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I have the quaternion $q=a+bi+cj+dk$ and wan't to compute $e^{q}$.
Since $i,j,k$ all behave similar one might assume the following formula works, but why doesn't it? $$e^{q}=e^{a}\cdot(\cos{b}+i\sin{b})\cdot(\cos{c}+j\sin{c})\cdot(\cos{d}+k\sin{d})$$

Edit: Of course this could be applied for any base $x$ using natural logarithms. (Thanks to Jonas Meyer for pointing this out)

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    What is the definition of $x^q$? – 2017-02-14
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    Do you mean $x^{a+bi+cj+dk}$ or do I get you wrong? – 2017-02-14
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    No, I mean, how is exponentiation defined? What is $x$, a positive real number or something else, and how is exponentiation defined when the exponent is quaternion? If it was $e^q$ I would imagine using the Taylor series would be a standard way. – 2017-02-14
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    I'm referring to the way it is defined for complex numbers and wan't to know why I can't apply the "same" logic. Also x can be any number (even $\in\mathbb{H}$) – 2017-02-14
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    That formula wouldn't apply for complex numbers, even if $x$ is positive. E.g., $2^{1+i}$ could be defined as $e^{(1+i)\log(2)}$, and if $\log(2)$ is the real logarithm of $2$ than this could be written as $2(\cos(\log(2))+i\sin(\log(2))$. You may need to consider logarithms if generalizing beyond $x=e$. What happens in that case? – 2017-02-14
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    Since $x^{bi}=\cos{b}+i\sin{b}$ I assume this could also apply to j and k, thus $x^q=x^a\cdot(\cos{b}+i\sin{b})⋅(\cos{c}+j\sin{c})⋅(\cos{d}+k\sin{d})$ – 2017-02-14
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    $x^{bi}=\cos(b)+i\sin(b)$ if $x=e$, not otherwise. – 2017-02-14
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    Oops, my bad, forgot to add the logarithm part, does this make the whole question obsolete? – 2017-02-14
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    It may be part of clarifying your question, not necessarily making it obsolete. – 2017-02-14
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    The formula would work if $i, j, k$ commuted, but they're very far from commuting. – 2017-02-14
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    @Qiauchu Thanks completely forgot about that. – 2017-02-14

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Start with $\;x=e=\exp(1)\,$ and note $\;I:=\dfrac{bi+cj+dk}{\rho}$ with $\;\rho:=\sqrt{b^2+c^2+d^2}\;$ then $\;I^2=-1\;$ and you may write : $$\tag{1}e^{a+\rho I}=e^{\,a}e^{\,\rho I}=e^a\left(\cos(\rho)+I\sin(\rho)\right)$$ The substitution of $\,I\,$ and $\,\rho\,$ giving you the final result.

Concerning the general power of $x$ you may use $\;\displaystyle x^{a+\rho I}=e^{\,\ln(x)(a+\rho I)}$ to conclude.
If $\,x\,$ is a quaternion you may have to define $\;\displaystyle x^q:=e^{\ln(x)\,q}\;$ (for example) first. $$-$$ Since this doesn't answer your question let's add that $(1)$ applies because $\,a\,$ and $\,u:=\rho I\,$ commute (this implies also that a similar method works for generalizations of quaternions like octonions)
so that : $$\tag{2}e^{\,a+u}=e^{\,a}\cdot e^{\,u}$$ Should $\,a\,$ and $\,u\,$ not commute then the Maclaurin expansion of the exponential at the left would for example contain the term $\;\displaystyle\frac {(a+u)^2}2=\frac{a^2+au+ua+u^2}2\,$ which differs from the corresponding terms $\;\displaystyle\frac {a^2}2+au+\frac {u^2}2=\frac{a^2+2au+u^2}2\,$ at the right and $(2)$ would not be true in general.

As indicated by Qiaochu Yuan since $i$, $j$ and $k$ don't commute you can't simply rewrite : $$e^{\,a+bi+cj+dk}=e^a\cdot e^{bi}\cdot e^{cj}\cdot e^{dk}$$ as you did to obtain a solution.

The same way $\,\ln(u\cdot v)=\ln(u)+\ln(v)\,$ is not true in general for quaternions since else you would have : $\,\ln(u\cdot v)=\ln(u)+\ln(v)=\ln(v)+\ln(u)=\ln(v\cdot u)\;$ even for $\;u\cdot v\neq v\cdot u$.

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    Thanks, so the two extensions are indeed quite similar? Could you also apply the same reasoning to octonions and so on? – 2017-02-14
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    Yes the same rule applies to octonions : $e^{\,\rho I}=\cos(\rho)+I\sin(\rho)$ as soon as $I^2=-1$ (see for example page $13$ of this [talk](http://people.oregonstate.edu/~drayt/talks/Purduepub.pdf) ). – 2017-02-14
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    Thanks, made things clear! – 2017-02-14
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    Glad it helped @Zonko. Fine continuation, – 2017-02-14