I have problem with Euler´s form of quaternion. My quaternion $q=\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j,$ so $q^2=-1$, because $$q^2=(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)=-\frac{1}{2}+\frac{1}{2}k-\frac{1}{2}k-\frac{1}{2}=-1.$$ Thus I can write $q$ in Euler´s form as $q=e^{q\frac{\pi}{2}}=\cos\frac{\pi}{2}+q \sin\frac{\pi}{2}=q.$ However, I can also write $$q=e^{q\frac{\pi}{2}}=e^{(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)\frac{\pi}{2}}=e^{\frac{\pi}{2\sqrt{2}}i+\frac{\pi}{2\sqrt{2}}j}=e^{\frac{\pi}{2\sqrt{2}}i}e^{\frac{\pi}{2\sqrt{2}}j}=\\=(\cos\frac{\pi}{2\sqrt{2}}+i \sin\frac{\pi}{2\sqrt{2}})(\cos\frac{\pi}{2\sqrt{2}}+j \sin\frac{\pi}{2\sqrt{2}})=\\=\cos^2\frac{\pi}{2\sqrt{2}}+(i+j) \sin\frac{\pi}{2\sqrt{2}}\cos\frac{\pi}{2\sqrt{2}}+k\sin^2\frac{\pi}{2\sqrt{2}},$$ but this is not $q$. Please can you somebody say me where is the mistake?
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3It is possible to define the exponential function for quaternions, but I would expect usually $e^{p+q} \neq e^p e^q$ – Will Jagy Mar 11 '16 at 19:15
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How do you have $q=e^{q\pi /2}$ ? – DanielWainfleet Mar 11 '16 at 19:39
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@user254665: when $q^2=-1$ we have for any real $x$ the relation $e^{,{x;q}}=\cos(x)+\sin(x)q$. – Raymond Manzoni Mar 12 '16 at 00:25
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For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$ the exponential is defined as: $$ e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right) $$ (see :Exponential Function of Quaternion - Derivation)
and in general, since quaternions are not commutative, we have:
$$ e^xe^y \ne e^ye^x \ne e^{x+y} $$
In your case: $q=\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j,$ is a pure imaginary quaternion $\mathbf {v} =q$ and $|q|=1$, so we have: $$ e^q= e^{\mathbf{v}}=\left( \cos (1) +\mathbf{v}\sin (1) \right)=\cos (1)+ \sin(1)\left(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j \right) $$
Emilio Novati
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