DQW

Not correct. Neutrons and protons are made of quarks, which are held together by gluons (not mesons - mesons themselves are made of quarks). Electrons are fundamental particles. They are not made up of quarks.

There a lot more confusion behind this question than just what is asked in the question itself. I recommend that you first understand the classification. Hadrons are all particles that interact through the strong force. This includes mesons like the pion (which are made up of pairs of quarks) and baryons (made up of 3 quarks) like the proton and neutron. Mesons, possessing integer spin values are also classified as bosons, while baryons are fermions. Particles that do not interact through the strong force are leptons, like the electron and its neutrino. What is the role played by the the short lived particles like the (who ordered that) muon ? I can't answer that question. I'd wait for one of the high-energy folks to show up.

I'm not sure I clearly understand the scheme outlined in post #7. The first line represents the polarizations transmitted by A. I imagine that 2 of these 4 polarizations represent a 0 bit and the other 2 represent a 1 bit. Now in the second line, you describe the filters that B uses - they can be either up/down/left/right (+) or diagonals (x). Okay, now I'm not sure how B interprets a photon that correctly goes through a filter. For example with the | photon (the second photon in A's message) and the x filter, B will observe 'transmission' or 'blocking' with equal probability. And in half those cases when the photon is transmitted, he will either measure \ or / (which can be determined by a second filter) with equal probability. Now one of these measurements WILL correspond to the correct bit, but the other will not. This brings me to line 3 : B does not know whether he is using the correct filter, so I imagine this line corresponds to knowledge that only A possesses. As for the 4th line, I've got no clue what it means. You haven't even told us the convention used to translate a polarization into a binary bit...so I can't make any sense of that line. In any case, you are asking about something that E does. You state that E has somehow (through a virus or other clever hacking trick) obtained a list of the correct filer sequence. But the correct filter sequence is essentially the key to the ciphertext. With the key, E can correctly determine all the photon polarizations. But this determination does not have to destroy the existing data. For instance, after E deciphers the photon polarizations, she merely prepares the same polarizations and passes the message onto B as though nothing happened along the way. In short, with a knowledge of the key, it is possible for E to decrypt the message and pass on an identical message to B. When B receives the message, it will be exactly what was sent by A.

First of all, it is not possible to measure the polarization of a single photon (of unknown polarization) using a polaroid filter. The probability that a photon of polarization angle [imath] \theta [/imath] with respect to the filter orientation, goes through the filter is [imath]cos^2 \theta[/imath]. Even in your scenario with 4 distinct polarizations, it is only possible to rule out one of the 4 polarizations by making a measurement. And once you make the measurement, you lose the knowledge of the original polarization because you have now polarized the photon along the filter orientation.

Psion, What you've shown in the picture IS [math]\frac{x^5 + 3x + 2}{x^3 + 2x+1}[/math] If you follow matt's procedure and go one step further, you will get the required answer. If you're not convinced, look at [imath]x^5/x^3[/imath] and [imath]x^3/x^5[/imath]. Which one gives you a quotient of [imath]x^2[/imath] ?

I don't understand what post 17 is saying, but here's post #16 without the LaTeX : "Isn't exp{i*theta /n} a constant number ?" So can you not pull it out of the summation ?

Isn't [imath]e^{i \theta/ n} [/imath] a constant number ?

No, please do not waste your time/money looking up Elton's book. You will find NOTHING of relevance to your question there. Take budu's advice instead. You can estimate molecular sizes from bond lengths and molecular geometries. Here's some bond lengths : http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html You can also estimate moolecular sizes in liquids from knowledge of density and molecular weight. This may give you a small overestimate (that depends on how you define molecular size).

Eureka !!

You fixed the first line but not the second : there should also be a 2a_0 in it. Remember that zero is also an even number.

Looks good so far. Any ideas on how to sum that series ? Does it look familiar ?

I couldn't comment on that unless I know how you got it.

Do you have ANY substantiation for any of this ?

From darkkazier's link : All is not lost. There is still hope... ...but on the other hand, there's this : http://politics.yahoo.com/s/ap/20050802/ap_on_go_pr_wh/bush_intelligent_design

Yes, but also, that body will be the only source of gravity. Remember, mass creates gravity. Not at all. The force of gravity on you would be completely unnoticeable. Yes, but since "you" are the only source of gravity, this force will be small. That's like bad sci-fi ! <help> The less stuff there is in the universe, the less gravity there is in the universe. 1. No, that's not true. Gravity does not distribute itself depending on "how many people there are, to eat the pie". This is known as the superposition principle. 2. If you remove objects from the universe, you also remove gravity from the universe. Finally, you are left with the tiny bit of gravity caused by the last remaining body. Gravity does not follow the laws of supply and demand. It does not switch on and off based on need. Outcome ? Nothing fancy. You'd have a very boring universe.

1. There's a small mistake there. Why don't you write it out step by step ? And can you write it in the exponential form ? 2. Does z change, if I replace [imath]\theta[/imath] with [imath] \theta + 2 \pi m [/imath], where m is any integer ? So, in the most general case, the argument should contain the latter, rather than the former.

(a) You have a tiny error in post #14. Why did a_0 change signs on the RHS ? (b) You are not asked for all the coefficients, right ? So, just write out the first few terms (I'll let you figure out how many you need to write), square it, and compare coefficients.

The last 2 places where you write "by definition of a subspace", you should write (if you want to write anything at all) "by definition of the intersection", because that's what you're using.

Yes, at the right temperatures and fields, you have ! See posts #3 and #6.

How do you show that the even terms are zero ? It does not matter what value of n you start from, for the even terms to vanish (but in any case, you start with n=0 because that term is part of a general power series).

You have z = a^{1/n}, where 'a' is some complex number. What is the most general polar representation for any complex number ? Use this representation for 'a'. What do you get ?

I was talking about part (a), not part (b) Apply the power series expansion to each side of the given equation (the "fact"). What do you get ?

(a) You are given 2 things : 1. tan(x) is analytic in (-pi/2, pi/2) 2. tan(x) is odd Substitute the expansion (from 1) in the equation for 2.