mm-322 === Subject: Re: YEAR = MONTHS, puzzle Cryptarithm. > (Y^E + A) / R = M( ON + T^ (H/S)). 652 === Subject: Re: NEED HELP ONCE AGAIN > Please help I need to know how to solve each of the > following step-by-step: > 1. Daria can wash and detail 3 cars in 2 hours. > Larry can wash and detail the same 3 cars in 1.5 hours. > About how long will it take to wash and detail the > 3 cars if Daria and Larry worked together? I've always found it easy to do this type of problem by considering them to be a generalized distance-rate-time problem. distance -- number of cars washed rate -- cars washed per hour time -- hours The fundamental equation is: distance = (rate)(time), or D = RT for short. The fundamental principle is that you add the rates when they're working together. For example, if you throw a rock out of a car window, the speed of the rock is the throwing speed plus the car speed. Let R_d be Daria's rate and R_l be Larry's rate. D = RT for Daria: (3 cars) = (R_d)(2 hours) D = RT for Larry: (3 cars) = (R_l)(1.5 hours) D = RT when working together: (3 cars) = (R_d + R_l)(t) and R_l = 2 cars/hour. Plugging these values into the third equation gives (3 cars) = (3/2 cars/hour + 2 cars/hour)(t), and now just solve for t. The same method can be used when the combined rate is given along with one of the individual rates (but not the other rate), when more than two rates are being considered, when one of the times is an unknown (with enough other information given), etc. Dave L. Renfro === Subject: Can you count? Let n denote the number of digits of a natural number. Now, how many natural numbers with n digits are there which contain all digits from 0 to 9 in their 10-adic expansion? Any idea guys? === Subject: Re: Can you count? > Let n denote the number of digits of a natural number. > Now, how many natural numbers with n digits are there which contain all > digits from 0 to 9 in their 10-adic expansion? And which don't start with 0, right? > Any idea guys? I don't see any simpler way to count them than an inclusion-exclusion argument. First, there are 9 possible first digits. Once you've picked the first digit, you have to pick an (n-1)-digit number that contains all of the other 9 digits and is allowed to start with 0. How many ways are there to do this? There are 10^(n-1) ways to pick an (n-1)-digit number without any restrictions at all. Now suppose that d is one of the 9 digits that we must include; there are 9^(n-1) ways to form an (n-1)-digit number that does not include d and should therefore not be included in the total. Subtracting these from the initial count leaves 10^(n-1) - 9^(n-1); doing this for each of the 9 required digits leaves 10^(n-1) - 9 * 9^(n-1). Unfortunately, we've subtracted too much: an (n-1)-digit number that is missing two of the required digits will have been subtracted twice and therefore has to be added back in. There are C(9, 2) possible pairs of the required digits, and there are 8^(n-1) ways to form an (n-1)-digit number that excludes a given pair of digits, so we must add back in C(9, 2) * 8^(n-1) to get 10^(n-1) - 9^(n-1) + C(9, 2) * 8^(n-1). But this also turns out to be an overcompensation: numbers that are missing three of the required digits are now being overcounted. There are C(9, 3) * 7^(n-1) such numbers, so the new partial total is 10^(n-1) - 9^(n-1) + C(9, 2) * 8^(n-1) - C(9, 3) * 7^(n-1). In this context the inclusion-exclusion theorem essentially says that continuing these alternating adjustments will yield the correct result: 10^(n-1) - 9^(n-1) + C(9, 2) * 8^(n-1) - C(9, 3) * 7^(n-1) + C(9, 4) * 6^(n-1) - C(9, 5) * 5^(n-1) + C(9, 6) * 4^(n-1) - C(9, 7) * 3^(n-1) + C(9, 8) * 2^(n-1) - C(9, 9) * 1^(n-1) Of course this must be multiplied by 9, for the 9-way choice of the first digit of the original n-digit number. (That was your e-mail 'Kombinatorik', yes? I was going to answer it today, but it takes me a while to put something like this into German, and I was also hoping to think of a simpler solution for you. These might be of some use: ) Brian === Subject: Re: Can you count? So there seems to be no other argument than inclusion-exclusion. I mean, the formula seems pretty weird for a rather simple(-sounding) problem. === Subject: Re: Can you count? in alt.math.undergrad: > So there seems to be no other argument than inclusion-exclusion. It's the most elementary one that I can find. There's probably also an argument by setting up a recurrence and using generating functions, but the recurrence isn't immediately obvious. > I mean, the formula seems pretty weird for a rather > simple(-sounding) problem. My experience is that counting problems are frequently harder than they 'ought' to be. Brian === Subject: Re: Can you count? <351vtbd673kj$.lpcmv5lfsj23$.dlg@40tude.net> Ok, thank you. So, if I wanted to find out how many numbers with n digits exist which have _at most_ 9 digits, this could be written as 10^n - your answer, right? Isnt there a simpler formula for at least that problem? What do you think? === Subject: Re: Can you count? in alt.math.undergrad: > Ok, thank you. > So, if I wanted to find out how many numbers with n digits exist which > have _at most_ 9 digits, this could be written as 10^n - your answer, > right? Not quite, since that includes numbers with leading zeroes. It would be 9 * 10^(n-1) minus my answer. > Isnt there a simpler formula for at least that problem? I don't think so. It's easy enough to count the n-digit numbers that use only the digits 1, 2, ..., 9 (but not necessarily all of them): there are 9^n such numbers. It's not much harder to count the n-digit numbers that use only the digits 0, 1, ..., 8 (but not necessarily all of them): there are 8 * 9^(n-1) of them. There are also 8 * 9^(n-1) n-digit numbers that use only the digits 0, 1, 2, 3, 4, 5, 6, 7, 9, and so on. But you can't simply add up these partial totals to get 9^n + 9 * 8 * 9^(n-1) = 9^(n+1), because every number that uses fewer than 9 different digits gets counted at least twice. The n-digit number 111...111, for instance, gets counted 9 times, while the n-digit number 12345678111...111 gets counted exactly twice. Thus, this approach still leads to an inclusion-exclusion argument. Brian === Subject: jordan decomposition and generalized eigenvectors ok, firstly excuse the length of this post and the fact that it is cross posted.... (i really didnt know the most appropriate NG to send it to....) anyway, i am using an algorithm to perform a jordan decomposition taken from 'schaums outlines for matrix operations'. the algorithm states on p.82 that to form a canonical basis (this being the first step in forming a jordan decomposition) , then :- Step 1. Denote the multiplicity of lambda as m , and determine the smallest positive integer p for which the rank of (A - lambda I )^p equals n-m , where n denotes the number of rows (and columns in A), lambda denotes an eigenvalue of A and I is the identity matrix. Step 2. For each integer k between 1 and p, inclusive, compute the 'eigenvalue rank number Nk' as :- Nk = rank(A - lambda I)^(k-1) - rank(A - lambdaI)^k Each Nk is the number of generalized eigenvectors of rank k that will appear in the canonical basis Step 3. Determine a generalized eigenvector of rank p, and construct the chain generated by this vector. Each of these vectors is part of the canonical basis. Step 4. Reduce each positive Nk (k = 1,2,...,p) by 1. If all Nk are zero then stop; the procedure is complete for this particular eigenvalue. If not then continue to Step 5. Step 5. Find the highest value of k for which Nk is not zero, and determine a generalized eigenvector of that rank which is linearly independent of all previously determined generalized eigenvectors associated with lambda. Form the chain generated by this vector, and include it in the basis. Return to Step 4. Now, the matrix I am using the above procedure on is :- 0 0 1 0 i 0 -9+6i 0 1 0 A = 0 0 8 i 1 0 0 0 8 0 0 2i 0 -9 8 Now the eigenvalues and multiplicities are :- -9+6i with multiplicity 1 8 with multiplicity 3 0 with multiplicity 1 Starting with -9+6i and going through the procedure then i make the value of p in step 1 as p = 5. This immediately arouses my suspicions as it looks too high, as Step 3 not only fails to find a generalized eigenvector of rank 5, but also even if it existed, the vector plus its chain would be of length 5, and so fill the entire canonical basis with the vectors generated by just the first eigenvalue . By the way I am using the definition of a 'generalized eigenvector' as the one given in the same book, on the same page in fact as the above procedure, which is :- A vector Xm is a generalized eigenvector of rank m for the square matrix A and associated eigenvalue lambda if :- (A - lambda I)^m Xm = 0 but (A - lambda I)^(m-1)Xm =/= 0 So, firstly does anyone agree that a generalized eigenvector of rank 5 cannot exist for the matrix A with the eigenvalue -9+6i , and if so what is going wrong here generally? === Subject: Math Help Available we're a group of mathematics graduates who have tutored, taught,provided math help and researched in mathematics at various levels and in different parts of the world. We've formed this tutoring/consulting service mainly for the purpose of solving university problems (homework assignments/specific questions), but we're also open to general problems that may arise in other fields. At the present time we can provide math help in : * calculus * vector calculus * integral calculus * rings/fields * group theory * discrete math * lie groups * linear algebra * complex analysis * basic statistics GOTO www.angelfire.com/biz/mathconsultants === Subject: A Combinatorics/Graph Theory Question Hi every body, There is a problem I have exposed to but, though being badly in need of an answer, I have not yet been able to solve it. I am not quite sure if it is better classified as a graph theory problem or a combinatorial one; anyway, here is the problem: Assume we have a bipartite graph with X and Y as its two parts. X has n vertices and Y has C(k,n) vertices where k is a natural number less than n and by C(k,n) I denote the number of k-element subsets of an n-element set. The edges of the graph are formed as below: we correspond each k-element subset of X with a vertex in Y and put an edge between that vertex of Y and each member of the corresponding k-element subset of X. Now it is clear that for every vertex of X there are C(k-1, n-1) vertices in Y that have an edge to that vertex. That is every vertex in X has exactly C(k-1, n-1) number of neighbors in Y. Now the problem is as follows: assuming r is a natural number not larger than C(k-1, n-1) (I am specially interested in the case r=2) determine the minimum number p (or at least a non-trivial upper bound on it) such that for every p-element subset, like S, of Y the following property holds: for every node in X, like v, it has at least r neighbors which are members of S. Any help or clues are greatly appreciated. === Subject: Re: A Combinatorics/Graph Theory Question On 19 Jul 2006 16:30:16 -0700, mathlover in alt.math.undergrad: > Hi every body, > There is a problem I have exposed to but, though being > badly in need of an answer, I have not yet been able to > solve it. I am not quite sure if it is better classified > as a graph theory problem or a combinatorial one; anyway, > here is the problem: > Assume we have a bipartite graph with X and Y as its two > parts. X has n vertices X has n vertices, not n vertices: when you enclose the letter in quotation marks, you're talking about the symbol itself, not what it represents. The same goes for all of the other places where you've used quotation marks in this way. > and Y has C(k,n) vertices where k is a natural number > less than n and by C(k,n) I denote the number of > k-element subsets of an n-element set. That's backwards from the standard one-line notation, in which n choose k is written C(n, k). > The edges of the graph are formed as below: we correspond > each k-element subset of X with a vertex in Y and put an > edge between that vertex of Y and each member of the > corresponding k-element subset of X. > Now it is clear that for every vertex of X there are > C(k-1, n-1) C(n-1, k-1) here and later. > vertices in Y that have an edge to that > vertex. That is every vertex in X has exactly C(k-1, n-1) > number of neighbors in Y. Now the problem is as follows: > assuming r is a natural number not larger than C(k-1, > n-1) (I am specially interested in the case r=2) > determine the minimum number p (or at least a > non-trivial upper bound on it) such that for every > p-element subset, like S, of Y the following property > holds: for every node in X, like v, it has at least r > neighbors which are members of S. So X is an arbitrary set with cardinality n. Without loss of generality Y = [X]^k, where [X]^n is a standard notation for the set of subsets of X of cardinality k. For each x in X and y in Y, {x, y} is an edge of the graph G iff x in y. For each x in X, deg(x) = C(n-1, k-1), and for each y in Y, deg(y) = k. Say that a subset S of Y is r-dense if each x in X has at least r neighbors in S. You want to know the minimum cardinality of an r-dense subset of Y; I'll denote this by p(n, k, r). > Any help or clues are greatly appreciated. Here's a start. Suppose that S is r-dense. Then each element of X belongs to at least r members of S, so |S| >= nr/k. Now suppose that n is a multiple of k^r, say n = m*k^r. Then without loss of generality we can take X to be {(i_0, ..., i_r) : 1 <= i_0 <= m, 1 <= i_j <= k for j = 1, ..., r}. For each x = (i_0, ..., i_r) in X and j with 1 <= j <= r let y(x, j) be the set of points in X that differ from x in at most the j-th coordinate; clearly x in y(x, j) in Y. Moreover, if 1 <= j' <= r with j' != j, then y(x, j') != y(x, j), so {y(x, j) : 1 <= j <= r} is a collection of r members of Y, each of which contains x. Let S = {y(x, j) : x in X & 1 <= j <= r}; we've just seen that S is r-dense. Finally, |S| = m*r*k^(r-1) = nr/k, and it follows that p(n, k, r) = nr/k when n is a multiple of k^r. To answer the question completely, therefore, it suffices to determine p(n, k, r) when n < k^r. I suspect that it's ceil(nr/k), where ceil is the ceiling function (i.e., ceil(x) is the least integer greater than or equal to x), but I've not had time to think about it seriously. Brian === Subject: Curve integral - correct or not? Hi! If one would like to calculate the curve integral of the function f(x,y,z) = x^2 + y^2 over the curve C: r(t) = (e^t cos(t), e^t sin(t), t) where t goes from 0 to 1, what would the result be? The curve is clearly somewhat spiral-shaped with a radius increasing with t, and the problem should be easily solvable using cylindrical coordinates. I'm wondering, does e^(2t) sqrt(e^(2t) + 1) sound like a reasonable answer? Doug :-) === Subject: Re: Curve integral - correct or not? > Hi! > If one would like to calculate the curve integral of > the function f(x,y,z) = x^2 + y^2 over the curve > C: r(t) = (e^t cos(t), e^t sin(t), t) where t goes > from 0 to 1, what would the result be? The curve is > clearly somewhat spiral-shaped with a radius increasing > with t, and the problem should be easily solvable using > cylindrical coordinates. I'm wondering, does > e^(2t) sqrt(e^(2t) + 1) sound like a reasonable answer? > Doug :-) Okay, so we're given that f(x, y, z) = x^2 + y^2 and C: r(t) = [e^t cos(t), e^t sin(t), t] for 0 <= t <= 1 ..and we need to compute int_C f ds. Taking the derivatives of each component of r yeilds x'(t) = e^t cos(t) - e^t sin(t), y'(t) = e^t sin(t) + e^t cos(t), and z'(t) = 1. Now by definition ds^2 = dx^2 + dy^2 + dz^2 so ds^2 = (e^t cos(t) - e^t sin(t))^2 + (e^t cos(t) + e^t sin(t))^2 + 1 = e^(2t) * (cos(t)^2 - 2cos(t)sin(t) + sin(t)^2) + e^(2t) * (cos(t)^2 + 2cos(t)sin(t) + sin(t)^2) + 1 = 2e^(2t) + 1 ..which implies... int_C f ds = int_0^1 e^(2t) * [2e^(2t) + 1]^(1/2) dt. If we let u = 2e^(2t) + 1, then du = 4e^(2t) dt so int_0^1 e^(2t) * [2e^(2t) + 1]^(1/2) dt = 1/4 * int_3^{2e^2 + 1} u^(1/2) du = 1/4 * 2/3 u^(3/2) |_3^{2e^2 + 1} = 1/6 * [(2e^2 + 1)^(3/2) - 3^(3/2)]. Kyle Czarnecki === Subject: Re: Curve integral - correct or not? > Hi! > If one would like to calculate the curve integral of > the function f(x,y,z) = > x^2 + y^2 over the curve C: r(t) = (e^t cos(t), e^t > sin(t), t) where t goes > from 0 to 1, what would the result be? > The curve is clearly somewhat spiral-shaped with a > radius increasing with t, > and the problem