mm-3779 === Subject: Re: Use of variable independence, core error > ... do the algebraic > integers indeed form a ring? > I've never said that algebraic integers don't form a ring. > What they form is a flawed ring, which doesn't include all the numbers > it must to prevent the possibility of appearing to prove two different > but opposite things. Well, that seems clear enough... but... is it possible that the flawed ring of algebraic integers includes too many numbers? Is that a possibility? === Subject: Re: JSH: About time > If your logic were correct it would apply > to other functions. > Yup. > Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Say Q(m) is factored in the form Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). Now note Q(0)/f^2 = x + f. That is, Q(0) = f * f * (x + f). Therefore when m = 0, we can say a1 = 0, a2 = 0, > and a3 = 1. Note that when m = 0, a1 and a2 are divisible by f. > Yup. > Now by your logic, for values of m other than 0, > a1 and a2 must be divisible by f and a3 is relatively > prime to f. Do you agree with this? This is a test of your > method. We need to know before we go to the next step. Nora B. > The answer is that yes, but with the qualification that m not equal 1, > but I have a better explanation for why than the wacky reply when I > initially freaked out. > Look again at your > Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). > Here the constant terms for the factors are revealed to be > f, f and x + f > where it just so happens that for your a3 you have something > like--yup, I know some will probably not like this, but it's the > reality--introducing h for the functions h(m), > a3 = h_1(m) - mx h_2(m) + x + f > where h_(1) has f^{2/3} as a factor and h_2(1)=1. > And yes, the same thing can happen with my argument, but it requires > mf^2 = 1 > but both m and f are integers in that argument, so that condition > doesn't occur. You have suggested elsewhere that I need to solve for the a's with expression. Let's do that. Recall that Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Let m = 1: [1] Q(1) = f^2*x^3 + f^3 = f^2*(x^3 + f). Assume this is factored in the form Q(1) = (a1*x + f)*(a2*x + f)*(a3*x + f). The roots of this polynomial are [2] r1 = -f/a1, r2 = -f/a2, and r3 = -f/a3. Of course from [1] I can compute the roots explicitly. They are: r1 = -f^{1/3} r2 = -f^{1/3}*(-1 + sqrt(3))/2 and r3 = -f^{1/3}*(-1 - sqrt(3))/2. and a3, since ai = -f/ri : a1 = f^{2/3}, a2 = f^{2/3} * (-1 - sqrt(3))/2, and a3 = f^{2/3} * (-1 + sqrt(3))/2. Note that a1, a2, and a3 are all algebraic integers. Note that each of them has a factor of f^{2/3}. Note that none of them have f as a factor. No surprises here. What has this got to do with your wanting m*f^2 to equal 1 ? > (If I didn't say they were before, well they are now.) > For the more adventurous, check Q(m) with m NOT equal to 1, and yes, > you will find that *two* of the a's have a factor that is f. See my other post on this. Nora B. PS: Oh yes, you also asked if I would be convinced if a computer verifies your proof. I would say *no*. I know exactly what is wrong with your proof already. If someone said a computer verified your proof, I would have to have a guarantee of at least three things: (1) That the computer program and the computing hardware were infallible. (2) That the translation of your proof into the program's language was done correctly. (3) That the person reporting the results could be trusted. All of these three conditions will be harder to ascertain than what I already know about your proof. I cannot see how (1) can be guaranteed at all. As for (2), checking that would probably considerably more tedious than what I have already done. And for (3), the most likely person to report positive results would be the least trustworthy. So no. But don't let me stop you. Go ahead and carry it out. Maybe you will convince somebody else. > It might be a fun exercise, assuming I didn't miss something. > If any of you out there think you can find fault with my conclusion > here, please try. > It is math after all. And it's also a lot of fun. > James Harris === === Subject: Re: Use of variable independence, core error ... > Yes, and so what? Algebraic integers are also roots of both monic and > non-monic polynomials. The definition of algebraic integers only states > that an algebraic number is an algebraic integer only if it is a root of > at least one monic polynomial. > However, as repeatedly shown by various posters on this board, > algebraic integers cannot be roots of non-monic polynomials with > integer coefficients that are irreducible over Q. (That is now what was shown, you should also state 'primitive'.) Normal integers cannot be roots of non-monic polynomials with integer coefficients that are primitive and irreducible over Q. > What asymmetry? Note that the distinction between integers and true > rationals is that rationals are *not* a root of a monic polynomial with > integer coefficients. The same distinction we have between algebraic > integers and other algebraic numbers. > However, as repeatedly shown by various posters on this board, > algebraic integers cannot be roots of non-monic polynomials with > integer coefficients that are irreducible over Q. See above. > This is nonsense. The definition of algebraic integers only label a > set of numbers with the name algebraic integers. How can such naming > lead to a contradiction? The name could also have been numbers that > are roots of monic polynomials. Would that name