Invariant - something that does not vary. Yes Covariant - if one changes the other changes. >Contravariant - if one increases the other decreases. > Simple but am I close? > No, not even in the right ball park. > Co/contra variant is about how vectors and tensors transform between > coordinates. Are the following covariant: sin(pi/2 - x) = cos(x) tan(pi/2 - x) = cot(x) sec(pi/2 - x) = csc(x) <40095f1f_2@127.0.0.1> ==== > Invariant - something that does not vary. >Yes > Covariant - if one changes the other changes. >> Contravariant - if one increases the other decreases. >> Simple but am I close? >No, not even in the right ball park. >Co/contra variant is about how vectors and tensors transform between >coordinates. Are the following covariant: sin(pi/2 - x) = cos(x) > tan(pi/2 - x) = cot(x) > sec(pi/2 - x) = csc(x) > No no. Those equations are identities. They're aren't even invariants. Constants are invariants. The length of a line segment over rigid transformations is an invariant. The end points of a line seqment aren't invariant over rigid transformations. 0 and 1 are invariants over squaring. 0 and -1 are invariants over cubing. conclusion. ==== > Co/contra variant is about how vectors and tensors transform between >> coordinates. examples? ==== >Invariant - something that does not vary. >> Yes >>Covariant - if one changes the other changes. >>Contravariant - if one increases the other decreases. >>Simple but am I close? >> No, not even in the right ball park. >> Co/contra variant is about how vectors and tensors transform between >> coordinates. > Are the following covariant: > sin(pi/2 - x) = cos(x) > tan(pi/2 - x) = cot(x) > sec(pi/2 - x) = csc(x) > No no. Those equations are identities. how to interpret it. >They're aren't even invariants. > Constants are invariants. The length of a line segment over rigid > transformations is an invariant. The end points of a line seqment aren't > invariant over rigid transformations. Ok. I understand that. >0 and 1 are invariants over > squaring. 0*0 = 0 1*1 = 1 >0 and -1 are invariants over cubing. 0*0*0 = 0 -1*-1*-1 = -1 ==== >Invariant - something that does not vary. >> Yes >>Covariant - if one changes the other changes. >>Contravariant - if one increases the other decreases. >>Simple but am I close? >> No, not even in the right ball park. >> Co/contra variant is about how vectors and tensors transform between >> coordinates. > Are the following covariant: > sin(pi/2 - x) = cos(x) > tan(pi/2 - x) = cot(x) > sec(pi/2 - x) = csc(x) > No no. Those equations are identities. They're aren't even invariants. > Constants are invariants. The length of a line segment over rigid > transformations is an invariant. The end points of a line seqment aren't > invariant over rigid transformations. 0 and 1 are invariants over > squaring. 0 and -1 are invariants over cubing. conclusion. http://mathforum.org/library/drmath/view/54027.html <40095f1f_2@127.0.0.1> <20040117220343.M35539@agora.rdrop.com> <400aa7b6_2@127.0.0.1> ==== > Covariant - if one changes the other changes. > Contravariant - if one increases the other decreases. >> Are the following covariant: >> sin(pi/2 - x) = cos(x) >> tan(pi/2 - x) = cot(x) >> sec(pi/2 - x) = csc(x) >No no. http://mathforum.org/library/drmath/view/54027.html > Bah, read it carefully. He made unique to him slang phrase co variant which means something about sin & cos, tan & cot, sec & csc. ==== [.snip.] >The theorem correctly states that, using these standard definitions and >>in ring of algebraic integers, or any other ring with identity, if 'f' >>divides 'a*b' and is coprime to 'a' then 'f' divides 'b'. I agree with you, but as Bill Dubuque has pointed out many times, >coprime does have other 'standard meanings' (though the one you use >is, as far as I am aware, the 'gold' standard in algebraic number >theory). We know that James uses coprime to mean any common >divisors are units; in the ring of algebraic integers the two notions >are equivalent, but that is a hard fact to establish. It is, of >course, provable; but it ->is<- hard to do so. > It's funny that so many posted as if something were so obvious only to > have Magidin come in and claim that it's not. > Since you are now arguing about whether the result is obvious or not, I assume that you agree that the result is true. [1] Where were we... There are two very similar statements. The first, Lemma 1, is false. The second, Lemma 1' is true. The only difference is that in Lemma 1', s and t are allowed to vary with m. It is, however, interesting to note that Lemma 1 does hold if g_1 and g_2 are polynomials. The importance is that James assumes that Lemma 1 holds for non-polynomial factors [2]. This, not surprisingly leads to a contradiction. Lemma 1: Let P(m) be a polynomial with coefficients in A, the algebraic integers. Let each coefficient be divisible (in A) by the algebraic integer f. Let P(m) = g_1(m) * g_2(m), with g_1(m) and g_2(m) functions from A to A. Then there exist s and t in A such that s*t=f; and for all m in A, s divides g_1(m) and t divides g_2(m). Lemma 1': Let P(m) be a polynomial with coefficients in A, the algebraic integers. Let each coefficient be divisible (in A) by the algebraic integer f. Let P(m) = g_1(m) * g_2(m), with g_1(m) and g_2(m) functions from A to A. Then there exist s(m) and t(m) in A such that s(m)*t(m)=f; and for all m in A, s(m) divides g_1(m) and t(m) divides g_2(m). - William Hughes [1] Actually, no one has claimed that this specific result (the equivalence between two definitions of coprime in the algebraic integers) is obvious. However, the result is very well known