# Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation

## Introduction

We derive here the important results that the Dispersion Relation for a Floating Elastic Plate can be written in the following form using the Mittag-Leffler expansion. This results is used to calculate the Free-Surface Green Function for a Floating Elastic Plate. The Mittag-Leffler expansion is a tool for expressing functions of a complex variable. We will use the Mittag-Leffler expansion to show that the function

$\displaystyle{ \hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) } }$

where

$\displaystyle{ d\left( \gamma,\omega\right) =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H}, }$

is the Dispersion Relation for a Floating Elastic Plate can be expressed by a linear sum of terms like $\displaystyle{ 1/\left( \gamma-a\right) }$, $\displaystyle{ a }$ being a zero of $\displaystyle{ d\left( \gamma,\omega\right) }$. We first remind ourselves of the Mittag-Leffler expansion that can be found in most text books on complex analysis, and then show that it can indeed be applied to $\displaystyle{ \hat {w}\left( \gamma\right) }$.

We will show that

$\displaystyle{ \hat{w}\left( \gamma\right) =\sum_{n=-2}^{\infty}\frac{2q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}} }$

where $\displaystyle{ R\left( q_n\right) }$ is the residue of $\displaystyle{ \hat{w}\left( \gamma\right) }$ at $\displaystyle{ \gamma=q_n }$ given by

$\displaystyle{ R\left( q_n\right) =\frac{\omega^{2}q_n}{\omega^{2}\left( 5q_n^{4}+u\right) +H\left[ \left( q_n^{5}+uq_n\right) ^{2}-\omega^{4}\right] }. }$

where $\displaystyle{ q_n }$ are the roots of the Dispersion Relation for a Floating Elastic Plate

$\displaystyle{ -q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \, }$

with $\displaystyle{ n=-1,-2 }$ corresponding to the complex solutions with positive imaginary part, $\displaystyle{ n=0 }$ corresponding to the imaginary solution with negative real part and $\displaystyle{ n\gt 0 }$ corresponding to the imaginary solutions with positive imagainary part.

## Mittag-Leffler expansion

Consider a function that is regular in the whole plane except at isolated points. A set of points is called isolated if there exists an open disk around each point that contains none other of the isolated points. Such a function is known as fractional function. We show that a fractional function that has an infinite number of poles can be expressed by infinite series of polynomials.

Let $\displaystyle{ f\left( \gamma\right) }$ be a fractional function that has an infinite number of poles. We note that a number of poles that are situated within a bounded region is always finite since the set of poles does not have limit-points. Indeed, if there is a limit-point $\displaystyle{ \gamma=c }$ then any small circle with centre at $\displaystyle{ \gamma=c }$ would contain an infinite number of poles. Once we have a finite number of poles in a confined part of the plane we can number them in the order of their non-decreasing moduli, so that denoting the poles by $\displaystyle{ a_{i} }$ we have

$\displaystyle{ \left| a_{1}\right| \leq\left| a_{2}\right| \leq\left| a_3\right| \leq..., }$

where $\displaystyle{ \left| a_{i}\right| \rightarrow\infty }$ as $\displaystyle{ i\rightarrow\infty }$. At every pole $\displaystyle{ \gamma=a_{i} }$ the function $\displaystyle{ f\left( \gamma\right) }$ will have a definite infinite part, which will be a polynomial with respect to the argument $\displaystyle{ 1/\left( \gamma-a_{i}\right) }$ without the constant term. We denote this polynomial term by

$\displaystyle{ G_{i}\left( \frac{1}{\gamma-a_{i}}\right) ,\,\,i=1,2,3,...\,. }$

We show that the fractional function $\displaystyle{ f\left( \gamma\right) }$ can be represented by a simple infinite series of $\displaystyle{ G_{i} }$ by making certain additional assumptions. Suppose that a sequence of closed contours $\displaystyle{ C_{n} }$ which surround the origin exists and satisfies following conditions.

