Recent content by Chirag B

I'm not sure I understand what you're saying Princeton Review is doing. Can you possibly scan a picture so others understand how this works?

Oh I see what you're saying. So they take up to be positive then, that's all.

This is more of a relative velocity question. As you said, with respect to the bottom mass, the top mass is not moving. The idea is the same as you driving inside a car. The car is moving, and with respect to the car, you are not moving. However, with respect to the ground the car is moving, so...

Haruspex is correct in his response. The idea is that the force given by Hooke's Law for Springs counters the motion. This is the reason for the negative sign. So here, the object is moving downward. Let us choose the downward direction for +\hat{i}. Then, mg is "positive", and the force that...

No. It will not always be from 0 to what makes the inside of the trigonometric function 2\pi. Here, 0 to \frac{2\pi}{3} will only give one loop. Since cos\left(n\theta\right) gives n loops when n is odd, there are three loops, or petals, here. Going from 0 to 2\pi will only give you one loop...

That's a good point. It appears I was mistaken. But all convergence/divergence tests appear to fail here. Is there perhaps another way to do it?

Samtouchdown, I would try something along these lines: We want to determine the convergence/divergence of the series \sum^{∞}_{n=1}\frac{sec(n)}{n}. You said writing out terms is of no help. Perhaps it is not? Let's try first. \sum^{∞}_{n=1}\frac{sec(n)}{n} = sec(1) + \frac{sec(2)}{2} +...

So I would have to use the Monotonic Sequence Theorem, I take it?

Unfortunately, I'm not familiar with Fixed Point Theorems. However, I am familiar with rudimentary sequences/series. Would I have to prove the convergence of the sequence as Office_Shredder has suggested, and only then use the formula that micromass gets? Thank you in advance.

Ah, that's a neat trick. So you would have to think a little outside of the square root, if you get what I mean. :-p Thanks!

I'm having a little bit of trouble figuring out how exactly to do this. Prove that \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}} = \frac{1\pm\sqrt{4n+1}}{2}. How exactly does one go about doing this? I mean, I understand it goes on infinitely, but doesn't that create an infinitely...

LOL. And for the poster who asked—you could try Wikipedia. I've heard that it has quite a bit of formulas as well as their proofs. …or you could always by a Trigonometry (Trigonometry only!) textbook online. I'll bet there's a decent one that covers everything from the basics to advanced...

That's a nice way to do it...interesting!

Makes sense. :smile:

Ah, well then I was wrong. But as far as I know, l'Hospital's Rule still applies. You were angry at the poster's making the mistake of taking the variable x outside of the limit, I presume?