mm-277 === Subject: Re: (n to the 5th power) mod 10 Since phi(10) = 4, the result follows trivially from Euler's theorem or Lagrange's threorem. Hint: look at a^(phi(N)) mod N === Subject: 27 straight lines on a cubic surface I was reading in a history text that Cayley and Salmon proved that any cubic surface has precisely 27 distinct straight lines lying completely in it. I'm having difficulty understanding the definition of these lines. Are they short sections of surface where the gradient is constant? Could someone give me a brief explanation? -- Alastair Rae, London, Europe. My opinions are not necessarily those of my employers. === Subject: Re: 27 straight lines on a cubic surface >I was reading in a history text that Cayley and Salmon proved that any >cubic surface has precisely 27 distinct straight lines lying >completely in it. >I'm having difficulty understanding the definition of these lines. Are >they short sections of surface where the gradient is constant? Nothing short about them. A cubic surface (for this purpose) is the set of points in projective 3-space with homogeneous coordinates (x:y:z:w) (i.e., x, y, z, and w are elements of the ground field [which you are welcome to assume is the complex numbers] which are not all 0, and for any non-zero t in the ground field, (tx:ty:tz:tw) represents the same point as (x:y:z:w)) such that, for some given homogenous polynomial G in four variables of degree 3, G(x,y,z,w)=0. We had better also assume that there are no points at which both G(x,y,z,w)=0 and the gradient of G is (0,0,0,0). Now the theorem is that, for generic such G at least (I'm surprised to see both any and distinct in the statement you give, and I really wonder if that's right), there are 27 distinct projective lines entirely contained in the surface. A projective line, in this case, is the intersection of two distinct projective planes, and a projective plane is the set of points with homogeneous coordinates (x:y:z:w) such that some linear form ax+by+cz+dw=0 (where a,b,c, and d aren't all 0). If your ground field isn't algebraically complete--e.g., if it's the real numbers instead of the complex numbers--I do believe that some of these 27 lines may not be defined over that field itself, but only over its algebraic closure (the complex numbers, if the ground field is the real numbers). In fact, I think it's quite a challenge to produce a non-singular real cubic surface on which all 27 lines are real (and thus visible in a physical model). But all I say here is === Subject to considerable correction and possibly contradiction. Exercise: try to find as many of the 27 lines as you can on the Fermat cubic, {(x:y:z:w) | x^3+y^3+z^3+w^3 = 0}. Hint: describe the subset of the Fermat cubic on which z = w = 0. === Subject: Re: 27 straight lines on a cubic surface > I was reading in a history text that Cayley and Salmon proved that any > cubic surface has precisely 27 distinct straight lines lying > completely in it. Add nonsingular and projective to cubic surface and work over an algebraically closed field. > I'm having difficulty understanding the definition of these lines. They're *lines* :-) > Are > they short sections of surface where the gradient is constant? What? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: 27 straight lines on a cubic surface > If your ground field isn't algebraically complete--e.g., if > it's the real numbers instead of the complex numbers--I do believe > that some of these 27 lines may not be defined over that field > itself, but only over its algebraic closure (the complex numbers, > if the ground field is the real numbers). In fact, I think it's > quite a challenge to produce a non-singular real cubic surface on > which all 27 lines are real (and thus visible in a physical model). I'm sure that you such surfaces do exist. If memory serves me right, there is a (clay ?) sculpture of one such example in the mathematical art collection at MSRI made by Mrs Hirzebruch. Cheers, === Subject: Re: 27 straight lines on a cubic surface >> If your ground field isn't algebraically complete--e.g., if >> it's the real numbers instead of the complex numbers--I do believe >> that some of these 27 lines may not be defined over that field >> itself, but only over its algebraic closure (the complex numbers, >> if the ground field is the real numbers). In fact, I think it's >> quite a challenge to produce a non-singular real cubic surface on >> which all 27 lines are real (and thus visible in a physical model). > I'm sure that you such surfaces do exist. If memory serves me right, > there is a (clay ?) sculpture of one such example in the mathematical > art collection at MSRI made by Mrs Hirzebruch. There used to be one in the bowels of DPMMS in Cambridge. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: 27 straight lines on a cubic surface If your ground field isn't algebraically complete--e.g., if >it's the real numbers instead of the complex numbers--I do believe >that some of these 27 lines may not be defined over that field >itself, but only over its algebraic closure (the complex numbers, >if the ground field is the real numbers). In fact, I think it's >quite a challenge to produce a non-singular real cubic surface on >which all 27 lines are real (and thus visible in a physical model). > I'm sure that you such surfaces do exist. If memory serves me right, >> there is a (clay ?) sculpture of one such example in the mathematical >> art collection at MSRI made by Mrs Hirzebruch. >There used to be one in the bowels of DPMMS in Cambridge. I didn't mean an insuperable challenge. My impression is that the all-27-lines-real locus in the space of real cubic surfaces, when compared to the various other non-empty loci defined by how many of the lines are real, is `small' in some non-precise sense (similar to the sense in which, I think, it's observed that among real affine cubic curves, one of has an oval, has no oval predominates--I don't remember which, and I won't even swear it's true, but I think it's been asserted in my hearing). Again, I may well be confused. Lee Rudolph === Subject: Re: 27 straight lines on a cubic surface > I was reading in a history text that Cayley and Salmon proved that any >> cubic surface has precisely 27 distinct straight lines lying >> completely in it. >Add nonsingular and projective to cubic surface and work over >an algebraically closed field. >> I'm having difficulty understanding the definition of these lines. >They're *lines* :-) >> Are >> they short sections of surface where the gradient is constant? >What? I did say that I was reading this in a *history* text so I was hoping for an simple explanation, not a ticking off. It is obvious that I don't understand what this theorem means but that was the point of my asking. I have done a reasonable amount of undergrad maths but a course in the history of maths doesn't expect you to know all the maths! I was just intrigued to know what Cayley-Salmon is about. I thought a cubic surface was a 3D object so I couldn't see how it could necessarily contain any infinite lines. Presumably the use of four homogenous coords projects a 4D object in to a 3D one. So is it correct to say that any fully 4D surface defined by a polynomial of order 3, has a projective 3D shadow with the 27 lines property? -- Alastair Rae, London, Europe. My opinions are not necessarily those of my employers. === Subject: Re: 27 straight lines on a cubic surface >I thought a cubic surface was a 3D object so I couldn't see how it >could necessarily contain any infinite lines. The placement of the words could and necessarily make it difficult to parse that sentence. Certainly there are plenty of surfaces in (ordinary, real) 3-dimensional space which contain infinite lines: for instance, any plane is such, as is any hyperboloid of one sheet, like {(x,y,z) | x^2+y^2=z^2+1 }, which contains the line {(x,y,z) | x=z, y=1 } and infinitely many more lines like it. (Calling a mathematical object X a surface, in other words, should not be taken to imply any guarantee that X is the surface of, i.e., boundary of, a bounded piece of space. It just means, not too roughly, that X is 2-dimensional.) >Presumably the use of >four homogenous coords projects a 4D object in to a 3D one. No. Or rather, yes, but that's not a very productive way to think of it. Given a homogeneous polynomial, one can always dehomogenize it by setting one of the variables equal to 1; for instance, you can dehomogenize the Fermat cubic x^3+y^3+z^3+w^3 to x^3+y^3+z^3+1. The set of points (x,y,z) in ordinary (not projective) 3-space at which the dehomogenized polynomial has the value 0 is now a so-called affine surface--for instance, {(x,y,z) | x^3+y^3+z^3+1=0} is the affine Fermat cubic surface. It contains almost all the points of the cubic surface I was talking about earlier; the only ones that are missing are the points at infinity where the homogeneous coordinate w equals 0. In this case (and typically), these points at infinity themselves form a (projective) curve in a (projective) plane at infinity; for the Fermat example in particular, this curve has the (again homogeneous!) equation x^3+y^3+z^3=0, and contains no (projective) lines at all. So we know that each of the 27 straight lines on the projective Fermat cubic surface has all but one of its points on the affine Fermat cubic surface. And a projective line with one point removes is, in fact, just an affine line! So there are 27 distinct straight lines in complex 3-space with the property that the coordinates of any point on any one of the lines satisfy the equation x^3+y^3+z^3+1=0. Taking my hint in the earlier post one step further, here is one of those lines: {(t,t,-1) | t is in C}. If you look only at the real points of the affine Fermat cubic surface, that is, the set of *real* triples (x,y,z) with x^3+y^3+z^3+1=0, then only *some* of the lines will be present; for instance, {(t,t,-1) | t is in R}. Lee Rudolph === Subject: Re: 27 straight lines on a cubic surface lrudolph@panix.com (Lee Rudolph) can't calculate. >And a projective line with one point removes is, in fact, >just an affine line! So there are 27 distinct straight lines >in complex 3-space with the property that the coordinates of >any point on any one of the lines satisfy the equation >x^3+y^3+z^3+1=0. Taking my hint in the earlier post one >step further, here is one of those lines: {(t,t,-1) | t is in C}. >If you look only at the real points of the affine Fermat cubic >surface, that is, the set of *real* triples (x,y,z) with >x^3+y^3+z^3+1=0, then only *some* of the lines will be present; >for instance, {(t,t,-1) | t is in R}. Of course (unless I'm compounding one blunder with another) those should be {(t,-t,-1) | t is in C} and {(t,-t,-1) | t is in R}. Sorry. Lee Rudolph === Subject: Re: 27 straight lines on a cubic surface > I thought a cubic surface was a 3D object so I couldn't see how it > could necessarily contain any infinite lines. A cone is a 3D object; can that not contain any infinite lines either? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: 2D geometry question I am looking for a way to determine container relationship of a point vs rectangle in 2 dimension. x, y coordinates of rectangle and of point are known. Thank you! Please let me know if this is not the right news group i am posting message to. - ap === Subject: Re: 2D geometry question If the vertices of the rectangle are A,B,C,D then the rectangle itself is {A+p(B-A)+q(C-A): 0<=p<=/B-A/ & 0<=q<=/C-A/}. | I am looking for a way to determine container relationship of a point | vs rectangle in 2 dimension. x, y coordinates of rectangle and of | point are known. Thank you! Please let me know if this is not the | right news group i am posting message to. | | - ap === Subject: Re: 3x+3 > I again, took Collatz and changed the algorithm to 3x+3 and founbd that all > of the numbers end with the integers of 12, 6, and 3. Right. Any 3x+b system ends at b. See discussion of loops below. > But the *interesting* thing is that any number that is a divisor of 8 ends > in a series like 3x+1 - 4, 2, 1. That's incorrect. The termination point in 3x+3 is 3 not 1. You must continue the sequence to the proper termination (or looping point) 4 2 1 6 3 (12 6 3 12 6 3 ...) The sequence 4 2 1 has no special meaning in 3x+3, it is just another side branch of the trunk whose root is 3. See http://members.aol.com/mensanator666/3x.htm for a comparison of 3x+1 and 3x+3. Here are some observations about loops in 3x+b systems: Take your numbers from any 3n+b sequence and create a sequence vector (where [1]=n/2 and [2]=3n+b). For the (3n+1) loop 1 4 2 1, the sequence vector is [2][1][1] Now graph this sybolically using down for [1] and right for [2]. k_l m n The Hailstone Function is k as a function of n and uses Inverse Rules. m = 2*n l = 2*2*n k = (2*2*n - b)/3 This simplifies to k = (4*n - b)/3 Every Hailstone Function simplifies to the form (X*n - Z*b)/Y In the above case X=4 Z=1 and Y=3. (Z*b)/(X-Y) If the Crossover Point is an integer, then the sequence vector is a loop in that 3n+b system. The [2][1][1] sequence vector's Crossover Point is (Z*b)/(X-Y) = (1*b)/(4-3) = b/1 = b Thus, 3n+1 has a loop at 1, 3n+3 has a loop at 3, 3n+5 has a loop at 5, etc. The sequence vector [2][1] has a Hailstone Function (2*n - b)/3 and a Crossover Point of b/(2-3) = b/(-1) = -b Thus, 3n+1 has a loop at -1, 3n+3 has a loop at -3, 3n+5 has a loop at -5, etc. The sequence vector [2][1][2][1][1] has a Hailstone Function (8*n - 5*b)/9 and a Crossover Point of (5*b)/(8-9) = 5*b/(-1) = -5*b Thus, 3n+1 has a loop at -5, 3n+3 has a loop at -15, 3n+5 has a loop at -25, etc. The sequence vector [2][1][2][1][2][1][2][1][1][2][1][2][1][2][1][1][1][1] has a Hailstone Function (2048*n - 2363*b)/2187 and a Crossover Point of (2363*b)/(2048-2187) = 2363*b/(-139) = -17*b Thus, 3n+1 has a loop at -17, 3n+3 has a loop at -51, 3n+5 has a loop at -85, etc. EVERY 3n+b system has a loop at b, -b, -5b and -17b. Now here's where it gets interesting. The sequence vector [2][1][1][1] has a Hailstone Function (8*n - b)/3 and a Crossover Point of (b)/(8-3) = b/5 This sequence vector is only a loop in systems where b is a multiple of 5. Thus, it is a loop in 3n+5, with root 1, but it is NOT a loop in 3n+1 or 3n+3. The Collatz Conjecture is therefore false in 3n+5 because there are two loops in the positive integers. Note that the 1 in this loop is not the same as the 1 in 3n+1 because it is derived from a different sequence vector. The 1 in 3n+1 is the root of the [2][1][1] sequence vector that is common to all systems. There at least two other loops in 3n+5, 19 and 23, but I haven't worked out their sequence vetors. I expect they will be fractions also and thus, not produce an integer crossover when b=1. So all one has to do to prove the Collatz Conjecture false for 3x+1 is find a sequence vector whose Crossover Point is a positive integer when b=1. === Subject: [Primes|Assymptotics] What is the order of Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p Hi to everyone, I have the following sum: S(n) = Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p where [a] is the integer part of a (that is the biggest integer r such that r <= a < r+1). I want to know if S(n) can be expressed as beautiful expression and what is it, and what is the assymptotics of S(n) Xan. === Subject: Re: [Primes|Assymptotics] What is the order of Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p What do you mean by beautiful? It can clearly be expressed as a Stieltjes integral. Then apply Euler-MacLauren summation. I will be happy to do this for you. However, I am currently unemployed and my consulting fee is $125/hr. However, if this is part of a more general mathematical problem and you wish to discuss it, I might be persuaded that it is worth my time to do so. === Subject: Re: a channel from God 1 ~ self, God 2 ~ couple, to 3 ~ child, free 4 ~ correct, for 5 ~ chaos 6 ~ sex 7 ~ Adam 8 ~ infinity 9 ~ no 10 ~ perfection, Eve 11 ~ yes 12 ~ step 15 ~ is 20 ~ access 21 ~ me 25 ~ nemesis 42 ~ for two I should have posted the list, the page I opened to asking about the pattern was opposite a big 21 a cursive 21 in Hiragana (Japanese) sounds like Ôme'. when I met Jen (Eve) she said I'm 21, its how ladies subconsciously introduce themselves. Herc what a dumb internet, god on an internet terminal for 3 years and everyone who understands shuts up to the voicestrous atheists and skeptics, even christnet just calls me a pagan for daring to say god is a man now. 100,000 against me face to face in townsville and 100,000 against me on the internet. 1000s say what the greatest moment in history would be, James Randi proving paranormal, god alive, telepathy, immortality, destiny, and when they actually see it its gossip fodder and nothing more. Herc === Subject: Re: a channel from God >1 ~ self, God >2 ~ couple, to >3 ~ child, free >4 ~ correct, for >5 ~ chaos >6 ~ sex >7 ~ Adam >8 ~ infinity >9 ~ no >10 ~ perfection, Eve >11 ~ yes >12 ~ step >15 ~ is >20 ~ access >21 ~ me >25 ~ nemesis >42 ~ for two >I should have posted the list, >the page I opened to asking about the pattern was opposite a big 21 >a cursive 21 in Hiragana (Japanese) sounds like Ôme'. >when I met Jen (Eve) she said I'm 21, its how ladies subconsciously >introduce themselves. >Herc >what a dumb internet, god on an internet terminal for 3 years and everyone >who understands shuts up to the voicestrous atheists and skeptics, >even christnet just calls me a pagan for daring to say god is a man now. >100,000 against me face to face in townsville and 100,000 against me >on the internet. >1000s say what the greatest moment in history would be, James Randi >proving paranormal, god alive, telepathy, immortality, destiny, and when they >actually see it its gossip fodder and nothing more. NO one is interested in your fantasies, because that's all they are. Get back on your prozac. -- Lt. General, Fanatic Legions. Commander of Southern Hemisphere Forces. Find out about Australia's most dangerous Doomsday Cult: http://users.bigpond.net.au/wanglese/pebble.htm === Subject: Re: a channel from God I worked it out, if you check google earlier today I was posting ascii art... _ ________ | | | _____ | | | | | | | | | | | | | | | | | | | | | | | | | | | | |____| | |_| |________| ________ | _____ | | | | | | |____| | |______ | | | _______| | |________| ________ | _____ | | | | | | |____| | | ____ | | | | | | |____| | |________| ________ |______ | | | | | | | | | | | |_| ________ | _______| | | | |______ | ____ | | | | | | |____| | |________| ________ | _______| | | | |______ |______ | | | ______| | |________| _ _ | | | | | | | | | |____| | |______ | | | | | |_| ________ |______ | | | ______| | |______ | | | ______| | |________| ________ |______ | | | ______| | | ______| | | | |______ |________| _ | | | | | | | | | | | | |_| ^ / / / | | | | | U | | | | S | | | | A | | | | | | | | | | | /| | / | | / | | / | | ---- -/_- ---- | /x ///x ////xxx /////xxx //////x /////x / / so 9 T is 9 10 lord was counting back up to 10, a la Bo Derek. which means finished! (These axioms sum up everything) I just channeled. channeling with a computer science book.!! I hope it means www.a1sites.com is finished as I started the final phase on promoting that today, and should just get it ready for my 2 months with no electricity. the quantum field always keeps a door open when others shut, but its always on the line. Herc === Subject: Re: a channel from God >1 ~ self, God >2 ~ couple, to >3 ~ child, free >4 ~ correct, for >5 ~ chaos >6 ~ sex >7 ~ Adam >8 ~ infinity >9 ~ no >10 ~ perfection, Eve >11 ~ yes >12 ~ step >15 ~ is >20 ~ access >21 ~ me >25 ~ nemesis >42 ~ for two 51 - idiot-boy --- John Hattan Grand High UberPope - First Church of Shatnerology john@thecodezone.com http://www.shatnerology.com === Subject: Re: a channel from God Why are you posting this crap ?? maybe I'll come and post a load of rubbish on alt.godnutters and see how you like that. > I worked it out, if