Recent content by danoonez

That's not what I was hoping to hear. I'll check the Dirac text.

Hmm, I have the Dirac text on my bookshelf. I will go back and look through it. Thanks!

Is anybody familiar with LCAO approximation? I'm working on my last problem for this assignment and the class text doesn't really cover it enough for me nor do my reference texts cover it enough. Does anybody know any good references so I can read more on LCAO approximation? Thanks.

The balls in the boxes is an interesting (new to me) way to look at it. I'll give that a try and see if I can make any progress. Thanks.

I'm having a problem with a proof I came across in one of my calculus books but it's not the calculus part of the proof that I'm having trouble with. Here's the actual proof: "Prove: The number of distinct derivatives of order n is the the same as the number of terms in a homogeneous...

They will all be mandrel bends.

Unfortunately, a full 180 degree bend is not possible b/c of space constraints.

Hmm, I guess it's not as simple as I imagined. The details you asked for: It will be a full 180 degree turn 2.5" diameter pipe, approximately 70 degrees F., approximately 8 lb/in^2, the straight pipes inbetween the bends would be about 6" max.

It's not so simple to me being that I have no background in fluid dynamics, but it seems like it would be fairly simple for most mechanical engineers; anyway: This is to help me with the design of my charge pipes in my custom turbocharger setup. Assuming a constant radius throughout, which...

I wasn't sure if my use of V_\textrm {group} = \frac {\Delta \omega} {\Delta k} was legal. But then I realized that the uncertainty principles are used for wave packets so it was fine.

Please check my work. Using the first uncertainty principle: \Delta x \Delta k \sim 1 derive the second uncertainty principle: \Delta \omega \Delta t \sim 1 My work: \Delta x \Delta k \sim 1 \frac {\Delta x} {\Delta t} = V \Rightarrow \Delta...

Excellent. Thanks for looking.

Thank you. I will do that for the paper I hand in. vsage: I can't tell which step you mean. Could you clarify? B/c it may not have been a typo. Anybody else see any mistakes?

From \int_{0}^{L} A^2 sin^2 \frac {n \pi x} {L} dx = 1 show that A = \sqrt {\frac {2} {L} Here's what I did: First bring A^2 out so that I have: A^2 \int_{0}^{L} sin^2 \frac {n \pi x} {L} dx = 1 Then I use u subsitution for the integral...