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Math Help - find m and b for min. expression.

  1. #1
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    find m and b for min. expression.

    Find all real values of m and b for which \sum_{j=1}^{50}(j^2+mj+b)^2 is Minimum.
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  2. #2
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    Quote Originally Posted by jacks View Post
    Find all real values of m and b for which \sum_{j=1}^{50}(j^2+mj+b)^2 is Minimum.
    First you will need to remember how to sum the the first n integers and their squares and the cubes.

    \displaystyle \sum_{j=1}^{n}j=\frac{n(n+1)}{2}

    \displaystyle \sum_{j=1}^{n}j^2=\frac{n(n+1)(2n+1)}{6}

    \displaystyle \sum_{j=1}^{n}j^3=\left[\frac{n(n+1)}{2}\right]

    Now the function you want to minimize is

    \displaystyle f(m,b)=\sum_{j=1}^{50}(j^2+mj+b)^2

    \displaystyle \frac{df}{dm}=\sum_{j=1}^{50}2(j^2+mj+b)j

    \displaystyle \frac{df}{db}=\sum_{j=1}^{50}2(j^2+mj+b)


    \displaystyle \frac{d^2f}{dm^2}=\sum_{j=1}^{50}2j^2


    \displaystyle \frac{d^2f}{db^2}=\sum_{j=1}^{50}2


    \displaystyle \frac{d^2f}{dmdb}=\frac{d^2f}{dbdm} = \sum_{j=1}^{50}2j

    Can you finish using the 2nd derivative test from here?
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    First you will need to remember how to sum the the first n integers and their squares and the cubes.

    \displaystyle \sum_{j=1}^{n}j=\frac{n(n+1)}{2}

    \displaystyle \sum_{j=1}^{n}j^2=\frac{n(n+1)(2n+1)}{6}

    \displaystyle \sum_{j=1}^{n}j^3=\left[\frac{n(n+1)}{2}\right]

    Now the function you want to minimize is

    \displaystyle f(m,b)=\sum_{j=1}^{50}(j^2+mj+b)^2

    \displaystyle \frac{df}{dm}=\sum_{j=1}^{50}2(j^2+mj+b)j

    \displaystyle \frac{df}{db}=\sum_{j=1}^{50}2(j^2+mj+b)


    \displaystyle \frac{d^2f}{dm^2}=\sum_{j=1}^{50}2j^2


    \displaystyle \frac{d^2f}{db^2}=\sum_{j=1}^{50}2


    \displaystyle \frac{d^2f}{dmdb}=\frac{d^2f}{dbdm} = \sum_{j=1}^{50}2j

    Can you finish using the 2nd derivative test from here?
    I should probably mention that all the derivatives involved are partial derivatives.
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