multivalued functions bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-02-03, 22:47 (This post was last modified: 2008-02-03, 22:48 by bo198214.) I am posting this as an excerpt of a private e-mail from GFR, who hopefully will join our forum soon: – We can build a “Theory of Numbers”, starting from the positive integers and trying to give them a “group” structure via the definition of a binary operation “+”, and a unit/neutral element “0”. We immediately (so to say) discover that, for justifying the group structure, we need to include within the postulates, the definition of “negative integers”, for allowing the inverse operation. If we go ... further, we may proceed defining the “multiplication group” structure around operation “x” and unit element “1” that, automatically, generates the exigency of defining the rational numbers (including the Z set), for allowing the inverse operation. For similar (but ... more complicated) reasons, the introduction of the “^” operation obliges us to define the set of real and complex numbers, even without reaching the “group structure”, for allowing the two inverse operations. In fact, the “^” operation is non-commutative and the “unit element” cannot be found. – Having said that (in a .... very informal way), we may observe that the implementation of any ... step (rank) in the hyper-operation hierarchy, dramatically increases the set of postulates of the theory, for justifying the return (inverse) operations. This postulate set is probably infinite-countable and Prof. Goedel would have been happy to read this. – My problem is that I was always convinced that hyper-operation rank s=3 (exponentiation), not only opened the door to complex (including real) numbers, but also opened ... the window to something that we might call “multiple” or “fibrated(?)” numbers. In fact, I am convinced that the square root of 4 is, indeed, a “double number”: -2 and +2. The fact of indicating it by plus/minus-sqrt(4) is orthodox (and perhaps also ... catholic, i.e. universal ) but does not satisfies my ... dirty mind. Moreover, we might write cbrt(8) = 2 and cbrt(-8) = -2, but we also know that we have, in both cases, two other well defined complex numbers, as solutions of this operation. In fact, the square roots of a number are two and the cubic roots of a number are three. Of course, this is very well known (the problem of the multiple complex roots of number 1). But this is often the subject of some superficially-deep specialist considerations, which we may very well leave out of the ... main window, if we limit ourselves to the ... reality. – Actually, equation x^n - a = 0 is a particular case of a general algebraic equation of n degree. The “General Theorem of Algebra” says that the multiplicity of solutions of an algebraic equation of n degree is n. In other words, the number of solutions of x^n = a must exactly be n. For example, the four solutions of the fourth root of 16 are: +2, +i2, -2, -i2, which we might put as 4th-rt(16) = {2,i2,-2,-i2}.The way of writing it is not relevant. The tricky (but not ... fuzzy!) idea behind is that, if the fourth root of 16 must be a number (one number), well, this number has to be “multiple”, with multiplicity 4. It is not a question of “choosing” (I know, this is the official point of view), there is nothing to be chosen, it’s just like that. But, this point of view implies the definition of multiple (fibrated? matrix?) numbers. Well ..., you know, a Civilization that was able to swallow number “i” can do much more than that. We just have to check if it is worth while! – Now (and here I risk to become ... crazy), we know that sqrt(4) = 4^(1/2). Then, we might put 4^(1/2) = {-2,2}, with multiplicity 2. But, how about 4^(1/e)? Or, better (worse?) we know that y = 4^x is a very civilized exponential function, with base 4, very “near” to the shape of y = e^x, mother and sister of all of us (analytic, fully derivable and with y = y’ = y”, and so on). Nevertheless (there is always a ... never-the-less), how to represent its complete behaviour, showing its multiplicities for x = 1/n? Or (again), if x is irrational transcendent (for instance, but not exclusively, < 