# An extension of the orthogonal derivative with adjustable precision

Enno Diekema 111email adress<EMAIL_ADDRESS>
###### Abstract

The orthogonal derivative is defined as a limit of an integral whose kernel
contains an orthogonal polynomial with its measure. When in practice no limit
is taken, it means that the accuracy of the derivative depends on the second
derivative of the given function. Liptaj shows that it is possible to define a
kernel in such a way that the accuracy depends on a higher derivative at your
own choice. The accuracy is therefore much greater than with the orthogonal
derivative. However Diekema and Koornwinder find a similar extension starting
directly from of the orthogonal derivative. The new kernel is not orthogonal
for order greater then one. The transfer function for this new derivative is
given.

##  1 Introduction

In many applications one needs to approximate the first or higher order
derivative of a function which is perturbed by noise. A good candidate for an
approximation of the first derivative $f^{\prime}(x)$ of a continuous function
$f(x)$ is given by

$f^{\prime}(x)\approx-\dfrac{3}{2\delta}\int_{-1}^{1}f(x+\delta\xi)\xi d\xi$

for small $\delta$. So part of the noise will be averaged out, because the
integral has an averaging effect. The history of this formula is described in
[2]. One can recognise that the function $\xi$ is the first Legendre
polynomial. In the paper [2] this idea is generalized to the so-called
orthogonal derivative. For the first order the derivative is called the
Lanczos’ derivative.

In this paper a modification of the orthogonal derivative is described. This
modification has the great advantage that the accuracy of the approximate
derivative can be set by means of an extra parameter. There are two methods to
go about this modification. The first method is based on an idea in some
papers of Liptaj [8, 9]. This method is based on the basic idea of integration
by differentiation, which originated from the method of the least squares. The
second method uses a formula from [2, Section 4]. This formula originates from
the use of orthogonal polynomials, which is also essentially based on the use
of the least squares method. Both methods derive a formula for the kernel of
an integral. The first method is more complicated than the second method.

The content of this paper is as follows. The orthogonal derivative is
discussed in section 2. This is a part of the first chapter from the author’s
thesis [3]. Section 3 derives the general formula for the kernel of the
approximate derivative of arbitrary order following the basis idea of Liptaj.
The result is a formula with a summation of a hypergeometric function. The
final polynomial is not orthogonal for the orders greater than one. So the
derivative cannot be an orthogonal derivative. In section 4 some special cases
for the basic kernel are described. In section 5 the second method using a
formula from [2, Section 4] is treated. Section 6 defines the general kernel
formula for the first derivative. It turns out that this kernel is an
orthogonal polynomial. The known properties of this polynomial are indicated.
The formula for the first derivative can be mentioned as an orthogonal
derivative or Lanczos’ derivative. In section 7 the transfer function of the
first derivative is determined. The transfer function is illustrated by a log-
log plot. A summary and conclusion is presented in section 8. In the
appendices an integral of a Gegenbauer polynomial is derived and an overview
is given for the basic kernel function for different values of the order and
the degree.

##  2 The orthogonal derivative

A well-known problem in signal analysis is to detect the signal that was
disturbed by noise. In his practice of signal analysis the author met the
problem of differentiating a signal that was disturbed by noise. Because
differentiating a signal will amplify the higher frequencies in the signal
more than the lower frequencies, the result of differentiating a signal with
noise becomes unstable. Therefore, an integrating factor must be added. Every
averaging has an integrating effect. The author choose in his work for the
method of the least square. (see Figure 1).

Figure 1: The definition of the least square derivative.

We approximate the function $f(x)$ by a linear function $g(x)$ with the
property that

$\int_{x}^{x+h}\big{[}f(u)-g(u)\big{]}^{2}du$ (2.1)

is minimal. This is the so-called method of the least squares. From Figure 1
it is seen that the noise will be averaged out. We take $g^{\prime}(x)$ as an
approximation of $f^{\prime}(x)$. With $g(u)=a+bu$ we have to minimize

$\int_{0}^{1}\big{[}f(x+uh)-a-b(x+uh)\big{]}^{2}du$ (2.2)

as a function of $a$ and $b$. This can be handled by putting the derivatives
of (2.2) with respect to $a$ and $b$ equal to zero and solving the resulting
system of linear equations in $a$ and $b$. The solution for $b$ is

$b=\dfrac{6}{h}\int_{0}^{1}f(x+hu)(2u-1)du.$

Then take $b$ as the approximation of $f^{\prime}(x)$. If we want to minimize
(2.1) for $g$ a quadratic function then we have to solve a system of three
linear equations. More generally, in order to minimize (2.1) for $g$ a
polynomial of degree $n$ we have to solve a system of $n+1$ linear equations.
When doing these computations for some successive low values of $n$ one will
see a structure arising in the expression for the approximation of the $n$-th
derivative. It involves an integral for which the integrand contains a so-
called shifted Legendre polynomial $P_{n}^{*}(x)$. Such polynomials are
orthogonal polynomials on the interval $[0,1]$ with respect to a constant
weight function. The resulting formula for the $n$-th derivative is

$\dfrac{d^{n}}{dx^{n}}f(x)=\dfrac{\Gamma(2n+2)}{\Gamma(n+1)}\lim\limits_{h\downarrow
0}\dfrac{1}{h^{n}}\int\nolimits_{x}^{x+h}f(u)P_{n}^{*}\left(\dfrac{u-x}{h}\right)du.$

Of course, this formula, obtained for low $n$ by tedious computations, cannot
be an accident. Indeed, a simple argument which involves the Taylor series
expansion of $f(u+h)$ up to the term with $h^{n}$ and some very basic
properties of orthogonal polynomials, gives a proof of the formula for general
$n$. Moreover, the proof of this formula does not depend on the special choice
of interval and weight function (or measure). We may start with minimizing

$\int_{\mathbb{R}}\big{[}f(x+hu)-g(u)\big{]}^{2}d\mu(u)$ (2.3)

for $g$ a polynomial of degree $\leq n$. $\mu$ is a positive measure on
$\mathbb{R}$, for instance $d\mu(u)=w(u)du$ on some interval and $=0$ outside
that interval, where $w$ is a positive weight function. Then it can be shown
[3, Ch. 2.3], [2] that

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim\limits_{h\downarrow
0}\dfrac{k_{n}n!}{h_{n}}\dfrac{1}{h^{n}}\int_{\mathbb{R}}f(x+hu)p_{n}(u)d\mu(u)$
(2.4)

where the polynomials $p_{n}$ are orthogonal polynomials with respect to the
measure $\mu$ i.e, $p_{n}$ is a polynomial of degree $n$ with coefficient of
highest degree term equal to $k_{n}$, and

$\int_{\mathbb{R}}p_{m}(x)p_{n}(x)d\mu(x)=h_{n}\delta_{m,n}.$

We call the right-hand side of (2.4), if the limit exists, the orthogonal
derivative of order $n$ of $f$ at $x$. This limit certainly exists if $f$ is
$n$ times differentiable at $x$, and then the orthogonal derivative equals the
ordinary derivative, but the limit exists much more generally, for instance,
it is equal to the $n$-th order Peano derivative of $f$ at $x$, if this
exists. The following items are some examples of orthogonal derivatives with
integration interval $[-1,1]$.

  * •

Legendre derivative

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim\limits_{h\downarrow
0}\dfrac{k_{n}n!}{h_{n}}\dfrac{1}{h^{n}}\int_{-1}^{1}P_{n}(u)f(x+hu)du.$ (2.5)

$P_{n}(x)$ are the Legendre polynomials with weight function $1$ on the
interval $[-1,1]$.

  * •

Gegenbauer derivative

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim\limits_{h\downarrow
0}\dfrac{k_{n}n!}{h_{n}}\dfrac{1}{h^{n}}\int_{-1}^{1}(1-u^{2})^{\alpha-1/2}C^{(\alpha)}_{n}(u)f(x+hu)du.$

$C^{(\alpha)}_{n}(x)$ are the Gegenbauer polynomials with weight function
$(1-x^{2})^{\alpha-1/2}$ on the interval $[-1,1]$.

  * •

Jacobi derivative

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim\limits_{h\downarrow
0}\dfrac{k_{n}n!}{h_{n}}\dfrac{1}{h^{n}}\int_{-1}^{1}(1-u)^{\alpha}(1+u)^{\beta}P^{(\alpha,\beta)}_{n}(u)f(x+hu)du.$

$P^{(\alpha,\beta)}_{n}(x)$ are the Jacobi polynomials with weight function
$(1-x)^{\alpha}(1+x)^{\beta}$ on the interval $[-1,1]$.

As an approximation of the orthogonal derivative we define the approximate
orthogonal derivative where we omit the limit for $h\downarrow 0$. This has
the property that the noise of a signal will be suppressed. We can illustrate
this property by working in the frequency domain. We look at the approximate
orthogonal derivative as a filter. This filter can be characterized by its
transfer function in the frequency domain.

A filter produces from an input signal an output signal. It has the property
that the quotient of the Fourier transform of the output signal and the
Fourier transform of the input signal is independent of the input. This
quotient, denoted by $H(\omega)$ with $\omega$ the frequency, is called the
transfer function of the filter. It is usually a complex-valued function of
the frequency. Both its modulus and its argument are of interest. In this
paper we use the modulus of the transfer function.

For a filter giving the $n$-th order ordinary derivative of the input we have
$H(\omega)=(i\omega)^{n}$, so for the modulus of the transfer function we can
draw the following log-log plots of Figure 2.

Figure 2: Moduli of the transfer functions of a differentiator of orders
$1$,$2$ and $5$.