1. None of poles of $\displaystyle{ f\left( \gamma\right) }$ are on the contours $\displaystyle{ C_{n},\,n=1,2,3,... }$
2. Every contour $\displaystyle{ C_{n} }$ lies inside the contour $\displaystyle{ C_{n+1} }$.
3. Let $\displaystyle{ l_{n} }$ be length of the contour $\displaystyle{ C_{n} }$ and $\displaystyle{ \delta_{n} }$ be its shortest distance from the origin then $\displaystyle{ \delta_{n}\rightarrow\infty }$ as $\displaystyle{ n\rightarrow\infty }$ , i.e., the contours $\displaystyle{ C_{n} }$ widen indefinitely in all directions as $\displaystyle{ n }$ increases.
4. A positive number $\displaystyle{ m }$ exists such that
$\displaystyle{ \frac{l_{n}}{\delta_{n}}\leq m\quad \mathrm{for}\quad n=1,2,3,.... }$

We now suppose that given such a sequence of contours, there exists a positive number $\displaystyle{ M, }$ such that on any contour $\displaystyle{ C_{n} }$ our fractional function $\displaystyle{ f\left(\gamma\right) }$ satisfies $\displaystyle{ \left| f\left( \gamma\right) \right| \leq M }$. Consider the integral

$\displaystyle{ \frac{1}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}-\gamma}d\gamma^{\prime}\,\,\, (1) }$

where the point $\displaystyle{ \gamma }$ lies inside $\displaystyle{ C_{n} }$ and is other than $\displaystyle{ a_{i} }$ (the poles inside $\displaystyle{ C_{n}. }$) We also consider the sum of the polynomials for the poles $\displaystyle{ a_{i} }$, inside $\displaystyle{ C_{n} }$,

$\displaystyle{ \omega_{n}\left( \gamma\right) =\sum_{\left( C_{n}\right) }G_{i}\left( \frac{1}{\gamma-a_{i}}\right) . \,\,\,(2) }$

The integrand of ((1)) has a pole $\displaystyle{ \gamma^{\prime}=\gamma }$ and poles $\displaystyle{ \gamma^{\prime}=a_{i} }$. We can calculate the residue at the pole $\displaystyle{ \gamma^{\prime}=\gamma }$ by

$\displaystyle{ \left. \frac{f\left( \gamma^{\prime}\right) }{\left( \gamma^{\prime }-\gamma\right) ^{\prime}}\right| _{\gamma^{\prime}=\gamma}=\left. f\left( \gamma^{\prime}\right) \right| _{\gamma^{\prime}=\gamma}=f\left( \gamma\right) . }$

The residues at the poles $\displaystyle{ \gamma^{\prime}=a_{i} }$ are, by the definition (2), the same as the residues of the function

$\displaystyle{ \frac{\omega_{n}\left( \gamma^{\prime}\right) }{\gamma^{\prime}-\gamma }.\,\,\, (3) }$

We note that all poles of this function are situated inside $\displaystyle{ C_{n} }$. We now show that the sum of residues of function (3) at the poles $\displaystyle{ a_{i} }$ is

$\displaystyle{ -\omega_{n}\left( \gamma\right) =-\sum_{\left( C_{n}\right) }G_{i}\left( \frac{1}{\gamma-a_{i}}\right) .\,\,\, (4) }$

Since the definition of $\displaystyle{ \omega_{n} }$ and $\displaystyle{ G_{i} }$ is a polynomial of $\displaystyle{ 1/\left( \gamma-a_{i}\right) , }$ the order of the denominator of function (3) is at least two units higher than that of the numerator of function (3). Hence, for a circle with a sufficiently large radius $\displaystyle{ R }$, we have

$\displaystyle{ 2\pi\mathrm{i}\sum_{\left( C_{n}\right) }Res _{\gamma^{\prime}=a_{i}}\frac{\omega_{n}\left( \gamma^{\prime}\right) }{\gamma^{\prime}-\gamma}=\oint_{C_{R}}\frac{\omega_{n}\left( \gamma^{\prime }\right) }{\gamma^{\prime}-\gamma}d\gamma^{\prime}. }$

The LHS of this does not change as the radius $\displaystyle{ R }$ increases, and the RHS$\displaystyle{ \rightarrow0 }$ as $\displaystyle{ R\rightarrow\infty }$. Indeed,