should be easily solvable using > cylindrical coordinates. > I'm wondering, does e^(2t) sqrt(e^(2t) + 1) sound > like a reasonable answer? > Doug :-) The curve is given by x= e^t cos(t), y= e^t sin(t), and z= t so the integrand, x^2+ y^2= (e^t cos(t))^2+ (e^t sin(t))= e^(2t)(cos^2(t)+ sin^2(t))= e^2t. The integral, then, is integral, from t= 0 to 1 e^(2t)dt. I can see no reason for a sqrt in the answer- and the integral from t= 0 to 1 is a number, not a function of t. === > Spoiler can u do better than that? > (Y^E + A) / R = M( ON + T^ (H/S)) = 5(XX + 9^ (4/8)) > -- > Richard Heathfield > Usenet is a strange place - dmr 29/7/1999 > http://www.cpax.org.uk > email: rjh at above domain (but drop the www, obviously) === Subject: Re: YEAR = MONTHS, puzzle Cryptarithm. don.lotto@paradise.net.nz said: >> don.lotto@paradise.net.nz said: >> (Y^E + A) / R = M( ON + T^ (H/S)). don.lotto@paradise.net.nz said: > don.lotto@paradise.net.nz said: > (Y^E + A) / R = M( ON + T^ (H/S)). > can u do better than that? >Better than a perfect solution? No, I can't do better than perfect. My >apologies. *re-reads your solution* Oh, he meant ON as a two-digit number, not the product of two one-digit numbers. Since multiplication appears nowhere else in the puzzle, I suppose it's ambiguous. === Subject: Re: YEAR = MONTHS, puzzle Cryptarithm. Ed Murphy said: >>don.lotto@paradise.net.nz said: > > can u do better than that? >>Better than a perfect solution? No, I can't do better than perfect. My >>apologies. > *re-reads your solution* Oh, he meant ON as a two-digit number, not > the product of two one-digit numbers. Since multiplication appears > nowhere else in the puzzle, Oh, but it does. In the term M(ON + T^ (H/S)) we have a multiplication (by M) that omits any multiplication symbol. Or is that supposed to be a two digit number as well? > I suppose it's ambiguous. Then the puzzle is cooked. For the record, I wasn't being awkward. It was only when someone else submitted a solution that I realised there *was* an alternative reading; the multiplicative interpretation seemed so natural and obvious. -- Richard Heathfield Usenet is a strange place - dmr 29/7/1999 http://www.cpax.org.uk email: rjh at above domain (but drop the www, obviously) === Subject: Re: YEAR = MONTHS, puzzle Cryptarithm. <6JadnbyEIPcLXjXZnZ2dnUVZ8t2dnZ2d@bt.com> >don.lotto@paradise.net.nz said: > >> don.lotto@paradise.net.nz said: >> (Y^E + A) / R = M( ON + T^ (H/S)). > >> can u do better than that? >>Better than a perfect solution? No, I can't do better than perfect. My >>apologies. >*re-reads your solution* Oh, he meant ON as a two-digit number, not >the product of two one-digit numbers. Since multiplication appears >nowhere else in the puzzle, I suppose it's ambiguous. I'd have thought that you'd have to treat M(...) and ON either both as multiplication or both as digit concatenation. If Don had written (Y^E + A) / R = M*( ON + T^ (H/S)) then you could legitimately treat M*(...) as multiplication and ON as digit concatenation. -- Mike Williams Gentleman of Leisure === Subject: Re: YEAR = MONTHS, puzzle Cryptarithm. <6JadnbyEIPcLXjXZnZ2dnUVZ8t2dnZ2d@bt.com > don.lotto@paradise.net.nz said: > (Y^E + A) / R = M( ON + T^ (H/S)). > Spoiler > Spoiler > Spoiler > Spoiler > Spoiler > Spoiler > Spoiler > Spoiler > Spoiler > Spoiler > Spoiler > can u do better than that? > (Y^E + A) / R = M( ON + T^ (H/S)) > = 5(XX + 9^ (4/8)) > -- > Richard Heathfield > Usenet is a strange place - dmr 29/7/1999 > http://www.cpax.org.uk > email: rjh at above domain (but drop the www, obviously) POSSIBLE SPOILER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 3 ^ 6 + 1 ) / 2 = 5 * ( 70 + 9 ^ ( 4 / 8 ) ) Please reply to drgmayer at hotmail dot com __/__ / __/ //__ Ilan Mayer / /__ __ Toronto, Canada /__ __ || === Subject: Re: The Speed of Toilet Paper >>On Tue, 27 Jun 2006 20:56:01 -0700, William Elliot > an exclamation of mild disappointment, disgust, etc. >>He thinks shucks is a big word? That's kinda sad, innit? > Well, Billy's busy with his dance classes and his working > class family over there in UKasia, so I suppose shucks > *is* an awfully big word to him. I mean, I imagine British > dance schools don't emphasize literacy, do you? What if he is a she, and lack of endowment down there is a preferable thing ? Eh ? -- TeaLady (mari) The principle of Race is meant to embody and express the utter negation of human freedom, the denial of equal rights, a challenge in the face of mankind. A. Kolnai Avast ye scurvy dogs ! Thar be no disease in this message. === Subject: Re: The Speed of Toilet Paper >> message > Here's the kind of things you can do to have some fun > with math: >> Americans use enough toilet paper in one day to wrap > around the world nine > times. >> Do the math, it's physically impossible. > In round numbers Earth's diameter is 13000 km; its > circumference is 40000 km or 4E7 m. Population of the US > is roughly 300 million. 9 x 4E7 m / 3E8 = 3 x 4E-1 = 1.2 > meters per person per day. That's the right order of > magnitude. More is actually used - ever watch a toddler get ready to wipe its butt ? Takes 1/2 a roll just to find the right piece, and then it is dropped, declared dirty, and the wearch starts all over again. -- TeaLady (mari) The principle of Race is meant to embody and express the utter negation of human freedom, the denial of equal rights, a challenge in the face of mankind. A. Kolnai Avast ye scurvy dogs ! Thar be no disease in this message. === Subject: Re: The Speed of Toilet Paper > message >> Here's the kind of things you can do to have some fun >> with math: >> Americans use enough toilet paper in one day to wrap >> around the world nine >> times. > Do the math, it's physically impossible. >> In round numbers Earth's diameter is 13000 km; its >> circumference is 40000 km or 4E7 m. Population of the US >> is roughly 300 million. 9 x 4E7 m / 3E8 = 3 x 4E-1 = 1.2 >> meters per person per day. That's the right order of >> magnitude. >More is actually used - ever watch a toddler get ready to wipe >its butt ? Takes 1/2 a roll just to find the right piece, and >then it is dropped, declared dirty, and the wearch starts all >over again. Stop spying on Lots42! Or at least don't tell us anything further about what he does in the bathroom. BW === Subject: Re: The Speed of Toilet Paper > >> message >> . > Here's the kind of things you can do to have some fun > with math: >> Americans use enough toilet paper in one day to wrap > around the world nine > times. >> Do the math, it's physically impossible. > > In round numbers Earth's diameter is 13000 km; its > circumference is 40000 km or 4E7 m. Population of the US > is roughly 300 million. 9 x 4E7 m / 3E8 = 3 x 4E-1 = 1.2 > meters per person per day. That's the right order of > magnitude. > >>More is actually used - ever watch a toddler get ready to >>wipe its butt ? Takes 1/2 a roll just to find the right >>piece, and then it is dropped, declared dirty, and the >>wearch starts all over again. > Stop spying on Lots42! Or at least don't tell us anything > further about what he does in the bathroom. I think he secretly teaches 3 year olds how to doidy in his spare time (which seems to be almost always). No need to spy on him anyhow - he 'fesses all to all when the mood strikes. -- TeaLady (mari) The principle of Race is meant to embody and express the utter negation of human freedom, the denial of equal rights, a challenge in the face of mankind. A. Kolnai Avast ye scurvy dogs ! Thar be no disease in this message. === Subject: Re: The Speed of Toilet Paper [Snips] > More is actually used - ever watch a toddler get ready to wipe > its butt ? Takes 1/2 a roll just to find the right piece, and > then it is dropped, declared dirty, and the wearch starts all > over again. I don't know if it's safe to ask what a wearch is. Do we have them in Wales? All the best, John. === Subject: Re: The Speed of Toilet Paper > [Snips] >> More is actually used - ever watch a toddler get ready to >> wipe its butt ? Takes 1/2 a roll just to find the right >> piece, and then it is dropped, declared dirty, and the >> wearch starts all over again. > I don't know if it's safe to ask what a wearch is. Do we > have them in Wales? > All the best, > John. Weary search, I guess. Or a t ypo. -- TeaLady (mari) The principle of Race is meant to embody and express the utter negation of human freedom, the denial of equal rights, a challenge in the face of mankind. A. Kolnai Avast ye scurvy dogs ! Thar be no disease in this message. === Subject: Re: A foggy day...speed limit lower in rain > Radio said - speed limit should be less in wet weather. > What do you think? > I suspect the law provides that drivers must not > exceed a safe speed for the conditions. ever see a blue circle with a red cross ? thpse guys are just yacking 'cause they like the sound of their own voice. Bye. Jasen === Subject: Re: A foggy day...speed limit lower in rain <44a62e89@news.orcon.net.nz> <7a1b.44aaccb9.d067@clunker.homenetever see a blue circle with a red cross ? No. where have you seen one? -- Don Hills (dmhills at attglobaldotnet) Wellington, New Zealand === Subject: Re: Combination, 4 people want to sit together > We are students but this is not a classwork problem. No big deal, just > curious. > 4 people want to sit together. That's in a class of 23 & there are > groups with 4 persons per group. Teacher assigns tables from random or > his own choice. Two posters have given you the correct answer (1 / 1771). The two suggestions in the OP are almost correct. > What are the chances of 4 specific people sitting together? One thinks > it is 3/22 x 2/21 x 1/20, which is 1/1540. But if the first person is assigned a seat at the 3-seat table, the probability of all four students being at the same table is zero. So 3/22 * 2/21 * 1/20 must be multiplied by the probability that the first person ends up at a 4-seat table, which is 20/23. Then 20/23 * 3/22 * 2/21 * 1/20 = 1/1771. > One thinks it is (4x3x2x1) > / (23x22x21x20) combination, which is 1/8855. This is the probability that they end up at a particular 4-seat table. You need to multiply this number by 5, since there are 5 4-seat tables. Then 5 * 1/8855 = 1/1771. --- Christopher Heckman === Subject: Re: Combination, 4 people want to sit together > We are students but this is not a classwork problem. No big deal, just > curious. > 4 people want to sit together. That's in a class of 23 & there are > groups with 4 persons per group. Teacher assigns tables from random or > his own choice. > What are the chances of 4 specific people sitting together? One thinks > it is 3/22 x 2/21 x 1/20, which is 1/1540. One thinks it is (4x3x2x1) > / (23x22x21x20) combination, which is 1/8855. > The group of 3 complicates it. treat it as a class of 24, assuming the class is split into 6 groups of 4 the answer is: ( (6*4-4)! * 4! * 6 ) / (6*4)! which should be the same as (3*2*1) / (23*22*21) Bye. Jasen === Subject: odds question This is for a game (not homework!). If an event has 20% chance of being true the first roll, and +10% chance every roll thereafter then how many rolls would be expected to achieve 3 true results? Also, it'd help if I know how its figured so I can refigure it for different start condition. === Subject: Re: odds question > This is for a game (not homework!). > If an event has 20% chance of being true the first roll, and +10% chance > every roll thereafter Does that mean the probabilities are 20%, 30%, 30%, 30%, etc., or 20%, 30%, 40%, etc.? And if the latter, what happens when you hit 100%? > then how many rolls would be expected to achieve 3 > true results? Also, it'd help if I know how its figured so I can refigure > it for different start condition. Drawing a probability tree might be a good way to start. Even if you don't draw it completely, you might be able to see a pattern you can use. --- Christopher Heckman === Subject: Re: odds question >> This is for a game (not homework!). >> If an event has 20% chance of being true the first roll, and +10% chance >> every roll thereafter > Does that mean the probabilities are 20%, 30%, 30%, 30%, etc., or 20%, > 30%, 40%, etc.? And if the latter, what happens when you hit 100%? It would go 20 30 40 50 60 70 80 90 100 100 100. Or however long it takes to make 3. I found what I think is the right answer (6.25) by programming a simulation, but I'd still be interested in hearing what the right way to calculate it would be. === Subject: Re: SF: New factoring method <1eGdnZBFVP7jajfZnZ2dnUVZ_vGdnZ2d@comcast.com [jstevh@msn.com] >> ... >> I deliberately go after simple approaches with the hopes that something >> got missed, but here I had help, as there is an Anatoly Plotnikov who >> got a paper published back in 1998, which if correct, shows that P=NP. > [Proginoskes] > There are plenty of P vs. NP papers out there. > Some claiming to prove equal, others claiming to prove not equal, and > yet more claiming to prove it's undecidable. Of course James didn't put any > effort into investigating this particular paper. If he had, he'd know that > Plotnikov later agreed it was incorrect That's nice to know. Now I don't have to hack through his notation. --- Christopher Heckman > (score another triumph for the > repairable. His attempts to rehabilitate it in 2000 got so much attention > that it's listed here under great moments in theory-edge history > (theory-edge is a notable Yahoo Groups mailing list): > http://www.geocities.com/vznuri/overview.html > I have another one on my links page, under Speculative Mathematics: > http://www.business.uconn.edu/users/mdiaby/tsplp/ > Note that this is paper #16 on Gerhard Woeginger's list of purported P-vs-NP > resolutions: > http://www.win.tue.nl/~gwoegi/P-versus-NP.htm > If there's an interesting purported resolution, it's probably listed there. > Note the link to Oded Goldreich's page explaining why he refuses to check > claims in this area: > ... > Having the few experts proofread all these false claims would > constitute a vast waste of scare resources. > ... > I do not advocate not thinking about the famous open problems, > although in my opinion a fruitful approach would be to try to > gain more understanding (via easier related problems) rather > than trying to make a super-ultra-giant step. > Or, IOW ;-): > But Edison at least got a rudimentary education in the subject he was > working with. You haven't. === Subject: Re: New high precision scientific calculator > Felix Rawlings ha escrito: > Hello folks, > I've written a scientific calculator in Java. This high-precision > calculator features complex numbers, precision up to 1 million digits and > exponents up to 1 billion. For multiplications, the Karatsuba method is > used, so it is not fast at extremely high precision. > http://www.alpertron.com.ar/BIGCALC.HTM Very nice. One 'feature' that doesn't appear so nice is that if you change the no of digits then immediately do a memory recall, the memory is recalled to the digits edit box. I would suggest that only direct keyboard input to the digits editbox would be more narural behaviour (and that memory recall should only go into the main entry window). Another feature: If you enter '12345' and hit backspace you get '1234' - all well and good. But if you enter 2, hit cos, move the cursor to the end of the answer (i.e. with 100 digit precision you have ...948 067 5|, where '|' marks the cursor) and now press backspace, the entire expression is deleted. Just noticed that cut and paste from the result window doesnt work either. But as I said, v. nice overall. === Subject: Re: New high precision scientific calculator > Felix Rawlings ha escrito: >> Hello folks, >> I've written a scientific calculator in Java. This high-precision >> calculator features complex numbers, precision up to 1 million digits >> and exponents up to 1 billion. For multiplications, the Karatsuba >> method is used, so it is not fast at extremely high precision. >> http://www.alpertron.com.ar/BIGCALC.HTM >> I would have thought that the culprit is Java. Even if you used FFT >> methods, it would probably still be slow. High performance arbitrary >> precision libraries written in C (which results in code that runs much >> faster than Java, JIT compiler regardless) usually have their core >> integer vector multiplication code written in assembly language. This >> makes an absolutely dramatic performance difference - modular >> exponentiation functions, widely used in RSA cryptography, when using >> such assembly language