We see that for high frequencies the modulus of the transfer function goes to
infinity. So the high frequencies will be stronger amplified than the lower
frequencies. This high frequency noise will disturb the derivative of the
original signal very much.

However, if we use for $n=1$ a filter corresponding to the approximation of
the first derivative implied by the right-hand side of (2.4) with $d\mu(u)=du$
on $[-1,1]$ (involving the first degree Legendre polynomial $P_{1}(x)=x$) then
we have a filter sending input function $f$ to output function $g$ by

$g(x)=\dfrac{3}{2h}\int\nolimits_{-1}^{1}f(x+hu)u\ du.$

The modulus of the corresponding transfer function is drawn in the log-log
plot of Figure 3.

Figure 3: Modulus of the transfer function of a first order Legendre
differentiator.

We see that $|H(\omega)|\approx\omega$ for lower frequencies but that it has
successive oscillations decreasing to zero for higher frequencies. Therefore,
higher frequency parts of the signal are suppressed by the filter.

##  3 First method for an alternative for the orthogonal derivative

Liptaj in a number of papers [8],[9] developed a method for the derivative by
integration of a function on an bounded interval with greater accuracy then
the approximate orthogonal derivative.

Looking for example to formula (2.5) we see that when expanding the function
$f(x+h\ t)$ into a series of powers of $h$, the first term we may neglect is
of order $n+1$ for a derivative of order $n$. Liptaj found that it is possible
to choose a function $k(t)$ in such a manner that the first term to neglect is
of higher order than $n+1$. He defined the following functions:

$k(t)=\dfrac{d^{n}}{dt^{n}}\omega(t).$
$\omega(t)=K(1-t^{2})^{n}\sum^{m}_{k=0}a_{2k}\,t^{2k}.$

He choose $n$ for the order of the derivative. His formula for the $n$-th
order derivative becomes

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim_{h\rightarrow
0}\left(-\dfrac{1}{h}\right)^{n}\int_{-1}^{1}k(t)f(x+h\,t)dt.$ (3.1)

We illustrate his method by the following example. We want to find the formula
for the third order derivative of the function $f(x)=x^{9}$. The exact answer
is known: $\dfrac{d^{3}}{dx^{3}}x^{9}=7*8*9*x^{6}$. For the given function we
use: $f(x+h\,t)=(x+h\,t)^{9}$. Theoretically, $h$ should go to zero. In
practice, an approximation is used. For the approximate derivative, the
following formula is used:

$k(t)=K\dfrac{d^{3}}{dt^{3}}\Big{(}(1-t^{2})^{3}(1+a_{2}t^{2}+a_{4}t^{4}+a_{6}t^{6})\Big{)}.$

It turns out that with this kernel, the first term to neglect is of order
$h^{8}$. We can use more or less terms, but there are only terms with even
power. Using (3.1) gives

$\dfrac{d^{3}}{dx^{3}}(x+h\,t)^{9}=K\lim_{h\rightarrow
0}\left(-\dfrac{1}{h}\right)^{n}\int_{-1}^{1}\dfrac{d^{3}}{dt^{3}}\Big{(}(1-t^{2})^{3}(1+a_{2}t^{2}+a_{4}t^{4}+a_{6}t^{6})\Big{)}(x+h\,t)^{9}dt.$
(3.2)

Omitting the limit and working out the integral we get

$\displaystyle\dfrac{d^{3}}{dx^{3}}(x+h\,t)^{9}$
$\displaystyle=K\dfrac{256}{230945}\big{(}4845+2261a_{2}+1197a_{4}+693a_{6}\big{)}h^{6}+$
$\displaystyle+K\dfrac{768}{2431}\big{(}663+255a_{2}+119a_{4}+63a_{6}\big{)}h^{4}x^{2}+$
$\displaystyle+K\dfrac{768}{143}\big{(}143+39a_{2}+15a_{4}+7a_{6}\big{)}h^{2}x^{4}+$
$\displaystyle+K\dfrac{256}{715}\big{(}1287+143a_{2}+39a_{4}+15a_{6}\big{)}x^{6}.$

If the first three terms are equal to zero and the factor for the last term is
equal to $7*8*9$, then the parameters $a_{2},\,a_{4},\,a_{6}$ and $K$ can be
determined. In that case, the inaccuracy of the derivative is proportional to
$h^{8}$. The equations are:

$\displaystyle 4845+2261a_{2}+1197a_{4}+693a_{6}=0.$ $\displaystyle
663+255a_{2}+119a_{4}+63a_{6}=0.$ $\displaystyle
143+39a_{2}+15a_{4}+7a_{6}=0.$ $\displaystyle
K\big{(}1287+143a_{2}+39a_{4}+15a_{6}\big{)}=7*8*9.$

It is evident that if we choose more unknown $a_{2k}$ we get more equations to
solve. The solution of this system of equations is:

$a_{2}=-15\qquad a_{4}=51\qquad a_{6}=-\dfrac{323}{7}\qquad
K=-\dfrac{105105}{32768}.$

Substitution in $\omega(t)$ gives

$\omega(t)=\dfrac{105105}{32768}(1-t^{2})^{3}\left(1-15t^{2}+51t^{4}-\dfrac{323}{7}t^{6}\right).$

Taking the third derivative results in

$k(t)=\dfrac{45045}{4096}\big{(}17765t^{9}-39780t^{7}+30030t^{5}-8580t^{3}+693t\big{)}.$

We can use this function in formula (3.1) and get the approximate third
derivative.

For each order of the derivative (which are up to order 4) and the number of
terms $k$ (up to $k=5$) Liptay in his paper [9] gives a method for deriving
the equations. For each case these equations should be solved. He does not
give a general solution for the function $k(t)$.

In the rest of this section a general formula for the function $k(t)$ is
derived. It is assumed that the given function is analytical in the given
domain. We call $k(t)$ the basic kernel function. Solving $m$ equations with
$m$ unknowns is then avoided. We start with

$\displaystyle\dfrac{d^{n}}{dx^{n}}f(x)$ $\displaystyle=\lim_{h\rightarrow
0}\left(-\dfrac{1}{h}\right)^{n}\int_{-1}^{1}f(x+h\,t)\dfrac{d^{n}}{dt^{n}}\left((1-t^{2})^{n}K\sum_{k=0}^{m}a_{2k}t^{2k}\right)dt$
$\displaystyle=\lim_{h\rightarrow
0}\left(-\dfrac{1}{h}\right)^{n}\int_{-1}^{1}f(x+h\,t)K\sum_{k=0}^{m}a_{2k}\dfrac{d^{n}}{dt^{n}}\left((1-t^{2})^{n}t^{2k}\right)dt.$
(3.3)

So we have $m+1$ terms in the summation. See (3.2). Because $a_{0}=1$ there
are $m$ unknown parameters to detect. The derivative can be taken by means of
Leibnitz’s rule.

$\dfrac{d^{n}}{dt^{n}}\big{(}f(t)g(t)\big{)}=\sum_{j=0}^{n}\binom{n}{j}D^{n-j}[f(t)]D^{j}[g(t)].$

Application of [7, 8.934(1)] gives

$D^{j}\big{[}(1-t^{2})^{n}\big{]}=(-2)^{j}\
j!\dfrac{(-n)_{j}}{(j-2n)_{j}}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t).$

$C^{(n-j+1/2)}_{j}(t)$ are the Gegenbauer polynomials. $(a)_{j}$ is the well-
known Pochhammer symbol.

$D^{n-j}\big{[}t^{2k}\big{]}=\dfrac{\Gamma(2k+1)}{\Gamma(2k+1-n+j)}t^{2k-n+j}.$

Substitution in (3.3) gives

$\displaystyle k(t)$
$\displaystyle=K\sum_{k=0}^{m}a_{2k}\dfrac{d^{n}}{dt^{n}}\left((1-t^{2})^{n}t^{2k}\right)$
$\displaystyle=K\sum_{k=0}^{m}a_{2k}\dfrac{\Gamma(2k+1)}{\Gamma(2k+1-n)}\sum_{j=0}^{n}\binom{n}{j}\dfrac{(-n)_{j}(-2)^{j}\
j!}{(2k+1-n)_{j}(j-2n)_{j}}t^{2k-n+j}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t).$
(3.4)

To find formulae for $a_{2k}$ and $K$ we can write the function $f(x+h\,t)$ as
a Taylor series. Then there arise all terms with $(x+h\,t)^{\lambda}$. We only
have to take the terms with $\lambda=2m+n$. Because the function $k(t)$
contains only terms with $\lambda\leq 2m+n$ we have only use the factor
$(x+h\,t)^{2m+n}$. The factor $(x+h\,t)^{2m+n}$ can be written as

$(x+h\,t)^{m+n}=\sum^{2m+n}_{i=0}\binom{2m+n}{i}x^{2m+n-i}(h\,t)^{i}.$

To compute the value of $a_{2k}$ it is sufficient to take only the term with
$x^{2k+n-i}$. Then we define with (3.1) and (3.4) the function $A(x)$ as

$A(x)=(-1)^{n}K\sum^{2m+n}_{i=0}\binom{2m+n}{i}\sum_{k=0}^{m}a_{2k}\dfrac{\Gamma(2k+1)}{\Gamma(2k+1-n)}\\\
\sum_{j=0}^{n}\binom{n}{j}\dfrac{(-n)_{j}(-2)^{j}\
j!}{(2k+1-n)_{j}(j-2n)_{j}}x^{2k+n-i}h^{i-n}\int_{-1}^{1}t^{i+2k-n+j}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t)dt.$
(3.5)

The integral is derived in Appendix A. Using the result and after some
manipulations with the Pochhammer symbols and the Gamma functions we obtain

$A(x)=(-1)^{n}K\dfrac{\Gamma(n+1)}{2}\sum^{2m+n}_{i=0}\binom{2m+n}{i}\big{[}1+(-1)^{i-n}\big{]}\sum^{m}_{k=0}a_{2k}x^{2k+n-i}h^{i-n}\\\
\dfrac{\Gamma\left(\dfrac{i+2k-n+1}{2}\right)}{\Gamma\left(\dfrac{i+2k+n+3}{2}\right)}\dfrac{\Gamma(2k+1)}{\Gamma(2k+1-n)}\sum^{n}_{j=0}\dfrac{(-n)_{j}(i+2k-n+1)_{j}}{(2k+1-n)_{j}}\dfrac{1}{j!}.$

The last summation is well-known.

