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Math Help - Increase of counter for differentiation of series?

  1. #1
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    Question Increase of counter for differentiation of series?

    In my notes, I am given that the differentiation of a geometry series is \frac{\mathrm{d} }{\mathrm{d} x}\sum_{n=0}^{\infty} x^{n} = \sum_{n=1}^{\infty} nx^{n-1}

    The counter n is increase by 1 after the differentiation. But can I generalise this to say that I need to increase the counter by 1 for any differentiation on any kind of series?

    Thanks.
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  2. #2
    Member HappyJoe's Avatar
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    I cannot see your LaTeX, because there currently is a general compilation error.

    The reason why you increase your counter by 1 when differentiating a geometric series is that the first term is constant, and constants vanish upon differentiation.

    You can picture yourself any series in which the first term is not a constant. For such a series, differentiation will not increase the counter by 1.
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  3. #3
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    This is my latex statement
    Code:
    \frac{d}{dx} \sum_{n=0}^{\infty} x^{n} = \sum_{n=1}^{\infty} n x^{n-1}
    and tried a few times but it still didn't show up. Weird.

    For a geometry series that starts from n=0, when differentiated, the first term vanishes because it is multiplied by 0. But if a series is starting from n=1 or n=2, then the first term will not vanish since the first term will multiply by 1 or 2, which will not make it to zero. And if I still increase the counter after differentiating, wouldn't the series have a missing first term?
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  4. #4
    Member HappyJoe's Avatar
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    You are quite right in that if your geometric series was truncated to begin with in the sense that the counter started from, say, 7, then you shouldn't increase it after differentiation. If you did, you would indeed have a missing term. It is only in the special case, where the first term is constant that the counter increases by 1.
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