I think you have to make sure that the limit is approaching 2 from below so that you can reject the answer of 4. Here is the proof for (2^0.5) (2^0.5)^(2^0.5)... < 2 for finite terms. 1. since (2^0.5) < 2 2. then (2^0.5)^(2^0.5) < (2^0.5)^2=2 3. and (2^0.5)^(2^0.5)^(2^0.5) < (2^0.5)^2 = 2 4. further more (2^0.5)^(2^0.5)^(2^0.5) ... < 2 for So answer 4 should be rejected by using your method, which is mathematically sound. |
it got an extra solution 4. If you have some idea, please let me know. I posted this one in wxc and Dr. math. I have not got good answers. |
The limit is infinite. |
the 2nd post says it is <=2. |
That proof does not seems to be right. 1,2,3 are OK, but how 4 is achieved?
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Can you think of 1,2,3 as steps in a "mathematical induction" procedure? And 4 is the conclusion. Good luck. |
Let me try to answer it. I guess we can blame the "irrationality" of the number for the strange solution of 4 you got. (2^0.5) is an irrational number. You are dealing with the irrational power of an irrational number. What is the meaning of an irrational power of an irrational number ? Can it be rational or irrational? I guess the answer is both. It is very strange. Very often, when we turn an infinite series into a finite algebraic equation to solve, we inevitably "added" something to the original problem. You have to rely on somthing else, being physical meaning or math reasoning, to get the right answer. I hope this is the case. Let us know you have something else to explain for the "4". |
You are right. I think I misunderstood the order of x^x^x^x.... -- I thought I had fell in such a 'pit' once before... |
2^0.5 is the only case to lead two solutions. If you try 3^(1/3) or 5^0.2, it will not introduce an extra solution. |
Maybe it has somthing to do with the fact that 2 is the only even prime number!Xiangdang's Conjecture! |
I think the method does not work. However, I do not know why. The reason the method does not work is the following, Consider x^x^x... (x>0). Replace x with 10^(1/10), using this method, we will get the result 10. But 10^(1/10) is smaller than 2^ (1/2). So the result should be smaller than 2, which was the solution of (2^0.5)^(2^0.5)^(2^0.5)..., by using this method. |
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