to James. When I stated that the issue has come up at least once before I was being ironic. The issue has come up many times. [2] The fact that the first thing that so many people say after looking at James' work is why do you use results at 0 to make claims for divisiblity not at 0 does not seem to bother James at all. (I think he suspects a conspiracy.) James has not offered any proof of his claims; he considers them obvious. He has given a number of examples, all of them involving polynomials. ==== Well, you can extract bezier curves out of edges of a patch, but you can just as well extract them anywhere on the surface (well, not exactly anywhere) to produce isoparms. Bezier patch is defined as a tensor product of two bezier curves and that produces 16 control points. Googling for tensor product and bezier patch should yield promising results. > hi all, I assume that a bezier patch has 16 control points and 4 corners control > points are attached onto the surface. It implies that there are 4 bezier > curves on the edges of the patch. My question is how can the middle 4 > control points change the shape of the surface? > newbie... > hi guys, 2^(x+1)+2^(3-x)=17 logging gives (x+1)lg2+(3-x)lg2=lg17 expanding gives lg2x+lg2+3lg2-lg2x=lg17 but lg2x-lg2x=0.....goinn mad...graphed it so i know x=3 but can't get there numerically.....please help i know i'm missing something simple shaun ==== <20040118120121.28470.00000202@mb-m15.aol.com>: >2^(x+1)+2^(3-x)=17 logging gives (x+1)lg2+(3-x)lg2=lg17 No. log(3+7) =/= log(3) + log(7) -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ For many Americans the danger of tyranny lies not in government but in employers or insurance companies or HMOs, from which we need government to protect us. To say that any worker is free to escape an oppressive employer by getting another job is about as realistic as to say that any citizen is free to escape an oppressive govern- ment by emigrating. -- Steven Weinberg, in /Facing Up/ (2001) ==== Hey to all, I need to solve the following equation, but not using trial and error any ideas? (1.2)^n > 1.6 + 0.4n I guess I fergot my basics Doug ==== I don't think this will help you (ProductLog). If n is assumed to be Real... equ = 1.2^n > 1.6 + 0.4*n Increase it's precission... equ = Rationalize[equ] (6/5)^n > 8/5 + (2*n)/5 List @@ Reduce[equ, Reals] n < (-4*Log[5] + 4*Log[6] + ProductLog[(3125*Log[5] - 3125*Log[6])/2592])/(Log[5] - Log[6]) and n > (-4*Log[5] + 4*Log[6] + ProductLog[-1, (3125*Log[5] - 3125*Log[6])/2592])/(Log[5] - Log[6]) Numerically... N[%] {n < -2.3801395955782705, n > 9.073517983127845} -- Dana > Hey to all, I need to solve the following equation, but not using trial and > error any ideas? (1.2)^n > 1.6 + 0.4n I guess I fergot my basics Doug ==== > Hey to all, I need to solve the following equation, but not using trial and > error > > any ideas? > > (1.2)^n > 1.6 + 0.4n > > I guess I fergot my basics > > Doug > > Firstly, it is not an equation unless you use > to mean is equal to. it should be called an inequality. Secondly, such equations or inequalities, with unknowns as exponents as well as in linear expressions, generally do not have exact solutions, but must be found by approximation methods. ==== >Hey to all, I need to solve the following equation, but not using trial and >error > any ideas? > (1.2)^n > 1.6 + 0.4n > I guess I fergot my basics > Doug > > Firstly, it is not an equation unless you use > to mean is equal to. it should be called an inequality. Secondly, such equations or inequalities, with unknowns as exponents as > well as in linear expressions, generally do not have exact solutions, > but must be found by approximation methods. One such method that I would use for this is the iteration method. 1. Rearrange to give an equation with n= on one side in terms of n on the other. E.g. n=((1.2)^n - 1.6)/0.4 2. Now add iterative subscripts. I usually use n for the subscripts - but since this is the variable, we'll use p instead. n(p+1) = ((1.2)^n(p) - 1.6)/0.4 3. Now set n(1) to a reasonable guess (say 1) and find n(2) using this formula. Now feed this into the equation again to give n(3), n(4), .... etc. The equation converges to a value of about -2.38. This will only give you an approximation though - unless of course you could repeat the process to find n(infinity). Also, Virgil was right - this is an inequality! This method is just giving one value for which the two sides are equal! If you want the other, just rearrange the original equation to give n= in terms of the other n. Hope this helps Vassago P.S. I know about initial conditions affecting convergence etc. but have skipped over this for the sake of simplicity. If anyone else feels they want to expand on the issue - feel free. ==== For n in [-2.38014;9.07352] is the equation wrong... PS: I don't really like cross-posting Max > Hey to all, I need to solve the following equation, but not using trial and > error any ideas? (1.2)^n > 1.6 + 0.4n I guess I fergot my basics Doug ==== > For n in [-2.38014;9.07352] is the equation wrong... > > PS: I don't really like cross-posting > > Max > > >Hey to all, I need to solve the following equation, but not using trial > and >error > any ideas? > (1.2)^n > 1.6 + 0.4n > I guess I fergot my basics > Doug > Cross posting beats multiple postings all hollow. It isn't an equation. I reqally don't like top posting. ==== Use misc.test -- that's what it's there for. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Thoroughness. I always tell my students, but they are constitutionally averse to painstaking work. -- Emma Thompson, in /Wit/ (2000)