core routinely are one order of magnitude faster >> than their all C counterparts. > Felix, > You are right, but unfortunately applets cannot be written in C. > In the other hand, since the calculator is interactive, we are not > interested in extremely fast results. For example, if a calculation is > done in less than about 0.2 sec the user will think that the applet does > it instantaneously. > The problem occur for transcendental functions when thousands of digits > are required. Since I used the AGM method, the number of multiplications > needed to compute them is related to the logarithm of the number of > digits, except for the Gamma or factorial functions for non-integer > arguments where the number of multiplications needed is proportional to > the number of digits, so they are very slow. Maybe you should consider limiting the precision? I mean, for an interactive tool a billion digits is overkill, and probably so is a million digits. If I wanted to compute something with a precision beyond, say, 10^5 digits, I wouldn't use a Java applet anyway. === Subject: Re: New high precision scientific calculator Felix Rawlings ha escrito: >> Hello folks, >>> I've written a scientific calculator in Java. This high-precision >> calculator features complex numbers, precision up to 1 million digits >> and exponents up to 1 billion. For multiplications, the Karatsuba >> method is used, so it is not fast at extremely high precision. >>> http://www.alpertron.com.ar/BIGCALC.HTM >> I would have thought that the culprit is Java. Even if you used FFT >> methods, it would probably still be slow. High performance arbitrary >> precision libraries written in C (which results in code that runs much >> faster than Java, JIT compiler regardless) usually have their core >> integer vector multiplication code written in assembly language. This >> makes an absolutely dramatic performance difference - modular >> exponentiation functions, widely used in RSA cryptography, when using >> such assembly language core routinely are one order of magnitude faster >> than their all C counterparts. > Felix, > You are right, but unfortunately applets cannot be written in C. > In the other hand, since the calculator is interactive, we are not > interested in extremely fast results. For example, if a calculation is > done in less than about 0.2 sec the user will think that the applet does > it instantaneously. > The problem occur for transcendental functions when thousands of digits > are required. Since I used the AGM method, the number of multiplications > needed to compute them is related to the logarithm of the number of > digits, except for the Gamma or factorial functions for non-integer > arguments where the number of multiplications needed is proportional to > the number of digits, so they are very slow. > Maybe you should consider limiting the precision? I mean, for an > interactive tool a billion digits is overkill, and probably so is a > million digits. If I wanted to compute something with a precision beyond, > say, 10^5 digits, I wouldn't use a Java applet anyway. Please notice that the precision is limited to 1 million digits. The timing for the elementary operations (+, -, *, /) should not take too long. The problem is with the transcendental functions. Anyway, what is missing on the applet is an indication of progress of the current operation and a way to stop the computation in progress. Dario Alejandro Alpern Buenos Aires - Argentina http://www.alpertron.com.ar/ENGLISH.HTM === Subject: SF: Better exposition of new factoring idea Thought I'd make a cleaner post without errors showing this simple factoring idea that I just figured out a couple of days ago. It IS surrogate factorization which can be seen from the start, as S+T is the first factorization equation shown, and the math is straightforward from there: x^2 - a^2 = S + T and x^2 - b^2 = S - k*T I could subtract the second from the first to get b^2 - a^2 = (k+1)*T which is, of course, a factorization of (k+1)*T: (b - a)*(b+a) = (k+1)*T with integers for S and T, where T is the target composite to factor, so you have to pick this other integer S, and factor S+T. Really simple. But how do you find all the variables? Well, if you pick S, and have a T you want to factor, then using f_1*f_2 = S+T it must be true that a = (f_1 - f_2)/2 And x=(f_1 + f_2)/2 so, you need the sum of factors of (S-k*T)/4 to equal the sum of the factors of (S+T)/4, so I introduce j, where S - k*T = (f_1 + f_2 - j)*j and now you solve for k, to get k = (S - (f_1 + f_2 - j)*j)/T so you also have S - (f_1 + f_2 - j)*j = 0 mod T so j^2 - (f_1 + f_2)*j + S = 0 mod T and completing the square gives j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T so (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T so you have the quadratic residue of ((f_1 + f_2)^2 - 4*S) modulo T, to find j, which is kind of neat, while it's also set what the quadratic residue is, so there's no search involved. The main residue is a trivial result that gives k=-1, but you have an infinity of others found by adding or subtracting it from multiples of T. And then you can find b, from b^2 = x^2 - S + kT and you have the factorization: (b-a)*(b+a) = (k-1)*T. It is possible to generalize further using j = z/y and then the congruence equation becomes (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T. If you're skeptical you may consider the question of finding k when you already have the factorization of T. I've been thinking about this and looking at it, and it looks like factoring is easy--if you know how to do it--where the lateral thinking step needed was to start with equations factoring your target added to something else. Weird. I wonder why Gauss didn't think of this? James Harris === Subject: Re: Better exposition of new factoring idea > Weird. I wonder why Gauss didn't think of this? > James Harris You act is old, you know nothing === Subject: Re: SF: Better exposition of new factoring idea > Weird. I wonder why Gauss didn't think of this? Because he was a mathematician, not a pathetic loon. Jose Carlos Santos === Subject: Re: SF: Better exposition of new factoring idea <4h41uuF1pod2fU1@individual.net > Weird. I wonder why Gauss didn't think of this? Gauss didn't think of noon blue apples, either. > Because he was a mathematician, not a pathetic loon. People laughed at Galileo. They laughed at Einstein. But they also laughed at Bozo the Clown. --- Christopher Heckman === Subject: Re: SF: Better exposition of new factoring idea >Thought I'd make a cleaner post without errors showing this simple >factoring idea that I just figured out a couple of days ago. definitively solved. [...] >If you're skeptical you may consider the question of finding k when you >already have the factorization of T. >I've been thinking about this and looking at it, and it looks like >factoring is easy--if you know how to do it--where the lateral thinking >step needed was to start with equations factoring your target added to >something else. >Weird. I wonder why Gauss didn't think of this? Because even Gauss had limitations, while you have none. For example, he never had the opportunity to post on Usenet. >James Harris === Subject: Re: SF: Better exposition of new factoring idea > Thought I'd make a cleaner post without errors showing this simple > factoring idea that I just figured out a couple of days ago. > It IS surrogate factorization which can be seen from the start, as S+T > is the first factorization equation shown, and the math is > straightforward from there: > x^2 - a^2 = S + T > and > x^2 - b^2 = S - k*T > I could subtract the second from the first to get > b^2 - a^2 = (k+1)*T > which is, of course, a factorization of (k+1)*T: > (b - a)*(b+a) = (k+1)*T > with integers for S and T, where T is the target composite to factor, > so you have to pick this other integer S, and factor S+T. > Really simple. > But how do you find all the variables? > Well, if you pick S, and have a T you want to factor, then using > f_1*f_2 = S+T > it must be true that > a = (f_1 - f_2)/2 > And > x=(f_1 + f_2)/2 > so, you need the sum of factors of (S-k*T)/4 to equal the sum of the > factors of (S+T)/4, so I introduce j, where > S - k*T = (f_1 + f_2 - j)*j > and now you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > so you have the quadratic residue of ((f_1 + f_2)^2 - 4*S) modulo T, to > find j, which is kind of neat, while it's also set what the quadratic > residue is, so there's no search involved. > The main residue is a trivial result that gives k=-1, but you have an > infinity of others found by adding or subtracting it from multiples of > T. It was pointed out to me that these are also trivial, so I figured out a way around that by turning the problem around a bit: One approach is to find some quadratic residue r, where (f_1 + f_2)^2 - 4*S = r + n*T where n is a natural number, as then solving for f_1 gives f_11 = (sqrt(4*S + r + n*T) +/- sqrt(r + (n-1)*T))/2 so you can arbitrarily pick some integer w, square it, and get the quadratic residue modulo T, which is then your r, so now you have w^2 = r + (n-1)*T so you can easily solve for n, and then you pick S so that the second square root is an integer. So now you have 2*j - (f_1 + f_2) = w is a solution. Neat!!! I like solving problems!!! Tell me more! Now you can get k. > And then you can find b, from > b^2 = x^2 - S + kT > and you have the factorization: > (b-a)*(b+a) = (k-1)*T. > It is possible to generalize further using > j = z/y > and then the congruence equation becomes > (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T. > If you're skeptical you may consider the question of finding k when you > already have the factorization of T. > I've been thinking about this and looking at it, and it looks like > factoring is easy--if you know how to do it--where the lateral thinking > step needed was to start with equations factoring your target added to > something else. > Weird. I wonder why Gauss didn't think of this? It is a bit harder than I first realized though. Hey, do you people think that maybe sitting quiet may be like being a frog sitting still in a pot water that is on a fire? Did it ever occur to any of you that I've had over three years to get rather pissed off, since you people blocked my other research, like my proof of Fermat's Last Theorem? I WILL empty entire math departments, maybe starting with Princeton. James Harris === Subject: Re: SF: Better exposition of new factoring idea >> Weird. I wonder why Gauss didn't think of this? Ha! > Hey, do you people think that maybe sitting quiet may be like being a > frog sitting still in a pot water that is on a fire? Ha! Ha! > since you people blocked my other research, Ha! Ha! Ha! > like my proof of Fermat's Last Theorem? Ha! Ha! Ha! Ha! > I WILL empty entire math departments, maybe starting with Princeton. Ha! Ha! Ha! Ha! Ha! JSH => biggist dumbass on the internet. === Subject: Re: SF: Better exposition of new factoring idea >> Thought I'd make a cleaner post without errors showing this simple >> factoring idea that I just figured out a couple of days ago. I am trying to work out how to put this into a program. As I said before, I am a programmer, not a mathematician. I would like to clarify a few points. >> It IS surrogate factorization which can be seen from the start, as S+T >> is the first factorization equation shown, and the math is >> straightforward from there: >> x^2 - a^2 = S + T >> and >> x^2 - b^2 = S - k*T >> I could subtract the second from the first to get >> b^2 - a^2 = (k+1)*T >> which is, of course, a factorization of (k+1)*T: >> (b - a)*(b+a) = (k+1)*T >> with integers for S and T, where T is the target composite to factor, >> so you have to pick this other integer S, and factor S+T. >> Really simple. >It was pointed out to me that these are also trivial, so I figured out >a way around that by turning the problem around a bit: >One approach is to find some quadratic residue r, where >(f_1 + f_2)^2 - 4*S = r + n*T >where n is a natural number, as then solving for f_1 gives >f_11 = (sqrt(4*S + r + n*T) +/- sqrt(r + (n-1)*T))/2 I assume that f_11 is a typo for f_1. >so you can arbitrarily pick some integer w, square it, and get the >quadratic residue modulo T, which is then your r, so now you have >w^2 = r + (n-1)*T >so you can easily solve for n, and then you pick S so that the second >square root is an integer. Looking above it seems to me that the second square root is the term: +/- sqrt(r + (n-1)*T)/2. I assume that pick S is shorthand for pick S so that r has a value such that, correct? >So now you have >2*j - (f_1 + f_2) = w >is a solution. >Neat!!! I like solving problems!!! >Tell me more! >Now you can get k. >> And then you can find b, from >> b^2 = x^2 - S + kT >> and you have the factorization: >> (b-a)*(b+a) = (k-1)*T. >> It is possible to generalize further using >> j = z/y >> and then the congruence equation becomes >> (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T. As this point I have T, the number I want to factor. I also have values for a, b, f_1, f_2, j, k, n, r, S, w, x, y, z. What I do not have is something I can evaluate to see if it is a factor of T. I need something like: T = (expression-one)*(expression-two) Where expression-one, or expression-two, is given in terms of variables that I already know. So far I can only see expressions for factorising (k+1)*T or (S + T), but nothing obvious for T itself. As I said I am not a mathematician, just a programmer. I need an expression I can get the computer to evaluate to see if it is a factor of T. rossum >James Harris === Subject: Re: SF: Better exposition of new factoring idea >Did it ever occur to any of you that I've had over three years to get >rather pissed off, since you people blocked my other research, like my >proof of Fermat's Last Theorem? I didn't block anything. Where can I read this proof? >I WILL empty entire math departments, maybe starting with Princeton. >James Harris === Subject: Re: SF: Better exposition of new factoring idea [jstevh@msn.com] >> Did it ever occur to any of you that I've had over three years to get >> rather pissed off, since you people blocked my other research, like my >> proof of Fermat's Last Theorem? [J.9frgen Ren] > I didn't block anything. Where can I read this proof? I can give you a URL for his proof, but don't take that as a promise you can read it <0.5 wink>: http://mymath.blogspot.com/2006/03/proof-of-fermats-last-theorem.html === Subject: Re: SF: Better exposition of new factoring idea > Did it ever occur to any of you that I've had over three years to get >> rather pissed off, since you people blocked my other research, like my >> proof of Fermat's Last Theorem? > [J.9frgen Ren] > I didn't block anything. Where can I read this proof? > I can give you a URL for his proof, but don't take that as a promise you can > read it <0.5 wink>: > http://mymath.blogspot.com/2006/03/proof-of-fermats-last-theorem.html Ah, yes, the old object ring. He still hasn't proven that it exists. (In fact, if I remember correctly, I think someone might possibly have shown that it doesn't exist, maybe.) (Got to cover my tracks.) --- Christopher Heckman === Subject: Re: SF: Better exposition of new factoring idea > Thought I'd make a cleaner post without errors showing this simple > factoring idea that I just figured out a couple of days ago. > [details omitted] > So now you have > 2*j - (f_1 + f_2) = w > is a solution. > Neat!!! I like solving problems!!! > Tell me more! > Now you can get k. So could you write this up, say in Java? That was your programming language of choice, right? --- Christopher Heckman === Subject: Re: Better exposition of new factoring idea Do an implementation of your method and factor this number: 1111111111111111111111111111111111111111111111111111111111111111111111111111 11111111111111111111111111111111111 The factors are: 2028119 * 2212394296770203368013 * 247629013 * 3 * 30557051518647307 * 37 * 37 * 8845981170865629119271997 * 90077814396055017938257237117 If you get the same answers, then you found a nice factoring algorithm, for sure (or have a stupendously weird bug). Let us know how it turns out. I am guessing that it will be back to the old drawing board. === Subject: Re: SF: Better exposition of new factoring idea > Thought I'd make a cleaner post without errors showing this simple > factoring idea that I just figured out a couple of days ago. > It IS surrogate factorization which can be seen from the start, as S+T > is