# Statistical Mechanics- moments/cumulants, log expansion

## Homework Statement

Using log taylor expansion to express cumulants in terms of moments

I have worked through the expansion- ##log(1+\epsilon)= ...## see thumbnail- and that's ok; my question is why does the expansion hold, i.e. all i can see is it must be that ##k## is small- how is this defined as so?

In all my studies on fourier transforms, I don't ever recall talking about the size of the transform variable being small- (minus the exception of perhaps a boundary condition, and then k is a function of L and then as L gets large k is small) but why here would we require k small, and why isn't this mentioned?

## Homework Equations

my lectures notes here:

## The Attempt at a Solution

as above

#### Attachments

• cm.png
171.4 KB · Views: 603

Ray Vickson
Science Advisor
Homework Helper
Dearly Missed

## Homework Statement

Using log taylor expansion to express cumulants in terms of moments

I have worked through the expansion- ##log(1+\epsilon)= ...## see thumbnail- and that's ok; my question is why does the expansion hold, i.e. all i can see is it must be that ##k## is small- how is this defined as so?

In all my studies on fourier transforms, I don't ever recall talking about the size of the transform variable being small- (minus the exception of perhaps a boundary condition, and then k is a function of L and then as L gets large k is small) but why here would we require k small, and why isn't this mentioned?

## Homework Equations

my lectures notes here:View attachment 232140

## The Attempt at a Solution

as above

The expansion
$$\ln(1+r) = \sum_{k=1}^{\infty} (-1)^{k-1} \frac{r^k}{k} \hspace{4ex}(1)$$ holds only for ##|r| < 1.## Thus, if you write
$$\rho(k) = 1 + \underbrace{\sum_{n=1}^{\infty} \frac{(-ik)^n}{n!} \langle x^n \rangle}_{=\;r}$$
you need ##|k|## small enough to ensure ##|r| < 1## in order to be able to apply expansion (1) to ##\tilde{\rho}(k) = \ln \rho(k).## (Since ##r = r(k)## is continuous in ##k## and vanishes when ##k=0##, there will be a ##k##-interval surrounding ##0## that ensures ##|r(k)| < 1## whenever ##k## lies in that interval.)

The expansion
$$\ln(1+r) = \sum_{k=1}^{\infty} (-1)^{k-1} \frac{r^k}{k} \hspace{4ex}(1)$$ holds only for ##|r| < 1.## Thus, if you write
$$\rho(k) = 1 + \underbrace{\sum_{n=1}^{\infty} \frac{(-ik)^n}{n!} \langle x^n \rangle}_{=\;r}$$
you need ##|k|## small enough to ensure ##|r| < 1## in order to be able to apply expansion (1) to ##\tilde{\rho}(k) = \ln \rho(k).## (Since ##r = r(k)## is continuous in ##k## and vanishes when ##k=0##, there will be a ##k##-interval surrounding ##0## that ensures ##|r(k)| < 1## whenever ##k## lies in that interval.)

Thank you for your reply, and where is it assumed in the notes that k is small enough ? Is this a common sort of assumption with Fourier transforms or not ? Ta.

StoneTemplePython
Science Advisor
Gold Member
any time you see a series with a constraint like ##\big \vert r \big \vert \lt 1## you should immediately ponder the geometric series as a building block.
- - - -
It looks like your author if defining the cumulant as the log of the characteristic function... I've always seen it the other way: cumulant is log of MGF.

It's worth contemplating the differences here.

Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
Thank you for your reply, and where is it assumed in the notes that k is small enough ? Is this a common sort of assumption with Fourier transforms or not ? Ta.

No: the Fourier transform may exist for all ##k##, and the series (1) may also be valid for all ##k##. However, as I said already (and will repeat here), the expansion ##\ln(1+r) = \sum_{k \geq 1} (-1) ^{k-1} r^k/k## is valid only if ##-1 < r \leq 1.## Therefore, when you write ##\rho(k) = 1 + r(k)## --- where ##r(k)## is the characteristic series for terms ##k^n, \; n \geq 1## --- we need ##|k|## small enough that we have ##|r(k)| < 1.## Just how large ##k## is allowed to be depends on the probability distribution, but for most distributions there will be at least a small ##k##-interval around ##0## that will work.

As for where in your notes it is assumed that ##k## is small enough: I cannot say. Maybe they do not say it anywhere---I cannot tell. All I can do is explain the situation to you as I have done: in typical case, having ##k## small enough ensures that ##|r(k)| < 1## and that allows your ##\ln##-function expansion.