Question

How to solve this equation?
\[z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 2\]

Update:

By solving \(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 2\) using the method we discussed in this post, we get \(z=\sqrt{2}\).
Now, when we solve \(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 4\) using the same method, we get \(z=\sqrt{2}\).
So, the left side of the two equations are the same as \(z=\sqrt{2}\) in both equations.
As a result, the right side should also be the same, which means we have 2=4.

The problem is... what's wrong?

Answer 1

Please post "z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 2" in the LaTeX equation editor at http://www.codecogs.com/latex/eqneditor.php to read the equation. Thanks!

Answer 2

can't get latex to work :C

Answer 3

One logarithm might suggest a solutions.

\(z^{z^{z^{.^{.{.}}}}}\cdot z = log(2)\)

Answer 4

wow

Answer 5

Whoops. That's log(z) on the left.

Answer 6

Callistro, you want this?
\[\huge\color{blue}{ z^{z^{z^z}} =2 }\]

Answer 7

@SolomonZelman No, not exactly. There are infinitely z's in the index part.

Answer 8

\(2\cdot \log(z) = \log(2)\)

Almost done.

Answer 9

LaTeX is working fine form me.

2*log(z) = log(2)

Answer 10

Sorry, got disconnected,

do you know why latex disconnects me so much?
\[\huge\color{blue}{ z^{z^{z^{.^{.^{.}}}}}=2 } \]
sorry for the late reply.

Answer 11

Ah!
So, here it is.
\[z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 2\]\[z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}}\log z = \log 2\]Replace \(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}}\) by 2 again in the equation, we get\[2\log z = \log 2\]\[\log z = \frac{1}{2} \log 2\]\[z = 2^{\frac{1}{2}}\] Is that right?

Answer 12

The arithmetic is good. Can you prove the sequence of exponents actually converges?

Answer 13

Would you mind giving me any hints to do it? :)

Answer 14

You cannot just keep raising a store number to the square root of 2. That is not the right structure.

I'm feeling like L.Euler, today. Let's just assume infinite sequences work and move on.

I did forget to ask. Is z a Real Number? Should we be looking for Complex Solutions?

Answer 15

10 Zs - 1.983668399
20 Zs - 1.999585623
40 Zs - 1.99999972847903
80 Zs - 1.99999999999988
88 Zs - 1.99999999999999 -- Ran out of that system's machine precision.

If I had to vote, I would vote for convergence.

Answer 16

Just real number. My teacher said we are mostly deal with real number system in this course. Though, finding the solution to the equation is actually the end of the question.

Answer 17

This should strike wonder to the deepest part of your mathematical soul.

An exponential tower of Irrational Numbers is a Rational Number???!!!!

Answer 18

And even worse, he used this to show 2=4...

Answer 19

Awesome. Maybe we were worrying about convergence from time to time.

Answer 20

Here is the problem he showed us in the first lecture:
Solving \(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 2\), we get \(z=\sqrt{2}\)
Solving \(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 4\), we get \(z=\sqrt{2}\)
So, the left side of the two equations are the same as \(z=\sqrt{2}\) in both equations.
As a result, the right side should also be the same, which means we have 2=4.

The problem is... what's wrong?

He said he hoped we could find out the problem at the end of the course, but I guess I can't wait that long.

Answer 21

First z cannot be > 1, because the tower of power of z will go to infinity.

Answer 22

If z =1, then the tower is on1

Answer 23

If z <1, then this tower stay less than one. What is wrong with my thinking?

Answer 24

It... seems right to me...

Answer 25

What was wrong in the reasoning to show that\( z=\sqrt 2 \) is that you use the fact
that this infinite tower of powers of z converges. It does not always.

Answer 26

@Callisto

Answer 27

I am sorry that I don't quite understand the step "ln(y_n)= z^n \ln(z)"

Answer 28

I think this is due to singularity at x=1 ...

Answer 29

Do it by induction

Answer 30

I am so confused...