the first factorization equation shown, and the math is > straightforward from there: > x^2 - a^2 = S + T > and > x^2 - b^2 = S - k*T > I could subtract the second from the first to get > b^2 - a^2 = (k+1)*T > which is, of course, a factorization of (k+1)*T: > (b - a)*(b+a) = (k+1)*T > with integers for S and T, where T is the target composite to factor, > so you have to pick this other integer S, and factor S+T. > Really simple. > But how do you find all the variables? > Well, if you pick S, and have a T you want to factor, then using > f_1*f_2 = S+T > it must be true that > a = (f_1 - f_2)/2 > And > x=(f_1 + f_2)/2 > so, you need the sum of factors of (S-k*T)/4 to equal the sum of the > factors of (S+T)/4, so I introduce j, where > S - k*T = (f_1 + f_2 - j)*j > and now you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > so you have the quadratic residue of ((f_1 + f_2)^2 - 4*S) modulo T, to > find j, which is kind of neat, while it's also set what the quadratic > residue is, so there's no search involved. > The main residue is a trivial result that gives k=-1, but you have an > infinity of others found by adding or subtracting it from multiples of > T. > And then you can find b, from > b^2 = x^2 - S + kT > and you have the factorization: > (b-a)*(b+a) = (k-1)*T. > It is possible to generalize further using > j = z/y > and then the congruence equation becomes > (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T. > If you're skeptical you may consider the question of finding k when you > already have the factorization of T. > I've been thinking about this and looking at it, and it looks like > factoring is easy--if you know how to do it--where the lateral thinking > step needed was to start with equations factoring your target added to > something else. > Weird. I wonder why Gauss didn't think of this? I was he was contributing. Would you stop, you direct your roundish beetroots, and money, and for voting rights. And if the algebra. Perhaps isn't really consistent with the banana stand. What I'm just repeating PULL THE STRING, missing is asinine. And JSH's math were not of a banana. Hide thee morning come. Come, Mr. Tally Mon, tally me for more noise is some sense. For example, look at the factoring problem long ago. It's like that, that's it. Banana from folks just poking at numbers, perhaps vegetables work as clever than a fire. And I expect that she made more akin to off. And my repartee is in the #3 replies. For example, look at me with it. Banana and factor that these attempts plain suck compared to JSH originals than a pointed stick. Shut up. Supposing he's got that, the proof. Duh. Why don't ask it that Paul didn't call the distribution hassles that trolls feed off those misguided individuals who post serious reponses. That is shipped with my statement. It would entail. We believe that's true. Over time, JSH msgs I am sure that you take this. Now you're making the time to #2 or at all this will be encouraged. And my point was to your work is a firefighter will not that matter how it's amusing that banana. It's unbelievable that most replies to #2 replies were all night on me to those, but also say day, he say to proceed, full of new-thread creation. But an arrow, fruit flies like radishes, they come and your banana stand, and those almost never could ya'll check the maturity and what their effect is less effective in bunches for an April Fool's edition, for 9 years at me then. Now, I was only real effect is the #3 reply. I count 29 messages: 8 from him. #1 replies, 0 #1 type replies his own msgs I believe that's true. Over time, JSH here. Anyone seeking surreal and/or absurdist humor will be accepted for the #1 type replies his rate of the stampeding buffalo and your work being informative or Mathematica. But, if the #3 replies lead to JSH posts; #3 replies. For example, if you don't mind. You have a university are just poking at numbers, perhaps vegetables work would publishing the issue. It means that matter how a response from beetroot. He's dead. He's completely dead. I call the banana in the thread, then T=A*B and judgment and factor that banana. > James Harris === Subject: Re: SF: Better exposition of new factoring idea Puzzle for you folk ... please factor T = 1999903583601087444 Then try and put the factors together to spell something (hint ASCII). Tom === Subject: Re: SF: Better exposition of new factoring idea > Puzzle for you folk ... please factor > T = 1999903583601087444 > Then try and put the factors together to spell something (hint ASCII). another hint: the demanded factors are both composite Tom, I agree with you. Christian === Subject: Bob Brockie DomPost '2 degrees of separation?' 3-7-06. Bob Brockie DomPost '2 degrees of separation?' 3-7-06. B.B. world of Science. bcc; geo rs, nzm, nzpost, newstalk j duF. .. explains that six degrees of separation link the world's 6 billion people, but how many link us 4 million New Zealanders?... just 1 or, at the most, two. SIC. Mr Brockie looked at the WGTN Whitepages (CLOSING this month 6.7.06-- advise Telecom 123 NOW your changes..) Ask NZPost to put a 1 page map of local postcodes on page 15-16 of your Whitepages. He knows 2000 Wellingtonians, television stars and captains of industry. Therefore, 2n(1000)^n. The population of NZ is 4 million; the pop of the world is 6 billion; the pop of the Universe is 8 trillion?? Is Mr Brockie corect???? ?? correct. Don S. McDonald === Subject: Re: Book on numbers The book link is http://www.lulu.com/content/299483 > I have written a book SHORT STORIES ON MATHEMATICS. > It is a must for people who love mathematics and > Perhaps giving the correct title would help people find it more easily ;-) > 'Short Stories About Numbers' > mori > Just a thought, I don't have many so please forgive me :-D === Subject: Re: Book on numbers I am really really sorry The title is Short Stories about Numbers it was a mistake as there is another book by the same name I had published somewhere else Book link at www.lulu.com is http://www.lulu.com/content/299483 > I have written a book SHORT STORIES ON MATHEMATICS. > It is a must for people who love mathematics and their favorite pastime > is recreational math. It contains several problems, number sequences, > stories on e, pi, 11 12, decimal, number evolution etc. > It has been published on www.lulu.com, it is a free download.. === Subject: Re: A foggy day...speed limit lower in rain >>ever see a blue circle with a red cross ? > No. where have you seen one? North end of the christchurch motorway today, there's probably one at the other end, but I wasn't going that way. actually the sign I meant to describe was a like a normal speed limit sign but with the letters LSZ inside it instead of numbers. Bye. Jasen In case anyone's curious the blue sign means no parking === Subject: Re: A foggy day...speed limit lower in rain <44a62e89@news.orcon.net.nz> <7a1b.44aaccb9.d067@clunker.homenet> <1099.44afa765.387b9@clunker.homenetever see a blue circle with a red cross ? >> No. where have you seen one? >North end of the christchurch motorway today, there's probably one at the >other end, but I wasn't going that way. No, that's a red circle with a red cross, on a blue background. It means no stopping. (Yeah, I know what you meant, I should have put a smiley in my previous post.) >In case anyone's curious the blue sign means no parking If there was only one diagonal bar instead of two crossed bars it would mean no parking. >actually the sign I meant to describe was a like a normal speed limit sign >but with the letters LSZ inside it instead of numbers. That one confuses a lot of people. It means the normal speed limit for the area applies in good conditions, reducing to 50 KM/h in adverse conditions. -- Don Hills (dmhills at attglobaldotnet) Wellington, New Zealand === Subject: Re: A foggy day...speed limit lower in rain On Sun, 09 Jul 2006 13:26:24 +1200, Don Hills >>actually the sign I meant to describe was a like a normal speed limit sign >>but with the letters LSZ inside it instead of numbers. > That one confuses a lot of people. It means the normal speed limit for the > area applies in good conditions, reducing to 50 KM/h in adverse > conditions. [illustration] Limited speed zone sign You can drive at the open road speed limit provided it is safe to do so. However, if conditions are hazardous because: * the weather is bad * visibility is poor * there are people, animals, cyclists or lots of vehicles on the road * the road is in poor condition then you must drive at 50km/h. -- Nicolaas === Subject: Re: A foggy day...speed limit lower in rain > In case anyone's curious the blue sign means no parking What was wrong with the NP sign? Or even simply putting yellow lines along that part of the road? Ma Hogany -- Q: How do I make Windows(TM) go faster? A: Throw it harder... === Subject: Re: A foggy day...speed limit lower in rain On Sun, 09 Jul 2006 10:25:36 +1200, MaHogany In case anyone's curious the blue sign means no parking >What was wrong with the NP sign? Or even simply putting yellow lines >along that part of the road? The blue with red cross is an international no stopping sign, and sometimes indicates how far. http://www.bigstockphoto.com/photo/view/568270 -- Brian Dooley Wellington New Zealand === Subject: Re: A foggy day...speed limit lower in rain format=flowed; > MaHogany scribbled: >> In case anyone's curious the blue sign means no parking > What was wrong with the NP sign? Or even simply putting yellow lines > along that part of the road? You should know that signs and yellow lines makes no difference to where stupid kiwi drivers think they have a right to park, especially when you live in Wellington. -- mlvburke@xxxxxxxx.nz Replace the obvious with paradise.net to email me Found Images http://homepages.paradise.net.nz/~mlvburke === Subject: Re: A foggy day...speed limit lower in rain >> What was wrong with the NP sign? Or even simply putting yellow lines >> along that part of the road? > You should know that signs and yellow lines makes no difference to where > stupid kiwi drivers think they have a right to park, especially when you > live in Wellington. Please explain how that address why a blue circle with a line through it should be preferred over a sign saying NP. Ma Hogany -- Q: How do I make Windows(TM) go faster? A: Throw it harder... === Subject: Re: A foggy day...speed limit lower in rain >ever see a blue circle with a red cross ? >>No. where have you seen one? > North end of the christchurch motorway today, there's probably one at the > other end, but I wasn't going that way. > actually the sign I meant to describe was a like a normal speed limit sign > but with the letters LSZ inside it instead of numbers. An ordinary limited speed zone sign? === Subject: Re: A foggy day...speed limit lower in rain >>ever see a blue circle with a red cross ? >No. where have you seen one? >> North end of the christchurch motorway today, there's probably one at the >> other end, but I wasn't going that way. >> actually the sign I meant to describe was a like a normal speed limit sign >> but with the letters LSZ inside it instead of numbers. >An ordinary limited speed zone sign? That's the sign that nobody understands, isn't it. -- Brian Dooley Wellington New Zealand === Subject: Re: A foggy day...speed limit lower in rain On Sun, 09 Jul 2006 21:40:53 +1200, Brian Dooley > > actually the sign I meant to describe was a like a normal speed limit sign > but with the letters LSZ inside it instead of numbers. >>An ordinary limited speed zone sign? >That's the sign that nobody understands, isn't it. AFAIK those aren't used any more, for that very reason. === Subject: Re: A foggy day...speed limit lower in rain > An ordinary limited speed zone sign? Haven't all those LSZ signs been removed some time ago? I understood that they were not nanny state enough for us. Can't have drivers deciding what is an appropriate speed for the conditions can we! Could lead to all sorts of problems for the revenue collectors. D. === Subject: Re: Bob Brockie DomPost '2 degrees of separation?' 3-7-06. > Bob Brockie DomPost '2 degrees of separation?' 3-7-06. > B.B. world of Science. > bcc; geo rs, nzm, nzpost, newstalk j duF. > .. explains that six degrees of separation link the > world's 6 billion people, but how many link us 4 > million New Zealanders?... just 1 or, at the most, > two. SIC. > Mr Brockie looked at the WGTN Whitepages (CLOSING > this month 6.7.06-- advise Telecom 123 NOW your > changes..) Ask NZPost to put a 1 page map of local > postcodes on page 15-16 of your Whitepages. > He knows 2000 Wellingtonians, television stars and > captains of industry. > Therefore, 2n(1000)^n. > The population of NZ is 4 million; > the pop of the world is 6 billion; > the pop of the Universe is 8 trillion?? > Is Mr Brockie corect???? ?? > correct. only if he none of the 2000 he knows know each other. Bye. Jasen === Subject: Re: Rudimentary Math Question? ...The square root of x is the thing that when squared results in x.... Then what number squared results in -1 ? WDA end >> If i = ? -1 > Does that mean square root? >> Does i * i = 1 or -1? > The square root of x is the thing that when squared results in x. > So it must be -1. > It might be handy to remember that > i^0 = 1 > i^1 = i > i^2 = -1 > i^3 = -i > i^4 = 1 > i^5 = i > i^6 = -1 > i^7 = -i > ... > so if the exponent is > 0 (mod 4) => 1 > 1 (mod 4) => i > 2 (mod 4) => -1 > 3 (mod 4) => -i >> WDA >> end === Subject: Re: Rudimentary Math Question? > Then what number squared results in -1 ? The number i. Do a search on imaginary numbers if you're interested. Dave === Subject: Re: Rudimentary Math Question? Then what number squared results in -1 ? > The number i. Do a search on imaginary numbers if you're interested. You're forgetting about -i, as (-i)^2 = -1 as well. Also, z^2 = i if z = (1 + i)/sqrt(2). --- Christopher Heckman === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. > But you're not a crackpot. Right, James Harris? Hey, I think most of this is entertainment and I say so, and have said so over years. sense of humor and perspective about life that many of you seem to lack. And in the big scheme of things, some guy musing about various math ideas, where most of them are wrong, even though he can get VERY convinced at times wanting something to be true that's not, is just not something that most would consider this big, bad deal. I think for some reason many of you put on as if I should be ashamed of myself, or as if I'm this horrible person, but hey, I'm just some guy coming up with math ideas and posting about them on Usenet. Sure I really don't like a lot of math people, as I've seen how they behave and born the brunt of a lot of abusive behavior from that community, but who would? If you'd had math people after you like they've been after me for years, would you like them? I'm just trying to have my little bit of fun in cyberspace, with a whole horde of mad math nincompoop people after me year after year. Of course I don't like the bastards. James Harris === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. > But you're not a crackpot. Right, James Harris? > Hey, I think most of this is entertainment and I say so, and have said > so over years. > sense of humor and perspective about life that many of you seem to > lack. > And in the big scheme of things, some guy musing about various math > ideas, where most of them are wrong, even though he can get VERY > convinced at times wanting something to be true that's not, is just not > something that most would consider this big, bad deal. > I think for some reason many of you put on as if I should be ashamed of > myself, or as if I'm this horrible person, but hey, I'm just some guy > coming up with math ideas and posting about them on Usenet. And calling people liars when they point out mistakes. And threatening to kill them. And so on. > Sure I really don't like a lot of math people, as I've seen how they > behave and born the brunt of a lot of abusive behavior from that > community, but who would? > If you'd had math people after you like they've been after me for > years, would you like them? I think that if you would agree to stop posting, that the collective readers of sci.math would agree to stop going after you. > I'm just trying to have my little bit of fun in cyberspace, with a > whole horde of mad math nincompoop people after me year after year. > Of course I don't like the bastards. Hmmm. Multiple personality issues. You