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

  • Thread starter jaobyccdee
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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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  • #2
HallsofIvy
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What is "Ri"?
 
  • #3
I like Serena
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Hi jaobyccdee! :smile:


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

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

Does that look familiar to you?
 
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er....actually that's what i start with....but i don't know that it should be familiar....
 
  • #5
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The average of the deviation squared is called "variance".
Its square root is called "standard deviation".
 
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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.
 
  • #7
I like Serena
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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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