Mixing
What's the power of quaternion $[e^{a} * (cos(r) + frac{sin(r)}{r}(bi+cj+dk)) ]^ 2$?
$begingroup$
Euler's identity extended into quaternions is:
$q = a + bi + cj + dk$ with a,b,c,d real numbers
for the below: $sqrt{b^2+ c^2 + d^2} = r > 0$, and $frac{bi+cj+dk}{r}$ = $sqrt{-1}$,
$e^q = e^{a + rsqrt{-1}} = e^ae^{rsqrt{-1}} = e^a(cos(r) + sqrt{-1} sin(r)) = e^a(cos(r) + frac{sin(r)}{r}(bi + cj + dk))$
Therefore, what is the higher power of $[e^{a} * (cos(r) + frac{sin(r)}{r} (bi+cj+dk)) ]^ 2$ ?
I found this site on the powers of quaternions, but it doesn't address quaternions in Euler's form.
http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/functions/power/index.htm
exponentiation quaternions eulers-method
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
$endgroup$
add a comment |
$begingroup$
Euler's identity extended into quaternions is:
$q = a + bi + cj + dk$ with a,b,c,d real numbers
for the below: $sqrt{b^2+ c^2 + d^2} = r > 0$, and $frac{bi+cj+dk}{r}$ = $sqrt{-1}$,
$e^q = e^{a + rsqrt{-1}} = e^ae^{rsqrt{-1}} = e^a(cos(r) + sqrt{-1} sin(r)) = e^a(cos(r) + frac{sin(r)}{r}(bi + cj + dk))$
Therefore, what is the higher power of $[e^{a} * (cos(r) + frac{sin(r)}{r} (bi+cj+dk)) ]^ 2$ ?
I found this site on the powers of quaternions, but it doesn't address quaternions in Euler's form.
http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/functions/power/index.htm
exponentiation quaternions eulers-method
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
$endgroup$
$begingroup$
Don't say a quaternion equals $sqrt{-1}$ if you just mean to say that its square is $-1$. If there is any standard choice for what $sqrt{-1}$ means (which I would say there is not), it is defintiely not the value in the question.
$endgroup$
– Marc van Leeuwen
Jan 22 at 13:16
$begingroup$
I have not stated the quaternion equals $sqrt{-1}$ but that for the following quaternion equations the written $sqrt{-1}$ is $frac{bi+ cj + dk}{r}$. This is the common reference point, as it's lack of clarification can lead to confusion due to it's many possibilities.
$endgroup$
– Du'uzu Mes
Jan 22 at 13:32
add a comment |
$begingroup$
Euler's identity extended into quaternions is:
$q = a + bi + cj + dk$ with a,b,c,d real numbers
for the below: $sqrt{b^2+ c^2 + d^2} = r > 0$, and $frac{bi+cj+dk}{r}$ = $sqrt{-1}$,
$e^q = e^{a + rsqrt{-1}} = e^ae^{rsqrt{-1}} = e^a(cos(r) + sqrt{-1} sin(r)) = e^a(cos(r) + frac{sin(r)}{r}(bi + cj + dk))$
Therefore, what is the higher power of $[e^{a} * (cos(r) + frac{sin(r)}{r} (bi+cj+dk)) ]^ 2$ ?
I found this site on the powers of quaternions, but it doesn't address quaternions in Euler's form.
http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/functions/power/index.htm
exponentiation quaternions eulers-method
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
$endgroup$
Euler's identity extended into quaternions is:
$q = a + bi + cj + dk$ with a,b,c,d real numbers
for the below: $sqrt{b^2+ c^2 + d^2} = r > 0$, and $frac{bi+cj+dk}{r}$ = $sqrt{-1}$,
$e^q = e^{a + rsqrt{-1}} = e^ae^{rsqrt{-1}} = e^a(cos(r) + sqrt{-1} sin(r)) = e^a(cos(r) + frac{sin(r)}{r}(bi + cj + dk))$
Therefore, what is the higher power of $[e^{a} * (cos(r) + frac{sin(r)}{r} (bi+cj+dk)) ]^ 2$ ?
