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标题: What is wrong with this method to find limit? [打印本页]

作者: yma16 时间: 2009-12-16 18:28
标题: What is wrong with this method to find limit?

[Question]
What is (2^0.5)^(2^0.5)^(2^0.5)...?

Here is a solution.

Let it be x, then we get x=(2^0.5)^x. It seems it will lead x=2 or 4.

The limit should be unique. What is wrong with this method?

Someone said that if x^x^x... (x>0) has a finite limit, x should
be <=e^(1/e). If he was right, then the maximum limit of x^x^x...
is e. So 4 cannot be the answer and 2 is. Is there a way to get
the answer directly (which is 2) and not use the maximum limit e to
reject 4?

作者: Xiangdang 时间: 2009-12-25 00:24
标题: 回复：What is wrong with this method to find limit?

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.

作者: yma16 时间: 2009-12-26 02:49
标题: thanks a lot. What I do not understand is how

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.

作者: HF: 时间: 2009-12-26 10:07
标题: 回复：What is wrong with this method to find limit?

The limit is infinite.

作者: yma16 时间: 2009-12-26 19:10
标题: 回复：回复：What is wrong with this method to find limit

the 2nd post says it is <=2.

作者: HF: 时间: 2009-12-26 19:59
标题: 回复：回复：回复：What is wrong with this method to find li

That proof does not seems to be right. 1,2,3 are OK, but how 4 is achieved?

原贴：

文章来源: yma16® 于 2009-12-26 11:10:30 (北京时间: 2009-12-27 0:10:30)
标题：回复：回复：What is wrong with this method to find limit

the 2nd post says it is <=2.

作者: Xiangdang 时间: 2009-12-28 21:22
标题: 回复：回复：回复：回复：What is wrong with this method to find

Can you think of 1,2,3 as steps in a "mathematical induction" procedure? And 4 is the conclusion. Good luck.

作者: Xiangdang 时间: 2009-12-28 21:50
标题: It is an "irrational" number!

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".

作者: HF: 时间: 2009-12-29 02:44
标题: 回复：回复：回复：回复：回复：What is wrong with this method to f

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...

作者: yma16 时间: 2010-1-1 04:41
标题: 回复：It is an "irrational" number!

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.

作者: Xiangdang 时间: 2010-1-6 00:33
标题: 回复：回复：It is an "irrational" number!

Maybe it has somthing to do with the fact that 2 is the only even prime number!

Xiangdang's Conjecture!

作者: yma16 时间: 2010-3-12 00:30
标题: 回复：What is wrong with this method to find limit?

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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