Exponential of hypercomplex numbers

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• friend
In summary, the exponential of a complex number is a complex number and the same applies to quaternions and octonions. The Taylor series for the exponential shows that the linear approximation for these numbers will also be a quaternion or octonion. However, the non-commutative nature of quaternions and octonions means that odd numbered exponents can be expressed in two different ways. The exponential of a quaternion can be written as a scalar plus a vector, and the exponential of an octonion is an octonion. This is supported by the infinite series definition of exponential.

friend

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.

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.