Remember this problem (Shown in Picture)?

Question

http://i2.photobucket.com/albums/y16/zorro1267/exponent.gif

It diverges because there are an infinite number of equations one can form from it.

Try doing this, then:

http://i2.photobucket.com/albums/y16/zorro1267/exp2.gif

Look at this too:

http://i2.photobucket.com/albums/y16/zorro1267/exp3.gif

Answer 1

let x = sqrt(2), your expression is x^x^x^x^...
but sqrt(2) = 2^0.5, thus expression is
(2^0.5)^(2^0.5)^...^(2^0.5)^(2^0.5)^(2^0.5)
Consider 0.5 = 2^(-1), and consider the last 2 brackets:
(2^(2^(-1)))^(2^0.5)
=2^(2^(-1)x2^0.5)
=2^(2^(-1+0.5))
=2^(2^(-0.5)), i.e. we get -0.5 from subtracting 1 from 0.5.
Thus the tail of the expression is now
...^(2^0.5)^(2^(2^(-0.5)))
Repeating, with (2^(-0.5)) as the power of the last term instead of 0.5, we will get (2^(-0.5)-1) as the expression in the last power.
Thus, one more time we will have the expression for the last power as 2^(2^(-0.5)-1)-1.
-1<-0.5<0 ==> 0<2^(-0.5)<1
==> -1<2^(-0.5)-1<0 ==> 0<2^(2^(-0.5)-1)<1
... and so on...

Got to stop here.. need to go home :)
next is to argue that (2^(-0.5)-1) approaches 0, and thus by the time the first 2 is considered, this power is approx. 2^0 ~ 1 and therefore the expression is 2^1 = 2.

Acknowledge: MSExcel was used to provide assurance.

Answer 2

I assume you want an answer to the infinite exponent sequence?

It diverges absolutely although it's significantly different than the divergence of products and sums.
Because the exponent is always the same and lies in integer proximity of 1, the divergence is agonizingly slow to be studied analytically...I reckon a regularization could make this finite, though I'm not all too sure how that'll be built up.

Answer 3

Well, it is easier to write: y = 2exp((1/2)1\2)........to infinity.

It is raised to the one half power to infinity means that at inifinity it approaches 1.

Well can you explain how you can form an infinite number of equations from the divergence?

Answer 4

just walk up the stais then choose the door u wanna enter

Answer 5

I am admirer of the people who can understand that kind of problems...

God created your brain special... :)

Answer 6

since the problem can be solved easily by x=sqrt(2)^ x than to solve it by x=sqrt(sqrt(2))^x . why to make things difficult?

Answer 7

Is there a question in there, somewhere, that I'm missing?


Doug

Answer 8

Can you please write a question next time. I feel ashamed to get two point without an answer.

Answer 9

i dint understand