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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$$Hrt_n(x,H_{n+1}(T,[z-1]))=T$$

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  • $\begingroup$ Can we take the definition of $x\uparrow^2 -[z]=y$ to be $y\uparrow^n+[z] = x$ ... ??? If that is the definition, and it makes sense, then this is a simple consequence. $\endgroup$ – GEdgar May 15 '13 at 21:41
  • $\begingroup$ If what you meant to say was $(x\uparrow^{n}-[z])\uparrow^{n}+[z]$, and $(y\uparrow^{n}+[z])\uparrow^{n}-[z]$, still assuming that $z$ is a constant and not a variable, then yes. But only as long as they satisfy $f(x)=x$ and $f(y)=y$ respectfully. $\endgroup$ – Moving Massive May 16 '13 at 2:10
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    $\begingroup$ @MovingMassive In my opinion I must tell you that your new notation is less clear that your first one, In fact it was not wrong (just not .. "offical" let's say, even if there is not really an official notation for the inverse like root and log en.wikipedia.org/wiki/Hyperoperation) but for example if $x^a=x \uparrow a=y$ we can write $\sqrt[a]{y}=y\uparrow -[a]$ or so we have that $y \uparrow^n -[a]$ is the $a$-th hyperroot of degree (level, or usually called rank) $n$ of $y$. $\endgroup$ – MphLee May 16 '13 at 17:23
  • $\begingroup$ @MovingMassive The thing that I find confusing is the sign $-$ that remember me the minus (It is a problem only for my mind, so don't worry). Anyways I find more clear use ,for example, $x^a=H_3(x,a)$ and $\sqrt[a]{y}=Hrt_3(y,a)$. but You can uses too other notation like $\sqrt[a]{y}=y \downarrow^3 a$ and this can make you free from the $-$ minus sign: $y \downarrow^n a$ is the $a$-th hyperrot of $y$, and $(x \uparrow^n a)\downarrow^n a=x$ and is pretty confortable at the beginning. Is becomes source of "problems" when you have hyperrots and hyperlogarithms in the same formulas. $\endgroup$ – MphLee May 16 '13 at 17:30
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    $\begingroup$ @MovingMassive For a nice notation about Hyperoperations chech this math.eretrandre.org/tetrationforum/attachment.php?aid=208 $\endgroup$ – MphLee May 16 '13 at 18:38
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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.

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  • $\begingroup$ @MovingMassive does this answer your question first question? $\endgroup$ – MphLee May 16 '13 at 21:21
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    $\begingroup$ Yes, thank you. Your notation is also very easy to understand, I might begin to use more letter symbols in my personal writing to avoid confusion. $\endgroup$ – Moving Massive May 17 '13 at 16:19

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