mm-298 == Subject: Re: Need help with a math problem>I've put a lot of time on this one and I can't come to the solution>yet. Here it is:>A gun shoot bullets at a given, regular interval. After it fire all of>it's bullets it need some time to reload until he can resume shooting.>Ammo : Number of bullets that the gun shoot before it need to reload.>RateOfFire : Time in seconds between each shot.>ReloadTime : Time in seconds required to reload.>I am searching for a formula that gives the number of shot for a given>number of seconds.>With no ammunition limits it would be :>BulletsShot = Time / RateOfFire>But I have to consider the time spent to reload the gun AND the fact>that the gun may be in the process of reloading, at a given second.>For example:>Ammo : 4>RateOfFire : 2>ReloadTime : 5>SecondsElapsed BulletsShot> 0 1> 1 1> 2 2> 3 2> 4 3> 5 3> 6 4> 7 Reloading> 8 Reloading> 9 Reloading> 10 Reloading> 11 Reloading> 12 5> 13 5> 14 6> 15 6> 16 7> 17 7> 18 8> 19 Reloading> 20 Reloading>and so onIf you are looking for the limit over a long time, it would be bulletsfired per interval times total time. If you load 4 bullets each time,and it takes 4 seconds to shoot them, plus 5 seconds to reload, thenyou would have 4 bullets per 9 seconds for large times. For shortertimes, you could use total time modulo (shoot time + reload time) tofind how many intervals. Then the ci formula applies to thecomplete intervals and a formula based on continuous fire applies tothe remainder:rateoffire defined as bullets per unit time without reloadingammo defined as one complete load of ammunitionshoottime = rateoffire * ammointervalsize = shoottime + reloadtimeintervals = totaltime mod (intervalsize)remainder = totaltime - intervalsthen,bulletsfired = (intervals * ammo ) + remainder * rateoffireThat's ignoring whether you start out loaded, and whether the barrelmelts.--john ==== Subject: Algebraic integer surpriseFor some time now I've been trying to explain an algebraic method foranalyzing polynomials that relies on non-polynomial factors. Myresearch is a natural extension in the polynomial domain of the ideaof irrationality in the integer domain. Basically, I look at theequivalent of irrational factors with polynomials.Recently Rick Decker, a professor at Hamilton College, apparentlytrying to refute my research came up with a quadratic example, which Ilike because it's a quadratic, and easier to manipulate than thecubics I've used before.If you wish to see his original post here are some headers which alsoshow that he is indeed at Hamilton College:== Subject: Re: Mathematical consistency, courageDecker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples ofnon-polynomial factors.Notice that despite not being polynomials they are algebraic integersif x is an algebraic integer because a_1(x) and a_2(x) are the tworoots ofa^2 - (x - 1)a + 7(x^2 + x).To my knowledge mathematicians have never done extensive research withnon-polynomial factors of polynomials, but instead most work in thearea has to do with finding roots to polynomials, or determining ifpolynomials with rational coefficients are reducible over rationals.The extension into the realm of non-polynomial factors has revealed aproblem with a previous understanding of mathematicians in the area ofthe ring of algebraic integers. It's easy to prove the problem.Consider that algebraically, factorizations multiply out in a certainway demonstra by(a+b)(c+d) = ac + ad + bc + bdand that's just a FACT which is not open to debate.Looking at(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) it is possible then to multiply out the factors on the left to obtain25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x + 2).Now subtracting 14 from both sides gives25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x)which reveals an imbalance as 35 is the constant left on the left,while 0 is what's on the right.A simple linear transformation helps balance the factorization, asusinga_2(x) = b_2(x) - 1, and making the substitution gives(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).Now multiplying out I get25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x + 2)and subtracting 14 from both sides gives25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)so as expec the constants 7 and 2 on the left are factors of 7(2)on the right.The mathematics is trivially obvious in that regard, but from thatsimple result, obvious from basic arithmetic, I have a truly profoundconclusion.That is, given that with(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)the 7 and 2 on the left are factors of 7(2) on the right, if I dividethat constant factor 7, visible as a factor of7(25x^2 + 30x + 2)from both sides then to remain in the ring of algebraic integers,there's only one way logically to do it:(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2as any other way, like factoring 7 into