How do I express a unit quaternion in exponential form? [closed]












1












$begingroup$


Let $t,u,v$ lie in the interval $(-pi, pi]$



If we assume that
$$ cos(t)cos(u)cos(v) + sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k = e^z$$ such that $z$ is also a quaternion, what does z equal and why?






quaternions







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$endgroup$



closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs Jan 23 at 21:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Maybe this can help: math.stackexchange.com/questions/1030737/…
    $endgroup$
    – Matti P.
    Jan 23 at 10:12
















1












$begingroup$


Let $t,u,v$ lie in the interval $(-pi, pi]$



If we assume that
$$ cos(t)cos(u)cos(v) + sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k = e^z$$ such that $z$ is also a quaternion, what does z equal and why?






quaternions







share|cite|improve this question











$endgroup$



closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs Jan 23 at 21:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Maybe this can help: math.stackexchange.com/questions/1030737/…
    $endgroup$
    – Matti P.
    Jan 23 at 10:12














1












1








1





$begingroup$


Let $t,u,v$ lie in the interval $(-pi, pi]$



If we assume that
$$ cos(t)cos(u)cos(v) + sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k = e^z$$ such that $z$ is also a quaternion, what does z equal and why?






quaternions







share|cite|improve this question











$endgroup$




Let $t,u,v$ lie in the interval $(-pi, pi]$



If we assume that
$$ cos(t)cos(u)cos(v) + sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k = e^z$$ such that $z$ is also a quaternion, what does z equal and why?






quaternions




quaternions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 10:43









John Hughes

64.5k24191




64.5k24191










asked Jan 23 at 10:03









Wire BowlWire Bowl

113




113




closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs Jan 23 at 21:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs Jan 23 at 21:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Maybe this can help: math.stackexchange.com/questions/1030737/…
    $endgroup$
    – Matti P.
    Jan 23 at 10:12


















  • $begingroup$
    Maybe this can help: math.stackexchange.com/questions/1030737/…
    $endgroup$
    – Matti P.
    Jan 23 at 10:12
















$begingroup$
Maybe this can help: math.stackexchange.com/questions/1030737/…
$endgroup$
– Matti P.
Jan 23 at 10:12




$begingroup$
Maybe this can help: math.stackexchange.com/questions/1030737/…
$endgroup$
– Matti P.
Jan 23 at 10:12










1 Answer
1






active

oldest

votes


















0












$begingroup$

The link provided by @MattiP shows this:




If $ mathbf{v} in mathbb{H}_P$ is an imaginary quaternion, putting $theta=|mathbf{v}|$ we have:
$$
e^mathbf{v}= costheta + mathbf{v};dfrac{sin theta}{theta}
$$




$$newcommand{bv}{{mathbf v}}
$$
In your case, we want to find $bv$ (which you've called "z"), but we know the right hand side. In particular, we can let
$$
theta = cos^{-1} (cos(t)cos(u)cos(v))
$$
and then $cos theta$ will be equal to the real part of your quaternion, as needed.



To determine $bv$, we multiply the vector part of your quaternion by $frac{theta}{sin theta}$ to get
$$
bv = frac{theta}{sin theta}
left( sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k right)
$$



In this expression, $sin theta$ can be simplified a little, because it's a sine of an arc-cosine, but it's probably not worth doing.



You can probably also get to all this by using Rodrigues' formula, but I doubt it's any simpler.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you John.
    $endgroup$
    – Wire Bowl
    Jan 23 at 11:18


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The link provided by @MattiP shows this:




If $ mathbf{v} in mathbb{H}_P$ is an imaginary quaternion, putting $theta=|mathbf{v}|$ we have:
$$
e^mathbf{v}= costheta + mathbf{v};dfrac{sin theta}{theta}
$$




$$newcommand{bv}{{mathbf v}}
$$
In your case, we want to find $bv$ (which you've called "z"), but we know the right hand side. In particular, we can let
$$
theta = cos^{-1} (cos(t)cos(u)cos(v))
$$
and then $cos theta$ will be equal to the real part of your quaternion, as needed.



