Finding the Missing X...

[sinCERA]

Our professor once gave us this problem, and none of us got this.
Could you please answer this with a solution.

PROBLEM: Solve for x if x raised to x raised to x raised to x and so on is equal to 3.

 

[albbg]

The general problem should be: x^x^x^x^x.....^x (infinite times) = k

if we call y=x^x^x^x^x.....^x "infinite-1" times then x^x^x^x^x.....^x (infinite times) = x^y

Since we are considering infinite then x^x^x^x^x.....^x (infinite times) = k ≈ y then we can write the approximation:

x^k=k from which x=k^(1/k)

the problem is that, as far as I remember, the function convergence is limited to "k=e" (that is k=2.71828...) or something like that, so k=3 cannot be reached.

 

[sinCERA]

The general problem should be: x^x^x^x^x.....^x (infinite times) = k

if we call y=x^x^x^x^x.....^x "infinite-1" times then x^x^x^x^x.....^x (infinite times) = x^y

Since we are considering infinite then x^x^x^x^x.....^x (infinite times) = k ≈ y then we can write the approximation:

x^k=k from which x=k^(1/k)

the problem is that, as far as I remember, the function convergence is limited to "k=e" (that is k=2.71828...) or something like that, so k=3 cannot be reached.

So why not just say x= 3^(1/3) ?

 

[zorro]

It is a tricky question.
(I hope I don't become confused.)

If you mean

x^(x^(x^(x^(x.....^x)))) = k

then the solution is x=k^(1/k), as you found.

But if you mean

((((x^x)^x)^x)^x.....)^x = k

then the solution should be the solution of k^x=k.
It seems that in this case there is a consistent solution only for k=1 (in which case x=1).
Regards

Z

 

[albbg]

OK: x^k=k is not exact but approximate. Since the function is strongly non linear infinitesimal increases of x can lead in very large variation in the solution.
Since we are out of the convergence 3^(1/3) will only give us an underestimate of x, that can be increase up to a value leading in k=exp(1). Higher values will cause non convergence.

zorro, you are write, but also k^x=k is an approximation, in effect I think the correct solution is given from the non-algebraic equation x^y=y^x that cannot be solved in closed-form. In any case k=3 is not reacheable.

 

[zorro]

Hi albbg

I see. Just, rather than say that "x^k=k is an approximation", we should say that "in the case a solution exist, it must satisfy x^k=k" , i.e. x=k^(1/k).

Let me explain. Consider the function f(k)=k^(1/k) for k positive. If it has a maximum for some k, Ln(f) has a maximum for the same k because Ln() is monotonic.

Equating the derivative of Ln(f(k))=Ln(k)/k to 0, we find that there is a maximum for k=e. So, we have the same solution of x=k^(1/k) for two values of k, say k1<e and k2>e.
For example, for k=2 we find the solution x=2^(1/2)=1.4142..., and for k=4 we find the same solution x=4^(1/4)=1.4142...

Obviously, if we put x=1.4142... into x^(x^(x^(x^(x.....^x)))) (infinite times) we can get only a single value for k, that is 2.
For x>=e^(1/e) the expression x^(x^(x^(x^(x.....^x)))) (N times) diverges as N->∞.
From the other side, the equation x^(x^(x^(x^(x.....^x))))=k (infinite times) has no solution for k>e.

Regards

Z