin(A-B): f(x) = sin(x) + cos(3x)*sin(2x) f(x) = sin(x) + (1/2)sin(5x) - (1/2)sin(x) f(x) = (1/2) [ sin(5x) + sin(x) ] This will continue oscillating forever, with a period of 2pi and [CapitalThorn]ve fast oscillations superimposed on every one slow oscillation. The amplitude is constant at (1/2)2 = 1. Now look at your second function, and use the de[CapitalThorn]nitions of asymptotes to answer the same question. -- Stan Brown, Oak Road Systems Cortland County, New York, USA http://OakRoadSystems.com You need any friends you can get. The only thing standing between you and a watery grave is your wits, and that's not my idea of adequate protection. -- /Beat the Devil/ (1954) === Subject: Urgent: need help for calculus Can anyone tell me how to get the integral of exp(-(x^2)/2). === Subject: Re: Urgent: need help for calculus > Can anyone tell me how to get the integral of exp(-(x^2)/2). You usually don't see such integrands until you learn double integrals. The best way to approximate this would be either numerical integration or possibly a power series. Dave === Subject: Re: Urgent: need help for calculus by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4IJrrg22373; >Can anyone tell me how to get the integral of exp(-(x^2)/2). Unfortunately, [exp(-(x^2)/2] is an extremely recalcitrant integrand, defying elementary integration techniques. For what it's worth, I looked this integral up in my handbook and the value is reported to be exactly sqrt(2pi) (assuming the limits are taken from -oo to +oo). To attain an approximate value, your best bet would be to try a numerical integration technique such as Simpson's rule or numerical quadrature. === Subject: Re: Urgent: need help for calculus > Can anyone tell me how to get the integral of exp(-(x^2)/2). No, nobody! It doesn't integrate except numerically. === Subject: Re: Urgent: need help for calculus > Can anyone tell me how to get the integral of exp(-(x^2)/2). > No, nobody! It doesn't integrate except numerically. Though for some limits of integration (like from 0 to oo or from -oo to +oo) you can get an exact value for it. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: problem in line drawing >To check whether a given point is on >line is used the below formula >y-y1 x-x1 >----- = ----- >y2-y1 x2-x1 >I'm not sure where that formula came from. It's point-slope form, just rearranged a bit. Point-slope form is usually given as y-y1 = m(x-x1) where m is the slope. But if two points on the line are known then m can be rewritten (y2-y1)/(x2-x1). That gives (y-y1) = [(y2-y1) / (x2-x1)] (x-x1) which is the same as (y-y1) / (y2-y1) = (x-x1) / (x2-x1) -- Stan Brown, Oak Road Systems Cortland County, New York, USA http://OakRoadSystems.com You need any friends you can get. The only thing standing between you and a watery grave is your wits, and that's not my idea of adequate protection. -- /Beat the Devil/ (1954) === Subject: problem about topology I need to solve this problem and I have the idea but I don't know how to write it: let (R^2,g) be a topological space where g is the standard topology on R^2. (R is the set of real numbers.) I have to prove that R^2(G1 U G2) never becomes a [CapitalThorn]nite set. G1 and G2 are nonempty disjunctiv sets, and they are elements of g. that is they are open sets in R^2. solution in words. if we let a point P be in G1 and a point Q in G2 then we can always make a linesegment PQ and [CapitalThorn]nd a point C that neither is in G1 nor in G2. But since the points G1 and G2 are arbitrary and since there are always an in[CapitalThorn]nite number of points in an open set and since the points don't lie in a straight line, then you will always have an in[CapitalThorn]nite number of points C. does the argument sound [CapitalThorn]ne? How can I write it with opensets? or how can I prove that there are always an in[CapitalThorn]nite number of points in a open set? === Subject: Re: problem about topology > let (R^2,g) be a topological space where g is the standard topology on R^2. > (R is the set of real numbers.) > I have to prove that R^2(G1 U G2) never becomes a [CapitalThorn]nite set. > G1 and G2 are nonempty disjunctiv sets, and they are elements of g. that is > they are open sets in R^2. This cannot be as G1 = G2 = R^2 - {(0,0)} is open set of R^2 and G1 / G2 = G1 U G2 = G1 union G2 = R^2 - {(0,0)} R^2 - (G1 / G2) = {(0,0)} So do you mean to show R^2 - G1xG2 is in[CapitalThorn]nite? Where G1xG2 = { (x,y) | x in G1, y in G2 } G1, G2 subset R > solution in words. > if we let a point P be in G1 and a point Q in G2 then we can always make a > linesegment PQ and [CapitalThorn]nd a point C that neither is in G1 nor in G2. But > since the points G1 and G2 are arbitrary and since there are always an > in[CapitalThorn]nite number of points in an open set and since the points don't lie in a > straight line, then you will always have an in[CapitalThorn]nite