$\displaystyle{ \left| \oint_{C_{R}}\frac{\omega_{n}\left( \gamma^{\prime}\right) } {\gamma^{\prime}-\gamma}d\gamma^{\prime}\right| \leq\oint_{C_{R}}\left| \gamma^{\prime}\frac{\omega_{n}\left( \gamma^{\prime}\right) } {\gamma^{\prime}-\gamma}\frac{1}{\gamma^{\prime}}d\gamma^{\prime}\right| \leq\max_{\left| \gamma^{\prime}\right| =R}\left| \gamma^{\prime} \frac{\omega_{n}\left( \gamma^{\prime}\right) }{\gamma^{\prime}-\gamma }\right| \frac{2\pi R}{R} }$

and the term $\displaystyle{ \left| \cdot\right| }$ tends to zero as $\displaystyle{ R\rightarrow\infty }$. Thus, the sum of residues at poles within a finite distance is zero. Since we know that the residue of (3) at $\displaystyle{ \gamma^{\prime}=\gamma }$ is $\displaystyle{ \omega_{n}\left( \gamma\right) }$, the sum of the rest is formula (4). Thus, we have an expression for the integral (1),

$\displaystyle{ \frac{1}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}-\gamma}d\gamma^{\prime}=f\left( \gamma\right) -\sum_{\left( C_{n}\right) }G_{i}\left( \frac{1}{\gamma-a_{i}}\right) . \,\,\,(5) }$

Also, when $\displaystyle{ \gamma=0 }$ we have

$\displaystyle{ \frac{1}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}}d\gamma^{\prime}=f\left( 0\right) -\sum_{\left( C_{n}\right) }G_{i}\left( -\frac{1}{a_{i}}\right) . (6) }$

Subtracting Equation (5) from Equation (6) gives

$\displaystyle{ \frac{\gamma}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}\left( \gamma^{\prime}-\gamma\right) } d\gamma^{\prime}=f\left( \gamma\right) -f\left( 0\right) -\sum_{\left( C_{n}\right) }\left[ G_{i}\left( \frac{1}{\gamma-a_{i}}\right) -G_{i}\left( -\frac{1}{a_{i}}\right) \right] . }$

We now prove that the integrand on the LHS of this expression tends to zero as $\displaystyle{ n\rightarrow\infty }$. Since, $\displaystyle{ \left| \gamma^{\prime}\right| \geq\delta _{n},\,\,\left| \gamma^{\prime}-\gamma\right| \geq\left| \gamma^{\prime }\right| -\left| \gamma\right| \geq\delta_{n}-\left| \gamma\right| , }$ we have

$\displaystyle{ \begin{matrix} \left| \int_{C_{n}}\frac{f\left( \gamma^{\prime}\right) }{\gamma^{\prime }\left( \gamma^{\prime}-\gamma\right) }d\gamma^{\prime}\right| & \leq \frac{Ml_{n}}{\delta_{n}\left( \delta_{n}-\left| \gamma\right| \right) }\\ & \lt \frac{Mm}{\delta_{n}-\left| \gamma\right| }.\,\,\, (7) \end{matrix} }$

Since $\displaystyle{ \delta_{n}\rightarrow\infty }$ as $\displaystyle{ n\rightarrow\infty }$ and {\bf condition 4}, the integral in inequality (7) tends to zero as $\displaystyle{ n }$ increases.

Finally, we have formula for $\displaystyle{ f\left( \gamma\right) }$,

$\displaystyle{ f\left( \gamma\right) =f\left( 0\right) +\lim_{n\rightarrow\infty} \sum_{\left( C_{n}\right) }\left[ G_{i}\left( \frac{1}{\gamma-a_{i} }\right) -G_{i}\left( -\frac{1}{a_{i}}\right) \right] . }$

Since, the contour $\displaystyle{ C_{n} }$ will widen indefinitely as $\displaystyle{ n }$ increases, the second term is a sum over all poles, so we have $\displaystyle{ f\left( \gamma\right) }$ in the form of an infinite series

$\displaystyle{ f\left( \gamma\right) =f\left( 0\right) +\sum_{i=1}^{\infty}\left[ G_{i}\left( \frac{1}{\gamma-a_{i}}\right) -G_{i}\left( -\frac{1}{a_{i} }\right) \right] . }$

For the expansion formula of $\displaystyle{ \hat{w}\left( \gamma\right) }$, the polynomial term is

$\displaystyle{ G_{i}\left( \frac{1}{\gamma-q_{i}}\right) =\frac{R\left( q_{i}\right) }{\gamma-q_{i}}. }$

## Expansion of the Dispersion Relation for a Floating Elastic Plate

Now we show that the function

$\displaystyle{ \hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) } }$

where

$\displaystyle{ d\left( \gamma,\omega\right) =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H}, }$

satisfies the

conditions for the Mittag-Leffler expansion.