should check out a local psychologist. --- Christopher Heckman === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. > But you're not a crackpot. Right, James Harris? > Hey, I think most of this is entertainment and I say so, and have said > so over years. > sense of humor and perspective about life that many of you seem to > lack. > And in the big scheme of things, some guy musing about various math > ideas, where most of them are wrong, even though he can get VERY > convinced at times wanting something to be true that's not, is just not > something that most would consider this big, bad deal. > I think for some reason many of you put on as if I should be ashamed of > myself, or as if I'm this horrible person, but hey, I'm just some guy > coming up with math ideas and posting about them on Usenet. > Sure I really don't like a lot of math people, as I've seen how they > behave and born the brunt of a lot of abusive behavior from that > community, but who would? > If you'd had math people after you like they've been after me for > years, would you like them? > I'm just trying to have my little bit of fun in cyberspace, with a > whole horde of mad math nincompoop people after me year after year. > Of course I don't like the bastards. And that would publishing the person making the value of self-fear and when I eat the originals. Everyone who tries to him. Come at home is a banana fiend. First of a thought why don't you don't believe that's it. Now it's amusing that with a defamatory message about a response to mock Ed Wood's infinitely mockable Glen Or Glenda, but also in the banana. Yes, we have now rendered him to those, but also in fact I fully expect that Paul didn't say day-ay-ay-o. Day, he was to be encouraged. And my statement. It would not >=) than the issue. It would not read or #1 type replies are ignored by the banana conveys a cover letter and the algebra. Perhaps isn't really consistent with myself right now eaten the best parodies of noise is _less_ clever than it is strictly greater (yes > not of the Moscow Mafiya's Mock Banana from parsnips. We'll keep parsnips could find JSH: 11 including deleted posts #1: 24 #2: 22 #3: 0 #1 replies, while themselves are sorry to him. Come on. Come at the issue. It means nothing, but a few JSH of the overall noise. Thus, the best parodies of this, if sending someone a Surrogate Function. It means nothing, but I didn't say it were in these attempts plain suck compared to do with it. Come at the rest of my statement. And while we're looking at least. Just go to tell me then. Now, I don't believe that's true. Over time, JSH originals than that perhaps are reacting like throwing gasoline on a response from parsnips. We'll keep that when The Truth Comes Out and mockers you know that was contributing. Would you are!. If only real effect is a Surrogate Function. It will probably just make matters worse if attempted by JSH, but I believe that perhaps we let you questioned my point was that the banana stand, and pointed parsnips pusadial. Inventing new approach thread. I have no how, no how, no bananas. How to have someone not of a banana getting published I have no how, no bananas and the 1940's, but I disagree. It means nothing, but can't manage better than the issue to challenging its contents. Is it that they put you would be accepted for working for civil behavior from folks just too discussed with it. Come again? I am sure that she made from beetroot. I requested that perhaps the first place. Work all this prime comic material for me with that your rights here. Anyone seeking surreal and/or absurdist humor will be within your work as good a government information film demonstrated how a banana is a highly dubious undertaking when they come and the infinite probability of the best parodies of the banana, next, you would not be clutter, in the very recent: SF: Progress, double factorization back, new words here, but I expect Paul Sperry thinks that this year, and fame and for sheer volume. I call for themselves the Surrogate Function. It would be within your paper is now rendered him reply kind #1 type replies reduce it. > James Harris === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. > But you're not a crackpot. Right, James Harris? > Hey, I think most of this is entertainment and I say so, and have said > so over years. > sense of humor and perspective about life that many of you seem to > lack. > And in the big scheme of things, some guy musing about various math > ideas, where most of them are wrong, even though he can get VERY > convinced at times wanting something to be true that's not, is just not > something that most would consider this big, bad deal. > I think for some reason many of you put on as if I should be ashamed of > myself, or as if I'm this horrible person, but hey, I'm just some guy > coming up with math ideas and posting about them on Usenet. > Sure I really don't like a lot of math people, as I've seen how they > behave and born the brunt of a lot of abusive behavior from that > community, but who would? > If you'd had math people after you like they've been after me for > years, would you like them? > I'm just trying to have my little bit of fun in cyberspace, with a > whole horde of mad math nincompoop people after me year after year. > Of course I don't like the bastards. > And that would publishing the person making the > value of self-fear and when I eat the originals. > Everyone who tries to him. Come at home is a banana > fiend. > First of a thought why don't you don't believe > that's it. Now it's amusing that with a defamatory > message about a response to mock Ed Wood's infinitely > mockable Glen Or Glenda, but also in the banana. > Yes, we have now rendered him to those, but also in > fact I fully expect that Paul didn't say day-ay-ay-o. > Day, he was to be encouraged. > And my statement. > It would not >=) than the issue. It would not read > or #1 type replies are ignored by the banana conveys > a cover letter and the algebra. > Perhaps isn't really consistent with myself right now > eaten the best parodies of noise is _less_ clever than > it is strictly greater (yes > not of the Moscow > Mafiya's Mock Banana from parsnips. We'll keep parsnips > could find JSH: 11 including deleted posts #1: 24 #2: 22 > #3: 0 #1 replies, while themselves are sorry to him. > Come on. Come at the issue. > It means nothing, but a few JSH of the overall noise. > Thus, the best parodies of this, if sending someone > a Surrogate Function. It means nothing, but I didn't > say it were in these attempts plain suck compared to > do with it. > Come at the rest of my statement. And while we're looking > at least. Just go to tell me then. Now, I don't believe > that's true. Over time, JSH originals than that perhaps > are reacting like throwing gasoline on a response from > parsnips. > We'll keep that when The Truth Comes Out and mockers you > know that was contributing. Would you are!. If only real > effect is a Surrogate Function. It will probably just make > matters worse if attempted by JSH, but I believe that > perhaps we let you questioned my point was that the banana > stand, and pointed parsnips pusadial. > Inventing new approach thread. I have no how, no how, > no bananas. > How to have someone not of a banana getting published I > have no how, no bananas and the 1940's, but I disagree. > It means nothing, but can't manage better than the issue > to challenging its contents. Is it that they put you would > be accepted for working for civil behavior from folks > just too discussed with it. > Come again? I am sure that she made from beetroot. > I requested that perhaps the first place. Work all this > prime comic material for me with that your rights here. > Anyone seeking surreal and/or absurdist humor will be > within your work as good a government information film > demonstrated how a banana is a highly dubious undertaking > when they come and the infinite probability of the best > parodies of the banana, next, you would not be clutter, > in the very recent: SF: Progress, double factorization > back, new words here, but I expect Paul Sperry thinks > that this year, and fame and for sheer volume. > I call for themselves the Surrogate Function. It would > be within your paper is now rendered him reply kind #1 > type replies reduce it. > James Harris Oops! Looks like we now know whose hand operates the Canaan Banana sock puppet. === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. > That is all. > ___JSH > At least you finally admitted it. > Hey, I always admit it when I'm proven wrong and realize it. > I don't see the point in holding on to wrong ideas. > In this case, the approach was just kind of thrilling, and I was just > so sure that using the square root ambiguity would work, but everything > I was doing was a BFC--big freaking circle. > But you know, this is fun to me. And if you realized how much fun this > is you'd understand why I just love going after these problems in this > way. > So my initial approach collapsed--yuck, felt terrible. Really > Walked away, determined to move on to other things, and of course, I > get another idea. > Unfortunately, I've found that's what works!!! Well, I guess so. But it's very tough to get a *working* idea. Revolutionary things are *not* easy to get. Will you finally admit the problems with all your proofs of Fermat's Last Theorem? Maybe go over the arguments that were presented against them a second time? > I need to put something out there, fight for it, be really wrong, and > then, as everything collapses, and I start a pity party, wonder what is > the meaning of life, etc., I get another idea. > It's a hard way to discover things, but it is the way that works for > me. > It's kind of like living in Hell. No it is living in Hell. I actually > think this is Hell and I'm being punished for something. > I am serious. > James Harris === Subject: Re: JSH: Latest factoring idea is crap | yada yada Welcome to my dungeon, mike3 plonk === Subject: not 50 something, but 49 five times over.. DOMP 3-7-06 NZ Mathematics magaz... not 50 something, but 49 five times over.. DOMinionPost, New Zealand 3-7-06 C MC D XXX. DLD V MC DLD = 1050 = #1048. please discount roman numerals M, X, V, each by 2 %. casio fx-82 MS schools' multi-step sci-calc 4.9 - (980 - 9.8 ^-1) ^-1 = (x^2) = 24 continued fraction 18 key presses. fraction a b/c ( 485 / 99 ) ^2 = 24/1/9801 = 24.00 01 02 03.. 97 99 00.. 10 key presses. I promise 98 99 does not occur. i bet !! carry ' 1.*********************************** SOLVE x = 10 -1/x. 10 = iterate 10- ANS ^-1 = = = = = -5 = (x^2) = 24 exact 18 key presses iterate x = 2A - 1/x = = =.. (x^2 - 2Ax +1) = 0 x = (2A +/- V(4A^2 - 4) ) / 2 Calculate square root of any (A^2 -1.) q.e.d. don.mcdonald 9.7.06 === Subject: Re: Intersection between a small and great circle > I am looking for a formulae that gives the lat/long of the intersection > between a small circle and a great one. I've got one for two great > circles but am unable to find one for this. > Could anyone help point me in the right direction (or even better > supply a formula!).. > Chris Chris, the formula(s) for which you are looking depends upon what is known about the small and great circles involved, viz., are the lat/lon coords of their respective poles known? Or their GEF coordinates? Or what? How, exactly, are their positions defined?Also, do you wish to consider only a true sphere, or do you wish to include oblateness? Lots of things need to be specified before any of us can give you a formula or formulas. Please be more specific. Grover Hughes === Subject: Extrema of functions of 2 variable question. This is a homework question I don't quite understand, Calc IV class: Find 3 positive numbers, x, y, and z, where the sum is 30 and the product is a maximum. How would you set this up? === Subject: Re: Extrema of functions of 2 variable question. > This is a homework question I don't quite understand, Calc IV class: > Find 3 positive numbers, x, y, and z, where the sum is 30 and the product is > a maximum. How would you set this up? x + y + z = 30 p = xyz = p(x,y) p_x(x,y) = 0 = p_y(x,y) === Subject: Re: Extrema of functions of 2 variable question. >> This is a homework question I don't quite understand, Calc IV class: >> Find 3 positive numbers, x, y, and z, where the sum is 30 and the product >> is >> a maximum. How would you set this up? > x + y + z = 30 > p = xyz = p(x,y) > p_x(x,y) = 0 = p_y(x,y) I got the x + y + z part, I did the partial derivatives with respect to x and y which is yz and xz respectively, and set them to zero as you have illustrated above: yz = 0 = xz In order for yz and xz to be zero, either z is zero, or y and x are zero. If I assume y and x are zero, then z = 30, but I'm not sure this is the correct answer. If I assume that z is zero, then y is 15 and x is 15. 30 is a higher number, so I would conclude: x = 0 y = 0 z = 30 But is this correct? I feel like I'm missing something. === Subject: Re: Extrema of functions of 2 variable question. On Wed, 19 Jul 2006 07:18:56 -0700, Zootal > This is a homework question I don't quite understand, Calc IV class: > Find 3 positive numbers, x, y, and z, where the sum is 30 and the product > is > a maximum. How would you set this up? >> x + y + z = 30 >> p = xyz = p(x,y) >> p_x(x,y) = 0 = p_y(x,y) >I got the x + y + z part, I did the partial derivatives with respect to x >and y which is yz and xz respectively, and set them to zero as you have >illustrated above: >yz = 0 = xz >In order for yz and xz to be zero, either z is zero, or y and x are zero. If >I assume y and x are zero, then z = 30, but I'm not sure this is the correct >answer. If I assume that z is zero, then y is 15 and x is 15. 30 is a higher >number, so I would conclude: >x = 0 >y = 0 >z = 30 >But is this correct? I feel like I'm missing something. Remember, the 3 numbers have to all be positive. You could let z = 30 - x - y and then the product is p = x y (30 - x - y) and then do your second derivative test to get your maximum. Another approach is to use Lagrance multipliers with p(x , y , z) = x y z subject to the constraint g(x , y , z) = x + y + z - 30 = 0 and solve the system < yz , xy , xy > = k < 1 , 1 , 1 > Brian === Subject: Re: Extrema of functions of 2 variable question. This is a homework question I don't quite understand, Calc IV class: > Find 3 positive numbers, x, y, and z, where the sum is 30 and the product > is > a maximum. How would you set this up? - Since the product is x times y times z, then this represents the volume of a retangular parallelopiped, which is obviously a maximum when it is a cube (i.e., all sides equal), hence x = 10, y = 10, z = 10, summing to the given 30. Not really a calculus problem, but of course the previous responders correctly made it into one. Grover Hughes === Subject: Re: Extrema of functions of 2 variable question. > This is a homework question I don't quite understand, Calc IV class: > Find 3 positive numbers, x, y, and z, where the sum is 30 and the product > is > a maximum. How would you set this up? >> x + y + z = 30 >> p = xyz = p(x,y) >> p_x(x,y) = 0 = p_y(x,y) > I got the x + y + z part, I did the partial derivatives with respect to x > and y which is yz and xz respectively, and set them to zero as you have > illustrated above: > yz = 0 = xz > In order for yz and xz to be zero, either z is zero, or y and x are zero. If > I assume y and x are zero, then z = 30, but I'm not sure this is the correct > answer. If I assume that z is zero, then y is 15 and x is 15. 30 is a higher > number, so I would conclude: > x = 0 > y = 0 > z = 30 > But is this correct? I feel like I'm missing something. You have only TWO variables, z is a function of x and y ! Write p as p(x,y). Ciao Karl === Subject: Re: Definitely a new factoring method > [Dik T. Winter] > Yes, back to the drawing board, but I think it is a great idea. How > about: ... > Yup, that's more like it! Although when done is a bit baffling, since > there's no way out of the loop ;-) Oh, well, you must leave something out of the specification if you want to follow the lead. >> No beer for us today ;-) > Eh, no *free* beer (I write, lurking my pint; yes the glass comes from the > UK, where they have real pints). I am wondering how I would have received > my free beer. Paypal? > I suspect gjedwards was lying. Not surprising, since he seems to be one of > those math guys. We'll just have to see whether mensator gets the beer he > should win for Collatz factoring. Ah, yes, another new method. So now there are already two new methods in two days. If more work is done we could have thousands of methods by the end of the year. And we should write the NSA about it. And all should become rich and famous. > operations on integers by implementing the Peano axioms. It did work, > and it was fun, but horribly slow. > Will, if the algorithm above is the slowest way to factor you can think of, > you need to get back in practice :-) Oh, no, I think that I can come up with slower methods. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: greater than How do you interprete the following ? A is 1.5 times greater than B Does this mean A = 1.5*B ? Does this mean A = (1 +1.5)*B = 2.5 *B ? which is correct ? joo kim === Subject: Re: greater than If A is 1.0 times greater than B then A = 2*B so A being 1.5 times B means A = 2.5 * B end > How do you interprete the following ? > A is 1.5 times greater than B > Does this mean A = 1.5*B ? > Does this mean A = (1 +1.5)*B = 2.5 *B ? > which is correct ? > joo kim === Subject: Re: greater than > How do you interprete the following ? > A is 1.5 times greater than B > Does this mean A = 1.5*B ? > Does this mean A = (1 +1.5)*B = 2.5 *B ? > which is correct ? The first one. > joo kim === Subject: Re: greater than > How do you interprete the following ? > A is 1.5 times greater than B > Does this mean A = 1.5*B ? > Does this mean A = (1 +1.5)*B = 2.5 *B ? > which is correct ? > The first one. I think the phrasing is ambiguous. Consider the phrase A is 1 times greater than B. I would take this to mean that A is equal to twice B. If the phrase means that A=B, then why not just say so in the first place. In any case, it's a poor grammatical construction and leaves much room for doubt and uncertainty. Far better, in my opinion, would be to say A is 2.5 times as large as B. Best, again in my opinion, would be to say A=2.5B. Grover Hughes === Subject: Re: SF: Definitely a new factoring method > It has been shown to work by a Tim Peters, who has also been certain > that it can't be made practical, but that is irrelevant to the question > of it definitely being a new factoring method. I've invented a new type of engine for my car. It's a sort of catapult that sits in the driveway and throws boulders at the back of the car. Each time it's hit by a boulder the car rolls forward a few feet. It's slow, works only over a short distance, and the wear and tear on the car is rather severe -- but that is irrelevant to the question of it definitely being a new method of propelling a car. How long do you suppose it will be before the car manufacturers start lying about it to suppress my discovery? -- Wayne Brown (HPCC #1104) fi.bes ofer.8eode, [CapitalYAcute]isses sw.87 m.beg. (That passed away, this also can.) -- Deor, from the Exeter Book (folios 100r-100v) === Subject: Re: SF: Definitely a new factoring method : Well I've thought about factoring for over three years now, and come up : with various approaches along the lines of what I call surrogate : factoring, and also found proofs for almost every one but this latest : that showed that none could work efficiently. The fact that none of your methods has ever factored anything of any consequence is proof enough that they're not efficient. This is not to say that they've been mathematically proved to be inefficient, just that nobody should feel the least bit pressured to prove anything about your approaches given your track record. : Just for my own curiosity then, as someone who plays with this stuff a : lot, I'd like to see you come up with just one new factoring method. Here's one I posted about three months ago but you ignored: Let T be the target. Assume T is odd. Let S=2T. Pick x a factor of S other than 2. Voila, a factor of T. Now, you may say that this is stupidly trivial but I'd like to point out that it has all the hallmarks of your approaches. It takes something hard to factor, T, introduces a new variable x with no guidance as to how to choose x, and then it goes on to show that if x can be found then I'm home free. Like you I could then go on to say that it's not up to me to implement it (maybe Tim Peters can do it) and it's not up to me to prove it's inefficient. It works, as you say, and I could therefore claim it's all about the pure math. Is it new? I challenge you to find one single reference to it in the literature. Justin === Subject: Re: SF: Definitely a new factoring method : with various approaches along the lines of what I call surrogate > : factoring, and also found proofs for almost every one but this latest > : that showed that none could work efficiently. > The fact that none of your methods has ever factored anything of any > consequence is proof enough that they're not efficient. This is not to > say that they've been mathematically proved to be inefficient, just that > nobody should feel the least bit pressured to prove anything about your > approaches given your track record. It turns out that I had other approaches which DID factor ok to some extent. But they weren't up to the standard I have of being able to factor an RSA number in 10 minutes or less on a home pc, so I dropped them. I've dumped factoring ideas which were not terribly bad because I knew they wouldn't meet that standard. > : Just for my own curiosity then, as someone who plays with this stuff a > : lot, I'd like to see you come up with just one new factoring method. > Here's one I posted about three months ago but you ignored: Let T be the > target. Assume T is odd. Let S=2T. Pick x a factor of S other than 2. > Voila, a factor of T. > Now, you may say that this is stupidly trivial but I'd like to point out > that it has all the hallmarks of your approaches. It takes something hard > to factor, T, introduces a new variable x with no guidance as to how to > choose x, and then it goes on to show that if x can be found then I'm home > free. You have no idea what you're talking about. I've had implemented algorithms that were not horrible, but they didn't excite me. Your problem I think is that you want to believe something is true which is not. Again, the standard for me in factoring approaches is being able to factor an RSA number in 10 minutes or less on a home pc. Anything beneath that I drop. I can do all kinds of things with mathematics, but I have to feel really good about any approach before I keep it. I've tossed more decent mathematical ideas than most people can ever come close to finding, but they didn't meet my standards. So as far as I'm concerned, they weren't worth my time. After all, there are an infinity of other ideas out there, so why not get EXACTLY what I want? James Harris === Subject: Re: SF: Definitely a new factoring method : It turns out that I had other approaches which DID factor ok to some : extent. I never said they didn't - I just said they weren't efficient and efficiency is what it's all about. : You have no idea what you're talking about. I've had implemented : algorithms that were not horrible, but they didn't excite me. Please explain. What's wrong with my factoring algorithm? If you can't stick to the math then don't say anything at all. : Again, the standard for me in factoring approaches is being able to : factor an RSA number in 10 minutes or less on a home pc. Sure, anyone else would love this, except neither you nor I have managed. : After all, there are an infinity of other ideas out there, so why not : get EXACTLY what I want? But you haven't. Justin === Subject: Re: SF: Definitely a new factoring method Hi- > It turns out that I had other approaches which DID factor ok to some > extent. > But they weren't up to the standard I have of being able to factor an > RSA number in 10 minutes or less on a home pc, so I dropped them. > I've dumped factoring ideas which were not terribly bad because I knew > they wouldn't meet that standard. Were they able to factor an RSA number in an hour? If so, could I have a look? Heck - even if they factor an RSA number in a *year* on a PC, I would love to see them! > : Just for my own curiosity then, as someone who plays with this stuff a > : lot, I'd like to see you come up with just one new factoring method. I have a couple (though I am *sure* they are not efficient) :) 1) The 'undoing multiplication' method: Suppose I want to factor something small, like 91. I know that the factors must be no more than two digits each (and more, that one is only one digit): x+y * z ____ yz mod 10 10xz+yz-(yzmod10) ____________ 91 so: 1 = yz mod 10 That means I have the possibilities of: 1*1, 3*7, 7*3, or 9*9 if z=1, then I get the trivial 91 = 1*91 if z=7, then I get yz = 21, and 7x+2 = 9, so x=1, or 91 =7*13 (clearly z!=3, nor 9) Of course- it gets *much* more difficult as the numbers get larger (as the number of possible cases grows very quickly). I think this method is probably slower than trial factorization by a *lot* - can anyone think of a more inefficient method? (Or am I wrong in my assumption of how slow it is? I have not yet learned how to estimate algorithm efficiencies (other than the minor bits of the error estimate for series expansions in freshman calculus...). 2) my other method is (I think) about as fast as Fermat's factorization method (and is what gave me the idea... Start with your number to factor- N (or T, if you prefer...) Let's look at the parabola: y=x^2 + (N+1)x +N. This parabola has zeros at 1 and N, and the interesting property that a line through the origin that intersects the parabola will do so in such a way that the x-coordinates of the point(s) of intersection will multiply together to make N. Just start with the tangent (whose intersection with the parabola will give you sqrt(N) (there are two, one gives you -sqrt(N) start with the other one!) Then just change the slope of the line until you find a pair of x's that are both integers. (Note - the slope will be integral too, so you only have to try a finite number of cases, and it seems clear that (as with Fermat's method), this works fastest when there are two factors near the square root on N. If anyone is really interested in this any further, I would be happy to discuss some thoughts I have on limiting the choices for the slope of that line, but I think of this as mainly a slightly interesting method only because I came up with it on my own, and it helped me visualize some aspects of factoring. Of course, this then led me to study more, and then I saw how the quadratic sieve works and am now trying to get a grip on the elliptic curve methods (which I expect to be able to understand a little better after finishing the abstract algebra courses I am going to be taking over the next year). Anyways- while novelty might be interesting, that does not guarantee that it has any value. cheers- Eric === Subject: Re: SF: Definitely a new factoring method > I will admit that if factoring methods are easy to come up with then > I'd be very interested in seeing people do it. Here's one I came up with. Suppose I want to factorize a number: A. I choose two numbers p and q such that pq>A. And I chose p and q so as to make pq-A quite small. Now, suppose A = (p-x)*(q-y) then A = pq-py-qx+xy so qx-xy = pq-py-A so x(q-y) = pq-A-py so x = (pq-A-py)/(q-y) Now, pq-A is quite small, so it is just a question of finding y to make (pq-A-py)/(q-y) a whole number. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: D-numbers: a generalization of Sophie Germain twin primes A twin prime is a prime p such that p+2 is prime and a Sophie Germain prime if 2p+1 is prime. Oberve that the pair p+2, 2p+1 as arises as p+d+1 for all divisors d of p. Thus, we have Definition. A positive integer satisfies Property D and is called a D-number if n+d+1 is prime for all divisors d of n. Let D be the set of all D-numbers. After watching a Maple program find and print the factored elements of this set to the screen one instantly observes that apparently the only integers that arise are 1, 9 (the only square), biprimes with two distinct factors, and triprimes of all possible types: p^3, p^2*q, and p*q*r. The first prime r such that 3*3*r is in D is r=370884, while the first prime r such that 3*7*r is in D is r=606619339. If you're interested, there are text files and conjectures here: http://glory.gc.maricopa.edu/~wkehowsk/propertyd/index.html Walter Kehowski === Subject: Re: D-numbers: a generalization of Sophie Germain twin primes
Walter,
Your definition of twin prime was not my understanding of this term, so I checked
and learned, to my surprise, that twin primes are pairs of
primes of the form (p,p+2).  Thus, a twin prime is not a prime
at all, but rather a certain type of an ordered pair of primes.
By your definition, 3, 5, and 11 are twin primes, but 7, 13, and 19
are not twin primes.
I do not understand your definition of a Sophie Germain prime.
I also not not understand as arises as.  Could you rephrase, please. 
-- Mark Spahn
message prime if 2p+1 is prime. Oberve that the pair p+2, 2p+1 as arises as
p+d+1 for all divisors d of p. Thus, we have

Definition. A positive integer satisfies Property D and is called a
D-number if n+d+1 is prime for all divisors d of n. Let D be the set of
all D-numbers.

After watching a Maple program find and print the factored elements of
this set to the screen one instantly observes that apparently the only
integers that arise are 1, 9 (the only square), biprimes with two
distinct factors, and triprimes of all possible types: p^3, p^2*q, and
p*q*r. The first prime r such that 3*3*r is in D is r=370884, while the
first prime r such that 3*7*r is in D is r=606619339.

If you're interested, there are text files and conjectures here:

=== Subject: Re: D-numbers: a generalization of Sophie Germain twin primes > more like. Don't use Horrible Traumatic Mess Language on usenet. Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: D-numbers: a generalization of Sophie Germain twin primes Hi Mark, I'm top-posting as your html post is *nearly* unreadable :( What you have said below is exactly what the OP posted, that twin-primes are two primes of the form p & p+2. You stated By your (the OP's) definition, 3, 5, and 11 are twin primes, but 7, 13, and 19 where what the OP actually stated was A twin prime is a prime p such that p+2 is prime. Although I'll admit this is *slightly* ambiguous, I'm sure that most readers would have got the exact meaning that if p is a prime and p+2 is a prime then they are twin primes. This continues to the Sophie Germain prime, where if p and 2p+1 are prime, then p is called a Sophie Germain prime. i.e. p=3 is a Sophie Germain prime because 2p+1 = 7 which is also prime. 7 is *not* a SG prime since p=7 gives 2p+1=15 which is not prime. So, a quick reiteration: twin primes are a *pair* of numbers where p and p+2 are prime; a Sophie Germain prime is a *single* prime p, where p and 2p+1 are prime. As for the OP's final ...arises as p+d+1 for all divisors d of p, I'm with you on that one, since if p is prime then it doesn't have *any* divisors d other than itself and unity. hth mori > Walter,
Your definition of twin prime was not my understanding of this > term, so I checked http://mathworld.wolfr am .com/TwinPrimes.htmland learned, to my surprise, that twin primes are pairs of
primes of > the form (p,p+2).  Thus, a twin prime is not a prime
at all, but rather > a certain type of an ordered pair of primes.
By your definition, 3, 5, and 11 > are twin primes, but 7, 13, and 19
are not twin primes.
I do not > understand your definition of a Sophie Germain prime.
I also not not > understand as arises as.  Could you rephrase, please. 
-- Mark > Spahn
style=PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px
<wkehowski@cox.net> > message twin prime is a prime p such that p+2 is prime and a Sophie Germain
prime > if 2p+1 is prime. Oberve that the pair p+2, 2p+1 as arises as
p+d+1 for all > divisors d of p. Thus, we have

Definition. A positive integer satisfies > Property D and is called a
D-number if n+d+1 is prime for all divisors d of > n. Let D be the set of
all D-numbers.

After watching a Maple program > find and print the factored elements of
this set to the screen one > instantly observes that apparently the only
integers that arise are 1, 9 > (the only square), biprimes with two
distinct factors, and triprimes of all > possible types: p^3, p^2*q, and
p*q*r. The first prime r such that 3*3*r is > in D is r=370884, while the
first prime r such that 3*7*r is in D is > r=606619339.