you check google earlier today I > was posting ascii art... > _ ________ > | | | _____ | > | | | | | | > | | | | | | > | | | | | | > | | | | | | > | | | |____| | > |_| |________| > ________ > | _____ | > | | | | > | |____| | > |______ | > | | > _______| | > |________| > ________ > | _____ | > | | | | > | |____| | > | ____ | > | | | | > | |____| | > |________| > ________ > |______ | > | | > | | > | | > | | > | | > |_| > ________ > | _______| > | | > | |______ > | ____ | > | | | | > | |____| | > |________| > ________ > | _______| > | | > | |______ > |______ | > | | > ______| | > |________| > _ _ > | | | | > | | | | > | |____| | > |______ | > | | > | | > |_| > ________ > |______ | > | | > ______| | > |______ | > | | > ______| | > |________| > ________ > |______ | > | | > ______| | > | ______| > | | > | |______ > |________| > _ > | | > | | > | | > | | > | | > | | > |_| > ^ > / > / > / > | | > | | > | U | > | | > | S | > | | > | A | > | | > | | > | | > | | > | | > /| | > / | | > / | | > / | | > ---- -/_- ---- > | > /x > ///x > ////xxx > /////xxx > //////x > /////x > / / > so 9 T is 9 10 lord was counting back up to 10, a la Bo Derek. > which means finished! (These axioms sum up everything) I just channeled. > channeling with a computer science book.!! > I hope it means www.a1sites.com is finished as I started the final phase > on promoting that today, and should just get it ready for my 2 months > with no electricity. the quantum field always keeps a door open when > others shut, but its always on the line. > Herc === Subject: Re: a channel from God > 1 ~ self, God > 2 ~ couple, to > 3 ~ child, free > 4 ~ correct, for > 5 ~ chaos > 6 ~ sex > 7 ~ Adam > 8 ~ infinity > 9 ~ no > 10 ~ perfection, Eve > 11 ~ yes > 12 ~ step > 15 ~ is > 20 ~ access > 21 ~ me > 25 ~ nemesis > 42 ~ for two I suspect this to be numerology, not mathematics ... David Ames === Subject: Re: a channel from God > I was just working on my space invaders game and blindly shooting > and looked up to notice a message on the screen. > A simple message but if you check my site www.adamskingdom.com > on numerology it is very straightforward. Numerology is for head-cases. Now go away with stuff your numbers where there sun don't shine... RS === Subject: Re: Aggregate of Two Geometric Series > David je v sporo.8filu novic > 4ce443a6.0308202008.4e97d615@posting.google.com napisal ... > I have a real-life math problem. > I have a product that customers subscribe to. Every year, on average, > a certain percentage of subscribers cancel. > For example, if I have 10 subscribers, and the cancellation rate is > 50% per year, the number of subscribers with time is a geometric > series: > 10, 5, 2.5, etc. > To calculate the average length of a subscription, I can apply the > formula: > Average Length of Subscription = 1/Cancellation Rate, which is a > modification of the formula for the sum of a geometric series. So in > the example above, the average length of subscription is 2 years > In formula for the sum of geometric series is s = a[1]/(1 - q); q is > qoutient (a[n+1] / a[n]) > To calculate the average lifetime value of the subscription, I simply > multiply this value by the price of subscription. So, if the > subscription costs $100, the average lifetime value of the > subscription in the example above will be $200. > Now for my problem. I want to factor-in price inßation. Every year, I > increase my prices by 25%. > # subscribers 10 5 2.5 1.25 > Here is q[a] = 0.5 and a[1] = 10 > Price $100 $125 $156 $195 > Here is q[b] = 1.25 and b[1] = $100 > Revenue $1000 $625 $391 $244 > r[n] = a[n] * b[n] > Theorem; q[r] = q[a]*q[b] > Prove: > q[r] = r[n+1]/r[n]=(a[n+1] * b[n+1])/(a[n] * b[n]) = (a[n+1] / a[n]) > *(b[n+1] / b[n]) =q[a]*q[b] Q.E.D. > I am assuming that since the series of revenue with time is a function > with only three factors: > - Initial number of subscribers > - Cancellation Rate > - Price Inßation > and that as long as the series revenue with time tends towards 0, it > seems to me to be an infinite geometric series, and a combination of > two other geometric series. > So, my problem is, how can I work out the formula to calculate the > average life time value, including price inßation? > Any suggestions? > LOOK UP > I ope it helps, Peter > David === Subject: Re: algrithm to list all possible partitioning of a network > Pardon me for ignorance. > Lets say, I want to generate all possible partition on a set networked > nodes. Is there an algrithm to do that? I understand that for a > general set, this would be related to Bell Number. But in my case, a > network relation is defined on the set. Theofore, using a set > partition algrithm is very ineffecient because it would generate a lot > of invalid partitioning. I assume a fairly large network of 50 nodes > but each node is connected to just a few (usually less than 3) other > nodes. > Does anyone know of an algrithm to list all partitioning of a network? > Many thanks in advance for the help. > The above sentence I want to generate all possible partition on a set > networked > nodes. should be I want to generate all possible partition on a set > of networked nodes. answer. It looks the question was not sufficiently clear though. Here is another way to put it. If a web of nodes is to be cut into pieces (two or more), is there an algorithm to list all the different ways to cut it (or all the outcome)? Consider each way of cutting to be the set of cuts needed to finish the job. === Subject: Re: algrithm to list all possible partitioning of a network > Pardon me for ignorance. Lets say, I want to generate all possible partition on a set networked > nodes. Is there an algrithm to do that? I understand that for a > general set, this would be related to Bell Number. But in my case, a > network relation is defined on the set. Theofore, using a set > partition algrithm is very ineffecient because it would generate a lot > of invalid partitioning. I assume a fairly large network of 50 nodes > but each node is connected to just a few (usually less than 3) other > nodes. Does anyone know of an algrithm to list all partitioning of a network? Many thanks in advance for the help. > The above sentence I want to generate all possible partition on a set > networked > nodes. should be I want to generate all possible partition on a set > of networked nodes. > answer. It looks the question was not sufficiently clear though. Here > is another way to put it. If a web of nodes is to be cut into pieces > (two or more), is there an algorithm to list all the different ways to > cut it (or all the outcome)? Consider each way of cutting to be the > set of cuts needed to finish the job. Does it matter if the web has or does not have a ßat graph? Kuratowski's theorem holds all sorts of (nasty) surprises. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: algrithm to list all possible partitioning of a network X-AUTHid: rubin > answer. It looks the question was not sufficiently clear though. Here > is another way to put it. If a web of nodes is to be cut into pieces > (two or more), is there an algorithm to list all the different ways to > cut it (or all the outcome)? Consider each way of cutting to be the > set of cuts needed to finish the job. I'm still not clear on how the web part fits in. Are you saying that the pieces need to be connected? Complete subgraphs? Some other restriction? Without any restriction on what happens to edges in the original graph (meaning that it is ok to delete them, and pieces need not be connected), the answer to what the different partitions might be is the set of all partitions (see the posting by Samik Raychaudhuri). I'm assuming that you are ruling some partitions out based on the edges, but it's not yet clear what your rules are. -- Paul === Subject: Re: a matrix problem related to eigevalues expand in minors across the top row | I have a matrix A(n), | where A(2)= [0 1/2 | 1 1/2] | A(3)= [0 0 1/3 | 0 1/2 1/3 | 1 1/2 1/3] | A(4)= [0 0 0 1/4 | 0 0 1/3 1/4 | 0 1/2 1/3 1/4 | 1 1/2 1/3 1/4] | And so on. | My purpose is to find the eigevalues. Through matlab simulation I | found they should be: | 1, -1/2, 1/3, -1/4,..., (-1)^n/n | But I cannot prove it. | | Climber === Subject: Re: a matrix problem related to eigevalues >expand in minors across the top row Ummm, that doesn't look so simple. For example, for det(A(4)-rI) the expansion by minors gives you [ r 1/3 1/4 ] [ 0 r 1/3 ] r det [ 1/2 1/3-r 1/4 ] - 1/4 det [ 0 1/2 1/3-r ] [ 1/2 1/3 1/4-r ] [ 1 1/2 1/3 ] So? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: a matrix problem related to eigevalues >expand in minors across the top row > Ummm, that doesn't look so simple. For example, for > det(A(4)-rI) the expansion by minors gives you > [ r 1/3 1/4 ] [ 0 r 1/3 ] > r det [ 1/2 1/3-r 1/4 ] - 1/4 det [ 0 1/2 1/3-r ] > [ 1/2 1/3 1/4-r ] [ 1 1/2 1/3 ] > So? Are there theorems on the effect of row operations (swapping rows, adding multiples of row i to row j) and eigenvalues? Because it's immediately clear that you can reduce R(n) to [0 0 .... 0 1/n] [0 0 ... 1/(n-1) 0] [ ... ] [0 1/2 ... 0 0] [1 0 ... 0 0] by subtracting each row from all the rows below it. What are the eigenvalues of an anti-diagonal matrix? - Randy === Subject: analytic/combinatorial number theory course notes Hi. I have placed online some notes on analytic/combinatorial number theory that Mathematics Program. Interested parties can find them at http://www.princeton.edu/~ppollack/notes/ I would be extremely grateful for notification of typos or other sorts of corrections, as well as general suggestions for improvement. Chapters 3 and up, in particular, could use a careful eye to proofread them, as they were never distributed in class. Paul === Subject: A non algebraic system of equations Once again I require the assistance of someone vastly more intelligent than myself. I'm playing around with the equations from www.numberspiral.com, and I wonder if this is possible. I have these equations: x=cos(sqrt(v)*2PI)*sqrt(v) y=-sin(sqrt(v)*2PI)*sqrt(v) is it possible to find v in terms of x and y? Mathematica pukes out on me when it tries to solve this system of equation. I tried everything I could think of to solve this, but I'm out of ideas. turned my infinite frustration into a mere (***frustration divided by 0***) -Paul Klemstine === Subject: Re: A non algebraic system of equations > I'm playing around with the equations from www.numberspiral.com, and I > wonder if this is possible. > I have these equations: > x=cos(sqrt(v)*2PI)*sqrt(v) > y=-sin(sqrt(v)*2PI)*sqrt(v) > is it possible to find v in terms of x and y? Mathematica pukes out > on me when it tries to solve this system of equation. I tried > everything I could think of to solve this, but I'm out of ideas. since you try to solve 2 equations for one unknown, Mathematica should encounter some difficulties. By squaring both equations and adding them you obtain x^2+y^2 = v, which coincides with the formula on http://www.numberspiral.com. What is the difficulty? Alois -- Alois Steindl, Tel.: +43 (1) 58801 / 32558 Inst. for Mechanics II, Fax.: +43 (1) 58801 / 32598 Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10 === Subject: Re: A non algebraic system of equations > Once again I require the assistance of someone vastly more intelligent > than myself. > I'm playing around with the equations from www.numberspiral.com, and I > wonder if this is possible. > I have these equations: > x=cos(sqrt(v)*2PI)*sqrt(v) > y=-sin(sqrt(v)*2PI)*sqrt(v) > is it possible to find v in terms of x and y? Mathematica pukes out > on me when it tries to solve this system of equation. I tried > everything I could think of to solve this, but I'm out of ideas. > turned my infinite frustration into a mere (***frustration divided by > 0***) > -Paul Klemstine How about the trivial v = x^2 + y^2, obtained by squaring bothe sides of each equation, adding equations and noting that sin(t)^2 + cos(t)^2 = 1 for every t. === Subject: Re: binomials over negative reals >... > If the m in binomial(n,m) (ie. n choose m) is a nonnegative integer, > then we can write binomial(n,m) as n*(n-1)*(n-2)*...(n-m+1)/m!, where n can be any complex number. > ... > Using this definition, then: binomial(-n,m) = (-n)*(-n-1)*(-n-2)*...(-n-m+1)/m! = (-1)^m n*(n+1)*(n+2)*...(n+m-1)/m! = (-1)^m binomial(n+m-1,m). >.....[a lot snipped] > binomial(-n,-m) = > (-1)^(m+n+1) ((n-m-1)/(-1))((n-m-2)/(-2))...(-m+1)/(-n+1) = > (-1)^(m+n+1) binomial(m-1,n-1). > (I think). > And, of course, binomial(-n,-n) = 1. > Hopefully, these are right. I multiplied by, or did not multiply by, -1 by accident somewhere above, apparently. (Do not have time to check where.) For, if m > n, m and n = positive integers, we get: binomial(-n,-m) = binomial(-n,m-n) (because binomial(j,k) = binomial(j,j-k)) = binomial(m-1,m-n)(-1)^(m+n) (because (m-n) is nonnegative integer and because binomial(-j,k) = (-1)^k binomial(j+k-1,k) ). And binomial(m-1,m-n)(-1)^(m+n) = binomial(m-1,n-1)(-1)^(m+n), the negative of the answer I got above. So, this works for all nonnegative integers m and n, however m and n compare to each other: binomial(-n,-m) = binomial(m-1,n-1)(-1)^(m+n). (So, as I originally suspected, binomial(-n,-m) = 0 IF n > m >= 1.) Right this time? Leroy Quet === Subject: Re: binomials over negative reals ... > I multiplied by, or did not multiply by, -1 by accident somewhere > above, apparently. > (Do not have time to check where.) no problem. > For, if m > n, m and n = positive integers, > we get: > binomial(-n,-m) = > binomial(-n,m-n) (because binomial(j,k) = binomial(j,j-k)) > = binomial(m-1,m-n)(-1)^(m+n) (because (m-n) is nonnegative integer > and because binomial(-j,k) = (-1)^k binomial(j+k-1,k) ). > And binomial(m-1,m-n)(-1)^(m+n) = binomial(m-1,n-1)(-1)^(m+n), > the negative of the answer I got above. > So, this works for all nonnegative integers m and n, however m and n > compare to each other: > binomial(-n,-m) = binomial(m-1,n-1)(-1)^(m+n). > (So, as I originally suspected, binomial(-n,-m) = 0 IF n > m >= 1.) > Right this time? Looks good. I am still curious: 1) why is this not the standard version (the standard version is bin(-n,-m) = 0, n,m pos ints (historically what choice was made since some choice you made above was different than the choice that leads to the standard version). That is, are any binomial identities broken by this? 2) Mathematica uses the standard version for negative ints. it seems (experimentally) to use the Gamma(n)/(Gamma(n-m)Gamma(m)) for negative reals. (I can't tell if yours is consistent with that or not). 3) is there an analytic formula either a hypergeometric, or an integral at the head for bin(n,m) that works for negatives. Mitch === Subject: Black-Scholes, standard normal I am working with a database that use the Black-Scholes formula, and within the function there is something that looks like the standard normal but I haven't figured out. If anyone could help me understand it, I would mighty appreciate it. The B-S is called with: Fnuction BS_CallPutOpt(OptionType,s,x,rf_rate,sigma,t,div) d_1=d_one(s,x,rf_rate,sigma,t,div) d_2=d_1-sigma*t^0.5 If OptionType=Call Then BS_CallPutOpt=s*Exp(-div*t)*SNorm(d_1) BS_CallPutOpt=BS_CallPutOpt-x*Exp(-rf_rate*t)*SNorm(d_2) Elseif ... ... ... Function d_one(s,x,rf_rate,sigma,t,div) d_one=Log(s/x)+(rf_rate-div+(sigma^2)/2)*t d_one=d_one/(sigma*t^0.5) End Function What I am having a problem understanding: Function SNorm(z) c1=2.506628 c2=0.3193815 c3=-0.3565638 c4=1.7814779 c5=-1.821256 c6=1.3302744 If z>0 or z=0 Then w=1 Else: w=-1 End If y=1/(1+0.231649*w*z) SNorm=0.5+w*(0.5-Exp(-z*z/2)/c1)*(y*(c2+y*(c3+y*(c4+y*(c5+y* c6)))))) End Function 2.506628 is sqrt(2*pi) But I am not sure what the other constants are, and to be honest, it's doesn't look much like the formula for the standard normal to me (where's the integral?) If anyone can tell me what SNorm does, I would be grateful. === Subject: Re: Black-Scholes, standard normal Yes, it should be the cumulative standard normal and SNorm is an approximation (i have not looked whether it is correct in what range). Note that rf_rate is for continous rates (ie: for 1 you get exp(rf_rate) after 1 year - while quotations often have other rules). And: working with dividends yields gives false values and general not a good idea. === Subject: Re: Byron and Fuller >> Anyway, now I'm very curious what their discussion of the >differences between closed and complete amounts to, >> because completeness and closedness (using their >> definition) are equivalent. Does their discussion of the >> differences end up by saying there are no differences? >Here is their discussion - I hope I haven't made any typos! >Definition 5.6 Let g(x) be any function in Hilbert space and let >{f_i(x)} be an orthonormal set of functions in Hilbert space. If there >exist constant coefficients {a_i} such that the sequence of partial sums >g_n(x) ~ SUM (i=1 to n) a_i f_i(x) converges in the mean to g(x), then >the set of functions {f_i} is a complete orthonormal set. Equivalently, >if the mean square error can be made arbitrarily small then the set >{f_i} is a complete orthonormal set of functions. >Definition 5.7. A set of orthonormal functions is said to be closed if >no nonzero function is orthogonal to every function in the set. >Theorem 5.2. A set of orthonormal functions in Hilbert space is complete >if and only if it is closed. ??? A minute ago you said that no, they didn't say there was no difference between closed and complete. But this theorem says exactly that there is no difference! ************************ === Subject: Re: Byron and Fuller Injector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061543452 X-URL: http://mygate.mailgate.org/mynews/sci/sci.physics/ caf61f5e52a78d994dced5b7cb2 b57ce.35661%40mygate.mailgate.org > ??? A minute ago you said that no, they didn't say there was no > difference between closed and complete. But this theorem says > exactly that there is no difference! Okay, thanks. Sounds like I'm getting confused. I'll ponder this over the weekend. But a couple of questions in the meantime. Having read their discussion of the difference between complete and closed, are you now happy with their definition of closed, and are you happy with their discussion generally? David. -- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG === Subject: Re: Byron and Fuller >> ??? A minute ago you said that no, they didn't say there was no >> difference between closed and complete. But this theorem says >> exactly that there is no difference! >Okay, thanks. Sounds like I'm getting confused. I'll ponder this over >the weekend. But a couple of questions in the meantime. Having read >their discussion of the difference between complete and closed, Are you just typing these posts because you have nothing better to do or what? There _is_ no discussion of the difference between closed and complete above. That's because there is no difference. This has already been made very clear. >are you >now happy with their definition of closed, And I've already addressed this quite clearly. Have you read any of the posts in this thread? > and are you happy with their >discussion generally? And I don't quite understand what you mean by this question. (But don't bother to explain - I'm not happy with _our_ discussion, because you don't even seem to be reading what people are writing.) >David. ************************ === Subject: Re: Byron and Fuller >??? A minute ago you said that no, they didn't say there was no >difference between closed and complete. But this theorem says >exactly that there is no difference! >>Okay, thanks. Sounds like I'm getting confused. I'll ponder this over >>the weekend. But a couple of questions in the meantime. Having read >>their discussion of the difference between complete and closed, >Are you just typing these posts because you have nothing >better to do or what? There _is_ no discussion of the difference >between closed and complete above. That's because there >is no difference. This has already been made very clear. >>are you >>now happy with their definition of closed, >And I've already addressed this quite clearly. Have you >read any of the posts in this thread? >> and are you happy with their >>discussion generally? >And I don't quite understand what you mean by this question. >(But don't bother to explain - I'm not happy with _our_ discussion, >because you don't even seem to be reading what people are >writing.) Perhaps the guy perused the posts rather than read them. