No, I definitely agree. But there's something amusing about the intersection of logic and math here, the fact that we can find a mathematical solution that ignores the physical realities, but because the problem is posed as a physical problem so its preconditions are impossible, logic dictates that any answer is as acceptable as the "correct" one. On Fri, Apr 8, 2011 at 4:53 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Tom — I think the original problem is interesting by itself. I just mean that if you graph the following there's no place where all three surfaces intersect. xyz=37 (x+1)(y+1)(z+1)=83 (x+2)(y+2)(z+2)=157
On Fri, Apr 8, 2011 at 4:30 PM, Tom Rokicki <rokicki@gmail.com> wrote:
So 42 cubic inches is a correct answer to the original question, correct? Or at least as valid as the other answers?
David, would you mark "42 cubic inches" as correct?
This reminds me of the little-known fact that if you hold up a guinea pig by its tail, its eyes fall out (although strangely pet stores seldom warn of this).
-tom
On Fri, Apr 8, 2011 at 4:12 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Ah - probably not your point, David, but this can't be a real box. No box would ever have volumes 37, 83, 156 as you increase the length of each side by 1. (At least one that's not non-imaginary.)
You can get close to your series — volumes 36, 83, 156, 263 — starting with a box 4.25 by 4.25 by 2, for example.
On Fri, Apr 8, 2011 at 12:10 AM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265.
Interesting that the error climbs to 5% then falls as the box gets more and more cube-like.
2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073
On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe <simon.plouffe@gmail.com wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
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