How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
My attempt:
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{partial}{partial a}sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}sum_{i=1}^N-a\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}a^N\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdot Ncdot a^{N-1}\
end{align}
Is this correct?
EDIT:
As per the accepted answer, this is how I got my final partial derivative.
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&=dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}Bigg]-dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}+dfrac{-a}{sigma_2^2}+cdots+dfrac{-a}{sigma_N^2}Bigg]\
&= -Bigg[dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}Bigg]+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_N^2}Bigg]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+cdots+dfrac{-1}{sigma_N^2}cdotdfrac{partial}{partial a}[a]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}+dfrac{-1}{sigma_2^2}+cdots+dfrac{-1}{sigma_N^2}Bigg]\
&= sum_{i=1}^{N}dfrac{1}{sigma_i^2}\
end{align}
partial-derivative
add a comment |
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
My attempt:
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{partial}{partial a}sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}sum_{i=1}^N-a\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}a^N\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdot Ncdot a^{N-1}\
end{align}
Is this correct?
EDIT:
As per the accepted answer, this is how I got my final partial derivative.
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&=dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}Bigg]-dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}+dfrac{-a}{sigma_2^2}+cdots+dfrac{-a}{sigma_N^2}Bigg]\
&= -Bigg[dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}Bigg]+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_N^2}Bigg]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+cdots+dfrac{-1}{sigma_N^2}cdotdfrac{partial}{partial a}[a]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}+dfrac{-1}{sigma_2^2}+cdots+dfrac{-1}{sigma_N^2}Bigg]\
&= sum_{i=1}^{N}dfrac{1}{sigma_i^2}\
end{align}
partial-derivative
1
Sorry, my mistake. (The sign is wrong.) The answer should be $-sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$. When you split the sum the sign is always held "+".
– kolobokish
Nov 26 at 17:40
add a comment |
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
My attempt:
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{partial}{partial a}sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}sum_{i=1}^N-a\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}a^N\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdot Ncdot a^{N-1}\
end{align}
Is this correct?
EDIT:
As per the accepted answer, this is how I got my final partial derivative.
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&=dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}Bigg]-dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}+dfrac{-a}{sigma_2^2}+cdots+dfrac{-a}{sigma_N^2}Bigg]\
&= -Bigg[dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}Bigg]+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_N^2}Bigg]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+cdots+dfrac{-1}{sigma_N^2}cdotdfrac{partial}{partial a}[a]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}+dfrac{-1}{sigma_2^2}+cdots+dfrac{-1}{sigma_N^2}Bigg]\
&= sum_{i=1}^{N}dfrac{1}{sigma_i^2}\
end{align}
partial-derivative
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
My attempt:
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{partial}{partial a}sum_{i=1}^Ndfrac{-a}{sigma_i^2}\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}sum_{i=1}^N-a\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdotdfrac{partial}{partial a}a^N\
&=sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-dfrac{1}{sigma_i^2}cdot Ncdot a^{N-1}\
end{align}
Is this correct?
EDIT:
As per the accepted answer, this is how I got my final partial derivative.
begin{align}
dfrac{partial}{partial a}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}-sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&=dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{D(x_i)-b(x_i)^2}{sigma_i^2}Bigg]-dfrac{partial}{partial a}Bigg[sum_{i=1}^Ndfrac{-a}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}+dfrac{-a}{sigma_2^2}+cdots+dfrac{-a}{sigma_N^2}Bigg]\
&= -Bigg[dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_1^2}Bigg]+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial a}Bigg[dfrac{-a}{sigma_N^2}Bigg]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+dfrac{-1}{sigma_1^2}cdotdfrac{partial}{partial a}[a]+cdots+dfrac{-1}{sigma_N^2}cdotdfrac{partial}{partial a}[a]Bigg]\
&= -Bigg[dfrac{-1}{sigma_1^2}+dfrac{-1}{sigma_2^2}+cdots+dfrac{-1}{sigma_N^2}Bigg]\
&= sum_{i=1}^{N}dfrac{1}{sigma_i^2}\
end{align}
partial-derivative
partial-derivative
edited Nov 26 at 17:37
asked Nov 26 at 17:09
kaisa
1019
1019
1
Sorry, my mistake. (The sign is wrong.) The answer should be $-sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$. When you split the sum the sign is always held "+".
– kolobokish
Nov 26 at 17:40
add a comment |
1
Sorry, my mistake. (The sign is wrong.) The answer should be $-sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$. When you split the sum the sign is always held "+".
– kolobokish
Nov 26 at 17:40
1
1
Sorry, my mistake. (The sign is wrong.) The answer should be $-sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$. When you split the sum the sign is always held "+".
