HP Forum Archive 21

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The answer to the second semi-mathematical puzzle Message #1 Posted by mike reed on 14 Feb 2013, 10:05 a.m. " int_(1)^(3^(1/3)) z^2 dz cos(3pi/9) = ln e^(1/3) (Note: Readers outside the US, this works better if you allow yourself the agony of pronouncing the letter “z” as “zee”.) I will post this answer one week from today" OK, the week has passed; here's the answer: The integral of z squared dz from 1 to the cube root of 3 times the cosine of 3 pi over 9 equals log of the cube root of e How did you do in solving it? mike

Re: The answer to the second semi-mathematical puzzle Message #2 Posted by Les Koller on 16 Feb 2013, 6:30 p.m., in response to message #1 by mike reed Not sure how to read it in Lymerick meter...is this close? The integral of z squared dz from one to cube root of 3 times the cos of 3pi divided by nine = the log of the cube root of three

Re: The answer to the second semi-mathematical puzzle Message #3 Posted by Marcus von Cube, Germany on 17 Feb 2013, 6:40 a.m., in response to message #2 by Les Koller The integral of z squared dz from one to cube root of 3 times the cos of 3pi divided by nine = the log of the cube root of three The formum software has wrapped the text. What you see here is enclosed in [pre]...[/pre] tags. As an alternative use [nl] at each line end.

Re: The answer to the second semi-mathematical puzzle Message #4 Posted by Derek Walker (UK) on 18 Feb 2013, 2:40 a.m., in response to message #3 by Marcus von Cube, Germany Shouldn't 'cosine' and 'nine' be at the ends of lines, to rhyme: The integral of z squared dz from 1 to the cube root of 3 times the cosine of 3 pi over 9 equals log of the cube root of e

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