Actually, it goes back to the question, how do you know if this equation can be solved or not. What is wrong with previous solution to solve z? Why does taking log on both sides use the fact that the infinite tower of powers of z converges (which is not true, but I need time to figure out.)?

I'm still trying to understand Dr's proof.

Sorry for being so stupid.

Answer 31

no no ... in analysis, you should prove that if you take infinitely many hyperpower ... you should show that it's limit exist before you find it's value. value come after you show that it exists. I don't understand either ... no latex here :(

Answer 32

lol .... my grammar!!

Answer 33

One of the problems is that I haven't taken any analysis courses yet. This questions was raised in my Linear Algebra course.

Answer 34

according to big bro ... the sequence of function only converges in the range .. (e^(-e), e^(1/e)) ...

Answer 35

I am not quite sure if I can understand the step "\(\ln(y_2)=z \ln (z^z)=z^2 \ln( z)\)"
For \(y_2\), isn't it \(y_2 = z^{z^{z}}}\)?
If it is, then wouldn't we get \(\ln(y_2)=z^z \ln (z)\) instead?

Answer 36

You are right, I am wrong @Callisto

Answer 37

let
\[

f(x) = x^{x^{ .. }} = a \\
x^{x^{ .. }} \log(x) = \log(a) \\
a \log(x) = \log(a) \hspace{1 cm} -(1)\\
\log(x) = \frac 1 a \log(a)
\]
differentiating both sides,
\[
\frac 1 x \frac{d x }{da} = \frac 1 {a^2 } (1 - \log(a))
\]
upon maximizing it, we get \(a = e\) hence the max value of \( x \) is \( e^{1/e} \) .... beyond that value, \( f(x) \) does not converge.


for \( x = \sqrt 2 \) , \( (1) \) obviously has two roots, http://www.wolframalpha.com/input/?i=Plot+y+Log%5Bx%5D+%3D+Log%5By%5D%3B+x%3DSqrt%5B2%5D+from+y%3D0+to+5
but we have restriction result from above. so it cannot be 4.

beyond the vertical tangent in this curve ... the function diverges.
http://www.wolframalpha.com/input/?i=Plot+y+Log%5Bx%5D+%3D+Log%5By%5D

Answer 38

\( \sqrt 2 < e^{1/e} \implies f(\sqrt 2 ) = 2 < e < 4 \)

@eliassaab prof i think this is true ... here is a short code that I wrote ... after 100 terms it shows that it's value is 2
http://ideone.com/28ZDlV

also I think we can also use monotone convergence theorem to show that it's limit point exits.

Answer 39

I am not so sure ... if it's the right proof ... let \( \displaystyle a_n = (\sqrt 2) ^ {a_{n-1}} \) with \( a_1 = \sqrt 2 \). let \( a_{n-1} < 2 \) , then \(a_n = (\sqrt 2) ^{a_{n-1}} < (\sqrt 2) ^{2} = 2\). Hence by induction, the sequence is bounded. for monotonicity ...

Answer 40

*for monotonicity ... it's obvious. \( a_{n} > a_{n-1} \) hence it must converge.

Answer 41

I just logged on to say that
\[
z= \sqrt 2
\]
then an infinite tower of z will converge to 2

Answer 42

Try this code with Mathematica with \( z=\sqrt 2\). We can construct the following table
with h[n_, z_] := Nest[z^# &, z, n]
h gives a tower of power of z of length n+1
{{10, 1.98871}, {11, 1.99219}, {12, 1.99459}, {13, 1.99626}, {14,
1.99741}, {15, 1.9982}, {16, 1.99876}, {17, 1.99914}, {18,
1.9994}, {19, 1.99959}, {20, 1.99971}, {21, 1.9998}, {22,
1.99986}, {23, 1.9999}, {24, 1.99993}, {25, 1.99995}, {26,
1.99997}, {27, 1.99998}, {28, 1.99998}, {29, 1.99999}, {30,
1.99999}, {31, 1.99999}, {32, 2.}, {33, 2.}, {34, 2.}, {35,
2.}, {36, 2.}, {37, 2.}, {38, 2.}, {39, 2.}, {40, 2.}, {41,
2.}, {42, 2.}, {43, 2.}, {44, 2.}, {45, 2.}, {46, 2.}, {47,
2.}, {48, 2.}, {49, 2.}, {50, 2.}}
@experimentX