$\,\mbox{}_{2}F_{1}\\!\left(\genfrac{}{}{0.0pt}{}{-n,b}{c};1\right)=\dfrac{(c-b)_{n}}{(c)_{n}}.$

Application gives

$A(x)=(-1)^{n}K\dfrac{\Gamma(n+1)}{2}\sum^{2m+n}_{i=0}\binom{2m+n}{i}\big{[}1+(-1)^{i-n}\big{]}\sum^{m}_{k=0}a_{2k}x^{2k+n-i}h^{i-n}\\\
\dfrac{\Gamma\left(\dfrac{i+2k-n+1}{2}\right)}{\Gamma\left(\dfrac{i+2k+n+3}{2}\right)}(-i)_{n}.$

For the last Pochhammer symbol we use

$(-i)_{n}=(-1)^{n}\dfrac{\Gamma(i+1)}{\Gamma(i-n+1)}.$

This factor gives only a contribution if $i\geq n$. Application gives

$A(x)=K\dfrac{\Gamma(n+1)}{2}\sum^{2m+n}_{i=n}\binom{2m+n}{i}\big{[}1+(-1)^{i-n}\big{]}\sum^{m}_{k=0}a_{2k}x^{2k+n-i}h^{i-n}\\\
\dfrac{\Gamma\left(\dfrac{i+2k-n+1}{2}\right)}{\Gamma\left(\dfrac{i+2k+n+3}{2}\right)}\dfrac{\Gamma(i+1)}{\Gamma(i-n+1)}.$

Replacing $i$ with $i+n$ and interchanging the summations (which is allowed
because all summations are bounded) gives

$A(x)=K\dfrac{\Gamma(n+1)}{2}\dfrac{\Gamma(2m+n+1)}{\Gamma(2m+1)}\sum^{m}_{k=0}a_{2k}x^{2k}\dfrac{\Gamma\left(\dfrac{1}{2}+k\right)}{\Gamma\left(\dfrac{3}{2}+n+k\right)}\\\
\sum^{m}_{i=0}\big{[}1+(-1)^{i}\big{]}\dfrac{(-2m)_{i}\left(\dfrac{1}{2}+k\right)_{i/2}}{\left(\dfrac{3}{2}+n+k\right)_{i/2}}\dfrac{1}{i!}\left(\dfrac{h}{x}\right)^{i}.$

The second summation gives a hypergeometric function

$\sum^{m}_{i=0}\big{[}1+(-1)^{i}\big{]}\dfrac{(-2m)_{i}\left(\dfrac{1}{2}+k\right)_{i/2}}{\left(\dfrac{3}{2}+n+k\right)_{i/2}}\dfrac{1}{i!}\left(\dfrac{h}{x}\right)^{i}=2\,\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-m,\dfrac{1}{2}-m,\dfrac{1}{2}+k}{\dfrac{1}{2},\dfrac{3}{2}+n+k};\dfrac{h^{2}}{x^{2}}\right).$

For $A(x)$ there rests

$A(x)=K\Gamma(n+1)\dfrac{\Gamma(2m+n+1)}{\Gamma(2m+1)}\sum^{m}_{k=0}a_{2k}x^{2k}\dfrac{\Gamma\left(\dfrac{1}{2}+k\right)}{\Gamma\left(\dfrac{3}{2}+n+k\right)}\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-m,\dfrac{1}{2}-m,\dfrac{1}{2}+k}{\dfrac{1}{2},\dfrac{3}{2}+n+k};\dfrac{h^{2}}{x^{2}}\right).$

This is a polynomial in $(x\,h)^{2}$. The factors before the powers of
$(x\,h)^{2}$ should be zero. So

$\displaystyle S$
$\displaystyle=\sum^{m}_{k=0}a_{2k}\dfrac{\Gamma\left(\dfrac{1}{2}+k\right)}{\Gamma\left(\dfrac{3}{2}+n+k\right)}\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-m,\dfrac{1}{2}-m,\dfrac{1}{2}+k}{\dfrac{1}{2},\dfrac{3}{2}+n+k};\dfrac{h^{2}}{x^{2}}\right)$
$\displaystyle=\sum^{m}_{k=0}a_{2k}\dfrac{\Gamma\left(\dfrac{1}{2}+k\right)}{\Gamma\left(\dfrac{3}{2}+n+k\right)}\sum_{i=0}^{m}\dfrac{(-m)_{i}\left(\dfrac{1}{2}-m\right)_{i}\left(\dfrac{1}{2}+k\right)_{i}}{\left(\dfrac{1}{2}\right)_{i}\left(\dfrac{3}{2}+n+k\right)_{i}}\dfrac{1}{i!}$
$\displaystyle=\dfrac{\Gamma\left(\dfrac{1}{2}\right)}{\Gamma\left(\dfrac{3}{2}+n\right)}\sum_{i=0}^{m}\dfrac{(-m)_{i}\left(\dfrac{1}{2}-m\right)_{i}}{\left(\dfrac{3}{2}+n\right)_{i}}\dfrac{1}{i!}\sum_{k=0}^{m}a_{2k}\dfrac{\left(\dfrac{1}{2}+i\right)_{k}}{\left(\dfrac{3}{2}+n+i\right)_{k}}=0$

with $a_{0}=1$. This gives $m$ equations to compute the $a_{2k}$. But it is
easy to check that the solution of these equations is

$a_{2k}=\dfrac{(-m)_{k}\left(m+n+\dfrac{3}{2}\right)_{k}}{\left(\dfrac{3}{2}\right)_{k}k!}.$
(3.6)

Using this formula for $\omega(t)$ together with [7, 8.932.3] gives the
polynomial

$\displaystyle\omega(t)=K(1-t^{2})^{n}\sum_{k=0}^{m}a_{2k}t^{2k}$
$\displaystyle=K(1-t^{2})^{n}\,\mbox{}_{2}F_{1}\\!\left(\genfrac{}{}{0.0pt}{}{-m,m+n+\dfrac{3}{2}}{\dfrac{3}{2}};t^{2}\right)$
$\displaystyle=-K(1-t^{2})^{n}\dfrac{\Gamma\left(-m-n-\dfrac{1}{2}\right)\Gamma(m+1)}{2\Gamma\left(\dfrac{1}{2}-n\right)}\dfrac{1}{t}C^{n+1/2}_{2m+1}(t).$
(3.7)

To compute the factor $K$ we set

$K\Gamma(n+1)\sum_{k=0}^{m}a_{2k}\dfrac{\Gamma\left(\dfrac{1}{2}+k\right)}{\Gamma\left(\dfrac{3}{2}+n+k\right)}=1.$

Substitution of $a_{2k}$ gives

$\displaystyle
K\Gamma(n+1)\sum_{k=0}^{m}a_{2k}\dfrac{\Gamma\left(\dfrac{1}{2}+k\right)}{\Gamma\left(\dfrac{3}{2}+n+k\right)}$
$\displaystyle=K\Gamma(n+1)\sum_{k=0}^{m}\dfrac{(-m)_{k}\left(m+n+\dfrac{3}{2}\right)_{k}}{\left(\dfrac{3}{2}\right)_{k}k!}\dfrac{\Gamma\left(\dfrac{1}{2}+k\right)}{\Gamma\left(\dfrac{3}{2}+n+k\right)}=$
$\displaystyle=K\dfrac{\Gamma\left(\dfrac{1}{2}\right)\Gamma(n+1)}{\Gamma\left(\dfrac{3}{2}+n\right)}\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-m,m+n+\dfrac{3}{2},\dfrac{1}{2}}{n+\dfrac{3}{2},\dfrac{3}{2}};1\right).$

The hypergeometric function is 1-balanced. So the summation is well-known.
After lots of manipulation with the Pochhammer symbols and the Gamma functions
we get

$K=\dfrac{2}{\pi}\dfrac{\Gamma\left(m+n+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{3}{2}\right)}{\Gamma(m+1)\Gamma(m+n+1)}.$
(3.8)

Substitution in (3.7) gives

$\omega(t)=\dfrac{(-1)^{m}}{\pi}\dfrac{\Gamma\left(m+\dfrac{3}{2}\right)\Gamma\left(n+\dfrac{1}{2}\right)}{\Gamma(m+n+1)}\dfrac{(1-t^{2})^{n}}{t}C_{2m+1}^{(n+1/2)}(t).$

For the kernel $k(t)$ we combine (3.4),(3.6) and (3.8). This results in

$k(t)=\dfrac{2}{\pi}\dfrac{\Gamma\left(m+n+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{3}{2}\right)}{\Gamma(m+1)\Gamma(m+n+1)}\sum_{k=0}^{m}\dfrac{(-m)_{k}\left(m+n+\dfrac{3}{2}\right)_{k}}{\left(\dfrac{3}{2}\right)_{k}}\dfrac{1}{k!}t^{2k}\\\
\sum_{j=0}^{n}(-1)^{n-j}\dfrac{(-2k)_{n-j}(-n)_{j}(-2n)_{j}}{2^{j}\left(\dfrac{1}{2}-n\right)_{j}}\left(\dfrac{1-t^{2}}{t}\right)^{n-j}C^{(n-j+1/2)}_{j}(t).$
(3.9)

Mathematica gives a hypergeometric function for the last summation. We get at
last for the basic kernel

$k(t)=\dfrac{2}{\pi}\dfrac{\Gamma\left(m+n+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{3}{2}\right)}{\Gamma(m+1)\Gamma(m+n+1)}\sum_{j=0}^{m}\dfrac{(-m)_{j}\left(m+n+\dfrac{3}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{j!}t^{2j-n}\\\
\dfrac{\Gamma(2j+1)}{\Gamma(2j+1-n)}\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-n,j+1,j+\dfrac{1}{2}}{j+1-\dfrac{n}{2},j+\dfrac{1}{2}-\dfrac{n}{2}};t^{2}\right).$
(3.10)

Attempts to simplify this formula have not yielded any results so far. A much
simpler formula is found in Section 5.