1), what can we do? Ggggghhhrrrrrhhhoooofff ... ! ;-) – Last but not least, let us consider y = [x->+oo] lim (sin(x)). Of course, it doesn’t exist, it is indeterminate. Nevertheless (ops ...) it must be a value limited between -1 and +1, i.e. something like y = [-1, 1], a ... continuou-multiple (or limited-undetermined number. What a mess, ... in my mind! Best regards. Gianfranco bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-02-03, 22:53 (This post was last modified: 2008-02-03, 23:06 by bo198214.) GFR Wrote:The tricky (but not ... fuzzy!) idea behind is that, if the fourth root of 16 must be a number (one number), well, this number has to be “multiple”, with multiplicity 4. It is not a question of “choosing” (I know, this is the official point of view), there is nothing to be chosen, it’s just like that. But, this point of view implies the definition of multiple (fibrated? matrix?) numbers. Well ..., you know, a Civilization that was able to swallow number “i” can do much more than that. We just have to check if it is worth while! Of course you can extend the number domain to contain sets of numbers. There you can then define sqrt to yield a set of two numbers. Next we have to define the operations on those sets, this could also be done in a natural way for example {1,5} + {3,4} := {1+3,1+4,5+3,5+4}={4,5,8,9}. How you have to handle multivalued functions for example sqrt. Perhaps you would proceed as in the following example: $\sqrt{\{4,9\}}=\{-2,2,-3,3\}$. But now - by constructing so - you get some really odd problems. I mean you surely want to have a * b / b = a, but have a look at the example: {1,5}*{2,3}/{2,3}={1*2,1*3,5*2,5*3}/{2,3} = {2/2,3/2,10/2,15/2,2/3,3/3,10/3,15/3} which is surely not {1,5}, though it contains {1,5}. Or even more simple you surely want to be $\sqrt{x^2}={sqrt{x}}^2=x$. But this is also not true: $\sqrt{\{4\}^2}=\sqrt{\{16\}}=\{+4,-4\}\neq\{4\}$ ${sqrt{\{4\}}}^2 = \{+2,-2\}^2 = \{(+2)^2,(-2)^2\} = \{4\}$ this time it is satisfied. And I really can not imagine any proper number system where this (i.e. that we can calculate with non-bijective functions as if they were bijective) would work. Quote:Now (and here I risk to become ... crazy), we know that sqrt(4) = 4^(1/2). Then, we might put 4^(1/2) = {-2,2}, with multiplicity 2. But, how about 4^(1/e)? Or, better (worse?) we know that y = 4^x is a very civilized exponential function, with base 4, very “near” to the shape of y = e^x, mother and sister of all of us (analytic, fully derivable and with y = y’ = y”, and so on). Nevertheless (there is always a ... never-the-less), how to represent its complete behaviour, showing its multiplicities for x = 1/n? Or (again), if x is irrational transcendent (for instance, but not exclusively, < 1), what can we do? Ggggghhhrrrrrhhhoooofff ... ! ;-) The function $a^x=e^{(\ln(a)+2\pi k)x}$ is (including $a=e$) rather a cloud than a function. It has infinite many values for $x$ being irrational. But the single valued function (as we know it, e.g. $e^x$, $4^x$) is the only continuous function through that cloud. For example for $a=4$ this continuous function goes through $4^{1/2}=2$ and not through $4^{1/2}=-2$. You could not make it continous if you would chose $4^{1/2}=-2$. GFR Junior Member Posts: 6 Threads: 0 Joined: Feb 2008 Reputation: 0 2008-02-06, 14:42 Hi Henryk! Concerning the multiplication of "multiple" numbers, of course, we have to better precise the "traffic rules". In my opinion, we should put: {1,5}*{2,3} = {{(1*2),(5*2)},{(1*3),(5*3)}} = {{2,10},{3,15}}. It is complicate to write it, but it is so. I am not teaching this to a Reihenalgebra Specialist. If we proceed like that, we have: {1,5}*{2,3}/{2,3} = {{(1*2),(5*2)},{(1*3),(5*3)}} / {2,3} = {{(1*2),(5*2)}/{2},{(1*3),(5*3)}/{3}} = {{1,5},{1,5}} = {1,5} So far so good. Nothing new under the sun. We live in the best possible world. Concerning the non-commutability of the sqrt and ^2 operators, the problem is neither depending on my proposals, nor on Reihenalgebra. It is just a thing due to a deep mysterious property of such operators and to a very old and tricky algebraic notation problem. What you are saying is normally (and