Define a sequence of square contours $\displaystyle{ C_{n} }$, square with its four corners at $\displaystyle{ \epsilon_{n}-\mathrm{i}\epsilon_{n} }$, $\displaystyle{ \epsilon_{n}+\mathrm{i} \epsilon_{n} }$, $\displaystyle{ -\epsilon_{n}+\mathrm{i}\epsilon_{n} }$ and $\displaystyle{ -\epsilon_{n}-\mathrm{i}\epsilon_{n} }$, where $\displaystyle{ \epsilon_{n}=\left( n+\frac{1}{2}\right) \pi/H,\,n=N,N+1,... }$. We start by showing that $\displaystyle{ \left| \hat{w}\left( \gamma\right) \right| }$ is bounded on any $\displaystyle{ C_{n} }$ in order to follow the proof of Mittag-Leffler expansion given in the previous subsection.

Before beginning we recall that $\displaystyle{ q_n }$ are the roots of the Dispersion Relation for a Floating Elastic Plate

$\displaystyle{ -q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \, }$

with $\displaystyle{ n=-1,-2 }$ corresponding to the complex solutions with positive imaginary part, $\displaystyle{ n=0 }$ corresponding to the imaginary solution with negative real part and $\displaystyle{ n\gt 0 }$ corresponding to the imaginary solutions with positive imagainary part.

For the sake of simplicity, write $\displaystyle{ u=1-m\omega^{2} }$. When $\displaystyle{ Im \gamma }$ is large the poles of $\displaystyle{ \hat{w} }$ are almost $\displaystyle{ \pm\mathrm{i} n\pi/H. }$ In fact, the poles $\displaystyle{ \left\{ \mathrm{i}q_{n}\right\} _{n=1,2,...} }$, $\displaystyle{ q_{n}\in\mathbb{R} }$ of $\displaystyle{ \hat{w} }$ satisfy

$\displaystyle{ \frac{1}{\left( q_{n}+u\right) q_{n}}=\tan\left( q_{n}H\right) , }$

so $\displaystyle{ \gamma_{n}\rightarrow\pm n\pi/H }$ as $\displaystyle{ n }$ increases. Thus, by choosing a large $\displaystyle{ N }$, the contour $\displaystyle{ C_{n} }$ is always a certain distance away from the poles for any $\displaystyle{ n\geq N }$. We prove the boundedness of $\displaystyle{ \left| \hat{w}\right| }$ by showing that $\displaystyle{ \left| \hat{w}\left( x+\mathrm{i}y\right) \right| }$ is bounded for $\displaystyle{ y=\pm\epsilon_{n} }$, $\displaystyle{ n=N,N+1,... }$, and $\displaystyle{ x,y\in\mathbb{R} }$, and then for $\displaystyle{ x=\pm\epsilon_{n},\,n=N,N+1,..., }$ $\displaystyle{ y\in\left[ -\epsilon_{n},\epsilon_{n}\right] }$.

For any $\displaystyle{ n\gt N }$ we have

$\displaystyle{ \left| \gamma^{4}+u\right| \gt \left| \gamma\right| ^{4}+C=\left| x+\mathrm{i}y\right| ^{4}+C }$
$\displaystyle{ \geq\epsilon_{n}^{4}+C\,\,\mathrm{for}\, \mathrm{any}\,\, x\in\mathbb{R},\,y=\epsilon _{n}, }$

where $\displaystyle{ C }$ is a constant determined by $\displaystyle{ u }$. When $\displaystyle{ y=\epsilon_{n} }$ we have

