Notation For limitation written on the cover an afterward circle box letter (①, ②, ③…I shall express) in entering circle parenthesis letter (n ).
I express (2) (a b = a (2) b ), 冪乗演算子 with the top by (1) (a + b = a (1) b ) with the top, a multiplication operator by an addition operator in (3) (a a (3) b ) with the top, and cock-up (n) (a (n ) b ) which generalized them in a non-negative integer general n is a hyper operator.
3 variable function hyper which assumed hypern , n expressing them in a function form a variable is defined. hyper1 multiplies addition, hyper2 by, and hyper3 is 冪乗, and, in hyper4, teth military ration (tetration), hyper5 are called ヘキセーション (hexation) ... by ペンテーション (pentation), hyper6 more.
The examples of n = 0-4 are as follows.
hyper0 ( a , b ) = hyper ( a , 0 , b ) = a ( 0 ) b = b + 1 hyper1 ( a , b ) = hyper ( a , 1 , b ) = a ( 1 ) b = a + b hyper2 ( a , b ) = hyper ( a , 2 , b ) = a ( 2 ) b = a b hyper3 ( a , b ) = hyper ( a , 3 , b ) = a ( 3 ) b = a b = a ↑ b hyper4 ( a , b ) = hyper ( a , 4 , b ) = a ( 4 ) b = b a = a ↑↑ b = a a ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a a ⏟ b copies of a {\displaystyle {\begin{aligned}\operatorname {hyper0} \left(a,b\right)=\operatorname {hyper} \left(a,0,b\right)=a^{\left(0\right)}b=&b+1\\\operatorname {hyper1} \left(a,b\right)=\operatorname {hyper} \left(a,1,b\right)=a^{\left(1\right)}b=&a+b\\\operatorname {hyper2} \left(a,b\right)=\operatorname {hyper} \left(a,2,b\right)=a^{\left(2\right)}b=&ab\\\operatorname {hyper3} \left(a,b\right)=\operatorname {hyper} \left(a,3,b\right)=a^{\left(3\right)}b=&a^{b}=a\uparrow b\\\operatorname {hyper4} \left(a,b\right)=\operatorname {hyper} \left(a,4,b\right)=a^{\left(4\right)}b=&\,^{b}a=a\uparrow \uparrow b=\underbrace {a^{a^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{a^{a}}}}}}}}}}}}}}}}}}}}}} _{b{\text{ copies of }}a}\end{aligned}}} hyper0 assumes it a latter function (an operand first a is ignored) of an operand second b . But I may use other definitions.
n > In the case of 4, I establish it as follows. This is n > In the case of 1, it is managed in all, but is not managed with n = 1.
hyper n ( a , b ) = hyper ( a , n , b ) = a ( n ) b = a ( n − 1 ) a ( n − 1 ) ⋯ ( n − 1 ) a ⏟ b copies of a {\displaystyle \operatorname {hyper} n\left(a,b\right)=\operatorname {hyper} \left(a,n,b\right)=a^{\left(n\right)}b=\underbrace {a^{\left(n-1\right)}a^{\left(n-1\right)}\cdots ^{\left(n-1\right)}a} _{b{\text{ copies of }}a}}
with other notation of relationships For n ≥3, the next relations are managed between arrow notation of Knuth and chain notation of Conway .
hyper ( a , n , b ) = a ( n ) b = a ↑ n − 2 b = a → b → ( n − 2 ) when n ≥ 3 {\displaystyle \operatorname {hyper} (a,n,b)=a^{(n)}b=a\uparrow ^{n-2}b=a\rightarrow b\rightarrow (n-2)\quad {\mbox{ when }}n\geq 3} In addition, for n ≥1, the next relations are managed between an expansion operator (Jonathan Bowers' Extended Operator) of Bower.
hyper ( a , n , b ) = a ( n ) b = a ⟨ n ⟩ b when n ≥ 1 {\displaystyle \operatorname {hyper} (a,n,b)=a^{(n)}b=a\langle n\rangle b\quad {\mbox{ when }}n\geq 1}
Recursive definition I can define it recursiv ely as follows. An exception handling at the age of b = 0 is careful about different things by n .
hyper ( a , n , b ) = a ( n ) b = { b + 1 , if n = 0 a , if n = 1 , b = 0 0 , if n = 2 , b = 0 1 , if n ≥ 3 , b = 0 a ( n − 1 ) ( a ( n ) ( b − 1 ) ) otherwise {\displaystyle \operatorname {hyper} (a,n,b)=a^{(n)}b={\begin{cases}b+1,&{\mbox{if }}n=0\\a,&{\mbox{if }}n=1,b=0\\0,&{\mbox{if }}n=2,b=0\\1,&{\mbox{if }}n\geq 3,b=0\\a^{(n-1)}(a^{(n)}(b-1))&{\mbox{otherwise}}\end{cases}}}
Expansion to a real number b which expanded 冪乗 in an exponential function , the natural expansion to a real number of n are not accomplished.
Hyper operator with the bottom Because it is more than n ≥3 (冪乗), and an associative law is not managed, priority from the right be determined,
hyper ( n + 1 ) ( a , b ) = a ( n + 1 ) b = a ( n ) a ( n ) ⋯ ( n ) a ( n ) a ⏟ b copies of a = a ( n ) ( a ( n ) ⋯ ( n ) ( a ( n ) a ) ⋯ ) ⏟ b copies of a {\displaystyle \operatorname {hyper} \left(n+1\right)\left(a,b\right)=a^{\left(n+1\right)}b=\underbrace {a^{\left(n\right)}a^{\left(n\right)}\cdots ^{\left(n\right)}a^{\left(n\right)}a} _{b{\text{ copies of }}a}=\underbrace {a^{\left(n\right)}\left(a^{\left(n\right)}\cdots ^{\left(n\right)}\left(a^{\left(n\right)}a\right)\cdots \right)} _{b{\text{ copies of }}a}} である.
In contrast, I can express operation from the left with priority by making a hyper operator lower with. In other words,
hyper n + 1 ( a , b ) = a ( n + 1 ) b = ( ⋯ ( a ( n ) a ) ( n ) ⋯ ( n ) a ) ( n ) a ⏟ b copies of a {\displaystyle \operatorname {hyper} _{n+1}\left(a,b\right)=a_{\left(n+1\right)}b=\underbrace {\left(\cdots \left(a^{\left(n\right)}a\right)^{\left(n\right)}\cdots ^{\left(n\right)}a\right)^{\left(n\right)}a} _{b{\text{ copies of }}a}} である.
But can easily express the hyper n +1 operator with the bottom using hyper n operator;, for example
a ( 4 ) b = a ( 3 ) a ( 3 ) ( b − 1 ) = a a ( b − 1 ) {\displaystyle a_{\left(4\right)}b=a_{\left(3\right)}a_{\left(3\right)}\left(b-1\right)=a^{a^{\left(b-1\right)}}} Not operation new essentially that is (from 冪乗法則), I do not have many uses of the hyper operator with the bottom.
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