Recent content by Feynman's fan

Not exactly homework in my case but I guess it fits the level. We know that a = v dv/dx holds when moving on a line. Does it hold when moving on a plane? If yes, what would be the exact formulation of the statement. I mean, I read a = v dv/dx as a(x) = v(x) dv/dx(x) (or are they functions of...

Here is the table of contents of Nonlinear Dynamics and Chaos (by Strogatz) Overview Flows on the Line Bifurcations Flows on the Circle Linear Systems Phase Plane Limit Cycles Bifurcations Revisited Lorenz Equations One-Dimensional Maps Fractals Strange Attractors Last quarter, there was a...

I'm looking for good examples of physical motivation for integrals over scalar field. Here is an example I've found: If you want to know the final temperature of an object that travels through a medium described with a temperature field then you'll need a line integral It appears to me that...

I'm going to take both eventually but I have only one free slot left in my bachelors program. How would you compare the difficulty level (assuming that Complex analysis is fully rigorous as well).

I have one slot to fill in in the coming term. The two candidates are Functional Analysis and Complex Analysis (both on the undergraduate level). Here are some questions: 1) Which one would you pick and why? 2) What other classes in the standard B.Sc. math curriculum rely on either of these...

Here is the problem statement: I thought it's a straightforward exercise on the divergence theorem, yet it looks like \operatorname{div} f = 0 . So the surface integral is zero? Am I missing some sort of a trick here? The exercise isn't supposed to be that easy. Any hints are very appreciated!

Simon Bridge, thank you, it's all clear now!

I'd like to show that if \alpha>\frac{1}{2} then (x^2+y^2)^\alpha is differentiable at (0,0). The usual way is to show that the partial derivatives are continuous at (0,0). Yet I am a little confused how to show that 2x\alpha(x^2+y^2)^{\alpha-1} is continuous at (0,0). I have tried working...

The problem statement: And here is the textbook solution: Note that the the simple calculation makes use of the Boole's inequality and the reasoning itself is the Probabilistic method And my questions are: Where do they use that those are cubes? What breaks down if we just go through the...

Thanks a lot for pointing out the direction! By the way, can we use the result from above, which is \lim\limits_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = \frac{1}{\sqrt{\pi}} e^{-x^2}as step functions to show that...

This approach sounds very interesting! Yet how did you arrive at the limit? I thought that if we want to drop \lfloor · \rfloor at the very beginning, we need to proceed as follows: \begin{align} \lim_{n \to \infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} &=...

I've tried the suggested things yet unfortunately I still cannot arrive at e^{-x^2} which supposedly should follow from something like (1+\frac{-x^2}{n})^n. But how do I proceed with \lfloor x\sqrt{n} \rfloor^2? Could you please elaborate on this?

Thank you for looking into it! Here is what I did. I was trying to squeeze the limit by using the Stirling's formula, which is: \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}<n!<\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12(n-1)}}. So I started with the RHS of the squeezing: For simplicity...

Here is the problem at hand: \lim\limits_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = \frac{1}{\sqrt{\pi}} e^{-x^2} I've tried to evaluate the limit using the Stirling's formula, many things canceled out yet I got stuck. Thanks a lot for any help or...