How to find $partialchi^2/partial b$ when $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} $?...
This question is an exact duplicate of:
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
1 answer
How do I find How to find $partialchi^2/partial b$ when $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} $?
My attempt:
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]+dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{1}{sigma_1^2}cdot2cdot b(x_1)-dfrac{1}{sigma_2^2}cdot2cdot b(x_2)-cdots-dfrac{1}{sigma_N^2}cdot2cdot b(x_N)\
&= -Bigg[dfrac{2}{sigma_1^2}cdot b(x_1)+dfrac{2}{sigma_2^2}cdot b(x_2)+cdots+dfrac{2}{sigma_N^2}cdot b(x_N)Bigg]\
&= -sum_{i=1}^Ndfrac{2b(x_i)}{sigma_1^2}
end{align*}
Is this correct?
EDIT: I previously asked this question. But the fact that $b$ is a function and changes with each different $x_i$ confuses me.
partial-derivative
asked Nov 26 at 17:50
kaisa
1019
marked as duplicate by callculus, KReiser, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:11
This question was marked as an exact duplicate of an existing question.
add a comment |
This question is an exact duplicate of:
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
1 answer
How do I find How to find $partialchi^2/partial b$ when $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} $?
My attempt:
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]+dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{1}{sigma_1^2}cdot2cdot b(x_1)-dfrac{1}{sigma_2^2}cdot2cdot b(x_2)-cdots-dfrac{1}{sigma_N^2}cdot2cdot b(x_N)\
&= -Bigg[dfrac{2}{sigma_1^2}cdot b(x_1)+dfrac{2}{sigma_2^2}cdot b(x_2)+cdots+dfrac{2}{sigma_N^2}cdot b(x_N)Bigg]\
&= -sum_{i=1}^Ndfrac{2b(x_i)}{sigma_1^2}
end{align*}
Is this correct?
EDIT: I previously asked this question. But the fact that $b$ is a function and changes with each different $x_i$ confuses me.
partial-derivative
asked Nov 26 at 17:50
kaisa
1019
marked as duplicate by callculus, KReiser, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:11
This question was marked as an exact duplicate of an existing question.
1
Does "$b(x)^2$" mean "$(b)cdot (x^2)$", or is $b$ a function so that it means "$(b(x))^2$"? If $b$ is a constant, then by linearity the derivative is just $-sum_i x_i^2/sigma_i^2$.
– MPW
Nov 26 at 17:54
@MPW $b(x_i)$ is a function.
– kaisa
Nov 26 at 17:57
add a comment |
This question is an exact duplicate of:
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
1 answer
How do I find How to find $partialchi^2/partial b$ when $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} $?
My attempt:
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]+dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{1}{sigma_1^2}cdot2cdot b(x_1)-dfrac{1}{sigma_2^2}cdot2cdot b(x_2)-cdots-dfrac{1}{sigma_N^2}cdot2cdot b(x_N)\
&= -Bigg[dfrac{2}{sigma_1^2}cdot b(x_1)+dfrac{2}{sigma_2^2}cdot b(x_2)+cdots+dfrac{2}{sigma_N^2}cdot b(x_N)Bigg]\
&= -sum_{i=1}^Ndfrac{2b(x_i)}{sigma_1^2}
end{align*}
Is this correct?
EDIT: I previously asked this question. But the fact that $b$ is a function and changes with each different $x_i$ confuses me.
partial-derivative
asked Nov 26 at 17:50
kaisa
1019
This question is an exact duplicate of:
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
1 answer
How do I find How to find $partialchi^2/partial b$ when $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} $?
My attempt:
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]+dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]+cdots+dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{1}{sigma_1^2}cdot2cdot b(x_1)-dfrac{1}{sigma_2^2}cdot2cdot b(x_2)-cdots-dfrac{1}{sigma_N^2}cdot2cdot b(x_N)\
&= -Bigg[dfrac{2}{sigma_1^2}cdot b(x_1)+dfrac{2}{sigma_2^2}cdot b(x_2)+cdots+dfrac{2}{sigma_N^2}cdot b(x_N)Bigg]\
&= -sum_{i=1}^Ndfrac{2b(x_i)}{sigma_1^2}
end{align*}
Is this correct?
EDIT: I previously asked this question. But the fact that $b$ is a function and changes with each different $x_i$ confuses me.
This question is an exact duplicate of:
How to find $partialchi^2/partial a$ where $chi^2=sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2}$?
