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How do I find How to find $\partial\chi^2/\partial b$ when $\chi^2=\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} $?

My attempt: \begin{align*} \dfrac{\partial}{\partial b}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}-\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}\Bigg]-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= 0-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{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)^2\Bigg]-\dfrac{1}{\sigma_2^2}\dfrac{\partial}{\partial b}\Bigg[b(x_2)^2\Bigg]-\cdots-\dfrac{1}{\sigma_N^2}\dfrac{\partial}{\partial b}\Bigg[b(x_N)^2\Bigg]\\ &= -\dfrac{1}{\sigma_1^2}\cdot2\cdot b(x_1)-\dfrac{1}{\sigma_2^2}\cdot2\cdot b(x_2)-\cdots-\dfrac{1}{\sigma_N^2}\cdot2\cdot 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}^N\dfrac{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.

StubbornAtom
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kaisa
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    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 '18 at 17:54
  • @MPW $b(x_i)$ is a function. – kaisa Nov 26 '18 at 17:57

1 Answers1

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I figured out the answer thanks to @MPW 's comment.

\begin{align*} \dfrac{\partial}{\partial b}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}-\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}\Bigg]-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= 0-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{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)^2\Bigg]-\dfrac{1}{\sigma_2^2}\dfrac{\partial}{\partial b}\Bigg[b(x_2)^2\Bigg]-\cdots-\dfrac{1}{\sigma_N^2}\dfrac{\partial}{\partial b}\Bigg[b(x_N)^2\Bigg]\\ &= -\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}^N\dfrac{(x_i)^2}{\sigma_i^2} \end{align*}

kaisa
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