lead to a > contradiction? > It's not the name; it's the exclusion. You have the exclusion wrong. > The only possibility is that in your contradiction you use a result > on algebraic integers that is wrong. Which is either a theorem you > use or your own logic. So you should identify the theorem you are > using that is wrong. The definition in and of itself is *not* wrong, > it just gives a labeling. > But my proof is very short, which means it is machine checkable. I have not yet seen a version that is machine chackable. > If the factors were polynomials, there'd be no argument, as who out > there believes that you can have a polynomial P(x), a polynomial > factor f(x) of P(x), where P(x) has 7 as a factor, for all algebraic > integers x, and f(0)=7, that 7 can be divided off such that P(0) is > coprime to 7, without f(x) having 7 as a factor? > It looks complicated, but *writing* it out with actual polynomials > makes it easy. > However, with non-polynomial factors you have to trust the > mathematical logic. That is your problem. You attempt to apply what holds for polynomial factors also does apply for non-polynomial factors. That is false. P(x) = (x + 1)(x + 2) is (whenever x is an integer) *always* divisible by 2. But 2 is not a polynomial factor of P(x), so it is never the same factor of P(x) that is divisible by 2, it is either (x + 1) or (x + 2), depending on the value of x. > One such theorem you use is that the algebraic integers form a ring. > Yes, that *is* a theorem that must be proven. To prove it is a ring > is to show the following (a, b and c algebraic integers): > R1: given a and b, a + b is also an algebraic integer (closed under > addition). > R2: given a and b, a * b is also an algebraic integer (closed under > multiplication) > the remaining requirements are trivial (a + b = b + a, etc) and follow > because algebraic integers are complex numbers. So to show that the > algebraic integers do not form a ring you have to show that either > R1 or R2 is false. The simplest proof of R1 and R2 is one where > given monic polynomials Pa(x) and Pb(x) of which a resp. b are roots, > two new monic polynomials Pc(x) and Pd(x) are constructed of which > a+b resp. a*b are roots. So either the construction is wrong, or > indeed the algebraic integers do form a ring. > > So what do you think? Is the construction wrong or do the algebraic > integers indeed form a ring? > I've never said that algebraic integers don't form a ring. > What they form is a flawed ring, which doesn't include all the numbers > it must to prevent the possibility of appearing to prove two different > but opposite things. In what way can a ring be flawed? A ring is something that satisfies some particular properties. If the something satisfies them, it must be a ring. Period. When you attempt to prove contradictionary things about this ring the possibilities are that either you are using a theorem about rings that is wrong, or that your own theorems are wrong. Or the ring properties that define a ring are contradictionary. Which is it? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: About time > If your logic were correct it would apply >to other functions. > Yup. > Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Say Q(m) is factored in the form Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). Now note Q(0)/f^2 = x + f. That is, Q(0) = f * f * (x + f). Therefore when m = 0, we can say a1 = 0, a2 = 0, >and a3 = 1. Note that when m = 0, a1 and a2 are divisible by f. > Yup. > Now by your logic, for values of m other than 0, >a1 and a2 must be divisible by f and a3 is relatively >prime to f. Do you agree with this? This is a test of your >method. We need to know before we go to the next step. Nora B. > The answer is that yes, but with the qualification that m not equal 1, > but I have a better explanation for why than the wacky reply when I > initially freaked out. > Look again at your > Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). > Here the constant terms for the factors are revealed to be > f, f and x + f > where it just so happens that for your a3 you have something > like--yup, I know some will probably not like this, but it's the > reality--introducing h for the functions h(m), > a3 = h_1(m) - mx h_2(m) + x + f > where h_(1) has f^{2/3} as a factor and h_2(1)=1. > And yes, the same thing can happen with my argument, but it requires > mf^2 = 1 > but both m and f are integers in that argument, so that condition > doesn't occur. > You have suggested elsewhere that I need to solve > for the a's with expression. > Let's do that. Recall that > Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. > Let m = 1: > [1] Q(1) = f^2*x^3 + f^3 = f^2*(x^3 + f). > Assume this is factored in the form > Q(1) = (a1*x + f)*(a2*x + f)*(a3*x + f). > The roots of this polynomial are > [2] r1 = -f/a1, r2 = -f/a2, and r3 = -f/a3. > Of course from [1] I can compute the roots > explicitly. They are: > r1 = -f^{1/3} > r2 = -f^{1/3}*(-1 + sqrt(3))/2 and > r3 = -f^{1/3}*(-1 - sqrt(3))/2. > and a3, since ai = -f/ri : > a1 = f^{2/3}, > a2 = f^{2/3} * (-1 - sqrt(3))/2, and > a3 = f^{2/3} * (-1 + sqrt(3))/2. > Note that a1, a2, and a3 are all algebraic integers. > Note that each of them has a factor of f^{2/3}. > Note that none of them have f as a factor. > No surprises here. What has this got to do > with your wanting m*f^2 to equal 1 ? > (If I didn't say they were before, well they are now.) > For the more adventurous, check Q(m) with m NOT equal to 1, and yes, > you