matt grime

who knows? you've only told us about its state at time t when t is greater than 0 and strictly less than 2 minutes, and that is the theoretical answer. practically it is not a possible situation.

You appear to be saying that it isn't obvious that "if 4 divides x then x is even" is true. i can't beleive you expect me to provide a proof of that... i didn't say you were stupid to not understand the reasoning behind "implies" being the way it is, i said that evidently you needed understandable examples to illustrate it where you can clearly see that a proposition is true despite it being "of the type" F => T ro F=>F. again, your pink elephant comment shows that you are using the common language maening of implies with its causality connotation instead of the mathematical one. the comical proposition you give is a true proposition (well, probably, depends what's acceptable as a pink elephant) albeit nonsensical. Thw wall papaer you refer to is the distinguished triangle in a traiangulated category that defines the homotompy colimit of a sequences of objects and morphisms X_1 --> x_2 ---> X_3---> X_4 ..... it is the triangulated version of the cokernel of an injection in an ableian category defining the colimit. There is no reason for you to know what this is supposed to be. i doubt that more than a handful of people would understand it. not because it is hard but because it is obscure.

erm, a la means in the style of. what is insulting abotu describing some articles written abuot elementary mathematics (first year undergraduate mathematics) as being in the style of polya?

There are only finitely many powers of two less than 509

Suppose for a minimal counter example. N, then there is a prime in the ragen N/2 to N, call it p, now think a little to get a contradiction,

THe empty set is not an exceptin. the cardinality of the emty set is zero, the cardinality of its power set is 1. restricted comprehension means that we can only specify sets in terms of other sets. Thus we can talk of the set of real numbers that have real square roots because we are restricting to "the set of real numbers" approximately, you shuold think of "a set theory" (for there are many) as a way of assigning objects with labels, some have the labe "set" some do not. we call this a model. sets in this model must obey the axioms of the theory, there is no way to create a model of ZF that has the collections of all things labelled with "set" also labelled with "set". more explicitly the way the restricted comprehension way to define a set in a model of ZF is: S={x in X:P(x)} that is it is the set of x's in some other set X satisfying some property P. It may appear circular on first reading but it isn't becuase it is only a consistency relation. the link can be fuond at http://www.dpmms.cam.ac.uk/~wtg10 follow it to his discussions of basic mathematics a la polya.

My name isn't Grimes. What on earth are you rambling about? this is a discussion of negation of propositions, nothing to do with set, please, don't confuse things pointlessly (and the set equivalent of not is complement). Moreobve, > and < are not usualy used for sets. containment of sets is round brackets. It would be corect for subgroup, not subset. Evidently you need some somple examples to see why the truth values of impliction are as they are. Simply put you are confusing "causality" with imlpication. For instance, thinking about natural numbers, take the propostion If 4 divides x then x is even. This is clearly a true proposition, though individually the case x=2 shows that we have "false implies true" ie the premise is false but the conclusion is true. x=7 shows that false impies false is true. Obvisoly we cannot find a case where the premise is true and the antecedent is false. That is the whole point of propositions. Now, why don't you write out the truth taable for not(A) or B and see what you get; hopefully now the mists will lieft and yuo will see why A=>B is the same as not(A) or B. I like the way you also think that there can be one and only one equivalent expression fro A implies B (not B implies not A). fortunately maths is quite a lot more varied than that, and you shuldn't think that only what you know is what is true.

why did you take 1 off the n's? I told you to take 1 from each term to go from y_i to Y_i-1. is the link between the terms of the series and the product not clear frmo the typesetting? we go from the product (1+x_1)(1+x_2).... to the series x_1+x_2+x_3... so going back subtract one off the terms, not the indices.