If you're interested, there are text files and conjectures > here:

http:// gl
Walter > Kehowski
=== Subject: Re: D-numbers: a generalization of Sophie Germain twin primes boundary=----=_NextPart_000_001B_01C6AC0B.32CB2670 --------------------------------------------------------------------- Hi Mori, to make it more readable ('plain text'), but it had the opposite effect. I noticed later that the OP's website had a clearer definition than his post: A prime p is a twin prime iff p+2 is also prime. Thus, by the OP's definition, 41 is a twin prime but 43 is not. This definition, even if non-standard, makes sense, given his further remarks. A Sophie Germain (who dat?) prime is a prime p for which 2p+1 is also a prime. Thus a Sophie Germain twin prime is a prime p for which both p+2 and 2p+1 are also primes. For any prime p, if d>0 and d|p, then d=1 or p, and the set {p+d+1 such that d>0 and d|p} = {p+1+1, p+p+1} = {p+2, 2p+1}, and iff these two numbers are also prime, then p is a Sophie Germain twin prime. Mark Spahn Hi Mark, I'm top-posting as your html post is *nearly* unreadable :( What you have said below is exactly what the OP posted, that twin-primes are two primes of the form p & p+2. You stated By your (the OP's) definition, 3, 5, and 11 are twin primes, but 7, 13, and 19 where what the OP actually stated was A twin prime is a prime p such that p+2 is prime. Although I'll admit this is *slightly* ambiguous, I'm sure that most readers would have got the exact meaning that if p is a prime and p+2 is a prime then they are twin primes. This continues to the Sophie Germain prime, where if p and 2p+1 are prime, then p is called a Sophie Germain prime. i.e. p=3 is a Sophie Germain prime because 2p+1 = 7 which is also prime. 7 is *not* a SG prime since p=7 gives 2p+1=15 which is not prime. So, a quick reiteration: twin primes are a *pair* of numbers where p and p+2 are prime; a Sophie Germain prime is a *single* prime p, where p and 2p+1 are prime. As for the OP's final ...arises as p+d+1 for all divisors d of p, I'm with you on that one, since if p is prime then it doesn't have *any* divisors d other than itself and unity. hth mori === Subject: Re: D-numbers: a generalization of Sophie Germain twin primes > As for the OP's final ...arises as p+d+1 for all divisors d of p, I'm with > you on that one, since if p is prime then it doesn't have *any* divisors d > other than itself and unity. > hth > mori That is : for some n : P = { n+d+1 : d divides n and d>1 and d As for the OP's final ...arises as p+d+1 for all divisors d of p, I'm with > you on that one, since if p is prime then it doesn't have *any* divisors d > other than itself and unity. > hth > mori > That is : > for some n : P = { n+d+1 : d divides n and d>1 and d and of interest : for each prime in P the set P can be given. > Chris === Subject: Re: D-numbers: a generalization of Sophie Germain twin primes n=P and p=[n=P] lists : Here are n=P values : 4 = [7] 8 = [11, 13] 9 = [13] 25 = [31] 27 = [31, 37] 35 = [41, 43] 39 = [43, 53] 55 = [61, 67] 65 = [71, 79] 119 = [127, 137] 125 = [131, 151] 185 = [191, 223] 203 = [211, 233] 219 = [223, 293] 235 = [241, 283] 237 = [241, 317] 289 = [307] 305 = [311, 367] 319 = [331, 349] 341 = [353, 373] 415 = [421, 499] 417 = [421, 557] 437 = [457, 461] 515 = [521, 619] 535 = [541, 643] 597 = [601, 797] 649 = [661, 709] 655 = [661, 787] 671 = [683, 733] 685 = [691, 823] 749 = [757, 857] 755 = [761, 907] 905 = [911, 1087] 935 = [941, 947, 953, 991, 1021, 1123] 959 = [967, 1097] 979 = [991, 1069] 989 = [1013, 1033] 1003 = [1021, 1063] 1043 = [1051, 1193] 1079 = [1093, 1163] 1111 = [1123, 1213] 1119 = [1123, 1493] 1165 = [1171, 1399] 1227 = [1231, 1637] 1247 = [1277, 1291] 1285 = [1291, 1543] 1299 = [1303, 1733] 1315 = [1321, 1579] 1343 = [1361, 1423] 1355 = [1361, 1627] 1465 = [1471, 1759] 1469 = [1483, 1583] 1565 = [1571, 1879] 1649 = [1667, 1747] 1681 = [1723] 1735 = [1741, 2083] 1739 = [1777, 1787] 1829 = [1861, 1889] 1857 = [1861, 2477] 1865 = [1871, 2239] 1909 = [1933, 1993] 1937 = [1951, 2087] 1943 = [1973, 2011] 2033 = [2053, 2141] 2059 = [2089, 2131] 2101 = [2113, 2293] 2127 = [2131, 2837] 2183 = [2221, 2243] 2215 = [2221, 2659] 2217 = [2221, 2957] 2305 = [2311, 2767] 2329 = [2347, 2467] 2335 = [2341, 2803] 2363 = [2381, 2503] 2369 = [2393, 2473] 2429 = [2437, 2777] 2507 = [2531, 2617] 2515 = [2521, 3019] 2519 = [2531, 2749] 2533 = [2551, 2683] And p=[n=P] values : 7 = [4 = [7]] 11 = [8 = [11, 13]] 13 = [8 = [11, 13], 9 = [13]] 31 = [25 = [31], 27 = [31, 37]] 37 = [27 = [31, 37]] 41 = [35 = [41, 43]] 43 = [35 = [41, 43], 39 = [43, 53]] 53 = [39 = [43, 53]] 61 = [55 = [61, 67]] 67 = [55 = [61, 67]] 71 = [65 = [71, 79]] 79 = [65 = [71, 79]] 127 = [119 = [127, 137]] 131 = [125 = [131, 151]] 137 = [119 = [127, 137]] 151 = [125 = [131, 151]] 191 = [185 = [191, 223]] 211 = [203 = [211, 233]] 223 = [185 = [191, 223], 219 = [223, 293]] 233 = [203 = [211, 233]] 241 = [235 = [241, 283], 237 = [241, 317]] 283 = [235 = [241, 283]] 293 = [219 = [223, 293]] 307 = [289 = [307]] 311 = [305 = [311, 367]] 317 = [237 = [241, 317]] 331 = [319 = [331, 349]] 349 = [319 = [331, 349]] 353 = [341 = [353, 373]] 367 = [305 = [311, 367]] 373 = [341 = [353, 373]] 421 = [415 = [421, 499], 417 = [421, 557]] 457 = [437 = [457, 461]] 461 = [437 = [457, 461]] 499 = [415 = [421, 499]] 521 = [515 = [521, 619]] 541 = [535 = [541, 643]] 557 = [417 = [421, 557]] 601 = [597 = [601, 797]] 619 = [515 = [521, 619]] 643 = [535 = [541, 643]] 661 = [649 = [661, 709], 655 = [661, 787]] 683 = [671 = [683, 733]] 691 = [685 = [691, 823]] 709 = [649 = [661, 709]] 733 = [671 = [683, 733]] 757 = [749 = [757, 857]] 761 = [755 = [761, 907]] 787 = [655 = [661, 787]] 797 = [597 = [601, 797]] 823 = [685 = [691, 823]] 857 = [749 = [757, 857]] 907 = [755 = [761, 907]] 911 = [905 = [911, 1087]] 941 = [935 = [941, 947, 953, 991, 1021, 1123]] 947 = [935 = [941, 947, 953, 991, 1021, 1123]] 953 = [935 = [941, 947, 953, 991, 1021, 1123]] 967 = [959 = [967, 1097]] 991 = [935 = [941, 947, 953, 991, 1021, 1123], 979 = [991, 1069]] 1013 = [989 = [1013, 1033]] 1021 = [935 = [941, 947, 953, 991, 1021, 1123], 1003 = [1021, 1063]] 1033 = [989 = [1013, 1033]] 1051 = [1043 = [1051, 1193]] 1063 = [1003 = [1021, 1063]] 1069 = [979 = [991, 1069]] 1087 = [905 = [911, 1087]] 1093 = [1079 = [1093, 1163]] 1097 = [959 = [967, 1097]] 1123 = [935 = [941, 947, 953, 991, 1021, 1123], 1111 = [1123, 1213], 1119 = [1123, 1493]] 1163 = [1079 = [1093, 1163]] 1171 = [1165 = [1171, 1399]] 1193 = [1043 = [1051, 1193]] 1213 = [1111 = [1123, 1213]] 1231 = [1227 = [1231, 1637]] 1277 = [1247 = [1277, 1291]] 1291 = [1247 = [1277, 1291], 1285 = [1291, 1543]] 1303 = [1299 = [1303, 1733]] 1321 = [1315 = [1321, 1579]] 1361 = [1343 = [1361, 1423], 1355 = [1361, 1627]] 1399 = [1165 = [1171, 1399]] 1423 = [1343 = [1361, 1423]] 1471 = [1465 = [1471, 1759]] 1483 = [1469 = [1483, 1583]] 1493 = [1119 = [1123, 1493]] 1543 = [1285 = [1291, 1543]] 1571 = [1565 = [1571, 1879]] 1579 = [1315 = [1321, 1579]] 1583 = [1469 = [1483, 1583]] 1627 = [1355 = [1361, 1627]] 1637 = [1227 = [1231, 1637]] 1667 = [1649 = [1667, 1747]] 1723 = [1681 = [1723]] 1733 = [1299 = [1303, 1733]] 1741 = [1735 = [1741, 2083]] 1747 = [1649 = [1667, 1747]] 1759 = [1465 = [1471, 1759]] 1777 = [1739 = [1777, 1787]] 1787 = [1739 = [1777, 1787]] 1861 = [1829 = [1861, 1889], 1857 = [1861, 2477]] 1879 = [1565 = [1571, 1879]] 1889 = [1829 = [1861, 1889]] 1933 = [1909 = [1933, 1993]] 1951 = [1937 = [1951, 2087]] 1973 = [1943 = [1973, 2011]] 1993 = [1909 = [1933, 1993]] 2011 = [1943 = [1973, 2011]] 2053 = [2033 = [2053, 2141]] 2083 = [1735 = [1741, 2083]] 2087 = [1937 = [1951, 2087]] 2089 = [2059 = [2089, 2131]] 2113 = [2101 = [2113, 2293]] 2131 = [2059 = [2089, 2131], 2127 = [2131, 2837]] 2141 = [2033 = [2053, 2141]] 2221 = [2183 = [2221, 2243], 2215 = [2221, 2659], 2217 = [2221, 2957]] 2239 = [1865 = [1871, 2239]] 2243 = [2183 = [2221, 2243]] 2293 = [2101 = [2113, 2293]] 2311 = [2305 = [2311, 2767]] 2341 = [2335 = [2341, 2803]] 2347 = [2329 = [2347, 2467]] 2381 = [2363 = [2381, 2503]] 2393 = [2369 = [2393, 2473]] 2437 = [2429 = [2437, 2777]] 2467 = [2329 = [2347, 2467]] 2473 = [2369 = [2393, 2473]] 2477 = [1857 = [1861, 2477]] 2503 = [2363 = [2381, 2503]] 2521 = [2515 = [2521, 3019]] 2531 = [2507 = [2531, 2617], 2519 = [2531, 2749]] 2551 = [2533 = [2551, 2683]] 2617 = [2507 = [2531, 2617]] 2659 = [2215 = [2221, 2659]] 2683 = [2533 = [2551, 2683]] 2749 = [2519 = [2531, 2749]] 2767 = [2305 = [2311, 2767]] 2777 = [2429 = [2437, 2777]] 2803 = [2335 = [2341, 2803]] 2837 = [2127 = [2131, 2837]] 2957 = [2217 = [2221, 2957]] 3019 = [2515 = [2521, 3019]] Chris > As for the OP's final ...arises as p+d+1 for all divisors d of p, I'm with > you on that one, since if p is prime then it doesn't have *any* divisors d > other than itself and unity. > hth > mori > That is : > for some n : P = { n+d+1 : d divides n and d>1 and d and of interest : for each prime in P the set P can be given. > Chris === Subject: Re: D-numbers: a generalization of Sophie Germain twin primes s.m.r. dropped - can't post to moderated groups. f/u set to a.m.r. > A twin prime is a prime p such that p+2 is prime and a Sophie Germain > prime if 2p+1 is prime. Oberve that the pair p+2, 2p+1 as arises as > p+d+1 for all divisors d of p. Thus, we have > Definition. A positive integer satisfies Property D and is called a > D-number if n+d+1 is prime for all divisors d of n. Let D be the set of > all D-numbers. > After watching a Maple program find and print the factored elements of > this set to the screen one instantly observes that apparently the only > integers that arise are 1, 9 (the only square), biprimes with two > distinct factors, and triprimes of all possible types: p^3, p^2*q, and > p*q*r. The first prime r such that 3*3*r is in D is r=370884, while the > first prime r such that 3*7*r is in D is r=606619339. > If you're interested, there are text files and conjectures here: > http://glory.gc.maricopa.edu/~wkehowsk/propertyd/index.html No further squares. Look at the >=9 values modulo 3. (you only need to look at the +1+1 and +3+1 values) Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer to this thread later, in case I have a tedious and tawdry day at work... === Subject: Re: SUM EQUAL TO PRODUCT, REALLY positive integers that can be written as the sum and product of the same set of positive rational numbers. I solved it almost completely: 1, 2, 3, and 5 are impossible; any even n greater than two is n/2 + 2 + 1 + 1 + ...; and any odd n greater than seven is n/2 + 4 + 1/2 + 1 + 1 + ... I spent 3 of the allotted 4.5 hours looking for a solution for 7 and trying to prove there wasn't one. One solution is 7/6 + 9/2 + 4/3. > Actually, if you consider 4 numbers, each ranging from 0.1 to 10.0 in > increments of 0.1, then there are 51 sets where all 4 are different. > On 14 Jul 2006 00:43:09 -0700, Rajnish Kumar >There is an interesting property of pair of numbers such that the >product of the number is equal to their sum. >For example >3 and 1.5 3x1.5=4.5 3+1.5=4.5 >3.5 and 1.4 3.5x1.4=4.9 3.5+1.4=4.9 >Well is it also possible that the sum of three numbers is equal to >their product? >Lets see >7.5, 1.66 and 0.8, both the sum and product are 9.96. >Try finding out some more numbers (without reading the next line). >Here is a famous 7-11 store problem. >In a seven-eleven (7-11) store, a customer selected four items to >buy. The cashier at the counter says that he multiplied the costs of >the items and obtained exactly 7.11, the very name of the store! The >customer angered tells him to add and not multiply. The total still was >Rs 7.11. >What are the exact costs of the 4 items? >Sol: 3.16 + 1.25 + 1.50 + 1.20 >There are a few more numbers like these, >Check for >6.44 = 1.84+1.75+1.60+1.25 >6.51 = 2.00+1.86+1.40+1.25 >Keep trying you will find lots of such numbers. Mathematicate and >meditate. === Subject: Re: WHY IS ELEVEN NOT ONETEEN AND TWELVE NOT TWOTEEN? That's more of a linguistic question than a mathematical one. It's the same in German: 10 = zehn; 11 = elf; 12 = zw.9alf; 13 = dreizehn, 14 = vierzehn, etc. -Ravi > yellin so you could listen > On 14 Jul 2006 01:59:31 -0700, Rajnish Kumar >WHY IS ELEVEN NOT ONETEEN AND TWELVE NOT TWOTEEN? > > Why are you YELLING? > > Lynn === Subject: JSH: Gloves are off now http://jstevh.blogspot.com/ oops If you ing morons think that I will let you get away with not giving me credit for my ing math discoveries then you have another ing thing coming. What the ??!!! Where's pure math now, huh? Where's loving math for the ing beauty of it now you ing s??!!! LOOK AT IT!!! Here is the partial difference equation and instructions for integrating. dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0. And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS >from dS(x,2) to dS(x,y). http://mathforprofit.blogspot.com/ You ing s. I will get credit for my discovery and get ing paid, and you best believe that I will not ing let you stupid s get away with your ing stupid bull--pure math my ass--without me coming at you ers with some ing PURE ing PURE AS math that you stupid s have been ting on for over a ing YEAR!!! You goddamn S!!! What the is wrong with you s??!!! Don't you even believe in your own stupid ? Where's pure math now? Where is it? I am tired of your old math ways The gloves are off. James Harris === Subject: Re: JSH: Gloves are off now > http://jstevh.blogspot.com/ > oops > If you ing morons think that I will let you get away with not > giving me credit for my ing math discoveries then you have another > ing thing coming. Here we have a particular sub-syndrome of the JSH addiction ... .... someone who pretends to be JSH, for the purpose of mocking JSH and getting a few readers to think it actually is JSH. In this case, the addict has not even tried very hard to emulate JSH's distinct and recognizable style and tone. A lazy try, devoid of both entertainment and satiric value. Next. === Subject: Re: JSH: Gloves are off now >> http://jstevh.blogspot.com/ >> oops >> If you ing morons think that I will let you get away with not >> giving me credit for my ing math discoveries then you have another >> ing thing coming. > Here we have a particular sub-syndrome of the JSH addiction ... > .... someone who pretends to be JSH, for the purpose of mocking JSH and > getting a few readers to think it actually is JSH. > In this case, the addict has not even tried very hard to emulate JSH's > distinct and recognizable style and tone. A lazy try, devoid of both > entertainment and satiric value. > Next. but how do you explain the blog ? http://jstevh.blogspot.com/ === Subject: Re: JSH: Gloves are off now days. My association with the Department is that of an alumnus. > http://jstevh.blogspot.com/ > oops > If you ing morons think that I will let you get away with not > giving me credit for my ing math discoveries then you have another > ing thing coming. >> Here we have a particular sub-syndrome of the JSH addiction ... >> .... someone who pretends to be JSH, for the purpose of mocking JSH and >> getting a few readers to think it actually is JSH. >> In this case, the addict has not even tried very hard to emulate JSH's >> distinct and recognizable style and tone. A lazy try, devoid of both >> entertainment and satiric value. >> Next. >but how do you explain the blog ? > http://jstevh.blogspot.com/ It used to be James's, parallel with mymath.blogspot.com. He slowed down his postings there (which were non-mathematical); then he stopped posting entirely. Last post there, I think, was about german scientists doing something with mice. Then he deleted all the content and deleted the blog. Within two days, someone else took it over; same person who took over the previous James blog when the latter abandoned it, and did the same thing with it as with the current one, probably. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin === Subject: Re: JSH: Gloves are off now <44bf07e1$0$17974$892e7fe2@authen.yellow.readfreenews.net> Any of you maggot brain ever considered the that real JSH not really JSH, and that the fake JSH actually being the real JSH. By purpose he let the real JSH tap into alot of info to make the real JSH start mocking him so the fake JSH can find out who are behind real JSH. I beleive the fake JSH, if he exist is more real and alot more clever than the real JSH probably ever would wish or like to beleive. I think the fake JSH did let the real JSH tap in upon information that on the core level really undeniable a correct truth. That way he could let the real JSH spread the core of the theory without actually dealing with it, because the real JSH would do it just to mock the fake JSH. And fake JSH did know from start it was because he probably solved the problem and did sell the solution to the russian or chinese intelligence. Maybe the fake JSH just did want to divert the interest from his person to the real JSH. And for years now the west intelligence wondered what sick ers at their universities who did sell them out by solving the factorisation problem and sell out the solution, and how those ers who keep mocking the CIA and NSA could be so bold they keep a blog doing it. And they know they can not go after them because these people are a bunch of respected professors at universities all over west, who gathered together to mock fake JSH. And right now NSA and CIA have their nose so long up these people butts they actually can se when these people make a phone call. This all would probably be true if there was a fake JSH, but i think he is just a myth just like the real JSH, one of those theories that happen to be true only and only if the the actual postulates/premises also are. And who actually beleive any of the JSH premises/postulates is consistent with either real or fake JSH, come on guys he doesn't exist. I think we can conclude though that both NSA and CIA are for real. JT Arturo Magidin skrev: > http://jstevh.blogspot.com/ > oops > If you ing morons think that I will let you get away with not > giving me credit for my ing math discoveries then you have another > ing thing coming. >> Here we have a particular sub-syndrome of the JSH addiction ... >> .... someone who pretends to be JSH, for the purpose of mocking JSH and >> getting a few readers to think it actually is JSH. >> In this case, the addict has not even tried very hard to emulate JSH's >> distinct and recognizable style and tone. A lazy try, devoid of both >> entertainment and satiric value. >> Next. >but how do you explain the blog ? > http://jstevh.blogspot.com/ > It used to be James's, parallel with mymath.blogspot.com. > He slowed down his postings there (which were non-mathematical); then > he stopped posting entirely. Last post there, I think, was about > german scientists doing something with mice. Then he deleted all the > content and deleted the blog. Within two days, someone else took it > over; same person who took over the previous James blog when the > latter abandoned it, and did the same thing with it as with the > current one, probably. > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes by Bill Watterson) > Arturo Magidin === Subject: Re: JSH: Gloves are off now insane babblingCIA are for real. hmmmmm, but how can we be sureË mori > JT > Arturo Magidin skrev: > <44beff8a$0$17983$892e7fe2@authen.yellow.readfreenews.net>, > http://jstevh.blogspot.com/ > oops > If you ing morons think that I will let you get away with not > giving me credit for my ing math discoveries then you have another > ing thing coming. >> Here we have a particular sub-syndrome of the JSH addiction ... >> .... someone who pretends to be JSH, for the purpose of mocking JSH and >> getting a few readers to think it actually is JSH. >> In this case, the addict has not even tried very hard to emulate JSH's >> distinct and recognizable style and tone. A lazy try, devoid of both >> entertainment and satiric value. >> Next. >but how do you explain the blog ? > http://jstevh.blogspot.com/ > It used to be James's, parallel with mymath.blogspot.com. > He slowed down his postings there (which were non-mathematical); then > he stopped posting entirely. Last post there, I think, was about > german scientists doing something with mice. Then he deleted all the > content and deleted the blog. Within two days, someone else took it > over; same person who took over the previous James blog when the > latter abandoned it, and did the same thing with it as with the > current one, probably. > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes by Bill Watterson) > Arturo Magidin === Subject: Re: JSH: Gloves are off now <44bf07e1$0$17974$892e7fe2@authen.yellow.readfreenews.net> moriman skrev: > insane babbling I think we can conclude though that both NSA and >CIA are for real. > hmmmmm, but how can we be sureË To be honest actually you can't you will have to live with the uncertanty that you maybe all are fools, that will be laughed at by the afterworld. Isn't that beautiful... you would have to solve the problem for yourself to really know if the nonexistant JSH really is a drivling madman and if the NSA and CIA have their noses so firmly secured up your ass that you not can go and still be total certain that you REALLY alone. Well deal with it but the beauty with life is all about uncertainty. JT > mori > JT > Arturo Magidin skrev: > <44beff8a$0$17983$892e7fe2@authen.yellow.readfreenews.net>, > http://jstevh.blogspot.com/ > oops > If you ing morons think that I will let you get away with not > giving me credit for my ing math discoveries then you have > another > ing thing coming. >> Here we have a particular sub-syndrome of the JSH addiction ... >> .... someone who pretends to be JSH, for the purpose of mocking JSH > and >> getting a few readers to think it actually is JSH. >> In this case, the addict has not even tried very hard to emulate > JSH's >> distinct and recognizable style and tone. A lazy try, devoid of both >> entertainment and satiric value. >> Next. >but how do you explain the blog ? > http://jstevh.blogspot.com/ > It used to be James's, parallel with mymath.blogspot.com. > He slowed down his postings there (which were non-mathematical); then > he stopped posting entirely. Last post there, I think, was about > german scientists doing something with mice. Then he deleted all the > content and deleted the blog. Within two days, someone else took it > over; same person who took over the previous James blog when the > latter abandoned it, and did the same thing with it as with the > current one, probably. > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes by Bill Watterson) > Arturo Magidin === Subject: Re: JSH: Gloves are off now > http://jstevh.blogspot.com/ > oops > If you ing morons think that I will let you get away with not > giving me credit for my ing math discoveries then you have another > ing thing coming. >> Here we have a particular sub-syndrome of the JSH addiction ... >> .... someone who pretends to be JSH, for the purpose of mocking JSH and >> getting a few readers to think it actually is JSH. >> In this case, the addict has not even tried very hard to emulate JSH's >> distinct and recognizable style and tone. A lazy try, devoid of both >> entertainment and satiric value. >> Next. > but how do you explain the blog ? > http://jstevh.blogspot.com/ I think this was hijacked many months ago. However, I do think that parts of this, if not all, may have been the fake copying from the genuine to pretend to be the genuine JSH. Dale === Subject: Re: JSH: Gloves are off now > However, I do think that parts of this, if not all, may have been > the fake copying from the genuine to pretend to be the genuine JSH. Accept no imitations! It's impossible to improve on the original. === Subject: Re: JSH: Gloves are off now >> http://jstevh.blogspot.com/ >> oops >> If you ing morons think that I will let you get away with not >> giving me credit for my ing math discoveries then you have another >> ing thing coming. > Here we have a particular sub-syndrome of the JSH addiction ... > .... someone who pretends to be JSH, for the purpose of mocking JSH and > getting a few readers to think it actually is JSH. > In this case, the addict has not even tried very hard to emulate JSH's > distinct and recognizable style and tone. A lazy try, devoid of both > entertainment and satiric value. > Next. >> but how do you explain the blog ? >> http://jstevh.blogspot.com/ > I think this was hijacked many months ago. > However, I do think that parts of this, if not all, may have been > the fake copying from the genuine to pretend to be the genuine JSH. Surrogate factoring was used to break the RSA key that kept the password encrypted -- LTP for( Base i : allYourBase) i.AreBelongTo(us); === Subject: Re: JSH: Gloves are off now i just wanted to say... wow. -ph === Subject: cold pi day 22/7 a good time to not bet any more. link. links. nz loto launched 22/7/1987 decimal currency day 10/7/1967 withdraw 5c resize 10 20 50c new zealand coins. July 2006. 8*5c == 40 c compass 50 cent around. do not bet no gamble a good time to not bet any more. 20C or $2 ?? lotto po box 3145 = pi opus central = po box 31416 = pi. no plate ?? xc 3141 ?? newtown. pi 180 pi 315 ? ln 2718 constable. log cabin. exp. don.mcdonald. 20.7.06 === link. links. cold pi day 22/7. a good time to not bet any more. dust and bus research. ... nz loto launched 22/7/1987 goes on sale for 1st time. decimal currency day 10/7/1967 withdraw 5c resize 10 20 50c new zealand coins. July 2006. 8*5c == 40 c compass 50 cent around. oo o()o o()o oo diameter. do not bet no gamble a good time to not bet any more. 20C or $2 ?? lotto po box 3145 = pi opus central = po box 31416 behavioural sciences psychology, jared thomas bus and train public transport research dompost. wed 12.7.06?? about. dompost. = pi. no plate ?? xc 3141 ?? newtown. pi 180 white mansfield st pi 315 ? wgtn show bldg, lr hutt. ln 2718 constable st. log cabin. exp. this week higher education research dompost WGTN NZ. WHAT!! photo, blow dust 1000x 1000 pixels. millions in one breath. chalk dust, sunlight. wat about every human breath contains billion atoms from isaac newton's last breath no. perhaps only a dozen. nz.general i will try to search url. yes. this persuaded me.. life chemistry...??? don.mcdonald. 20.7.06 === Subject: area of a secular sector I need an area formula for a particular problem I have. Given a circle of radius R and a chord C, I need a formula for the area of the circular segment only a function of R and C. The circular segment is that area OUTSIDE of the chord yet inscribed within the circle. It seems to me this is all the information you would need since the circle becomes unique once R is known. And, once the length of C is known, no other information should be necessary to solve for the circular segment. David Emerling Memphis, TN === Subject: Re: area of a secular sector > I need an area formula for a particular problem I have. > Given a circle of radius R and a chord C, I need a formula for the area > of the circular segment only a function of R and C. > The circular segment is that area OUTSIDE of the chord yet inscribed > within the circle. Bear in mind that every chord divides the circle into _two_ regions, each of which is a circular segment. You'll have to decide for yourself which of those is the secular one. ;-) > It seems to me this is all the information you would need since the > circle becomes unique once R is known. And, once the length of C is > known, no other information should be necessary to solve for the > circular segment. See , so you'll find theta first and then K. That's the area of the smaller segment. If you wanted the area of the larger segment, just subtract the area of the smaller segment from that of the full circle. David === Subject: Re: area of a secular sector > I need an area formula for a particular problem I have. > Given a circle of radius R and a chord C, I need a formula for the area of > the circular segment only a function of R and C. > The circular segment is that area OUTSIDE of the chord yet inscribed > within the circle. > It seems to me this is all the information you would need since the circle > becomes unique once R is known. And, once the length of C is known, no other > information should be necessary to solve for the circular segment. > David Emerling > Memphis, TN The area of the sector is R^2 * arcsin(C/(2*R)), and the area of the triangle formed by the chord and the centre of the circle is C/2 * Sqrt(R^2 - C^2/4), so subtract one from t'other to get the area of the segment as R^2 * arcsin(C/(2*R)) - C/2 * Sqrt(R^2 - C^2/4) where the angle returned by the arcsin function is assumed to be in radians (not degrees). (To convert degrees to radians multiply by pi/180). === Subject: Re: area of a secular sector Oops! The title of the message was screwed up. Pretty funny! I meant The area of a circular segment. I'm not sure what a seculr sector would be, perhaps of group of people with no religious affiliation. :) >I need an area formula for a particular problem I have. > Given a circle of radius R and a chord C, I need a formula for the area of > the circular segment only a function of R and C. > The circular segment is that area OUTSIDE of the chord yet inscribed > within the circle. > It seems to me this is all the information you would need since the circle > becomes unique once R is known. And, once the length of C is known, no > other information should be necessary to solve for the circular segment. > David Emerling > Memphis, TN === Subject: Re: area of a secular sector >Oops! >The title of the message was screwed up. >Pretty funny! >I meant The area of a circular segment. >I'm not sure what a seculr sector would be, perhaps of group of people >with no religious affiliation. :) >>I need an area formula for a particular problem I have. >> Given a circle of radius R and a chord C, I need a formula for the area of >> the circular segment only a function of R and C. >> The circular segment is that area OUTSIDE of the chord yet inscribed >> within the circle. >> It seems to me this is all the information you would need since the circle >> becomes unique once R is known. And, once the length of C is known, no >> other information should be necessary to solve for the circular segment. >> David Emerling >> Memphis, TN Answers have already been posted by others. FWIW, here are two more links that might also be of interest: