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== 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
<center><math>
\hat{w}\left(  \gamma\right)  =\frac{1}{d\left(  \gamma,\omega\right)  }
</math></center>
where
<center><math>
d\left(  \gamma,\omega\right)  =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2}
}{\gamma\tanh\gamma H},
</math></center>
is the [[Dispersion Relation for a Floating Elastic Plate]]
can be expressed by a linear sum of terms like <math>1/\left(  \gamma-a\right)  </math>,
<math>a</math> being a zero of <math>d\left(  \gamma,\omega\right)  </math>. 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 <math>\hat
{w}\left(  \gamma\right)  </math>.

We will show that 
<center><math>
\hat{w}\left(  \gamma\right)  =\sum_{n=-2}^{\infty}\frac{2q_nR\left(
q_n\right)  }{\gamma^{2}-q_n^{2}} 
</math></center>
where <math>R\left(  q_n\right)  </math> is the residue of <math>\hat{w}\left(  \gamma\right)  </math> at <math>\gamma=q_n</math>
given by
<center><math>
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]  }.
</math></center>
where <math>q_n</math> are the roots of the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math> 
-q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = 
-\omega^2 \cosh(kH)  \, 
</math></center>
with <math>n=-1,-2</math> corresponding to the complex solutions with positive imaginary part, 
<math>n=0</math> corresponding to the imaginary solution with negative real part and
<math>n>0</math> 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 <math>f\left(  \gamma\right)  </math> 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 <math>\gamma=c</math> then any small
circle with centre at <math>\gamma=c</math> 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 <math>a_{i}</math> we have
<center><math>
\left|  a_{1}\right|  \leq\left|  a_{2}\right|  \leq\left| a_3\right|
\leq...,
</math></center>
where <math>\left|  a_{i}\right|  \rightarrow\infty</math> as <math>i\rightarrow\infty</math>. At
every pole <math>\gamma=a_{i}</math> the function <math>f\left(  \gamma\right)  </math> will have a
definite infinite part, which will be a polynomial with respect to the
argument <math>1/\left(  \gamma-a_{i}\right)  </math> without the constant term. We
denote this polynomial term by
<center><math>
G_{i}\left(  \frac{1}{\gamma-a_{i}}\right)  ,\,\,i=1,2,3,...\,. 
</math></center>
We show that the fractional function <math>f\left(  \gamma\right)  </math> can be
represented by a simple infinite series of <math>G_{i}</math> by making certain
additional assumptions. Suppose that a sequence of closed contours <math>C_{n}</math>
which surround the origin exists and satisfies following conditions.

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

We now suppose that given such a sequence of contours, there exists a positive
number <math>M,</math> such that on any contour <math>C_{n}</math> our fractional function 
<math>f\left(\gamma\right)  </math> satisfies <math>\left|  f\left(  \gamma\right)  \right|  \leq M</math>.
Consider the integral
<center><math>
\frac{1}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left(  \gamma^{\prime
}\right)  }{\gamma^{\prime}-\gamma}d\gamma^{\prime}\,\,\, (1)
</math></center>
where the point <math>\gamma</math> lies inside <math>C_{n}</math> and is other than <math>a_{i}</math> (the
poles inside <math>C_{n}.</math>) We also consider the sum of the polynomials
for the poles <math>a_{i}</math>, inside <math>C_{n}</math>,
<center><math>
\omega_{n}\left(  \gamma\right)  =\sum_{\left(  C_{n}\right)  }G_{i}\left(
\frac{1}{\gamma-a_{i}}\right)  . \,\,\,(2)
</math></center>
The integrand of ((1)) has a pole <math>\gamma^{\prime}=\gamma</math> and poles
<math>\gamma^{\prime}=a_{i}</math>. We can calculate the residue at the pole
<math>\gamma^{\prime}=\gamma</math> by
<center><math>
\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)  .
</math></center>
The residues at the poles <math>\gamma^{\prime}=a_{i}</math> are, by the definition
(2), the same as the residues of the function
<center><math>
\frac{\omega_{n}\left(  \gamma^{\prime}\right)  }{\gamma^{\prime}-\gamma
}.\,\,\, (3)
</math></center>
We note that all poles of this function are situated inside <math>C_{n}</math>. We now
show that the sum of residues of function (3) at the poles <math>a_{i}</math>
is
<center><math>
-\omega_{n}\left(  \gamma\right)  =-\sum_{\left(  C_{n}\right)  }G_{i}\left(
\frac{1}{\gamma-a_{i}}\right)  .\,\,\, (4)
</math></center>
Since the definition of <math>\omega_{n}</math> and <math>G_{i}</math> is a polynomial of <math>1/\left(
\gamma-a_{i}\right)  ,</math> 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 <math>R</math>, we
have
<center><math>
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}.
</math></center>
The LHS of this does not change as the radius <math>R</math> increases, and the
RHS<math>\rightarrow0</math> as <math>R\rightarrow\infty</math>. Indeed,
<center><math>
\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}
</math></center>
and the term <math>\left|  \cdot\right|  </math> tends to zero as <math>R\rightarrow\infty</math>.
Thus, the sum of residues at poles within a finite distance is zero. Since we
know that the residue of (3) at <math>\gamma^{\prime}=\gamma</math> is
<math>\omega_{n}\left(  \gamma\right)  </math>, the sum of the rest is formula
(4). Thus, we have an expression for the integral (1),
<center><math>
\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)
</math></center>
Also, when <math>\gamma=0</math> we have
<center><math>
\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)
</math></center>
Subtracting Equation (5) from Equation (6) gives
<center><math>
\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]  .
</math></center>

