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\title{A Remark on Power Series Estimation via Boundary Corrections
with Parameter}
\author{W. Solak}

\pagestyle{myheadings}

\markright{APIC'95, El Paso,
Extended Abstracts,
A Supplement to the international journal of {\rm Reliable
Computing}\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ }

\maketitle

  
\auffil{The author is with the Institute of Mathematics, Academy of
Mining and Metallurgy, Al. Mickiewicza 30, 30-059 Krak\'ow, Poland.}

\section{Formulation of the Problem}
Many numerical methods are based on the fact that the 
desired result $s$ can be represented as
a sum of an infinite series$$s:=\sum_{n=1}^\infty a_n:\eqno{(1)}$$ 
Taylor series, Fourier series, etc. Such series are typical, e.g., in
physical applications. Due to this representation, we can
approximate the desired result 
$s$ with a finite sum $$\sum_{n=1}^N a_n;$$ the larger $N$,
the better the approximation. The majority of the existing methods are
{\it heuristic:} they either fix $N$ from the very beginning, or
choose $N$ for which $a_N$ becomes sufficiently small. These heuristic
methods, however, do not guarantee any accuracy of the resulting
estimate. It is therefore desirable to design estimates with a
guaranteed estimate. 

If we use a finite sum with some $N$ as an estimate for $s$, then the
error of this estimate can also be represented as a sum of the
infinite series:
$$\sum_{n=N+1}^\infty a_n.$$ Therefore, if we have a method of
estimating the sum of an infinite series, then this same method will
enable us to estimate the error of the $N-$term approximation. 
One way to estimate the sum of the series is to take into
consideration that a series can be viewed as a quadrature for an
integral over an infinite domain: 
$$I(f)=\int_1^\infty f(x)\,dx\eqno{(2)}$$for some
function $f:[1,\infty)\to R$ 
for which $f(n)=a_n$ for all $n$. Therefore, if for a
given series, we know an explicitly integrable function $f(x)$ with
this property, then we can take the value $I(f)$ of the integral 
as an estimate for $s$. 
How accurate is this estimate?

\section{Main Idea}
In this paper, we describe a way to estimate the accuracy of this
estimate $I(f)$. This method (based on formulas from \cite{2,3,4}) 
will be applicable for the case when the
function $f(x)$ satisfies the following properties: 
\begin{itemize}
\item $f(x)$ is {\it monotonic}. Since the sum (1) converges, it means that
$f$ is either positive and decreasing, or
negative and increasing; 
\item 
$f$ must be {\it smooth}, namely, for some $\bar\beta>0$, $f$ is four times
differentiable on the semi-line $[1-\bar\beta,\infty)$; 
\item the fourth derivative
$f^{IV}(x)$ of the function $f$ 
must be different from 0 for all $x$ from this semi-line; and
\item the third derivative $f^{III}(x)$ must 
have a finite limit as $x\to \infty$. 
\end{itemize}

\section{Main Result}
If $f^{IV}(x)>0$ for all $x\in[1-\bar\beta,\infty)$, then for every
$\bar\beta\in (0,1]$ and for every $\beta\in [\sqrt{6}/8,1]$, we get
the following estimates:
$${a_1\over 2}+\int_1^\infty f(x)\,dx-$$
$${1\over 24\beta}[-3f(1)+4f(1+\beta)-f(1+2\beta)]\le s\le$$
$${a_1\over 2}+\int_1^\infty f(x)\,dx-{1\over
24\bar\beta}[f(1+\bar\beta)-f(1-\bar\beta)].\eqno{(3)}$$

If $f^{IV}(x)<0$, then we get a similar inequality, but with the
right-hand side instead of the left-hand side and vice versa. 

\section{Proof}
The desired formula (3) follows from the following two quadrature
formulas: 

The first formula was proven in \cite{2,3}:
\newpage 
$$\int_a^b f(x)\, dx=T_{N+1}[f]+$$
$${h\over 24\bar\beta}
[f(a+\bar\beta h)-f(a-\bar\beta h)-f(b+\bar\beta h)+f(b-\bar\beta h)]$$
$$+R^{\bar\beta}_{N+5}[f],$$
where $$R^{\bar\beta}_{N+5}[f]={1+10\bar\beta^2\over
720}(b-a)h^4f^{IV}(\xi)$$ for some $\xi\in(a,b)$. 