$\displaystyle{ \left| \frac{1}{\gamma\tanh\left( \gamma H\right) }\right| =\left| \frac{e^{2xH}e^{\mathrm{i}2yH}+1}{\left( x+\mathrm{i}y\right) \left( e^{2xH}e^{\mathrm{i}2yH}-1\right) }\right| }$
$\displaystyle{ =\frac{\left| e^{2xH}e^{\mathrm{i}2yH}+1\right| }{\left| x+\mathrm{i}y\right| \left| e^{2xH}e^{\mathrm{i}2yH}-1\right| } }$
$\displaystyle{ =\frac{\left| e^{2xH}-1\right| }{\left| x+\mathrm{i}y\right| \left| e^{2xH}+1\right| }\leq\frac{1}{\left| x+\mathrm{i}y\right| }\leq\frac{1}{\epsilon_{n}} }$

for any $\displaystyle{ x\in\mathbb{R} }$. (We used $\displaystyle{ \exp\left( \mathrm{i}\left( 2n+1\right) \pi\right) =-1 }$ and

$\displaystyle{ \left| \frac{e^{2xH}-1}{e^{2xH}+1}\right| \leq1 }$

to show this.) For large $\displaystyle{ \left| \gamma\right| }$ we have

$\displaystyle{ \left| \gamma^{4}+u-\frac{\omega^{2}}{\gamma\tanh\left( \gamma H\right) }\right| \geq\left| \gamma^{4}+u\right| -\left| \frac{\omega^{2}} {\gamma\tanh\left( \gamma H\right) }\right| . }$

Since the RHS of this inequality is positive from the previous equations,

$\displaystyle{ \left| \hat{w}\left( \gamma\right) \right| \leq\frac{1}{\left| \gamma ^{4}+u\right| -\left| \frac{\omega^{2}}{\gamma\tanh\left( \gamma H\right) }\right| }\leq\frac{1}{\epsilon_{n}^{4}+C-\frac{\omega^{2}}{\epsilon_{n}} } }$

for any $\displaystyle{ n\geq N }$. Note that the same relationship holds for $\displaystyle{ y=-\epsilon_{n} }$.

For $\displaystyle{ \gamma }$ on the line segment $\displaystyle{ \epsilon_{n}-\mathrm{i}\epsilon_{n} }$ to $\displaystyle{ \epsilon_{n}+\mathrm{i}\epsilon_{n} }$ we use the fact that

$\displaystyle{ \frac{\left| e^{2xH}e^{\mathrm{i}2yH}+1\right| }{\left| x+\mathrm{i}y\right| \left| e^{2xH}e^{\mathrm{i}2yH}-1\right| }\leq\frac{1}{\left| x+\mathrm{i}y\right| }\frac{1+\left| e^{-2xH}\right| }{1-\left| e^{-2xH}\right| }\leq\frac{E_{N}}{\epsilon_{N}} }$

for any $\displaystyle{ y }$, $\displaystyle{ n\geq N }$, where $\displaystyle{ E_{N} }$ is defined as

$\displaystyle{ \frac{1+\left| e^{-2xH}\right| }{1-\left| e^{-2xH}\right| }\leq \frac{1+\left| e^{-2\epsilon_{N}H}\right| }{1-\left| e^{-2\epsilon_{N} H}\right| }=E_{N}\,\,=, = \frac{1}{\left| x+\mathrm{i}y\right| }\leq\frac{1}{\epsilon_{N}}. }$

Therefore

$\displaystyle{ \frac{1}{\left| \gamma^{4}+u\right| -\left| \frac{\omega^{2}}{\gamma \tanh\left( \gamma H\right) }\right| }\leq\frac{1}{\epsilon_{N}^{4} +C-\frac{\omega^{2}E_{N}}{\epsilon_{N}}} }$

for any $\displaystyle{ n\geq N }$. The same proof can be applied for the line segment $\displaystyle{ -\epsilon_{n}-\mathrm{i}\epsilon_{n} }$ to $\displaystyle{ -\epsilon_{n} +\mathrm{i}\epsilon_{n} }$. We have proved that $\displaystyle{ \left| \hat{w}\left( \gamma\right) \right| }$ is bounded on all sides of the contours $\displaystyle{ C_{n},\,n\geq N }$ where $\displaystyle{ N }$ is chosen to be large so that the contours are a certain distance away from all the poles of $\displaystyle{ \hat{w} }$.

