Has this phenomenon been discovered and named?

If $$x-\frac{x}{2}=\frac{x}{2},$$ and $$\frac{x}{\sqrt{x}}=\sqrt{x},$$ and $$x-\uparrow(x-\uparrow^22)=x-\uparrow^22$$ when $(x\uparrow^n-[A])\uparrow^nA=x$, where $A$ is some constant, and one uses standard Knuth up-arrow notation[where ($\uparrow^{-1}x$)($-\uparrow^{-1}x$), ($\uparrow^0x$)($-\uparrow^{0}x$), ($\uparrow{x}$)($-\uparrow{x}$), ($\uparrow^2x$), and ($-\uparrow^2x$) all mean addition(and subtraction), multiplication(and division), exponentiation(and roots), and tetration(and super-roots) in that order],will $$x-\uparrow^n(x-\uparrow^{n+1}2)$$ always yield $$x-\uparrow^{n+1}2$$

for every degree $n$? If so, could someone please let me know where I can find an explanation of this identity?

Another, possibly more mathematically concise definition could be given by $$x-\uparrow^n\sum_{k=1}^{z-1}{x-\uparrow^{n+1}z}=x-\uparrow^{n+1}z$$

where the sigma could signify the sum, product, etc. depending on the degree of the up-arrow. $x\uparrow^az$ could be the $z$-th hyperpower($a$-power) of $x$ ($x$ $(n+2)$-ated to the $z$-th), or inversely, $x-\uparrow^az$ is the $z$-th hyperroot of x. Or in a different notation (MphLee's),

$$Hrt_n(x,\sum_{k=1}^{z-1}{Hrt_{n+1}(x,z))}=Hrt_{n+1}(x,z)$$

Where the $\sum$ still doesn't nessecarily represent the sum, but $z-1$ iterations of hyperoperations of degree $n$, and rank $z$.

Ex. $$Hrt_{1}(x,\sum_{k=1}^{2}{Hrt_2(x,3)})$$

Here, $Hrt_1$ represents a difference between $x$ and $\sum_{k=1}^{2}{Hrt_2(x,3)}$. Sigma ($\sum$) is operating in the first degree; addition, or $H_1$ (note that although we are using $Hrt$ and not $H$, only the subscript or 'degree' matters when defining Sigma ($\sum$)), and is the sum of 2 ($\sum_{}^{2}$) identical terms: $$Hrt_2(x,3)$$ that is $\frac{x}{3}$. Therefore,$$Hrt_{1}(x,\sum_{k=1}^{2}{Hrt_2(x,3)})=x-((\frac{x}{3})_1+(\frac{x}{3})_2)=x-2(\frac{x}{3})=\frac{x}{3}$$.

I am fairly certain that this equation is 'clumsy' to some extent, but if anyone could suggest a better form of notation, I would greatly appreciate it.

Here is a better rendering: $$Hrt_n(x,H_{n+1}(Hrt_{n+1}(x,z),[z-1]))=Hrt_{n+1}(x,z)$$ and if $Hrt_{n+1}(x,z)=T$, then$$Hrt_n(x,H_{n+1}(T,[z-1]))=T$$

Solution 1:

Sorry but I'm really getting confused with your up-arrow's notation of the inverses (root-type inverse), I'll try to write this problem with this notation:

$H_n$ is the $n$-th hyperoperation and $Hrt_n$ is its hyper-root

$H_1(x,a)=x+a=y$ and its root is $$Hrt_1(y,a):=y-a=x$$

$H_2(x,a)=x\times a=y$ and its root is $$Hrt_2(y,a):=y/a=x$$

$H_3(x,a)=x^a=y$ and its root is $$Hrt_3(y,a):=\sqrt[a]{y}=x$$

$H_4(x,a)=x\uparrow \uparrow a=y$ (that is tetration) and its root is $$Hrt_4(y,a):=srrt_a(y)=x$$ ($a$-th super-root)

and if we have $H_n(x,a)=y$ than $$Hrt_n(y,a):=x$$

$\mathcal Y$our statement:

$$x-\uparrow^n(x-\uparrow^{n+1}2)=x-\uparrow^{n+1}2$$ or $$x-\uparrow^nr=r$$ and $r=x-\uparrow^{n+1}2$

It can now be written as

$$Hrt_n(x,Hrt_{n+1}(x,2))=Hrt_{n+1}(x,2)$$ or $$Hrt_n(x,r)=r$$ and $r=Hrt_{n+1}(x,2)$

Now, we know that for the definiton of hyper-root must hold $H_n(Hrt_n(x,a),a)=x$, where $H_n$ is the $n$-th hyperoperation.

so if we apply this to your formula we get

$Hrt_n(x,r)=r$

$H_n(Hrt_n(x,r),r)=H_n(r,r)$

$x=H_n(r,r)$

Since for all Hyperoperations ($n\ge 1$) we have $H_n(r,r)=H_{n+1}(r,2)$ and $r=Hrt_{n+1}(x,2)$

$x=H_n(r,r)=H_{n+1}(r,2)=H_{n+1}(Hrt_{n+1}(x,2),2)=x$

$\mathcal O$bservation:

Here in the proof of your statement I'm using some theorems that I have not proved here, and I have instead assumed them to be true:

I assumed that $H_n(r,r)=H_{n+1}(r,2)$ for $n\ge 1$ that is same as $H_n(r,H_{n+1}(r,0))=H_{n+1}(r,1)=r$ and we can see that holds:

$r+(r\times 0)=r$

$r\times(r^0)=r$

$r^{(r\uparrow \uparrow 0)}=r$

and that $H_n(Hrt_n(x,a),a)=Hrt_n(H_n(x,a),a)=x$

the last affermation we know that does not hold always, for example it does't hold for the sqare-root or the $a$-th root ($\sqrt[a]{y}$) when we are working with negative (or complex) numbers because the function $f(x)=H_n(x,a)=y$ is not invertible than $f^{-1}(y)=Hrt_n(y,a)$ is not a fucntion, but it works for the naturals.