Travelling at twice the sound of speed
Feb. 10th, 2004 06:28 pmAMC today! Score: 126½. 19 right, 1 stupid mistake, 5 blank. Chris got 4 wrong and just barely passed with 101½. I don't remember any other scores right now.
EDIT: I had forgotten about Fergie's epsilon convention. Using that, I got 126½+9½ε. (I got a probability of 5/8 for number 20! I seriously did!)
I dropped a parabola. Have you seen it?
Chandi: No, I've only seen ellipses.
Marshall: What was the distance between the directrix and the focus?
Me: About half a foot.
Dr. Ostfeld: I think that's a bit of hyperbole.
Avi: What was it made out of?
Me: Blue.
Dr. Nevard: It shattered into 3 hyperbolae.
Narratives came today. All good, yay!
I knew there was a final case of integration by partial fractions which we didn't cover. Now I know why. *shudder*. Let me quote Apostol on it:
The case m>1 may be reduced to the case m=1 by repeated application of the recursion formula
∫du/(u²+α²)m=(1/(2α²(m-1)))(u/(u²+α²)m-1)+((2m-3)/(2α²(m-1)))∫du/(u²+α²)m-1,
which is obtained by integration by parts.
Now *there's* a nasty recursion for you.
EDIT: Oh, and don't forget, everyone, tomorrow's the 42nd day of the year!
-Sniffnoy
--
"REPAINT VERSION
This is not a new product at all. The Edition of us
whom it is already on market were painted again in the new method,
hi-skill and a point of view."
-engrish.com
EDIT: I had forgotten about Fergie's epsilon convention. Using that, I got 126½+9½ε. (I got a probability of 5/8 for number 20! I seriously did!)
I dropped a parabola. Have you seen it?
Chandi: No, I've only seen ellipses.
Marshall: What was the distance between the directrix and the focus?
Me: About half a foot.
Dr. Ostfeld: I think that's a bit of hyperbole.
Avi: What was it made out of?
Me: Blue.
Dr. Nevard: It shattered into 3 hyperbolae.
Narratives came today. All good, yay!
I knew there was a final case of integration by partial fractions which we didn't cover. Now I know why. *shudder*. Let me quote Apostol on it:
The case m>1 may be reduced to the case m=1 by repeated application of the recursion formula
∫du/(u²+α²)m=(1/(2α²(m-1)))(u/(u²+α²)m-1)+((2m-3)/(2α²(m-1)))∫du/(u²+α²)m-1,
which is obtained by integration by parts.
Now *there's* a nasty recursion for you.
EDIT: Oh, and don't forget, everyone, tomorrow's the 42nd day of the year!
-Sniffnoy
--
"REPAINT VERSION
This is not a new product at all. The Edition of us
whom it is already on market were painted again in the new method,
hi-skill and a point of view."
-engrish.com