##  4 A closer look at the basic kernel function

In this section some of the features of the basic kernel function are
discussed. The basic kernel depends on two parameters. The first parameter is
$n$. This is the order of the derivative. The case $n=1$ is so special that is
will be discussed in the next section. The second parameter is $m$. This
parameter represents the number of terms used to approximate the derivative.
This gives a direct measure of the accuracy of the derivative.

The first attribute that stands out, is that the hypergeometric function is
zero-balanced. However, this attribute does not lead to a known function here.
Next it is seen that

if $n=even$ the function $k(t)=even\qquad k(t)=k(-t)$.

if $n=odd\ \ $ the function $k(t)=odd\qquad\ k(t)=-k(-t)$.

Using (3.10) for computing the function $k(t)$ there arises a problem. The
fraction before the hypergeometric function is not convergent if $2k+1-n\leq
0$. But the corresponding lower term in the hypergeometric function becomes
also negative. The summation should be split into two summations. The first
summation concerns the positive values of the relevant factor and the second
sum concerns the negative values of the relevant factor. In that case the
limit should be taken. First we rewrite the basic kernel function and get

$k(t)=\dfrac{2^{n+1}}{\pi}\dfrac{\Gamma\left(m+n+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{3}{2}\right)}{\Gamma(m+1)\Gamma(m+n+1)}\sum_{j=0}^{m}\dfrac{(-m)_{j}\left(m+n+\dfrac{3}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{j!}t^{2j-n}\\\
\dfrac{\Gamma(j+1)\Gamma\left(j+\dfrac{1}{2}\right)}{\Gamma\left(j+\dfrac{1-n}{2}\right)\Gamma\left(j+\dfrac{2-n}{2}\right)}\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-n,j+1,j+\dfrac{1}{2}}{j+\dfrac{1-n}{2},j+\dfrac{2-n}{2}};t^{2}\right).$
(4.1)

We have to distinguish between the case where n is even or n is odd. For the
case $n$ even the term $j+\dfrac{2-n}{2}$ is an integer. Then we have to split
the summation for two cases: $j+\dfrac{2-n}{2}>0$ and $j+\dfrac{2-n}{2}\leq 0$
dependent of the value of $m$. When $n$ odd the term $j+\dfrac{1-n}{2}$ is an
integer. Then we have to split the summation for two cases:
$j+\dfrac{1-n}{2}>0$ and $j+\dfrac{1-n}{2}\leq 0$ and also dependent of the
value of $m$. In both cases we can use the following formula

$\dfrac{1}{\Gamma(-M)}\
_{p+1}F_{p}\left(\begin{array}[]{l}a_{0},\dots,a_{p}\\\
-M,b_{2},\dots,b_{p}\end{array};x\right)=\\\
=\dfrac{x^{M+1}(a_{0})_{M+1}\dots(a_{p})_{M+1}}{\Gamma(M+2)(b_{2})_{M+1}\dots(b_{p})_{M+1}}\
_{p+1}F_{p}\left(\begin{array}[]{l}a_{0}+M+1,\dots,a_{p}+M+1\\\
M+2,b_{2}+M+1,\dots,b_{p}+M+1\end{array};x\right)$ (4.2)

with $M$ a positive integer. For the proof of this formula see [4, Lemma 2].
Both cases are not discussed further here and are left to the reader.

A very special case occurs when $m=0$. We distinguish again the cases with $n$
even and $n$ odd. Starting with $n$ even, equation (4.1) becomes

$k(t)=\dfrac{\Gamma\left(n+\dfrac{3}{2}\right)}{\Gamma(n+1)\Gamma\left(\dfrac{1-n}{2}\right)}\dfrac{1}{t^{n}}\lim_{m\rightarrow
0}\dfrac{1}{\Gamma\left(m+\dfrac{2-n}{2}\right)}\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-n,1,\dfrac{1}{2}}{m+\dfrac{2-n}{2},\dfrac{1-n}{2}};t^{2}\right).$

Application of (4.2) gives

$k(t)=\dfrac{2^{n}\Gamma\left(n+\dfrac{3}{2}\right)}{\Gamma\left(\dfrac{1}{2}\right)}\,\mbox{}_{2}F_{1}\\!\left(\genfrac{}{}{0.0pt}{}{-\dfrac{n}{2},\dfrac{n+1}{2}}{\dfrac{1}{2}};t^{2}\right).$

The hypergeometric function is equal the Legendre polynomial. At last we get

$k(t)=(-1)^{n}\dfrac{\Gamma(2n+2)}{2^{n+1}\Gamma(n+1)}P_{n}(t).$

For $n$ odd we get the same function. This is the expected kernel for the
Legendre derivative.

##  5 Second method for an alternative for the orthogonal derivative

Diekema and Koornwinder derive formula for the approximation of higher
derivatives in [2, Section 4]. With a slight adaptation their formula [2,
(4.6)] becomes

$h^{-n}\sum_{j=0}^{m}\left(\int_{-1}^{1}f(x+ht)P_{n+2j}(t)dt\right)\dfrac{P_{n+2j}^{(n)}(0)}{h_{n+2j}}=\\\
=f^{(n)}(x)+(-1)^{m}\dfrac{\left|P_{n+2m+2}^{(n)}(0)\right|f^{(n+2m+2)}(x)}{\left|k_{n+2m+2}\right|(n+2m+2)!}h^{2m+2}+o(h^{2m+2}).$

In this formula $P_{j}^{(n)}(t)$ denotes the $n$-th derivative of the Legendre
polynomial $P_{n}(t)$. We use the notations $P_{n}(t)=k_{n}t^{n}+\dots$ and
$h_{n}=\int_{-1}^{1}P_{n}(t)^{2}dt=2/(2n+1)$. We can write the left-hand side
as

$(-h)^{-n}\int_{-1}^{1}f(x+\delta t)k(t)dt$

with

$k(t)=(-1)^{n}\sum_{j=0}^{m}\dfrac{P_{n+2j}(t)P^{(n)}_{n+2j}(0)}{h_{n+2j}}.$
(5.1)

With this formula we can derive a simple formula for the function $\omega(t)$.
For the Legendre polynomials we have the following formula

$\displaystyle
P_{n+2j}(t)=\dfrac{(-1)^{n}}{2^{n}(2n+1)_{n}}\dfrac{d^{n}}{dt^{n}}\left((1-t^{2})^{n}P^{(n,n)}_{2j}(t)\right)$
$\displaystyle\dfrac{d^{n}}{dt^{n}}P_{n+2j}(t)=2^{-n}(n+2j+1)_{n}P^{(n,n)}_{2j}(t).$
(5.2)

$P^{(n,n)}_{j}$ are the Jacobi polynomials with constants $h_{j}^{(n,n)}$ and
$k_{j}^{(n,n)}$. Using these formula we get

$\displaystyle k(t)$
$\displaystyle=2^{-2n}\sum_{j=0}^{m}\dfrac{(n+2j+1)_{n}}{(2j+1)_{n}}\dfrac{1}{h_{n+2j}}\dfrac{d^{n}}{dt^{n}}\left((1-t^{2})^{n}P^{(n,n)}_{2j}(t)P^{(n,n)}_{2j}(0)\right)$
$\displaystyle=\dfrac{d^{n}}{dt^{n}}\left(2^{-2n}\sum_{j=0}^{m}\dfrac{(n+2j+1)_{n}}{(2j+1)_{n}}\dfrac{1}{h_{n+2j}}\left((1-t^{2})^{n}P^{(n,n)}_{2j}(t)P^{(n,n)}_{2j}(0)\right)\right).$

Remembering $k(t)=\omega^{(n)}(t)$ gives for $\omega(t)$

$\omega(t)=2^{-2n}(1-t^{2})^{n}\sum_{j=0}^{m}\dfrac{(n+2j+1)_{n}}{(2j+1)_{n}}\dfrac{1}{h_{n+2j}}P^{(n,n)}_{2j}(t)P^{(n,n)}_{2j}(0).$

Using

$h_{n+2j}\dfrac{2^{2n}(2j+1)_{n}}{(n+2j+1)_{n}}=h_{2j}^{(n,n)}$

gives

$\omega(t)=(1-t^{2})^{n}\sum_{j=0}^{m}\dfrac{P^{(n,n)}_{2j}(t)P^{(n,n)}_{2j}(0)}{h_{2j}^{(n,n)}}=(1-t^{2})^{n}\sum_{k=0}^{2m}\dfrac{P^{(n,n)}_{k}(t)P^{(n,n)}_{k}(0)}{h_{k}^{(n,n)}}.$

For the summation we can use the formula of Christoffel-Darboux and get

$\displaystyle\omega(t)$
$\displaystyle=(1-t^{2})^{n}\dfrac{k_{2m}^{(n,n)}}{h_{2m}^{(n,n)}k_{2m+1}^{(n,n)}}\dfrac{P_{2m+1}^{(n,n)}(t)P_{2m}^{(n,n)}(0)-P_{2m}^{(n,n)}(t)P_{2m+1}^{(n,n)}(0)}{t-0}$
$\displaystyle=(1-t^{2})^{n}\dfrac{k_{2m}^{(n,n)}}{h_{2m}^{(n,n)}k_{2m+1}^{(n,n)}}\dfrac{1}{t}P_{2m+1}^{(n,n)}(t)P_{2m}^{(n,n)}(0).$

The Jacobi polynomials can be written as Gegenbauer polynomials.

