Cumulant Expansion: How Does It Work?

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The discussion focuses on the mechanics of cumulant expansion, specifically addressing the logarithmic transformation of the expansion in equation (29-26). Participants express confusion regarding the transition from equation (29-26) to (29-27) and emphasize the importance of verifying the derivatives of the logarithm of the generating function, G(k), at k=0. The derivative is confirmed as , aligning with the expected outcome for the cumulants Ξr. This highlights the necessity of understanding the underlying mathematical principles for accurate application.

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Can anyone tell me how a cumulant expansion, like the one on the picture, works? If I take the logarithm of the expansion in eq. (29-26) I don't get what is stated below. What should I do to get that?
 

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aaaa202 said:
If I take the logarithm of the expansion in eq. (29-26) I don't get what is stated below. What should I do to get that?
I also don't agree with the transfer from (29-26) to (29-27). Either there is something very wrong with the statement or there is something going on backstage that we are not shown.
 
You need to verify that the derivatives of ##\log(G(k))## evaluated at ##k=0## are exactly the Ξr
For example: ##\frac{d}{dk}\log(G(k)) = \frac{G^\prime(k)}{G(k)}##. Remember that ##G(0) = 1## and ##G^\prime(0) = <Y>##. Hence the derivative evaluated at ##0## is ##<Y>##, as desired. I will let you compute the other Ξr.
 

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