Certainly it converges for all |z|<1 since it is a subseries of the series z^n and certainly it diverge when z=1. But a more interesting exercise: show it diverges for a dense set of points on the circle |z|=1, thus it can how no continuation beyond the unit circle. This is what makes it an intersting function.

nope, think that the sum of the gp 1/2+1/4+1/8+... is 1. there are two other = signs but they don't need changing either. i really do'nt know anything about lebesgue integration; this is all stuff i've cobbled together f from various sources that feels like a reasonable idea of what it is. i've had to use measures for probability but i never really understood them except to konw that it can safely be ignored. it is a useful tool where, once you develop a certain level of intuition, what yuo want to be true is true.

lebesgue integration is best ignored, as the book states, but let me give you the definition of "almost" that this is talking about. Let R be the real numbers. A set S in R has measure zero if given e>0 there is a collectuion of open intervals that cover S that has total length less than e. example: let S be any coutnable set, and let s_r be the elements in S with r in N, whcihch we can do sinvce S is countable. about the element s_r place an open interval of width e2^{-r} then these discs cover S and their total width we can easily compute is the geometric series e(1/2+1/4+1/8+...)=e so any coutnable set has measure zero. what lesbegue measure does is this: it assigns the size b-a to the interval [a,b) and then extends this measure to all sets that can be obtained by intersecting, complementing, taking the union of these basic sets, and do so additively. notice that (a,b) [a,b] and (a,b] all have measure b-a since we are adding or subtracting one point at a time and a point p is contained in the set [p,p+e) and has measure less than e for all e>0 hence it must be zero. the idea then is loosely that to work out an integral of f(x) we look at the points where f(x)=[t,t+e) and work out the measure of this set of x's and assign a contribution of e times the measure (a little like adding up lots of rectangles of area) and work out the limits and see what happens. It is all very messy. the upshot of it is that if we take a fucntion like f(x)=0 if x is rational and 1 if x is irrational, then the integral over the intercval [0,1] doesn't exist in the riemann sense but does in the lesbegue sense. f is only not zero on a set of measure zero (the rationals which are countable) and 1 on the rest, thus the area must be 1. it is usually hard to tell what the measure of a set is, and is very hard to find a set that is not measurable (there are some but they are patholgically bad in a technical sense) the moral of the story is that behaviour on a set of measure zero can be ignored. if we ignore a set of measure zero then we are still saying "almost everywhere" ie that last function is almost everywhere equal to 1 since the the places it isn't 1 are a set of measure zero. if two functions agree everywhere except a set of measure zero (ie almost everywhere) then they have the same measure theoretic features, that is the integrals are the same over any given region.

your first piece of latex is once more not a series, and you even then call it a sequence, and no that isn't what i said the terms of the series ought to be. secondly things can't diverge to 8.7 that would be a typo, tough. read it agina carefulyl: prod(1+x_i) converges if sum x_i converges notice that the terms in the product are different from the ones in the sum. i#ve no idea if the associated sereis converge.

yes, the entries are called binomial coefficients. nCr is the r'th number in the n'th row as r goes from 0 to n i think you should figure out the formual yourself by thinking how when you expand (1+x)(1+x)(1+x)...(1+x) you can obtain x^r: hint it's lik picking the x from r brackets and the 1 from the remaining brackets.

that isn;t a series, that is product. you were supposed to work out the series to sum and see if it converged thus telling you if the product converged.

the product (1+x_1)(1+x_2)(1+x_3)... converges if and only if the series sum x_i converges. or in this case the product (y_1)(y_2)(y_3)... converges if and only if the series sum(y_i - 1) converges so translate it to a sum, what is the series?

