SERVING THE QUANTITATIVE FINANCE COMMUNITY

EStealth
Topic Author
Posts: 48
Joined: July 9th, 2004, 6:56 pm

### "Equation"

QuoteOriginally posted by: MirceaEstealth,Let’s define the series y_k = x^(x^(x^..) k times. We know how to compute y_kThen, we define x^x^x^… to be its limit (ie lim_(k->infty) y_k ). If (when) the limit exists, it’s a valid question. So, the problem is well defined, as well as the fun Yes, we know how to compute y_k, but we don't know how to compute lim_(k->infty) y_k and if you can then what is it in case sqrt(2)^sqrt(2)^sqrt(2)^... 2 or 4?

zerdna
Posts: 3856
Joined: July 14th, 2002, 3:00 am

### "Equation"

another look at the issue that Temnik pointed out is that the limit is not defined for this function properly.if f(n) is the n-th power, it is possible that f(n+1)=x^f(n) and f(n+1)=f(n)^x. Once you define which of these definitions you use to take the limit, the answer to the original problem becomes certain. In particular, the sqrt(2) solution will be a valid one for the first definition and not valid for the second. Basically, the problem is not well-defined.
Last edited by zerdna on July 12th, 2004, 10:00 pm, edited 1 time in total.

temnik
Posts: 112
Joined: September 17th, 2003, 6:17 pm

### "Equation"

There's no need to take a limit of n - note that L.H.S. is an infinite expression. Let's call the whole L.H.S. of the equation Y. Note that in addition to Y=2 we have the following:Y^x = 2 and x^Y = 2.There's no solution to these three equations in the domain of either real or complex numbers - can easily see by taking logarithm.Even if we assume that x is a vector in an infinitely-dimensional space with a "measure" (Banach?) - we keep bumping against the contradiction that ||x|| has to equal 1 and SQRT(2) at the same time.So, x cannot be a "number". But, if x is a logical "true" and you choose to designate "true" by symbol "2" - then the equation works out and you, EStealth, are a weirdo.OTOH, a physics problem with an infinite resistor chain is good clean well-defined fun...

EStealth
Topic Author
Posts: 48
Joined: July 9th, 2004, 6:56 pm

### "Equation"

Logical true is not fun unless in addition we have at least false and even then we have to define true^false and everything… Considering that the truth is more then false (if we are not talking VB, are we? ) we can choose false=0 and true=1 then the original equation reads true^true = true and I’m a weirdo.

EStealth
Topic Author
Posts: 48
Joined: July 9th, 2004, 6:56 pm

### "Equation"

QuoteOriginally posted by: zerdnaanother look at the issue that Temnik pointed out is that the limit is not defined for this function properly.if f(n) is the n-th power, it is possible that f(n+1)=x^f(n) and f(n+1)=f(n)^x. Once you define which of these definitions you use to take the limit, the answer to the original problem becomes certain. In particular, the sqrt(2) solution will be a valid one for the first definition and not valid for the second. Basically, the problem is not well-defined.There is no need to define the power operation. At least I didn't mean to redifine it in the original equation.Consider f(n) =x_1^x_2…^x_n (all x_i are different) then f(n+1) is not equal to f(n)^x_(n+1) neither it’s equal to x_(n+1)^f(n). The confusion comes form the fact that all x_i are equal in the original equation. The limit is not difined for the same reason. I've already pointed it out. 2^3^4 is not equal to (2^3)^4, it is equal to 2^(3^4).Y=2, Y^x = 2 and x^Y = 2 -> wrong!

tristanreid
Posts: 441
Joined: May 12th, 2004, 6:58 pm

### "Equation"

QuoteOriginally posted by: EStealthLogical true is not fun unless in addition we have at least false and even then we have to define true^false and everything… Considering that the truth is more then false (if we are not talking VB, are we? ) we can choose false=0 and true=1 then the original equation reads true^true = true and I’m a weirdo.If true=1 and false=0,true+false = truetrue*false = falsetrue^false = truetrue^true = truefalse^false = truefalse^true = falseYou can take the true root (true identity) of something, but not the false root. As the limit of the root approaches false, the operand becomes false.false = true^(1/false) is not a legal statement but false=false^(1/true) is valid and true If you add true and false, you are left with only true. If you mix true and false, you will create false. Taking something to the false power creates truth. The true root of something is true, even if that thing is false, but it is impossible to find the false root of anything. How zen.For more explorations of software truth:The oreilly nutshell book for the unix 'true' command-t.