I found this site on the powers of quaternions, but it doesn't address quaternions in Euler's form.
http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/functions/power/index.htm
exponentiation quaternions eulers-method
exponentiation quaternions eulers-method
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
asked Jan 22 at 13:00
Du'uzu MesDu'uzu Mes
135
135
$begingroup$
Don't say a quaternion equals $sqrt{-1}$ if you just mean to say that its square is $-1$. If there is any standard choice for what $sqrt{-1}$ means (which I would say there is not), it is defintiely not the value in the question.
$endgroup$
– Marc van Leeuwen
Jan 22 at 13:16
$begingroup$
I have not stated the quaternion equals $sqrt{-1}$ but that for the following quaternion equations the written $sqrt{-1}$ is $frac{bi+ cj + dk}{r}$. This is the common reference point, as it's lack of clarification can lead to confusion due to it's many possibilities.
$endgroup$
– Du'uzu Mes
Jan 22 at 13:32
add a comment |
$begingroup$
Don't say a quaternion equals $sqrt{-1}$ if you just mean to say that its square is $-1$. If there is any standard choice for what $sqrt{-1}$ means (which I would say there is not), it is defintiely not the value in the question.
$endgroup$
– Marc van Leeuwen
Jan 22 at 13:16
$begingroup$
I have not stated the quaternion equals $sqrt{-1}$ but that for the following quaternion equations the written $sqrt{-1}$ is $frac{bi+ cj + dk}{r}$. This is the common reference point, as it's lack of clarification can lead to confusion due to it's many possibilities.
$endgroup$
– Du'uzu Mes
Jan 22 at 13:32
$begingroup$
Don't say a quaternion equals $sqrt{-1}$ if you just mean to say that its square is $-1$. If there is any standard choice for what $sqrt{-1}$ means (which I would say there is not), it is defintiely not the value in the question.
$endgroup$
– Marc van Leeuwen
Jan 22 at 13:16
$begingroup$
Don't say a quaternion equals $sqrt{-1}$ if you just mean to say that its square is $-1$. If there is any standard choice for what $sqrt{-1}$ means (which I would say there is not), it is defintiely not the value in the question.
$endgroup$
– Marc van Leeuwen
Jan 22 at 13:16
$begingroup$
I have not stated the quaternion equals $sqrt{-1}$ but that for the following quaternion equations the written $sqrt{-1}$ is $frac{bi+ cj + dk}{r}$. This is the common reference point, as it's lack of clarification can lead to confusion due to it's many possibilities.
$endgroup$
– Du'uzu Mes
Jan 22 at 13:32
$begingroup$
I have not stated the quaternion equals $sqrt{-1}$ but that for the following quaternion equations the written $sqrt{-1}$ is $frac{bi+ cj + dk}{r}$. This is the common reference point, as it's lack of clarification can lead to confusion due to it's many possibilities.
$endgroup$
– Du'uzu Mes
Jan 22 at 13:32
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083138%2fwhats-the-power-of-quaternion-ea-cosr-fracsinrrbicjdk%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083138%2fwhats-the-power-of-quaternion-ea-cosr-fracsinrrbicjdk%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Don't say a quaternion equals $sqrt{-1}$ if you just mean to say that its square is $-1$. If there is any standard choice for what $sqrt{-1}$ means (which I would say there is not), it is defintiely not the value in the question.
$endgroup$
– Marc van Leeuwen
Jan 22 at 13:16
$begingroup$
I have not stated the quaternion equals $sqrt{-1}$ but that for the following quaternion equations the written $sqrt{-1}$ is $frac{bi+ cj + dk}{r}$. This is the common reference point, as it's lack of clarification can lead to confusion due to it's many possibilities.
$endgroup$
– Du'uzu Mes
Jan 22 at 13:32