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Byron and Fuller Injector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061543452 X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 16055123d6547b9b802215fce5f5f0 95.35661%40mygate.mailgate.org > Are you just typing these posts because you have nothing > better to do or what? There _is_ no discussion of the difference > between closed and complete above. Okay, I get the picture, but I don't think you explained it very well - the two definitions amount to the same thing - they are both definitions of complete and they should never have confused the issue by using the term closed when they should have used complete. So what we have, in fact, is two ways of describing the same property. Sorry if I've been a bit slow with this - my area of experise is organic chemistry - come and talk to me about organic chemistry and it will be you who is shown to be the fool :-) David. -- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG === Subject: Re: Byron and Fuller >> Are you just typing these posts because you have nothing >> better to do or what? There _is_ no discussion of the difference >> between closed and complete above. >Okay, I get the picture, but I don't think you explained it very well - >the two definitions amount to the same thing - they are both definitions >of complete and they should never have confused the issue by using the >term closed when they should have used complete. So what we have, in >fact, is two ways of describing the same property. The reason I never explained the point you make in that last sentence is that it's wrong. It's not true that they should never use two different words for the same property - if they weren't allowed to do that then they couldn't say closed is equivalent to complete. >Sorry if I've been a bit slow with this - my area of experise is organic >chemistry - come and talk to me about organic chemistry and it will be >you who is shown to be the fool :-) I didn't call you a fool. If we were talking about chemistry I might ask what H2O was. Someone might explain that it was water. _After_ that was explained I wouldn't _repeatedly_ ask questions about the difference between H2O and water. If I were studying chemistry I might not know at first that H2O and water were the same thing. But I wouldn't ask what the difference was, _quote_ the book as saying H2O is water and then ask _again_ what the difference was. Or: After I quoted the book saying H2O is water, and _two_ people pointed out that the book was saying there is no difference between H2O and water, I wouldn't reply asking someone whether he was happy with the book's discussion of the difference between H2O and water. I mean really: your not knowing some bit of math is no problem, that's the point to sci.math. And slow is no problem. But when it _really_ seems like you're simply not even _glancing_ at the replies you get then it gets a little irritating when you repeat the question - why should someone bother to answer it a fourth time, given that you didn't read the first three replies? >David. ************************ === Subject: Re: Byron and Fuller Injector-Info: news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1061543452 X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 50f8c9525040b9abf046feb6d05a80 d0.35661%40mygate.mailgate.org > I mean really: your not knowing some bit of math is no > problem, that's the point to sci.math. And slow is no > problem. But when it _really_ seems like you're simply > not even _glancing_ at the replies you get then it gets > a little irritating when you repeat the question - why > should someone bother to answer it a fourth time, > given that you didn't read the first three replies? Well, it's simply not true that I didn't read the posts - but it _is_ true that I didn't properly comprehend them (my apologies). My main problem was that I didn't appreciate that the conclusion of the discussion in Byron and Fuller was that the two definitions were equivalent as it was couched in heavy mathematical language. And yes, I know you pointed this out! As for the chemistry bit, sorry, you've lost me - I don't know anything about water or H2O - that's inorganic rather than organic chemistry (I clearly stated _organic_ chemistry in my post). David. -- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG Author-Supplied-Address: bds ipp mpg de === Subject: Re: Byron and Fuller Mail-To-News-Contact: abuse@dizum.com | |> |>> ??? A minute ago you said that no, they didn't say there was no |>> difference between closed and complete. But this theorem says |>> exactly that there is no difference! |> |>Okay, thanks. Sounds like I'm getting confused. I'll ponder this over |>the weekend. But a couple of questions in the meantime. Having read |>their discussion of the difference between complete and closed, | |> Are you just typing these posts because you have nothing |> better to do or what? There _is_ no discussion of the difference |> between closed and complete above. That's because there |> is no difference. This has already been made very clear. No need to get snappy... I also think the authors deliberately took a weaker definition of the concepts closed and complete (which is how I understand them) and then went on to show that they are in fact equivalent. Not that they start out this way, just that with orthogonal functions they can be _shown_ to _end_ this way. -- cu, Bruce drift wave turbulence: http://www.rzg.mpg.de/~bds/ === Subject: Re: Calculus is irrational? Hamish Reid > Anyway, back here, you then explictly redefined human rationality to > mean something it doesn't currently mean (and your definition is much > more restrictive than the standard one), and once again seemed surprised > when no one else thought that that was a good idea. I admit that I made a mistake in trying to use mathematical defintions when I originally started this thread. I had expected resistence to the lack of math in my post and that was the way I countered it (ineffectivly, I agree). But the above is not a redefintion. It's a clarification. I suspect once it has been understood that it will lead to a great deal of reßection on discussions we had thought were all but wrapped up. I'm also willing to accept the possibility that I'm delusional... but who of us isn't? :-) Mike Helland === Subject: Re: Calculus is irrational? > Acid Pooh > You're reasoning about imaginary unicorns right there. Since > according to your definition of rationality, you cannot do this, and > you just did, your definition is obviously wrong. > How does my defintion of rationality say that I cannot reason about > unicorns? > My defintion of rationality says: reasoning about something you've > observed. > It doesn't say that it's impossible to reason about something you > haven't observed. It just says that reasoning about something you > haven't observed isn't rational. It can still be reasonable. Here's a pretty good definition of rationality: imagination bounded > by logic. If you can imagine something, you can reason about it. > I don't disagree with the second sentence. But how about imagination > bounded by logic and observation as rationality? Why resistence to > that suggestion? Why resist the suggestion that one may be rational about non-physical (and hence, inobservable) objects? Clearly, you understand that we can reason about non-physical objects. I'll make a leap here and suggest that you further understand that mathematics uses logic to form a (semi-) consistent whole. (That semi- might be provocative, but it's irrelevant to the issue at hand. It is only included for accuracy.) To quote William Burroughs, Words mean what you want them to mean. Though true, if you wish to have intelligent discussions with others, you had better use the conventional definition. So, if by Calculus is irrational, you mean that Calculus uses the methods of logic to study non-physical objects, fine. I doubt anyone would disagree with your intended meaning. But irrational has many strong connotations--and what it denotes is not what you think it does. Think of it this way--how much more productive could your time here have been if you had just said, Calculus uses infinity, a non-physical non-object, to form its solutions. People would have immediately told you, Calculus does not require infinity to be used, and would explain why. You could have learned from this. Instead, you chose to argue about a definition. Ôcid Ôooh === Subject: Re: Calculus is irrational? > Gregory L. Hansen >Yes, thats exactly my point. A solution can only be as rational as the >premises assumed to find that solution. > If the solution can be expressed as a ratio of integers, then the solution > is rational. Otherwise it's not. > Sorry, you're still missing my point. What you said should have been: If the _result of a_ solution can be expressed as a ratio of > integers, then the _result of a_ solution is rational. Otherwise it's > not. > However that says nothing about the solution itself or its > rationality. For example, we both know of solutions to problems that > use division by zero that could lead to a result that can be expressed > rationally. However, despite the result being a rational number, we > both agree that the solution itself is not rational as it requires > irrationality to come to its result. Agreed? So if I have... (2 sqrt 2) / (sqrt 2), the solution itself is somehow irrational, because I had irrational numbers in the original problem, though the result of the solution is rational? === Subject: Re: Calculus is irrational? Acid Pooh > Why resist the suggestion that one may be rational about non-physical > (and hence, inobservable) objects? Clearly, you understand that we > can reason about non-physical objects. I resist because I think there is a subtle but important difference in being rational and being reasonable. That's why I clarified my defintion. > I'll make a leap here and > suggest that you further understand that mathematics uses logic to > form a (semi-) consistent whole. (That semi- might be provocative, > but it's irrelevant to the issue at hand. It is only included for > accuracy.) Yes, here's the thing semi-consistent. The problem isn't that its semi-consistent, the problem is when this system is used to discuss something like Time. Time is the boundry of what we observe, so I think it's fair to say that Time is the boundy or our rationality. Along comes Peter Lynds mathematician wishes, he can prove Lynds wrong by showing how calculus can be used to determine what Lynds says does not exist. The issue is that calculus isn't talking about the real world, where Lynds is. You say to be accurate we need to call Calc semi-consistent. Well of course a semi-consistent notion will contradict a consistent notion. I don't see why we just don't use the word irrational to describe things that lie outside the real world. If we're clear on what is rational and what is not, it allows us to sort through sticky === Subjects like Time much easier. > But irrational has many > strong connotations--and what it denotes is not what you think it > does. Depending on the degree of irrationality. Calculus starts with some irrationality (a ball that bounces an infinite number of times) and then follows a rational process to solve. So it's not completely irrational. THere is quite a bit of rationality to it, but it's not completely rational either. Mike Helland === Subject: Re: Calculus is irrational? > Acid Pooh > You're reasoning about imaginary unicorns right there. Since > according to your definition of rationality, you cannot do this, and > you just did, your definition is obviously wrong. > How does my defintion of rationality say that I cannot reason about > unicorns? > My defintion of rationality says: reasoning about something you've > observed. > It doesn't say that it's impossible to reason about something you > haven't observed. It just says that reasoning about something you > haven't observed isn't rational. It can still be reasonable. Here's a pretty good definition of rationality: imagination bounded > by logic. If you can imagine something, you can reason about it. > I don't disagree with the second sentence. But how about imagination > bounded by logic and observation as rationality? Why resistence to > that suggestion? > Why resist the suggestion that one may be rational about non-physical > (and hence, inobservable) objects? Clearly, you understand that we > can reason about non-physical objects. I'll make a leap here and > suggest that you further understand that mathematics uses logic to > form a (semi-) consistent whole. (That semi- might be provocative, > but it's irrelevant to the issue at hand. It is only included for > accuracy.) > To quote William Burroughs, Words mean what you want them to mean. > Though true, if you wish to have intelligent discussions with others, > you had better use the conventional definition. > So, if by Calculus is irrational, you mean that Calculus uses the > methods of logic to study non-physical objects, fine. I doubt anyone > would disagree with your intended meaning. But irrational has many > strong connotations--and what it denotes is not what you think it > does. > Think of it this way--how much more productive could your time here > have been if you had just said, Calculus uses infinity, a > non-physical non-object, to form its solutions. People would have > immediately told you, Calculus does not require infinity to be used, > and would explain why. You could have learned from this. Instead, > you chose to argue about a definition. Nothing uses infinity if calculus doesn't. You're obviously talking about numerical infinity. Which calculus and group theory heads don't know exist, which is why mathematicans are obviously are always given low-bugdet intergral jobs. And we save the logic for people with brains. > Ôcid Ôooh === Subject: Changing symbols, explaining problem I've seen consternation with what I see as a simple proof of a problem with the ring of algebraic integers. However, it occurs to me that many of you are used to the convention of capital letters for constants, and may prefer x as a variable, so I'll make some changes. A problem with the ring of algebraic integers can be seen by focusing on the expression P(x), in the ring of algebraic integers, where P(x) = A^2(( A^4 x^3 - 3 A^2 x^2 + 3x) B^3 - 3(-1+A^2 x )B C^2 + C^3 A) = (B y_1 + CA)(B y_2 + CA)(B y_3 + CA) as one of the y's must be coprime to A, when A is coprime to 3. Can't you see it? Come on, can't you understand *basic* algebra??!!! It can be seen by considering the constant term P(0), as you have P(0)/A^2 = C^2(3B + CA). So the expression is special in that while it's a polynomial with x as the variable, you can factor it with respect to *constants*, which is easy. Can't you see it? Now with a polynomial P(x) given that the constant term is coprime to A, and given that two of the y's equal 0, when x equals 0, it makes sense they're related to x, for instance, possibly each has sqrt(x) as a factor? Do you know? Can't you tell by looking? However, I have faced challenges to that position as it shows a problem with the ring of algebraic integers, as for values like x=1, A=sqrt{5}, it is possible to prove that the y's cannot have a factor that is sqrt{5} in the ring of algebraic integers. Faced with the problem in the ring, some may wish to hold on to the belief that it is fine by believing that A^2 divides off in varying ways dependent on the value of x. But you can't see it? Can you? You're all lost and instead of looking at the math going to the words. So desperate to reach for the familiar in a world of PURE mathematics, which is possibly too pure, eh? I've given the argument to you in a way that you CAN understand, but possibly you wish values? Then let A=sqrt(5), C=1, so you have P(x) = 5(( 25 x^3 - 3(5) x^2 + 3x) B^3 - 3(-1+5 x )B + sqrt(5)) = (B y_1 + sqrt(5))(B y_2 + sqrt(5))(B y_3 + sqrt(5)) and is that better for you? Ok then, let x=1, B=z, as then you have P(1) = 65z^3 - 60z + 5sqrt(5) and is THAT easier? Why don't you let w=sqrt(5), as then you have P(1) = 65z^3 -12w^2z + w^3 and is that finally familiar? It amazes me how few of you understand your basic limitations and how your brains are limited by symbols, when supposedly you know algebra. But you see, I tested what you *really* knew about algebra, versus what you've learned through parroting in your *belief* that you understood. Most of you don't understand basic mathematics. You simply have rote processes, in your preferred ways of doing things, and you can't even follow the expressions in this post, until the end, though they were given in a way you may have *thought* you preferred. I need at least some of you to show some intellectual maturity. I think you know who you are. The rest are walking dead, and soon even the illusion that they're alive will be taken. There is not much time left. Death has been incarnated and now She walks the earth. And She's hungry for human souls. So some of you *must* quickly learn how to be more than human. Some of you will learn. The earth is going through an extinction and the Universe has a sense of humor. There's no telling how creatively She has rolled the dice this time, and I begin already to hear Her laughter. James Harris === Subject: Re: Changing symbols, explaining problem .. I begin already to hear Her laughter. That laughter you hear ... do you think they're laughing with you, or at you? Gib === Subject: Re: Changing symbols, explaining problem >I've seen consternation with what I see as a simple proof of a problem >with the ring of algebraic integers. [Rest of post deliberately cut] This many-times repeated assertion of Mr Harris that there is a Ôproblem with the ring of algebraic integers' irritates me, because I do not know what it means. Without going into details of the maths, what is Mr Harris claiming? Is it perhaps: 1. There is something inconsistent or ambiguous about the definition of an algebraic integer ? 2. The definition is OK, but the set of algebraic integers do not form a ring ? 