40A05-AbelsLemma.tex

\documentclass[12pt]{article}
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\pmcanonicalname{AbelsLemma}
\pmcreated{2013-03-22 13:19:49}
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\pmowner{mathcam}{2727}
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\pmtitle{Abel's lemma}
\pmrecord{14}{33843}
\pmprivacy{1}
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\pmtype{Theorem}
\pmcomment{trigger rebuild}
\pmclassification{msc}{40A05}
\pmsynonym{summation by parts}{AbelsLemma}
\pmsynonym{Abel's partial summation}{AbelsLemma}
\pmsynonym{Abel's identity}{AbelsLemma}
\pmsynonym{Abel's transformation}{AbelsLemma}
\pmrelated{PartialSummation}

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{\bf Theorem 1 } Let 
$\{a_i\}_{i=0}^N$ and $\{b_i\}_{i=0}^N$ be sequences of 
real  (or complex) numbers with $N\ge 0$. 
For $n=0,\ldots, N$, let $A_n$ be the  partial sum
$A_n=\sum_{i=0}^na_i$.
Then 
$$\sum_{i=0}^N a_i b_i = \sum_{i=0}^{N-1}A_i(b_i-b_{i+1})+A_N b_N.$$

In the trivial case, when $N=0$, then sum on the right hand side
should be interpreted as identically zero. In other words, 
if the upper limit is below the lower limit, there is no summation. 

An inductive proof can be found \PMlinkname{here}{ProofOfAbelsLemmaByInduction}. 
The result can be found in \cite{guenther} (Exercise 3.3.5).

If the sequences are indexed from $M$ to $N$, we have the following 
variant:

{\bf Corollary}
Let $\{a_i\}_{i=M}^N$ and $\{b_i\}_{i=M}^N$ be sequences of 
real  (or complex) numbers with $0\le M \le N$. 
For $n=M,\ldots, N$, let $A_n$ be the  partial sum
$A_n=\sum_{i=M}^na_i$.
Then 
$$\sum_{i=M}^N a_i b_i = \sum_{i=M}^{N-1}A_i(b_i-b_{i+1})+A_N b_N.$$

\emph{Proof.} By defining
$a_0=\ldots =a_{M-1}=b_0=\ldots =b_{M-1} =0$, we can apply Theorem 1
to the sequences $\{a_i\}_{i=0}^N$ and $\{b_i\}_{i=0}^N$.
$\Box$
 
\begin{thebibliography}{9}
\bibitem{guenther}
R.B. Guenther, L.W. Lee,
\emph{Partial Differential Equations of Mathematical Physics and Integral Equations},
Dover Publications, 1988.
\end{thebibliography}
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