Zeta function generalized to quaternions?

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Has the $zeta(s)$ function, $sum_n 1/n^s$, been generalized
to quaternions, so $zeta(q)$ for $q$ a quaternion?
Euler defined it for $s$ integers, Chebyshev for $s$ real,
Riemann for $s$ complex. So it is natural to explore $s$ a quaternion. But perhaps this does not lead to an interesting Quaternion Hypothesis?
complex-analysis number-theory riemann-zeta quaternions
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up vote
4
down vote
favorite
Has the $zeta(s)$ function, $sum_n 1/n^s$, been generalized
to quaternions, so $zeta(q)$ for $q$ a quaternion?
Euler defined it for $s$ integers, Chebyshev for $s$ real,
Riemann for $s$ complex. So it is natural to explore $s$ a quaternion. But perhaps this does not lead to an interesting Quaternion Hypothesis?
complex-analysis number-theory riemann-zeta quaternions
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Has the $zeta(s)$ function, $sum_n 1/n^s$, been generalized
to quaternions, so $zeta(q)$ for $q$ a quaternion?
Euler defined it for $s$ integers, Chebyshev for $s$ real,
Riemann for $s$ complex. So it is natural to explore $s$ a quaternion. But perhaps this does not lead to an interesting Quaternion Hypothesis?
complex-analysis number-theory riemann-zeta quaternions
Has the $zeta(s)$ function, $sum_n 1/n^s$, been generalized
to quaternions, so $zeta(q)$ for $q$ a quaternion?
Euler defined it for $s$ integers, Chebyshev for $s$ real,
Riemann for $s$ complex. So it is natural to explore $s$ a quaternion. But perhaps this does not lead to an interesting Quaternion Hypothesis?
complex-analysis number-theory riemann-zeta quaternions
edited Aug 25 at 11:37
asked Aug 25 at 0:52
Joseph O'Rourke
17.3k248104
17.3k248104
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1 Answer
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You can define $zeta(q)$ for any quaternion with real part $ne 1$ by plugging it into the Taylor series for $zeta(x)$ around a real point far enough from $1$ in the appropriate direction.
This is not very exciting, though, because every quaternion $q$ decomposes as $a+bell$ where $a$ and $b$ are real and $ell$ is a quaternion satisfying $ell^2=-1$. So for every fixed $q$, all terms in the Taylor series will lie in the complex plane spanned by $1$ and $ell$, and therefore no really new phenomena show up.
This shows that it doesn't matter which center for the Taylor series we choose, as long as it is closer to $q$ than it is to $1$. It also shows that we can extend the function continuously to quaternions with real part $1$ (other that $1$ itself).
Thanks for this response. If $q=2+0i+0j+0k$, will $zeta(q)$ will still equal $pi^2/6 =zeta(2)$? In other words, would the extension agree with standard $zeta$ for real $q$, and for complex $q$?
â Joseph O'Rourke
Aug 25 at 12:18
@JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $mathbb Hsetminus1$ in one go).
â Henning Makholm
Aug 25 at 12:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You can define $zeta(q)$ for any quaternion with real part $ne 1$ by plugging it into the Taylor series for $zeta(x)$ around a real point far enough from $1$ in the appropriate direction.
This is not very exciting, though, because every quaternion $q$ decomposes as $a+bell$ where $a$ and $b$ are real and $ell$ is a quaternion satisfying $ell^2=-1$. So for every fixed $q$, all terms in the Taylor series will lie in the complex plane spanned by $1$ and $ell$, and therefore no really new phenomena show up.
This shows that it doesn't matter which center for the Taylor series we choose, as long as it is closer to $q$ than it is to $1$. It also shows that we can extend the function continuously to quaternions with real part $1$ (other that $1$ itself).
Thanks for this response. If $q=2+0i+0j+0k$, will $zeta(q)$ will still equal $pi^2/6 =zeta(2)$? In other words, would the extension agree with standard $zeta$ for real $q$, and for complex $q$?
â Joseph O'Rourke
Aug 25 at 12:18
@JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $mathbb Hsetminus1$ in one go).
â Henning Makholm
Aug 25 at 12:26
add a comment |Â
up vote
4
down vote
accepted
You can define $zeta(q)$ for any quaternion with real part $ne 1$ by plugging it into the Taylor series for $zeta(x)$ around a real point far enough from $1$ in the appropriate direction.
