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mathwonk

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try the infinite series definition of exponential. what happens?

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The Taylor series for the exponential is,try the infinite series definition of exponential. what happens?

[itex]{e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + ...[/itex]

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.

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[itex]q = {x_0} + i{x_1} + j{x_2} + k{x_3}[/itex],

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

Its conjugation is written,

[itex]{q^*} = {x_0} - i{x_1} - j{x_2} - k{x_3}[/itex].

Its norm can be written,

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

But

[itex]q = {x_0} + \vec v[/itex],

where,

[itex]\vec v = i{x_1} + j{x_2} + k{x_3}[/itex].

Then the exponential of a quaternion,

[itex]{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\|)[/itex].

So since the [itex]{\vec v}[/itex] carries the

I assume the same construction leads to the fact that the exponential of a octonion is an octonion.

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