1 answer
partial-derivative
partial-derivative
asked Nov 26 at 17:50
kaisa
1019
asked Nov 26 at 17:50
kaisa
1019
edited Nov 26 at 17:56
asked Nov 26 at 17:50
kaisa
1019
asked Nov 26 at 17:50
kaisa
1019
asked Nov 26 at 17:50
kaisa
1019
1019
marked as duplicate by callculus, KReiser, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:11
This question was marked as an exact duplicate of an existing question.
marked as duplicate by callculus, KReiser, max_zorn, Cyclohexanol., José Carlos Santos Nov 27 at 7:11
This question was marked as an exact duplicate of an existing question.
1
Does "$b(x)^2$" mean "$(b)cdot (x^2)$", or is $b$ a function so that it means "$(b(x))^2$"? If $b$ is a constant, then by linearity the derivative is just $-sum_i x_i^2/sigma_i^2$.
– MPW
Nov 26 at 17:54
@MPW $b(x_i)$ is a function.
– kaisa
Nov 26 at 17:57
add a comment |
1
Does "$b(x)^2$" mean "$(b)cdot (x^2)$", or is $b$ a function so that it means "$(b(x))^2$"? If $b$ is a constant, then by linearity the derivative is just $-sum_i x_i^2/sigma_i^2$.
– MPW
Nov 26 at 17:54
@MPW $b(x_i)$ is a function.
– kaisa
Nov 26 at 17:57
1
1
Does "$b(x)^2$" mean "$(b)cdot (x^2)$", or is $b$ a function so that it means "$(b(x))^2$"? If $b$ is a constant, then by linearity the derivative is just $-sum_i x_i^2/sigma_i^2$.
– MPW
Nov 26 at 17:54
Does "$b(x)^2$" mean "$(b)cdot (x^2)$", or is $b$ a function so that it means "$(b(x))^2$"? If $b$ is a constant, then by linearity the derivative is just $-sum_i x_i^2/sigma_i^2$.
– MPW
Nov 26 at 17:54
@MPW $b(x_i)$ is a function.
– kaisa
Nov 26 at 17:57
@MPW $b(x_i)$ is a function.
– kaisa
Nov 26 at 17:57
add a comment |
1 Answer
1
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I figured out the answer thanks to @MPW 's comment.
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]-dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]-cdots-dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{(x_1)^2}{sigma_1^2}dfrac{partial}{partial b}[b]-dfrac{(x_2)^2}{sigma_2^2}dfrac{partial}{partial b}[b]-cdots-dfrac{(x_N)^2}{sigma_N^2}dfrac{partial}{partial b}[b]\
&= -dfrac{(x_1)^2}{sigma_1^2}cdot1-dfrac{(x_2)^2}{sigma_2^2}cdot1-cdots-dfrac{(x_N)^2}{sigma_N^2}cdot1\
&= -Bigg[dfrac{(x_1)^2}{sigma_1^2}+dfrac{(x_2)^2}{sigma_2^2}+cdots+dfrac{(x_N)^2}{sigma_N^2}Bigg]\
&= -sum_{i=1}^Ndfrac{(x_i)^2}{sigma_i^2}
end{align*}
answered Nov 26 at 18:05
kaisa
1019
add a comment |
1 Answer
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1 Answer
1
active
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active
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active
oldest
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I figured out the answer thanks to @MPW 's comment.
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]-dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]-cdots-dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{(x_1)^2}{sigma_1^2}dfrac{partial}{partial b}[b]-dfrac{(x_2)^2}{sigma_2^2}dfrac{partial}{partial b}[b]-cdots-dfrac{(x_N)^2}{sigma_N^2}dfrac{partial}{partial b}[b]\
&= -dfrac{(x_1)^2}{sigma_1^2}cdot1-dfrac{(x_2)^2}{sigma_2^2}cdot1-cdots-dfrac{(x_N)^2}{sigma_N^2}cdot1\
&= -Bigg[dfrac{(x_1)^2}{sigma_1^2}+dfrac{(x_2)^2}{sigma_2^2}+cdots+dfrac{(x_N)^2}{sigma_N^2}Bigg]\
&= -sum_{i=1}^Ndfrac{(x_i)^2}{sigma_i^2}
end{align*}
answered Nov 26 at 18:05
kaisa
1019
add a comment |
I figured out the answer thanks to @MPW 's comment.