– kolobokish
Nov 26 at 17:40
Sorry, my mistake. (The sign is wrong.) The answer should be $-sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$. When you split the sum the sign is always held "+".
– kolobokish
Nov 26 at 17:40
add a comment |
1 Answer
1
active
oldest
votes
Nope. If $D$, $b$ and $sigma$ are independet of $a$, then the partial differential of their combination should be zero. Second mistake is the sign you used. (between sum signs it should be "+"). However later you corrected it. Next you can't take our $frac{1}{sigma_{i}^{2}}$ out of the sum sign (as $sigma_i$-s depends on $i$). And the last mistake is that sum of $a$-s is not $a^{N}$, but rather $aN$. So as to me the answer should be $sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$ (because partial of $frac{a}{sigma_{i}^{2}}$ is $frac{1}{sigma_{i}^{2}}$).
If $D$, $b$ and $sigma$ depends on $a$(As it may when the initial problem comes from statistics), then the dependence form is crucial. (Without it you can't explicitly tell what the partial will be.)
Sorry(too long to type it as comment).
add a comment |
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Nope. If $D$, $b$ and $sigma$ are independet of $a$, then the partial differential of their combination should be zero. Second mistake is the sign you used. (between sum signs it should be "+"). However later you corrected it. Next you can't take our $frac{1}{sigma_{i}^{2}}$ out of the sum sign (as $sigma_i$-s depends on $i$). And the last mistake is that sum of $a$-s is not $a^{N}$, but rather $aN$. So as to me the answer should be $sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$ (because partial of $frac{a}{sigma_{i}^{2}}$ is $frac{1}{sigma_{i}^{2}}$).
If $D$, $b$ and $sigma$ depends on $a$(As it may when the initial problem comes from statistics), then the dependence form is crucial. (Without it you can't explicitly tell what the partial will be.)
Sorry(too long to type it as comment).
add a comment |
Nope. If $D$, $b$ and $sigma$ are independet of $a$, then the partial differential of their combination should be zero. Second mistake is the sign you used. (between sum signs it should be "+"). However later you corrected it. Next you can't take our $frac{1}{sigma_{i}^{2}}$ out of the sum sign (as $sigma_i$-s depends on $i$). And the last mistake is that sum of $a$-s is not $a^{N}$, but rather $aN$. So as to me the answer should be $sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$ (because partial of $frac{a}{sigma_{i}^{2}}$ is $frac{1}{sigma_{i}^{2}}$).
If $D$, $b$ and $sigma$ depends on $a$(As it may when the initial problem comes from statistics), then the dependence form is crucial. (Without it you can't explicitly tell what the partial will be.)
Sorry(too long to type it as comment).
add a comment |
Nope. If $D$, $b$ and $sigma$ are independet of $a$, then the partial differential of their combination should be zero. Second mistake is the sign you used. (between sum signs it should be "+"). However later you corrected it. Next you can't take our $frac{1}{sigma_{i}^{2}}$ out of the sum sign (as $sigma_i$-s depends on $i$). And the last mistake is that sum of $a$-s is not $a^{N}$, but rather $aN$. So as to me the answer should be $sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$ (because partial of $frac{a}{sigma_{i}^{2}}$ is $frac{1}{sigma_{i}^{2}}$).
If $D$, $b$ and $sigma$ depends on $a$(As it may when the initial problem comes from statistics), then the dependence form is crucial. (Without it you can't explicitly tell what the partial will be.)
Sorry(too long to type it as comment).
Nope. If $D$, $b$ and $sigma$ are independet of $a$, then the partial differential of their combination should be zero. Second mistake is the sign you used. (between sum signs it should be "+"). However later you corrected it. Next you can't take our $frac{1}{sigma_{i}^{2}}$ out of the sum sign (as $sigma_i$-s depends on $i$). And the last mistake is that sum of $a$-s is not $a^{N}$, but rather $aN$. So as to me the answer should be $sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$ (because partial of $frac{a}{sigma_{i}^{2}}$ is $frac{1}{sigma_{i}^{2}}$).
If $D$, $b$ and $sigma$ depends on $a$(As it may when the initial problem comes from statistics), then the dependence form is crucial. (Without it you can't explicitly tell what the partial will be.)
Sorry(too long to type it as comment).
answered Nov 26 at 17:31
kolobokish
40438
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Sorry, my mistake. (The sign is wrong.) The answer should be $-sum_{i=1}^{N}frac{1}{sigma_{i}^{2}}$. When you split the sum the sign is always held "+".
– kolobokish
Nov 26 at 17:40