two other non-unit factorswill require that you have numbers not available in the ring ofalgebraic integers.For instance, assume there exists some value of x within the ring ofalgebraic integers where the factors (5a_1(x) + 7) and (5b_2(x) + 2)each have sqrt(7) as a factor (in fact, x=1 will work), then you'dhave(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2which reveals the two factors of 2 on the right, as being sqrt(7) and2/sqrt(7) on the left.Remember, algebra gives a rigid format for multiplying out, as Ipoin out before with(a+b)(c+d) = ac + ad + bc + bdso there shouldn't be debate here from people who accept algebra.Notice that works in the field of algebraic numbers, but not in thering of algebraic integers.But it turns out that it's possible to prove that even with thefactorization(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2that (5a_1(x)/7 + 1) is not in general an algebraic integer.The problem here is that the natural follow through from the algebrais that there exists b_1(x), where a_1(x) = 7b_1(x), so that youfinally have(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2and while b_1(x) does exist, as it must, it's rather easy to show thatit's not an algebraic integer.What makes this result so surprising is that the necessary conclusionis that you cannot operate completely in the ring of algebraicintegers if you divide both sides of(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)by 7, as then *necessarily* from basic algebra, you're forced out ofthe ring as you then must have factors of 2 that are outside the ringof algebraic integers, as I demonstra with sqrt(7) as *necessarily*then the factors of 2 are sqrt(7) and 2/sqrt(7) which are not in thering of algebraic integers.Extensions of mathematical knowledge tend to produce surprisingresults, and some people can't handle a world where areas once thoughtsettled are revealed to contain new results. But my hope is that someof you are actually mathematical researchers versus seeing yourselvesas mere caretakers of dogma.For more on my research in this area and a broader overview, pleasesee my blog archives: ==== Subject: hmm by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0PDEfV25090; === so where does x^y = y^x when x > 0 and y > 0? joe ==== Subject: Re: hmm> so where does x^y = y^x when x > 0 and y > 0?> joex = y = 1 ? ==== Subject: Re: hmm> so where does x^y = y^x when x > 0 and y > 0?x = y = 1 ?Hmm. x = y = _any_ constant > 0 obviously works.Such trivial solutions aside, the only positive integer solutions for x andy are given by (x,y)=(2,4) and (x,y)=(4,2).For _all_ real solutions of the equation x^y = y^x, see my sci.math.David Cantrell ==== Subject: Chair in Statistics - VacancyVacancy for a Chair in Statistics at Eindhoven University of Technology,Eindhoven, The Netherlands ==== ============================================== ==== ========================================================== The Department of Mathematics and Computer Science has a vacancy for a fullprofessorship in StatisticsGeneral Information:-------------------------------------------------- -------------------------------------------------------------- -------------------------------------------------------------- ---------------------------The Department of Mathematics and Computer Science provides undergraduate,MSc and PhD programs in Industrial and Applied Mathematics and in ComputerScience.The Department has research collaborations with other Departments at theTechnical University Eindhoven, as well as with a large number of otheruniversities and companies,both at home and abroad.The Department has approximately 300 employees and over 700 students. Thechair of Statistics is one of the nine chairs in Mathematics, and one ofthe four chairs in the section Statistics and Operations Research. The othertwo sections in Mathematics cover Analysis and Discrete Mathematics.For the future, the Department of Mathematics and Computer Science envisionsimportant new opportunities for research on biological, biomedical,industrial and engineering applications. The statistics group is activelyinvolved with the activities at EURANDOM, the European institute forresearch in stochastics.What are your duties?