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Is it possible to use sigma notation for non-integer steps, for example I want to sum $\ln(x)^2$ from $2$ to $20$ with steps of $0.5$, is there a way I could write this in sigma notation or some other form of notation.

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4 Answers 4

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In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like $$\sum_{s\in S}f(s).$$ In particular you might have $S=\{2,2.5,3,\dots,20\}$.

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  • $\begingroup$ Arbitrary finite set (or else invoke topology and convergence conditions). $\endgroup$ Feb 11, 2019 at 10:17
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    $\begingroup$ You could also index over the set $S$, to get $\sum_{i=0}^nf(s_i)$, where $S=\{s_0,s_1,\ldots,s_n\}$. $\endgroup$
    – Jam
    Feb 11, 2019 at 10:20
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Does $$\sum_{n = 0}^{36}\ln\left(2+\dfrac{1}{2}n\right)^2$$ satisfy your requirements?

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You could change $\ln(x)^2$ into $\ln\left(\frac{x}{2}\right)^2$ to achieve the steps of $0.5$ in this case. You want $\frac{x}{2}$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$\sum_\limits{x = 4}^{40}\ln\left(\frac{x}{2}\right)^2$.

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Just use $\frac{x}{2}$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.

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