Hari Seldon’s proof is good enough for the purposes of solving the OP’s problem, but one can obtain more relaxed bounds on the set of real numbers for which the sequence converges; see for example this page by math professor Lou Talman (Fun Stuff, Item 5).
No, first you define it for any n, then you allow n to approach infinity, and find a limit. So it’s more like, try it with x=2, then let n=4, now n=5, now n=6, and what limit are you getting closer and closer and closer too? That’s the limit at infinity.
That’s a counterexample for silverfish’s claimed equality. I know what a limit is, thank you very much.
Even proving that the sequence cnverges is not enough. You actually have to show it converges to 2 and not, say to 4.
Suppose you were given the equation x^x^x^x^… = 4. You could go through the same argument I gave to show that if x#n < 4 then x#(n+1) < 4 so it is a bounded increasing sequence and therefore converges. You could see that x^4 = 4 and then conclude that x = sqrt(2). But that is not the solution; there is none. The rest of my post was devoted to showing that when x = sqrt(2), then the only possible values for x^x^x^… are 2 and 4 and then the fact that x#n < 2 for all n implies the limit is 2.
“It is not enough to succeed; others must fail” (I think this was attributed to some Hollywood director.)
Just for fun and interest, here’s a similar question with an interesting answer.
Solve for X:
X = sqrt ( 1 + sqrt (1 + sqrt (1 + sqrt (1 + …
x = sqrt(1+x)
x^2 = 1+x
x^2 - x - 1 = 0
x = (1+sqrt(5))/2
That’s the nested radical form of the Golden Ratio, as Polerius notes.
The problem with the method Bryan Ekers and a few others suggested is that power towers are not associative like addition and multiplication; they have to be evaluated right-to-left.
Frylock, you have to enable scientific mode, then exponentiation is in the lower-left center, as “x^y”.
[QUOTE=Frylock]
I’m looking at Windows Calculator and don’t see a button for exponentiation. Where is it?
-FrL-[/QUOTE
Change the View on your calculator from Standard to Scientific.
Thanks, av8rmike and cmrdough. I didn’t realize there was a half-decent calculator to be had using that application!
So anyway, now I’ve tried out sqrt2 as an answer to the question, and as far as I can tell, it doesn’t converge to 2. It hits two on step three, then continues right on past it.
But maybe I’m doing it wrong.
I’m hitting the following buttons:
2
x^y
.5
M+
then
MR
x^y
MR
then
x^y
MR
over and over again.
This, it seems to me, raises sqrt2 to the power of sqrt2, then raises the result to the power of sqrt2, then raises the new result to the power of sqrt2, and so on.
Is that not the right way to understand the equation in question?
-FrL-
No, it’s not. It’s like I said, you have to evaluate the tower from right to left, what you’re doing is equivalent to:
(((((x^x)^x)^x)^x)…), which is not the same as (x^(x^(x^(x^x)…)))). Write out a few terms and you’ll see what I mean. To do it on the calculator, you’ll need to compute:
sqrt(2)^sqrt(2),
store that result,
take sqrt(2), raise that to the value of the stored number,
store that result,
take sqrt(2), raise that to the value of the stored number,
etc…
Again, you have to be more careful. It is easy to see that the golden ratio, I will call it g, is the only possible solution. What is harder is showing it is. Let x_n be the nth interate, so x_{n+1} = sqrt(1+x_n). If we suppose x_n < g, then 1+x_n < 1+g. Then x_{n+1} = sqrt(1+x_n) < sqrt(1+g) = g and so we see that all x_n < g. It is not quite evident that the sequence is monotone. So I will show that x_n < x_{n+1}. In fact from x_n < g, we get x_n+1 < g+1 and then multiply the inequalities (incidentally, it is clear that all terms are positive so we can take positive square roots and multiply inequalities as lib), we get x_n^2 - x_n < g^2 - g and then x_n^2 - x_n -1 < g^2 - g -1 = 0. But then x_n^2 < x_n +1 and, taking square roots, x_n < sqrt(x_n+1) = x_{n+1}.
What Hari Seldon is distinguishing here is the difference between existence and uniqueness proofs. For both of the problems mentioned, it’s easy to show that if a solution exists, it must be some particular value. Showing that a solution does exist is a little harder.
Sqrt2^(Sqrt2^(Sqrt2…))) seems to come to both 2 and 4.
Its clear it comes to 2 from comments on this thread. But by a similar argument, it seems like we can show that it comes to 4 since if we set x^(x^(x…)))=4, we get x^4=4 which means x=Sqrt2.
So what I presently don’t understand/don’t know is:
- Why does 2 get to be the limit and not 4?
- Is it true that Sqrt2^(Sqrt2^(Sqrt2…))) in some way really “equals” (or anyway “comes to” in some sense) both 2 and 4? Or is the 4 result somehow deceptive?
I have a feeling Hari Seldon’s posts answer this issues I’ve asked about, but I can’t follow them. He skips too many steps for my level of knowledge. And I dont’ know what “monotone” means in this context.
Is there a free graphing utility? Thinking about this has made me curious as to what the graph for y = the “xth root” of x would look like.
-FrL-
The argument that x^(x^(x^(…))) = 4 implies x = sqrt(2) shows that if the original equation has a solution, the only solutions (in R, anyway) are +sqrt(2). What still needs to be shown is that there is a solution to that equation, and that’s what Hari Seldon has disproved.
We say that a function f is monotone increasing iff x > y implies that f(x) > f(y).
Right. As I tried to indicate, I figured this was what Hari was up to. I was just hoping there could be a more complete explanation as to how his proof works. But I recognize this may not be possible (reasonably feasible) in a venue such as this.
-FrL-
Playing around with this, it appears that x^(x^(x^(…))) = y only has a solution if y is less than (or equal?) to e (i.e. 2.71828…).
Can anyone offer a proof? I’m headached out.
x^(x^(x^(…))) converges for (1/e)^e <= x <= e^(1/e). If x=e^(1/e) the sequence converges to e and satifies the equation for y=e.