To determine $bv$, we multiply the vector part of your quaternion by $frac{theta}{sin theta}$ to get
$$
bv = frac{theta}{sin theta}
left( sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k right)
$$



In this expression, $sin theta$ can be simplified a little, because it's a sine of an arc-cosine, but it's probably not worth doing.



You can probably also get to all this by using Rodrigues' formula, but I doubt it's any simpler.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you John.
    $endgroup$
    – Wire Bowl
    Jan 23 at 11:18
















0












$begingroup$

The link provided by @MattiP shows this:




If $ mathbf{v} in mathbb{H}_P$ is an imaginary quaternion, putting $theta=|mathbf{v}|$ we have:
$$
e^mathbf{v}= costheta + mathbf{v};dfrac{sin theta}{theta}
$$




$$newcommand{bv}{{mathbf v}}
$$
In your case, we want to find $bv$ (which you've called "z"), but we know the right hand side. In particular, we can let
$$
theta = cos^{-1} (cos(t)cos(u)cos(v))
$$
and then $cos theta$ will be equal to the real part of your quaternion, as needed.



To determine $bv$, we multiply the vector part of your quaternion by $frac{theta}{sin theta}$ to get
$$
bv = frac{theta}{sin theta}
left( sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k right)
$$



In this expression, $sin theta$ can be simplified a little, because it's a sine of an arc-cosine, but it's probably not worth doing.



You can probably also get to all this by using Rodrigues' formula, but I doubt it's any simpler.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you John.
    $endgroup$
    – Wire Bowl
    Jan 23 at 11:18














0












0








0





$begingroup$

The link provided by @MattiP shows this:




If $ mathbf{v} in mathbb{H}_P$ is an imaginary quaternion, putting $theta=|mathbf{v}|$ we have:
$$
e^mathbf{v}= costheta + mathbf{v};dfrac{sin theta}{theta}
$$




$$newcommand{bv}{{mathbf v}}
$$
In your case, we want to find $bv$ (which you've called "z"), but we know the right hand side. In particular, we can let
$$
theta = cos^{-1} (cos(t)cos(u)cos(v))
$$
and then $cos theta$ will be equal to the real part of your quaternion, as needed.



To determine $bv$, we multiply the vector part of your quaternion by $frac{theta}{sin theta}$ to get
$$
bv = frac{theta}{sin theta}
left( sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k right)
$$



In this expression, $sin theta$ can be simplified a little, because it's a sine of an arc-cosine, but it's probably not worth doing.



You can probably also get to all this by using Rodrigues' formula, but I doubt it's any simpler.






share|cite|improve this answer









$endgroup$



The link provided by @MattiP shows this:




If $ mathbf{v} in mathbb{H}_P$ is an imaginary quaternion, putting $theta=|mathbf{v}|$ we have:
$$
e^mathbf{v}= costheta + mathbf{v};dfrac{sin theta}{theta}
$$




$$newcommand{bv}{{mathbf v}}
$$
In your case, we want to find $bv$ (which you've called "z"), but we know the right hand side. In particular, we can let
$$
theta = cos^{-1} (cos(t)cos(u)cos(v))
$$
and then $cos theta$ will be equal to the real part of your quaternion, as needed.



To determine $bv$, we multiply the vector part of your quaternion by $frac{theta}{sin theta}$ to get
$$
bv = frac{theta}{sin theta}
left( sin(t)cos(u)cos(v)i + sin(u)cos(v)j + sin(v)k right)
$$



In this expression, $sin theta$ can be simplified a little, because it's a sine of an arc-cosine, but it's probably not worth doing.



You can probably also get to all this by using Rodrigues' formula, but I doubt it's any simpler.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 23 at 10:52









John HughesJohn Hughes

64.5k24191




64.5k24191












  • $begingroup$
    Thank you John.
    $endgroup$
    – Wire Bowl
    Jan 23 at 11:18


















  • $begingroup$
    Thank you John.
    $endgroup$
    – Wire Bowl
    Jan 23 at 11:18
















$begingroup$
Thank you John.
$endgroup$
– Wire Bowl
Jan 23 at 11:18




$begingroup$
Thank you John.
$endgroup$
– Wire Bowl
Jan 23 at 11:18



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