matt grime

I see the standard of mathematics here hasn't improved. Just look at the groups of order what ever. It is trivial to find counter examples.

Well done, everyone. You've just convinced me that this is not a site ever worth visiting again in the hope that there was some interesting and informed mathematical discussion about anything. The sheer idiocy on this thread alone is amazing. Goodbye.

Yes, really you ought to finding the integral [math]\int_0^t v(x)dx[/math]

Good, I'm glad this had a decent answer. It ought to have done. The reason I was disappointed earlier was that I thought you had this nice proof in your head and were going to tell us it. Thank God shmoe was around to point you in the right direction straight away. I was thinking about asking you to consider that phi(p^n)=p^n-p^{n-1} and multiplicative and then trying to rewrite the sum, which would have led nowhere fast.

So when you said it was a surprising value you didn't actually mean it? As in you don't know what it is as some interesting function of some well known natural constants.

Juststuit: The difference is very important, juststuit. I imagine solving when the integrand is sec(x^3) is impossible in elementary functions. Woelen. I haven't seen that question before but the answer is e. It is always e. Or pi. Or possibly log(2), or something. (The limit seems to be 0.307...something, though I've not bothered to figure out what that is or why at the moment). Real calc, since someone asked, would at least need the definition that f is continuous if and only if f^-1 sends open sets to open sets.

Two comments: 1. is that (sec x)^3 or sec(x^3)? 2. if this is what people call calculus then I agree with insane_alien. here's an interesting calculus problem: work out the sum [math] \sum_{n=1,3,5,\ldots} \frac{ (-1)^{(n-1)/2}}{n^2}[/math] You will need to learn about fourier series (why is this interesting: because it is surprising that you can work out this sum at all, but you can, similarly you could try proving that the sum 1/n^2 is whatever fraction of pi it is supposed to be.)

This is 10-adic arithmetic. Look up p-adic and find out all about it.

That can't be correct. Why would they meet at a point when x=y=z necessarily? And you can solve that for t and get t=1 but that doesn't have anything to do with s and the second line, does it? You just need to find a point, if any exists, that lies on both lines, so do it. When are the x coordinates the same? And the y and z coordinates. Those are conditions on s and t. Now can they be simultaneously satisfied?

You cannot do that integral. It does not exist. You have integrated over a singularity at x=0.

It's not necessary to try to 'see that the GD can't get bigger'. If d divides x and y it divides ax+by for integers a and b. So in particular the divisors of n and m are the same as the divisors of n-m and m (you don't actually need the restriction on n=>m: divisors are well defined for negative numbers). So the set of divisors of n and m is the same as the set of divisors of n-m and m, hence the largest element in each of those two sets must be the same.

Rather unusual that someone decides that they should start a thread on this rather than its negation. And somewhat pleasing.

It just reflects a choice (one that is more natural to humans, it seems) to start counting from zero. You haven't seen 'negative' number of geese becuase you choose to start counting a set with no geese as 0. This doesn't hold with temperatures. You have seen negative temperatures (in all likelihood). We could just as easily have decided that the normal amount of geese to have is 100, so 1 goose in the old money is -99 in the new. 1 and -1 are just as real as each other, not that I'm saying how real that is. It is just that it normally makes more sense to scale your world so that 0 corresponds to nothing. Note, you've not seen a 1 ever, either, just something with a property someone labels oneness.

yes, and you should avoid going into these things for a while. there is no need to be at all concerned with manifolds, wedge products and the like, since you're only doing this with surfaces in 3 space. The motivation behind that is because you're doing stuff that is 'real life'. What woelen is talking about is the generalization to arbitrary dimensions, and not relevant to you.

Well, who's to say that he turned it down because he didn't want recognition (in whatever sense)? I'm sure he has many reasons for declining things, and I'm sure that none of us here knows what they (all) are. Mathematicians do things for many reasons, and recognition is almost never one of them. Some people do actively want to win the Fields medal, though, and one person I spoke to about these things actually said he refused to go to the ICM because there were too many of these self-publicists bound to be there. Perhaps Perelmen is just at the extreme end of those who find it bemusing to be recognised for doing something you enjoy.

Stick with it. Summation convention absolutely throws everyone the first time they see it (and boasting 'oh I got it straight away replies' are not wanted), but once you've got used to it you'll not understand how you survived without it and you'll have no problem using it. I mean, I can still remember that [math]\epsilon_{ijk}\epsilon_{ipq}= \delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}[/math] and I haven't used it for 8 years or more. (It may seem odd that I say you will not understand how you got by without it whilst saying I've not used it for 9 years, but I chose pure maths where summation convention is rarely used. For instance it is common to let v_i be an eigenvector with eigenvalue k_i and we want Av_i=k_iv_i, and we definitely don't want to sum that.)

The above comment is appropriate: S is not bounded above, and you don't need Lagrange multipliers to see this, it is also, correspondingly, not bounded below (z=1, x=-y, and let x tend to infinity or minus infinity).

Why can't you believe it? He turned down a prize from the ECM, he resigned his position in the Stekhlov. It was entirely inkeeping with what we have come to expect from him. (And is your caps lock key playing up?)

you are confusing two different uses of the word scalar. In ordinary analysis, when one talks of scalars one means constants, fixed unvarying elements of the real numbers (or complex numbers). Thus in this sense, if k is a scalar (a real number) then the derivative of kx is k, and the derivative of k is zero. However, in this context a scalar means a one dimensional vector. It is not a constant, just something that takes values in the basefield. Think of it as a 'scalar quantity' rather than a 'scalar', and to think of those old 'scalars' whose derivative vanishes as 'constants', if it helps. |vxr| is a scalar quantity: it is an element of R, but it is not a constant: it changes with r, which is a function of (x,y,z).

It does, but your r's go from minus something to plus something, and negative radii are not used.

*may* have solutions of the form, not *does* have. In anycase, it is badly written, but that is not uncommon in textbooks (it for instance states the solution in the degree one case is a^nx_0 and then immediately asserts x_n=a^n is a solution. It is not unless x_0=1

there are at least two factual errors in there, not to mention a matter of opinion that is very moot... ok, down right wrong. the value of a function at zero is incredibly useful, and is the whole basis of Taylor/Maclaurin series. Some folk don't half not help things.

We do not assume that the solution is b^n at all. We assume that the solution is of the form c_1*b_1^n + c_2*b_2^n: the equation is linear. And then from this knowledge we know that b must satisfy a quadratic equation based upon the information given. All you're doing is taking sloppy shorthand for absolute truth.

Well, that isn't what happens. Plenty of thngs around here do tell you what is going on. Please read them.

The probability is what the probability is. It's that *you* do not know what it is that matters. Note, *you* are assuming that each of the distribution of balls, from 1 to 99 of one colour is equally likely: why are you doing this? (and awhy not allow for 0 balls of one colour?)