$\omega(t)=\dfrac{(1-t^{2})^{n}}{t}\dfrac{k_{2m}^{(n,n)}P_{2m}^{(n,n)}(0)}{h_{2m}^{(n,n)}k_{2m+1}^{(n,n)}}\dfrac{(n+1)_{n}}{(2m+n+2)_{n}}C_{2m+1}^{(n+1/2)}(t).$

For the parameters we get

$\displaystyle
k_{2m}^{(n,n)}=\dfrac{1}{2^{2m}}\binom{4m+2n}{2m}\qquad\qquad\qquad\qquad\qquad\qquad
k_{2m+1}^{(n,n)}=\dfrac{1}{2^{2m+1}}\binom{4m+2+2n}{2m+1}$ $\displaystyle
h_{2m}^{(n,n)}=\dfrac{2^{2n+1}}{(4m+2n+1)}\dfrac{\Gamma(2m+n+1)^{2}}{(2m)!\Gamma(2m+2n+1)}\qquad\quad
P_{2m}^{(n,n)}(0)=\dfrac{(-1)^{m}}{2^{2m}}\binom{2m+n}{m}.$

Substitution gives after some manipulation with the Pochhammer symbols and the
Gamma functions

$\displaystyle\omega(t)$
$\displaystyle=\dfrac{(-1)^{m}}{\pi}\dfrac{\Gamma\left(m+\dfrac{3}{2}\right)\Gamma\left(n+\dfrac{1}{2}\right)}{\Gamma(m+n+1)}\dfrac{(1-t^{2})^{n}}{t}C_{2m+1}^{(n+1/2)}(t)$
$\displaystyle=\dfrac{(-1)^{m}}{2^{2m+2n+1}}\dfrac{\Gamma(2m+2)\Gamma(2n+1)}{\Gamma(m+1)\Gamma(n+1)\Gamma(m+n+1)}\dfrac{(1-t^{2})^{n}}{t}C_{2m+1}^{(n+1/2)}(t)$

We found this formula already in section 3. Now we have to compute the basic
kernel function $k(t)$. The parameters are

$h_{n+2j}=\dfrac{2}{2n+4j+1}$

and using (5.2)

$P_{n+2j}^{(n)}(0)=2^{-n}(n+2j+1)_{n}P_{2j}^{(n,n)}(0)=(n+2j+1)_{n}\dfrac{(-1)^{j}}{2^{2j+n}}\binom{2j+n}{j}.$
(5.3)

Substitution in (5.1) gives after some manipulation with the Pochhammer
symbols and the Gamma functions

$k(t)=(-1)^{n}\dfrac{2^{n-1}}{\sqrt{\pi}}\sum_{j=0}^{m}(2n+4j+1)\Gamma\left(n+j+\dfrac{1}{2}\right)\dfrac{1}{j!}(-1)^{j}P_{n+2j}(t).$
(5.4)

##  6 The first derivative

Looking at formula (3.10) we see that the hypergeometric function is a
polynomial of degree $n$. $n$ is also the order of the derivative. In this
section we will discuss the case where $n=1$ so we look for the kernel $k(t)$
for the first derivative.

We get from (3.10) with $n=1$

$k_{m}(t)=\dfrac{4}{\pi}\dfrac{\left(m+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{5}{2}\right)}{\Gamma(m+1)\Gamma(m+2)}\sum_{j=0}^{m}\dfrac{(-m)_{j}\left(m+\dfrac{5}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{j!}\big{[}j-(j+1)t^{2}\big{]}t^{2j-1}.$
(6.1)

The first few functions (see appendix B), which we call $k_{m}(t)$ are

$\displaystyle k_{0}(t)=-\dfrac{3}{2}t$ $\displaystyle
k_{1}(t)=-\dfrac{15}{8}t(5-7t^{2})$ $\displaystyle
k_{2}(t)=-\dfrac{105}{128}t(35-126t^{2}+99t^{4})$ $\displaystyle
k_{3}(t)=-\dfrac{315}{512}t(105-693t^{2}+1287t^{4}-715t^{6}).\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$

It is easy to check that the function $k_{m}(t)$ satisfies the following
difference equation

$k_{m+1}(t)=\left(1+\dfrac{C_{m}}{A_{m}}-\dfrac{1}{A_{m}}t^{2}\right)k_{m}(t)-\dfrac{C_{m}}{A_{m}}k_{m-1}(t)$

with

$\dfrac{1}{A_{m}}=\dfrac{(4m+5)(4m+7)}{(2m+2)^{2}}\qquad\text{ and
}\qquad\dfrac{C_{m}}{A_{m}}=\dfrac{(2m+3)^{2}(4m+7)}{(2m+2)^{2}(4m+3)}.$

Then we may conclude that the function $k_{m}(t)$ is orthogonal. Looking at
equation (6.1) it turns out that the summation can be performed. It is
possible to write the function $k_{m}(t)$ in a highly simplified form.

We start with (6.1) and split the right hand side.

$k_{m}(t)=-\dfrac{4}{\pi}\dfrac{\Gamma\left(m+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{5}{2}\right)}{\Gamma(m+1)\Gamma(m+2)}\sum_{j=0}^{m}\dfrac{(-m)_{j}\left(m+\dfrac{5}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{j!}t^{2j+1}+\\\
+(1-t^{2})\dfrac{4}{\pi}\dfrac{\left(m+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{5}{2}\right)}{\Gamma(m+1)\Gamma(m+2)}\sum_{j=1}^{m}\dfrac{(-m)_{j}\left(m+\dfrac{5}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{(j-1)!}t^{2j-1}.$

Changing the summation in the second term we get

$k_{m}(t)=-\dfrac{4}{\pi}\dfrac{\Gamma\left(m+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{5}{2}\right)}{\Gamma(m+1)\Gamma(m+2)}\sum_{j=0}^{m}\dfrac{(-m)_{j}\left(m+\dfrac{5}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{j!}t^{2j+1}+\\\
-(1-t^{2})\dfrac{8}{3\pi}\dfrac{\left(m+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{7}{2}\right)}{\Gamma(m)\Gamma(m+2)}\sum_{j=0}^{m-1}\dfrac{(1-m)_{j}\left(m+\dfrac{7}{2}\right)_{j}}{\left(\dfrac{5}{2}\right)_{j}}\dfrac{1}{j!}t^{2j+1}.$
(6.2)

The summations can be done and there appears Jacobi functions.

$\sum_{j=0}^{m}\dfrac{(-m)_{j}\left(m+\dfrac{5}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{j!}t^{2j+1}=t\dfrac{\sqrt{\pi}\Gamma(m+1)}{2\Gamma\left(m+\dfrac{5}{2}\right)}P_{m}^{(1/2,1)}(1-2t^{2}).$
$\sum_{j=0}^{m-1}\dfrac{(1-m)_{j}\left(m+\dfrac{7}{2}\right)_{j}}{\left(\dfrac{5}{2}\right)_{j}}\dfrac{1}{j!}t^{2j+1}=t\dfrac{3\sqrt{\pi}\Gamma(m)}{4\Gamma\left(m+\dfrac{3}{2}\right)}P_{m-1}^{(3/2,2)}(1-2t^{2}).$

Substitution in (6.2) gives

$k_{m}(t)=-\dfrac{2\Gamma\left(m+\dfrac{5}{2}\right)}{\sqrt{\pi}\Gamma(m+2)}\left[t\,P_{m}^{(1/2,1)}(1-2t^{2})+\left(m+\dfrac{5}{2}\right)(1-t^{2})t\,P_{m-1}^{(3/2,2)}(1-2t^{2})\right].$
(6.3)

To continue there are two possibilities. The first possibility transforms the
Jacobi polynomials to Gegenbauer polynomials. These can be transformed into
Legendre polynomials. We get a sum of two Legendre polynomials.

The second possibility uses transform formulas for the Jacobi polynomials and
at last we get a single Jacobi polynomial.

First possibility.

