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I am curious about simplifying the following expression:

$$\log \left(\sum_\limits{i=0}^{n}x_i \right)$$

Is there any rule to simplify a summation inside the log?

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    $\begingroup$ No there is no such standard rule as far as I can remember. $\endgroup$ – SchrodingersCat Nov 20 '15 at 14:44
  • $\begingroup$ Is there anything to prevent you from taking the exponential and work with the sum directly $\endgroup$ – Chinny84 Nov 20 '15 at 14:44
  • $\begingroup$ this is one of my way to solve my own problem, just asking if this can work, thanks! $\endgroup$ – KennyYang Nov 20 '15 at 14:48
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Sometimes we have asymptotics with most significant term $M \to \infty$ we do this $$ \log(M+A) = \log(M\cdot(1+S)) $$ so that $S=A/M$ is "small" in the sense $S=o(1)$, and then $$ \log(M\cdot(1+S)) = \log M + \log(1+S) = \log M + S - \frac{1}{2}S^2+\frac{1}{3}S^3+\cdots $$

reference
G. A. Edgar, Transseries for beginners, Real Anal. Exchange 35 (2010), no. 2, 253--309.

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  • $\begingroup$ this sounds good, but $x_i$ is i.i.d.(identical independent ) in my case.So maybe I should trt another way..? $\endgroup$ – KennyYang Nov 20 '15 at 15:07
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You may note that this is equivalent to trying to solve $$\log(a+b+c)$$You realize you can't do much about $a,b,$ or $c$.

The only way this can be simplified, is if you can factor something out and then apply log properties.

A more interesting question might concern $$\log\left(\Pi_{i=0}^na_i\right)=\sum_{i=0}^n\log(a_i)$$

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