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]-dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]-cdots-dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{(x_1)^2}{sigma_1^2}dfrac{partial}{partial b}[b]-dfrac{(x_2)^2}{sigma_2^2}dfrac{partial}{partial b}[b]-cdots-dfrac{(x_N)^2}{sigma_N^2}dfrac{partial}{partial b}[b]\
&= -dfrac{(x_1)^2}{sigma_1^2}cdot1-dfrac{(x_2)^2}{sigma_2^2}cdot1-cdots-dfrac{(x_N)^2}{sigma_N^2}cdot1\
&= -Bigg[dfrac{(x_1)^2}{sigma_1^2}+dfrac{(x_2)^2}{sigma_2^2}+cdots+dfrac{(x_N)^2}{sigma_N^2}Bigg]\
&= -sum_{i=1}^Ndfrac{(x_i)^2}{sigma_i^2}
end{align*}
answered Nov 26 at 18:05
kaisa
1019
add a comment |
I figured out the answer thanks to @MPW 's comment.
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]-dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]-cdots-dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{(x_1)^2}{sigma_1^2}dfrac{partial}{partial b}[b]-dfrac{(x_2)^2}{sigma_2^2}dfrac{partial}{partial b}[b]-cdots-dfrac{(x_N)^2}{sigma_N^2}dfrac{partial}{partial b}[b]\
&= -dfrac{(x_1)^2}{sigma_1^2}cdot1-dfrac{(x_2)^2}{sigma_2^2}cdot1-cdots-dfrac{(x_N)^2}{sigma_N^2}cdot1\
&= -Bigg[dfrac{(x_1)^2}{sigma_1^2}+dfrac{(x_2)^2}{sigma_2^2}+cdots+dfrac{(x_N)^2}{sigma_N^2}Bigg]\
&= -sum_{i=1}^Ndfrac{(x_i)^2}{sigma_i^2}
end{align*}
answered Nov 26 at 18:05
kaisa
1019
I figured out the answer thanks to @MPW 's comment.
begin{align*}
dfrac{partial}{partial b}sum_{i=1}^Ndfrac{D(x_i)-a-b(x_i)^2}{sigma_i^2} &= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}-sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{D(x_i)-a}{sigma_i^2}Bigg]-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= 0-dfrac{partial}{partial b}Bigg[sum_{i=1}^Ndfrac{b(x_i)^2}{sigma_i^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}+dfrac{b(x_2)^2}{sigma_2^2}+cdots+dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{partial}{partial b}Bigg[dfrac{b(x_1)^2}{sigma_1^2}Bigg]-dfrac{partial}{partial b}Bigg[dfrac{b(x_2)^2}{sigma_2^2}Bigg]-cdots-dfrac{partial}{partial b}Bigg[dfrac{b(x_N)^2}{sigma_N^2}Bigg]\
&= -dfrac{1}{sigma_1^2}dfrac{partial}{partial b}Bigg[b(x_1)^2Bigg]-dfrac{1}{sigma_2^2}dfrac{partial}{partial b}Bigg[b(x_2)^2Bigg]-cdots-dfrac{1}{sigma_N^2}dfrac{partial}{partial b}Bigg[b(x_N)^2Bigg]\
&= -dfrac{(x_1)^2}{sigma_1^2}dfrac{partial}{partial b}[b]-dfrac{(x_2)^2}{sigma_2^2}dfrac{partial}{partial b}[b]-cdots-dfrac{(x_N)^2}{sigma_N^2}dfrac{partial}{partial b}[b]\
&= -dfrac{(x_1)^2}{sigma_1^2}cdot1-dfrac{(x_2)^2}{sigma_2^2}cdot1-cdots-dfrac{(x_N)^2}{sigma_N^2}cdot1\
&= -Bigg[dfrac{(x_1)^2}{sigma_1^2}+dfrac{(x_2)^2}{sigma_2^2}+cdots+dfrac{(x_N)^2}{sigma_N^2}Bigg]\
&= -sum_{i=1}^Ndfrac{(x_i)^2}{sigma_i^2}
end{align*}
answered Nov 26 at 18:05
kaisa
1019
answered Nov 26 at 18:05
kaisa
1019
answered Nov 26 at 18:05
kaisa
1019
answered Nov 26 at 18:05
kaisa
1019
1019
add a comment |
add a comment |
1
Does "$b(x)^2$" mean "$(b)cdot (x^2)$", or is $b$ a function so that it means "$(b(x))^2$"? If $b$ is a constant, then by linearity the derivative is just $-sum_i x_i^2/sigma_i^2$.
– MPW
Nov 26 at 17:54
@MPW $b(x_i)$ is a function.
– kaisa
Nov 26 at 17:57