Find the real value of √2 ^ (√2 ^ (√2 ^ (√2 ^ (√2 ^ (√2 ^ (√2 ^ (√2 ^ ......?

Question

EASY....
^_^

Answer 1

As previous respondants have noted, if this expression does, indeed, have a value, it is either 2 or 4. Dougy brought up the good point that one needs to be sure that these exponents don't cause the value of the expression to blow up, so that approximations to this number by computing it part way out do not head to infinity. While he usually gives good responses, in this case, he was careless (as was one other respondant) and assumed it would by glancing at it. We can show that in fact, it does stay bounded. The other point that people brought up is that one needs to be sure that it doesn't converge to 4, another possible solution of x^2=2^x.

To see rigorously that it does, indeed, converge to 2, let us define by a_n the n'th approximation to the expression, i.e.

a_n = sqrt(2^sqrt(2^sqrt(2^. . .sqrt(2)))

where this expression has n square roots in it.

We will first show by induction that for each n, a_n < 2. Notice that a_1 = sqrt(2) < 2, so the first case is true. Now assume that a_n < 2 for some n. Then a_(n+1) = sqrt(2^(a_n)) < sqrt(2^2) = 2, where the < sign follows from our assumption that a_n < 2 and the fact that the functions sqrt(x) and 2^x are increasing. Thus, by induction, a_n < 2 for all n, so the expression certainly isn't diverging to infinity.

In fact, we should be sure that it does converge to some value, and doesn't oscillate around, say between the values of 2 and 4. To see this, we can show by induction that the sequence a_n is increasing. In fact, it is clear that each a_n > 0 since the range of sqrt(x) and 2^(x) consist only of positive numbers when you plug in positive values for x. So we know from above that 0 < a_n < 2 holds for all n. For 0 < x < 2, one can check that 2^x > x^2, so that 2^(a_n) > (a_n)^2 for all n. Then for any n, a_(n+1) = sqrt(2^(a_n)) > sqrt((a_n)^2) = a_n.

Thus, the sequence of the a_n's is increasing. As it is also bounded above by 2, it converges to a real number that is at most 2. Since it has to satisfy x^2 = 2^x by previous arguments, the expression you gave is, in fact, equal to 2.

Answer 2

I'm going to assume that each exponent is within the radical to its left -- i.e., that we're talking about √[ 2^( √[ 2^( ... ) ] ) ]

Let x be the entire expression.

Square both sides. This gives us x² = 2^{√[ 2^( √[ 2^( ... ) ] ) ]}

or x² = 2^x.

We can eyeball the solutions of this equation as x = 2 and x = 4.

Now oddly enough, if you try to work the value out on your calculator (by starting with some x_0 value and iteratively computing x_n+1 = √[2^x_n] ), the result seems to converge on the x=2 solution for initial guesses < 4, and diverge for initial guesses > 4; the only way to get it to converge on 4 is if you start with 4 as your initial guess. And a graph of the curve f(x) = x² - 2^x, which is obviously continuous, seems to have three zeroes, but one of them is negative and clearly can't be the value of the original expression -- it must be a false solution that was introduced when we squared both sides. Perhaps the same is true of the x = 4 solution.

Hope that helps!

Answer 3

its 1

Answer 4

Consider the sequence defined recursively by a_1 = sqrt(2), and a_{n+1} = a_n^sqrt(2) for all natural numbers n. Then what you seek in your question is lim n->inf a_n.

This sequence is an increasing sequence, furthermore, it's increasingly increasing, and so therefore it diverges - i.e. goes to infinity.

Note on previous answer: While this number does satisfy x = sqrt(2)^x, and by squaring both sides gives the result x^2 = 2^x, you need to consider that x=2, 4 and infinity each satisfy this equation, and therefore this equation does not give a unique solution.

Edit: I now see that my definition from the sequence that I answered a different question. The sequence should be defined with a_{n+1} = sqrt(2)^a_n instead to correspond with the bracketing in the question. I believe that mathbear77's proof is sound.

Answer 5

It is easy to see this will diverge if one rewrites sqrt(2) =1.414

1.414^1.414 is certainly larger than 1.414

continuing this recursively will cause it to diverge.


Note, mistakes in previous answers have assumed that sqrt(2)^(sqrt(2) etc
reduces to 2^(0.5*2*0.5*2*0.5....)

This is NOT the case.

Answer 6

2?
(2^.5)^(2^.5)..=2^1^1...=2?

Answer 7

Let x = root( 2 raised-to ( root ... ....
If you just observe this equation for a while, you'll note -

x = root ( 2 raised-to (x) )... Right?
Now square both sides...
x^2 = 2^x

Answer: x = 2

Answer 8

Amey P's equation also allows x = 4.

Answer 9

Stack error

Answer 10

Amey P is right.