Welcome to Anagrammer Crossword Genius! Keep reading below to see if hyperope is an answer to any crossword puzzle or word game (Scrabble, Words With Friends etc). Scroll down to see all the info we have compiled on hyperope.
hyperope
Searching in Crosswords ...
The answer HYPEROPE has 0 possible clue(s) in existing crosswords.
Searching in Word Games ...
The word HYPEROPE is VALID in some board games. Check HYPEROPE in word games in Scrabble, Words With Friends, see scores, anagrams etc.
Searching in Dictionaries ...
Definitions of hyperope in various dictionaries:
noun - a person with hyperopia
noun - a farsighted person
HYPEROPE - In mathematics, the hyperoperation sequence is an infinite sequence of arithmetic operations (called hyperoperations) that starts with the unary oper...
Word Research / Anagrams and more ...
Keep reading for additional results and analysis below.
Possible Dictionary Clues |
---|
A farsighted person, a hyperopic person, a person with hyperopia. |
a person with hyperopia a farsighted person |
Hyperope might refer to |
---|
In mathematics, the hyperoperation sequence is an infinite sequence of arithmetic operations (called hyperoperations) that starts with the unary operation of successor (n = 0), then continues with the binary operations of addition (n = 1), multiplication (n = 2), and exponentiation (n = 3), after which the sequence proceeds with further binary operations extending beyond exponentiation, using right-associativity. For the operations beyond exponentiation, the nth member of this sequence is named by Reuben Goodstein after the Greek prefix of n suffixed with -ation (such as tetration (n = 4), pentation (n = 5), hexation (n = 6), etc.) and can be written as using n − 2 arrows in Knuth's up-arrow notation. * Each hyperoperation may be understood recursively in terms of the previous one by:* * * * a * [ * n * ] * b * = * * * * * a * [ * n * − * 1 * ] * ( * a * [ * n * − * 1 * ] * ( * a * [ * n * − * 1 * ] * ( * ⋯ * [ * n * − * 1 * ] * ( * a * [ * n * − * 1 * ] * ( * a * [ * n * − * 1 * ] * a * ) * ) * ⋯ * ) * ) * ) * * ⏟ * * * * * b * * * copies of * * * a * * * * , * * n * ≥ * 2 * * * {\displaystyle a[n]b=\underbrace {a[n-1](a[n-1](a[n-1](\cdots [n-1](a[n-1](a[n-1]a))\cdots )))} _{\displaystyle b{\mbox{ copies of }}a},\quad n\geq 2} * It may also be defined according to the recursion rule part of the definition, as in Knuth's up-arrow version of the Ackermann function: * * * * * a * [ * n * ] * b * = * a * [ * n * − * 1 * ] * * ( * * a * [ * n * ] * * ( * * b * − * 1 * * ) * * * ) * * , * * n * ≥ * 1 * * * {\displaystyle a[n]b=a[n-1]\left(a[n]\left(b-1\right)\right),\quad n\geq 1} * This can be used to easily show numbers much ... |