We now prove that the integrand on the LHS of this expression tends to zero as
<math>n\rightarrow\infty</math>. Since, <math>\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|  ,</math> we
have
<center><math>\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)
}\\
& <\frac{Mm}{\delta_{n}-\left|  \gamma\right|  }.\,\,\, (7)
\end{matrix}</math></center>
Since <math>\delta_{n}\rightarrow\infty</math> as <math>n\rightarrow\infty</math> and
{\bf condition 4}, the integral in inequality (7) tends to zero
as <math>n</math> increases.

Finally, we have formula for <math>f\left(  \gamma\right)  </math>,
<center><math>
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]  .
</math></center>
Since, the contour <math>C_{n}</math> will widen indefinitely as <math>n</math> increases, the
second term is a sum over all poles, so we have <math>f\left(  \gamma\right)  </math> in
the form of an infinite series
<center><math>
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]  .
</math></center>

For the expansion formula of <math>\hat{w}\left(  \gamma\right)  </math>, the polynomial
term is
<center><math>
G_{i}\left(  \frac{1}{\gamma-q_{i}}\right)  =\frac{R\left(  q_{i}\right)
}{\gamma-q_{i}}.
</math></center>

==Expansion of the [[Dispersion Relation for a Floating Elastic Plate]]==

Now we show that the function
<center><math>
\hat{w}\left(  \gamma\right)  =\frac{1}{d\left(  \gamma,\omega\right)  }
</math></center>
where
<center><math>
d\left(  \gamma,\omega\right)  =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2}
}{\gamma\tanh\gamma H},
</math></center>satisfies the
conditions for the Mittag-Leffler expansion.

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

Before beginning we recall that
<math>q_n</math> are the roots of the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math> 
-q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = 
-\omega^2 \cosh(kH)  \, 
</math></center>
with <math>n=-1,-2</math> corresponding to the complex solutions with positive imaginary part, 
<math>n=0</math> corresponding to the imaginary solution with negative real part and
<math>n>0</math> corresponding to the imaginary solutions with positive imagainary part.

For the sake of simplicity, write <math>u=1-m\omega^{2}</math>. When <math>Im 
\gamma</math> is large the poles of <math>\hat{w}</math> are almost <math>\pm\mathrm{i}
n\pi/H.</math> In fact, the poles <math>\left\{  \mathrm{i}q_{n}\right\}
_{n=1,2,...}</math>, <math>q_{n}\in\mathbb{R}</math> of <math>\hat{w}</math> satisfy
<center><math>
\frac{1}{\left(  q_{n}+u\right)  q_{n}}=\tan\left(  q_{n}H\right)  ,
</math></center>
so <math>\gamma_{n}\rightarrow\pm n\pi/H</math> as <math>n</math> increases. Thus, by choosing a
large <math>N</math>, the contour <math>C_{n}</math> is always a certain distance away from the
poles for any <math>n\geq N</math>. We prove the boundedness of <math>\left|  \hat{w}\right|
</math> by showing that <math>\left|  \hat{w}\left(  x+\mathrm{i}y\right)
\right|  </math> is bounded for <math>y=\pm\epsilon_{n}</math>, <math>n=N,N+1,...</math>, and
<math>x,y\in\mathbb{R}</math>, and then for <math>x=\pm\epsilon_{n},\,n=N,N+1,...,</math>
<math>y\in\left[  -\epsilon_{n},\epsilon_{n}\right]  </math>.