3. The set of algebraic integers as defined do form a ring, but there is a valid mathematical argument concerning this ring which leads to a contradiction ? I request enlightenment. Derek Holt. === Subject: Re: Changing symbols, explaining problem >I've seen consternation with what I see as a simple proof of a problem >with the ring of algebraic integers. How would you know? You made it abundantly clear recently that you have no idea what the word ring means. [...] >Can't you see it? Come on, can't you understand *basic* algebra??!!! You're insisting that Z[pi] is a field, by arguments that actually imply that the field of rationals is equal to the field of reals, and you ask _us_ whether we understand basic algebra? It is to laugh. >[...] >Most of you don't understand basic mathematics. Right. Just like most of us are so stupid we insist that decidable means convergent and in support of that assertion we give links to definitions that show exactly the opposite. >You simply have rote >processes, in your preferred ways of doing things, and you can't even >follow the expressions in this post, until the end, though they were >given in a way you may have *thought* you preferred. >I need at least some of you to show some intellectual maturity. I >think you know who you are. The rest are walking dead, and soon even >the illusion that they're alive will be taken. >There is not much time left. Death has been incarnated and now She >walks the earth. And She's hungry for human souls. Look. If you don't like it when people call you crazy, which evidently you don't, then don't rave this way. >So some of you *must* quickly learn how to be more than human. Some >of you will learn. The earth is going through an extinction and the >Universe has a sense of humor. There's no telling how creatively She >has rolled the dice this time, and I begin already to hear Her >laughter. Also please avoid such easy setups. >James Harris ************************ === Subject: Re: Changing symbols, explaining problem >>I've seen consternation with what I see as a simple proof of a problem >>with the ring of algebraic integers. > [Rest of post deliberately cut] > This many-times repeated assertion of Mr Harris that there is a > Ôproblem with the ring of algebraic integers' irritates me, because > I do not know what it means. Without going into details of the maths, > what is Mr Harris claiming? Is it perhaps: > 1. There is something inconsistent or ambiguous about the definition of > an algebraic integer ? > 2. The definition is OK, but the set of algebraic integers do not form a ring ? > 3. The set of algebraic integers as defined do form a ring, but there is > a valid mathematical argument concerning this ring which leads to a > contradiction ? > I request enlightenment. > Derek Holt. As I understand it, his problem is that he believes he has found algebraic integers a,b,c with product 65, where GCF(a,5), GCF(b,5), GCF(c,5) are units. The a,b,c he is talking about are based on factoring his polynomial. The problem with his proof is: explicit calculation shows that none of the GCFs are units. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Changing symbols, explaining problem > I've seen consternation with what I see as a simple proof of a problem > with the ring of algebraic integers. However, it occurs to me that > many of you are used to the convention of capital letters for > constants, and may prefer x as a variable, so I'll make some changes. You still fail to realize that your method is ßawed. Your derivations are NOT derivations, and your conclusions are in error. Instead of paying more attention at the evidence, the clear computations that show that you are concluding something that is false, you redouble your efforts at trying to get people to understand this very ßawed technique that you seem to care about. How's that for application of the scientific method? Run an experiment, and when the results are shown to be at variance with what your ideas, throw that data out and rerun the experiment. After all, you have a degree in PHYSICS, and that makes you invincible. Those of us with degrees and decades of experience in mathematics can't be expected to know how to do our jobs, but you, with (all) your BS, can somehow see so much more clearly than everyone else. Why is it that your credentials, meager as they are, are more valid than our credentials? Why is it that your untold thousands of errors that you have defended, crying from the mountaintops, proclaiming should not be taken into consideration? Why should your method, which is the fount of all this (non-)wisdom, not be held accountable for such folly? I say that until you can produce a coherent argument that is accessible to others, your failures *should* be your legacy, and your credibility *should* be based on a total absence of correct result. That is, make the argument accessible to others. Show that it works for a specific example. *THEN* ask others to believe it. As of now, your argument is inaccessible. I'll show you: > A problem with the ring of algebraic integers can be seen by focusing > on the expression P(x), in the ring of algebraic integers, where > P(x) = > A^2(( A^4 x^3 - 3 A^2 x^2 + 3x) B^3 - 3(-1+A^2 x )B C^2 + C^3 A) > = > (B y_1 + CA)(B y_2 + CA)(B y_3 + CA) > as one of the y's must be coprime to A, when A is coprime to 3. No explanation of this set of formulas. You say P(x) is in the ring of algebraic integers, but I believe [correct me if I'm wrong] x is supposed to be a variable, as are the y's. Algebraic integers are *NOT* variables, they are constants. Further, no variable shares any non-unit factors with any constant (except for the constant zero). > Can't you see it? Come on, can't you understand *basic* algebra??!!! This does not constitute a proof. I know, you think people are being obtuse. Fine, think whatever you like, but there is no proof there. > It can be seen by considering the constant term P(0), as you have > P(0)/A^2 = C^2(3B + CA). > So the expression is special in that while it's a polynomial with x as > the variable, you can factor it with respect to *constants*, which is > easy. There's no x there. Did you neglect to put x in there? Or, are you claiming that when you put x = 0, you get the above expression? Why are you saying it's a polynomial in x? > Can't you see it? > Now with a polynomial P(x) given that the constant term is coprime to > A, and given that two of the y's equal 0, when x equals 0, it makes > sense they're related to x, for instance, possibly each has sqrt(x) as > a factor? A polynomial in x doesn't have any y's, does it? What is y supposed to be? > Do you know? Can't you tell by looking? The claim that something follows by looking can work if people all agree that it follows by looking. If anyone claims it *doesn't* follow by looking, the person promoting an argument has the obligation to make the argument explicit. So far, you have not done that. > However, I have faced challenges to that position as it shows a > problem with the ring of algebraic integers, as for values like x=1, > A=sqrt{5}, it is possible to prove that the y's cannot have a factor > that is sqrt{5} in the ring of algebraic integers. Are you saying that the y's cannot be divisible by sqrt(5), or are you saying that the y's are coprime to sqrt(5), or are you saying that the y's cannot share non-unit factors with sqrt(5) (all in the ring of algebraic integers)? > Faced with the problem in the ring, some may wish to hold on to the > belief that it is fine by believing that A^2 divides off in varying > ways dependent on the value of x. you *assume* a problem in the ring of algebraic integers. That has not been proven. > But you can't see it? Can you? You're all lost and instead of > looking at the math going to the words. So desperate to reach for the > familiar in a world of PURE mathematics, which is possibly too pure, > eh? Again, the technique of proof by ßatulence is not accepted. The idea is to present the argument, not to accuse people of incompetence or cowardice in their unwillingness to buy this kettle of fish. > I've given the argument to you in a way that you CAN understand, but > possibly you wish values? Then let A=sqrt(5), C=1, so you have > P(x) = 5(( 25 x^3 - 3(5) x^2 + 3x) B^3 - 3(-1+5 x )B + sqrt(5)) = > (B y_1 + sqrt(5))(B y_2 + sqrt(5))(B y_3 + sqrt(5)) > and is that better for you? Three points. 1. It is a person's prerogative to understand what he can. 2. It is unspeakably rude to give an argument as full of crap as this one, and then to blame the reader for failing to accept it. 3. What is the point here? I mean, is that better for you? is hardly a mathematical argument, is it? > Ok then, let x=1, B=z, as then you have > P(1) = 65z^3 - 60z + 5sqrt(5) > and is THAT easier? Is WHAT easier? What are you hoping to conclude? Are you concluding anything at all? > Why don't you let w=sqrt(5), as then you have > P(1) = 65z^3 -12w^2z + w^3 > and is that finally familiar? Of course, this polynomial is the one you have claimed factors like this P = (a1 z + w)(a2 z + w)(a3 z + w) where the a's are coprime to 5. I have shown you common factors between each of the a's and 5 (algebraic integer factors, that is), have shown that those factors CANNOT be units, and you couldn't accept it. Here's what it came down to: This from Google: > === Subject: Re: How I know, linchpin of my FLT proof > Original Format >> ... stuff deleted ... Synopsis of what I've presented in the cited thread: Given any root a of the equation p(x) = x^3 - 12 x + 65 the following factorizations hold: q(a)r(a) = 5 r(a)s(a) = a where the polynomials q,r,s are given as q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 Further, the minimal polynomial of r(a) is this: x^3 - 969 x^2 + 315 x + 5 showing that r(a) is an algebraic integer and not a unit. Now, we'll continue with my protestation to JSH: >> What? Are you saying that the factorization I've posted and provided >> all the necessary calculation to verify that, is somehow inadequate? > Please explain. Show me the error. > That is, show how one of these is not correct: > 1. 5 = q(a)*r(a) >> 2. a = r(a)*s(a) >> 3. r(a) is an algebraic integer >> 4. r(a) is NOT a unit. > You have not given proof that r(a) cannot be coprime to 5. And there you have it. I have r(a) a *DIVISOR* of 5, not a unit, yet JSH claims that I have not shown r(a) cannot be coprime to 5. coprime to 5 implies r(a) is a unit, JSH complains: > Since a = rs, and 5 = qr, this equation can be written: > (rs) u + (qr) v = 1 > factoring, we have: > r (su + qv) = 1. > Thus if w = su + qv, we have an algebraic integer w, >> for which > r w = 1. > Therefore, r must be a unit in the ring of algebraic >> integers. > You went in a circle. That is, I make the following argument: Given: r divides 5. r is not a unit If r is coprime to 5, then r must be a unit. Conclusion r cannot be coprime to 5. JSH's response: You went in a circle. > It amazes me how few of you understand your basic limitations and how > your brains are limited by symbols, when supposedly you know algebra. You go off hiding after waving the bogus Dale's a racist ßag, and now return with the same old garbage. It amazes me how very little you understand your basic limitations, and how your brain is limited by your hatred of the readers of sci.math. > But you see, I tested what you *really* knew about algebra, versus > what you've learned through parroting in your *belief* that you > understood. Can you show me how I've parroted *anything*? Can you find the work I've produced here, *anywhere else*? In *any* textbook? I'm sure the material is covered there (in many textbooks, in fact), in more than enough detail, but to be successful in such a canard, you must be prepared to illustrate an act of plagiarism. Put up or shut up. > Most of you don't understand basic mathematics. You simply have rote > processes, in your preferred ways of doing things, and you can't even > follow the expressions in this post, until the end, though they were > given in a way you may have *thought* you preferred. Yeah, we're all such babies. We learn *by rote* how to do Galois theory, what ideals are, why the rationals are NOT the reals, why Z[1/3] is simply the smallest ring that contains both Z and a multiplicative inverse for 3, and how and why it's isomorphic to Z[x]/<3x-1>. Further, it's *by rote* that we understand how Z[q] is isomorphic to Z[x] whenever q is transcendental over Q. > I need at least some of you to show some intellectual maturity. I > think you know who you are. The rest are walking dead, and soon even > the illusion that they're alive will be taken. Oh, intellectual maturity, is it? JSH always like to put the fear of the Devil into his writing, so here he goes on his Dat Ole Debbil Gonna Gitcha tirade: > There is not much time left. Death has been incarnated and now She > walks the earth. And She's hungry for human souls. > So some of you *must* quickly learn how to be more than human. Some > of you will learn. The earth is going through an extinction and the > Universe has a sense of humor. There's no telling how creatively She > has rolled the dice this time, and I begin already to hear Her > laughter. Oh, I'm sorry, that was me laughing. I couldn't help myself. > James Harris Dale === Subject: Circulant matrices over GF(2) What is the number of n X n invertible circulant matrices over GF(2) ? === Subject: Re: Data analysis software > It plots and analyses any x-y data for peak location, peak height, peak > width, semi-derivative, derivative, integral, semi-integral, convolution, > deconvolution, curve fitting, and separating overlapped peaks and > background. > www.chemSoftware.com === Subject: Dumbasses, complaints, etc. If you have thoughts regarding how anti virus/anti spam software should behave when confronted with virus spam emails, I don't see a problem with you crossposting your divined from above knowledge to news.admin.net-abuse.email and some other groups. But, if you want to keep your rants secret, that's ok too. === Subject: Re: Dumbasses, complaints, etc. > If you have thoughts regarding how anti virus/anti spam software should > behave when confronted with virus spam emails, I don't see a problem with > you crossposting your divined from above knowledge to > news.admin.net-abuse.email and some other groups. This will be my only comment in this thread seen in the crossposted NGs. I am currently the victim; apparently Sobig.F is being distributed with from all over. I am reasonably sure my linux system is not virused, or in any way distributing this material. My thought regarding how anti-virus/anti-spam software should behave when confronted with virus spam e-mails is that it shouldn't try replying to what it thinks is the source unless it has some magical way of determining beyond a shadow of a doubt who the source actually is. Currently the poorly-thought-out notifications are helping a net abuser abuse my inbox. I am not the only victim. I'm sure I speak on behalf of all the other victims when I say that this isn't right. -- All relevant people are pertinent. All rude people are impertinent. Therefore, no rude people are relevant. -- Solomon W. Golomb === Subject: Re: Dumbasses, complaints, etc. X-Posting-Server: RumorMill 1.3.4 > If you have thoughts regarding how anti virus/anti spam software should > behave when confronted with virus spam emails, I don't see a problem with > you crossposting your divined from above knowledge to > news.admin.net-abuse.email and some other groups. It's really very simple: drop the mail completely and log the event. In some cases (such as NAV) the antiviral product is actually the external SMTP gateway and so can choose to shun the sending IP address. Many sites would want to do just that, and it seems to me that with Sobig (which sends from infected machines, which are by definition under the control of buffoons and the worms they've welcomed) it would be a very good idea to stop accepting any mail at all from the source IP and to perhaps report it in some way to be blocked in a broader fashion. (I guess I can't recommend that an antiviral use the sooper-sekrit CPU-exploding trick as a response, eh?) -- Now where did I hide that website... === Subject: Empirical Proof of NP=P I was wondering if current state of the art SAT solvers provide an empirical proof of NP=P. It can solve a 100 variable SAT problem in a few seconds. My program counts the number of clauses examined. This is the inner most loop of the program and I think this is a valid measure of the number of operations performed to solve a problem. competition. The average number of operations to solve each problem was around 3,000,000 clauses examined. The worst case was 7,523,876 clauses examined. This is approximately 100^3.4. There are existing state of the art solvers are much better than mine. Has anyone found a solvable problem that can't be solved in polynomial time by these programs? Russell - 2 many 2 count === Subject: Re: Empirical Proof of NP=P : Has anyone found a solvable problem that can't be solved : in polynomial time by these programs? These solvers can solve anyi given problem in constant time. The constant might be quite large, though. Or did you have a particular polynomial in mind? -Greg === Subject: Re: Empirical Proof of NP=P 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > I was wondering if current state of the art SAT solvers > provide an empirical proof of NP=P. I am under the impression that current state of the art SAT solvers take an amount of time that is exponential (but a very small exponential) when applied to random instances near the satisfiable-unsatisfiable threshhold. You can not test this empirically by looking only at problems of a certain fixed size, large though it may be (and 100 variables is well within the range these algorithms can solve) -- you need to plot average runtime against problem size for different problem sizes and fit a curve to it. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Empirical Proof of NP=P Interesting: Within 24 hours, newsgroup sci.math receives two posts, one with a prospective method to prove P=NP and the other to prove the opposite! === Subject: Empirical Proof of NP=P === Subject: one-way function (VMPC) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Empirical Proof of NP=P Distribution: world > competition. The average number of operations to > solve each problem was around 3,000,000 clauses examined. > The worst case was 7,523,876 clauses examined. > This is approximately 100^3.4. So how long does it take with 200 variables? 300? Running something with 100 variables might be interesting in some ways, but doesn't tell you much of anything about the complexity of your solution. I'm betting that you wouldn't be able to handle 200 variables.... And you're still talking about a pretty small number of variables. But if you get it working with more variables, I have some very interesting things you could do with it (very large formulas, but still relatively few variables -- say 2000-4000 variables). -- Steve Tate - srt[At]cs.unt.edu | A computer lets you make more mistakes faster Dept. of Computer Sciences | than any invention in human history with the University of North Texas | possible exceptions of handguns and tequila. Denton, TX 76201 | -- Mitch Ratliffe, April 1992 === Subject: Re: Empirical Proof of NP=P | > I was wondering if current state of the art SAT solvers | > provide an empirical proof of NP=P. | You can not test this empirically by looking only at problems of a | certain fixed size, large though it may be (and 100 variables is | well within the range these algorithms can solve) -- you need to | plot average runtime against problem size for different problem | sizes and fit a curve to it. Even that's wouldn't be sufficient. An algorithm that runs in O(n^5 + 2^{n/10000}) steps would exhibit polynomial behavior for any realistic value of n. On the other hand, an algorithm that ran in O(n^{100}) time would be effectively indistinguishable from exponential. (Add zeros as necessary.) The most you can hope for is an empirical proof that it doesn't matter whether or not P=NP. -- Jeff Erickson jeffe@cs.uiuc.edu Computer Science Department http://www.uiuc.edu/~jeffe University of Illinois at Urbana-Champaign === Subject: Re: Empirical Proof of NP=P How about getting 100 places for the value of f(1/2) where exp(f(x))=f(x+1) with f(0)=0? It starts f(1/2)=.49856328794... It would require 270 equations in 270 unknowns with partition(270) terms in some equations!! === Subject: Re: Empirical Proof of NP=P > competition. The average number of operations to > solve each problem was around 3,000,000 clauses examined. > The worst case was 7,523,876 clauses examined. > This is approximately 100^3.4. > So how long does it take with 200 variables? 300? I solved ten 50 variable problems. The average number of clauses examined was about 63,500. This is around 50^2.8. 300 and 500 variables. > Running something with 100 variables might be interesting in some > ways, but doesn't tell you much of anything about the complexity of > your solution. I'm betting that you wouldn't be able to handle 200 > variables.... 