This is not very exciting, though, because every quaternion $q$ decomposes as $a+bell$ where $a$ and $b$ are real and $ell$ is a quaternion satisfying $ell^2=-1$. So for every fixed $q$, all terms in the Taylor series will lie in the complex plane spanned by $1$ and $ell$, and therefore no really new phenomena show up.
This shows that it doesn't matter which center for the Taylor series we choose, as long as it is closer to $q$ than it is to $1$. It also shows that we can extend the function continuously to quaternions with real part $1$ (other that $1$ itself).
Thanks for this response. If $q=2+0i+0j+0k$, will $zeta(q)$ will still equal $pi^2/6 =zeta(2)$? In other words, would the extension agree with standard $zeta$ for real $q$, and for complex $q$?
â Joseph O'Rourke
Aug 25 at 12:18
@JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $mathbb Hsetminus1$ in one go).
â Henning Makholm
Aug 25 at 12:26
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You can define $zeta(q)$ for any quaternion with real part $ne 1$ by plugging it into the Taylor series for $zeta(x)$ around a real point far enough from $1$ in the appropriate direction.
This is not very exciting, though, because every quaternion $q$ decomposes as $a+bell$ where $a$ and $b$ are real and $ell$ is a quaternion satisfying $ell^2=-1$. So for every fixed $q$, all terms in the Taylor series will lie in the complex plane spanned by $1$ and $ell$, and therefore no really new phenomena show up.
This shows that it doesn't matter which center for the Taylor series we choose, as long as it is closer to $q$ than it is to $1$. It also shows that we can extend the function continuously to quaternions with real part $1$ (other that $1$ itself).
You can define $zeta(q)$ for any quaternion with real part $ne 1$ by plugging it into the Taylor series for $zeta(x)$ around a real point far enough from $1$ in the appropriate direction.
This is not very exciting, though, because every quaternion $q$ decomposes as $a+bell$ where $a$ and $b$ are real and $ell$ is a quaternion satisfying $ell^2=-1$. So for every fixed $q$, all terms in the Taylor series will lie in the complex plane spanned by $1$ and $ell$, and therefore no really new phenomena show up.
This shows that it doesn't matter which center for the Taylor series we choose, as long as it is closer to $q$ than it is to $1$. It also shows that we can extend the function continuously to quaternions with real part $1$ (other that $1$ itself).
edited Aug 25 at 1:32
answered Aug 25 at 1:26
Henning Makholm
229k16295526
229k16295526
Thanks for this response. If $q=2+0i+0j+0k$, will $zeta(q)$ will still equal $pi^2/6 =zeta(2)$? In other words, would the extension agree with standard $zeta$ for real $q$, and for complex $q$?
â Joseph O'Rourke
Aug 25 at 12:18
@JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $mathbb Hsetminus1$ in one go).
â Henning Makholm
Aug 25 at 12:26
add a comment |Â
Thanks for this response. If $q=2+0i+0j+0k$, will $zeta(q)$ will still equal $pi^2/6 =zeta(2)$? In other words, would the extension agree with standard $zeta$ for real $q$, and for complex $q$?
â Joseph O'Rourke
Aug 25 at 12:18
@JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $mathbb Hsetminus1$ in one go).
â Henning Makholm
Aug 25 at 12:26
Thanks for this response. If $q=2+0i+0j+0k$, will $zeta(q)$ will still equal $pi^2/6 =zeta(2)$? In other words, would the extension agree with standard $zeta$ for real $q$, and for complex $q$?
â Joseph O'Rourke
Aug 25 at 12:18
Thanks for this response. If $q=2+0i+0j+0k$, will $zeta(q)$ will still equal $pi^2/6 =zeta(2)$? In other words, would the extension agree with standard $zeta$ for real $q$, and for complex $q$?
â Joseph O'Rourke
Aug 25 at 12:18
@JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $mathbb Hsetminus1$ in one go).
â Henning Makholm
Aug 25 at 12:26
@JosephO'Rourke: Yes, because all the terms in the Taylor series would be unchanged, and convergence works the same. (You can also use the Laurent series around $1$ to get values for all of $mathbb Hsetminus1$ in one go).
â Henning Makholm
Aug 25 at 12:26
add a comment |Â
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