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CROSSWORD
ANSWER

superadd

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The answer SUPERADD has 3 possible clue(s) in existing crosswords.

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The word SUPERADD is VALID in some board games. Check SUPERADD in word games in Scrabble, Words With Friends, see scores, anagrams etc.

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Definitions of superadd in various dictionaries:

verb - to add further

SUPERADD - In mathematics, a sequence { an }, n ≥ 1, is called superadditive if it satisfies the inequality a ...

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Keep reading for additional results and analysis below.

Possible Crossword Clues
Drink time (two days) put on as extra
Swallow terrible dread and apply to be another extra?
Combine with total already reached

Last Seen in these Crosswords & Puzzles
Oct 26 2013 The Times - Concise
Jun 16 2010 The Times - Cryptic
Jan 16 2003 The Guardian - Cryptic crossword

Superadd might refer to
In mathematics, a sequence { an }, n ≥ 1, is called superadditive if it satisfies the inequality* * * * * a * * n * + * m * * * ≥ * * a * * n * * * + * * a * * m * * * * * {\displaystyle a_{n+m}\geq a_{n}+a_{m}} * for all m and n. The major reason for the use of superadditive sequences is the following lemma due to Michael Fekete.Lemma: (Fekete) For every superadditive sequence { an }, n ≥ 1, the limit lim an/n exists and is equal to sup an/n. (The limit may be positive infinity, for instance, for the sequence an = log n!.) * Similarly, a function f is superadditive if * * * * * f * ( * x * + * y * ) * ≥ * f * ( * x * ) * + * f * ( * y * ) * * * {\displaystyle f(x+y)\geq f(x)+f(y)} * for all x and y in the domain of f. * For example, * * * * f * ( * x * ) * = * * x * * 2 * * * * * {\displaystyle f(x)=x^{2}} * is a superadditive function for nonnegative real numbers because the square of * * * * ( * x * + * y * ) * * * {\displaystyle (x+y)} * is always greater than or equal to the square of * * * * x * * * {\displaystyle x} * plus the square of * * * * y * * * {\displaystyle y} * , for nonnegative real numbers * * * * x * * * {\displaystyle x} * and * * * * y * * * {\displaystyle y} * . * The analogue of Fekete's lemma holds for subadditive functions as well. * There are extensions of Fekete's lemma that do not require the definition of superadditivity above to hold for all m and n. There are also results that allow one to deduce the rate of convergence to the limit whose existence is stated in Fekete's lemma if some kind of both superadditivity and subadditivity is present. A good exposition of this topic may be found in Steele (1997).If f is a superadditive function, and if 0 is in its domain, then f(0) ≤ 0. To see this, take the inequality at the top. * * * * f * ( * x * ) * ≤ * f * ( * x * + * y * ) * − * f * ( * y * ) * * * {\displaystyle f(x)\leq f(x+y)-f(y)} * . Hence * * * * f * ( * 0 * ) * ≤ * f * ( * 0 * + * y * ) * − * f * ( * y * ) * = * 0 * * * {\displaystyle f(0)\leq f(0+y)-f(y)=0} * * The negative of a superadditive function is subadditive.

Superadd might refer to

In mathematics, a sequence { an }, n ≥ 1, is called superadditive if it satisfies the inequality*
*
*
*
* a
*
* n
* +
* m
*
*
* ≥
*
* a
*
* n
*
*
* +
*
* a
*
* m
*
*
*
*
* {\displaystyle a_{n+m}\geq a_{n}+a_{m}}
* for all m and n. The major reason for the use of superadditive sequences is the following lemma due to Michael Fekete.Lemma: (Fekete) For every superadditive sequence { an }, n ≥ 1, the limit lim an/n exists and is equal to sup an/n. (The limit may be positive infinity, for instance, for the sequence an = log n!.)
* Similarly, a function f is superadditive if
*
*
*
*
* f
* (
* x
* +
* y
* )
* ≥
* f
* (
* x
* )
* +
* f
* (
* y
* )
*
*
* {\displaystyle f(x+y)\geq f(x)+f(y)}
* for all x and y in the domain of f.
* For example,
*
*
*
* f
* (
* x
* )
* =
*
* x
*
* 2
*
*
*
*
* {\displaystyle f(x)=x^{2}}
* is a superadditive function for nonnegative real numbers because the square of
*
*
*
* (
* x
* +
* y
* )
*
*
* {\displaystyle (x+y)}
* is always greater than or equal to the square of
*
*
*
* x
*
*
* {\displaystyle x}
* plus the square of
*
*
*
* y
*
*
* {\displaystyle y}
* , for nonnegative real numbers
*
*
*
* x
*
*
* {\displaystyle x}
* and
*
*
*
* y
*
*
* {\displaystyle y}
* .
* The analogue of Fekete's lemma holds for subadditive functions as well.
* There are extensions of Fekete's lemma that do not require the definition of superadditivity above to hold for all m and n. There are also results that allow one to deduce the rate of convergence to the limit whose existence is stated in Fekete's lemma if some kind of both superadditivity and subadditivity is present. A good exposition of this topic may be found in Steele (1997).If f is a superadditive function, and if 0 is in its domain, then f(0) ≤ 0. To see this, take the inequality at the top.
*
*
*
* f
* (
* x
* )
* ≤
* f
* (
* x
* +
* y
* )
* −
* f
* (
* y
* )
*
*
* {\displaystyle f(x)\leq f(x+y)-f(y)}
* . Hence
*
*
*
* f
* (
* 0
* )
* ≤
* f
* (
* 0
* +
* y
* )
* −
* f
* (
* y
* )
* =
* 0
*
*
* {\displaystyle f(0)\leq f(0+y)-f(y)=0}
*
* The negative of a superadditive function is subadditive.

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