EStealth
Topic Author
Posts: 48
Joined: July 9th, 2004, 6:56 pm

### "Equation"

QuoteOriginally posted by: tristanreidIf true=1 and false=0,true+false = truetrue*false = falsetrue^false = truetrue^true = truefalse^false = truefalse^true = falseYou can take the true root (true identity) of something, but not the false root. As the limit of the root approaches false, the operand becomes false.false = true^(1/false) is not a legal statement but false=false^(1/true) is valid and true If you add true and false, you are left with only true. If you mix true and false, you will create false. Taking something to the false power creates truth. The true root of something is true, even if that thing is false, but it is impossible to find the false root of anything. How zen.For more explorations of software truth:The oreilly nutshell book for the unix 'true' command-t.Cool. What about VB? Must be the other way around If true=0 and false=1.

Mircea
Posts: 19
Joined: July 9th, 2004, 6:04 pm

### "Equation"

Estealth,I guess we got lost on notations. Here is how I understand them. 1. When I see a notation like x^y^z, I understand x^(y^z)2. It is not common to use the notation (x^y)^z since this is equal to x^(y*z) which is a lot more simple. Thus when you posted the problem x^x^x^…=2, I understood x^(x^(x^(..))) = 2. The other one (..(x^x)^x)^…) = 2 is trivial since (..(x^x)^x)^..) = lim_{n->infty}x^(x^n) =2. For this one there is no solution. The proof is straightforward. This problem is in the second formulation from zerdna’s message (below) given by f(n+1)=f(n)^x. Then it follows, if the sequence {f(n)} has a limit, say F=2, then the recursive relation becomes F=F^x. This equation has no solution for F != 1 and for F=1 the solution is trivial x=1.In contrast the first problem x^x^x^…=2 (or x^(x^(x^(..))) = 2 if you wish) is far from being trivial. This in zerdna’s notations is f(n+1) = x^f(n). The problem becomes: if the sequence {f(n)} has a limt F=2 then what is x? My conjecture is a bit more general and is like this:The equation x^x^x^…=b with b > 0 has a solution if and only if b is in (0,e], then the solution is x=b^(1/b). If b>e then there is no solution to this equation and for any x > e^(1/e) we have x^x^x^…=inftyIn particular for b=2 < e the solution is x=sqrt(2). For b=4 > e there is no solution. The confusion comes from the wrong manipulation of eq f(n+1) = x^f(n). In general we say: if the sequence {f(n)} is convergent with the limit b, then we have b=x^b and so x=b^(1/b). Then we take b=2 and we say that x=sqrt(2) is the solution. Moreover, we take b=4 and again we say that x=4^(1/4)=sqrt(2) is a solution. Finally we get the absurd consequence that 2=4. Why is that? Our weak point is in the assumption that {f(n)} is convergent no mater what!!! Well, that is wrong. In fact it can be prove that for x=sqrt(2) the sequence {f(n)} is convergent with the limit 2 and moreover there is no real x such that the sequence {f(n)} is convergent with limit 4. The problem of convergence was bothering ScilabGuru in her very first message, then was somehow forgotten.If nobody will do it before, tomorrow I will post my proof to the above conjecture. It is relatively long.PS I don’t understand what is the confusion when you thought that the power operator was redefined. Given the infinite sequence {x_k} we can build the infinite sequence {f(n)=x1^x^2^..^x_n} the limit of the {f(n)} is formally x1^x2^x3^….. Indeed here f(n+1)!=f(n)^x_(n+1)!=x1^f(n). Never the less the sequence {f(n)} exist and any of its elements can be computed (in principal). Then the question if the sequence {f(n)} is convergent is a valid one. In particular for 2^3^4^…=infty. You can prove that, assuming my conjecture true and finding a diverging minor sequence for {f(n)} (one of the Caushy theorems).

temnik
Posts: 112
Joined: September 17th, 2003, 6:17 pm

### "Equation"

Last edited by temnik on July 13th, 2004, 10:00 pm, edited 1 time in total.

