Geek Joke of the Day
Posted in Personal Stuff by John E. Pannell on November 7th, 2008 at 1:00 pm In early 2007, I posted on this blog a mathematician’s limerick. It’s been a long time since I’ve tested the geekiness of this blog’s readers, or my friends on Face Book who also see this blog. So, I offer this limerick:
Mathematical Limerick
Who will expose their inner geekiness by supplying the translation of this. Try to solve it first without “cheating” by looking up the answer. It’s harder than the last one I posted, but not unreasonably so. Remember, it’s a limerick.
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Within a couple of hours two geeks made themselves known to me by answering this riddle. One got it right. The second was very close and just needed to brush up a bit on his calculus.
The limerick was easy… just read the equation aloud:
Integral t squared dt
from 1 to the cube root of three
times the cosine
of 3Pi over nine
equals log of the cube root of e.
Of course you need some sort of math background to know what any of that was, and how to “read” it. I’ve not attempted to verify it, but I’m told the equation is correct.
Ok… It works… Now I’m going to show how much of a geek I am. Those who aren’t math students should skip the rest of this comment.
The calculus is basic “freshman calculus” stuff. The trigonometry and natural logs are topics I was taught in high school. (I’m kind of embarrassed I had to look up cos(pi/3). )
1. Integrate t^2dt over the interval of 1 to 3^(1/3)
== 1/3 ( (3^(1/3)^3 – 1) == 1/3 ( 3 – 1) == 2/3
2. Compute the cosine(3 pi /9) = cos(pi/3) = 1/2
3. Left side of the equation == 2/3 * 1/2 == 1/3
4. Simplify right side (very easy) = ln(e^1/3) = 1/3 ln(e) = 1/3
Eureka!