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 Height of Zero Tetration Problem moejoe Junior Fellow Posts: 2 Threads: 1 Joined: Jun 2010   06/27/2010, 09:50 PM 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. 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. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 06/28/2010, 03:26 AM (This post was last modified: 06/28/2010, 03:46 AM by bo198214.) (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 $x^{m\cdot n} = (x^m)^n$ for integer numbers. If we extend this law to rational numbers we can set $x=x^{\frac{1}{n} n} = (x^{\frac{1}{n}})^n$ which means that $x^{\frac{1}{n}$ is the inverse of $x^n$, i.e. $y=x^{\frac{1}{n}}$ is the number such that $y^n = x$. All this consideration fails for tetration because we dont have the rule x^^(m*n)=(x^^m)^^n. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 06/28/2010, 06:24 AM Also this thread discusses a similar issue. moejoe Junior Fellow Posts: 2 Threads: 1 Joined: Jun 2010 06/28/2010, 11:34 PM Thank you for your reply and for the thread link. It has been obvious to me that for tetrations: ${^a(^bx)}\neq{^{ab}x}$ is a correct statement. But it has not been obvious to me that: ${^{1/n}}({^{n}x)\neq x$ 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 ${^{1/n}}({^{n}x)=x$. Therefore this means that you can not define ${}^{1/n}x$ to be the "super-root of order 'n' " and consequentially one doesn't even know what ${}^{1/n}x$ 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. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 06/29/2010, 03:44 AM (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:. « Next Oldest | Next Newest »

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