will find that *two* of the a's have a factor that is f. > See my other post on this. > Nora B. > PS: Oh yes, you also asked if I would be convinced if > a computer verifies your proof. > I would say *no*. I know exactly what is wrong with your > proof already. If someone said a computer verified > your proof, I would have to have a guarantee of at least > three things: > (1) That the computer program and the computing hardware > were infallible. > (2) That the translation of your proof into the program's > language was done correctly. > (3) That the person reporting the results could be trusted. > All of these three conditions will be harder to ascertain > than what I already know about your proof. I cannot see how > (1) can be guaranteed at all. As for (2), checking that would > probably considerably more tedious than what I have already done. > And for (3), the most likely person to report positive results > would be the least trustworthy. > So no. But don't let me stop you. Go ahead and carry it out. > Maybe you will convince somebody else. Not to mention that unless someone had the code to look at against the output, it wouldn't tell you a thing. > It might be a fun exercise, assuming I didn't miss something. > If any of you out there think you can find fault with my conclusion > here, please try. > It is math after all. And it's also a lot of fun. > James Harris David Moran === Subject: Help I really don't understand algebra at all. I am real stupid when it comes to letters and numbers. I have 2 problems that I need solved and don't really understand them. I am hoping someone will help me. for i = squareroot -1, If 3i (2 + 5i) = x + 6i, then x = ? if f(4) = 0 and f(6) = 6, then what would represent f(x)? === Subject: Re: Help > I really don't understand algebra at all. I am real stupid when it comes to > letters and numbers. I have 2 problems that I need solved and don't really > understand them. I am hoping someone will help me. > for i = squareroot -1, If 3i (2 + 5i) = x + 6i, then x = ? This is an identity. This is the same as 6i-5=x+6i therefore x=-5. > if f(4) = 0 and f(6) = 6, then what would represent f(x)? This means (4,0) and (6,6) are on the curve. Since two points are given, I have to assume it's a linear function. m=6/2=3 0=3(4)+b b=-12 Therefore f(x)=3x-12 David Moran === Subject: Re: Help > I really don't understand algebra at all. I am real stupid when it comes to > letters and numbers. I have 2 problems that I need solved and don't really > understand them. I am hoping someone will help me. > for i = squareroot -1, If 3i (2 + 5i) = x + 6i, then x = ? First, subtract 6*i from both sides and, lo and behold, there's x: x = 3*i*(2 + 5*i) - 6*i. I suppose you would like to (or are required to) simplify. 3*i*(2 +5*i) - 6*i = 3*i*2 + 3*i*5*i - 6*i = 6*i + 15*i^2 - 6*i = -15 [i^2 = -1] > if f(4) = 0 and f(6) = 6, then what would represent f(x)? A lot things. I hope I don't seen too grumpy, but this is probably why you are having difficulties: You didn't realize that the rest of the problem was important. I'll bet the problem really said something like: f(x) is a linear function, f(4) = 0 and f(6) = 6. Find an equation for f(x). You are supposed to think: Hmm. Linear function means y = m*x + b where m is the slope. The slope is the change in y's divided by the change in x's. OK. Here we go. Change in y's divided by change in x's is (0 - 6)/(4 - 6) = -6/-2 = 3. So far, y = 3*x + b. How to get b? Well, y = 0 when x = 4 so 0 = 3*4 + b so b = -12 and f(x) = y = 3*x - 12. Check f(6) = 3*6 - 12 = 6. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Tricks >> p is a prime >> n is a natural number >> set A = { a_0, a_1, , a_j} where n = a_0 * 10^0 + ... + a_j * 10^j >> For each p there exists a relation, R_p, between p and one or more >> elements in A so that n = (p R_p A) (mod p). >> (btw is this a correct use of a relation ???) > Makes no sense. (mod p) is a relation between two numbers. > Tho n is a number, p R_p A isn't a number, it's statement. > In addition R_p is sloppy, in need of clearer description. > Ok, > well say then instead that there exists a function f from A^i, where > i <= j, to N so that n = f (mod p). Would that be more correct? I don't know how to untangle your line of thought. How about a rewrite instead of a patch? > Or maybe just defining the function to be f: N -> N, and leaving the a_j > stuff to be defined in the function. > Better?? I'm not pcychi. How about a complete rewrite instead of a guess? === Subject: Re: Michael Gordon Nagus - February 18th 1978 > I'm not the slightest bit scared of you nor your threats. And Michael, > you keep in mind that you can beat the crap out of me and it's very > unlikely that the Saskatoon City Police will ever charge you with > assault, for I only have the right to be assaulted and brutalized. If I thought that were true, I'd look up Michael's address and mail him a copy of this myself. In fact, I think I will anyway. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Help connie ckaeser@fuse.net says and requests: >I really don't understand algebra at all. I am real stupid when it comes to >letters and numbers. I have 2 problems that I need solved and don't really >understand them. I am hoping someone will help me. Before you are permitted to say that about yourself, you should have studied algebra twice as an adult without really trying; otherwise you just may be one of those people whose nerves run for things other than written symbolic notation. Maybe you could study at your own pace, outside of any semester structure. For some people, trying to fit introductory algebra inside of 4 months is too difficult. >for i = squareroot -1, If 3i (2 + 5i) = x + 6i, then x = ? You are intending to use i*i = -1; in this case, i is a constant (like a variable, but now it HAS a value). Use distributive property on the left side of your equation. Use additive inverse of 6i (or of 6*i) in order to isolate x. You must apply the additive inverse to BOTH sides of the equation. Combine like terms; all the real numbers and all the imaginary numbers (those which are shown combined with 'i' using multiplication.) Your value for x should be a complex number. >if f(4) = 0 and f(6) = 6, then what would represent f(x)? Not enough information for that one. f(x) is a function (or at least, a relation). Maybe, tell us your mathematical background. G C === Subject: Re: Help from plsperry@sc.rr.com: >I suppose you would like to (or are required to) simplify. >3*i*(2 +5*i) - 6*i = 3*i*2 + 3*i*5*i - 6*i = 6*i + 15*i^2 - 6*i = -15 I was not watching too closely when I gave my earlier advice. This one comes out to be a real number. (the coefficient for the i component is 0). G C === Subject: Re: Algebra problem > 1/ Prove (4^47 + 2^12)^14 == 4 (mod 13) <=> ( 1 + 4 )^2 == 12 via multiply prior by 4^2 == 3 and use x^12 == 1 if x !== 0 > 2/ Prove 11^2n + 5^(2n+1) == 6 (mod 24) Calculate 11^2, 5^2 (mod 24) -Bill Dubuque === Subject: Help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9C49JM18254; I really don't understand algebra at all. I am real stupid when it comes to letters and numbers. I have 2 problems that I need solved and don't really understand them. I am hoping someone will help me. for i = squareroot -1, If 3i (2 + 5i) = x + 6i, then x = ? if f(4) = 0 and f(6) = 6, then what would represent f(x)? === Subject: Partial Sums? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9C49IY18216; (n-0)(n-1) + (n-1)(n-2) + (n-2)(n-3)+....+ I am thinking maybe to use a partial sum, but could someone please explain how to approach this? === Subject: Re: help with velocity > Needing help with velocity word problem can anyone help me come up > with the equasion? Here's the question: > A ball is thrown upward from the top of a 196 ft building with an > initial velocity of 64 ft per second. Approximate the maximum height > of the ball. > What height of the ball am I looking for? I don't think I get the > question and that'd why the equasion is so hard for me. Can anyone > please help. In addition to the other answers you got, there's an easier way: at the maximum height, the velocity is zero, and the velocity varies as v(t) = v_0 - g * t where g is the acceleration due to gravity. So, the ball takes v_0 / g to reach the max height, and therefore it will have travelled d = v_0 * t - g * t^2 / 2 = v_0^2 / g - v_0^2 / 2g = = v_0^2 / 2g from it starting position. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: Tricks <5cface913fd5dc576acfa5885b5993d0@news.teranews.com> <6b37b08dcb4b642121a698ccb7a3e876@news.teranews.com I'm not pcychi. How about a complete rewrite instead of a guess? n and m are natural numbers p is a prime A is a set { a_0, ... , a_j} where n = a_0*10^0 +...+ a_j*10^j For every p there exists a relation, R_p, beetween m and one or more elements in A so that n = m (mod p). Examples R_2 is m = a_0 (mod 2) R_11 is m = sum(k=0,j) (-1)^k*a_k The important part being that it is an existance conjecture. If the relation exists one can for instance reduce the numeber of calculations when factoring primes. Of course it's an impossible task to find R_p for every p but still it would be fun if it where so. If nothing else having this kind of relation for p < 50 aids every day manual calculations. My question was if there has been any research done of these relations or if R_p is known for any more p than 2,3,5,11 P.S. Hope it makes sense this time :) -- Sigblock empty. By choice. === Subject: Re: Partial Sums? >(n-0)(n-1) + (n-1)(n-2) + (n-2)(n-3)+....+ >I am thinking maybe to use a partial sum, but could someone please >explain how to approach this? as a finite series the sum is well known and is: 1*2 + 2*3 + 3*4+ ... + (n-1)(n) = n(n+1)(n+2)/3 === Subject: Re: Group Theory to be able to understand how to do the problem myself. I really Holly === Subject: Re: Help >I really don't understand algebra at all. I am real stupid when it >comes to letters and numbers. I have 2 problems that I need solved and Please don't post the same question multiple times. Please _do_ show us what you tried to do on your own. Simply giving up is usually not effective or necessary -- there is almost always an example in your textbook that throws some light on a homework problem. And don't call yourself stupid. Not well prepared might be more accurate, perhaps even lazy -- but there's something you can do about both of those. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Address munging may or may not reduce the spam you get; it surely reduces the number of useful answers you get. http://www.cs.tut.fi/~jkorpela/usenet/laws.html === Subject: Re: f(x) > I wanna know a history about Function( in simplified form, f(x)) > please.... Victor J. Katz, A History of Mathematics: an Introduction. The index entry Functions - defined leads to a few pages covering just what you want. Ken Pledger. === Subject: System Of Differential Equations Having trouble solving this system of non homogeneous differential equations. If anyone could offer some help, it would be much appreciated. d/dx[y1 (t)] = -7*y1(t) + 4*y2(t) d/dx[y2 (t)] = -8*y1(t) + 1*y2(t) + t === Subject: Re: System Of Differential Equations > Having trouble solving this system of non homogeneous differential > equations. If anyone could offer some help, it would be much appreciated. > d/dx[y1 (t)] = -7*y1(t) + 4*y2(t) > d/dx[y2 (t)] = -8*y1(t) + 1*y2(t) + t As y1, y2 are independent of x, d/dx y1(t) = d/dx y2(t) = 0 and system of equations becomes algebra problem. Is the problem stated correctly? === Subject: Re: System Of Differential Equations >> Having trouble solving this system of non homogeneous differential >> equations. If anyone could offer some help, it would be much >> appreciated. >> d/dx[y1 (t)] = -7*y1(t) + 4*y2(t) >> d/dx[y2 (t)] = -8*y1(t) + 1*y2(t) + t > As y1, y2 are independent of x, d/dx y1(t) = d/dx y2(t) = 0 > and system of equations becomes algebra problem. > Is the problem stated correctly? Sorry, that should read d/dt, not d/dx Here's the corrected version: d/dt[y1 (t)] = -7*y1(t) + 4*y2(t) d/dt[y2 (t)] = -8*y1(t) + 1*y2(t) + t === Subject: Re: System Of Differential Equations > d/dt[y1 (t)] = -7*y1(t) + 4*y2(t) > d/dt[y2 (t)] = -8*y1(t) + 1*y2(t) + t 1st eq --------- 2nd eq y1 = -1/8 ( y2' - y2 - t) and y1' = -1/8 y2'' + 1/8 y2' + 1/8 use this in the 1st eq and get y2'' + 6 y2' - 25 y2 = 7t - 1 solve this, get y2(t) and use the solution to get y1(t) === Subject: Re: System Of Differential Equations === Subject: Re: System Of Differential Equations >d/dt[y1 (t)] = -7*y1(t) + 4*y2(t) >d/dt[y2 (t)] = -8*y1(t) + 1*y2(t) + t x = x(t); y = y(t) x' = -7x + 4y y' = -8x + y + t x' = -7x + 4y y' = -8x + y Is a homogeneous equation in x, y. One solution is x = 0 = y x = at + b, y = rt + s a = -7at - 7b + 4rt + 4s r = -8at - 8b + rt + s + t -7a + 4r = 0 -8a + r + 1 = 0; r = 8a - 1 -7a - 4 + 32a = 0; a = 4/25; r = 7/25 4/25 = -7b + 4s 7/25 = -8b + s, etc ---- === Subject: sign function === Subject: Re: sign function Please do not post the same question _separately_ to different newsgroups. If you feel that you _must_ post the same question to more than one group, crossposting is much better. I already responded to your question in k12.ed.math . Interested readers may see the response there, once its moderator posts my message. David Cantrell === Subject: A funny situation re combination/permutation Hi all, I have another interesting problem. What is the probability of getting exactly one pair in a hand of 5 cards? One of the approach is 52/52 * 3/51 * 48/50 * 44/49 * 40/48 *C(5, 2) = 352/833, where C(n, r) = nCr. But the following approach sounds perfectly okay to me: the number of ways of choosing the pair = C(13, 1) After choosing the pair, there are C(4, 2) ways to choose which two from the four suites. The third card must be chosen from the remaining 52 - 4 = 48 card, so and so forth. Hence the number of combinations of getting exactly one pair = 13 * 6 * 48 * 44 * 40 = 6589440. BUT the number in the sample space is C(52, 5) = 2598960 which is way less than 6589440!!! It sounds so plausible yet impossible! Any hint? Anthony === Subject: Re: A funny situation re combination/permutation >Hi all, >I have another interesting problem. What is the probability of getting >exactly one pair in a hand of 5 cards? >One of the approach is 52/52 * 3/51 * 48/50 * 44/49 * 40/48 *C(5, 2) = >352/833, where C(n, r) = nCr. >But the following approach sounds perfectly okay to me: >the number of ways of choosing the pair = C(13, 1) >After choosing the pair, there are C(4, 2) ways to choose which two from the >four suites. The third card must be chosen from the remaining 52 - 4 = 48 >card, so and so forth. >Hence the number of combinations of getting exactly one pair = 13 * 6 * 48 * >44 * 40 = 6589440. >BUT the number in the sample space is C(52, 5) = 2598960 which is way less >than 6589440!!! >It sounds so plausible yet impossible! Any hint? Your mistake comes in calculating the number of ways of choosing the last three cards: you've counted separately the six different orders in which the same three cards can be chosen. Divide your 6589440 by 6 to get 1098240 = 352 * 3120, and note that 2598960 = 833 * 3120. Brian === Subject: Re: sign function I won't be so rude to you. The expressions you're looking for are: sin(x) = (exp(jx)-exp(-jx))/j2 cos(x) = (exp(jx)+exp(-jx))/2 If you are a mathematician, and not an engineer, you may know j as i = sqrt(-1). You can work this out from Euler's formula (this is a German name, pronounced Oiler), which is: exp(jx) = cos(x) + j.sin(x) This formula is especially amazing when you use x = pi. e ( = 2.7182818 . . .), j ( = sqrt(-1)) and pi ( = 3.1415926535 . . .) are all constants we obtained from other places, but we can show that they have a peculiar relationship: e^(j*pi) = 1. Yep - one. *Exactly* one. MadJock function, === Subject: help with a trigonometric limit This really _shouldn't_ be any trouble, but it's grating at me, and I'd appreciate some help. The question (forgive me if my notation's not standard): limit((((cos(3*x)^2)-1)/x^2), x=0) It's a given that the limit(((cos(x)-1)/x), x=0) = 0. What I've tried to do is factor the top term and separate the entire thing into two fractions. I got: 9*limit(((cos(3*x)-1)/(3*x))*((cos(3*x)+1)/(3*x)), x=0) I know that 3*limit(((cos(3*x)-1)/(3*x)), x=0) = 0, so shouldn't the limit equal 0? It doesn't; I've plugged it into Maple and multiplied with a calculator, and it really does seem to approach -9 as x approaches 0. Could someone tell me where I'm