What is the average of (Ri-1/N*Sum[Ri])^2 The sum is from i=1 to i=N.

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The discussion centers on calculating the average of the expression (Ri - 1/N * Sum[Ri])^2, where the sum is taken from i=1 to N. Participants clarify that 1/N * Sum[Ri] represents the average, denoted as &bar;R, and that the expression simplifies to the variance, which is the average of the squared deviations from the mean. The conversation also touches on the relationship between the square root of the average squared radius, sqrt, and its equivalence to sqrt(Lζ/3), linking contour length to the radius of gyration.

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What is the average of (Ri-1/N*Sum[Ri])^2 The sum is from i=1 to i=N.
 
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What is "Ri"?
 


Hi jaobyccdee! :smile:


1/N*Sum[Ri] is the average \bar R of R_i.

So you're asking about {1 \over N}\sum (R_i - \bar R)^2

Does that look familiar to you?
 


er...actually that's what i start with...but i don't know that it should be familiar...
 


The average of the deviation squared is called "variance".
Its square root is called "standard deviation".
 


Sorry, i should have mentioned that the question starts with sqrt<R^2>, and asks to prove that it is equal to sqrt(Lζ/3). This is the relationship between contour length and the radius of gyration. So i was trying to evaluate it to something that looks like sqrt(Na/6) where Na^2 is the variance.
 


jaobyccdee said:
Sorry, i should have mentioned that the question starts with sqrt<R^2>, and asks to prove that it is equal to sqrt(Lζ/3). This is the relationship between contour length and the radius of gyration. So i was trying to evaluate it to something that looks like sqrt(Na/6) where Na^2 is the variance.

I can't make chocolate of all these symbols you just introduced here...
 

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