# Homework Help: A number raise to it self infinitely

1. Jul 29, 2007

### bigjoe5263

hi I have encountered this...

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Code (Text):
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X
X
X
X    =   3

I tried to solve it using properties of logarithims however I always end up with the initial equation... how will I find the value of X? ( x is raised to it self endlessly ).

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Last edited: Jul 29, 2007
2. Jul 29, 2007

### Dick

You have to parenthesize that exponential tower or it's ambiguous in meaning. 3^(3^3) is not equal to (3^3)^3. And as it's infinite, you need to express the meaning in terms of an infinite sequence. So start with a_0=x. Is a_n=x^(a_(n-1)) or is a_n=(a_(n-1))^x?

Last edited: Jul 30, 2007
3. Jul 30, 2007

### Curious3141

The generally understood meaning of the left hand side of that equation is the infinite power tower function

$$f(x) = x^{(x^{(x^{(x^...$$

or simply, work "downwards" from "infinity".

The "trick" solution to f(x) = N is to see that this implies x^f(x) = N hence x^N = N when a solution exists, giving x = N^(1/N).

The problem is that there is an upper bound on N for a solution to exist. I'll leave it to the OP to find that bound and hence deduce that f(x) = 3 has no solution.

4. Jul 31, 2007

### HallsofIvy

If $x^{(x^{(x^{(x^...}}} = 3$, Then
x3= $$x^[{(x^{(x^{(x^...]}}}}= x^{(x^{(x^{(x^...}}}$$. What is that equal to?

Curios3141, I don't see why that "has no solution". Am I missing something?

5. Jul 31, 2007

### D H

Staff Emeritus
Halls, try calculating $$f(x)=x^{x^{x^\cdots}}$$ when $$x=3^{1/3}$$. It does not converge to 3. It does not converge, period. The upper bound on N is between 2 and 3 and is a very ubiquitous number.

6. Aug 1, 2007

### HallsofIvy

How are you calculating that? If, for example, you calculate 31/3, store it in "x" in a calculator, do x^x, then "ans"^x, repeatedly, it does not converge.

But that is NOT $$f(x)=x^{x^{x^\cdots}}$$! To calculate that you need to repeatedly do x^"ans" and that does converge, slowly, to 3.

7. Aug 1, 2007

### D H

Staff Emeritus
I misspoke. It does converge, just not to the value you expect. Let

$$f_1(x) = x$$
$$f_k(x) = x^{f_{k-1}(x)}$$

The desired function $f(x)$ is

$$f(x) = x^{x^{x^{\cdots^x}} = \lim_{k\to\infty}f_k(x)$$

Now further define

$$g(x) = x^{1/x}$$

THe question at hand is $f(g(x))=x$? For $x \in [ 1/e,e ]$, $f(g(x))=x$. The convergence is very slow as x approaches 1/e or e.

Things get a bit interesting when x>e. For example, $f(g(3)) = 2.478053\cdots$ rather than 3. What's going on here? Note that $g(3) = 1.44225\cdots$. However, $g(2.478053\cdots) = 1.44225\cdots$ also. The inverse of $g(x)$ has two branches, with a branch point at $$e^{1/e}$$. $f(g(x))$ converges to $$g^{-1}(g(x))$$, where $$g^{-1}$$ refers to the lower branch.

Things get even more interesting when x<1/e. The sequence $\{f_k(x)\}$ doesn't converge. Instead, the sequence toggles between a pair of values as $k\to\infty$, one close to zero and the other close to one.

8. Aug 2, 2007

### HallsofIvy

I'm a bit confused as to what converges, "converge, just not to the value you expect". The original problem was to solve
[tex\ x^{x^x^{\cdot\cdot\cdot}}= 3[/tex].
IF the sequence $a_1= x^x$, $a_2= x^{a_1}, ... converges to 3, then x must be equal to 31/3. 9. Aug 2, 2007 ### Dick That's true. But the converse (if x=3^(1/3) then the sequence converges to 3) isn't true. The sequence does converge to a limit L that you can find by solving log(L)/L=log(3)/3 (not the L=3 root, the other root L=2.47805...). 10. Aug 2, 2007 ### HallsofIvy I am going to have to meditate upon this! 11. Aug 2, 2007 ### D H Staff Emeritus The problem is that no x exists such that the sequence [itex]\{a_k(x)\}$ converges to 3.

12. Aug 3, 2007

### Curious3141

Halls, it's simple. The radius of convergence of the hyperpower function $$f(x) = ^{\infty}x$$ (using the tetration notation) is $$({(1/e)}^e, e^{(1/e)})$$

For x > e^(1/e), the function does not coverge.

For a value like 3^(1/3), the function *does* converge. But it does not converge to 3.

The function $$g(x) = x^{(\frac{1}{x})}$$ reaches a maximum at x = e. In other words, you cannot find a value of x that makes g(x) > e^(1/e), so g(f(x)) will always converge for any real positive x > e.

So for N > e, f([N^(1/N)]) will never equal N, even though convergence will hold. Hence f(x) = N has no solution for x > e.

Hope that clarifies things. I haven't explored the lower bound this carefully, so I shan't comment on it.

Last edited: Aug 3, 2007
13. Aug 3, 2007

### HallsofIvy

Yes, thanks. I showed that "if the sequence of powers converges to 3, then x= 31/3. The converse, "if x= 31/3, then the sequence converges to 3" is not true.

14. Aug 4, 2007

### Curious3141

Of course, the first condition (convergence to 3) will never actually be met.

15. May 14, 2008

### erszega

D_H's f(g(x)) appears to provide solutions for x^y = y^x, x<>y, x>e.
For instance y=f(g(3)) is the other solution for 3^y = y^3, besides the obvious y=3.
When x=e, the only solution is y=e.
However, such solutions also exist when 1 < x < e, eg 2^4 = 4^2, but f(g(x)) does not help then, eg f(g(2)) = 2.

Last edited: May 14, 2008