------------------------------------------------------- -------------------------------------------------------------- -------------------------------------------------------------- --------------------------You are expec to stimulate and coordinate the fundamental and appliedresearch of the group, to initiate new research directions, to establishlinks with other researchprograms at our university and to become actively involved with EURANDOM.You give and coordinate courses in Statistics and Probability, and you areresponsible for updating these courses.You advise MSc and PhD students.You fulfill key management functions in the section and the faculty.Your skills are:---------------------------------------------------------- -------------------------------------------------------------- -------------------------------------------------------------- -------------------------A deep and broad insight into statistics, as is reflec in a stronginternational reputation in the field and a large number of - also recent -publications in the international literature.Much affinity with, and knowledge of stochastics.Interest in application-orien research, experience and expertise inacquisition and consultancy.Excellent didactical qualities; leadership and management qualities.What we have to offer:-------------------------------------------------------- -------------------------------------------------------------- -------------------------------------------------------------- ----------------------------A prominent leading position in a stimulating scientific environment, inwhich you will work with a strong group in stochastics, enthusiasticstudents,trainee design engineers, and PhD students, as well as postdocs fromEURANDOM.A full-time appointment in accordance with the Collective Labour Agreementfor Dutch Universities.A maximum salary of euro 7.875,00 gross per month depending on yourexperience.An extensive package of fringe benefits.Inquiries:------------------------------------------- -------------------------------------------------------------- -------------------------------------------------------------- ----------------------------------------For more information, please contact Prof.dr. W.T.F. den Hollander, e-mail:denhollander@eurandom.tue.nl, tel. +31 40 2478100.For information concerning job conditions, please contact Mr. W.C.J.Verhoef, head human resources, e-mail: w.c.j.verhoef@tue.nl, tel. +31 402472321.How to respond:------------------------------------------------------ -------------------------------------------------------------- -------------------------------------------------------------- -----------------------------Please submit a written letter of application accompanied by a recentcurriculum vitae to:Mrs. Drs. S. Udo,Managing Director of the Department of Mathematics and Computer Science,Eindhoven University of Technology,HG 6.22,P.O. Box 513,5600 MB Eindhoven,The Netherlands,mentioning the number of the vacancy V32854 in your letter and on theenvelope. ==== Subject: does ANYONE has ANY math problem?I'm a high-school student, I came across a applied maths test thatrequires me write something to solve an applied maths problem. It'snearly 4000 - 5000 words. I had a classmate to do it with me. But tillnow( I have thought since Chinese Spring Festival ) I haven't decidedwhat to do. Does anyone has math problem?PS: we are only high school students && we're not good at ENGLISH. ==== Subject: Re: does ANYONE has ANY math problem?>I'm a high-school student, I came across a applied maths test that>requires me write something to solve an applied maths problem. It's>nearly 4000 - 5000 words. I had a classmate to do it with me. But till>now( I have thought since Chinese Spring Festival ) I haven't decided>what to do. Does anyone has math problem?>PS: we are only high school students && we're not good at ENGLISH.You don't say much about what you know so it is hard to predict whatwill be an interesting problem. It is also hard to define applied. Forexample is computing the area of a triangle an applied problem? Ifso Heron's solution of computing the area of a triangle given 3 sidesis something that only requires high school geometry. SeeWilliam Dunham's Journey through Genius isbn 014014739X (about $15),which is very accessible.-- http://www.math.fsu.edu/~bellenotbellenot math.fsu.edu +1.850.644.7189 (4053fax) ==== Subject: Maple problemIs there a newsgroup for discussing problems in using Maple?Thanks.-- David G. CantorCenter for Communications ResearchSan Diego, CA ==== Subject: Re: Maple problem> Is there a newsgroup for discussing problems in using Maple?Yes, comp.soft-sys.math.maple . Also, there is a Yahoo group dedica tothat,http://groups.yahoo.com/group/maple-assist/Alec Mihailovshttp://webpages.shepherd.edu/amihailo/ ==== Subject: help with subsI have a function E = kx + cx'(x' = time derivatve of variable x calcula using diff(x(t),t). x isonly a function of time)now, i need to get differential of E wrt x'. any ideas on how i can dothis ?of course, this is a very simple function, but i need to do this forcomplica multivariable functions,(trying to solve mechanics problemsusing Lagrange's eqns)can anyone give me tips on how to go about it ?Vin.