Hence, the expansion of $\displaystyle{ \hat{w}\left( \gamma\right) }$ becomes, from $\displaystyle{ \hat{w}\left( 0\right) =0 }$,

$\displaystyle{ \hat{w}\left( \gamma\right) =\sum_{n=-2}^{\infty}\left[ \frac{R\left( q_n\right) }{\gamma-q_n}+\frac{R\left( q_n\right) }{q_n}\right] =\sum_{n=-2}^{\infty} \left[ \frac{2q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}}+\frac{2R\left( q_n\right) }{q_n}\right] . }$

Note that the summation on the first line is over all poles of $\displaystyle{ \hat{w}\left( \gamma\right) }$. Note that $\displaystyle{ R\left( q\right) =-R\left( -q\right) }$, since $\displaystyle{ \hat{w}\left( \gamma\right) }$ is an even function and

$\displaystyle{ -\left( \gamma-q\right) \hat{w}\left( \gamma\right) =\left( -\gamma+q\right) \hat{w}\left( -\gamma\right) =\left( \gamma+q\right) \hat{w}\left( \gamma\right) . }$

Note that the term $\displaystyle{ \sum2R\left( q\right) /q }$ is zero. Indeed, expansion of the function $\displaystyle{ \hat{w}\left( \gamma\right) \gamma }$ which has the same analytic properties and poles as the function $\displaystyle{ \hat{w} }$ and residues $\displaystyle{ R\left( q\right) q }$ at $\displaystyle{ \gamma=q }$. Hence, $\displaystyle{ \hat{w}\left( \gamma\right) \gamma }$ is expanded as,

$\displaystyle{ \hat{w}\left( \gamma\right) \gamma=\sum_{n=-2}^{\infty}\left[ \frac{q_nR\left( q_n\right) }{\gamma-q_n}+\frac{q_nR\left( q_n\right) }{q_n}\right] = \sum_{n=-2}^{\infty} \frac{2\gamma q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}}. }$

The fact that $\displaystyle{ \sum2R\left( q\right) /q }$ is zero can also be confirmed by using the contour integration of the function $\displaystyle{ \hat{w}\left( \gamma\right) /\gamma }$. The function $\displaystyle{ \hat{w}\left( \gamma\right) /\gamma }$ is an odd function and has the same poles as the function $\displaystyle{ \hat{w}\left( \gamma\right) }$ with the residues $\displaystyle{ R\left( q\right) /q }$. Notice that $\displaystyle{ \gamma=0 }$ is not a singular point of $\displaystyle{ \hat{w}\left( \gamma\right) /\gamma }$. Hence, the integration over the real axis is zero and $\displaystyle{ \hat{w}\left( \gamma\right) /\gamma\rightarrow0 }$ on the semi-arc with order of $\displaystyle{ A^{-3} }$ as $\displaystyle{ A\rightarrow\infty }$.

The residues $\displaystyle{ R\left( q\right) }$ can be calculated using the usual formula. Since each of the poles of $\displaystyle{ \hat{w}\left( \gamma\right) }$ is simple, the residue $\displaystyle{ R\left( q\right) }$ at a pole $\displaystyle{ q }$ can be found using the expression

$\displaystyle{ \begin{matrix} R\left( q\right) & =\left[ \left. \frac{d}{d\gamma}d\left( \gamma ,\omega\right) \right| _{\gamma=q}\right] ^{-1}\\ & =\left[ 4q^{3}+\omega^{2}\left( \frac{qH+\tanh qH-qH\tanh^{2}qH} {q^{2}\tanh^{2}qH}\right) \right] ^{-1}. \end{matrix} }$

As each pole $\displaystyle{ q }$ is a root of the dispersion equation, we may substitute $\displaystyle{ \tanh qH=\omega^{2}/\left( q^{5}+uq\right) }$, where for brevity we have defined $\displaystyle{ u=\left( 1-m\omega^{2}\right) }$. The residue may then be given as the rational function of the pole

$\displaystyle{ R\left( q\right) =\frac{\omega^{2}q}{\omega^{2}\left( 5q^{4}+u\right) +H\left[ \left( q^{5}+uq\right) ^{2}-\omega^{4}\right] }. }$

This form avoids calculation of the hyperbolic tangent which becomes small at the imaginary roots. This causes numerical round-off problems since $\displaystyle{ q_{n}H }$ tends to $\displaystyle{ n\pi }$ as $\displaystyle{ n }$ becomes large, which makes $\displaystyle{ \tan q_{n}H }$ become small.