quickly) written like that: $\sqrt{4^2}=\sqrt{16}=+4,-4$ ${sqrt{4}}^2 = 4$ Normally (so to say) the problem is bypassed by using the plus/minus sign in front of the sqrt. It is practical, like in the famous formula for finding the two roots of a second degree equation. But it is without any possible serious meaning. Like if someboy was saying: you have two solutions x_1 and x_2, choose among them! There is nothing to choose. The solutions are just two (real and different, real and coincident or complex-conjugate). Concernin the old good exponential, are you meaning that we have two interpretations of te exponential function?. If I understand what you say, there should be the "legal" function, which is the one that we know from the textbooks and another one, including it, but also another "hidde one" including an infinite number of infinite-countable values of y (for x rational) and of infinite-uncountable values of y (for x irrational). What a ... cloudy world ! Do you also mean that x = -2 is not a solution of y = sqrt(4)? How can we then explain that (-2)x(-2) = 4? Here I really need a cup of coffee. Don't worry, I passed all my mathematical exams, with the satisfaction of all my professors. I can easily ... simulate, if I have to. My Father was always saying: "La matematica non è un'opinione" (Mathematics is not an opinion). He was wrong!! Ciao Henryk! See you soon. GFR PS - By the way, how many stars do you have? bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-02-07, 17:11 GFR Wrote:In my opinion, we should put: {1,5}*{2,3} = {{(1*2),(5*2)},{(1*3),(5*3)}} = {{2,10},{3,15}}. It is complicate to write it, but it is so. I am not teaching this to a Reihenalgebra Specialist. If we proceed like that, we have: {1,5}*{2,3}/{2,3} = {{(1*2),(5*2)},{(1*3),(5*3)}} / {2,3} = {{(1*2),(5*2)}/{2},{(1*3),(5*3)}/{3}} = {{1,5},{1,5}} = {1,5} So far so good. Nothing new under the sun. We live in the best possible world. Hm, but this is probably not what we want, as multiplication and addition is no more commutative here: {1,5}*{2,3}={{2,10},{3,15}} {2,3}*{1,5}={{2*1,3*1},{2*5,3*5}}={{2,3},{10,15}} Constructing new algebraic systems is a tricky thing, I have my experiences with it Quote:Concernin the old good exponential, are you meaning that we have two interpretations of te exponential function?. If I understand what you say, there should be the "legal" function, which is the one that we know from the textbooks and another one, including it, but also another "hidde one" including an infinite number of infinite-countable values of y (for x rational) and of infinite-uncountable values of y (for x irrational). What a ... cloudy world ! Yes, we have the good old function $b^x$, but we also have the functions $x^a$. Though uniquely defined on the reals the latter functions have branches on the sliced complex plane. $x^{1/3}$ has 3 branches $x^{\sqrt{2}}$ has infinitely many branches. $b^x$ is the only continuous function through the cloud of argument-value-pairs $\{ (x,{b^x}_k) : k\in\mathbb{Z},x\in\mathbb{C}\}$, where ${b^x}_k$ denotes the $k$-th branch of $f(b)=b^x$. Quote:Do you also mean that x = -2 is not a solution of y = sqrt(4)? How can we then explain that (-2)x(-2) = 4? I mean that x=-2 is not a value of the function sqrt and hence is $-2\neq \sqrt{4}$. However of course is $x=-2$ a solution of $x^2=4$. And it is well known that $\{\sqrt{y},-\sqrt{y}\}$ is the set of solutions of $x^2=y$. Quote:My Father was always saying: "La matematica non è un'opinione" (Mathematics is not an opinion). He was wrong!! But this only regards foundation of mathematics and definitions! Quote:PS - By the way, how many stars do you have? 7? GFR Junior Member Posts: 6 Threads: 0 Joined: Feb 2008 Reputation: 0 2008-02-08, 09:39 bo198214 Wrote:Hm, but this is probably not what we want, as multiplication and addition is no more commutative here: {1,5}*{2,3}={{2,10},{3,15}} {2,3}*{1,5}={{2*1,3*1},{2*5,3*5}}={{2,3},{10,15}}Yes, of course! But it is rare to have to multiply among them, for instance, ... the roots of two second degree equations, without saying in which order you have to do that. My proposal would suppose an order, unless ... otherwise