Rainman DX Posted December 11, 2006 Report Share Posted December 11, 2006 If I were sticking with my math major, I'd be taking Abstract Algebra I next semester, along with Numerical Analysis I... But a math minor'll have to do instead. Quote Link to post Share on other sites
Steben Posted December 13, 2006 Report Share Posted December 13, 2006 Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway. Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x. Quote Link to post Share on other sites
Bahamut Posted December 19, 2006 Report Share Posted December 19, 2006 Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x. f is any old function? And what properties does H have? Connectedness? Quote Link to post Share on other sites
Steben Posted December 19, 2006 Report Share Posted December 19, 2006 Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x. f is any old function? And what properties does H have? Connectedness? You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H. Quote Link to post Share on other sites
Bahamut Posted December 19, 2006 Report Share Posted December 19, 2006 Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x. f is any old function? And what properties does H have? Connectedness? You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H. I guess since x is in the interior of H, one can take a ball around it that is contained in the interior of H. Using the fact that a function can only have a countable number of discontinuities, this means that the function must be piecewise continuous, and so must be continuous about x. This argument seems stronger than needed though. Quote Link to post Share on other sites
Steben Posted December 21, 2006 Report Share Posted December 21, 2006 Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x. f is any old function? And what properties does H have? Connectedness? You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H. I guess since x is in the interior of H, one can take a ball around it that is contained in the interior of H. Using the fact that a function can only have a countable number of discontinuities, this means that the function must be piecewise continuous, and so must be continuous about x. This argument seems stronger than needed though. x isn't in the interior of H. For example, 0 is not a limit point of the integers, as the open set (-.5,.5) about 0 does not contain a point of the integers distinct from 0. The interior of the integers is empty. Here's some useful characterizations to apply for this problem. x is a limit point of a set H iff for every open set S about x, there is a point y in S distinct from x such that y is in H. (It follows that x is not a limit point of a set H iff there exists an open set S about x such that S does not intersect any point of H besides x.) A fuction f:H->K is said to be continuous at x in H iff for every open set T about f(x), there is an open set S about x such that f is a subset of T. On an unrelated note, it occurs to me that we need to have some problems that require more cleverness than upper-level math experience. I should look for some of my high school math competition materials so that more people can participate. Quote Link to post Share on other sites
Bahamut Posted December 22, 2006 Report Share Posted December 22, 2006 Hm...I'm so rusty with the undergrad real analysis . Guess I'll get to review it in the first 3 weeks of next semester in the actual grad real analysis. So, here's a math question for anyone with knowledge of complex analysis - does there exist a non-constant function that is harmonic on the complex plane and is 0 everywhere on the real and imaginary axis? Oh and math problems are very easy to find - one can search for any AMC, AIME, USAMO, and IMO problems and post them here. There is a good chance I will be able to answer them. Quote Link to post Share on other sites