> restart:> time_dpndt := {x=x(t)}; time_dpndt := {x = x(t)}> time_indpt := {x(t)=x}; time_indpt := {x(t) = x}> E := (1/2)*(K*x^2 + C*diff(subs(time_dpndt,x),t)); 2 K x /d E := ---- + 1/2 C |-- x(t)| 2 dt /> v1 := diff(subs(time_dpndt,x),t); d v1 := -- x(t) dt> diff(E,v1); ==== Subject: Re: help with subs===> I have a function E = kx + cx'(x' = time derivatve of variable x calcula using diff(x(t),t). x is> only a function of time)now, i need to get differential of E wrt x'.... ^^^^^^^^^^^^ derivative?Remember (dE/dx') = (dE/dt)/(dx'/dt). Ken Pledger. ==== Subject: Re: help with subsoops...i forgot to add that i am trying to that that calculation inMaple.I have a function E = kx + cx'(x' = time derivatve of variable x calcula using diff(x(t),t). x is> only a function of time)now, i need to get differential of E wrt x'. any ideas on how i can do> this ?of course, this is a very simple function, but i need to do this for> complica multivariable functions,(trying to solve mechanics problems> using Lagrange's eqns)can anyone give me tips on how to go about it ?> Vin.restart:> time_dpndt := {x=x(t)};> time_dpndt := {x = x(t)}> time_indpt := {x(t)=x};> time_indpt := {x(t) = x}E := (1/2)*(K*x^2 + C*diff(subs(time_dpndt,x),t)); 2> K x /d > E := ---- + 1/2 C |-- x(t)|> 2 dt /v1 := diff(subs(time_dpndt,x),t); d> v1 := -- x(t)> dt> diff(E,v1); ==== Subject: Questions about the usage of mupadpolynomials together with the coefficients of Bezout identity. It is thefollowing:bezout := proc(a:Type::PolyExpr(x,Type::Rational), b:Type::PolyExpr(x,Type::Rational)) local u,x0,x1,y0,y1,v,r,q; begin //Special case: at least one polynomial is 0 if a*b=0 then if (a=b) then error(gcd(0,0) is not defined); end_if; u := a+b; return(a+b,1,1); end_if; x0 := 1; x1 := 0; y0 := 0; y1 := 1; u := a; v := b; r := divide(u,v)[2];// while ( r <> poly(0,[x]) ) do while ( r <> 0 ) do q := divide(u,v)[1]; x2 := x0-q*x1; y2 := y0-q*y1; x0 := x1; x1 := x2; y0 := y1; y1 := y2; u := v; v := r; r := divide(u,v)[2] ; end_while; //gcd(a,b)=v=x1*a+y1*b return(v,x1,y1); end_proc;I have some questions about the syntax and usage, in order to constructa function which should be as close as possible to the state of art:1) the usage of the function varies with respect to the input: it works ifI insert the polynomials simply as expressions (eg, bezout(x^2-1,x^2+2*x+1)). It doesn't work if I insert them using thefunction poly. For this to work, I have to change the while line with thecommen one, and similarly in other places (and also I need to take careof when, in the multiplication q*x1, q is a constant: quite boring). Howcan I define it so that it works in any case it make sense? (I.e.: Iprovide either an expression which can be conver to a polynomial ordirectly a polynomial)2) As everybody knows, the preceding algorithm works every time the input(a,b) are in an euclidean domain, provided the system is able to computea function divide which returns quotient and reminder. Is there a wayto implement the algorithm so that the euclidean domain is not specifieda priori but only in the moment when the function is used? I ask thisbecause I am in the need to use the function in various environments(Z,Q[x],Z_p[x]) and I would like to define only one function.TIAFabio ==== Subject: Re: Questions about the usage of mupadI do not know much about mupad, but can't you pass an extraargument in to your program, the name of the divide function?If Mupad does not allow some level of polymorphism, e.g. notallowing the coefficient type of a and b to vary, then it isa poor design in this day and age. Mimicking Maple, as it seemsto do, is unfortunate.e.g. bezourt:=proc (a,b,main_variable,coefficient_type, divide_program) ....Also, it seems to me you should be doingx0:= poly(1,[x]), and divide should return a poly form.Also, testing a*b=0 is a wasteful way to test (a=0)or(b=0).If neither is zero, you will be doing a pointless polynomialmultiplication.RJFpolynomials together with the coefficients of Bezout identity. It is the> following:bezout := proc(a:Type::PolyExpr(x,Type::Rational),> b:Type::PolyExpr(x,Type::Rational))> local u,x0,x1,y0,y1,v,r,q;> begin > //Special case: at least one polynomial is 0> if a*b=0 then> if (a=b) then> error(gcd(0,0) is not defined);> end_if;> u := a+b; > return(a+b,1,1);> end_if;> x0 := 1; x1 := 0; y0 := 0; y1 := 1;> u := a; v := b; r := divide(u,v)[2];> // while ( r <> poly(0,[x]) ) do> while ( r <> 0 ) do> q := divide(u,v)[1]; > x2 := x0-q*x1;> y2 := y0-q*y1;> x0 := x1; x1 := x2;> y0 := y1; y1 := y2;> u := v; v := r; r := divide(u,v)[2] ;> end_while;> //gcd(a,b)=v=x1*a+y1*b> return(v,x1,y1);> end_proc;> I have some questions about the syntax and usage, in order to construct> a function which should be as close as possible to the state of art:> 1) the usage of the function varies with respect to the input: it works if> I insert the polynomials simply as expressions > (eg, bezout(x^2-1,x^2+2*x+1)). It doesn't work if I insert them using the> function poly. For this to work, I have to change the while line with the> commen one, and similarly in other places (and also I need to take care> of when, in the multiplication q*x1, q is a constant: quite boring). How> can I define it so that