• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 exponential distributivity bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/22/2008, 12:36 PM quickfur Wrote:I forgot to note, what I had in mind here is $(ab)^c = a^{c}b^c$ and $(a+b)c = ac + bc$.Oh right, misinterpreted that. Another idea would then be to investigate the distribution law for exponentiation, i.e. whether there is an operation # such that (a^b)#c=(a#c)^(b#c). As a side node I give a proof that there are only trivial operations # satisfying this property. Note however that on the just discussed hierarchy of operations {n}, there is always distributivity from {n+1} over {n}, but exponentiation does not belong to this hierarchy. Now to the proof, if there would be such an operation then we can consider the function f(x)=x#c (fixed c) which satisfies f(a^b)=f(a)^f(b). Then: f(a^(b^n))=f((...(a^b)^b...)^b)=f(a)^(f(b)^n) but also f(a^(b^n))=f(a)^f(b^n)=f(a)^(f(b)^f(n)) hence f(a)^(f(b)^f(n))=f(a)^(f(b)^n) then for f(a)!=1, f(a)>0 f(b)^f(n)=f(b)^n and again for f(b)!=1, f(b)>0 f(n)=n f(a)^m=f(a^m)=f((a^(m/n))^n)=(f(a)^f(m/n))^n=f(a)^(f(m/n)n), then for f(a)!=1, f(a)>0: m=f(m/n)n f(m/n)=m/n If we now suppose that f is continuos we extend this law from the fractional to the positive real numbers: f(x)=x for all positive real x. This was however under the assumption that there is an $a>0$ with $f(a)\neq 1$. So the other possibility is: f(x)=1 for all positive real x. Summarizing Proposition: Any operation # defined on the positive real numbers such that (a^b)#c=(a#c)^(b#c) (for each a,b,c) and such that f(x)=x#c is continuous (for each c), is either given by x#c=x or by x#c=1 for all x,c. quickfur Junior Fellow Posts: 22 Threads: 1 Joined: Feb 2008 02/22/2008, 06:51 PM bo198214 Wrote:[...] Another idea would then be to investigate the distribution law for exponentiation, i.e. whether there is an operation # such that (a^b)#c=(a#c)^(b#c). As a side node I give a proof that there are only trivial operations # satisfying this property. [...] Now to the proof, if there would be such an operation then we can consider the function f(x)=x#c (fixed c) which satisfies f(a^b)=f(a)^f(b). [...] Summarizing Proposition: Any operation # defined on the positive real numbers such that (a^b)#c=(a#c)^(b#c) (for each a,b,c) and such that f(x)=x#c is continuous (for each c), is either given by x#c=x or by x#c=1 for all x,c.Hmm, interesting. I was going to ask what happens if we consider f(x)=c#x instead, but it doesn't change the conclusion because we assume the same operator #. So we established that a higher operation cannot distribute over exponentiation without becoming trivial. I wonder what is the root cause of this... maybe it has something to do with the non-associativity of exponentiation? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/22/2008, 07:17 PM quickfur Wrote:Hmm, interesting. I was going to ask what happens if we consider f(x)=c#x instead, but it doesn't change the conclusion because we assume the same operator #. So we established that a higher operation cannot distribute over exponentiation without becoming trivial. I wonder what is the root cause of this... maybe it has something to do with the non-associativity of exponentiation? Who knows, but the proof can be similarly done for other distributivity type functional equations, for example $f(xy)=f(x)f(y)$ with the only continuous solutions $f(x)=x^c$ or $f(x)=0$ $f(x+y)=f(x)+f(y)$ with the only continuous solutions $f(x)=cx$. quickfur Junior Fellow Posts: 22 Threads: 1 Joined: Feb 2008 02/23/2008, 12:29 AM That's interesting, because analysing f(x+y) = f(x)f(y) leads to the exponential function $e^x$. Unfortunately, we don't seem to have such a handy property involving tetration that we can use, except for $f(x+1)=b^{f(x)}$, which isn't very helpful for