The Jacobi polynomials can be transformed to Gegenbauer polynomials. Using [1,
22.5.25, 22.5.26] gives

$\displaystyle
t\,P_{m}^{(1/2,1)}(1-2t^{2})=-\dfrac{1}{(2m+3)}C_{2m+1}^{(3/2)}(t)$
$\displaystyle
t\,P_{m-1}^{(3/2,2)}(1-2t^{2})=\dfrac{1}{2(2m+3)\left(m+\dfrac{5}{2}\right)}\dfrac{d}{dt}\left[C_{2m+1}^{(3/2)}(t)\right].$

For the derivative of the Gegenbauer function we use [1, 22.9.2] and get

$\dfrac{d}{dt}\left[C_{2m+1}^{(3/2)}(t)\right]=\dfrac{3}{t}C_{2m}^{(5/2)}(t)-\dfrac{1}{t^{2}}C_{2m+1}^{(3/2)}(t).$

Substitution in (6.3) gives

$k_{m}(t)=-\dfrac{\Gamma\left(m+\dfrac{3}{2}\right)}{\sqrt{\pi}\Gamma(m+2)}\left[\left(-1-\dfrac{1-t^{2}}{2t^{2}}\right)C_{2m+1}^{(3/2)}(t)+\dfrac{3}{2}\left(\dfrac{1-t^{2}}{t}\right)C_{2m}^{(5/2)}(t)\right].$

For the second Gegenbauer function we use [1, 22.7.3] and get

$3(1-t^{2})C_{2m}^{(5/2)}(t)=(4+2m)t\,C_{2m+1}^{(3/2)}-(2m+2)C_{2m+2}^{(3/2)}(t).$

Substitution gives

$k_{m}(t)=-\dfrac{\Gamma\left(m+\dfrac{3}{2}\right)}{\sqrt{\pi}\Gamma(m+2)}\left[\left(\dfrac{3}{2}+m-\dfrac{1}{2t^{2}}\right)C_{2m+1}^{(3/2)}(t)-\dfrac{1}{t}(m+1)C_{2m+2}^{(3/2)}(t)\right].$

With [1, 22.7.3]

$C_{n-2}^{(\lambda+1)}(t)=\dfrac{(2\lambda+1)t}{2\lambda(1-t^{2})}C_{n-1}^{(\lambda)}(t)-\dfrac{n}{2\lambda(1-t^{2})}C_{n}^{(\lambda)}(t).$

we can decrease the upper indexes of the Gegenbauer functions with one. For
$\lambda=1/2$ the Gegenbauer polynomial becomes a Legendre polynomial. Using
the standard recurrence equation for the Legendre polynomials [1, 22.7.10]

$P_{n+1}(t)=\dfrac{2n+1}{n+1}t\,P_{n}(t)-\dfrac{n}{n+1}P_{n-1}(t)$

a lot of times we get at last

$\displaystyle k_{m}(t)$
$\displaystyle=\dfrac{1}{B(m+1,m+2)2^{2m+2}}(-1)^{m+1}\,\dfrac{1}{t^{2}}\Big{[}(2m+2)t\,P_{2m+2}(t)+P_{2m+1}(t)\Big{]}$
$\displaystyle=\dfrac{(2m+3)}{B(m+1,m+2)(4m+5)2^{2m+2}}(-1)^{m+1}\dfrac{1}{t^{2}}\Big{[}(2m+2)P_{2m+3}(t)+(2m+3)P_{2m+1}(t)\Big{]}$
(6.4)

where $B(m+1,m+2)$ is the Beta function. We conclude that the kernel function
$k_{m}(t)$ can be written as a sum of two Legendre functions.

Second possibility.

We repeat here (6.3).

$k_{m}(t)=-\dfrac{2\Gamma\left(m+\dfrac{5}{2}\right)}{\sqrt{\pi}\Gamma(m+2)}t\left[P_{m}^{(1/2,1)}(1-2t^{2})+\left(m+\dfrac{5}{2}\right)(1-t^{2})P_{m-1}^{(3/2,2)}(1-2t^{2})\right].$
(6.5)

We define the form between the square brackets as

$K=P_{m}^{(1/2,1)}(1-2t^{2})+\left(m+\dfrac{5}{2}\right)(1-t^{2})\,P_{m-1}^{(3/2,2)}(1-2t^{2}).$

For the Jacobi polynomials there are a lot of transformation formula [1,
10.8]. The goal of these transformations is to bring the first order to $3/2$
and the second order to zero. First of all we use [5, 10.8.(33)] a lot of
times

$2\left(m+\dfrac{\alpha+\beta}{2}+1\right)P_{m}^{(\alpha,\beta+1)}(1-2t^{2})=\dfrac{(m+\beta+1)}{(1-t^{2})}P_{m}^{(\alpha,\beta)}(1-2t^{2})+\dfrac{(m+1)}{(1-t^{2})}P_{m+1}^{(\alpha,\beta)}(1-2t^{2})$

and get at last

$\displaystyle K$
$\displaystyle=\dfrac{2(m+1)}{(4m+5)}\dfrac{1}{(1-t^{2})}\left(P_{m}^{(1/2,0)}(1-2t^{2})+P_{m+1}^{(1/2,0)}(1-2t^{2})\right)+$
$\displaystyle+\dfrac{2m(m+1)(2m+5)}{(4m+3)(4m+5)}\dfrac{1}{(1-t^{2})}\left(P_{m-1}^{(3/2,0)}(1-2t^{2})+P_{m}^{(3/2,0)}(1-2t^{2})\right)+$
$\displaystyle+\dfrac{2m(m+1)(2m+5)}{(4m+5)(4m+7)}\dfrac{1}{(1-t^{2})}\left(P_{m}^{(3/2,0)}(1-2t^{2})+P_{m+1}^{(3/2,0)}(1-2t^{2})\right).$

Only the polynomials with first order $1/2$ should be converted into
polynomials with order $3/2$. We use [5, 10.8.(35)]

$P_{m}^{(\alpha-1,0)}(1-2t^{2})=\dfrac{(m+\alpha)}{(2m+\alpha)}P_{m}^{(\alpha,0)}(1-2t^{2})-\dfrac{m}{(2m+\alpha)}P_{m-1}^{(\alpha,0)}(1-2t^{2}).$

After gathering the different terms there remains

$\displaystyle K$
$\displaystyle=\dfrac{2m(m+1)(2m+3)}{(4m+3)(4m+5)}\dfrac{1}{(1-t^{2})}P_{m-1}^{(3/2,0)}(1-2t^{2})+$
$\displaystyle+\dfrac{2(m+1)(4m^{2}+10m+3)}{(4m+3)(4m+7)}\dfrac{1}{(1-t^{2})}P_{m}^{(3/2,0)}(1-2t^{2})+$
$\displaystyle+\dfrac{2(2m+5)(m+1)^{2}}{(4m+5)(4m+7)}\dfrac{1}{(1-t^{2})}P_{m+1}^{(3/2,0)}(1-2t^{2}).$

and there rests all Jacobi polynomials with the same orders. Only the index of
the polynomials are different. We use the standard recurrence equation for the
Jacobi polynomials [5, 10.8.(11)]

$\displaystyle
2(m+\alpha)(m+\beta)(2m+\alpha+\beta+2)P_{m-1}^{(\alpha,\beta)}(x)=$
$\displaystyle=(2m+\alpha+\beta+1)\left[(2m+\alpha+\beta)(2m+\alpha+\beta+2)x+\alpha^{2}-\beta^{2}\right]P_{m}^{(\alpha,\beta)}(x)-$
$\displaystyle-2(m+1)(m+\alpha+\beta+1)(2m+\alpha+\beta)P_{m+1}^{(\alpha,\beta)}(x).$

Application gives at last

$K=(m+1)P_{m}^{(3/2,0)}(1-2t^{2})=(m+1)(-1)^{m}P_{m}^{(0,3/2)}(2t^{2}-1).$

Then we get for $k_{m}(t)$

$k_{m}(t)=(-1)^{m+1}\dfrac{2\Gamma\left(m+\dfrac{5}{2}\right)}{\sqrt{\pi}\Gamma(m+1)}t\,P_{m}^{(0,3/2)}(2t^{2}-1).$
(6.6)

######  Remark 6.1.

Combining (6.4) and (6.6) results in a relation between a special Jacobi
polynomial and two Legendre polynomials.

$P_{m}^{(0,3/2)}(2t^{2}-1)=\dfrac{(m+2)2^{m+3}}{(4m+5)}\dfrac{2}{t^{3}}\Big{[}(2m+2)P_{2m+3}(t)+(2m+3)P_{2m+1}(t)\Big{]}.$
(6.7)

######  Remark 6.2.

For the second order derivative ($n=2$) the formula for $k(t)$ becomes

$k_{m}(t)=(-1)^{m}\dfrac{4\Gamma\left(m+\dfrac{7}{2}\right)}{\sqrt{\pi}\Gamma(m+1)(4m+7)}\Big{[}(2m+4)P_{m}^{(0,3/2)}(2t^{2}-1)+(2m+3)P_{m+1}^{(0,3/2)}(2t^{2}-1)\Big{]}.$

With (6.7) we can express this function $k(t)$ as a summation of Legendre
functions of odd orders.

As we noted earlier, the function $k_{m}(t)$ for $n=1$ is orthogonal. Here
follows an overview of the orthogonality properties of this function.

Overview

\- The orthogonal polynomial

$k_{m}(t)=(-1)^{m+1}\dfrac{\Gamma(2m+4)}{\Gamma(m+1)\Gamma(m+2)2^{2m+2}}t\,P_{m}^{(0,3/2)}(2t^{2}-1)\qquad\qquad-1\leq
t\leq 1.$

\- The basic orthogonality property

$\int_{-1}^{1}k_{m}(t)k_{n}(t)\,w(t)dt=h_{m}\delta_{mn}$

with

$h_{m}=\dfrac{\Gamma(2m+4)^{2}}{2^{4m+3}(4m+5)\Gamma(m+1)^{2}\Gamma(m+2)^{2}}.$

\- The weight function

$w(t)=\left(1+\sum_{k=1}^{\infty}c_{k}\,t^{2k+1}\right)t^{2}.$

$c_{k}$ are arbitrary parameters (and can also be omitted).