going wrong? How would I proceed from here? Best, James === Subject: Re: help with a trigonometric limit > limit((((cos(3*x)^2)-1)/x^2), x=0) Use L'Hospital's rule, since both the numerator and the denominator converge to 0. lim {x -> 0} (((cos 3x)^2 - 1)/x^2) = lim {x -> 0} (-6 cos 3x sin 3x / 2x) = lim {x -> 0} (-3 sin 6x / 2x) = lim {x -> 0} (-18 cos 6x / 2) = lim {x -> 0} (-9 cos 6x) = -9 hth meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Which math Software to buy? I am an undergraduate student working toward a computer science degree and taking first year calculus courses. I am considering purchasing MATH software (Maple, Matlab, or Mathmatica). I Would like to know if anyone has a favorite and why? === Subject: Re: help with a trigonometric limit > This really _shouldn't_ be any trouble, but it's grating at me, and > I'd appreciate some help. The question (forgive me if my notation's > not standard): > limit((((cos(3*x)^2)-1)/x^2), x=0) Hint: cos^2 + sin^2 = 1, so (cos(3x))^2 - 1 = ?? (I would avoid L'Hospital, because no one learns anything by using it.) === Subject: Re: Which math Software to buy? > I am an undergraduate student working toward a computer science degree > and taking first year calculus courses. > I am considering purchasing MATH software (Maple, Matlab, or > Mathmatica). > I Would like to know if anyone has a favorite and why? I'd suggest checking with your department before buying anything, as they might have a site or multi-user license. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: help with a trigonometric limit alt.math.undergrad, James Cunningham I'd appreciate some help. The question (forgive me if my notation's >not standard): >limit((((cos(3*x)^2)-1)/x^2), x=0) >It's a given that the limit(((cos(x)-1)/x), x=0) = 0. What I've tried >to do is factor the top term and separate the entire thing Why not just use l'Hopital's rule? lim[x->0, 2*cos(3x)*[-sin(3x)]*3/2x] lim[x->0, -3*sin(6x)/2x] lim[x->0, -9*sin(6x)/6x] = -9 Or are you constrained not to use l'Hopital's rule? -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Address munging may or may not reduce the spam you get; it surely reduces the number of useful answers you get. http://www.cs.tut.fi/~jkorpela/usenet/laws.html === Subject: Re: Which math Software to buy? > I am an undergraduate student working toward a computer science degree > and taking first year calculus courses. > I am considering purchasing MATH software (Maple, Matlab, or > Mathmatica). > I Would like to know if anyone has a favorite and why? I haven't tried all you mention, but I use Derive (www.derive.com) for my symbolic math needs, and it has always done what I needed. I also like Geometer's Sketchpad. (I am a computer science engineer with recreational math intereste.) Rick http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: help with a trigonometric limit > limit((((cos(3*x)^2)-1)/x^2), x=0) > It's a given that the limit(((cos(x)-1)/x), x=0) = 0. lim(x->0) ((cos (3x)^2) - 1)/x^2 = 9.lim(x->0) ((cos (3x)^2) - 1)/(3x)^2 Now use theorem, when g is continuous lim(x->a) f(g(x)) = lim(y->g(a)) f(y) Thus to continue, using g = ??, f = ?? = 9.lim(y->0) ((cos y) - 1)/y = 9 * 0 = 0 === Subject: Re: Which math Software to buy? > I am an undergraduate student working toward a computer science degree > and taking first year calculus courses. > I am considering purchasing MATH software (Maple, Matlab, or > Mathmatica). > I Would like to know if anyone has a favorite and why? Two points: 1) There are free packages on the net that will probably meet your needs. MuPad and Maxima come to mind. 2) If you *need* software to solve your calculus problems, you are in big trouble. You are better off learning it without. The software should only be used to check your answers. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: help with a trigonometric limit >> limit((((cos(3*x)^2)-1)/x^2), x=0) >> It's a given that the limit(((cos(x)-1)/x), x=0) = 0. >lim(x->0) ((cos (3x)^2) - 1)/x^2 > = 9.lim(x->0) ((cos (3x)^2) - 1)/(3x)^2 >Now use theorem, when g is continuous > lim(x->a) f(g(x)) = lim(y->g(a)) f(y) >Thus to continue, using g = ??, f = ?? > = 9.lim(y->0) ((cos y) - 1)/y > = 9 * 0 = 0 Only problem with that is, L'Hopital's rule gives an answer of -9, which matches the textbook according to the OP. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Address munging may or may not reduce the spam you get; it surely reduces the number of useful answers you get. http://www.cs.tut.fi/~jkorpela/usenet/laws.html === Subject: Re: Tricks <5cface913fd5dc576acfa5885b5993d0@news.teranews.com> <6b37b08dcb4b642121a698ccb7a3e876@news.teranews.com> <75027c9483fe2ce63537dd27c238ee86@news.teranews.com> n and m are natural numbers >> p is a prime >> A is a set { a_0, ... , a_j} where n = a_0*10^0 +...