• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Attempt to formally generalize log, exp functions to 3,4,5..(n,m) log exp Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/01/2008, 01:54 PM (This post was last modified: 06/01/2008, 02:22 PM by Ivars.) Logarithms were invented in response to the need to deal numerically and later analytically with numbers that were too big/too small for calculation abilities of that time. Similar situation arises when dealing with tetration and higher operations today. One can try to generalize the usefulness of exp, log functions to increase/reduce order of operations by 1. We denote summation with [1], multiplication with [2], exponentiation with [3] and tetration with [4] etc. Let us call normal logarithm 2_log, and normal exponentiation 2_exp. Then: 2_exp(a+b) = 2_exp(a[1]b)=2_exp(a)*2_exp(b) = 2_exp(a)[2]2_exp(b) 2_log (a*b) = 2_log(a)+2_log(b) = (2_log(a))[1](2_log(b)) We can form now more logarithms and exponetiations, requiring: 3_log(a^b) =3_log(a[3]b)= 3_log(a)*3_log(b) = 3_log(a)[2]3_log(b) 4_log(a[4]b)= 4_log(a)[3]4_log(b) = (4_log(a))^(4_log(b)) n_log(a[n]b)=n_log(a)[n-1] n_log(b) This 3 log will have the same value for a^b, and b^a which would only be true for roots of equation a^b=b^a. Otherwise, 3_log must have either sign difference or index to indicate order of a and b. Inversely, 3_exp, 4_exp and n_exp: 3_exp(a*b) =3_exp(a[2]b) = (3_exp(a))^(3_exp(b)) = 3_exp(a)[3]3_exp(b) 4_exp(a^b) = 4_exp(a[3]b)=4_exp(a)[4]4_exp(b) n_exp(a[n-1]b)=n_exp(a)[n]n_exp(b) Since operations above and including tetration are all too fast for numerical analysis, n_log defined in such way as above would not help much, but we could look for (3, 2) - log which reduces the order of infinity by 2 and generally for log that reduces order of infinity by as much as we need (e.g. - m). So far we stick to integer (n,m) values of orders of infinity since iteration to intermediate should be possible once the properties of functions are figured out. (3, 2)_log (a^b) = (3, 2)_log (a[3]b)= (3,2)_log (a)+(3,2)_log (b)= (3,2)_log(a)[1](3,2)_log(b) (4, 2)_log (a[4]b) = (4, 2)_log (a[4]b)= (4,2)_log (a)*(4,2)_log (b)= (4,2)_log(a)[2](4,2)_log(b) In general: (n,m)_log(a[n]b)=(n,m)_log(a) [n-m] (n,m)_log(b) (n,m)_exp(a[n]b)=(n,m)_exp(a)[n+m] (n,m)_exp(b) We may ask if there exists natural basis for such logarithms and exponentials, series expansion of some kind, is it related to n-factorials, and how to deal with noncommmutative/non-associative and multivalue properties of these functions arising from that.The development of the nature of such basis itself, (even if only to jump over one order of infinity, but more so over 2 or more) should be very informative. After that we may form a 2-d table of few available values arising from (n,m) as corresponding natural basis and some values of numbers in those basis along 3rd d. But most useful would probably be the possibility to have rules for dealing with those functions analogous to rules dealing with log ,exp based on calculus. Which might mean most likely generalization of calculus and infinitesimal analysis-similarly how its basic form appeared soon after logarithms were invented. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/01/2008, 03:01 PM (This post was last modified: 06/01/2008, 03:08 PM by bo198214.) Lots of ideas however mostly not working, for example: Ivars Wrote:3_log(a^b) =3_log(a[3]b)= 3_log(a)*3_log(b) = 3_log(a)[2]3_log(b) Assume there would be a function $f$ with $f(x^y)=f(x)f(y)$ Then surely $f(x)f(x^n)=f(x^{x^n})=f((...