For any <math>n>N</math> we have
<center><math>
\left|  \gamma^{4}+u\right|    >\left|  \gamma\right|  ^{4}+C=\left|
x+\mathrm{i}y\right|  ^{4}+C
</math></center>
<center><math>
\geq\epsilon_{n}^{4}+C\,\,\mathrm{for}\, \mathrm{any}\,\, x\in\mathbb{R},\,y=\epsilon
_{n}, 
</math></center>
where <math>C</math> is a constant determined by <math>u</math>. When <math>y=\epsilon_{n}</math> we have
<center><math>
\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| 
</math></center>
<center><math>
=\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|
}
</math></center>
<center><math>
 =\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}} 
</math></center>
for any <math>x\in\mathbb{R}</math>. (We used <math>\exp\left(  \mathrm{i}\left(
2n+1\right)  \pi\right)  =-1</math> and
<center><math>
\left|  \frac{e^{2xH}-1}{e^{2xH}+1}\right|  \leq1
</math></center>
to show this.) For large <math>\left|  \gamma\right|  </math> we have
<center><math>
\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|  .
</math></center>
Since the RHS of this inequality is positive from the previous equations,
<center><math>
\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}}
} 
</math></center>
for any <math>n\geq N</math>. Note that the same relationship holds for <math>y=-\epsilon_{n}</math>.

For <math>\gamma</math> on the line segment <math>\epsilon_{n}-\mathrm{i}\epsilon_{n}</math>
to <math>\epsilon_{n}+\mathrm{i}\epsilon_{n}</math> we use the fact that
<center><math>
\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}}
</math></center>
for any <math>y</math>, <math>n\geq N</math>, where <math>E_{N}</math> is defined as
<center><math>
\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}}.
</math></center>
Therefore
<center><math>
\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}}}
</math></center>
for any <math>n\geq N</math>. The same proof can be applied for the line segment
<math>-\epsilon_{n}-\mathrm{i}\epsilon_{n}</math> to <math>-\epsilon_{n}
+\mathrm{i}\epsilon_{n}</math>. We have proved that <math>\left|  \hat{w}\left(
\gamma\right)  \right|  </math> is bounded on all sides of the contours
<math>C_{n},\,n\geq N</math> where <math>N</math> is chosen to be large so that the contours are a
certain distance away from all the poles of <math>\hat{w}</math>.

Hence, the expansion of <math>\hat{w}\left(  \gamma\right)  </math> becomes, from
<math>\hat{w}\left(  0\right)  =0</math>,
<center><math>
\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]  . 
</math></center>
Note that the summation on the first line is over all poles of <math>\hat{w}\left(
\gamma\right)  </math>. Note that <math>R\left(  q\right)  =-R\left(  -q\right)  </math>, since
<math>\hat{w}\left(  \gamma\right)  </math> is an even function and
<center><math>
-\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)  .
</math></center>
Note that the term <math>\sum2R\left(  q\right)  /q</math> is zero. Indeed, expansion of
the function <math>\hat{w}\left(  \gamma\right)  \gamma</math> which has the same
analytic properties and poles as the function <math>\hat{w}</math> and residues <math>R\left(
q\right)  q</math> at <math>\gamma=q</math>. Hence, <math>\hat{w}\left(  \gamma\right)  \gamma</math> is
expanded as,
<center><math>
\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}}.
</math></center>

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

The residues <math>R\left(  q\right)  </math> can be calculated using the usual formula.
Since each of the poles of <math>\hat{w}\left(  \gamma\right)  </math> is simple, the
residue <math>R\left(  q\right)  </math> at a pole <math>q</math> can be found using the expression
<center><math>\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}</math></center>
As each pole <math>q</math> is a root of the dispersion equation, we may substitute
<math>\tanh qH=\omega^{2}/\left(  q^{5}+uq\right)  </math>, where for brevity we have
defined <math>u=\left(  1-m\omega^{2}\right)  </math>. The residue may then be given as
the rational function of the pole
<center><math>
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]  }. 
</math></center>
This form avoids calculation of the hyperbolic tangent which becomes small at the imaginary roots. This causes numerical round-off problems since <math>q_{n}H</math> tends to <math>n\pi</math> as <math>n</math> becomes large, which makes <math>\tan q_{n}H</math> become small. 

[[Category:Floating Elastic Plate]]