Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove: $$\sum_{s=1}^S\left(\frac{y_s-x_s}{x_s}\right)=\bigtriangledown J_\vec x\cdot(\vec y-\vec x)$$ with: $$J_\vec x=\sum_s\ln(x_s)$$

share|cite|improve this question

1 Answer 1

up vote 1 down vote accepted

With $$ J_x = \sum_s \ln(x_s)$$ you have \begin{align} \nabla J_x &= (\frac{\partial}{\partial x_1} (\sum_s \ln(x_s)),\frac{\partial}{\partial x_2} (\sum_s \ln(x_s)), ...\frac{\partial}{\partial x_S} (\sum_s \ln(x_s)) \bigr)^T \\&=( \frac{1}{x_1},\frac{1}{x_2},...,\frac{1}{x_S})^T \end{align} furthermore \begin{align} \vec x-\vec y = (x_1-y_1, x_2-y_2,...,x_S-y_S)^T \end{align} If you now take the dot product of the two terms you get \begin{align} \nabla J_x \cdot (\vec x-\vec y) & = ( \frac{1}{x_1},\frac{1}{x_2},...,\frac{1}{x_S})^T \cdot (x_1-y_1, x_2-y_2,...,x_S-y_S)^T \\ & =\frac{x_1-y_1}{x_1} + \frac{x_2-y_2}{x_2}+ ... +\frac{x_S-y_S}{x_S}\end{align} Which is your left hand side.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.