((x^x)^x)\dots)^x)=f(x)^{n+1}$ So if there is an $x$ with $f(x)\neq 0$ we have $f(x^n)=f(x)^n$. for $y=x^n$ we also have: $f(y^{1/n})=f(x)=f(y)^{1/n}$ and both together we get $f(x^{m/n})=f(x)^{m/n}$ If we assume that $f$ is continuous then this is valid not only for rationals but also for reals $\alpha$: $f(x^\alpha)=f(x)^\alpha$. No let $x=c$ constant and let $\alpha$ be variable: $f(c^t)=f( c)^t$ $f(x)=f(c^{\log_c(x)})=f( c)^{\log_c(x)}=x^{\log_c(f( c))}=x^d$ So $f(x)=x^d$ for some constant $d$. But then: $(x^y)^d=f(x^y)=f(x)f(y)=x^d y^d$ if $d\neq 0$: $x^y=xy$ which can not be satisfied for all $x$, $y$. Hence either $d=0$ or $f(x)=0$ which we excluded in our previous considerations. Proposition. Let $f$ be a continuous function defined on the positive reals such that $f(x^y)=f(x)f(y)$ for all $x,y$, then either $f(x)=0$ for all $x$ or $f(x)=1$ for all $x$. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/01/2008, 05:02 PM (This post was last modified: 06/02/2008, 10:22 AM by Ivars.) Yes.. good exercise to understand that it can not be so simple if these are normal continuos functions, which they of course may be not, or continuity idea itself needs to be changed to include some (un)limited jumps; But then of course I have no idea how to do it right now. Full stop, and much better idea after good training how to work with hyperop notation Either f(x) = 0 or f(x) = 1 for all x.... hmm. If that function one could oscillate between those 2 values, it would than be 1 non-continuous function that does the job. Such infinitely oscillating function would also be interesting in terms of providing ultimate discretization for any continuous function since it should oscillate faster than there is difference between 2 real numbers, meaning the period is biggest real number, (or, perhaps, straight to imaginary frequency) ? So that every continuous real function when multiplied (composed?) by this function would be turned in 2 valued function, like bifurcation. Then, I would like to extend this notion to functions oscillating fast between n constant values and applying such function to continuous function would make it n-furcating, etc. Also, we were looking with Gottfried at behaviour of convergence of certain iterations that exibit simple oscillating, triangular, polygonial apprach to limit point instead of smooth spiralling. There should be a link between those beahviours , n-furcations and oscillating functions. Anyway, if there is a reason for chaos, it lies in higher speed operations invisible with normal mathematics being hidden in dx and continuity -which is questionable (the continuity ) in my opinion.It is only continuity defined within a set of certain operations. If we move outside those operations, continuity can not be taken for granted, as what seemed continuos when acting on it with slow operations, might prove discontinuos when amplified fast enough. If tetration transforms some function let us say y=x into discontinuous domains, than the question was y=x truly continuous in all range of x is open. ( just considering variuos behaviours of h(x) at various domains of x)) . If function h(x) which is just a function applied to x and established certain relationship between sets of values of h(x) and R, and changes its behaviour with jumps just because x changes values "continuosly" , then perhaps x itself (reals) is not continuous, or , at least, should not be considered linearly growing. So, the function or relationship is continuous, but numbers are not. But I have to read more about Fourier transform and some chaos theory to see where is the clue. Ivars andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 06/02/2008, 04:46 PM (This post was last modified: 06/02/2008, 04:46 PM by andydude.) Ivars Wrote:3_log(a^b) =3_log(a[3]b)= 3_log(a)*3_log(b) = 3_log(a)[2]3_log(b) This is where it starts to break down, because this is not a normal logarithm. Normal logarithms satisfy $\log(a^b) = b\log(a)$ Ivars Wrote:3_exp(a*b) =3_exp(a[2]b) = (3_exp(a))^(3_exp(b)) = 3_exp(a)[3]3_exp(b) Since this is not the definition of logarithms, we should really separate out the 'log' notation from the equations: $\log(a^b) = \log(a) b$ $f(a^b) = f(a) f(b)$ Using these equations we can evaluate at the following points: If $a=b^{1/b}$: $f(b) = f(b^{1/b}) f(b)$ $1 = f(b^{1/b})$ then we find that f(a) is a constant. If a=1: $f(1) = f(1) f(b)$ then f(b) is a constant. If b=1: $f(a) = f(a) f(1)$ $1 = f(1)$ then we find the point f(1)=1. If b=0: $f(1) = f(a) f(0)$ $1 = f(a) f(0)$ which means f(a) is a constant. So, not only can we show that f(x) is a constant, but we can show it in multiple ways, which is a very bad sign. I think this is proof that no such function can exist (or the only solution to all of these equations is f(x)=1). Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/02/2008, 07:15 PM (This post was last modified: 06/02/2008, 08:59 PM by Ivars.) So the solution to f(x^y) =f(x)*f(y) is that there does not exist a contionuos real function defined on reals>=0 such that f(x) is not 1 or 0 for all x. Right? Then the only option is to look for more interesting solution outside ( or rather, in between) real numbers. If between any 2 real numbers there is a certain (e.g. infinite) amount of e.g. hyperreal numbers, than, for the purposes of this function, one can try to decide what function of hyperreals would satisfy this equation. What is required is that such function has to have values 0 and 1 for all reals. So for x=real , x>=0 f(x) = 1 or 0. We can choose 2 hyperreal values infinitely close to any real number x- e.g on both sides of it- so that one of them h1(x) < x would give value f(h1(x))=0, while other h2(x)>x would give value of function f(h2(x))=1. Then we have a function f( h1(x), h2(x)) = 0 if arg=h1(x), 1 if arg=h2(x).) Since h1(x) and h2(x) is infinitely close to x, in reals this function falls apart into 2 separate functions, or one that is not quite continuous. In this case, f becomes a function only of hyperreals as it is 2 constant functions reals. We can use either discrete function which oscillates between 1 and 0 like f(h(x)) = 1,0,1,0,1,0,1 for any 2 consequtive hyperreals in interval between 2 reals, or we can form a one to one correspondence between all hyperreal interval between 2 neighbouring real numbers and Pi/2 and then f(h(x)) = sin (h(x)) where h1(x) = pi/2 and h2(x) = 0. This correspondence between hyperreals and pi/2 is something I can not handle yet. Has it ever been shown what is the cardinality of hyperreals and what is the cardinality of pi/2? Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/03/2008, 07:14 AM (This post was last modified: 06/03/2008, 07:18 AM by Ivars.) To put it shorter: Can we construct a function defined on all set of hyperreals or subset of them ( or superreals )which would preserve the property: REAL f( subset of hyperreals) = such that f(h(x)^h(y)) = f( h(x)) * f(h(y)) for all x, y>=0 Where h(x), h(y) is the subset of hyperreals relevant to the problem and the same for all x, y- what ever x, y, the subset of hyperreals we can use to construct such function does not change. This requires that it is possible to establich one to one correspondence between subset of hyperreals and real numbers, or, at least, 2 real numbers 1 and 0. In my simple thinking, it requires only 2 hyperreals besides each real to do that- as indicated in previous post. I wonder also how to extend this to hyper imaginaries as there is not reason they can not exist, so they must exist. If true, such function would perform the necessary transformation required from 3_log. If so , that would allow simple generalization to n_log, (n,m)_log by just adding more scales of numbers , like 2_hyperreals which are formed by 2 neighbouring hyperreals , between them, 3_hyperreals formed from 2 neighbouring 2_hyperreals etc. The nice thing is, surreal numbers of Conway are constructed exactly in such way, so that field has been already covered very deeply, with all ordinalities , cardinalities calculated and proved. Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/03/2008, 08:09 AM (This post was last modified: 06/03/2008, 08:15 AM by bo198214.) andydude Wrote:So, not only can we show that f(x) is a constant, but we can show it in multiple ways, which is a very bad sign. I think this is proof that no such function can exist (or the only solution to all of these equations is f(x)=1). Yeah, I see its even much simpler to prove than I did, and we can also drop the demand of continuity. However your proof(s) omits the case of $f(x)=0$. So I will add the detail: If $f(x^y)=f(x)f(y)$ for f being defined on $[1,\infty)$, then $f(x)=f(x)f(1)$ for $y=1$. So if there is at least one $x$ with $f(x)\neq 0$ (i.e. $f$ is not identically 0) then $f(1)=1$. Let now $x=1$ in our original equation: $f(1)=f(1)f(y)$. With $f(1)=1$ we get $f(y)=1$ for all $y$. So either $f=1$ or $f=0$. @Ivars: this proof also works for hyperreals. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/03/2008, 08:49 AM (This post was last modified: 06/03/2008, 08:51 AM by Ivars.) bo198214 Wrote:@Ivars: this proof also works for hyperreals. Based on "language analogy" between real and hyperreal sets? I will look into this deeper before making any new statements. At least, as a first step, is it possible to construct such discontinuos function f(x) on real numbers which in each point $x_o$ is 0 if the point is approached from left, and 1 if approached from right( or vice versa)? So: $\lim_{x\to\\{-x_o}} f(x_o)= 1,$ $\lim_{x\to\\{+x_o}} f(x_o)= 0,$ Alternatively : $\lim_{x\to\\{-x_o}} f(x_o)= 0,$ $\lim_{x\to\\{+x_o}} f(x_o)= 1,$ Can we define a discontinuous function in such manner? The one use of such function as a function of $x_o$ would be for any $x_o$ to discern from which side we are approaching it. If we approach $x_o$ from left, this function $f(x_o)$ has value e.g. 0 (alternatively 1) if we approach $x_o$ from right ,this function $f(x_o)$ has value e.g. 1 (alternatively 0) Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/03/2008, 01:13 PM Ivars Wrote:At least, as a first step As a first step to what? The proof shows that there is no such function, not even a discontinuous one. Quote:$\lim_{x\to\\{-x_o}} f(x_o)= 0,$ $\lim_{x\to\\{+x_o}} f(x_o)= 1,$ Thats not difficult if it is only for one point $x_0$. Just define $f(x)=0$ for $x and $f(x)=1$ for $x>x_0$ and $f(x_0)$ as you like. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/03/2008, 01:22 PM (This post was last modified: 06/03/2008, 01:35 PM by Ivars.) bo198214 Wrote:Thats not difficult if it is only for one point $x_0$. Just define $f(x)=0$ for $x and $f(x)=1$ for $x>x_0$ and $f(x_0)$ as you like. No, for all real points $x_0$. $f(x_0)$ remains arbitrary for time being, just a. May be some a is better than other. If existing , I wonder if such function has connections to Kronecker Delta, Heaviside Step, and Dirac Delta functions. Ivars « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post The AB functions ! tommy1729 0 1,650 04/04/2017, 11:00 PM Last Post: tommy1729 the inverse ackerman functions JmsNxn 3 6,146 09/18/2016, 11:02 AM Last Post: Xorter Look-alike functions. tommy1729 1 2,135 03/08/2016, 07:10 PM Last Post: hixidom Inverse power tower functions tommy1729 0 1,957 01/04/2016, 12:03 PM Last Post: tommy1729 [2014] composition of 3 functions. tommy1729 0 1,858 08/25/2014, 12:08 AM Last Post: tommy1729 Intresting functions not ? tommy1729 4 5,324 03/05/2014, 06:49 PM Last Post: razrushil generalizing the problem of fractional analytic Ackermann functions JmsNxn 17 22,856 11/24/2011, 01:18 AM Last Post: JmsNxn Discrete-analytic functions Ansus 4 5,855 07/30/2011, 04:46 PM Last Post: tommy1729 product functions tommy1729 5 6,646 06/01/2011, 05:38 PM Last Post: tommy1729 Periodic functions that are periodic not by addition JmsNxn 0 2,837 04/17/2011, 09:54 PM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)