log of summation expression

Question

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?

Is there anything to prevent you from taking the exponential and work with the sum directly — Chinny84, Nov 20 '15 at 14:44
this is one of my way to solve my own problem, just asking if this can work, thanks! — KennyYang, Nov 20 '15 at 14:48

score 18 · Answer 1 · answered Nov 20 '15 at 15:03

18

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.

answered Nov 20 '15 at 15:03

GEdgar

117,296

1

this sounds good, but $x_i$ is i.i.d.(identical independent ) in my case.So maybe I should trt another way..? – KennyYang Nov 20 '15 at 15:07
1

@KennyYang Are you doing any asymptotics? There's not much that can be said generally, but maybe for your specific problem there is? Like, do the $x_{i}$ have finite mean/variance, and are you taking $n$ large? – Sherwin Lott Sep 09 '20 at 22:37

score 14 · Answer 2 · edited Jul 09 '20 at 14:57

14

There's this: $$\log\left(\sum_{i=0}^n x_i\right) = \log(x_0) + \log\left(1+\sum_{i=1}^n\left(\frac{x_i}{x_0}\right)\right)$$ $x_0$ must be the biggest in the series. It didn't help me with what I'm trying to solve, but I think it answers the question.

edited Jul 09 '20 at 14:57

Especially Lime

46,692

answered Jul 09 '20 at 14:11

tlloyd87

151

3

Actually, the expression is true, irrespective of whether $x_0$ is the biggest in the series or not. – G Frazao Sep 29 '22 at 16:18

score 11 · Answer 3 · answered Dec 23 '15 at 13:26

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 $$\log\left(\Pi_{i=0}^na_i\right)=\sum_{i=0}^n\log(a_i)$$

log of summation expression

3 Answers3

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