matt grime

I hope, when you go to these dances, that you wear some badge indicating that you're a bigot who does not feel comfortable with other people expressing themselves by wearing whatever clothes they feel like so that these prospective girlfriends know what kind of person you are too: the kind who confuses transvestism with sexual preference (what on earth makes you think this person would even want to dance with someone like you?) and feels it acceptable to refer to other people by 'thing' and 'it'. Keep it up and you will get banned. Heck, I'd ban you anyway. Grow up.

I have a simple rule of thumb: if it is mechanics and I can do it it's easy. In general I am useless at applied maths having no intuition at all for it.

they mean nothing: names are not universal, the syllabi are not universal. Beyond the presumption that Math I is done before Math II there is nothing more that anyone can possibly say. The syllabi should be available for you to look at. Questions beyond that should be directed at the person at the college responsible for answering these questions (and there will be someone to do that).

that is not a formula, it is an equation. (a+b)^2 expanded might be what you want. try doing multiplying out: it's long multiplication and nothing more scary (and, no, before you say something, that is not belittling you it is trying to point out that you're thinking this is more complicated than it is. If you can multiply 37 by 23 be thinking of it is (20+3)*(30+7) which is 20*30 +3*30+20*7+3*21 like we were doing in our first years at school you can do this kind of expansion too).

It's just simple harmonic motion. It's not just a 'cambridge interview question', it is a very standard simple exercise in mechanics.

They're just names, people, they don't actually have any universal meaning. If you want to find out how two things are different just look at the syllabi, or talk to the lecturers/admin people who might know what is going on at your school.

google for rotation matrices hit the first link you get given.

this contains the two digit numbers 48, 10, 32, and 76, and none of these is prime. All but one of the 2 digit numbers is prime, that means that exactly one is composite, you appear to have it back to front.

1 is not prime

But YT you do (or did) know how to do this question. It is just long division that you are doing on that bit of paper with a pencil. X/D=P + R/D Y/D=Q + R/D Is all that the question is saying. Well, where you go from there depends on the person I guess. It's quite a nice question, and if you scribble a few musings it becomes clear what to do - this is somewhere where a lot of students go wrong: they look at the question, throw their hands up and say it's too hard and don't attempt it because they don't 'see' the answer instantly. If they just played around with a few symbols they'd soon realize the questions aren't that hard. OK, you might not have spotted the fact that D must divide the difference between X and Y, but this is one of the equivalent ways of defining modulo arithmetic. Say that X~Y if D divides X-Y. This is equivalent to saying that X and Y have the same remainder on division by D, or that there is an integer E such that X=Y+DE. It's good for people to work that one through the first time they meet it. It's also a shame that teachers don't tend to emphasize that this modulo arithmetic thing is just doing long division again. I have this weird observation about maths: you start off doing long division with remainder until you learn about fractions and then remainders go out the window. Then you learn about the evil decimal expansion and forget about fractions too (I even saw a text book for final year highschool students write sqrt(2)=1.4, come on people!). But then you start to learn that actually you should use 'exact' symbols like pi, e, sqrt and fractions rather than decimals so you start going backwards. By the time you come round to doing maths at university in your first number theory course you're doing, in some sense, long division with remainder again.

You need to take more care. 3/2 is far more preferable than any decimal (never use decimals). And in those limits you're using n in the n->2 part but x inside. It is only a small thing but you need to keep these things in check so you don't develop any bad habits. 0/0 at the limit point does not mean undefined as a limit: that is why you're trying to work out the the limit: you don't just put x equal to where you're letting it tend to. If you get 0/0 doing that then you can't draw any conclusions abuot behaviour. x^2/x, sin(x)/x and x/x^2 are all "0/0" symbols at zero yet the limits are 0,1, and undefined respectively. Try to avoid writing 1/0 or 0/0, especially since 1/0 does not equal 1/x, again don't get into bad habits: only use equals signs when things are equal, and use full sentences. Do you know l'hopital's rule?

Presumably for the same reason you're dividing f(b) by two: the area of a trapesium is one half the sum of the two sides times the base. All 'sides' are used twice except the first and the last in the approximation. And no there's no reason to suppose that trapezia should have evenly sized bases, especially if you're going to claim a general formula.

Pleaes don't take this the wrong way, but he used elementary/primary/junior school mathematics. Dressing it up we know that we have two numbers X and Y and that divided by some D they have the same remainder R, or as we learnt at elementary/primary/junior school and then forgot when we were given calculators, this means that there are numbers P and Q satisfying: X=DP+R Y=DQ+R and R is between 0 and D-1 (quotient plus remainder arithemetic, long division, the old fashoined stuff) So, using some little algebraic manipluations it follows that D divides X-Y, or D divides 7*41*11 and is three digits long as is R. This extra information allows you to work out which three digit D and R it is: presumably dividing by 7*41=287 leaves only a 1 or two digit remainder.

what generality? You assumed a=0 at one point (and not at another) and that f(a)=0, so it was not very general, you also assumed that there are 3 points defining the subdivision into trapezia and that they are equally spaced, again that is not very 'general', and n=4 so why not put that in? You also need to divide f(a) by 2.

It is still not clear if you're doing this in general of for some specific function and interval: you are mixing and matching, so why not put a=0, b=2, use 4 trapezoids (n=4) instead? b can't have meant 'the same thing', since it was the upper limit of the integral (b-a) and a subscript (x_b). Since b was in this example 2 you'd have x_2 appearing twice which presumably you don't want. Since you're doing equal sized subintervals why not use that fact too?

Slightly puzzled: you use b twice for different things, a generic 'n' for the number of trapezoids but you only use 4 values of f, and one of them is divided by 2 and none of the others is, and surely f(a) and f(b) ought to appear somewhere too.

if you differentiate your solution you will find out. not necessarily a good idea to revive dead threads that have nothing to do with your question, by the way.

If you actually read the paper that google leads you to you see that all the author has done is (naturally) redefine the word 'identity' so that it only has to act as 1 on some subset of the underlying set of the 'beta group'. Normally we would call that an idempotent. It is rather trivial to construct formal examples of them, whether or not they are of any use is another question entirely, or if they are indeed internally consistent. Noticeably there are no concrete examples of them given, nor am I entirely happy with his demonstration of internal consistency. http://uosis.mif.vu.lt/%7ejavtokas/files/betagroups.pdf BG3 contains the worrying phrase for e_i there are e_j such that e_ie_j=e_j for all i,j in N. which is confusingly written. I suspect he is trying to say that there are a countable number of e_j satisfying e_i=e_j. In anycase it looks as though it is trivial to generate potential beta algebras using infinite ordinals, though he doesn't appear to have noticed this.

It perhaps wasn't the best use of the word 'definition'. Better would be, perhaps: it is a simple consequence of the axioms of a group that the identity is unique. Or indeed any other algebraic structure that has elements i and j that satisfy ix=x=xi = jx=xj for all x must have that i=j.

a, k, and r were all explained in post 7.

no, since a group, by definition has a unique identitiy element. You might want to notice that whatever you refer to is strictly not called a group even by the person who discovered/invented it.

So, you're not learning this from any notes? How did you come to have to do the questions? I am just curious: if you're supposed to be doing these at school then you've been let down by them, and if yuo're just doing it for fun then you/we need to start from the beginning. I don't know of any intro calc books, so perhaps someone else can help out there. If you know how to do the finite geometric sum just take limits, and if you read post 7 you'll see that Phil explained what the answer is.

which would be exactly the same answer given by the previous poster....