# 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

• 171.4 KB Views: 450

Related Calculus and Beyond Homework Help News on Phys.org
Ray Vickson
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
Gold Member
2019 Award
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
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.