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?







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    up vote
    4
    down vote

    favorite
    1













    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?







    share|cite|improve this question
























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1






      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?







      share|cite|improve this question















      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?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 25 at 11:37

























      asked Aug 25 at 0:52









      Joseph O'Rourke

      17.3k248104




      17.3k248104




















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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).






          share|cite|improve this answer






















          • 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 $q$?
            – 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).
            – Henning Makholm
            Aug 25 at 12:26











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          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).






          share|cite|improve this answer






















          • 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 $q$?
            – 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).
            – Henning Makholm
            Aug 25 at 12:26















          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).






          share|cite|improve this answer






















          • 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 $q$?
            – 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).
            – Henning Makholm
            Aug 25 at 12:26













          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).






          share|cite|improve this answer














          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).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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 $q$?
            – 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).
            – 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 $q$?
            – 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).
            – 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 $q$?
          – 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 $q$?
          – 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).
          – 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).
          – Henning Makholm
          Aug 25 at 12:26


















           

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