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Difference between revisions of "Abel inequality"

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An estimate for the sum of products of two numbers. If sets of numbers $(a_k)$ and $(n_k)$ are given such that the absolute values of all sums $B_k = b_1 + \cdots + b_k$, $k=1,2,\ldots$, are bounded by a number $B$, i.e. $|B_k| \le B$, and if either $a_i \ge a_{i+1}$ or $a_i \le a_{i+1}$, $i = 1,2,\ldots,n-1$, then
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An estimate for the sum of products of two numbers. If sets of numbers $(a_k)$ and $(b_k)$ are given such that the absolute values of all sums $B_k = b_1 + \cdots + b_k$, $k=1,\ldots,n$, are bounded by a number $B$, i.e. $|B_k| \le B$, and if either $a_i \ge a_{i+1}$ or $a_i \le a_{i+1}$, $i = 1,2,\ldots,n-1$, then
 
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\left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B(|a_1| + 2|a_n|)
 
\left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B(|a_1| + 2|a_n|)
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[[Category:Real functions]]

Latest revision as of 12:13, 9 November 2014

An estimate for the sum of products of two numbers. If sets of numbers $(a_k)$ and $(b_k)$ are given such that the absolute values of all sums $B_k = b_1 + \cdots + b_k$, $k=1,\ldots,n$, are bounded by a number $B$, i.e. $|B_k| \le B$, and if either $a_i \ge a_{i+1}$ or $a_i \le a_{i+1}$, $i = 1,2,\ldots,n-1$, then $$ \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B(|a_1| + 2|a_n|) $$

If the $a_k$ are non-increasing and non-negative, one has the simpler estimate: $$ \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B a_1 \ . $$


Abel's inequality is proved by means of the Abel transformation.

How to Cite This Entry:
Abel inequality. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Abel_inequality&oldid=34410
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article