Mac OS 10.4.10 Under Development

[oober_freak]

Mother, I'd like to ****.

Q. While I'm thinking about it... what starts with an f and ends in uck?

A. Firetruck. you dirty minded b******s.
.

Q. What starts with F and ends with K, and if a man doesn't get it, then he uses his hands?

A. Fork!

I couldn't resist posting this.

 

[ryaxnb]

I can tell you those extra 67MHz don't make a lot of difference. First Santa Rosa MBP and I'm buying - was going to wait until leopard, but... well, you know...

I'm living on a two-year old Mac Mini with a gig of RAM and it's still slow. In October we're getting a Mini or Macbook w/ intel.

 

[jsalzer]

Hope this cleared things up for you, jsalzer.

Thanks, bud! I'm sure I knew this at some point during my undergrad, but when you don't use it, you lose it.

Just like ".10" version numbers. (To attempt to tie back in to the original topic.)

 

[slayr-cxf50]

10.4.10 oo look, its a mnemonic.
Hmm... perhaps this patch will introduce some basic support for some Leopard features? As in Tiger being able to detect them..

 

[The Roc 1217]

I don't know, Clive, my friend. A friend asked me what the amount was, and that's not what I came up with.

We have:

0.002 + e^i + 23.14 + 1

(Breaking up e^(iPi))

This gave me 24.142 plus an imaginary amount. And since imaginary money doesn't go through banks, I figure the final answer was $24.14.

I guess my question is, how did you get that e^(iPi) = -1?

That would mean that e^i = -1/23.14. Where did that come from?

I'm confuzzled.

Polar coordinates. R x e^(iΦ) gives the point on a circle with radius R, rotated Φ radians counterclockwise from horizontal right. that's because e^iΦ=cosΦ + i sinΦ. and when you split a power into factors, you don't add the two new ones or multiply them, , like e^ab, it's not e^a+e^b or e^ae^b, it's (e^a)^b. in that example, Φ=π and R = 1, so you get the value -1+0i.

 

[jsalzer]

Yup

Polar coordinates. R x e^(iΦ) gives the point on a circle with radius R, rotated Φ radians counterclockwise from horizontal right. that's because e^iΦ=cosΦ + i sinΦ. and when you split a power into factors, you don't add the two new ones or multiply them, , like e^ab, it's not e^a+e^b or e^ae^b, it's (e^a)^b. in that example, Φ=π and R = 1, so you get the value -1+0i.

Thus the importance of reading to the end of the thread before replying. Clive already taughteth, and I already learndeth.

Thanks for trying, though. And welcome to the board.