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Height of Zero Tetration Problem
#1
Question 
Hello,

I have a question that I am really confused by that I hope someone can explain to me thoroughly. Also I am a Physics student so forgive my lack of understanding of mathematical rigor.

I am going to use the ASCII notation of tetration for explaining: a^^n

So everywhere I am reading I see that "a^^0 = 1". Intuitively, nothing seems wrong here because "a^^1 = a" and "a^^2 =a^a" and so forth so "a^^0 = 1" seems to be a logical result. Also there's a similarity with the exponentiation "a^0 = 1". Now, I thought why this is the case for exponentiation and I realized:
1) This makes the function a^x to be continuous around and at x=0 (at least for real 'a').
2) If you were to consider "a^k = y" and by algebraic manipulation "a = y^(1/k)" and take the limit of 'k' to 0 then you get "a^0=y" and "a = y^∞". You can see that the 2nd equation is satisfied for when y is approximately equal to 1 (from the right if a>1 and from the left if a<1) and "a^0=y≈1" follows from that. This means you can practically evaluate a^0 by taking an infinite root 'a'.

Now onto Tetration:
Here's the reason why a^^0=1 makes sense and this is because if you take the property of tetration: "a^^n = a^(a^^(n-1))" and set n=1 then you get "a^^1 = a^(a^^0)" and you want the answer to be "a^^1=a" and this requires that "a^^0=1".
BUT, taking my 2nd example for why a^0=1 in exponentiation - this doesn't work at all with tetration because if you do the following:
"a^^k = y" and "a = y^^(1/k)" and do the limit of k to zero to get "a^^0 = y" and "a = y^^∞". then we all know that the infinite tetration doesn't give infinity for all numbers greater than 1 and 0 for numbers less than one (like y^∞ would) but it converges to "y^^∞ = W(-ln(y))/ln(y)" for the range [e^(-e) , e^(1/e)]. Using that you get "a = W(-ln(y))/ln(y)" and whose inverse is written as "a^(1/a) = y". So this means that "a^^0=y=a^(1/a)" but not "a^^0=1". This can be verified by taking a very large super-root (i.e. a^^(1/∞)) and see that it does not converge to 1 for all numbers like the normal root (a^(1/∞)=1) would.

If we were to say that the property "a^^n = a^(a^^(n-1))" applies for non-integer values of 'n' (which according to Wikipedia says is necessary for an extension to real heights) then if we substitute n=1.5 for example then we have "a^^1.5 = a^(a^^(1.5-1)) = a^(a^^(0.5))" which CAN be evaluated by evaluating a^^(1/2) first. This can be generalized if we substitute "n = I + 1/x" where I and x are both integers and now we can evaluate 'almost' any positive non-integer range. It will look like this:
"a^^(I + 1/x) = a^a^..^a^(a^^(1/x))". Now if we set x=∞ so that "n = I+1/∞ = I" we get "a^^(I + 1/∞) = a^^(I + 0) = a^^n = a^a^..^a^(a^^(1/∞)) = a^a^..^a^(a^(1/a)) ≠ a^^n". So it becomes apparent that a^^n is a non-continuous function because there are discontinuities at every integer value (see image). This is of course unless the property "a^^n = a^(a^^(n-1))" doesn't hold for non-integer 'n' and that it is NOT a requirement for an extension to real heights. Note that using this method I was unable to calculate a^^(1.75) for example because a^(a^^(0.75)) and specifically a^^(0.75) is incalculable using Newton's method.

[Image: 4739312447_096718df48_b.jpg]

The blue line connects all the integer values of 'n' and the square data points are the ones evaluated using the above mentioned method. The dotted line is a guide. I used "a = 1.444" because it doesn't blow up to infinity for large values of 'n'.

So to me this all seems like some kind of paradox is happening but I hope that someone will be able to put this in a clear light.

Thank you.
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#2
(06/27/2010, 09:50 PM)moejoe Wrote: BUT, taking my 2nd example for why a^0=1 in exponentiation - this doesn't work at all with tetration because if you do the following:
"a^^k = y" and "a = y^^(1/k)"
...
So to me this all seems like some kind of paradox is happening but I hope that someone will be able to put this in a clear light.

I think your paradox emerges from the wrong assumption in your equations.
You dont get the inverse of f(x)=x^^k (i.e. the tetration root) by taking x^^(1/k).
This is the case for powers but not for tetration.

It works for x^k because we have for integer numbers.
If we extend this law to rational numbers we can set which means that is the inverse of , i.e. is the number such that .

All this consideration fails for tetration because we dont have the rule x^^(m*n)=(x^^m)^^n.
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#3
Also this thread discusses a similar issue.
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#4
Thank you for your reply and for the thread link.
It has been obvious to me that for tetrations: [Image: mimetex.cgi?{^a(^bx)}\neq{^{ab}x}] is a correct statement. But it has not been obvious to me that: [Image: n}}({^{n}x)\neq%20x] which from the thread is said to be the case because the above property (1st eq.) isn't an equality then you can not assume that [Image: n}}({^{n}x)=x]. Therefore this means that you can not define [Image: n}x] to be the "super-root of order 'n' " and consequentially one doesn't even know what [Image: n}x] defines (i.e. evaluates to).
ARE non-integer tetrations defined (not in interpolative methods that is)? Also without making any extra assumptions.

Btw, is there a better way to write TeX in here? I have to keep using "Insert Image" and referring to mimetex.cgi to do this.
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#5
(06/28/2010, 11:34 PM)moejoe Wrote: ARE non-integer tetrations defined (not in interpolative methods that is)? Also without making any extra assumptions.
Yes, this is the main discussion topic of this forum. Have a look at the last post in the FAQ thread in the "general" section (though they are not proper introductions into the topic, you may find it helpful).

Quote:Btw, is there a better way to write TeX in here? I have to keep using "Insert Image" and referring to mimetex.cgi to do this.

Ya, enclose it into
Code:
[tex][/tex]
.
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