number of points C. > does the argument sound [CapitalThorn]ne? It's not to my taste. > How can I write it with opensets? or how can I prove that there are always > an in[CapitalThorn]nite number of points in a open set? Now that's an soothing approach. Let U = { u1,..uj } be a [CapitalThorn]nite open subset of R^2. Show { u1,.. u_(j-1) } = Uuj = U / R^2uj is open set. / intersection. By induction demostrate { u1 } is open. Can you show that is impossible and conclude proof? === Subject: Re: problem about topology > let (R^2,g) be a topological space where g is the standard topology on R^2. > (R is the set of real numbers.) > I have to prove that R^2(G1 U G2) never becomes a [CapitalThorn]nite set. > G1 and G2 are nonempty disjunctiv sets, and they are elements of g. that is > they are open sets in R^2. > This cannot be as G1 = G2 = R^2 - {(0,0)} is open set of R^2 and > G1 / G2 = G1 U G2 = G1 union G2 = R^2 - {(0,0)} > R^2 - (G1 / G2) = {(0,0)} [...] You seem to have overlooked the requirements for G1 and G2 : non-empty, disjoint and open. -- Paul Sperry Columbia, SC (USA) === Subject: non modular lattices by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4IKxBT28726; does anyone knows an example of a subgroup lattice which is not modular and a subring lattice, also not modular??? please help === Subject: Re: non modular lattices days. My association with the Department is that of an alumnus. >does anyone knows an example of a subgroup lattice which is not >modular and a subring lattice, also not modular??? please help Obviously, the subgroup lattice would have to be of a nonabelian group. The subgroup lattice of A_4 includes the following full sublattice: A_4 / / H K | / M / / / 1 where H is the subgroup of order 4 generated by (1,2)(3,4) and (1,3)(2,4); M is the subgroup of order 2 generated by (1,2)(3,4); and K is the subgroup of order 3 generated by (1,2,3). So the lattice cannot be modular. For subrings, take the group algebra F_2[A_4]. Then you have subrings corresponding to F_2[H], F_2[M], and F_2[K] as above. The intersection of F_2[H] and F_2[M] each with F_2[K] is just F_2 = F_2[{e}]. The smallest subring of F_2[A_4] that contains F_2[M] and F_2[K] must contains F_2[] = F_2[A_4], and so be the same as the smallest one containing F_2[H] and F_2[K]. So you get the same sublattice in subrings. -- 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 === Subject: TI-89 Programs for Calculus III Anyone recommend/have some good TI-89 programs for help with verifying Calculus III problems? Or multivariable problems like tangent plane to a surface, green's theorem, line integrals, etc? Netsniper === Subject: How to Find Limit Function f(x) = lim dx->0 of g(x,x+dx) To All, Could someone out there point me in the right direction in solving/proving the following? Kevin J. Brewer Given a funtion of 2 variables g(x,y), how does one [CapitalThorn]nd the limit function f(x) for the situation in which y=x ? (Having no Greek characters, let dx represent delta x in the following.) Asked more elegantly, how does one [CapitalThorn]nd f(x) = lim g(x,x+dx) dx->0 In one hemodialysis paper (call it paper 1) that I am studying, it states that / K ( 1 - -- ) Qb | Qd | Ko*A = ------ * ln| ------- | eqn1 Qb | K | 1 - -- ( 1 - -- ) Qd Qb / when Qd <> Qb and K Ko*A = ------ eqn2 K 1 - -- Qb when Qd = Qb where: Ko is the mass transfer coef[CapitalThorn]cient of the dialyzer A is the mass transfer area of the dialyzer Ko*A is the mass transfer area coef[CapitalThorn]cient of the dialyzer Qb is the blood ßow rate through the dialyzer Qd is the dialysate ßow rate through the dialyzer K is the urea clearance coef[CapitalThorn]cient of the dialyzer without any physical or mathematical proof. Another hemodialysis paper (call it paper 2), which did not was able to follow the derivation of eqn1. *If* eqn2 is true, possibly without knowing it, in essence, paper 1 was stating that the limit of eqn1 when Qd = Qb is eqn2. Via numerical methods, I have found that this appears to be true by setting Qd = Qb + 1 mL/min, etc. Obviously, eqn1 has a singularity at Qb = Qd where it (eqn1) is indeterminate because its 1st term tends toward in[CapitalThorn]nity and its 2nd term tends toward 0. As a result, this singularity plane must be approached via limiting. Thinking of eqn1 and eqn2 as / K ( 1 - -- ) Qb | Qd | g(Qb,Qd) = ------ * ln| ------- | eqn3 Qb | K | 1 - -- ( 1 - -- ) Qd Qb / and K f(Qb) = ------ eqn4 K 1 - -- Qb Since both eqn3 and eqn4 are equal to Ko*A, in essence, paper 1 was stating that f(Qb) = lim g(Qb,Qb+dQ) dQ->0 where Qd = Qb + dQ How does one [CapitalThorn]nd the limit function of eqn1 as Qd goes to Qb? Or from another point of view, how does one prove that eqn2 is the limit function of eqn1 as Qd goes to Qb? === Subject: Re: How to Find Limit Function f(x) = lim dx->0 of g(x,x+dx) Kevin, I answered this over in alt.math. In