\- The recurrence equation.

$k_{m+1}(t)=\left(1+\dfrac{C_{m}}{A_{m}}-\dfrac{1}{A_{m}}t^{2}\right)k_{m}(t)-\dfrac{C_{m}}{A_{m}}k_{m-1}(t)$

with

$\dfrac{C_{m}}{A_{m}}=\dfrac{(2m+3)^{2}(4m+7)}{(2m+2)^{2}(4m+3)}\text{\qquad
and \qquad}\dfrac{1}{A_{m}}=\dfrac{(4m+5)(4m+7)}{(2m+2)^{2}}$

and

$k_{-1}(t)=0\text{\qquad and \qquad}k_{0}(t)=-\dfrac{3t}{2}.$

\- The differential-difference equation

$t(1-t^{2})\dfrac{d}{dt}k_{m}(t)=\left(\dfrac{4m^{2}+4m+3}{4m+3}-(2m+1)t^{2}\right)k_{m}(t)-\dfrac{(2m+3)^{2}}{4m+3}k_{m-1}(t).$

\- The differential equation

$t^{2}(1-t^{2})^{2}\dfrac{d^{2}}{dt^{2}}k_{m}(t)+A_{m}t(1-t^{2})\dfrac{d}{dt}k_{m}(t)-B_{m}k_{m}(t)=0$

with

$A_{m}=2(1-2t^{2})\text{\qquad and
\qquad}B_{m}=2(m+2)(2m+1)t^{4}-2(2m+3)(m+1)t^{2}+2.$

##  7 Transfer function of the new derivative

Liptaj in his paper [9] shows that the formula for the derivatives are much
more accurate then the standard Legendre derivatives. He uses some test
functions to demonstrate this effect. The derivative can be regarded as a
filter. This filter is characterized by his transfer function. The author in
his thesis [3] argues that it is recommended to use the transfer function
because then you can see

\- the effect of the noise disturbed input signal on the output signal

\- the effect of the frequency content of the input signal on the output
signal.

In that case you are independent of the test functions. Liptaj uses test
functions with very low different frequency contents. In this way it is
difficult to decide how accurate the filter is.

In this section we investigate the transfer function of the filter equation
(3.1) of the derivative which will be repeat here

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim_{h\rightarrow
0}\left(-\dfrac{1}{h}\right)^{n}\int_{-1}^{1}k(t)f(x+h\,t)dt.$

Let us call the transfer function of the left hand side $H(\omega)$. The
transfer function follows from (5.4) and is then given by (omitting the limit)

$H(h,\omega)=\left(\dfrac{1}{h}\right)^{n}\dfrac{2^{n}}{\sqrt{\pi}}\sum_{j=0}^{m}(2n+4j+1)\Gamma\left(n+j+\dfrac{1}{2}\right)\dfrac{1}{j!}(-1)^{j}\int_{-1}^{1}P_{n+2j}(t)e^{-ih\omega
t}dt.$ (7.1)

The integral is known [6, 3.3(1)]

$\displaystyle\int_{-1}^{1}P_{n}(t)e^{-i\omega t}dt$
$\displaystyle=(-1)^{n}i^{n}(2\pi)^{1/2}\omega^{-1/2}J_{n+1/2}(\omega)$
$\displaystyle=(-1)^{n}i^{n}j_{n}(\omega)$

$J_{n}(\omega)$ is the Bessel function of the first kind. $j_{n}(\omega)$ is
the spherical Bessel function of the first kind. Substitution gives

$H(h,\omega)=\left(-\dfrac{1}{h}\right)^{n}i^{n}\dfrac{2^{n}}{\sqrt{\pi}}\sum_{j=0}^{m}(2n+4j+1)\Gamma\left(n+j+\dfrac{1}{2}\right)\dfrac{1}{j!}j_{n+2j}(h\omega)$

In the next figure we show the modulus of the transfer function for $n=2$ and
different values for $m$. For $m=0$ we get the transfer function of the second
order Legendre derivative. This is also given in [3, section 2.6]. When using
another order of the derivative, the shape of the figures does not change,
only the axes are adjusted. The same for the value of $h$. The product
$(h\omega)$ determines the adjustment of the horizontal axe.

Figure 4: Moduli of the absolute value of the transfer function of a second
order differentiator for $m=0$, $m=2$ and $m=5$. $h=0.01$. For $m=0$ this is
the transfer function of the second order Legendre differentiator.

We see in the figure that with higher values of $m$ the bandwidth of the
derivative will increase. This is in line with what we expect. But there is a
also a limit for the bandwidth of the transfer function.

As an approximation of the maximum of the transfer function we get

$\omega_{nax}\approx\dfrac{2}{h}\sqrt{2m+n+\dfrac{5}{2}}.$

For practical reasons it is recommended using a signal with its highest
frquency component is lower then 10 times this maximum. This means that if we
want to use test functions, carefully attention must be paid to the frequency
content of those test functions. After all the accuracy of the differentiator
depends on the frequency content of the test signal. With a test signal with
low frequencies the approximation will be better than with a test signal with
high frequencies.

##  8 Summary and conclusion

This paper describes an extension of the orthogonal derivative which first
appears under this name in the thesis of the author [3]. In [2] there is a
survey of this derivative which is already known for long time under different
names. The standard formula is

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim\limits_{h\downarrow
0}\dfrac{k_{n}n!}{h_{n}}\dfrac{1}{h^{n}}\int_{\mathbb{R}}f(x+hu)p_{n}(u)d\mu(u)$

with the necessary conditions (2.4). Liptay in his papers [8],[9] developed a
new formula for the derivative by integration of a function on an bounded
interval with greater accuracy then the approximate orthogonal derivative. His
formula is

$\dfrac{d^{n}}{dx^{n}}f(x)=\lim_{h\rightarrow
0}\left(-\dfrac{1}{h}\right)^{n}\int_{-1}^{1}k(t)f(x+h\,t)dt$

with $k(t)$ is a kernel which can be computed by solution of a number of
equations. He claims that the accuracy of his formula is much better then the
standard Lanczos’ derivative which is the first order Legendre derivative. He
did not derive a general formula for $k(t)$. In this paper a general solution
for $k(t)$ is derived. We found

$k(t)=\dfrac{2}{\pi}\dfrac{\Gamma\left(m+n+\dfrac{3}{2}\right)\Gamma\left(m+\dfrac{3}{2}\right)}{\Gamma(m+1)\Gamma(m+n+1)}\sum_{j=0}^{m}\dfrac{(-m)_{j}\left(m+n+\dfrac{3}{2}\right)_{j}}{\left(\dfrac{3}{2}\right)_{j}}\dfrac{1}{j!}t^{2j-n}\\\
\dfrac{\Gamma(2j+1)}{\Gamma(2j+1-n)}\,\mbox{}_{3}F_{2}\\!\left(\genfrac{}{}{0.0pt}{}{-n,j+1,j+\dfrac{1}{2}}{j+1-\dfrac{n}{2},j+\dfrac{1}{2}-\dfrac{n}{2}};t^{2}\right).$

Diekema and Koornwinder [2] found

$k(t)=(-1)^{n}\dfrac{2^{n-1}}{\sqrt{\pi}}\sum_{j=0}^{m}(2n+4j+1)\Gamma\left(n+j+\dfrac{1}{2}\right)\dfrac{1}{j!}(-1)^{j}P_{n+2j}(t).$

As a special case we developed a formula for the kernel for the first
derivative. We get

$\displaystyle k_{m}(t)$
$\displaystyle=\dfrac{1}{B(m+1,m+2)2^{2m+2}}(-1)^{m+1}\,\dfrac{1}{t^{2}}\Big{[}(2m+2)t\,P_{2m+2}(t)+P_{2m+1}(t)\Big{]}$
$\displaystyle=(-1)^{m+1}\dfrac{2\Gamma\left(m+\dfrac{5}{2}\right)}{\sqrt{\pi}\Gamma(m+1)}t\,P_{m}^{(0,3/2)}(2t^{2}-1).$

In the first formula there are Legendre polynomials and in the second formula
there is a Jacobi polynomial. This kernel function is orthogonal. For higher
order the kernel function is not orthogonal. For all orders of the derivative
it turns out that k$(t)$ can be written as a sum of two Legendre polynomials
where each polynomial is multiplied by a polynomial of $t$. Liptaj comes to
the same conclusion [9, Appendix A].

In [3] formula are developed for the transfer functions of the approximate
orthogonal derivative. These are used to see how the accuracy of the function
behaves in the frequency domain. In this paper we found

$H(h,\omega)=\left(-\dfrac{1}{h}\right)^{n}i^{n}\dfrac{2^{n}}{\sqrt{\pi}}\sum_{j=0}^{m}(2n+4j+1)\Gamma\left(n+j+\dfrac{1}{2}\right)\dfrac{1}{j!}j_{n+2j}(h\omega)$

Finally, we get the following conclusions:

  1. –

The formula for the approximate derivatives we use in this paper depends on a
parameter $h$. When $h$ goes to zero the exact derivative appears.

  2. –

The idea of Liptaj for his system of deriving formulas for the derivative by
integration is a good idea. Only his method to compute the kernel is not very
useful in practice.

  3. –

The formula for the basic kernel in this paper is much more useful to compute
this kernel.

  4. –

The formula for the kernel for the first order is orthogonal. The formula for
the kernel for higher order derivatives are not.