specified. But, you are a champion of that stuff! bo198214 Wrote:I mean that x=-2 is not a value of the function sqrt and hence is $-2\neq \sqrt{4}$. However of course is $x=-2$ a solution of $x^2=4$. And it is well known that $\{\sqrt{y},-\sqrt{y}\}$ is the set of solutions of $x^2=y$.Say it again, ... please ?!?! So, indeed, you have seven stars. How lucky you are .... ! GFR bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-02-08, 14:57 And as I now see you rascal even propose obvious nonsense! : Quote:{{(1*2),(5*2)}/{2},{(1*3),(5*3)}/{3}} = {{1,5},{1,5}} = {1,5} So we have neither commutativity nor a*b/b=a. There are not much usable laws left in this structure, are there? Quote:bo198214 Wrote:I mean that x=-2 is not a value of the function sqrt and hence is $-2\neq \sqrt{4}$. However of course is $x=-2$ a solution of $x^2=4$. And it is well known that $\{\sqrt{y},-\sqrt{y}\}$ is the set of solutions of $x^2=y$. Say it again, ... please ?!?! Wot? Sqrt is a (singlevalued) function and with the help of this function you can express the set of solutions of an quadratic equation. Interestingly if you consider a certain set polynomials they are bijective on the positive numbers. This set of polynomials $P$ is built recursively by 1. We put $x$ into $P$. 2. Whenever there are two functions $f$ and $g$ in $P$ then we also put the function $f+g$ and $f\cdot g$ into $P$. For example $x,2x,nx,x^2,nx+x^2\in P$ where $n$ is a positive integer. Generally this set is the set of all polynomials with integer coefficients $\ge 0$ where the first coefficient must be nonzero. The cool thing is now if you add or multiply two strictly increasing functions they remain strictly increasing. This means we can add a 3rd rule in our creation process: 3. Whenever $f\in P$ then also $f^{-1}\in P$ And still keep all resulting functions bijective. For example $x\in P$ then by rule 2 $f(x)=x+x^2+x^3+x^4+x^5$ is in $P$. This function is strictly increasing on the positive reals, and hence bijective. So there exist $f^{-1}$ again by rule 2, we have also $g(x)=x^2+ f^{-1}(x)$ in $P$ which is again strictly increasing, continuous and heads towards 0 for $x\to 0$ and hence has again an inverse function, and so on. We see that on those polynomials on the positive real numbers the topic of multiple roots does not even occur. Quote:So, indeed, you have seven stars. How lucky you are .... ! Yes, very lucky, if you can prescribe yourself 7 stars, instead of laboriously earn them *ggg* (I mean its not even prescribing, its just the default!). But be reassured that I left out my stars on the tetration forum, so that you can have more of them there *gg*. GFR Junior Member Posts: 6 Threads: 0 Joined: Feb 2008 Reputation: 0 2008-02-09, 13:53 Ref. your kind and amusing observation: bo198214 Wrote:And as I now see you rascal even propose obvious nonsense! : {{(1*2),(5*2)}/{2},{(1*3),(5*3)}/{3}} = {{1,5},{1,5}} = {1,5} So we have neither commutativity nor a*b/b=a. There are not much usable laws left in this structure, are there?Sorry! I didn't mean that. Let us see: {(1*2),(5*2)}/{2}} = {(2*1),(2*5)}/{2}} = {(1*2)/2,(5*2)/2} = {(2*1)/2,(2*5)/2} = {1,5} as well as all other commutations in the multiplications. Then: {{(1*2),(5*2)}/{2},{(1*3),(5*3)}/{3}} = {{(1*2)/2,(5*2)/2},{(1*3)/3,(5*3)3} = {{1,5},{1,5}} = {1,5} Actually, operation "(...*...)" i.e. multiplication, will remain commutative. Notation (operation ?) "{... , ...}" will not be commutative. The pairs are "oriented". So: (a*b) / a = (b*a) / a = a but: {a,b} / a = {1, b/a}, while {b,a} / a = {b/a, 1}. In fact, I supposed that {a,b} would determine an order in the pair. But, I don't insist. It was just an idea in the air. Bye, dear Bo(ss) ! GFR bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-02-09, 15:09 Now i am really confused, lets see whether we can fix this. I thought you use the {,} in the common way, i.e. as finite sets. If you mean an ordered pair or list you should use (,). However I am not sure what should be the order of solutions of an equation. Another possibility would also be the use of multisets, i.e. where one element can occur multiple times, as corresponding to the multiplicities of