I need to know how to get the summation of a constant (c) from i=0 to log(n) of a constant
what is the summation from i=0 to log(n)
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$\begingroup$
summation
logarithms
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4$\;\log n-1\;$ won't usually be a natural number, so what do you *exactly* mean? – 2017-02-12
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0Maybe there's a ceiling or flooring operation on $\log(n) - 1$, for $n>0$? – 2017-02-12
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0Your question is not enough clear ! – 2017-02-12
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0Welcome to MSE! It's important that posts are clearly written in TeX and explained so users can help as much as possible! – 2017-02-12
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0sorry its a typo, it is without -1 @DonAntonio – 2017-02-12
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0$C\log(n)$ when $\log(n)$ is natural, otherwise it is sometimes considered abuse of notation unless rigorously defined; where it still equals $C\log(n)$. – 2017-02-12
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1@Alexandralopez That doesn't matter: the number's still not a natural one. – 2017-02-12
1 Answers
2
$$\sum_{i = 0}^{\ln(n)} C = C \sum_{i = 0}^{\ln(n)} 1 = C\cdot (\ln(n)+1) = C\ln(n) + C $$
"More rigorously"
$$\lim_{N\to \ln(n)} \sum_{i = 0}^{N} C = C\cdot (N+1) = C\ln(n) + C$$
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1there are $\log(n) + 1$ indices when starting from zero. So it's $C(\log(n)+1)$. The sum is $\sum_{j=0}^{\log(n)-1} C = C\log(n)$. – 2017-02-12
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0@james.nixon Whoops, thank you! – 2017-02-12