  5. –

The accuracy of the derivative depends on the first term with the parameter
$h$ in the Taylor series of the given function. This term can be chosen
freely.

  6. –

To test the accuracy of the formula for the derivative, it is recommended to
look at the behaviour of the given function in the frequency domain.

  7. –

As expected the bandwidth of the transfer function increases when the number
of terms without the parameter $h$ in the Taylor series of the input function
increases.

Acknowledgements

The author is very grateful to T.H. Koornwinder and W. Van Assche for their
comments on section 6.

Appendices

##  Appendix A An integral with a Gegenbauer polynomial

In this Appendix we compute the integral

$I=\int_{-1}^{1}t^{i+2k-n+j}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t)dt.$

The integral is not known, but we can convert this integral into an integral
that is known. We use

$C^{(n-j+1/2)}_{j}(-t)=(-1)^{j}C^{(n-j+1/2)}_{j}(-t).$ (A.1)

Define:

$p(t)=t^{i+2k-n+j}(1-t^{2})^{n-j}.$

Then for the integral we get

$\displaystyle\int_{-1}^{1}p(t)C^{(n-j+1/2)}_{j}(t)dt$
$\displaystyle=\int_{-1}^{0}p(t)C^{(n-j+1/2)}_{j}(t)dt+\int_{0}^{1}p(t)C^{(n-j+1/2)}_{j}(t)dt=$
$\displaystyle=(-1)^{i+2k-n+j}\int_{0}^{1}p(t)C^{(n-j+1/2)}_{j}(-t)dt+\int_{0}^{1}p(t)C^{(n-j+1/2)}_{j}(t)dt=$
$\displaystyle=\big{[}1+(-1)^{i-n}\big{]}\int_{0}^{1}t^{i+2k-n+j}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t)dt.$

This integral is known [10, 2.21.1(1)]

$\int_{0}^{1}x^{\alpha-1}(1-x^{2})^{\lambda-1/2}C^{(\lambda)}_{2r}(x)dx=\dfrac{(-1)^{r}}{2(2r)!}(2\lambda)_{2r}\Gamma\left(\lambda+\dfrac{1}{2}\right)\dfrac{\Gamma\left(\dfrac{\alpha}{2}\right)}{\Gamma\left(\dfrac{\alpha+1}{2}+\lambda+r\right)}\left(\dfrac{1-\alpha}{2}\right)_{r}$

with $\lambda>-\dfrac{1}{2}$ and $\alpha>0$. Application of this integral with
$\alpha=i+2k-n+j+1$, $\lambda=n-j+\dfrac{1}{2}$ and $r=\dfrac{j}{2}$ gives

$\int_{0}^{1}t^{i+2k-n+j}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t)dt=\\\
=\dfrac{(-1)^{j/2}}{2j!}(2n+1-2j)_{j}\Gamma(n+1-j)\dfrac{\Gamma\left(\dfrac{i+2k-n+1+j}{2}\right)}{\Gamma\left(\dfrac{n+i+2k+3}{2}\right)}\left(\dfrac{n-i-2k-j}{2}\right)_{j/2}.$

After some manipulations with the Pochhammer symbols and the Gamma functions
we obtain

$\int_{0}^{1}t^{i+2k-n+j}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t)dt=\\\
=(-1)^{j}\dfrac{\Gamma(n+1)2^{j}\left(\dfrac{1}{2}-n\right)_{j}(i+2m-n+1)_{j}}{(-2n)_{j}(i+2k+n+1)j!}\dfrac{\Gamma\left(\dfrac{i+2k-n+1}{2}\right)}{\Gamma\left(\dfrac{i+2k+n+1}{2}\right)}.$

Then we get

$\int_{-1}^{1}t^{i+2k-n+j}(1-t^{2})^{n-j}C^{(n-j+1/2)}_{j}(t)dt=\\\
=\left[1+(-1)^{i-n}\right](-1)^{j}\dfrac{\Gamma(n+1)2^{j}\left(\dfrac{1}{2}-n\right)_{j}(i+2m-n+1)_{j}}{(-2n)_{j}(i+2k+n+1)j!}\dfrac{\Gamma\left(\dfrac{i+2k-n+1}{2}\right)}{\Gamma\left(\dfrac{i+2k+n+1}{2}\right)}.$

##  Appendix B Overview of the kernel functions

In this appendix an overview is given of the kernel functions $k_{m}(t)$ for
different values of the order $n$ and the parameter $m$ using (3.10).

\- First order $n=1$

$\displaystyle k_{0}(t)=-\dfrac{3}{2}t$ $\displaystyle
k_{1}(t)=-\dfrac{15}{8}t(5-7t^{2})$ $\displaystyle
k_{2}(t)=-\dfrac{105}{128}t(35-126t^{2}+99t^{4})$ $\displaystyle
k_{3}(t)=-\dfrac{315}{512}t(105-693t^{2}+1287t^{4}-715t^{6})$ $\displaystyle
k_{4}(t)=-\dfrac{3465}{32768}t(1155-12012t^{2}+38610t^{4}-48620t^{6}+20995t^{8})$
$\displaystyle
k_{5}(t)=-\dfrac{9009}{131072}t(3003-45045t^{2}+218790t^{4}-461890t^{6}+440895t^{8}-156009t^{10}.\qquad\qquad\qquad$

\- Second order $n=2$

$\displaystyle k_{0}(t)=\dfrac{15}{4}(-1+3t^{2})$ $\displaystyle
k_{1}(t)=-\dfrac{105}{32}(5-42t^{2}+45t^{4})$ $\displaystyle
k_{2}(t)=\dfrac{315}{256}(-35+567t^{2}-1485t^{4}+1001t^{6})$ $\displaystyle
k_{3}(t)=-\dfrac{3465}{4096}(105-2772t^{2}+12870t^{4}-20020t^{6}+9945t^{8})$
$\displaystyle
k_{4}(t)=\dfrac{45045}{65536}(-231+9009t^{2}-64350t^{4}+170170t^{6}-188955t^{8}+74613t^{10})$
$\displaystyle
k_{5}(t)=-\dfrac{45045}{524288}(3003-162162t^{2}+1640925t^{4}-6466460t^{6}+11904165t^{8}-10296594t^{10}+$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\
\ \ +3380195t^{12}).\qquad\ \ \ $

\- Third order $n=3$

$\displaystyle k_{0}(t)=-\dfrac{105}{4}t(-3+5t^{2})$ $\displaystyle
k_{1}(t)=\dfrac{945}{32}t(21-90t^{2}+77t^{4})$ $\displaystyle
k_{2}(t)=-\dfrac{10395}{256}t(-63+495t^{2}-1001x^{4}+585x^{7})$ $\displaystyle
k_{3}(t)=\dfrac{45045}{4096}t(693-8580t^{2}+30030t^{4}-39780t^{6}+17765t^{8})$
$\displaystyle
k_{4}(t)=-\dfrac{135135}{65536}t(-9009+160875t^{2}-850850t^{4}+1889550t^{6}-1865325t^{8}+676039t^{10})$
$\displaystyle
k_{5}(t)=\dfrac{2297295}{524288}t(9009-218790t^{2}+1616615t^{4}-5290740t^{6}+8580495t^{8}-6760390t^{10}+$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+2064825t^{12}).\qquad\
\ \ $

\- Fourth order $n=4$

$\displaystyle k_{0}(t)=\dfrac{945}{16}(3-30t^{2}+35t^{4})$ $\displaystyle
k_{1}(t)=-\dfrac{10395}{64}(-7+135t^{2}-385t^{4}+273t^{6})$ $\displaystyle
k_{2}(t)=\dfrac{135135}{2048}(63-1980t^{2}+10010t^{4}-16380t^{6}+8415t^{8})$
$\displaystyle
k_{3}(t)=-\dfrac{135135}{8192}(-693+32175t^{2}-250250t^{4}+696150t^{6}-799425t^{8}+323323t^{10})$
$\displaystyle
k_{4}(t)=\dfrac{2297295}{262144}(3003-193050t^{2}+2127125t^{4}-8817900t^{6}+16787925t^{8}-14872858t^{10}+$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\
\ +4970875t^{12})\ \ \ $ $\displaystyle
k_{5}(t)=-\dfrac{43648605}{1048576}(-1287+109395t^{2}-1616615t^{4}+9258795t^{6}-25741485t^{8}+37182145t^{10}-$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\
-26842725t^{12}+7653825t^{14}).$

\- Fifth order $n=5$

$\displaystyle k_{0}(t)=-\dfrac{10395}{16}t(15-70t^{2}+63t^{4})$
$\displaystyle k_{1}(t)=\dfrac{135135}{64}t(-45+385t^{2}-819t^{4}+495t^{6})$
$\displaystyle
k_{2}(t)=-\dfrac{675675}{2048}t(1485-20020t^{2}+73710t^{4}-100980t^{6}+46189t^{8})$
$\displaystyle
k_{3}(t)=\dfrac{11486475}{8192}t(-1287+25025t^{2}-139230t^{4}+319770t^{6}-323323t^{8}+119301t^{10})$
$\displaystyle
k_{4}(t)=-\dfrac{218243025}{262144}t(6435-170170t^{2}+1322685t^{4}-4476780t^{6}+7436429t^{8}-5965050t^{10}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\
\ \ +1847475t^{12})$ $\displaystyle
k_{5}(t)=\dfrac{43648605}{1048576}t(-328185+11316305t^{2}-116660817t^{4}+540571185t^{6}-1301375075t^{8}+$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \ \
+1691091675t^{10}-1125112275t^{12}+300540195t^{14}).\qquad$

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