Recent content by susskind_leon

Damn, I guess you're right ;-)

Imagine |a>=(1,0,0), |b>=(0,1,0), |c>=(0,0,1) What would the square matrix be then? Once you've got that, how can you relate your "hypothetical" square matrix to the correct one?

You cannot solve this at all, because neither of the two limits exist. Thus you cannot carry out the substraction.

Keep in mind \arctan(1/x)=sgn(x)\pi/2 - \arctan(x)

Can you use complex numbers? If so, you can rewrite your cosine in term of complex exponentials which will make things quite easy.

Check out Dixon's identity and the rules for the Gamma function (basically that it's just factorial for integer arguments) and you will find your way from eq.3 to eq.1

I still think you can definitely do it by multiplying with the denominators, you just need to be very persistent and the calculation is very lengthy.

In principal, I would say it's a good idea to exploit contour integration, but keep in mind that when you have a removable singularity, the residue is 0, so the straight-forward way where you just consider the residue and argue that the arc doesn't contribute won't work.

Well, you probably should have done the steps the other way around, starting with 0 ≤ √x√y and arriving at √x+y ≤ √x + √y

What do you get when you multiply by all the denominators? Can you do that and isolate |a-b| on the lhs and |a-c|+|c-b| on the rhs?

It obviously doesn't work when an is not bounded. I am understanding correctly that an does not converge to 0, right? if so, just consider bn=1/n and an=n^2. an*bn=n does not convergen. Bam!

If I'm not mistaking, Ar has only one Element for each r, so what does that tell you about the intersection? What is the condition for an element to be in the intersection of sets? That keeping in mind, what does that tell you about the intersection of Br. As for Cr, well, can you imagine what...

Please write down your entire calculation then I can help you with it.

You're weeeeeeeeeeeeeeeeeelcooooooooooooooooooooooome ;-)

How about you do the calculation and see what you end up with? There is a possibility that L cancels out... Btw.: Which one is the second formula for you?