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Math Help - Differentiation of Summation operators

  1. #1
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    Differentiation of Summation operators

    Hello all,

    I am having a hard time understanding how to differentiate summation operators. More specifically, I was given the following solution that I do not understand. Any step by step advice (or a link) would be fantastic....

    $\dfrac{d}{dx}$$\sum\nolimits_{i}p_{i}x_{i}=p_{i}$ I do not understand why this is equal to p, not the sum of p

    Thank you kindly for the help,

    Nick
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  2. #2
    MHF Contributor chisigma's Avatar
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    May be that the formula is...

    \displaystyle \frac{d}{d x_{i}} \sum_{i} p_{i}\ x_{i} = p_{i} (1)

    ... and if the x_{i} are independent variables that's true...

    Kind regards

    \chi \sigma
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  3. #3
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    What if \dfrac{d}{dx_i}$$\sum\nolimits_{i}p_{i}x_{i}=p_{i} ?
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  4. #4
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    Quote Originally Posted by salohcin View Post
    Hello all,

    I am having a hard time understanding how to differentiate summation operators. More specifically, I was given the following solution that I do not understand. Any step by step advice (or a link) would be fantastic....

    $\dfrac{d}{dx}$$\sum\nolimits_{i}p_{i}x_{i}=p_{i}$
    As the others are implying, what you have written makes no sense. The sum depends on many different variables, labeled x_1, x_2, ..., x_n but you are differentiating with respect to "x" which is not any of the variables!

    What you have is something like \frac{d(p_1y+ p_2z)}{dx} where you are differentiating a function of y and z with respect to "x". That derivative is, in fact, 0!

    But the partial derivatives of \frac{\partial (p_1x_1+ p_2x_2+ p_3x_3)}{\partial x_1}= p_1, \frac{\partial (p_1x_1+ p_2x_2+ p_3x_3)}{\partial x_2}= p_2, and \frac{\partial (p_1x_1+ p_2x_2+ p_3x_3)}{\partial x_3}= p_3

    I do not understand why this is equal to p, not the sum of p

    Thank you kindly for the help,

    Nick
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