EStealth
Topic Author
Posts: 48
Joined: July 9th, 2004, 6:56 pm

### "Equation"

QuoteOriginally posted by: MirceaEstealth,...My conjecture is a bit more general and is like this:The equation x^x^x^…=b with b > 0 has a solution if and only if b is in (0,e], then the solution is x=b^(1/b). If b>e then there is no solution to this equation and for any x > e^(1/e) we have x^x^x^…=inftyIn particular for b=2 < e the solution is x=sqrt(2). For b=4 > e there is no solution. The confusion comes from the wrong manipulation of eq f(n+1) = x^f(n). In general we say: if the sequence {f(n)} is convergent with the limit b, then we have b=x^b and so x=b^(1/b). Then we take b=2 and we say that x=sqrt(2) is the solution. Moreover, we take b=4 and again we say that x=4^(1/4)=sqrt(2) is a solution. Finally we get the absurd consequence that 2=4. Why is that? Our weak point is in the assumption that {f(n)} is convergent no mater what!!! Well, that is wrong. In fact it can be prove that for x=sqrt(2) the sequence {f(n)} is convergent with the limit 2 and moreover there is no real x such that the sequence {f(n)} is convergent with limit 4. The problem of convergence was bothering ScilabGuru in her very first message, then was somehow forgotten.If nobody will do it before, tomorrow I will post my proof to the above conjecture. It is relatively long.Sounds like fun is getting serious now. Looking forward to see your proof. Hope in your proof f(n+1)!=f(n)^x_(n+1)!=x1^f(n).

karakfa
Posts: 139
Joined: May 25th, 2002, 5:05 pm

### "Equation"

This reminds me the use of 1/(1-x) expansion 1+x+x^2+x^3+... at x=-1. Therefore:1/2 = 1-1+1-1+1....What is name of the famous mathematician who asserted this?

Mircea
Posts: 19
Joined: July 9th, 2004, 6:04 pm

### "Equation"

My conjecture wasThe eq x^x^x^…=b has a solution if and only if b is in (0,e] and that solution is x=b^(1/b).The proof is in four steps. First we will show that the conjecture is true in (0,1], second that for x=e^(1/e) the limit x^x^x^… exist and is equalt to e, third that the conjecture is true for b in (1,e), and finally fourth that there are no solutions for b in (e,infty).As usual we denote by f(n)=x^..^x n-times, x^x^x^… is the formal limit for the sequence {f(n)}.Observation: for any positive x we have f(n+1) > f(n). The sequence is monotonically increasing.Step 1. b in (0,1] (this is the trivial)For any positive a < 1 and positive b we have a^b < 1. Then, it is clear that f(2)=x^x < 1. By induction f(n+1)=x^f(n) < 1.Thus the sequence {f(n)} is bounded. Since it is also monotonic, {f(n)} is convergent and has a limit. Obviously, that limit is b^(1/b).Step 2. x = e^(1/e)We have:f(1) = e^(1/e) < ef(2) = (e^(1/e))^f(1) = e^(f(1)/e) < eby induction assuming f(n-1) < e we havef(n)=(e^(1/e))^f(n-1)=e^(f(n-1)/e) < eSo the sequence {f(n)} is bounden, thus convergent, and its limit is e.Step 3. If b in (1,e) then x in (1,e^(1/e))We have: f(1)=x < g(1)=e^(1/e)f(2)=x^x < (e^(1/e))^g(1) = g(2)by induction assuming f(n-1) < g(n-1) we havef(n) = x^f(n-1) < (e^(1/e))^g(n-1) = g(n)Since {g(n)} is bounden (see Step 2) it follows that {f(n)} is bounden. So it has limit. Step 4. b in (e,infty)The only candidate for a solution is x=b^(1/b). Since b is in (e,infty) we have x in (e^(1/e),infty).On other hand the function F(t)=t^(1/t) has the following propertiesi. F(t)->0 when t->0ii. F(1) = 1.iii. F(t) has one maximum for t=e (the max value is e^(1/e))iv. F(t)->1 as t->inftyThen for any b > e^(1/e) there is 1 < b1 < e^(1/e) such that b^(1/b)=b1^(1/b1). Form Step 2 we know that for x=b1^(1/b1) with 1 < b1 < e^(1/e) we have x^x^x^…=b1. Thus x=b^(1/b) cannot be the solution.QEDObs. x^x^x^…=2 has the solution sqrt(2). x^x^x^…=4 has no solutionSomebody mention a C code in order to verify this eq for sqrt(2). Before to do that, one must ensure that the recursive expression f(n+1)=x^f(n) is stable relative to truncation errors.Assume w is the truncation error in the estimation of f(n). Then we have ff(n+1)=x^(f(n)+w) where ff is the numerical approximation of f(n+1). Using the first order Teylor expansion on w we have ff(n+1)= x^f(n) [1+w*ln(x)]= f(n)+w*ln(x)*x^f(n) < f(n+1) +w*e*1/e = f(n+1)+w. Thus the error in estimating f(n+1) is the same order with the error in f(n). The recursive relation is stable relative to the numerical truncations. Now you can do a C program where you can loop b = pow (x,b) with initial b=1 and x=sqrt(2). You will see that b will approach 2 from below. In this way you can verify that math.h is working properly

