Undergrad Exponential of hypercomplex numbers

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The discussion explores whether the exponential of hypercomplex numbers, specifically quaternions and octonions, yields results within the same number system. It confirms that the exponential of a quaternion, derived from its Taylor series, results in a quaternion, as the series retains the quaternion structure. The non-commutative nature of quaternions and octonions complicates the behavior of odd-numbered exponents, which can be expressed in two ways depending on multiplication order. The exponential of an octonion is similarly assumed to remain within the octonion set, following analogous reasoning. Overall, the exponential function extends to hypercomplex numbers while maintaining their respective identities.
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The exponential of a complex number is a complex number. Does this extent to the quaternions and the octonions? Does the exponential of a quaternion give a quaternion? Does the exponential of an octonion give an octonion? Thanks.
 
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try the infinite series definition of exponential. what happens?
 
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mathwonk said:
try the infinite series definition of exponential. what happens?
The Taylor series for the exponential is,

{e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + ...

So obviously the linear approximation for the exponential will be the same as the exponent. If the exponent is a quaternion, then the linear approximation will be a quaternion, and same goes for the octonions. However, the quaternions and octonions do not commute. I think this means that you can still get a quadratic approximation (correct me if I'm wrong), but odd numbered exponents will have two ways of being expressed, depending on whether you multiply from the left or right. So I don't know what to say about this.
 
I learn from Wikipedia that a quaternion can be written generally as

q = {x_0} + i{x_1} + j{x_2} + k{x_3},

where {x_0}, {x_1}, {x_2}, {x_3} are real numbers. And {i^2} = {j^2} = {k^2} = - 1

Its conjugation is written,

{q^*} = {x_0} - i{x_1} - j{x_2} - k{x_3}.

Its norm can be written,

\left\| q \right\| = \sqrt {q{q^*}} = \sqrt {{x_0}^2 + {x_1}^2 + {x_2}^2 + {x_3}^2}.

But q can also be written as a scalar plus a vector as,

q = {x_0} + \vec v,

where,

\vec v = i{x_1} + j{x_2} + k{x_3}.

Then the exponential of a quaternion, q, can be written as,

{e^q} = \sum\limits_{n = 0}^\infty {\frac{{{q^n}}}{{n!}} = {e^{{x_0}}}(\cos \left\| {\vec v} \right\|} + \frac{{\vec v}}{{\left\| {\vec v} \right\|}}\sin \left\| {\vec v} \right\|).

So since the {\vec v} carries the i, j, k components, it does indeed seem that the exponential of a quaternion is a quaternion.

I assume the same construction leads to the fact that the exponential of a octonion is an octonion.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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