general, it's not a good idea to post the same problem to different groups separately. People here can't see what is written over there, and vice-versa, so someone may go to all the effort to duplicate what's already been done on another group. === Subject: Re: Ran accross this forum, need help if possible by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4JF3Ux23692; >I'm not a student or anything, but needed a probability [CapitalThorn]gured out if >anyone could be so kind. It would take me days to [CapitalThorn]gure out and you >guys maybe a minute or two. >Anyway here goes, say X has a 99% success rate. What would be the >odds of X failing exactly 4 out of 5 occurances? I know it'd be >really low : P but would like to know the exact number if possible? Hi Aaron, if X counts the number of failings during n ocurrences, then X is binomially distributed with p = 0.01. Thus P(X = i) = C(n,i)*(0.01)^i*(1-0.01)^(n-i) with C(n,i) = n!/(i! * (n-i)!) For your special case (n=5 and i=4): P(X = 4) = C(5,4)*(0.01)^4*(0.99)^(5-4) = 4.95e-8 . Best wishes Torsten. === Subject: Re: Sequences and Subsequences by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4JF6VF23929; >Hi everyone, >I've run into a problem in my discrete maths. book which says >that the following is true: >A sequence a_1, a_2, ..., a_k where k = n^2 + 1 of distint numbers >contains either an increasing subsequence of length n + 1 or >a decreasing subsequence of length n + 1. >Suppose the sequence is 1, 3, 2, 5, 4 so k = 5 and n = 2. I see no >increasing or decreasing subsequence of length 3. All the increasing >or decreasing subsequences are of length 1 or 2. I looked at the >errata sheet of this book, but it doesn't say anything about this.] >I'm confuzzled here. >Bernd Hi Bernd, 1,3,5 or 1,3,4 or 1,2,5 or 1,2,4 are monotonically increasing subsequences of length 3 ... Best wishes Torsten. === by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4JJJPm18581; I am an American High School math teacher from Colorado, I just returned from a world elementary math education forum in Shanghai, National College Entrance Exam. All Chinese HS students in their senior year take this exam, along with 5 other subjects. And their total score determine how good of a college they go to, or they can go to college at all. Boy, if American students have to take this kind of math test to get to college, none of us would have a college degree. I know I wouldn't... I translated the entire test here: http://www.geocities.com/gzhzhx/test.html It's non-calculus, but that does not mean it's easy.... Enjoy.... === I am an American High School math teacher from Colorado, I just returned from a world elementary math education forum in Shanghai, National College Entrance Exam. All Chinese HS students in their senior year take this exam, along with 5 other subjects. And their total score determine how good of a college they go to, or they can go to college at all. Boy, if American students have to take this kind of math test to get to college, none of us would have a college degree. I know I wouldn't... I translated the entire test here: http://www.geocities.com/gzhzhx/test.html It's non-calculus, but that does not mean it's easy.... Enjoy.... === I know this drifts off the subject signi[CapitalThorn]cantly; but, what kind of comparable test and high school level mathematics instruction occurs in VietNam? G C === Subject: Re: My paper passed peer review by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4JLl8Z01180; >They plotted their conspiracy *openly* in posts, as if it were >nothing. Now they're assaulting the journal process, assaulting the >peer review process, and doing it PUBLICLY because that's Usenet. >The last word really is that my paper passed peer review. Others >will come here in this thread and mouth off with their rage because >that is Usenet. The long overdue publication of your paper, while something of a pyrrhic victory, represents a milestone in mathematics. I would like to offer you a genuine and heartfelt handshake - Congratulations, James! And, no, you won't have to endure any disparaging remarks from me. Far from it. I, for one, took the time to read -- no, scrutinize -- your papers. Their contents so lucidly describe the framework for a radically innovative approach to the way we view algebra. Our research company, specializing in the re[CapitalThorn]nement and applications of multidimensional hyperbolic annihilators, has already made explicit use of those very same seminal concepts delineated in your work with overwhelming success. To all those who wrongfully scorn James, I offer the following admonition: Einstein himself was initially ridiculed for his seemingly zany and far-fetched views of our universe. Just as nature abhors a vacuum, so does the average mind abhor genius. J.P. === Subject: Re: My paper passed peer review Let us know when it's published.