it works in any case it make sense? (I.e.: I> provide either an expression which can be conver to a polynomial or> directly a polynomial)> 2) As everybody knows, the preceding algorithm works every time the input> (a,b) are in an euclidean domain, provided the system is able to compute> a function divide which returns quotient and reminder. Is there a way> to implement the algorithm so that the euclidean domain is not specified> a priori but only in the moment when the function is used? I ask this> because I am in the need to use the function in various environments> (Z,Q[x],Z_p[x]) and I would like to define only one function.TIAFabio ==== Subject: Re: Questions about the usage of mupad> I do not know much about mupad, but can't you pass an extra> argument in to your program, the name of the divide function?> If Mupad does not allow some level of polymorphism, e.g. not> allowing the coefficient type of a and b to vary, then it is> a poor design in this day and age. Mimicking Maple, as it seems> to do, is unfortunate.e.g. bezourt:=proc (a,b,main_variable,coefficient_type, divide_program) > ....> Also, it seems to me you should be doing> x0:= poly(1,[x]), and divide should return a poly form.In any case, this way I only get a generalization to K[x], with K field, not a true generalization to euclidean domain; for example, I needto redefine it for Z or gauss integers.> Also, testing a*b=0 is a wasteful way to test (a=0)or(b=0).> If neither is zero, you will be doing a pointless polynomial> multiplication.I was wondering if it was better to do {just one test and amultiplication} or {(maybe) two tests}, but thinking it better it seemsto me that you are right: the second alternative should be faster.Fabio ==== Subject: Re: Questions about the usage of mupad / defining extended gcd, domains, programmingIt is probably the case that there are only about 3 or 4 domains ofinterest, and a generic program will be slower on the simpledomains. I suggest that if you want to have an extendedEuclidean algorithm for integers, you write one that is specificto that. And you might look to see if mupad has a program alreadydefined. Macsyma has such a program, but it is good only for univariateZ[w,x,y,z,...] with one distinguished main variable.extended_gcd(u,v,x):= block([u1,u2,u3,v1,v2,v3,t1,t2,t3,q], u: rat(u,x), v: rat(v,x), [u1,u2,u3]:[rat(1),rat(0),u], [v1,v2,v3]:[rat(0),rat(1),v], while v3#0 do (q: quotient(u3,v3,x), [t1,t2,t3]:[u1,u2,u3]-q*[v1,v2,v3], [u1,u2,u3]:[v1,v2,v3],[v1,v2,v3]:[t1,t2,t3]), [u1,u2,u3])$I don't actually check for u=v=0, since the answer it comes outwith 1*0+0*0 = 0is acceptable to me. The only time the gcd is 0 is if bothinputs are 0.Macsyma does not require type declarations, though there havebeen various attempts to include such information, and forcompilation to numerical code, certain declarations help. Butthe Macsyma language dates back to about 1966 or so.RJF ==== Subject: Nedd helpThis is the problem:You are planning a party and want to have two types of snacks:Stuffed mushroom and cheese sticks. Each mushroom costs about $0.25and each cheese stick will cost about $0.15. You want enough to makeso that each personcan have at least two stuffed mushroom and three cheese sticks. Thereare going to be at most 40 people at the party and you don't want tospend more than $45.Write a system of linear equations that shiows the various numbers ofstuffed mushrooms and cheese sticks that you could make.I would really appreciate anybody who can help and tell me what thosetwo equations are. I came up with the first one2x(0.25)+ 3y(0.15) <= 45I need help with the second one. Thanks. ==== Subject: Non-standard analysisConsider the collection of sets I1, I2, I3, ... , In, ...Where I1 = {1}, I2 = {1,2}, I3 = {1,2,3},...,In = {1,2,3,...,n},...for natural numbers n. That is, Ij is the set of natural numbers lessthan or equal to j.Define a set S to be finite if there is a 1-1 mapping of S onto In forsome natural number n.Define a finite set S to be larger than a natural number n, if thereis a natural number m and a 1-1 map of S onto Im with m > n. Consider further the following countable collection of sentences:1. There exists a finite set S1 larger than 1.2. There exists a finite set S2 larger than 2....n. There exists a finite set Sn larger than n....Clearly any finite subset of this collection is consistent. [let j bethe largest index of a sentence in the finite subset, then, forexample, Ij+1 provides the model] By compactness, the whole collection is consistent and therefore thereis a model with the property that there exists a finite set S which islarger than n for any n. This is clearly a contradiction.[S finite implies by definition that there exists some natural numbern, such that S can be mapped 1-1 onto In. But S larger than n implies,again by definition, that S can be mapped 1-1 onto Im for some naturalnumber m > n]