Recent content by shakgoku

I have seen four two qubit entangled states of the form: $ \frac{1}{\sqrt{2}} \left | 00 \right > \pm \left | 11 \right >$ $ \frac{1}{\sqrt{2}} \left | 01 \right > \pm \left | 10 \right >$ I want to write a most general two qubit entangled state. I presume it can be of the form: $...

1. There are x distinguishable ants and there are x pots full of food. Due to the smell, the ants arrange themselves in a circle on the circular rim of each pot. In how many ways can they do this? note: Any number of pots can be free of ants. For example all the ants can be on the circular...

Multiply and divide by (n+2)^{\frac{1}{2}}+n^{\frac{1}{2}} After simplification you will get, \frac{2}{n^{\frac{1}{2}}[1+(1+\frac{2}{n})]} Now use the comparision test, by comparing with \sum \frac{1}{n^{\frac{1}{2}}} which diverges by P-test.

I just realized I made a blunder in calculation. That substitution yields \int \frac{\sin^2\theta}{cos \theta}d \theta=\int \frac{1-\cos^2\theta}{\cos \theta} d\theta and finally it is equal to, \int \sec \theta d\theta-\int \cos\theta d\theta = \ln...

I don't get it. if we take x^3 inside we get, \int \sqrt{\frac{x^4-1}{x^6}}dx=\int \sqrt{\frac{1}{x^2}-\frac{1}{x^6}}dx

Homework Statement \int \frac{\sqrt{x^4-1}}{x^3}dx Homework Equations The Attempt at a Solution tried substituting x^4 = \sec^2 \theta to get rid of square root but it was of no use because, I got another complex integral \int \frac{\sin^2\theta}{\cos^{\frac{3}{4}}...

hi shashankac655! :smile: Most straightforward way to solve it is using partial fractions.

Hey tiny-tim! Thank you , now everything in the page makes sense to me! :)

Homework Statement This is an extract from elements of mechanics by giovanni. I'[m unable to understand the meaning of R^d and C^\infty http://flic.kr/p/9K8P7p Homework Equations The Attempt at a Solution

Thanks , its so obvious now. The slope is at x = 0 is beta. So, if beta<1 the slope keeps decreasing and never gets a chance to intersect y = x again. But beta > 1, tanh(beta*x) stays > (y = x) but due to decreasing slope, intersects y = x again at two places x = + and - r (if r is a root)...

hi tiny-tim, The shape of tanh(βx) is nearly straight line near origin and has horizontal asymptotes at y = 1 and y = -1 The shape of x is straight line through origin , slope = 45 degrees. so x = 0 is definitely a root. how to find it there are more roots?

You can find the formulas yourself by suitable substitution in integral for x-bar and y-bar. and using from both of them try to get separate integrals for r and theta

Homework Statement what happens to the number of solutions of the equation x = \tanh(\beta x) When \beta is varied from \frac{1}{2} to \frac{3}{2} [/tex] a) unchanged b) increase by 1 c) increase by 2 d) increase by 3 Homework Equations...

:biggrin: I get the same answer by considering, F = -kx and using k = \frac{dF}{dx} ==> k = \frac{2Gm_{1}m_{2}}{x^3} and T = {2 \pi} \sqrt { \frac{\mu}{k}} At \frac{T}{4} Collison occurs, so t = \frac{T}{4} = {\frac{\pi...

Thanks. I had doubts but I just got carried away because its easy to solve after making that wrong step :D I finally made the substitution you suggested and got an integral of form A \int_{0}^{d} \sqrt{\frac{x}{d-x}} dx =t And I solved it to get t = \frac{\pi d}{2}...