Recent content by eigenperson

If you're going to ignore both the original poster's notation (##\theta## has a specific meaning, and it's a vector integral) and the domain of integration, you can't expect to just type the integral into Wolfram Alpha and expect to get the right result.

Well, I didn't check the ones involving the hyperbolic functions, but it is the same thing that is going on in the first and last sets in the above post. EDIT: Obviously it has nothing to do with the identity in the original post, though -- I don't think there's any clear relationship between...

I think there is a genuine, though not precisely defined, duality here. For example, the dual of the identity ##2\sin^2 x = 1 - \cos 2x## is ##\arccos(1-2y^2) = 2\arcsin(\left|y\right|).## (Note: the absolute value comes into the picture because ##\arccos(\cos(x)) = \left|x\right|## -- at least...

Please tell us some more: What mathematical background do you have? Are you planning to study pure mathematics, or applied mathematics?

I'm surprised to hear that you are getting an odor from the saltpeter, because potassium nitrate really shouldn't produce much of an odor. There might be a small amount of nitrogen dioxide, but I can't imagine it would be enough for you to be able to smell it. Maybe someone sold you something...

Ideally the impurity will be much less soluble than the compound you are trying to isolate (let's say >100 times less soluble). If this is the case, then at first, when you add a little bit of solvent, a noticeable amount of the compound will dissolve (so the quantity of solid will decrease...

To count the number of k-element multisubsets of an n-element set using balls and bins, first make a "bin" for each element in the set. (In this case, there are n bins. Your version of the problem uses the notation "k" for the number of bins, which I think is the source of the confusion.) Then...

No. The number of multisets is ##\left(\!\left(n \atop k\right)\!\right)##, not ##\left(\!\left(k \atop n\right)\!\right)##.

Yes, the loop around which you are computing the emf must always be the boundary of the surface. If you are confused about this, I would suggest that you watch Walter Lewin talk about it (start at 5:00) if you haven't already. As for an example, basically any Faraday's Law problem will do the...

You are allowed to choose any surface* as long as its boundary is the loop around which you are computing the emf. In general you choose a flat surface if that's possible, so as to make calculations easier, but you could choose whatever surface you like as long as it has the right boundary. *...

EDIT: Removed due to your edit. EDIT 2: Bah. I deleted my post by accident. Stay tuned... Obviously my second post was not helpful, so let's get back to the main issue, which is this paragraph in your original argument: My point is that this is an invalid argument. In Sentence 1, you...

OK, I just thought of another way to say this which might (or might not) be clearer. Let's say I have a set S of positive integers with the following properties: 1. Exactly 1/2 of all the numbers in S are divisible by 2. 2. Exactly 1/3 of all the numbers in S are divisible by 3. 3...

No, that doesn't really address the point I'm trying to make. Let's look, for clarity, at what happens to the numbers between 1 and 600. According to your procedure, I start by removing all the numbers divisible by 2. This takes away 300 of the numbers, so that 300 of the numbers are left...

For integrals involving products of cosines and sines, the general principle is as follows: 1. If one of the trig functions has an odd exponent, substitute for the other function. If they are both odd, substitute for either one -- it's your choice. (Example: if the problem is to integrate...

One issue I have with this argument is this line: Even assuming you formalize this... how do you know? Sure, ##p_i## divides 1 out of every ##p_i## numbers, but that is not the same as saying it divides 1 out of every ##p_i## remaining numbers. Maybe the primes before ##p_i## have already...