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### "Equation"

Posted: July 12th, 2004, 5:35 pm
Solve the equation:P.S. The real fun is not to find the “solution”, but to discuss this “equation”. Imagine, that you were asked to find a solution of this “equation” on the interview. What would you say after having been thinking for a while? Originally this task was given to hi school students on a mathematical contest to see how do they react and for a fun.

### "Equation"

Posted: July 12th, 2004, 6:13 pm
My answer is sqrt(2)"Solution"Define y_n = x ^(x^...) - n_timesThin Lim y_n = 2;then log lim y_n = log(x) lim y_{n-1} = log(x) *2 = log(2) -> x =sqrt(2)Of course, problem of converging y_n is omitted here

### "Equation"

Posted: July 12th, 2004, 6:18 pm

### "Equation"

Posted: July 12th, 2004, 6:46 pm
You are right. The "solution" is quite obvious. But it’s not a fun is it? So let me ask you:1. Do you believe it’s a solution? Why or why not? 2. To write a function on a plain C or whatever to verify that solution is like a couple of minutes… Well and what is the result?

### "Equation"

Posted: July 12th, 2004, 8:45 pm
Similar question:solve the following "ODE":f = sum_k=1^\infty f^(k)where f^(k) is the k-th derivative of f.

### "Equation"

Posted: July 12th, 2004, 9:17 pm
this stuff belongs in the discussion on the difference between an econometrician and a physicist that people flog in the general topics. function has to be infinite for any x>1, that was my first thought, it took me a minute to understand where people are coming with sqrt(2) from.

### "Equation"

Posted: July 12th, 2004, 9:49 pm
Generalization:x^x^x... = n -> x =n^(1/n)It does converge for x >1 and for arbitary x > 1 there exists n > 0 such that the equation holds true.

### "Equation"

Posted: July 12th, 2004, 11:17 pm
Corollary:Find the max of x^x^.... for convergent series.

### "Equation"

Posted: July 13th, 2004, 2:46 am
About to the generalization: if n=2 then the solution is x=sqrt(2). If n=4 then the solution is again x=sqrt(2), thus 2=4!!! Let’s analyze the function x^(1/x). It is zero for x->0, 1 for x=1, it has one max for x=e at e^(1/e) >1, and it is 1 for x->+infty (horizontal asymptote). It is easy to draw the plot. So it can be inverted only on the domain (0,1]+{e}. Note that only in this domain the eq a=x^(1/x) has an unique solution in rest there are x1 and x2!=x1 such that a=x1^(1/x1)=x2^(1/x2). Thus, the eq x^x^x…=2 dose not have any solution for real numbers, since 2 is not in the (0,1]+{e}.

### "Equation"

Posted: July 13th, 2004, 12:28 pm
Let’s add to the confusion: We are looking for the solution of eq x^x^x…= n.Let’s call (in the spirit of a previous comment, no offences to the actual professions)1. “physicist solution” : n must be in the domain (0,e] (because is working!), then sqrt(2) is a solution for n=22. “econometrician solution”: n must be in (0:1], then sqrt(2) is not a solution for n=2In both case the general solution is x=n^(1/n).And let’s have a democratic vote for the next 48 hours. One can vote independently on each point (eg 1yes 2yes, or 1no 2no etc). There is only one requirement (initially posted with the problem) have fun.(We can try a point 3. for n in (0,infty) but is to obviouse)

### "Equation"

Posted: July 13th, 2004, 2:13 pm
QuoteOriginally posted by: MirceaLet’s add to the confusion: We are looking for the solution of eq x^x^x…= n.Let’s call (in the spirit of a previous comment, no offences to the actual professions)1. “physicist solution” : n must be in the domain (0,e] (because is working!), then sqrt(2) is a solution for n=22. “econometrician solution”: n must be in (0:1], then sqrt(2) is not a solution for n=2In both case the general solution is x=n^(1/n).And let’s have a democratic vote for the next 48 hours. One can vote independently on each point (eg 1yes 2yes, or 1no 2no etc). There is only one requirement (initially posted with the problem) have fun.(We can try a point 3. for n in (0,infty) but is to obviouse)I go for sqrt(2) is not a solution for n=2. Anyway, I thought that was the physicists' solution not the way round you have it.

### "Equation"

Posted: July 13th, 2004, 2:19 pm
Excellent Mircea! I was waiting for another solution 2=4 and this leads to the statement that all numbers are the same and there are no numbers. What is wrong with the equation? As to your voting stuff… why bother? You have already pointed out that this is not the real equation and it doesn’t have a solution even though it seams pretty easy to find one…What is wrong with the equation (to add an oil into the fire)? How would you calculate something like 2^3^4^5? How would you put parentheses there? 2^(3^(4^5)) meaning that you have to start from 4^5 and go down. When we are talking about something like x^x^x^... we have to start from infinity which is a nonsense. We can write lim(x->1){1/(1-x)}, but we cannot consider 1/(1-x) where x=1.

### "Equation"

Posted: July 13th, 2004, 2:36 pm
FV,No offence there, the classification was done below by zerdna.

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Posted: July 13th, 2004, 3:21 pm
Estealth,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

### "Equation"

Posted: July 13th, 2004, 3:50 pm
QuoteOriginally posted by: MirceaAbout to the generalization: if n=2 then the solution is x=sqrt(2). If n=4 then the solution is again x=sqrt(2), thus 2=4!!! /q]Mmmmm, .. cool, MirceaGood question to fail somebody on an interview