Abel’s lemma
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 $n=0,\mathrm{\dots},N$, let ${A}_{n}$ be the partial sum ${A}_{n}={\sum}_{i=0}^{n}{a}_{i}$. 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 here (http://planetmath.org/ProofOfAbelsLemmaByInduction). The result can be found in [1] (Exercise 3.3.5).
If the sequences are indexed from $M$ to $N$, we have the following variant:
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 $n=M,\mathrm{\dots},N$, let ${A}_{n}$ be the partial sum ${A}_{n}={\sum}_{i=M}^{n}{a}_{i}$. Then
$$\sum _{i=M}^{N}{a}_{i}{b}_{i}=\sum _{i=M}^{N-1}{A}_{i}({b}_{i}-{b}_{i+1})+{A}_{N}{b}_{N}.$$ |
Proof. By defining ${a}_{0}=\mathrm{\dots}={a}_{M-1}={b}_{0}=\mathrm{\dots}={b}_{M-1}=0$, we can apply Theorem 1 to the sequences ${\{{a}_{i}\}}_{i=0}^{N}$ and ${\{{b}_{i}\}}_{i=0}^{N}$. $\mathrm{\square}$
References
- 1 R.B. Guenther, L.W. Lee, Partial Differential Equations^{} of Mathematical Physics and Integral Equations, Dover Publications, 1988.
Title | Abel’s lemma |
---|---|
Canonical name | AbelsLemma |
Date of creation | 2013-03-22 13:19:49 |
Last modified on | 2013-03-22 13:19:49 |
Owner | mathcam (2727) |
Last modified by | mathcam (2727) |
Numerical id | 14 |
Author | mathcam (2727) |
Entry type | Theorem |
Classification | msc 40A05 |
Synonym | summation by parts^{} |
Synonym | Abel’s partial summation |
Synonym | Abel’s identity |
Synonym | Abel’s transformation |
Related topic | PartialSummation |