non-integer x. Now that I think of it, I think the root of the problem is that exponentiation is non-associative, so there's no easy way to algebraically "access the top of the exponential tower", so to speak. In order to derive any useful relations, we need to have some way of "reaching the top of the tower". For example, given b[4]n, if we can somehow reach the top of the tower and add another tower of height m, then we can state the property that J(b[4]n, b[4]m) = b[4](n+m), where the hypothetical J operator attaches the second tower to the top of the first tower. But I doubt that J is expressible as an algebraic operation (I'm not even sure if it can be consistently defined if b is not fixed!). And so even if we could make such a statement, it wouldn't be of the same utility as $f(x+1)=b^{f(x)}$ w.r.t. exponentiation. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 09/22/2011, 03:27 PM (This post was last modified: 09/22/2011, 03:34 PM by JmsNxn.) (02/23/2008, 12:29 AM)quickfur Wrote: That's interesting, because analysing f(x+y) = f(x)f(y) leads to the exponential function $e^x$. Unfortunately, we don't seem to have such a handy property involving tetration that we can use, except for $f(x+1)=b^{f(x)}$, which isn't very helpful for non-integer x. Now that I think of it, I think the root of the problem is that exponentiation is non-associative, so there's no easy way to algebraically "access the top of the exponential tower", so to speak. In order to derive any useful relations, we need to have some way of "reaching the top of the tower". For example, given b[4]n, if we can somehow reach the top of the tower and add another tower of height m, then we can state the property that J(b[4]n, b[4]m) = b[4](n+m), where the hypothetical J operator attaches the second tower to the top of the first tower. But I doubt that J is expressible as an algebraic operation (I'm not even sure if it can be consistently defined if b is not fixed!). And so even if we could make such a statement, it wouldn't be of the same utility as $f(x+1)=b^{f(x)}$ w.r.t. exponentiation. I'd like to respectfully disagree. consider: $\mu (x) =\, ^x \mu$ $a \,\otimes_\mu\,b = \mu(\mu^{-1}(a) + \mu^{-1}(b))$ it's very easy to see that: $^c \mu \, \otimes_\mu\,^d \mu = \,^{c+d} \mu$ I think even, this operator may give some short cuts for algebra involving tetration. I did a little research into this under the following thread: http://math.eretrandre.org/tetrationforu...hp?tid=699 The only rule is if we create: $\phi (x) = \,^x \phi$ $a \,\otimes_\phi \,b = \phi(\phi^{-1}(a) + \phi^{-1}(b))$ $\otimes_\mu \neq \otimes_\phi$ whereas when we create the same operator with exponentiation instead of tetration, the operators are equivalent and both are multiplication. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Math overflow question on fractional exponential iterations sheldonison 4 3,740 04/01/2018, 03:09 AM Last Post: JmsNxn An exponential "times" table MikeSmith 0 1,829 01/31/2014, 08:05 PM Last Post: MikeSmith exponential baby Mandelbrots? sheldonison 0 2,048 05/08/2012, 06:59 PM Last Post: sheldonison help with a distributivity law JmsNxn 3 3,777 09/22/2011, 03:32 PM Last Post: JmsNxn Base 'Enigma' iterative exponential, tetrational and pentational Cherrina_Pixie 4 9,749 07/02/2011, 07:13 AM Last Post: bo198214 Two exponential integrals Augustrush 2 4,401 11/10/2010, 06:44 PM Last Post: Augustrush HELP NEEDED: Exponential Factorial and Tetrations rsgerard 5 8,357 11/13/2009, 02:27 AM Last Post: rsgerard Exponential factorial mike3 3 5,964 10/07/2009, 02:04 AM Last Post: andydude Additional super exponential condition bo198214 4 5,380 10/21/2008, 03:40 PM Last Post: martin exponential polynomial interpolation Gottfried 3 7,028 07/16/2008, 10:32 PM Last Post: andydude

Users browsing this thread: 1 Guest(s)