What is wrong with this method to find limit?

yma16 · 发表于 2009-12-16 18:28:08

[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?

yma16 · 发表于 2010-3-12 00:30:14

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.

Xiangdang · 发表于 2010-1-6 00:33:10

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

Xiangdang's Conjecture!

yma16 · 发表于 2010-1-1 04:41:35

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.

HF: · 发表于 2009-12-29 02:44:10

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...

Xiangdang · 发表于 2009-12-28 21:50:48

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".

Xiangdang · 发表于 2009-12-28 21:22:23

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

HF: · 发表于 2009-12-26 19:59:06

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.

yma16 · 发表于 2009-12-26 19:10:30

the 2nd post says it is <=2.

HF: · 发表于 2009-12-26 10:07:40

The limit is infinite.

yma16 · 发表于 2009-12-26 02:49:28

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.

Xiangdang · 发表于 2009-12-25 00:24:05

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.

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What is wrong with this method to find limit?

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It is an "irrational" number!

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thanks a lot. What I do not understand is how

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