The second formula
was proven in \cite{3,4}:
$$\int_a^b f(x)\, dx=T_{N+1}[f]+$$ $${h\over 24\beta}
[-3f(a)+4f(a+\beta h)-f(a+2\beta h)-3f(b)+$$ $$4f(b-\beta h)-f(b-2\beta
h)]$$ 
$$+{\tilde R}^{\beta}_{N+5}[f],$$
where $${\tilde R}^{\beta}_{N+5}[f]={h^5\over
720}[30\beta^3+N(1-20\beta^2)]f^{IV}(\delta)$$ for some $\delta\in
(a,b)$ (this formula is only true if $\beta\in [\sqrt{6}/8,1]$). 

We apply these formulas for $a=1$, $b=N$, and $h=1$. Then, the integral
sum $T_{N+1}(f)$ is equal to $a_1+a_2+...+a_N$.

For our choice of $\beta$, we have $1-20\beta^2\le 0$. Therefore,  
if the fourth derivative is always positive, then $R\ge 0$ and (for
sufficiently large $N$), $\tilde R\le 0$. So, if we let $N$ go to
$\infty$, and take into consideration that $f(x)\to 0$ as
$x\to\infty$, we get the desired inequalities. 

If $f^{IV}<0$, then we get the opposite inequalities. 
\smallskip

\noindent{\it Comment.} We can also get the explicit expressions for
the difference between $s$ and its estimates from (3). 
Namely, the Peano kernels $K_{...}^{...}$ 
of these quadrature formulas are of opposite signs,
and $f^{IV}\ne 0$. Therefore, we can apply the mean value theorem to
the integrals that describe the remainders of these formulas: 
$$R^{\bar\beta}_{N+5}[f]=\int_{a-\bar\beta h}^{b+\bar\beta
h}K^{\bar\beta}_4(x)\,dx=$$
$$=K^{\bar\beta}_4(\alpha)[f^{III}(b+\bar\beta h)-f^{III}(a-\bar\beta
h)]$$ for some $\alpha\in (a-\bar\beta h,b+\bar\beta h)$. Similarly, 
$${\tilde R}^{\beta}_{N+5}[f]=\int_{a}^{b}
{\tilde K}^{\beta}_4(x)\,dx=$$
$$={\tilde K}^{\bar\beta}_4(\gamma)[f^{III}(b)-f^{III}(a)]$$
for some $\gamma\in (a,b)$. When $N\to\infty$, these terms turn into 
$$R^{\bar\beta}_{\infty}[f]=
{\nu\over 24}\big({1\over 16}+{\bar\beta^2\over 3}\big)
[f^{III}(1-\bar\beta)-\lim_{x\to\infty}f^{III}(x)]\eqno{(4)}$$ for some $\nu\in
[0,1]$, and 
$${\tilde R}^{\bar\beta}_\infty[f]=
-\mu{\beta^2\over 36}
[f^{III}(1)-\lim_{x\to\infty}f^{III}(x)]\eqno{(5)}$$ for some $\mu\in
[0,1]$. These terms are always non-positive and non-negative.
Therefore, 
in the limit $N\to\infty$, the above-described quadrature formulas
lead to the desired inequalities.

\section{Examples}
\subsection{Example 1}
Let us estimate the value $\zeta(3)$ of the Riemann zeta function
$$\zeta(3)=\sum_{n=1}^\infty {1\over n^3}$$ (cf. \cite{1}, p. 232). 
As a first approximation to the sum, we will take the sum of
its first nine terms. Then, to get the estimate for the sum itself,
we must estimate the remainder $$s=\sum_{n=10}^\infty {1\over
n^3}=\sum_{n=1}^\infty{1\over (9+n)^3}.$$ From (3), we conclude that 
$${1\over 2\cdot 10^3}+\int_1^\infty{dx\over (9+x)^3}-$$ $${1\over
24\beta}[-{3\over 10^3}+{4\over (10+\beta)^3}-{1\over
(10+2\beta)^3}]\le s\le $$
$${1\over 2\cdot 10^3}+\int_1^\infty{dx\over (9+x)^3}-{1\over
24\bar\beta}[{1\over 10+\bar\beta)^3}-{1\over 10-\bar\beta)^3}].$$
If we choose  
$\beta=0.5$ and $\bar\beta=0.1$, the for 
$$\zeta(3)=s+\sum_{n=1}^9{1\over n^3}$$ we get an estimate $$1.2020567\le
\zeta(3)\le 1.2020571,$$ i.e., an estimate with an accuracy $\le
4\cdot 10^{-7}$ (actually, in this case,
$${\tilde R}^\beta_\infty=\mu\cdot 4.16\cdot 10^{-7}$$ and 
$$R^{\bar\beta}_\infty=-\nu\cdot 1.65\cdot 10^{-7}.)$$ 