Steben Posted December 24, 2006 Report Share Posted December 24, 2006 Just got out of my Analysis exam. Here's a problem that was pretty easy, but I thought it was fun anyway.Let f:H->K be a function. Show that if x is in H but isn't a limit point of H, then f is continuous at x. f is any old function? And what properties does H have? Connectedness? You shouldn't need anything else. You're just considering f's continuity at a point of H which is not a limit point of H. I guess since x is in the interior of H, one can take a ball around it that is contained in the interior of H. Using the fact that a function can only have a countable number of discontinuities, this means that the function must be piecewise continuous, and so must be continuous about x. This argument seems stronger than needed though. x isn't in the interior of H. For example, 0 is not a limit point of the integers, as the open set (-.5,.5) about 0 does not contain a point of the integers distinct from 0. The interior of the integers is empty. Here's some useful characterizations to apply for this problem. x is a limit point of a set H iff for every open set S about x, there is a point y in S distinct from x such that y is in H. (It follows that x is not a limit point of a set H iff there exists an open set S about x such that S does not intersect any point of H besides x.) A fuction f:H->K is said to be continuous at x in H iff for every open set T about f(x), there is an open set S about x such that f is a subset of T. On an unrelated note, it occurs to me that we need to have some problems that require more cleverness than upper-level math experience. I should look for some of my high school math competition materials so that more people can participate. Alrighty, here's the proof. x is not a limit point of H, so let S be an open set such that it intersects no point of H distinct from x. Consider an open set T about f(x). f=f[{x}]={f(x)} which is a subset of T. Therefore, f is continuous at x. Quote Link to post Share on other sites
Bahamut Posted December 24, 2006 Report Share Posted December 24, 2006 Once you gave all that info, I figured it out quite easily. It's not nice handing over the tools if the tools are only like 2 things . Quote Link to post Share on other sites
Red Shadow Posted March 7, 2007 Report Share Posted March 7, 2007 im finding myself too dumb to prove this, send help: Tan^2x(sec^2x)=tan^2x+cos^2x-1 Quote Link to post Share on other sites
Bahamut Posted March 7, 2007 Report Share Posted March 7, 2007 So you know (sin x)^2 + (cos x)^2 = 1, so just divide everything by (cos x)^2. Quote Link to post Share on other sites
Red Shadow Posted March 7, 2007 Report Share Posted March 7, 2007 for some reason im still not seeing it aggravates the hell out of me because these are damned easy i end up getting sin^2xcos^2x on the right wth Quote Link to post Share on other sites
Bahamut Posted March 7, 2007 Report Share Posted March 7, 2007 Oh wait I misread the problem ha. Working on it atm Edit: I seem to be getting that this is impossible. So, by moving the (tan x)^2 terms to the left side, you get (tan x)^2 * [ (sec x)^2 - 1 ] = (cos x)^2 - 1 ==> (tan x)^4 = - (sin x)^2 ==> (sin x)^4 = - (sin x)^2 * (cos x)^4 = -(sin x)^2 * (1 - (sin x)^2 )^2 ==> (forget about the possible division by 0, since we're looking backwards and for arbitrary x) (sin x)^2 = - (1 - 2 * (sin x)^2 + (sin x)^4 ) ==> (sin x)^4 - (sin x)^2 + 1 = 0 So, set t = sin x. Note that t^4 - t^2 + 1 = 0 is an even function, and the range of t is [-1, 1], so it is enough to look at [0, 1]. However, this equation is always greater than 0 for the possible values of t, so I think there's a typo here or your teacher gave a false problem. Quote Link to post Share on other sites
Hadriel Posted March 7, 2007 Report Share Posted March 7, 2007 Linear algebra is making me want to shoot myself in the face. One of my relatives died, so I missed several class periods and horribly bombed the first exam. The really depressing part is that there's only that exam, one more in April, and a comprehensive final, with our homework counting as one collective exam grade. Not only that, he hasn't even told us the grading scheme, although he's mentioned that the final will probably count for more than the other tests. Unless the final counts for like 40% of my grade, I think a B is the highest possible grade I can get right now; I need better if I want to transfer to UT next spring. Considering dropping so it doesn't go on my transcript. Also considering taking a differential equations class this summer at another nearby university, though the costs and driving times behind that would be troublesome. If people would have signed up for the damn class over here they wouldn't have axed it. Quote Link to post Share on other sites
Red Shadow Posted March 7, 2007 Report Share Posted March 7, 2007 yeah it should have been sin instead of sec >:[ Quote Link to post Share on other sites