the roots of polynomials. I would propose to write multisets like [a,b,c]. Ok, but for now I still assume finite sets {a,b,c}. Quote:{(1*2),(5*2)}/{2}} = {(2*1),(2*5)}/{2}} = {(1*2)/2,(5*2)/2} = {(2*1)/2,(2*5)/2} = {1,5} Here is an unmatched bracket (red)! And you dont obey your own rule which was {a,b}x{c,d}={{axc,bxc},{bxc,bxd}} (where x is an arbitrary operation). According to this principle we have {a,b}x{c}={{axc,bxc}}. I need some explanations from you at this point GFR Junior Member Posts: 6 Threads: 0 Joined: Feb 2008 Reputation: 0 2008-02-09, 22:02 bo198214 Wrote:Now i am really confused ..... ...... Another possibility would also be the use of multisets, i.e. where one element can occur multiple times, as corresponding to the multiplicities of the roots of polynomials. I would propose to write multisets like [a,b,c]. Ok, but for now I still assume finite sets {a,b,c}.I agree for [...,...] notation. I must try to better formalizing these vague ideas. Nevertheless, my first need was to indicate by {...,...} a "multiple solution" of an equation or a "multiple result" of an operation. I didn't think of an actual Multivalued Numbers Algebra. Why not. If you can (or wish to) help! bo198214 Wrote:Here is an unmatched bracket.Ooops, .... sorry! bo198214 Wrote:And you dont obey your own rule which was {a,b}x{c,d}={{axc,bxc},{bxc,bxd}} (where x is an arbitrary operation). According to this principle we have {a,b}x{c}={{axc,bxc}}...,... I need some explanations from you at this point But ... I just wanted to say: {a,b}x{c,d}={{axc,bxc},{axd,bxd}}, and, of course: {a,b}x{c}={axc,bxc} I need your approval of that, asap By the way, how to indicate continuous sets, like: [x->+oo]lim(sin(x)) = ???? , i.e. indeterminate between 0 and 1 ? Don't shoot me, please. I have honest intentions. Ciao, for the moment. GFR bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-02-19, 18:39 GFR Wrote:Nevertheless, my first need was to indicate by {...,...} a "multiple solution" of an equation or a "multiple result" of an operation. I didn't think of an actual Multivalued Numbers Algebra. Why not. If you can (or wish to) help!If you think of solutions to polynomes, the multiset is probably the more appropriate variant, as the solutions of a polynome has multiplicities. But a theory of multivalued functions, should at least make things easier to work with (if it doesnt solve some problems otherwise). And I really dont see where this would be the case. What would be an application of multivalued functions, where it would be very cumbersome to work with proper functions? Quote:But ... I just wanted to say: {a,b}x{c,d}={{axc,bxc},{axd,bxd}}, and, of course: {a,b}x{c}={axc,bxc} I need your approval of that, asap No, no approval. You want to handle singletons differently from other sets? That would make things really difficult to handle (you always have to distinguish whether something is a singleton or not) and I dont see where it would help. Or did you rather mean: {a,b}xc={axc,bxc}, i.e. to have a mixture of sets of numbers and ordinary numbers? Quote:By the way, how to indicate continuous sets, like: [x->+oo]lim(sin(x)) = ???? , i.e. indeterminate between 0 and 1 ? I have no idea, I guess you want to have [x->+oo]lim(sin(x)) = [0..1]? Quote:Don't shoot me, please. I have honest intentions. Look, the idea to handle multiple roots as a single outcome is quite apparent. But if you want to make it more than just a convenient idea (and you want it to revolutionize mathematics ), then there must be some substance behind: a theory or whatever with which you can do things that you couldnt do before, or things that were cumbersome before, etc. For example it is very cumbersome to make the distinctions of different inverse functions for a non-injective function. So everyone would be happy if there would be a system of multi-valued function that would allow to calculate with an arbitrary function as if it were a bijective function, i.e. just to take the inverse function that then possibly could be multivalued. However as far as I can see multivalued function can not help with this, even if your intentions are quite honerable. « Next Oldest | Next Newest »