Derivative of a function with summation












1












$begingroup$


Can someone please help me to verify whether the derivative of the following function:



$$z(zeta)=frac{R}{2}left[aleft(frac{1}{zeta}+sum_{k=1}^{N}m_kzeta^kright)+bleft(zeta+sum_{k=1}^{N}frac{m_k}{zeta^k}right)right]$$



with respect to $zeta$



is this one below?
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$



in which $m_k$, $R$, $a$ and $b$ are all constants.



Can someone please help me to verify? I am just not so sure about my own result.






derivatives







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  • 1




    $begingroup$
    Yes, this is right.
    $endgroup$
    – Yves Daoust
    Jan 24 at 18:41










  • $begingroup$
    Thanks @YvesDaoust
    $endgroup$
    – BeeTiau
    Jan 26 at 23:00
















1












$begingroup$


Can someone please help me to verify whether the derivative of the following function:



$$z(zeta)=frac{R}{2}left[aleft(frac{1}{zeta}+sum_{k=1}^{N}m_kzeta^kright)+bleft(zeta+sum_{k=1}^{N}frac{m_k}{zeta^k}right)right]$$



with respect to $zeta$



is this one below?
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$



in which $m_k$, $R$, $a$ and $b$ are all constants.



Can someone please help me to verify? I am just not so sure about my own result.






derivatives







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, this is right.
    $endgroup$
    – Yves Daoust
    Jan 24 at 18:41










  • $begingroup$
    Thanks @YvesDaoust
    $endgroup$
    – BeeTiau
    Jan 26 at 23:00














1












1








1


1



$begingroup$


Can someone please help me to verify whether the derivative of the following function:



$$z(zeta)=frac{R}{2}left[aleft(frac{1}{zeta}+sum_{k=1}^{N}m_kzeta^kright)+bleft(zeta+sum_{k=1}^{N}frac{m_k}{zeta^k}right)right]$$



with respect to $zeta$



is this one below?
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$



in which $m_k$, $R$, $a$ and $b$ are all constants.



Can someone please help me to verify? I am just not so sure about my own result.






derivatives







share|cite|improve this question











$endgroup$




Can someone please help me to verify whether the derivative of the following function:



$$z(zeta)=frac{R}{2}left[aleft(frac{1}{zeta}+sum_{k=1}^{N}m_kzeta^kright)+bleft(zeta+sum_{k=1}^{N}frac{m_k}{zeta^k}right)right]$$



with respect to $zeta$



is this one below?
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$



in which $m_k$, $R$, $a$ and $b$ are all constants.



Can someone please help me to verify? I am just not so sure about my own result.






derivatives




derivatives






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 18:40









Yves Daoust

130k676229




130k676229










asked Jan 24 at 18:17









BeeTiauBeeTiau

758




758








  • 1




    $begingroup$
    Yes, this is right.
    $endgroup$
    – Yves Daoust
    Jan 24 at 18:41










  • $begingroup$
    Thanks @YvesDaoust
    $endgroup$
    – BeeTiau
    Jan 26 at 23:00














  • 1




    $begingroup$
    Yes, this is right.
    $endgroup$
    – Yves Daoust
    Jan 24 at 18:41










  • $begingroup$
    Thanks @YvesDaoust
    $endgroup$
    – BeeTiau
    Jan 26 at 23:00








1




1




$begingroup$
Yes, this is right.
$endgroup$
– Yves Daoust
Jan 24 at 18:41




$begingroup$
Yes, this is right.
$endgroup$
– Yves Daoust
Jan 24 at 18:41












$begingroup$
Thanks @YvesDaoust
$endgroup$
– BeeTiau
Jan 26 at 23:00




$begingroup$
Thanks @YvesDaoust
$endgroup$
– BeeTiau
Jan 26 at 23:00










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$begingroup$

The above derivation is correct.



$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$






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    $begingroup$

    The above derivation is correct.



    $$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The above derivation is correct.



      $$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The above derivation is correct.



        $$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$






        share|cite|improve this answer









        $endgroup$



        The above derivation is correct.



        $$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 23:01









        BeeTiauBeeTiau

        758




        758






























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