Recent content by khari

Sorry I haven't checked back since, I got caught up in an Optics lab and forgot. Yes, the proof you gave is essentially the same as was given given in class:) I don't think anyone got it right, or if they did I don't think anyone proved it. Thanks for your help. And yes, I am interested in the...

It sure does seem to be. I've much to think about, and not much time to think it in. I'm going to check back with this thread in the morning to see if anything conclusive has turned up. Otherwise I guess I'm going to hand it my answer as is, confused or not. Thankfully this assignment is just...

Well, it did correct a misunderstanding I had about the completeness theorem, which I had added to my explanation after I posted this thread, but before you brought it up. I had it stated that because |sin n| > 0, it must have a positive lower bound, which is apparently incorrect, because it can...

Well it's been interesting for me so far, and has given me a little doubt about my conclusion or at least that my statement of my conclusion is quite lacking.

Yeah, that's basically where I'd gotten to. My problem is that I'm not sure if it can be simply stated like that or if there's a more formal way to prove it. Even if it is obvious enough to state, I'd still like to see a formal proof, just because I need to get as familiar with proofs as...

Homework Statement Is it true or false that \exists c>0 s.t. \forall n \in N, |sin n|> c, Justify your answer Homework Equations The Attempt at a Solution I figure it is true, because n != (k-1)\pi \forall n,k \in N, so |sin n|>0 \forall n It seems fairly obvious to me that if |sin n|>0...

I got it. Thank you all, you've been a big help.

I must be missing something obvious, because I've tried that as well with pretty much no success. :frown:

Sorry if it seems stupid, but I'm horrible at proofs, and am still coming up a bit short on proving n^2 \leq {(\frac{x}{2})}^n I mean it seems fairly obvious that this is true to me, and more obvious still when I begin writing it out term by term for increasing n, but is this good enough to...

Problem: Find the limit of the sequence a_{n}=\frac{n^{2}2^{n}}{n!} My first thought was to say that 0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!} and by squeeze theorem, since \frac{x^n}{n!}=0 for all real x, my original limit must be 0 as well. Now all I need to do is prove...

Is Halliday Resnick and Krane the same as Halliday Resnick and Walker, only a different edition?

Has anyone ever used Modern Physics by Serway? Is it any good?

I browsed through but couldn't find it. Could you link me to it?