200^4 would be 1,600,000,000 operations. This would take a long time on my computer. > And you're still talking about a pretty small number of variables. > But if you get it working with more variables, I have some very > interesting things you could do with it (very large formulas, but > still relatively few variables -- say 2000-4000 variables). I would be interested in your problems. Special problems with millions of variables have been solved. Russell - 2 many 2 count === Subject: Exp-Generating-Function for x(exp(x(1+e^x))) I am guessing that.... If a(m) = sum{k=1 to m} binomial(m,k) k^(m+1-k), then we have a(m) is the m_th derivative of x*exp(x(1+e^x)) at x = 0, ie. sum{k=1 to oo} a(k) x^k /k! = x(exp(x(1+e^x))). This sequence {a(k)} begins: 1, 4, 18, 92, ... Does this sequence have a simpler representation? Now a(m)/m is simply sequence A080108 in the EIS. But I am unsure if my particular sequence is in the EIS. By the way, x(exp(x(1+e^x))) = f(f(x)) if f(x) = x*exp(x), ie. sum{k=1 to oo} a(k) (W(W(x))^k /k! = x. (W(x) is the Lambert function, of course.) Is there an easy way to get a(n,m), where sum{k=1 to oo} a(n,k) x^k /k! = f(f(...f(x)..)) (with n f's) if f(x) = x exp(x)? Leroy Quet === Subject: Re: Face facts, many Americans *like* idea of killing > Ha. Yeah, they did sort of tinker with the details a bit. Guess > that's how it has to be when you only have 128 pages or so to work > with. > Ed Howdershelt - Abintra Press > Science Fiction and Semi-Fiction >> A little off topic from this particular thread, but when did >> Semi-Fiction officially become a recognized genre? Is it restricted >> to science specifically or can the mix include a combination of any >> or all genre's? Just curious. > Genre? Recognized? By definition, perhaps. > Don't know if a ÔSemi-Fiction' genre exists. > Don't really care, either. :) > I use the term ÔSemi-Fiction' to describe those of my works in which > only names of people and certain places have been changed to avoid > lawsuits. See titles Kim, Mindy, Anne, and Field Decision. > Ed Howdershelt - Abintra Press > Science Fiction and Semi-Fiction > http://abintrapress.tripod.com > of it? It seems that changing the names of people and places isn't always > enough. I was going to jokingly mention a couple of semi-celebs in my book > but now I'm not sure if it's worth the risk. > http://www.publishlawyer.com/carousel4.htm > 2poor The segment about privacy issues would pertain to my stuff. I've cleared Kim, Mindy, and Anne with the women involved and accepted many of their suggested revisions, and none of them have called their lawyers. Field Decision's key female character died in 2001, but she had no objections to what I sent her for review. Guess I'll wait and see if anyone sues. If nothing else, there may be a book in it. Ed Howdershelt - Abintra Press Science Fiction and Semi-Fiction http://abintrapress.tripod.com === Subject: FLT and the Barcelona conjecture The Barcelona conjecture: Let c=(x+y+z)^p/(pxyz2^p) for integer c,x,y,z and p prime greater than or equal to 5, the Barcelona conjecture is that no solutions exist with gcd(c,xyz)=1 (no c exist that shares no factor with x or y or z). While this is a very interesting problem in itself (the heart of it is the relation between the factors of a sum versus the product of the addends - ie (x+y+z)^p/(xyz)), and may at first seem to be easy to resolve, it is in fact intimately related to Fermat's Last Theorem. In fact, proving the Barcelona conjecture also proves FLT (no integer ABC that satisfy A^p+B^p=C^p for p prime greater than 2) for prime exponents greater than 3. [note that letting c=1 for p=3 also solves this exponent for FLT] There are integer solutions to the above, but all found to date have a factor in common for c and xyz. === Subject: Re: FLT and the Barcelona conjecture >In fact, proving the Barcelona conjecture also proves FLT Now, to me, that sounds like some fun. Unfortunately I am working with p=3 right now, and not on FLT. Yours, Doug Goncz, Replikon Research, Seven Corners, VA The hormones work at different speeds: In a fight-or-ßight scenario, glucocorticoids are the ones drawing up blueprints for new aircraft carriers; epinephrine is the one handing out guns. === Subject: fraction function Has any considered the neat definition for a fraction function of f(x)=a+b*x? It would be defined as f^u(x)=a*((1-b^u)/(1-b))+(b^u)*x unless a=b=1 in which case f^u(x)=u+x. We would have f^(0)(x)=x, f^(1)(x)=a+b*x, f^(-1)=-a+(1/b)*x the inverse function of x, f^(2)(x)=f(f(x))=a*(1+b)+b^2*x, f^(1/2)(x)=a/(1+sqrt(b))+sqrt(b)*x, and finally the inverse fuction of f^u(x) would be just f^(-u)(x). === Subject: Re: fraction function > Has any considered the neat definition for a fraction function of f(x)=a+b*x? > It would be defined as f^u(x)=a*((1-b^u)/(1-b))+(b^u)*x unless a=b=1 in which > case f^u(x)=u+x. We would have f^(0)(x)=x, f^(1)(x)=a+b*x, f^(-1)=-a+(1/b)*x > the inverse function of x, f^(2)(x)=f(f(x))=a*(1+b)+b^2*x, > f^(1/2)(x)=a/(1+sqrt(b))+sqrt(b)*x, and finally the inverse fuction of f^u(x) > would be just f^(-u)(x). More generally, you might use (ax+b)/(cx+d), so that you are talking about 2x2 matrices, of which you want fractional powers. === Subject: Re: Fraud in Computer Science Publishing > 1. Do you consider Predicate Calculus wffs to be programs? Some Predicate Calculus wffs are programs. > 2. How do you define a program? A program is an machine-readable specification of acceptable (computer) behavior. > 3. Do you know of a better way to sepcify the largest proper factor > of a given number? No. > 4. Do you know of a way to specify it that isn't programming? No. > 5. Do you know of a simpler way to specify it? No. > 6. How do you think that a Mathematician would specify it? No comment. > 7. Did you know that the state-of-the-art in Program Synthesis is to > specify the program requirement as a Predicate Calculus wff? Lambda Calculus is also used. > 8. Do you see a qualitative difference between a Predicate Calculus > wff and a computer program? No. > 9. How about the fact that a Predicate Calculus wff has no assignment, > conditional execution, loops or the possibility of not terminating? Those are all properties of the implementation. Nevertheless, the intention behind all of those ideas can be captured in predicate calculus, as Hehner shows in _A Practical Theory of Programming_ (no doubt others have demonstrated this as well.). As for the possibility of not terminating, consider a boolean function of two arguments that is true just when its first argument corresponds to a Turing machine that halts when given the second argument as input. > 10. Do you think that Predicate Calculus wffs and computer programs > are at the same level of abstraction? Predicate Calculus's heavy use of quantifiers makes it more abstract at the language level than many of today's computer programming languages. > 11. Do you think that Predicate Calculus wffs and computer programs > are in a one-to-many relationship in that one wff can be implemented > by multiple programs based on different algorithms, but for one > program there is essentially only one wff that represents the > functionality that it provides (ignoring permutations of the > conjuncts/disjuncts and other logical redundancy)? No. They are in a many-to-many relationship. Just there is no algorithm to decide whether two computer programs are equivalent, there is also no algorithm to detect whether two predicate calculus formulas are equivalent. > 12. Do you think that computer programs have to be analyzed to > determine what function they compute, and in general you cannot do > that? Yes. > 13. Do you think that there is a corresponding process of analysis to > detemine what a Predicate Calculus wff is doing, or is it the final > word as to the definition that it is conveying? Yes, there is a corresponding process of analysis for P. C. wffs. > 14. Do you think that a simple wff may require complex programs to > implement it (defining simple and complex informally or > intuitively)? Yes. > 15. Do you see value in being able to determine the wff that a > particular program computes? Yes. > 16. Do you see value in being able to determine computer programs that > implement a given predicate calculus wff? Yes. > 17. Does my system determmine programs that compute a given wff? Not if those programs are required to halt for all inputs. > 18. Do you know of any system other than mine that determines programs > that compute a given wff? False premise. === Subject: Re: Functional analysis question > Hi all, > This must be a triviality to some of you, but not (yet) to me > unfortunately. Here's the question: > Suppose C[0,1] is our space, the space of continuous functions on the > closed interval [0,1]. We have two norms on that space: the integral > norm, N_1(f)=int_0^1 |f(x)|dx and the, well, let's call it moderated > sup-norm, N_2(f)=||xf||_oo, i.e. sup{|x*f(x)|:x in [0,1]}. Now if we > take f_n(x)=x^n we see that this is a sequence of functions converging > to zero in the integral norm, but not in the moderated sup-norm, since > N_2(f_n)=1 for all n, while N_1(f_n)=(n+1)^{-1}->0 if n->oo. This has > led me to the strong belief that if a sequence goes to zero in the > N_2-norm, that it also goes to zero in the N_1-norm. Can you confirm my > intuition or am I way off here? If I'm correct, can you give a hint (so > that I can still try to complete it myself) how to prove this? > Micha Here's a (big) hint: If f_n ---> 0 in the sup norm then for any e > 0 there is an N such that if n > N then |f_n(x)| < e for all x in [0,1]. What is the L_1 norm of such f_n ? === Subject: Re: Functional analysis question >>Hi all, >>This must be a triviality to some of you, but not (yet) to me >>unfortunately. Here's the question: >>Suppose C[0,1] is our space, the space of continuous functions on the >>closed interval [0,1]. We have two norms on that space: the integral >>norm, N_1(f)=int_0^1 |f(x)|dx and the, well, let's call it moderated >>sup-norm, N_2(f)=||xf||_oo, i.e. sup{|x*f(x)|:x in [0,1]}. Now if we >>take f_n(x)=x^n we see that this is a sequence of functions converging >>to zero in the integral norm, but not in the moderated sup-norm, since >>N_2(f_n)=1 for all n, while N_1(f_n)=(n+1)^{-1}->0 if n->oo. This has >>led me to the strong belief that if a sequence goes to zero in the >>N_2-norm, that it also goes to zero in the N_1-norm. > I can't imagine why you think this is evidence of that... Well, it was not only that that gave me the idea. It was also the relation between the integral norm and the (ordinary) sup norm. Since the integral norm (where the integration is over [0,1]) is bounded by the sup norm and that thus for every sequence converging in sup norm it is true that it is also convergent in the integral norm, I thought this would be the case. However, your help and of the others has been illuminating in the sense that I see I was way off with my idea. Cheers, Micha === Subject: Re: Functional analysis question > Hi Micha, >>Suppose C[0,1] is our space, the space of continuous functions on the >>closed interval [0,1]. We have two norms on that space: the integral >>norm, N_1(f)=int_0^1 |f(x)|dx and the, well, let's call it moderated >>sup-norm, N_2(f)=||xf||_oo, i.e. sup{|x*f(x)|:x in [0,1]}. Now if we >>take f_n(x)=x^n we see that this is a sequence of functions converging >>to zero in the integral norm, but not in the moderated sup-norm, since >>N_2(f_n)=1 for all n, while N_1(f_n)=(n+1)^{-1}->0 if n->oo. This has >>led me to the strong belief that if a sequence goes to zero in the >>N_2-norm, that it also goes to zero in the N_1-norm. Can you confirm >>my intuition or am I way off here? > You are off ! > Take, for each integer n>=1, > f_n(x)= 1/(xLog(Log(n))), if 1/n<=x<=1, > and f_n lineal between (0,0) and (1/n, f_n(1/n)). > Then N_2(f_n)-> 0 and N_1(f_n)->+ infty > So these 2 norms give non comparable metrics on C[0,1]. > Is this O.K. for you? Very O.K. :-) This is a very nice couterexample of my belief. I see I'm Cheers, Micha === Subject: Re: Functional analysis question > Let f_n(x) = b_n/a_n for 0 <= x < a_n, b_n/x for a_n <= x <= 1, for some > choice of sequences {a_n}, {b_n} of positive numbers (with a_n in [0,1]). > This is in C[0,1] and N_2(f_n) = b_n, while N_1(f_n) = b_n(1 - ln a_n). > So any choice such that b_n -> 0 while b_n(1 - ln a_n) does not will give > a counterexample to your claim, e.g. b_n = 1/n, a_n = e^(1-n). Cheers, Micha === Subject: Re: Functional analysis question >>Hi all, >>This must be a triviality to some of you, but not (yet) to me >>unfortunately. Here's the question: >>Suppose C[0,1] is our space, the space of continuous functions on the >>closed interval [0,1]. We have two norms on that space: the integral >>norm, N_1(f)=int_0^1 |f(x)|dx and the, well, let's call it moderated >>sup-norm, N_2(f)=||xf||_oo, i.e. sup{|x*f(x)|:x in [0,1]}. Now if we >>take f_n(x)=x^n we see that this is a sequence of functions converging >>to zero in the integral norm, but not in the moderated sup-norm, since >>N_2(f_n)=1 for all n, while N_1(f_n)=(n+1)^{-1}->0 if n->oo. This has >>led me to the strong belief that if a sequence goes to zero in the >>N_2-norm, that it also goes to zero in the N_1-norm. Can you confirm my >>intuition or am I way off here? If I'm correct, can you give a hint (so >>that I can still try to complete it myself) how to prove this? >>Micha > Here's a (big) hint: If f_n ---> 0 in the sup norm then for any e > 0 > there is an N such that if n > N then |f_n(x)| < e for all x in [0,1]. > What is the L_1 norm of such f_n ? Gowan, Did you notice I wasn't talking about the regular sup norm but ||xf||_oo. f multiplied by the identity function x |--> x. Otherwise you're of course right. Cheers, Micha === Subject: Re: Functional analysis question Note that if ||xf||_oo is replaced with ||sqrt(x)f||_oo, then your original guess is correct. === Subject: Re: GRE Math === Subject Test Well, I have been working out of a book titled GRE Math === Subject test by The Princeton review. I have heard that the problems in the book are a little tougher than what is on the actual test, but it is suppossed to be good preparation. I think for text books you should just refer to the ones from which you learned. No sense in wasting your time getting used to a new author's style. You can go to the gre.org site and they will explain to you the rescaling. At least they had info on it a while ago. Lurch > I want to do MS/MA in Math. Are there some standard text books that I > can refer to? Besides, what is rescaled in the test? > kjr === Subject: HELP! DCT is not good enough. How to design my own transform basis matrix for image compression? Dear all, I want to design a my own transform basis matrix. For 2D case, forward transform is: Y=A*X*B, inverse transform is: Z=C*Y*D... let's say X, Y, Z, A, B, C, D are 8x8 matrices... Can anybody tell me what relationship should A, B, C, D have? 1)Orthognality? 2)A=B'? 3)C=D'? 4)A=B^(-1)? 5)C=D^(-1)? 6)B=C? 7)A=D? ... Thank you very much if you can also point me to some resources for reference? -Walala === Subject: Re: Help with complex exponentiation Hi all! >Please help me with this (not homework): >n^(p+qi) = a + bi >a = f(n,p,q)= ? >b = g(n,p,q)= ? >N. >n is positive? If so, then: >n^(p+qi) = exp((p+qi) log n) = exp(p log n) exp(i q log n) > = exp(p log n) cos(q log n) + i exp(p log n) sin(q log n) > which simplifies, a little: a = n^p cos(q log n) b = n^p sin(q log n) -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Help with complex exponentiation Ôexp','log',[Capi talOTilde]sin' and Ôcos'? i.e.: a = sqrt(p^2+q^2)/2pq + .... N. A N Niel escribi.97 en el mensaje > Hi all! > Please help me with this (not homework): > n^(p+qi) = a + bi > a = f(n,p,q)= ? > b = g(n,p,q)= ? > N. > n is positive? If so, then: > n^(p+qi) = exp((p+qi) log n) = exp(p log n) exp(i q log n) > = exp(p log n) cos(q log n) + i exp(p log n) sin(q log n) === Subject: Re: Help with complex exponentiation Content-transfer-encoding: 8bit > Ôexp','log',[Capit alOTilde]sin' and Ôcos'? 2^i = cos(log 2) + i sin(log 2), so I would think not. === Subject: Re: Help with complex exponentiation > Hi all! > Please help me with this (not homework): > n^(p+qi) = a + bi > Much easier if n is in polar form. > Let N = the magnitude of n, |n|, and let > theta = the phase of n. That is, > n = N exp(i*theta). > a = f(n,p,q)= ? > b = g(n,p,q)= ? > n^(p+qi) = [N*exp(i*theta)]^(p+qi) > = [exp(log N) exp(i*theta)]^(p+qi) > = exp(log N)^p exp(logN)^(qi) exp(i*theta)^p exp(i*theta)^(qi) > = N^p exp(logN * qi) exp(i*p*theta) exp(-theta*q) > = [N^p* exp(-theta*q)] * exp[i*(logN*q + theta*p)] I assumed n was a general complex number. If I work with n being positive real as others have, then N = n and theta = 0. Then this reduces to n^(p+qi) = n^p *exp(i*q log n) = n^p * cos (q log n) + i * n^p * sin(q log n) which agrees with the solution provided by others. - Randy === Subject: Help with exercise in Dunford & Schwartz Linear Operators charset=iso-8859-1 I am self-studying Dunford & Schwartz Linear Operators and have been having trouble with one of the exercises. If you have the book, it is VII.10.3. In case you don't, here is the statement: (I'll write infty for infinity, and R for (-infty,infty)) In the spaces L p(R), 1 <= p <= infty, let T be the operator (Tx)(t) = x'(t) with domain D(T) = {x | x is absolutely continuous on each finite interval, x' is in L p(R)}. Show that (a) T is a closed unbounded linear operator whose domain is dense for p < infty, and (b) the spectrum of T is the imaginary axis, and R(l; T)(x, t) = int 0^infty e^{-ls} x(t+s) ds when Real(l) > 0, and =-int {-infty}^0 e^{-ls) x(t+s) ds for Real(l) < 0. I have no trouble doing part a. I also have no trouble showing that the expressions given for the resolvent, R(l; T) are correct. I also have no trouble showing that the spectrum is the imaginary axis in the case p = infty. The trouble I have is in showing that the spectrum is the imaginary axis for 1 <= p < infty. I can show that 0 is in the spectrum. by showing that the characteristic function of the interval [0,1] (which is in L p(R)) cannot be mapped onto by -T. It is clear to me where some of the difficulties come from (namely that e^{irt} is not in L p(R)) and how the solutions for points off the imaginary axis fail to work for points on the imaginary axis. It would seem there are three ways that (irI-T) could fail to have an inverse. First it could fail to be one-to-one, although I think that it actually is one-to-one. Second it could fail to be onto all of L p(R). Finally, it could be onto all of L p(R), but the inverse function could fail to be bounded (as an operator). I have not been able to make any progress on showing which is true. Any suggestions? === Subject: Re: Help with exercise in Dunford & Schwartz Linear Operators >I am self-studying Dunford & Schwartz Linear Operators and >have been having trouble with one of the exercises. If you >have the book, it is VII.10.3. In case you don't, here is the >statement: (I'll write infty for infinity, and R for (-infty,infty)) > In the spaces L_p(R), 1 <= p <= infty, let T > be the operator (Tx)(t) = x'(t) with domain D(T) = {x | x is > absolutely continuous on each finite interval, x' is in L_p(R)}. > Show that (a) T is a closed unbounded linear operator whose domain > is dense for p < infty, and (b) the spectrum of T is the imaginary > axis, and > R(l; T)(x, t) = int_0^infty e^{-ls} x(t+s) ds when Real(l) > 0, and > =-int_{-infty}^0 e^{-ls) x(t+s) ds for Real(l) < 0. >I have no trouble doing part a. I also have no trouble showing that >the expressions given for the resolvent, R(l; T) are correct. I also >have no trouble showing that the spectrum is the imaginary axis in the >case p = infty. The trouble I have is in showing that the spectrum >is the imaginary axis for 1 <= p < infty. >I can show that 0 is in the spectrum. by showing that the characteristic >function of the interval [0,1] (which is in L_p(R)) cannot be mapped onto >by -T. It is clear to me where some of the difficulties come from (namely >that e^{irt} is not in L_p(R)) and how the solutions for points off the >imaginary axis fail to work for