To repeat myself, when i said you cannot generalize the cross product I was talking about a bilinear "multiplication" of vectors. not a multilinear form. your epsilon with the subscripts is an unnecessarily complicated way of writign the sign of a permutation, but you have discovered the idea of the exterior algebra if you didnt' know it already. The hodge dual is in spirit this: to integrate over a volume in R^n i need a priori to have picked a convetion about which direction is "positive", right, ie in 1 dimensions i take the integral in the usual sense and we know what we mean by a positive and negative area, and if we reverse the limits in the integral we reverse the sign of the integral, similarly in R^2 and so on. so, we pick our infinitesimal volume dxdydzd...... with n letters in some order and so on. Now, roughly, given a set of r vectors we look at the r-dimensional volume they span and thin how our infintesimal volume dxdydz.... restricts to this volume, and then try and add in other vectors until we make someting that roughly agrees with dxdyz.. on the infintesimal level. The other vectors we pick and the orientation we tahe for them give us the "hodge dual" of our original r vectors. this is not very accurate but it is right in spirit. Example. Suppose we pick our standard example in R^3 and we take our integrals with respect to dxdydz in the usual sense. No let me take, for simplicity, the unit vectors i,j and let me work out the hodge dual, well, the i and j span the dx dy bits of the volume so i need to take a vector in the dz bit to make it a proper volume, that is the vector k. if i were to do it for j,i in that order i've switched the orientation of the span so i need to take -j. extend this linearly and so on. so the "hodge dual" takes in r vectors and spews out n-r other vectors that allow us to do integrals over R^n with the correct sign (and zero if the r vectors are not linerly independent) i am glossing over some facts. such as the fat that the hodge dual is actually defined on /\^r but then /\^r can be thought of as the r-tuples of vectors with a + or - sign modulo some equivalence relation. the equivalence realtion is related to the reording and sign changes that you've come up with your self.

i meant the vector product as an operation on two vectors in simple terms of its components. Of course, given any two vectors (or n-1) i can find some vector orthogonal to both (or all), but that isn't the generalization i meant. to be technical it's because (deep breath): /\^2(R^3) is isomorphic as a vector space to R^3, and this is a perculiarity of the fact that 3=2+1. Tecnhically the isomorphism is via the hodge dual, and indeed given any n-1 linearly independent vectors in R^n then there is a nonzero element of /\(R^n) labelled y=x_1/\../\x_{n-1} which yields a unique vector x_n such that y/\x_n is the unique voume form. The so called hodge star duality (x_n is written as *y). Phew, ok?

that is a very simplistic gross anatomy of it, and not correct in general. for instance a scalar can be recovered as a component of a vector (eg use of the vector product might yield the vector (1,0,0) from whcih the scalar 1 can be read off). that said, it isn't a bad "rough" idea to remember. there is more than one way to do any question. example: find a vector orthogonal to a=(1,-1,0) and b=(2,0,1) can be fouund simply as axb, and is useful for finding the equation of planes) or you can solve the two equations a.x=0 and b.x=0 simultaneoulsy (there are infintely many solutions for x). by inspection (1,1,-2) will do it, and is infintely easier than using the cross product) one important note is that the dot prodcut can be generalized to any dimension, but the vector prodcut is perculiar to the 3 d world.

physics is full of good examples, and was indeed the inspiration behind the basis of vector algebra. so throw in plenty of examples. the only person who can say if they're good or not is yourdadonapogostick.

the title "vector multiplication" is at best a "title by analogy" and at worst misleading. With very few exceptions can you mutliply vectors in a sense that makes the name reasonable and accurate in its suggestiivity. The reason I dislike it is because it immediately makes people think that you can divide vectors, and you cannot, in general. here's another use for th dot product. suppose that i want to describe all points lying in some plane P. I can do it like this. Let r be some point in the plane, and let n be a vector orthogonal to the plane. imagine drawing the vector n with its tail at r. then you can see that a point x is in the plane exactly when its displacement vector from r is orthogonal to n. that is P={x : (x-r).n=0} is the description of the plane. you can do the same thing with lines. let L be a line. suppose that p is a displacement vector parallel to L, and that s is a point on the line. then some point y is also on the line exactly when the displacement vector from s to y is parallel with p, that is L={y: (y-s)xp=0}

and that n.n=1 as well, sicne i said n was a unit vector.

you do know that u.(v+w)=u.v+u.w, right?

what are you dot prodcucting with what? think about it. you want to kill the m, so how do you kill the m if dotting with orthoginal vectors gives zero?

or should we go more basic: if A=B then doing the same thing to both sides leaves them still equal since that is what an eqaulity means,

i've already in this thread pointed out how to do this question, it has nothign to do with dividing (we'venot stated ahwt multiplication is, never mind division). use the fact that the dot product of orthogonal vectors is zero.