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 March 28th, 2018, 02:40 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 394 Thanks: 27 Math Focus: Number theory Way to transcendence Can both infinite, rational terms of series -- and finite, irrational coefficients of polynomials -- generate and eventually identify transcendental numbers?
 March 28th, 2018, 04:45 PM #2 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves Identifying Trancendental Numbers ok the first link I would recommend is the work of the late Alan Baker: https://en.wikipedia.org/wiki/Baker%27s_theorem And a good paper by Michel Waldschmidt entitled "Transcendence of Periods: The State of the Art" that has a lot of good theorems for deciding whether a number is transcendental or not that follow from Baker's theorem of linear independence. This file size is too large for the limit on this forum, so if you cannot find a copy on the net, pm me and I will email you a copy. Last edited by Adam Ledger; March 28th, 2018 at 04:48 PM. Reason: meh
March 28th, 2018, 05:06 PM   #3
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Quote:
 Originally Posted by Adam Ledger This file size is too large for the limit on this forum ....
Fermat tried that dodge in 1637.

 March 28th, 2018, 06:54 PM #4 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves Well sometimes the old ways are the best. But we can all rest easy now that a teenager with what appears to be a severe Adderall induced psychosis has shown us that simple elementary proof in a recent thread
 March 28th, 2018, 09:52 PM #5 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 394 Thanks: 27 Math Focus: Number theory i^i is actually a real transcendental number. Here is how you can compute the value of i^i = 0.207879576... 1. Since e^(ix) = Cos x + i Sin x, then let x = Pi/2. 2. Then e^(iPi/2) = i = Cos Pi/2 + i Sin Pi/2; since Cos Pi/2 = Cos 90 deg. = 0. But Sin 90 = 1 and i Sin 90 deg. = (i)*(1) = i. 3. Therefore e^(iPi/2) = i. 4. Take the ith power of both sides, the right side being i^i and the left side = [e^(iPi/2)]^i = e^(-Pi/2). 5. Therefore i^i = e^(-Pi/2) = .207879576... From Cliff Pickover, The 15 Most Famous Transcendental Numbers - Cliff Pickover
 March 28th, 2018, 09:57 PM #6 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 394 Thanks: 27 Math Focus: Number theory Baker's theorem deserves more contemplation.
March 28th, 2018, 11:13 PM   #7
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 Originally Posted by Loren From Cliff Pickover, The 15 Most Famous Transcendental Numbers - Cliff Pickover
Great page, thanks for the link. I love Chaitin's number. It's an easily understood example of a real number that can't possibly be computable.

I looked up Alan Baker, I'd never heard of him before. Won the Fields medal in 1970 for his work on transcendentals.

I was fascinated to discover that it's unproven whether $\pi^e$ is transcendental.

Last edited by Maschke; March 28th, 2018 at 11:21 PM.

 March 29th, 2018, 11:50 AM #8 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves One great theorem that arises from Schwarz's Lemma in the work of Schneider on Abelian functions is this: Let a and b be rational numbers which are not integers and such that a + b is not an integer. Then B(a,b) is transcendental https://en.wikipedia.org/wiki/Beta_function

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