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I've been messing around with microtonal scales and I came up with this number and I was wondering if it is Transcendental.

U = 17.312340490667609 to 15 decimal places

U = 17.31234049066756088831909617202270564911975144983583120962539288317331063017422095862631947472207613396397021971583053106711722024602031571264794589868175203529444843861662192356841665196

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closed as unclear what you're asking by 6005, Daniel W. Farlow, Shailesh, Leucippus, G. Sassatelli Jul 22 at 1:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

No number for which you can give a finite decimal expansion is even irrational, let alone transcendental. Is $U$ some sort of approximation of a number that you can express in some other way? – Aaron Jun 3 '11 at 14:22
Since it is impossible to list infinitely many digits, you need to specify a mathematical expression for $U$. Was there a formula that you were using that you got $U$ out of? What was it? – Zev Chonoles Jun 3 '11 at 14:29
@Ginger Bill: I don't understand what "trial and error" means here. We need to know a mathematical expression, i.e. formula, for $U$, in order to answer this question. Presumably you used a computer when producing this number $U$; can you describe the process that you used to produce it? Perhaps started with an integer, then added smaller and smaller fractions? Also, as I said before, "only specifying a finite number of decimal places doesn't suffice to determine whether or not $U$ is transcendental", so while I appreciate your providing more digits of $U$, I'm afraid it doesn't help. – Zev Chonoles Jun 3 '11 at 14:43
@Ginger Bill: At this moment the "short" and "long" versions differ at the $13$th place after the decimal point, where presumably the $5$ is the long version is a typo. You must have at least an excellent algorithm for computing $U$. Is it possible to have a clear description of this? – André Nicolas Jun 3 '11 at 14:45
@Ginger Bill: The statement that a real number is transcendental requires a usually-complicated proof. Mathematicians took a long time to prove that $\pi$ is transcendental, and the proof is not so obvious! (See here). Can you describe why you think $U$ is transcendental? – Zev Chonoles Jun 3 '11 at 14:56
up vote 19 down vote accepted

It looks as if the number is $12/(\ln 2)$. So it is transcendental.

It was the connection to music, and hence to $2^{1/12}$, that did it.

The transcendence of $\ln 2$ follows from the Lindemann-Weierstrass Theorem.

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$12/\ln(2) = 17.312340490667562...$ which disagrees with the number above. It is possible that the OP miscalculated the constant, but I'm not sure it is what you have written (although I would like it to be!). – JavaMan Jun 3 '11 at 15:06
@DJC: The much longer version added in the post actually agrees with what you wrote, as I pointed out the short and long versions disagree in the $13$th place after the decimal. – André Nicolas Jun 3 '11 at 15:13
@AndréNicolas How did you discover that 12/ln2 came close to the OP's number? – zerosofthezeta Aug 8 '13 at 1:17
Connection to music, $12$-tone scale made me suspect. The agreement to many places clinches it, unless something very weird is happening. – André Nicolas Aug 8 '13 at 1:23

No. Any transcendental number must have an infinite decimal expansion; your number has a finite decimal expansion, so it is in fact a rational number, specifically $$\small U=\frac{1731234049066756088831909617202270564911975144983583120962539288317331063017422095862631947472207 \atop 613396397021971583053106711722024602031571264794589868175203529444843861662192356841665196}{10^{185}}.$$

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This is only to 15 decimal places. I'll add a note. – Ginger Bill Jun 3 '11 at 14:24
+1: This answer is really funny. – Eric Naslund Jun 3 '11 at 16:09
Very nice. And I've never come across \atop before. Cool. – mixedmath Jun 3 '11 at 21:17
@Zev: I was just thinking about this answer and wishing I would could vote it up again! – JavaMan Jul 20 '11 at 3:29

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