Recent content by rpf_rr

yes, you are right, 2i, I wrote wrong, but the problem for me is the integral "from 0 to 2i", i can't find a simple expression, while Mathematica do (only a cotangent and something like 1/p, or similar). The other methods, for me (maybe I'm not too experienced) gives too "complex" results...

I found this integral in a book ("A course of modern analysis", Whittaker): \int_{0}^{\infty} \frac{sin(bx)}{e^{\pi x}-1} dx I tried to use residue theorem in the rectangular domain [0,R]x[0,i], with R-> \infty , but i couldn't do the integral in [0,i]

Officeshredder is right. There are two possible situations, you pick 2 chinese 3 british and 2 indians or you pick 2 chinese 2 british and 3 indians, so you have 2C2*4C3*5C2+2C2*4C2*5C3=100

for nCk you mean n!/(k!*(n-k)!) ? If yes we computed the same thing, or better, you are right, I've done an error in the third factor, it is actually 5C2=5!/(3!2!), for me it's 600, but i made the same reasoning as you did, and i think it's right, if we understand the problem correctly.

you have to choose two chinese forced, at least two british and two indians, and one between british or indian. so you have 1*\frac{4!}{2!2!}*\frac{4!}{2!2!}*(3+2)=180 the last term: 3 indians+2 british

yes you are right, if a and b are primes

It's the line created by the vectors that have their component perpendicular to a equal to x*sin(theta) (it's not exact, the vectors must stay in a certain sense at right of a, if not they created two lines), the vector product is for definition the product of a times the component of the other...

you should write 6

You are right, you can see why in another way, there are no corrents in vc and there is a short circuit beetween the nodes near vc, i.e. voltage=0, so vc must be zero

Integrate the wikipedia equation with the conditions x(0)=0, v(0)=15.24 (cauchy problem); the system is underdamped, you will find a solution like exponential*sin, for the maximum displacement set the condition v(t)=0, i found it at t=0.01572s, if you want further help i can give you the...

my result is correct for relativity (at least for special as far i know), it is even reported in my textbook

Integrate, you find p=Ft, substitute p=mv/sqrt(1-v^2/c^2), some arithmetics and you fininshed, your solution is wrong, is valid for v^2

a, b real numbers with binomial a(a-b)+b(a-b)=(a-b)(a+b)=a2-b2 or with computation a(a-b)+b(a-b)=a2-ab+ab-b2=a2-b2 it's a basic binomial formula: (a+b)(a-b)=a2-b2