points on the imaginary axis. >It would seem there are three ways that (irI-T) could fail to have an >inverse. First it could fail to be one-to-one, although I think that >it actually is one-to-one. Second it could fail to be onto all of L_p(R). >Finally, it could be onto all of L_p(R), but the inverse function could >fail to be bounded (as an operator). I have not been able to make >any progress on showing which is true. Well, first I should say that never having studied unbounded operators I'm really not quite sure what the definition of the spectrum is. But taking your word for the fact that the above conditions suffice, it's easy to see that any point of the imaginary axis is in the spectrum. Note that I'm going to talk about the Fourier transform, assuming that Pi = 2Pi = 1; no point in trying to get the formulas actually right in a context like the present because where the Pi's go varies from book to book... For example, given r, choose f infinitely differentiable with compact support such that f^(r) <> 0. Then f cannot be in the range of irI - T. Suppose to the contrary that g is an element of D and g' - ir g = f. If we knew that g was, say, in L^1 and actually absolutely continuous then we could conclude that (g' - ir g)^(r) = 0, giving a contradiction. We don't know that g is that nice but this argument convinces me that g' - ir g cannot equal f regardless... Ah. g' - ir g = f shows that (exp(-irt) g(t))' = exp(-irt) f(t). But the integral of (exp(-irt) g(t))' from -infinity to infinity must be zero, since exp(-irt) g(t) is in D, while the integral of exp(-irt) f(t) is nonzero. >Any suggestions? ************************ === Subject: Re: Help with exercise in Dunford & Schwartz Linear Operators charset=iso-8859-1 > For example, given r, choose f infinitely differentiable with compact > support such that f^(r) <> 0. Then f cannot be in the range of > irI - T. Suppose to the contrary that g is an element of D and > g' - ir g = f. If we knew that g was, say, in L^1 and actually > absolutely continuous then we could conclude that > (g' - ir g)^(r) = 0, giving a contradiction. We don't know > that g is that nice but this argument convinces me that > g' - ir g cannot equal f regardless... > Ah. g' - ir g = f shows that > (exp(-irt) g(t))' = exp(-irt) f(t). > But the integral of (exp(-irt) g(t))' from -infinity to infinity > must be zero, since exp(-irt) g(t) is in D, while the integral > of exp(-irt) f(t) is nonzero. So then I just need to remember that g(t) must go to zero at plus and minus infinity to conclude that the first integral you mention above vanishes, right? Since you explained that one to me several years ago, I guess I'm Jeff Rubin === Subject: How to calculate ÔVideo Plus' codes. Hi Does anyone out there know how to calculate the ÔVideo Plus' codes that appear beside each programme on the TV listing pages in the newspapers? An example for one that appears today is: and the code is 652627 Johno === Subject: Re: How to calculate ÔVideo Plus' codes. >Does anyone out there know how to calculate the ÔVideo Plus' codes >that appear beside each programme on the TV listing pages in the >newspapers? A Google search for video-plus+ algorithm turns up lots of pages. I have a vague recollection of a posting stating that the 8-digit version was now known, but I can't remember any more about it. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: how to make a (fired) bullet stay put > .... The only horizontal forces acting on the bullet are the initial > velocity of the bullet and the velocity of the aircraft as the bullet leaves > the barrel of the gun. Forces are not velocities. They are not even proportional to one another. -- G.C. === Subject: Re: how to make a (fired) bullet stay put NNTP-Proxy-Relay: library2.airnews.net charset=iso-8859-1 > How fast would an aircraft have to ßy to make a > bullet fall straight down if shot from the rear of the aircraft? Say > the bullet travels at 400mph. Then set your autopilot for 400mph, and the bullet will be stationary with respect to the earth as the bullet and the barrel part company. Kindof reminds me of a magician whisking the tablecloth out from under a formal dinner setting... === Subject: Re: how to make a (fired) bullet stay put > There seem to be quite a few riße shells that provide > 3300-3800 fps (2250-2591 mph) muzzle velocity (see e.g. > http://www.chuckhawks.com/25mag.htm ). An 87 grain bullet out of a 270 case should certainly smoke a 22-250... > The 400 mph bullet would be about 587 fps and I think that > even a 45 acp would beat that. > I think you are saying it would be difficult to find a slow- > enough bullet to get it to drop straight down when fired > out the back of a plane ßying 400 mph. However, a pellet > gun like the Crossman 2200B (one third of the way through > http://www.gungarage.com/crossmanr.htm) would be close, > at 595 feet per second or 405.7 miles per hour. Note that 22 pellet guns are up to 1000 fps and that 17 caliber pellet guns are up to 1200 fps... > Regarding the 45 acp, you probably mean .45, with 820 fps. > I think 45's are more typically 1800-2525 fps, like the one > I saw on a recent trip (http://pat7.com/jp/45caliber.jpg). A 45 caliber is .45 inch diameter (but defined in some convenient increment)...and so Ô45' by itself does identify the cartridge except within context. ACP is Ôautomatic caliber pistol' so the word Ôcaliber' or the letter Ôc' is precisely correct usage with Ô45'. WWII military loads of 45 acp were less than 820 fps... === Subject: How to transform spherical/polar coordinates to Cartesian for n-dimensions? I am looking for the method to transform spherical coordinates of arbitrarily high dimensions (n) to Cartesian coordinates (of the same number of dimensions of course). I have found how to go from polar (2d) to Cartesian, and spherical (3d) to Cartesian, but I have not yet found the n-dimension solution. It is my intent to implement this in Java for a program I am writing. Any help or links would be greatly appreciated. Craig craiggarrett@yahoo.com === Subject: Re: How to transform spherical/polar coordinates to Cartesian for n-dimensions? in message <2rh1b.31872$0u4.12349@news1.central.cox.net>: > I am looking for the method to transform spherical coordinates of > arbitrarily high dimensions (n) to Cartesian coordinates (of the same > number > of dimensions of course). I have found how to go from polar (2d) to > Cartesian, and spherical (3d) to Cartesian, but I have not yet found the > n-dimension solution. It is my intent to implement this in Java for a > program I am writing. Any help or links would be greatly appreciated. The usual scheme, modulo renumbering of indices, is: x_1 = r sin(a_1) x_2 = r cos(a_1) sin(a_2) x_3 = r cos(a_1) cos(a_2) sin(a_3) ... x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) Of course, r >= 0, -pi/2 <= a_i <= pi/2 except -pi < a_{n-1} <= pi -- Jim Heckman === Subject: Re: How to transform spherical/polar coordinates to Cartesian for n-dimensions? Was that last line supposed to be: x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) or x_n = r cos(a_1) cos(a_2) cos(a_3) ... sin(a_{n-1}) or x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) sin(a_n) ? It looked like in the x_1 through x_3 that the last factor was always going to be sin of a_n... Just checking... Craig > in message <2rh1b.31872$0u4.12349@news1.central.cox.net>: > I am looking for the method to transform spherical coordinates of > arbitrarily high dimensions (n) to Cartesian coordinates (of the same > number > of dimensions of course). I have found how to go from polar (2d) to > Cartesian, and spherical (3d) to Cartesian, but I have not yet found the > n-dimension solution. It is my intent to implement this in Java for a > program I am writing. Any help or links would be greatly appreciated. > The usual scheme, modulo renumbering of indices, is: > x_1 = r sin(a_1) > x_2 = r cos(a_1) sin(a_2) > x_3 = r cos(a_1) cos(a_2) sin(a_3) > ... > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) > Of course, r >= 0, -pi/2 <= a_i <= pi/2 except -pi < a_{n-1} <= pi > -- > Jim Heckman === Subject: impulse response and green's function This is probably a naive question. I have little background on diff equations. How do you prove that a system characterised by a linear differential equation is representable by what is known as impulse response ? ( proof should stay within time domain. dont go to freq domain. ) The impulse response is the green's function of the system. right ? Is the stability of the system deducible from the diff equation itself ? How ? thanks shankar === Subject: Re: impulse response and green's function > This is probably a naive question. I have little background on diff > equations. > How do you prove that a system characterised by a linear differential > equation is representable by what is known as impulse response ? > ( proof should stay within time domain. dont go to freq domain. ) I'm going to let this alone because I'm rusty in the theory of distributions. > The impulse response is the green's function of the system. right ? > Is the stability of the system deducible from the diff equation itself ? How ? For a linear differential equation with constant coefficients, the answer is yes. You only need to look at the real part of the roots of the characteristic equation. This is pretty easy to see from the solution to the homogeneous equation. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === Subject: Re: impulse response and green's function > How do you prove that a system characterised by a linear differential > equation is representable by what is known as impulse response ? Let Lu=f, where L=linear differential operator. (1) Let Lh=delta, where h=impulse response. (2) u=f*h, where * represents convolution. (3) See if you recover (1). === Subject: Re: impulse response and green's function >This is probably a naive question. I have little background on diff >equations. >How do you prove that a system characterised by a linear differential >equation is representable by what is known as impulse response ? Using the so-called convolution integral, one can determine the output of the system given only the input and the impulse response function. >( proof should stay within time domain. dont go to freq domain. ) >The impulse response is the green's function of the system. right ? >Is the stability of the system deducible from the diff equation itself ? How ? of the system, one can either compute the roots and examine the real parts, or apply the Routh Criterion. > thanks > shankar -- John E. Prussing University of Illinois at Urbana-Champaign Department of Aerospace Engineering http://www.uiuc.edu/~prussing === Subject: Re: Is there a simple way to see if a matrix can be factorized ieee std > Is there a simple way to see if a matrix can be factorized? > Example, A is a matrix > A = |23, 34 | > |77, 46 | > Is there a simple way to know if A can be factorized as product > of two matrix A= B*C, where each element of B and C is positive integer ? > If we're talking only of square matrices, det(A) will have to be equal to > det(B)det(C). Better, the characteristic polynomial of A will be the product > of the c.p.'s of B and C. > But you might want to clarify just what is meant by factorization. E.g. if > B is invertible (e.g. a 0-1 matrix representing a permutation) then we can > always factor A as A=BDB^{-1} simply by setting > D=B^{-1}AB. > LH element of B, or DB^(-1) are positive integer. === Subject: Re: Is there someone to help me PLEASE... > 1-The first qustion is to find the largest set > of natural numbers that if we choose two members a and b then ab+1 is square > of a natural number.(is there any limit for the number of members of the > set?) > 4 or 5. See http://www.math.hr/~duje/dtuples.html - summary is, there > are lots of ways of getting 4, it's impossible to get 6, no one has > ever gotten 5, and if there are any 5s there are only finitely many. > The reference given is A. Dujella, There are only finitely many > Diophantine quintuples, J. Reine Angew. Math., to appear. I vaguely remembered a very similar problem that was solved by Baker and Davenport, and running a Google search on Baker and Davenport threw up the link Gerry gave, and also http://at.yorku.ca/cgi-bin/amca/cakl-71 and http://arxiv.org/PS_cache/math/pdf/0304/0304242.pdf among others. A Dujella certainly has been busy on these problems over the last few years! Cheers ------------------------------------------------------------- -------------- John R Ramsden (jr@adslate.com) ------------------------------------------------------------- -------------- I thank God I never read a book, nor never will. It was merrie in England afore all this new learning came among us. Yea, I would all things were as they hath been in times past. Duke of Norfolk, 1532 === Subject: Re: James Harris >> Or perhaps everything is a gigantic theatrical spam. >I've skimmed a few of the posts and recently noticed he is not getting any benefit of doubt. >He makes several mathematical statements, people rebuke the maths, then >James corrects them and that's the end of the thread. Noone admits when >their counters to his proof were incorrect, and James has to repeatedly address >the same attacks. It won't go anywhere until the sequences of posts are as >systematic as the posts should be themselves. Try a new tactic sci.math isolate >where his proof if ~right~ first. It is considered impolite to reply to a post without including at least some of the original text. >http://tinyurl.com/ko46 heard of the hammer?, this is KO >Herc Larry (this space unintentially left blank ..... make obvious deletion for email === Subject: Re: James Harris Here's a summary of the JSH lamebrain's FLT postings to sci.math: ************1996**************** *Feb. 27th: first post, and it contains a mild hint that he already solved FLT. *Feb to April: strongly suggests he proved FLT. *April: realizes he's a bonehead and blowhard, apologizes for being an idiot, and unsubscribes from sci.math ------------------------------------------------------------- ----- ...obviously my attempt at a proof isn't even close. ...obviously, I was suffering from a bit of delusion.... ...the mild release making outrageous claims has given me is finally surmounted by a sense of shame; so I'll quit. ....My apologies to anyone concerned or in any way interested. ------------------------------------------------------------- ------ *June: rested for a month and a half, he's back *August: decides Pierre done beat him this time: My last post was an apology for annoying you with my attempts at proving FLT with simple algebra which is obviously an impossible task, and such attempts have been a source of irritation for mathematicians for hundreds of years. I was forcing myself off because I was scared that I was insane and hopelessly obsessed with a Quixotic approach, and was just making a complete fool of myself.... *September: posts 4 different proofs *October: on two consecutive days, posts a Correct Proof, a Corrected Proof, and a Completed Proof, each one including an apology for the previous post being incorrect. His errors included ...a brain fart. ....I reworked everything, including taking out some embarrassing errors or unimportant statements that had stayed in most of the previous posts. *November: one post *December: no posts ************end 1996****************** * 1997: 65 posts (approx 1.25 per week) * 1998: 517 posts (approx. 1.4 per day) One post includes: I shouldn't be suprised that every time I think I'm finished with FLT there seems to be a little bit to be added. * 1999: 1,030 posts (approx. 2.8 per day) * 2000: 393 posts (approx. 1 per day) * 2001: 1,140 posts (approx. 3 per day) * First 49 days of 2002: 411 posts, or an avg of 8.4 per day > The more I view the postings of/from James Harris, the replies and > counterreplies, I have a certain thought that this is more than a > competition amongst mathemathicans and amateurs. I think this is a survey! > No one can be as dumb as James. Everything is laid out to him, either as > definitions or as proofs. He can't refute them. Still he disagrees, > sometimes unpolitely, somtimes unscholardly. > I think that this is some survey that some of the mathematisions has > started, and it has gone wild. Can they stop it before I die in laughs! > Karl-Olav Nyberg === Subject: Re: James Harris |> I think that this is some survey that some of the mathematisions has |> started, and it has gone wild. Can they stop it before I die in laughs! | |I do think James harris is relatively stupid, but not _that_ stupid. |He certainly lacks quite a lot of _social_ intelligence Don't we all? ;-) Well, okay, I guess some of us don't. |since he |stubbornly repeats and repeats his random babble. However, I think |his _motive_ is not his self-proclaimed genious, but just to make a |lot of noise in the sci.math community. It really is about his need |for strokes or feedback. I guess he has a poor social life. | |Or perhaps everything is a gigantic theatrical spam. I don't understand why people feel the need for some of these explanations. I don't think lack of intelligence has much to do with any of it. There are plenty of people of average intelligence who find mathematics of the kind being discussed confusing. They just don't usually conclude that their own confusions are truths missed by the rest of the world. Certainly he seems to want attention. But do we have any real reason to think he's merely pretending to think so highly of himself just for the attention or for kicks? People getting the idea that they're misunderstood geniuses is a common phenomenon. You might think he occasionally lets slip some sign of not being confident of the solidity of his view of the world, and thus that he actually knows that his reasoning doesn't work. This may be. I suspect a lot of people who cling very tightly to beliefs have a certain underlying anxiety about the reliability of those beliefs. A lot of true believers can be seen vigorously attempting to shore up their belief systems, and seem often to be worried that the truth they've seen will lose out if not vigorously pushed. On the other hand I would say that this is a different story from intentionally maintaining a facade. They say human beings generally tend to be mildly anxious, and I think the insecurity of the true believer is only an aggravation of the ordinary human condition. Keith Ramsay === Subject: Re: James Harris > Here's a summary of the JSH lamebrain's FLT postings to sci.math: Well, for the bulk of my postings that isn't necessarily a bad assessment. I keep mentioning that my degree is in PHYSICS in the hopes that some of you might show a recognition of the scientific method. Other times, figuring some of you might hate science, I hope that at least you appreciate the benefits, like lights, so I mention Thomas Edison. > ************1996**************** > *Feb. 27th: first post, and it contains a mild hint that he already solved > FLT. > *Feb to April: strongly suggests he proved FLT. > *April: realizes he's a bonehead and blowhard, apologizes for being an > idiot, and unsubscribes from sci.math > ------------------------------------------------------------- ----- ...obviously my attempt at a proof isn't even close. ...obviously, I was suffering from a bit of delusion.... ...the mild release making outrageous claims has given me > is finally surmounted by a sense of shame; so I'll quit. ....My apologies to anyone concerned or in any way interested. > ------------------------------------------------------------- ------ > *June: rested for a month and a half, he's back > *August: decides Pierre done beat him this time: My last post was an > apology for annoying you with my attempts at proving FLT with simple algebra > which is obviously an impossible task, and such attempts have been a source > of irritation for mathematicians for hundreds of years. I was forcing > myself off because I was scared that I was insane and hopelessly obsessed > with a Quixotic approach, and was just making a complete fool of myself.... Of course, posters on sci.math weren't so nice as to *suggest* but would not only repeatedly call me insane, they'd talk about ways that I should disappear, or call me inhuman! Now why the math community would find posts about mathematics--albeit often ßawed--evidence of a *lack* of humanity is something on which I've wondered. > *September: posts 4 different proofs > *October: on two consecutive days, posts a Correct Proof, a Corrected > Proof, and a Completed Proof, each one including an apology for the > previous post being incorrect. His errors included ...a brain fart. ....I reworked everything, including taking out some embarrassing errors or > unimportant statements that had stayed in most of the previous posts. > *November: one post > *December: no posts > ************end 1996****************** > * 1997: 65 posts (approx 1.25 per week) > * 1998: 517 posts (approx. 