EStealth
Topic Author
Posts: 48
Joined: July 9th, 2004, 6:56 pm

### "Equation"

QuoteOriginally posted by: MirceaObservation: for any positive x we have f(n+1) > f(n). The sequence is monotonically increasing.Wrong for 0<b<1.For Step 1. you have proven that x<1, but "Obviously, that limit is b^(1/b)" seams to be weak. You'd probably should show that there is such n1 that f(n1) is in \epsilon range from e and for any n2>n1 and so on...Step 2. The same thing. For any n you have: f(n)=(e^(1/e))^f(n-1)=e^(f(n-1)/e) < e. But what about the limit? ...Good but not perfect...

EStealth
Topic Author
Posts: 48
Joined: July 9th, 2004, 6:56 pm

### "Equation"

As to computer calculations it’s not only about truncation errors in computation. There is no way to represent sqrt(2) precisely dew to irrational nature of sqrt(2). All numbers n^(1/n) are irrational with the exeption n=1.Who knows, may be one day we’ll be seating in front of the computer with some sort of quantum processor where it’s possible. In expression for probability to go down or up one step for quantum oscillator there is sqrt(2), well this one is just an observation and what about all the rest of “irrationality” after all. (the power of set of rational numbers)/(the power of set of irrational numbers) = 0 if you will. That is how the computer computations weak. Well not really weak, there is a good deal on numerical calculation error estimation and everything…QuoteOriginally posted by: MirceaMy conjecture wasSomebody mention a C code in order to verify this eq for sqrt(2). Before to do that, one must ensure that the recursive expression f(n+1)=x^f(n) is stable relative to truncation errors.Assume w is the truncation error in the estimation of f(n). Then we have ff(n+1)=x^(f(n)+w) where ff is the numerical approximation of f(n+1). Using the first order Teylor expansion on w we have ff(n+1)= x^f(n) [1+w*ln(x)]= f(n)+w*ln(x)*x^f(n) < f(n+1) +w*e*1/e = f(n+1)+w. Thus the error in estimating f(n+1) is the same order with the error in f(n). The recursive relation is stable relative to the numerical truncations. Now you can do a C program where you can loop b = pow (x,b) with initial b=1 and x=sqrt(2). You will see that b will approach 2 from below. In this way you can verify that math.h is working properly
Last edited by EStealth on July 14th, 2004, 10:00 pm, edited 1 time in total.

Mircea
Posts: 19
Joined: July 9th, 2004, 6:04 pm

### "Equation"

EStealth,You miss the point. So again, if the sequence {f(n)} is convergent with limit b then x=b^(1/b). That was proved since the very first replay.Then the sequence {f(n)} is convergent if and only if b is in (0,e] (that implies x is in (0,e^(1/e)]. This is what is proved.I cannot help you more.PS The difference between a real number and its finite precision representation in a computer is called numerical truncation. That “w” was counting for that.QuoteOriginally posted by: EStealthAs to computer calculations it’s not only about truncation errors in computation. There is no way to represent sqrt(2) precisely dew to irrational nature of sqrt(2). All numbers n^(1/n) are irrational with the exeption n=1.Who knows, may be one day we’ll be seating in front of the computer with some sort of quantum processor where it’s possible. In expression for probability to go down or up one step for quantum oscillator there is sqrt(2), well this one is just an observation and what about all the rest of “irrationality” after all. (the power of set of rational numbers)/(the power of set of irrational numbers) = 0 if you will. That is how the computer computations weak. Well not really weak, there is a good deal on numerical calculation error estimation and everything…