\subsection{Example 2}
Estimate $$s=\sum_{n=1}^\infty{(-1)^n\over\sqrt{n}}\eqno{(6)}$$ (cf.
\cite{1}, p. 243). In this case, the values $a_n$ are not monotonic.
So, to apply our method, we must represent the sum of this series in
the desired form. This can be done by combining together neighboring
terms of opposite sign. 

We can start this combination with the very
first term, and thus get
$$s=\sum_{n=1}^\infty{(-1)^n\over \sqrt{n}}=\sum_{n=1}^\infty
\big(-{1\over \sqrt{2n-1}}+{1\over\sqrt{2n}}\big),$$ so that 
$$a_n=-{1\over \sqrt{2n-1}}+{1\over\sqrt{2n}},$$ 
$$f(x)=-{1\over \sqrt{2x-1}}+{1\over\sqrt{2x}},$$ 
and $$f^{IV}(x)=-3\cdot 5\cdot
7\cdot(2x-1)^{-9/2}+3\cdot 5\cdot 7\cdot (2x)^{-9/2}>0$$ for all
$x>1/2$. 

We can also
start combining with the second term, in which case we end up with 
$$s=\sum_{n=1}^\infty{(-1)^n\over \sqrt{n}}=-1+
\sum_{n=1}^\infty\big({1\over \sqrt{2n}}-{1\over\sqrt{2n+1}}\big),$$ so that 
$$a_n={1\over \sqrt{2n}}-{1\over\sqrt{2n+1}},$$ 
$$f(x)={1\over \sqrt{2x}}-{1\over\sqrt{2x+1}},$$ 
and $f^{IV}(x)=3\cdot 5\cdot
7\cdot(2x)^{-9/2}-3\cdot 5\cdot 7\cdot (2x+1)^{-9/2}>0$ for all $x>0$. 

Applying an estimate with $\bar\beta=0.01$ to both formulas, 
we get $$-0.6164982\le s\le -0.6037183.$$ Applying an estimate with
$\beta=0.5$, we get $$-0.6048987\le s\le -0.59707716.$$ Combining
these two inequalities, we get the following estimate for $s$:
$$-0.6048987\le s\le -0.6037183.$$ 

To get a better estimate, we compute the first nine terms separately, and
estimate the remainder:
$$s=\sum_{n=1}^\infty{(-1)^n\over\sqrt{n}}=\sum_{n=1}^9{(-1)^n\over\sqrt{n}}+
\sum_{n=5}^\infty\big({1\over \sqrt{2n}}-{1\over\sqrt{2n+1}}\big).$$
For the remaining sum, for $\bar\beta=0.1$ and $\beta=\sqrt{6}/8$, we
get 
$${\tilde R}^\beta_\infty=\mu\cdot 1.4304291\cdot 10^{-6}$$ and 
$$R^{\bar\beta}_\infty=-\nu\cdot 3.690692\cdot 10^{-6}.$$ 

\begin{thebibliography}{99}
\bibitem{1}
G. Schmeisser and H. Schirmeier, {\bf Practiche Mathematik}, De
Gruger, Berlin, N.Y., 1976.

\bibitem{2}
W. Solak, ``On quadrature formulas with end correction'', {\it Journal
of Computational and Applied Mathematics}, 1989, Vol. 25, pp.
373--377.

\bibitem{3}
W. Solak, ``Boundary corrections for numeric integration formulas'',
{\it Opuscula Math.}, 1991, Vol. 8. 

\bibitem{4}
W. Solak and Z. Szydelko, ``Quadrature rules with Gregory-Laplace end
correction'', {\it Journal of Computational and Applied Mathematics},
1991, Vol. 30, pp. 251--253. 

    
\end{thebibliography}


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