Bahamut Posted March 7, 2007 Report Share Posted March 7, 2007 Oh then that makes sense. So, (tan x)^2 * (sin x)^2 = (tan x)^2 + (cos x)^2 - 1 is equivalent to (tan x)^2 * (sin x)^2 = (tan x)^2 - (sin x)^2 (using (sin x)^2 + (cos x)^2 = 1) (sin x)^2 + (tan x)^2 * (sin x)^2 = (sin x)^2 * (1 + (tan x)^2 ) = (tan x)^2 (sin x)^2 * (sec x)^2 = (sin x)^2 / (cos x)^2 = (tan x)^2, which is certainly true (Note: I used (tan x)^2 + 1 = (sec x)^2 here) Quote Link to post Share on other sites
Red Shadow Posted March 7, 2007 Report Share Posted March 7, 2007 yeah i went ahead on mine and changed tan^2 on the left to sec^2-1 and worked from there etc etc Quote Link to post Share on other sites
Steben Posted March 9, 2007 Report Share Posted March 9, 2007 Oh then that makes sense.So, (tan x)^2 * (sin x)^2 = (tan x)^2 + (cos x)^2 - 1 is equivalent to (tan x)^2 * (sin x)^2 = (tan x)^2 - (sin x)^2 (using (sin x)^2 + (cos x)^2 = 1) (sin x)^2 + (tan x)^2 * (sin x)^2 = (sin x)^2 * (1 + (tan x)^2 ) = (tan x)^2 (sin x)^2 * (sec x)^2 = (sin x)^2 / (cos x)^2 = (tan x)^2, which is certainly true (Note: I used (tan x)^2 + 1 = (sec x)^2 here) Usually teachers prefer starting with the left side and manipulating it to get the right. tan^2*sin^2 = (sec^2 - 1)*sin^2 [by tan^2 = sec^2 - 1] = tan^2 - sin^2 [by distributive law] = tan^2 + cos^2 - 1 [by -sin^2 = cos^2 - 1] Quote Link to post Share on other sites
Escariot Posted March 9, 2007 Report Share Posted March 9, 2007 Those numbers don't mean a thing to me. MATH111 here is calculus.But yeah, I'm having the same trouble in spanish. I'm having to withdraw from it entirely. We don't have a Math 111, It's Math 110, which is Basic Calc, and Math 140, which is Calc w/ Analytical Geometry Quote Link to post Share on other sites
Broken Posted March 10, 2007 Report Share Posted March 10, 2007 Will someone help me with this Trig. identity, please? cos(A)/\4 + 2cos(A)/\2 sin(A)/\2 + sin(A)/\4. and to clarify, that's cosine A to the 4th plus two times cosine A squared times sine A squared plus sine A to the 4th. The furthest I could get on this was to re-arrange the expression so that it was: cosA/\4 + sinA/\4 + 2cosA/\2 sinA/\2 And then apply the pythag. identity to the terms cosA/\4 + SinA/\4, making that expression equal to 1. So, then I had: 1 + 2cosA/\2 sinA/\2. Am I going about this the right way? If anyone could help me out, that would be excellent. Quote Link to post Share on other sites
Bahamut Posted March 11, 2007 Report Share Posted March 11, 2007 So, (cos A)^4 + 2 * (cos A * sin A)^2 + (sin A)^4 = ( (cos A)^2 + (sin A)^2 )^2 = 1 ^ 2 = 1 Quote Link to post Share on other sites
Broken Posted March 11, 2007 Report Share Posted March 11, 2007 Oh, so the whole problem was just a factored polynomial basically? And the two factors end up being (sinA^2 + cosA^2) and (sinA^2 + cosA^2) or 1X1. Did I understand that correctly? Quote Link to post Share on other sites
Steben Posted March 11, 2007 Report Share Posted March 11, 2007 Oh, so the whole problem was just a factored polynomial basically? And the two factors end up being (sinA^2 + cosA^2) and (sinA^2 + cosA^2) or 1X1. Did I understand that correctly? Exactly. I'm tutoring for Precal with Trig, and that's something they like to pull - requiring you to factor trig functions in order to simplify them. Here's a similar example: (note that I use sin^n as shorthand for sin^n(x)) (sin^3 + cos^3)/(sin + cos) = 1 - sin*cos (sin^3 + cos^3)/(sin + cos) = (sin + cos)(sin^2 - sin*cos + cos^2)/(sin + cos) {by the factorization of the sum of cubes} = sin^2 - sin*cos + cos^2 {by the cancellation of (sin + cos) over itself - note that this does not hold true for angle values of 3pi/4 or 7pi/4, which cause sin(x) + cos(x) to equal zero} = 1 - sin*cos {by the Pythagorean identity} Quote Link to post Share on other sites
Bahamut Posted March 11, 2007 Report Share Posted March 11, 2007 Well, there is a long way to do the problem too I think, but it's more trouble than it's worth (by making use of (cos x)^2 = 1 - (sin x)^2 ) Quote Link to post Share on other sites
Red Shadow Posted December 9, 2008 Report Share Posted December 9, 2008 hey kids pay attention somebody is probably going to be in here with a question soon Quote Link to post Share on other sites
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