Undergrad Question about simplifying Sigma notation

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The discussion centers on simplifying the expression ∑i=1log(n) 1(log(n) - i), which the original poster struggles to understand. Participants clarify that the sum involves adding the value 1 a total of log(n) times. This leads to the realization that the simplified result is log(n), as repeatedly adding 1 log(n) times yields log(n). The conversation also touches on the general properties of summation and exponentiation, emphasizing that raising 1 to any power remains 1. Ultimately, the poster gains clarity on the simplification process.
RoboNerd
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Hello everyone!

I have this expression which I have to simplify:
∑i=1log(n) 1(log(n) - i)

And my book apparently simplifies it to being log(n).
I am struggling to figure out why this is the case. Could anyone help?

Thanks!
 
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What is ##e_i=1^{\text{ insert any integer }}## and what is ##\sum_{i=1}^{n} a_i## if you write it out with dots? And finally set ##a_i=e_i##.
 
Firstly, how many terms are there in the sum - ie how many things are being added together?

Secondly, what is the value of each term? What do you get when you raise 1 to the power of any other real number?

EDIT: Uh oh - Jinxed again!
 
Ohh, I see thanks!

We add 1 log(n) times over and over again!

Thanks for the help everyone!
 

Thread 'Hypothesis testing: Defining H0, HA hypotheses so that ( H_A)_A' makes sense'

The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...
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