+ a_j*10^j > Let A = { a0,.. aj } and explain what a0,.. aj are. Decimal digets? > Let n = a0 + a1 10 + ... + aj 10^j Ok, p is a prime n, m and j are natural numbers m < n (otherwise this conjecture is rather pointless) A is a set containing all the decimal digits of n so that A = { a0 + a1 + ... + a(j-1) + aj } and n = a0 + a1*10 + .. + a(j-1)*10^(j-1) + aj*10^j >> For every p there exists a relation, R_p, beetween m and one >> or more elements in A so that n = m (mod p). > 'Between one or more elements' is weird usage of the notion of relation. Why? In this case R_p is a relation between A^j and m for a given p. Lets say R_p is function f_p : A^j -> N and in the case of p = 2 write f_2(a0, a1, ... , aj) = a0 + 0*(a1 + a2 + ... + aj) the same for 11 f_11(a0, a1, ... , aj) = a0 - a1 + a2 - .... > You seem to be saying. > For all primes p, there's some m in A with n = m (mod p) What I mean is that there exists a relation like the function above between a number m and the elements in A. The number n is then congruent with n mod p. >> Examples >> R_2 is m = a_0 (mod 2) >> R_11 is m = sum(k=0,j) (-1)^k*a_k > That m isn't a member of A and the rest of R_11 > is known only within your mind. Like where's the modulus p? There is no mod inside the relation, the R_2 case is special for 2 and 5. Hope it makes more sense this time.... -- Sigblock empty. By choice. === Subject: Re: Tricks === Subject: Re: Tricks >p is a prime >n, m and j are natural numbers >m < n (otherwise this conjecture is rather pointless) >A is a set containing all the decimal digits of n so that >A = { a0 + a1 + ... + a(j-1) + aj } You stated A is the singleton set containing just the sum of the digets. >n = a0 + a1*10 + .. + a(j-1)*10^(j-1) + aj*10^j To summarize: let A = { a0,.. aj } be a collection of decimal digets, and n = a0 + a1 10 + ... + aj 10^j >Lets say R_p is function f_p : A^j -> N and Let's just say f_p is a function and dismiss R_p >in the case of p = 2 write >f_2(a0, a1, ... , aj) = a0 + 0*(a1 + a2 + ... + aj) >the same for 11 >f_11(a0, a1, ... , aj) = a0 - a1 + a2 - .... >What I mean is that there exists a relation like the function above >between a number m and the elements in A. The number n is then >congruent with n mod p. Oh sure, and there exists a relation between me and the lady in the grocery store that's checking out at the other cash register. You'll have to be more explicit. Like, the lady flew a paper airplane into my shopping basket. >There is no mod inside the relation, the R_2 case is special for 2 >and 5. Hope it makes more sense this time.... ---- === Subject: Re: Tricks <5cface913fd5dc576acfa5885b5993d0@news.teranews.com> <6b37b08dcb4b642121a698ccb7a3e876@news.teranews.com> <75027c9483fe2ce63537dd27c238ee86@news.teranews.com> === > Subject: Re: Tricks >p is a prime >n, m and j are natural numbers >m < n (otherwise this conjecture is rather pointless) >A is a set containing all the decimal digits of n so that >A = { a0 + a1 + ... + a(j-1) + aj } > You stated > A is the singleton set containing just the sum of the digets. Ahh, sorry.. >n = a0 + a1*10 + .. + a(j-1)*10^(j-1) + aj*10^j > To summarize: > let A = { a0,.. aj } be a collection of decimal digets, > and n = a0 + a1 10 + ... + aj 10^j >Lets say R_p is function f_p : A^j -> N and > Let's just say f_p is a function and dismiss R_p >in the case of p = 2 write >f_2(a0, a1, ... , aj) = a0 + 0*(a1 + a2 + ... + aj) >the same for 11 >f_11(a0, a1, ... , aj) = a0 - a1 + a2 - .... >What I mean is that there exists a relation like the function above >between a number m and the elements in A. The number n is then >congruent with n mod p. > Oh sure, and there exists a relation between me and the lady in the > grocery store that's checking out at the other cash register. You'll have > to be more explicit. Like, the lady flew a paper airplane into my > shopping basket. For every p there exist an f_p : A^j->N so that f_p < n n = f_p (mod p) For every prime there exists a function that takes the decimal digits of a natural number n and outputs a number smaller than n and congruent with n modulus p. I cant be more specific than this. Does or does not this generalised relation(function) exist and is anything know anything about it, i.e. my original question. Can it even be proved? -- Sigblock empty. By choice. === Subject: Re: Tricks > n and m are natural numbers > p is a prime > A is a set { a_0, ... , a_j} where n = a_0*10^0 +...+ a_j*10^j Let A = { a0,.. aj } and explain what a0,.. aj are. Decimal digets? Let n = a0 + a1 10 + ... + aj 10^j > For every p there exists a relation, R_p, beetween m and one > or more elements in A so that n = m (mod p). 'Between one or more elements' is weird usage of the notion of relation. You seem to be saying. For all primes p, there's some m in A with n = m (mod p) > Examples > R_2 is m = a_0 (mod 2) > R_11 is m = sum(k=0,j) (-1)^k*a_k That m isn't a member of A and the rest of R_11 is known only within your mind. Like where's the modulus p? === Subject: Re: help with a trigonometric limit > limit((((cos(3*x)^2)-1)/x^2), x=0) >> It's a given that the limit(((cos(x)-1)/x), x=0) = 0. lim(x->0) ((cos (3x)^2) - 1)/x^2 > = 9.lim(x->0) ((cos (3x)^2) - 1)/(3x)^2 Now use theorem, when g is continuous > lim(x->a) f(g(x)) = lim(y->g(a)) f(y) Thus to continue, using g = ??, f = ?? > = 9.lim(y->0) ((cos y) - 1)/y > = 9 * 0 = 0 > Only problem with that is, L'Hopital's rule gives an answer of -9, > which matches the textbook according to the OP. > They made a mistake. No, they didn't. (cos^2(3x) - 1)/x^2 = - sin^2(3x)/x^2 = -9 {sin(3x)/(3x)}^2 which --> -9 as x --> 0 (using that sin(u)/u --> 1 as u --> 0). > Using l'Hopital > lim(x->0) ((cos (3x)^2) - 1)/x^2 > = lim(x->0) ((-sin (3x)^2) * 2(3x) * 3)/2x > = lim(x->0) -9 sin (3x)^2 = 0 If f(x) = cos^2(3x) - 1 and g(x) = x^2, then f'(x) = -6 cos(3x) sin(3x), g'(x) = 2x so L'Hopital's rule doesn't apply (yet). You'd need to consider lim_{x-->0} f''(x)/g''(x) = lim_{x-->0} -9(cos^2(3x) - sin^2(3x)} = lim_{x-->0} -9 cos(6x) = -9 I *think* that the problem here is that you're reading 'cos(3x)^2' as 'cos(9x^2)' ... which would be a very idiosyncratic reading ... > As in today's hyper-commercial society, profits are more important than > learning, money is saved by publishing math books that haven't been > 'proof' read it, that is without checking proofs and problems. > That's why in math newsgroups, errors in a text are not an uncommon topic. === Subject: Re: help with a trigonometric limit > They made a mistake. Using l'Hopital > lim(x->0) ((cos (3x)^2) - 1)/x^2 > = lim(x->0) ((-sin (3x)^2) * 2(3x) * 3)/2x > = lim(x->0) -9 sin (3x)^2 = 0 > As in today's hyper-commercial society, profits are more important than > learning, money is saved by publishing math books that haven't been > 'proof' read it, that is without checking proofs and problems. > That's why in math newsgroups, errors in a text are not an uncommon topic. Actually, I think you've got the problem wrong, which you can chalk up to my inability to write in accepted notation. It's really: lim(x->0) (cos^2(3x)-1)/x^2. What I was supposed to do is recognize that cos^2(3x)-1 = -sin^2(3x), which I didn't 'cause I'm quite the maroon sometimes. Then I could evaluate it easily using the proven lim(x->0) sin(x)/x = 1; the answer is definitely -9. thank-you from the news server I usually use, but it never showed up. === Subject: Re: help with a trigonometric limit === Subject: help with a trigonometric limit >The question (forgive me if my notation's not standard): >limit((((cos(3*x)^2)-1)/x^2), x=0) limit( (( (cos(3*x)^2) -1 )/x^2), x=0 ) >> lim(x->0) ((cos (3x)^2) - 1)/x^2 >> = lim(x->0) ((-sin (3x)^2) * 2(3x) * 3)/2x >> = lim(x->0) -9 sin (3x)^2 = 0 >Actually, I think you've got the problem wrong, which you can chalk >up to my inability to write in accepted notation. It's really: >lim(x->0) (cos^2(3x)-1)/x^2. Agreed. ---- === Subject: Re: help with a trigonometric limit >> limit((((cos(3*x)^2)-1)/x^2), x=0) >> It's a given that the limit(((cos(x)-1)/x), x=0) = 0. >lim(x->0) ((cos (3x)^2) - 1)/x^2 > = 9.lim(x->0) ((cos (3x)^2) - 1)/(3x)^2 >Now use theorem, when g is continuous > lim(x->a) f(g(x)) = lim(y->g(a)) f(y) >Thus to continue, using g = ??, f = ?? > = 9.lim(y->0) ((cos y) - 1)/y > = 9 * 0 = 0 > Only problem with that is, L'Hopital's rule gives an answer of -9, > which matches the textbook according to the OP. They made a mistake. Using l'Hopital lim(x->0) ((cos (3x)^2) - 1)/x^2 = lim(x->0) ((-sin (3x)^2) * 2(3x) * 3)/2x = lim(x->0) -9 sin (3x)^2 = 0 As in today's hyper-commercial society, profits are more important than learning, money is saved by publishing math books that haven't been 'proof' read it, that is without checking proofs and problems. That's why in math newsgroups, errors in a text are not an uncommon topic. === Subject: Calculus 3 Test Questions 1) Ok, when doing tangential and normal components of acceleration, we have a position equation of R(t)=(cos(2t))i+(sin(2t))j. The general equation of finding tangential and normal components is acceleration=(d^2s/dt^2)T(hat)+k(ds/dt)^2(N(hat)). If that makes any sense T(hat) is tangential component and N(hat) is normal component. k is curvature. The question is...how do i show that the tangential component of acceleration is zero using the position equation above, and how do i show that the normal component is constant? === Subject: Re: Calculus 3 Test Questions > 1) Ok, when doing tangential and normal components of acceleration, > we have a position equation of R(t)=(cos(2t))i+(sin(2t))j. The > general equation of finding tangential and normal components is > acceleration=(d^2s/dt^2)T(hat)+k(ds/dt)^2(N(hat)). If that makes any > sense T(hat) is tangential component and N(hat) is normal component. > k is curvature. The question is...how do i show that the tangential > component of acceleration is zero using the position equation above, > and how do i show that the normal component is constant? You don't need the equation for acceleration; you just need the definitions of T(hat) and N(hat) (I'm dropping hats from T & N in what follows to declutter the notation). By definition, T = (dR/dt) / |dR/dt|. Also by definition, N = (dT/dt) / |dT/dt| So: T = ( -sin(2t), cos(2t) ) N = ( -cos(2t), -sin(2t) ) = -R Now express d^2R/dt^2 in terms of these. -- P.A.C. Smith 'If the Apocalypse comes, beep me.' <*> http://www.srcf.ucam.org/~pas51 === Subject: Re: Planning to buy a calculus textbook! You might try Alt.binaries.e-books.technical I've gotten a few Calculus books from that site. In pdf format. What is the title of the Stewart book? My suspicion is that more and more math texts are going to be converted to pdf format to be made available to anyone with a good newsreader. >> I'm a senior in high school, and I'm taking AP calculus. The book I'm >using >> is by Larson, ISBN 0669327093, which is the 5th edition, and it sucks! I'm >> looking for a better book (to buy for my own) to use, which would clearly >> explains much wider types of problems. The one I'm using only shows about >> examples; on the excercise, I have to sit there for hours trying to even >> begin understanding how they even got the answer (answer to odd questions >> back). Usually the next day my teacher show us how that certain problem >was >> done AFTER he assigns the homework. >I have 2 editions of a book by James Stewart and it is one of the best books >I've seen. === Subject: Re: Adam Joseph Shebelski - July 30th 1981 > Continuing in Newton and Liebnitz' footsteps? Are primes any more > mystical than non-primes? More like Charles Manson's footsteps, probably. He may not be mystical, but he's a prime candidate for a lunatic asylum (again). -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock