Wikipedia:Reference desk/Archives/Mathematics/2016 March 7

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March 7

Quaternion addition

The quaternion article says that quaternion multiplication is not commutative. But what about quaternion addition?? Given the properties of logarithms, if quaternion multiplication is not commutative, numbers that we can create as logarithms of quaternions must form a set where addition is not commutative, right?? Georgia guy (talk) 16:58, 7 March 2016 (UTC)

Quaternion addition is commutative.

You can think of quaternions as matrices (see Quaternion#Matrix representations), so what's true for matrices is true for quaternions. In general, $\exp(A+B)\neq \exp(A)\exp(B)$, so your intuitions on how addition, multiplication, exponentiation and logarithms all interact may be incorrect. If A and B commute, that is, $AB=BA$, then things start to behave in a more familiar way ($\exp(A+B)=\exp(A)\exp(B)$ and so on). It's also true that every matrix (or quaternion) commutes with any matrix found by applying a polynomial or power series on it. -- Meni Rosenfeld (talk) 17:25, 7 March 2016 (UTC)

For general (possibly noncommutative) algebras, $\exp(A)\exp(B)=\exp(A+B+{\tfrac {1}{2}}[A,B]+{\tfrac {1}{12}}([A,[A,B]]+[B,[B,A]])+\cdots )$ where $[x,y]=xy-yx$ is the commutator (see Baker–Campbell–Hausdorff formula). -- BenRG (talk) 21:28, 7 March 2016 (UTC)

(To put it another way, $\log ab=\log a\oplus \log b$ where $p\oplus q=p+q+{\tfrac {1}{2}}[p,q]+\cdots$. So there is a kind of noncommutative addition; it's just not the "official" addition of the algebra.) -- BenRG (talk) 23:24, 7 March 2016 (UTC)

Interesting question. If you look at complex numbers the log is not unique - one can add k2πi where k is any integer. The same is true of the quaternions, the exponential function is unique but the logarithm is not - and for the quaternions the non-unique bit is k2πn where n is a unit vector of ai+bj+ck - and when you add two of these up with different unit vectors the result when expressed as a multiple of a unit vector may not be k2π with k an integer. Thus multiplying quaternions by taking the log adding and exponentiating can give all sorts of different results depending on the non-unique factors of the log. Dmcq (talk) 17:36, 7 March 2016 (UTC)

How about octonion, sedenion, or duo-tricenion addition?? Georgia guy (talk) 18:38, 7 March 2016 (UTC)

All of these number systems (though I've never heard of "duo-tricenion") are given by defining some multiplication operation on top of the usual addition operation in Rⁿ. These additions are all commutative, though the multiplications are not. Staecker (talk) 19:41, 7 March 2016 (UTC)

What did you think the name of the next member after the sedenions was?? It has to be derived from 32, right?? Georgia guy (talk) 20:21, 7 March 2016 (UTC)

Google hasn't heard of "duo-tricenion" either, though I'm sure someone else somewhere has imagined a 32-dimensional noncommutative and nonassociative algebra over the reals obtained by applying the Cayley–Dickson construction to the sedenions. It seems we can't have an article yet. Dbfirs 20:56, 7 March 2016 (UTC)

trigintaduonions T

• The Basic Subalgebra structure of the Cayley-Dickson Algebra of Dimension 32 (Trigintaduonions) http://arxiv.org/pdf/0907.2047.pdf
• Compounding Fields and Their Quantum Equations in the Trigintaduonion Space http://arxiv.org/pdf/0704.0136.pdf
• Wikipedia:Articles for deletion/Trigintaduonion

(To get us started)Naraht (talk) 21:21, 7 March 2016 (UTC)

And also Wikipedia:Articles for deletion/Bitredeconion.--JohnBlackburne^words_deeds 00:09, 8 March 2016 (UTC)

Powerful numbers

In the article Powerful numbers, at the start of the section Mathematical properties, it says

and so the sum [emphasis added] of reciprocals of powerful numbers converge

but the equation on the following line shows a product symbol, and not a summation symbol

Maybe it should it say "product of reciprocals of powerful numbers converge"?

Or maybe the equation should have a summation symbol?

Or maybe something has been accidentally omitted?

Or maybe I am imagining problems where none exist?

Bh12 (talk) 22:50, 7 March 2016 (UTC)

What is written is correct: the sum converges, and the value it converges to is also the value of that (convergent) infinite product. I agree that this is not as clear as it could be. --JBL (talk) 23:00, 7 March 2016 (UTC)

Thank you. I leave it to you math savants to make the article clearer.Bh12 (talk) 01:50, 8 March 2016 (UTC)

I have made an attempt; is it clearer now? --JBL (talk) 14:15, 9 March 2016 (UTC)