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Let's say that $X$ is a hypercomplex number from one of the Cayley-Dickson algebras, possibly CD(32).
I was looking for general formulas for exp() and ln() of any $X$.

I found:
https://math.stackexchange.com/a/1047246/301741
https://math.stackexchange.com/a/2554/301741

These formulas are given for quaternions but it seems to me that they should hold for any of the higher algebras, right?

Also, please notice that in both formulas we multiply $Im(X)$ by a scalar, i.e. a number from $R$, i.e. a number from CD(1). In general I cannot rely on commutativity and associativity as these are lost. However, if I see this correctly, operations with $X$ and scalars would still hold these properties and in the end I can juggle the order of operations however I want?

Also, it is not necessary that I perform $X \times (y,0,0,0...)$ according to CD definition of multiplication, right? That would be embedding the scalar in a higher algebra and then carry out a tedious multiplication by recursive decomposition into $(a,b)\times(c,d)$. I should use the multiplication property from a vector space and notice that a product of $X$ and a scalar is just $X$ with all the components multiplied by $y$, right?

Last, but not least: I assume that exp and ln are still inverses, even for the higher algebras(?) Therefore exp(ln(X)) = X = ln(exp(X))?

maciek
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  • In order for the Taylor series to work, you need to have that powers are well-defined, i.e. that when you talk about $x^4$ it doesn't matter if you interpet it as $x(x(xx))$ or $(xx)(xx)$ or $x((xx)x)$ or etc. This is called 'power-associtive'. I'm not sure if all the higher CD-algebras are power associtaive – Vincent Nov 24 '20 at 20:42
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    Wiki says they are: https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction#Further_algebras – maciek Nov 24 '20 at 20:47
  • Cool. Also there is a notion of norm, so it makes sense to talk about convergence – Vincent Nov 24 '20 at 21:08
  • Still you will likely loose the properties for which log's were initially invented: $\log(ab) = \log(a) + \log(b)$ and conversely $\exp(a+b) = \exp(a)\exp(b)$ – Vincent Nov 24 '20 at 21:10

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