1.4 per day) > One post includes: I shouldn't be suprised that every time I think I'm finished with FLT there > seems to be a little bit to be added. > * 1999: 1,030 posts (approx. 2.8 per day) > * 2000: 393 posts (approx. 1 per day) > * 2001: 1,140 posts (approx. 3 per day) > * First 49 days of 2002: 411 posts, or an avg of 8.4 per day Wow. I've been quite dedicated. Still I see posters confidently, and proudly making posts that apparently are bent on attacking me for following one of the greatest ideas in human history: the scientific method. Apparently to these people you see *mathematics* has nothing to do with thinking, so advanced problem solving techniques aren't necessary, and in fact are forbidden. But you see words like brain-storming, creativity, problem solving, and perseverance DO actually point to activities that society finds appealing. But then again, it's been well-known that mathematicians pride themselves on NOT being practical. They may feel that electic lights, and maybe electricity itself is a personal affront. James Harris === Subject: Re: James Harris >> Here's a summary of the JSH lamebrain's FLT postings to sci.math: >Well, for the bulk of my postings that isn't necessarily a bad >assessment. >[...] > *August: decides Pierre done beat him this time: My last post was an >> apology for annoying you with my attempts at proving FLT with simple algebra >> which is obviously an impossible task, and such attempts have been a source >> of irritation for mathematicians for hundreds of years. I was forcing >> myself off because I was scared that I was insane and hopelessly obsessed >> with a Quixotic approach, and was just making a complete fool of myself.... >Of course, posters on sci.math weren't so nice as to *suggest* but >would not only repeatedly call me insane, Just today, at the end of a post that was ostensibly about mathematics, you said this: There is not much time left. Death has been incarnated and now She walks the earth. And She's hungry for human souls. So some of you *must* quickly learn how to be more than human. Some of you will learn. The earth is going through an extinction and the Universe has a sense of humor. There's no telling how creatively She has rolled the dice this time, and I begin already to hear Her laughter. That does not sound like a sane person talking. It might sound sane in some other context, but as part of a post that's supposed to be explaining some bit of mathematics it sounds totally wacky. You make that sort of statement a _lot_. That's why people draw those unpleasant conclusions. >they'd talk about ways that >I should disappear, or call me inhuman! >Now why the math community would find posts about mathematics--albeit >often ßawed--evidence of a *lack* of humanity is something on which >I've wondered. That post I quote above was about mathematics? >> *September: posts 4 different proofs > *October: on two consecutive days, posts a Correct Proof, a Corrected >> Proof, and a Completed Proof, each one including an apology for the >> previous post being incorrect. His errors included ...a brain fart. ....I reworked everything, including taking out some embarrassing errors or >> unimportant statements that had stayed in most of the previous posts. > *November: one post > *December: no posts > ************end 1996****************** >> * 1997: 65 posts (approx 1.25 per week) > * 1998: 517 posts (approx. 1.4 per day) > One post includes: >I shouldn't be suprised that every time I think I'm finished with FLT there >> seems to be a little bit to be added. > * 1999: 1,030 posts (approx. 2.8 per day) > * 2000: 393 posts (approx. 1 per day) > * 2001: 1,140 posts (approx. 3 per day) > * First 49 days of 2002: 411 posts, or an avg of 8.4 per day >Wow. I've been quite dedicated. >Still I see posters confidently, and proudly making posts that >apparently are bent on attacking me for following one of the greatest >ideas in human history: the scientific method. Uh, no. There's nothing scientific about refusing to learn the basics of the field you're attempting to revolutionize, about ignoring your critics, or refusing to learn enough material to understand their objections. >Apparently to these people you see *mathematics* has nothing to do >with thinking, so advanced problem solving techniques aren't >necessary, and in fact are forbidden. >But you see words like brain-storming, creativity, problem >solving, and perseverance DO actually point to activities that >society finds appealing. >But then again, it's been well-known that mathematicians pride >themselves on NOT being practical. They may feel that electic lights, >and maybe electricity itself is a personal affront. >James Harris ************************ === Subject: Re: JSH: Testing failure, once again [...] > That's interesting. I looked at the NNTP-Posting-Host. In the > thread, it was: 67.192.37.140. > In an earlier discussion on JSH forgeries, I had made a list of > so it seemed). > One of these was: 67.192.37.160 , which is close to the one above. > It seems that my IP address > is 67.68.134.54 right now. > Let's see if it shows up > in the NNTP-Posting-Host > How often does JSH > include the JSH tag in > his === Subject lines? > David Bernier > I think that only tells us he's posting from Google. Regarding JSH, I think he sees himself as a mathematical genius and an amateur writer. So maybe he was in fiction writer-mode over the top. My IP address is now 67.68.137.19 David Bernier posting from Google === Subject: Re: leech lattice generator matrix > Amrani, O. et al, The Leech Lattice and the Golay Code: > Bounded-Distance Decoding and Multilevel Constructions, > IEEE Transactions on Information Theory, > vol. 40, No. 4, pp. 1030-1043, Jul. 1994. Sorry, I don't have easy access to this paper. > The leech lattice is defined as > a) It is either A-type or B-type > b) It consists either of only even columns or only odd columns > c) The overall k-parity is even if the array is A-type, and odd > otherwise > d) The overall h-parity is even if the array columns are even, and odd > otherwise > e) The projection of the array is a codeword of H6. > This defintion seems very similiar to the MOG defintion > on SPLAG p304, the defintion for Golay code is very similiar. Sounds like > p133 gives the generator matrix Tables of sphere packings and spherical codes > Sloane, N.; > Information Theory, IEEE Transactions on , Volume: 27 Issue: 3 , May > 1981 > Page(s): 327 -338 > It defines the lattice lattice as > a(0 +2c + 4x) > a(1 + 2c + 4y) > a = 1/sqrt(8) c is any codword in the ginary extended Golay code of > lenght 24, and > x in Z^24, y in Z^24 saitisfy > sum x_i = 0 mod(2) > sum y_i = 1 mod(2) Same as contrsuction in COnway and Sloane mentioned above. > In the paper it gives the generator matrix If you multiply this by 2 you get an integer matrix rather similar to the first snipped matrix. They aren't the same and don't generate the same lattice, but the two lattices differ only by permuting the coordinates. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: leech lattice generator matrix On paper > Tables of sphere packings and spherical codes > Sloane, N.; > Information Theory, IEEE Transactions on , Volume: 27 Issue: 3 , May > 1981 > Page(s): 327 -338 > It defines the lattice lattice as > a(0 +2c + 4x) > a(1 + 2c + 4y) > a = 1/sqrt(8) c is any codword in the ginary extended Golay code of > lenght 24, and > x in Z^24, y in Z^24 saitisfy > sum x_i = 0 mod(2) > sum y_i = 1 mod(2) > Same as contrsuction in COnway and Sloane mentioned above. If they are the same, why the generator matrix are different? > In the paper it gives the generator matrix > the first snipped matrix. They aren't the same and don't generate > the same lattice, but the two lattices differ only by permuting the > coordinates. I don't think so, it is not only permuting. If you run 2*G/N you will get 0.5 in the matrix. I don't have strong math background in this area, I am trying to apply leech lattice to engineering, thanks a lot for discussing with me. === Subject: Re: Leibniz and a mechanical binary calculator > Could you tell which paper/book claimed that a binary > --A church-school McCrusade (Blair's ideals?): > Harry-the-Mad-Potter want's US to kill Iraqis?... > http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) > http://www.rwgrayprojects.com/synergetics/plates/plates.html > http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX Binary system was invented in China more than two thousand years before Leibniz: the hexagram of I Ching was a binary system. martin === Subject: Linear Equation Could someone with more experience in math then I please show me how to solve this equation algebraic? (n^2 + n)/2 = 15 I realise this might be a very simple problem. But any help would be greatly appreciated! LepeL === Subject: Re: Linear Equation n^2+n=30,n^2+n-30=0,(n+6)(n-5)=0, n={-6,5} > Could someone with more experience in math then I please show me how to > solve this equation algebraic? > (n^2 + n)/2 = 15 > I realise this might be a very simple problem. But any help would be > greatly appreciated! > LepeL === Subject: Re: Linear Equation > Could someone with more experience in math then I please show me how to > solve this equation algebraic? > (n^2 + n)/2 = 15 1) Multiply both sides by 2: n^2 + n = 30 2) Subtract 30 from both sides: n^2 + n - 30 = 0 3) Recognize that the left side has this factorization: (n + 6)(n - 5) = 0 How would you recognize this factorization? The trick is to see that you're looking for two numbers whose product is -30, and whose sum is 1. If you didn't happen to recognize the factorization, you would instead use the quadratic formula. 4) Therefore the solutions are n = -6 and n = 5. 5) Check both solutions by plugging them back into the original equation: (5^2 + 5)/2 = 30/2 = 15 ok and ((-6)^2 - 6) / 2 = 30/2 = 15 ok === Subject: Re: Linear Equation > Could someone with more experience in math then I please show me how to > solve this equation algebraic? > (n^2 + n)/2 = 15 > I realise this might be a very simple problem. But any help would be > greatly appreciated! > LepeL Others have shown you how to solve it, but it is NOT a linear equation, it is a quadratic equation. === Subject: Re: Linear Equation >>Could someone with more experience in math then I please show me how to >>solve this equation algebraic? >>(n^2 + n)/2 = 15 > 1) Multiply both sides by 2: > n^2 + n = 30 > 2) Subtract 30 from both sides: > n^2 + n - 30 = 0 > 3) Recognize that the left side has this factorization: > (n + 6)(n - 5) = 0 > How would you recognize this factorization? The trick is to see that > you're looking for two numbers whose product is -30, and whose sum is 1. > If you didn't happen to recognize the factorization, you would instead > use the quadratic formula. > 4) Therefore the solutions are n = -6 and n = 5. > 5) Check both solutions by plugging them back into the original equation: > (5^2 + 5)/2 = 30/2 = 15 ok > and > ((-6)^2 - 6) / 2 = 30/2 = 15 ok Thank you very much for your complete and educational explanation! You really helped me out. Tnx! LepeL === Subject: Re: Linear Equation ANother method, called complete the square... Multiply both sides by 2: n^2 + n = 30 add the third term for a perfect square on the left*: n^2 + n + 1/4 = 30 + 1/4 so it is (n+1/2)^2 = 121/4 take square-root n+1/2 = 11/2 or n+1/2 = -11/2 subtract 1/2 from both sides n = 10/2 or n = -12/2 simplify n = 5 or n = -6 *[ (x+a)^2 = x^2 + 2ax + a^2, so in this case we must have a=1/2 to match the first two terms. ] === Subject: Re: line - surface intersection >Given line inside 3d surface, >the line given by >[x y z] = [x0 y0 z0] + [nx ny nz] * t >where [x0 y0 z0] , [nx ny nz] are known and what's the surface? >I would like to find the intersection points between this line and the 3d >surface >(in case the surface is convex there are supposed to be two such points) Maybe, maybe not... >note that the 3d surface is a bunch of points - 3 x n array No, a 3 x n array is one way to specify n points. A surface has infinitely many points. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: LOGIC ^ ATHEISM > This is a sci.logic post that may hold some interest here, > No. > -- Am, m E sci.math, ~interest(m, LOGIC^ATHEISM) ? Herc === Subject: Re: LOGIC ^ ATHEISM This is a sci.logic post that may hold some interest here, >> No. >> -- >Am, m E sci.math, ~interest(m, LOGIC^ATHEISM) ? T. f(u*k) 0ff --- Merlyn LeRoy === Subject: Map two triangles by transforming bounding rectangle triangle into a new triangle by distorting the bounding rectangle of the first rectangle into a parallelogram/rectangle. The problem goe slike this: I have a triangle with corners A, B, & C and the locations of the corners of the rectangle that surrounds that surrounds it (such that the bounding rect is always on the horizontal plane) e.g.: _____A_____________ | x | | x x | | x x | | x x | | x x |C |x x | B x------------------- What I now need to do is to work out the new locations for the corners of the original bounding rectangle such that I can map the points A, B & C onto any given new locations to create a new triangle of known coordinates. I believe this will turn the bounding rect into a parallelogram in most cases and maybe a rectangle in some. There are a few issues that i can be sure of. These include that one of the corners of the triangle will always lie on one corner of the rectangle although sometimes 2 and 3 also. Is there any simple solution to this problem or is it going to be very complex? electron-dot-cloud are galaxies === Subject: Re: math answers on a daily rise percentage Re: 100 free SBC; VonNeumann Gametheory how to play StockMarket Apparently I should have taken advantage of yesterdays rise in Qwest to 4.55 with SBC at 23.80, because today Qwest has fallen back to 4.35. It appears that whenever a stock rises 8% or more in one day that the probability of it continuing to rise is 33% and the probability of it falling to the level of where it was before the 8% rise is 67% Some weeks ago I bought 15,000 shares of Qwest at 4.30 and so if I had sold them at 4.55 would have been a gain of $3,750. and SBC at 23.80 would have given me approx 157 free shares of SBC before taxes. But I need to confirm whether Probability theory approves of those numbers. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Mathematica: LCM for variable length list? [this question originally posted to comp.soft-sys.math.mathematica] What is the syntax to have Mathematica calculate the LCM for a list? For example: if myList = {2, 3, 4, 5, 6}, I want to get 60 back. The list length varies at runtime so I can't list the elements explicitly in the code like LCM[ arg1, arg2, arg3, ... ]. I tried passing the list like this: LCM[myList], but it returns the original list instead of a single number (perhaps the LCM is operating on each element of the list individually). Any help greatly appreciated. Mark -- Mark Lookabaugh mlookabaugh (at) cox.net USS Brewton FF-1086 Home Page http://www.ussbrewton.com === Subject: Re: Mathematica: LCM for variable length list? Why not recursion? If lcm returns the least common multiple of two integers, define LCM as returning the sole element of a list of length 1, or lcm(head(list),LCM(tail(list))) if list has several elements. | [this question originally posted to comp.soft-sys.math.mathematica] | | What is the syntax to have Mathematica calculate the LCM for a list? | | For example: if myList = {2, 3, 4, 5, 6}, I want to get 60 back. | | The list length varies at runtime so I can't list the elements | explicitly in the code like LCM[ arg1, arg2, arg3, ... ]. | | I tried passing the list like this: LCM[myList], but it returns the | original list instead of a single number (perhaps the LCM is operating | on each element of the list individually). | | Any help greatly appreciated. | | Mark | | -- | Mark Lookabaugh | mlookabaugh (at) cox.net | USS Brewton FF-1086 Home Page | http://www.ussbrewton.com === Subject: Re: Mathematica: LCM for variable length list? > [this question originally posted to comp.soft-sys.math.mathematica] > What is the syntax to have Mathematica calculate the LCM for a list? > For example: if myList = {2, 3, 4, 5, 6}, I want to get 60 back. Apply[LCM,myList] or LCM @@ myList -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Mathematica: LCM for variable length list? Elaine Jackson a dit : > Why not recursion? If lcm returns the least common multiple of two > integers, define LCM as returning the sole element of a list of length > 1, or lcm(head(list),LCM(tail(list))) if list has several elements. If it is a list it's smart. If it's an array or a vector, it's cerebral masturbating. And I think that the writting of two functions is not necessary. > | [this question originally posted to comp.soft-sys.math.mathematica] > | > | What is the syntax to have Mathematica calculate the LCM for a list? > | > | For example: if myList = {2, 3, 4, 5, 6}, I want to get 60 back. > | > | The list length varies at runtime so I can't list the elements > | explicitly in the code like LCM[ arg1, arg2, arg3, ... ]. > | > | I tried passing the list like this: LCM[myList], but it returns the > | original list instead of a single number (perhaps the LCM is > | operating on each element of the list individually). > | > | Any help greatly appreciated. > | > | Mark > | > | -- > | Mark Lookabaugh > | mlookabaugh (at) cox.net > | USS Brewton FF-1086 Home Page > | http://www.ussbrewton.com -- Alexandre Charitopoulos mailto:a.charito@wanadoo.fr Em6 / Eb7(5b) / Dm7 / Db7(5b, 9b) / Cmaj7 === Subject: Re: Mathematica: LCM for variable length list? >Why not recursion? If lcm returns the least common multiple of two integers, >define LCM as returning the sole element of a list of length 1, or >lcm(head(list),LCM(tail(list))) if list has several elements. Excellent idea, thanks. That sounds like it might be expensive in terms of memory and execution speed (the list may get very long). I was hoping Mathematica might have some optimized list operator for situations like this. Any other options available? -- Mark Lookabaugh mlookabaugh (at) cox.net USS Brewton FF-1086 Home Page http://www.ussbrewton.com === Subject: Re: Math Trick > Here is a good Math Trick that I invented. > Check this web site with a very unique Math Game > http://w4u.sytes.net/MagicMath > You can send me your comments at gonzalep@.84scientist.com don't forget to > remove the .84 from the e-mail. My old boss at BGC name was Pablo Gonzalez.?? Herc === Subject: Mean Reverting Process This is a question regarding stochastic processes. For the mean reverting process, I've always thought that it is reßected at 0. But I've been told recently that whether it is reßected or absorbed at 0 depends on the parameters. Can someone explain this to me, or point me to a book or website that explains this well? electron-dot-cloud are galaxies === Subject: muon catalyzed fusion: stickyness + muon radii = tokamak Coulomb Re: Coulomb barrier becomes Fusion Barrier Principle; compounding of Maxwell Equations A curious note occurred last night while looking into the Nagamine muon catalyzed fusion. I notice that the muonic atom radii which is much smaller than the radii of DT, some 200 times closer packed nuclei. And it is important in my theory that the Fusion Barrier Principle is the same for all forms of fusion. Implying that muon fusion equals tokamak fusion. So where is muon fusion equal to tokamak fusion since both have a Coulomb barrier. The Coulomb barrier for tokamak fusion is straightforward and call it Tokamak-coulomb. The Coulomb barrier for muon fusion is 1/200 tokamak coulomb. So if my theory is correct then something has to diminish for muon catalyzed fusion to make this equation proper: Tokamak-coulomb = Muon-coulomb X detractor What is the detractor that would equalize muon catalyzed fusion and regain its equality with Tokamak fusion. So the upshot is that muon fusion decreases the radii for fusion events and thus decreases the Tokamak-coulomb. But this smaller term is compensated for, by the stickyness of muons. And thus we have overall this equation: Tokamak-coulomb = muonic radii coulomb X stickyness My theory says that all forms of controlled fusion in a machine, all have a 2/3 breakeven limit. So that if the Coulomb barrier for muons is made less because of muonic radii, then that lessening must be compensated for by another aspect of muon fusion-- Stickyness. However, I must also allow for the compensator of the fact that the muon decays in a few microseconds. A full equation which equates tokamak fusion to muon fusion would have the decay rate of muonic atoms. If my theory is correct and obviously counter to the prevailing physics communities dream/wish of successful fusion, then my theory would show equations of Equality or Equivalence between the various fusion machines whether they be tokamaks or muon or inertial confinement etc etc. All are stopped at 2/3 breakeven and thus all are Equivalent once they get closer and closer to the 2/3 mark. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: My (Un)Originality (was Re: Reminder: Wages, Employment Not Determined By Supply, Demand) >> 1. Being in a position of corporate power (control over production >> factors) often allows one to extract what are effectively economic >> rents. IMO there is an enormous amount of rent seeking going on at >> the top of productive organizations, especially big ones. This is >> much the same in socialist societies as capitalist ones. >I agree with that, though could you elaborate with some (perhaps >hypothetical) examples? Just as a most obvious way it is done, consider what often happens in cases of merger/acquisition: the CEOs and perhaps other top execs cut themselves in for huge slices of options, etc. by effectively holding the deal to ransom. Often it is not so obvious as that, but it seems to me that almost everything I read about specific deals in the business press includes some factor that will allow hefty rent seeking by those in positions of power. It sure isn't happening by accident. >> 2. The reward conferred by the market for creating a temporary >> illusion of value is often just as great as the reward for creating a >> lasting reality, and the former is much easier for some people to do. >I assume you're referring to pumping up stock values temporarily. >That certainly is an issue, though the phenomenon I'm thinking about >is more general. I am also thinking of the more general phenomenon of salesmanship/marketing/PR. >I think it's difficult to for firms to estimate the value of a >particular employee, whether managerial or otherwise. I agree very strongly with this. Everywhere I have worked, I have seen some workers very overpaid for what they were doing, and others equally underpaid. Yet managers were always reluctant to can the former, and mystified when the latter left for greener pastures. >Essentially, most non-managerial >employees are labor commodities---an electrical engineer is just >another EE. IMO this is just ßat wrong, even for assembly line workers, and labor unions have a lot to answer for in having foisted the consequences of their idiotic Marxist ideology on their members as well as society. >As with most commodities, price competition will keep >wages down. High-level managers, however, are viewed as having >special skills which are usually not fully commoditized, and as such >their compensation is higher. Right. But it is an error to believe that the high value placed on these individuals is strongly related to any kind of contribution to production. It's more skill in self-promotion, negotiating ability (a la Morgan), awareness of rent-seeking opportunities, lack of conscience, etc. >A particular example is that of people who are paid a commission basis >for their work in trading securities or commodities, working out >financial deals, etc. I would maintain that total commissions paid >should usually be a sublinear function of the total traded (i.e., the >marginal %age commission should be a monotone decreasing function). >Often this isn't the case. I suppose this is a form of capturing >rents. Yes, especially as competition in these fields is often strictly limited: you need a license to sell real estate, you need a seat on the stock exchange to deal in stocks, etc. >Finally, I find it difficult to believe that compensation levels for >various occupations aren't inßuenced by sociological/socio-biological >factors. Managers and some salesmen tend to be alpha male types and >might be rewarded partly on that basis, regardless of their actual >contribution to the firm. Right. IME just being pushy and demanding often gains extravagant rewards that mere merit can only dream of. One of the greatest advantages enjoyed by those born to wealth is that they grow up giving other people orders which are obeyed. They don't think its rude to treat others like servants, because they have always done exactly that. -- Roy L === Subject: Re: My (Un)Originality (was Re: Reminder: Wages, Employment Not Determined By Supply, Demand) >> 1. Being in a position of corporate power (control over production >> factors) often allows one to extract what are effectively economic >> rents. IMO there is an enormous amount of rent seeking going on at >> the top of productive organizations, especially big ones. This is >> much the same in socialist societies as capitalist ones. >I agree with that, though could you elaborate with some (perhaps >hypothetical) examples? > Just as a most obvious way it is done, consider what often happens in > cases of merger/acquisition: the CEOs and perhaps other top execs cut > themselves in for huge slices of options, etc. by effectively holding > the deal to ransom. Often it is not so obvious as that, but it seems > to me that almost everything I read about specific deals in the > business press includes some factor that will allow hefty rent seeking > by those in positions of power. It sure isn't happening by accident. I've known several high level managers through mergers. They all pretty much assume half the managemnt staff will be whacked after the merger completes, so a lot of that is preemptive looting in anticipation of later RIFs. Some of this is rather like piracy. But not playing the pirate-game is very dangerous. Ask the top guys at Phillips Petroleum when T. Boone Pickens was after Ôem. That took that company a long way down. Phillips was a very kindly- fatherly company, lifetime employer and they got decimated by the greenmail phenonemon. >> 2. The reward conferred by the market for creating a temporary >> illusion of value is often just as great as the reward for creating a >> lasting reality, and the former is much easier for some people to do. >I assume you're referring to pumping up stock values temporarily. >That certainly is an issue, though the phenomenon I'm thinking about >is more general. > I am also thinking of the more general phenomenon of > salesmanship/marketing/PR. For whatever reason, those phenomena are attributed as *the* top level skills in this culture, at this instant. Regardless of actual value, these things are widely beleived to be what feeds the bulldog. >I think it's difficult to for firms to estimate the value of a >particular employee, whether managerial or otherwise. > I agree very strongly with this. Everywhere I have worked, I have > seen some workers very overpaid for what they were doing, and others > equally underpaid. Yet managers were always reluctant to can the > former, and mystified when the latter left for greener pastures. Over/underpaid by standards of whom? I presume folks can make/break employment contracts as they see fit. In the airline seatback magazines is the timeless phrase you get what you negotiate. I beleive it's mostly true. And some of this is analogous to high-school clique politics - people in power tend to reward what they like, not what can be objectively demonstrated as best. I know of at least one manager who hired a crony purely to accept abuse on his behalf by the board. He was a professional whipping boy. Anythign that was likely to cause these problems was handled by the whipping boy. It worked - the manager was quite popular. >Essentially, most non-managerial >employees are labor commodities---an electrical engineer is just >another EE. > IMO this is just ßat wrong, even for assembly line workers, and labor > unions have a lot to answer for in having foisted the consequences of > their idiotic Marxist ideology on their members as well as society. No, labor unions sometimes act as pre-management. They provide some additional services, generally, once they're a mature union like the UAW. It of course varies, but especially in assembly line work, a union can act to add value for what the labor base does - training, some measure of rule enforcement, standardizaiton of the workforce. There have to be negative effects from unions, but they at least try to PR a value-added component these days. >As with most commodities, price competition will keep >wages down. High-level managers, however, are viewed as having >special skills which are usually not fully commoditized, and as such >their compensation is higher. > Right. But it is an error to believe that the high value placed on > these individuals is strongly related to any kind of contribution to > production. It's more skill in self-promotion, negotiating ability (a > la Morgan), awareness of rent-seeking opportunities, lack of > conscience, etc. Yes, but Morgan also once locked all the principals on Wall Street in a room and averted a depression. I see this backwards - a Morgan adds value *because* people behave that way around him. At the time, a really huge bank was a good thing ( or so I guess - it's hard to say otherwise ). To get really goofy about it, there is a certain symmetry to most people's behavior en masse that causes a lot of friction. A Morgan type person will break a bunch of that symmetry and cause Things to happen. In analogy, Grant ended up being the guy who would trade bodies for victory, and won the Civil War. A real cold blooded guy, but the strategy worked. As a really busted comparison, which was worse, the seige of Vicksburg or slavery? I don't think humans made it to the top of the food chain by out-nice-ing the other animals - I think we out-predated them, and we bear the mark of that evolutionary history today. >A particular example is that of people who are paid a commission basis >for their work in trading securities or commodities, working out >financial deals, etc. I would maintain that total commissions paid >should usually be a sublinear function of the total traded (i.e., the >marginal %age commission should be a monotone decreasing function). >Often this isn't the case. I suppose this is a form of capturing >rents. > Yes, especially as competition in these fields is often strictly > limited: you need a license to sell real estate, you need a seat on > the stock exchange to deal in stocks, etc. But isn't some of that the whole idea of them being a registered practitioner? You break the rules, you lose the seat? >Finally, I find it difficult to believe that compensation levels for >various occupations aren't inßuenced by sociological/socio-biological >factors. Managers and some salesmen tend to be alpha male types and >might be rewarded partly on that basis, regardless of their actual >contribution to the firm. > Right. IME just being pushy and demanding often gains extravagant > rewards that mere merit can only dream of. Absolutely. And somehow, human behavior makes that so. In general, I beleive things are they way they are for a reason. Sometimes the reaosn is just because nobody thought to change things, but experience tells me that the vast majority of these things are due to what turn out to be good reasons. This doesn't work 100% of the time, but it works quite often. > One of the greatest > advantages enjoyed by those born to wealth is that they grow up giving > other people orders which are obeyed. They don't think its rude to > treat others like servants, because they have always done exactly > that. And some people will all but instinctively submit to the show of authority. > -- Roy L -- Les Cargill === Subject: Re: My (Un)Originality (was Re: Reminder: Wages, Employment Not Determined By Supply, Demand) [...] | I don't think humans made it to the top of the food chain by | out-nice-ing the other animals - I think we out-predated them, | and we bear the mark of that evolutionary history today. I think this view is somewhat one-sided. Our success has a lot to do with our forming tribes and being able to talk (which I consider primarily nice features). If we're good at ganging up on each other and on other animals, it's largely because we're also good at teamwork, including gathering a shared repository of knowledge and other less aggressive activities. Some of the rarest cognitive abilities we have are abilities to model each other-- to put ourselves in each other's shoes as it were. I also don't find animals to be so nice. It used to be for instance that we thought chimpanzees lived in peace (and wondered why we were so aggressive by comparison), but then someone saw them fight a war, complete with all the usual atrocities. Dogs seem to have a tendancy similar to ours toward being nice to friends and nasty to enemies, only with a more stupid idea on average of who to consider an enemy. I can be talking with the owner of a dachshund away from their shared home turf; the owner may be being entirely friendly to this enemy, and the enemy may be ten times larger than it, but it still has the idea that an attack is in order. Male domestic cats seem often to have an endless greed for each other's territories, and they fight dirty. Keith Ramsay === Subject: Re: My (Un)Originality (was Re: Reminder: Wages, Employment Not Determined By Supply, Demand) >I think it's difficult to for firms to estimate the value of a >particular employee, whether managerial or otherwise. > I agree very strongly with this. Everywhere I have worked, I have >> seen some workers very overpaid for what they were doing, and others >> equally underpaid. Yet managers were always reluctant to can the >> former, and mystified when the latter left for greener pastures. >Over/underpaid by standards of whom? Me, as their co-worker. My impressions were quite consistently supported by others in the same workplaces. >As with most commodities, price competition will keep >wages down. High-level managers, however, are viewed as having >special skills which are usually not fully commoditized, and as such >their compensation is higher. > Right. But it is an error to believe that the high value placed on >> these individuals is strongly related to any kind of contribution to >> production. It's more skill in self-promotion, negotiating ability (a >> la Morgan), awareness of rent-seeking opportunities, lack of >> conscience, etc. >Yes, but Morgan also once locked all the principals on Wall Street >in a room and averted a depression. Which would otherwise have resulted from the sort of shenanigans he and they were profiting from... >I see this backwards - a >Morgan adds value *because* people behave that way around him. No, he just gets to _keep_ the value because people act that around him. >In analogy, Grant ended up being the guy who would trade bodies >for victory, and won the Civil War. A real cold blooded guy, but >the strategy worked. As a really busted comparison, which was >worse, the seige of Vicksburg or slavery? The seige was to right the wrong of slavery. >I don't think humans made it to the top of the food chain by >out-nice-ing the other animals - I think we out-predated them, >and we bear the mark of that evolutionary history today. No. We got to where we are _precisely_ by out-nice-ing other animals: the cooperation in a modern human society dwarfs that in a beehive. The most successful human lineages have not been those that excelled in predation, but in cooperation and production. While we still have the atavistic ape genes that make us reward alpha male antics like J.P. Morgan's beyond all reason, they are more accurately considered handicaps than advantages. >A particular example is that of people who are paid a commission basis >for their work in trading securities or commodities, working out >financial deals, etc. I would maintain that total commissions paid >should usually be a sublinear function of the total traded (i.e., the >marginal %age commission should be a monotone decreasing function). >Often this isn't the case. I suppose this is a form of capturing >rents. > Yes, especially as competition in these fields is often strictly >> limited: you need a license to sell real estate, you need a seat on >> the stock exchange to deal in stocks, etc. >But isn't some of that the whole idea of them being a registered >practitioner? You break the rules, you lose the seat? That's the rationalization, anyway.... >In general, I beleive things are they way they are for a reason. >Sometimes the reaosn is just because nobody thought to change things, >but experience tells me that the vast majority of these things >are due to what turn out to be good reasons. This doesn't work >100% of the time, but it works quite often. Unfortunately, all the progress ever achieved by the human race has been achieved exclusively by those who realized there _wasn't_ a good reason for things to be the way they were. >> One of the greatest >> advantages enjoyed by those born to wealth is that they grow up giving >> other people orders which are obeyed. They don't think its rude to >> treat others like servants, because they have always done exactly >> that. >And some people will all but instinctively submit to the show of >authority. True. But is that really their nature, or just their training? I have had first-person accounts of the school systems in authoritarian countries like Germany and Japan, and it was very obvious to me that the entire public-school populations were being trained to unquestioning obedience. You can bet the guys at the top weren't training their own kids for lives of servitude. -- Roy L === Subject: Re: notion of a Ring Peter > Hi guys, > My problem is why a Ring is defined as it is defined. What is the purpose of > such a definition. A ring has to me an association with something infinite, > no beginning no end. > Peter === Subject: Re: number of trees >Does anyone know how to find : >The total number of binary trees with N nodes and L leaves. A binary tree with N nodes always has exactly (N+1)/2 leaves. This is because each non-leaf has 2 children, and each node except the root is a child. >Does this problem have a solution with general k-ary trees. Similarly, a k-ary tree with N nodes always has N - (N-1)/k leaves. >The total number of binary trees(and k-ary) can be written using the >catalan number, but there is no constraint on the number of leaves here. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2