### Saddle point method and end-to-end distribution of polymer models

**Introduction**

Saddle point methods are widely used in estimation of integrals with form

$$ I = \int \exp\left(-f(x)\right)\mathrm{d}x $$

where function $f$ can be approximated by first 2 terms of its Taylor series around some $x_0$, i.e.

$$ f(x)\approx f(x_0) + f^\prime (x_0)(x-x_0) + \frac{1}{2}f^{\prime\prime}(x_0)(x-x_0)^2 $$

The integral is thus approximated by its saddle point, where $f^\prime (x_0)=0$ and $f^{\prime\prime}(x_0)>0$:

$$ \begin{align} I&\approx \int \exp\left(-f(x_0) - \frac{1}{2}f^{\prime\prime}(x_0)(x-x_0)^2\right) \mathrm{d}x\\ &=\exp(-f(x_0))\sqrt{\frac{2\pi}{f^{\prime\prime}(x_0)}} \end{align}$$

**Examples**

- Stirling's formula:

$$N!=\int_0^\infty \exp(-x)x^N\mathrm{d}x$$

let $f(x):=x-N\ln(x)$, with large $N$, the negative part is negligible, solving $f^\prime (x) = 0$, we have:

$$N!\approx\exp(-N+N\ln(N))\sqrt{2\pi{}N}=\sqrt{2\pi{}N}\left(\frac{N}{e}\right)^N$$

- Partition function:

$$Z = \int \exp(-\beta U(\mathbf{x})) \mathrm{d}\mathbf{x}$$

with

$$U(\mathbf{x})\approx U(\mathbf{x}_0) + \frac{1}{2}(\mathbf{x}-\mathbf{x}_0)^T H[U](\mathbf{x}-\mathbf{x}_0)$$

where $H$ represents hessian matrix.

**End-to-end distribution function of random walk model of polymer chains**

For an $N$-step random walk model, the exact end-to-end vector distribution is

$$ \begin{align} P(\mathbf{Y})&=\frac{1}{(2\pi)^3}\int \mathrm{d}\mathbf{k} \exp(-i\mathbf{k}\cdot\mathbf{Y})\tilde{\phi}^N\\ &=\int_{0}^{\infty} k\sin(kY) \left(\frac{\sin(kb)}{kb}\right)^N \mathrm{d}k \end{align} $$

with $\phi(\mathbf{x})=\frac{1}{4\pi b^2}\delta(|\mathbf{x}|-b)$ is the distribution of one step vector (length=$b$) and $\tilde{\phi}$ is the characteristic function of $\phi$; $\mathbf{Y}:=\sum_{i=1}^N \mathbf{x}_i$ is the end-to-end vector. Let $s=kb$ and $f(s):=i\frac{Y}{Nb}s -\ln\frac{\sin(s)}{s}$ then we have:

$$ \begin{align} P&=\frac{i}{4\pi^2 b^2 Y}\int_{-\infty}^{+\infty} s \exp\left(-is\frac{Y}{b}\right)\left(\frac{\sin(s)}{s}\right)^N\mathrm{d}s\\ &=\frac{i}{4\pi^2 b^2 Y} \int s \exp(-Nf(s)) \mathrm{d} s\end{align}$$

in this step, the integral is extened to $(-\infty, +\infty)$ due to the symmetry of sin/cos function: the first sin function is replaced with form of $\exp(-ix)$ by Eular's equation: $\exp(ix)=\cos(x)+i\sin(x)$.

Solving for $f^\prime(s)=0$, one could find that $is$ satisfies

$$\coth(is)-\frac{1}{is}=\frac{Y}{Nb}$$

i.e. the Langevin function, $is_0=L^{-1}(\frac{Y}{Nb})$. We therefore have:

$$\begin{align} P &\approx\frac{s_0}{4\pi^2b^2Y}\sqrt{\frac{2\pi}{Nf^{\prime\prime}(s_0)}}e^{-Nf(s_0)}\\ &=\frac{1}{(2\pi{}Nb^2)^{3/2}}\frac{L^{-1}(x)^2}{x(1-(L^{-1}(x)\csc\left(L^{-1}(x)\right))^2)^{1/2}}\\&\times\left(\frac{\sinh\left(L^{-1}(x)\right)}{L^{-1}(x)\exp\left(xL^{-1}(x)\right)}\right)^N\end{align}$$

with $x:=\frac{Y}{Nb}$.

nice

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