Mixing
Derivative of a function with summation
$begingroup$
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
edited Jan 24 at 18:40
Yves Daoust
130k676229
asked Jan 24 at 18:17
BeeTiauBeeTiau
758
$endgroup$
add a comment |
$begingroup$
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
edited Jan 24 at 18:40
Yves Daoust
130k676229
asked Jan 24 at 18:17
BeeTiauBeeTiau
758
$endgroup$
1
$begingroup$
Yes, this is right.
$endgroup$
– Yves Daoust
Jan 24 at 18:41
$begingroup$
Thanks @YvesDaoust
$endgroup$
– BeeTiau
Jan 26 at 23:00
add a comment |
$begingroup$
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
edited Jan 24 at 18:40
Yves Daoust
130k676229
asked Jan 24 at 18:17
BeeTiauBeeTiau
758
$endgroup$
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
edited Jan 24 at 18:40
Yves Daoust
130k676229
asked Jan 24 at 18:17
BeeTiauBeeTiau
758
edited Jan 24 at 18:40
Yves Daoust
130k676229
asked Jan 24 at 18:17
BeeTiauBeeTiau
758
edited Jan 24 at 18:40
Yves Daoust
130k676229
edited Jan 24 at 18:40
Yves Daoust
130k676229
edited Jan 24 at 18:40
Yves Daoust
130k676229
130k676229
asked Jan 24 at 18:17
BeeTiauBeeTiau
758
asked Jan 24 at 18:17
BeeTiauBeeTiau
758
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
add a comment |
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
add a comment |
1 Answer
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$begingroup$
The above derivation is correct.
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The above derivation is correct.
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
$endgroup$
add a comment |
$begingroup$
The above derivation is correct.
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
$endgroup$
add a comment |
$begingroup$
The above derivation is correct.
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
$endgroup$
The above derivation is correct.
$$frac{dz}{dzeta}=frac{R}{2}left[aleft(-frac{1}{zeta^2}+sum_{k=1}^{N}km_kzeta^{k-1}right)+bleft(1-sum_{k=1}^{N}km_kzeta^{-k-1}right)right]$$
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
answered Jan 26 at 23:01
BeeTiauBeeTiau
758
758
add a comment |
add a comment |
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$begingroup$
Yes, this is right.
$endgroup$
– Yves Daoust
Jan 24 at 18:41
$begingroup$
Thanks @YvesDaoust
$endgroup$
– BeeTiau
Jan 26 at 23:00