Answer 43

The code to generate the above code is
Table[{i, h[i, N[Sqrt[2]]]}, {i, 10, 50}]

Answer 44

Looking at general case:\[
\begin{array}{rcl}
z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} &=& c \\
z^{\left(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}}\right)} &=& z^c \\
z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} &=& z^c \\
c &=& z^c \\
z^c-c &=& 0 \\
\end{array}
\]
Seems like you'd need to use Newton's method to find a solution.

Answer 45

Would you mind elaborating it? By the way, I haven't learnt Newton's method yet.

Answer 46

It's an iterative root finding method. You guess \(x_0\), then you use:\[
x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)}
\]To get closer to the root. In this case \(f(x) = x^c-c\).

Answer 47

Wait nevermind what I said.

Answer 48

Take both sides to the power of \(1/c\). \[
z = c^{1/c}
\]

Answer 49

so if you had \(=4\) then it would be \(4^{1/4}\)

Answer 50

Which is \(\sqrt{2}\)?

Answer 51

not all infinities are the same

Answer 52

Hmmm, this is a tough question.

Answer 53

So far, no one has elaborated on the fallacy.

Answer 54

For one, is there a value for \(z\) which will even give us 4? It seems \(\sqrt{2}\) doesn't do this.

Answer 55

Hmm, but isn't it what we get when we solve the equation? Why is it not correct? What's wrong when we solve the equation?

Answer 56

Well, everything we did when we solved to get \(z=c^{1/c}\) assumed that the tower converged.

Answer 57

certain assumptions were made. I'm not sure which assumption was wrong.

Answer 58

I have a question. When your teacher solved for c, did he use my method, or the log method?

Answer 59

Here's what he wrote:
\[z^{z^{z^{z^{.^{.^{.}}}}}}=2\]\[z^2=2\]\[z=\sqrt{2}\]

Answer 60

http://www.wolframalpha.com/input/?i=x%5E%28x%5E%28x%5E%28x%5E%28x%29%29%29%29%2C+x+%3D+0+to+5

Answer 61

This is a rough approximation.

Answer 62

When we consider the series @experimentX talked about, we could find out where it hypothetically stops being convergent.

Answer 63

All I can really think to say is that the flaw with:
\[z^{z^{z^{z^{.^{.^{.}}}}}}=4\]\[z^4=4\]\[z=\sqrt{4}\]
lies with the assumption that \(z\) exists.

Answer 64

Consider if we assume for example that and \(x\) exists such that:\[
0x = 4
\]

Answer 65

Now, I am very confused.
How do you define and equation? How do you know if there exist a solution to the equation? How are convergence and divergence related to solving an equation?

Answer 66

convergence and divergence are related to anything that involve infinity.

Answer 67

There is no way to know solutions exist.

Answer 68

Unless some other math person has found solutions

Answer 69

Okay think of the equation:\[
0x = 4
\]once more. If \(x=\infty\) then we can possibly make this equation work.

Answer 70

From now on let's define \[
h(z) = z^{z^{z^{z^{.^{.^{.}}}}}}
\]

Answer 71

Consider we tried \(\lim_{z\to -\infty} h(z)\)
It would swap between 0 and an indeterminate form.

Answer 72

If we tried \(\lim_{z\to 0}h(z)\) we would get an indeterminate form

Answer 73

What if \(h(z)\) isn't actually a function?

For example consider \(\pm x=2\).
We let \(x=2\) and get \(-2\) or \(2\) and then proceed to say \(-2=2\).

Yet in the case of \(x=0\), we know \(\pm x=0\) and is a function.

Answer 74

Remember than indeterminate forms such as \(0/0\) are problematic because they can give you any number.

Answer 75

What if the solution to the equation is a complex number?
In that case, our exponent rules would not work.

Answer 76

I wish I had a better answer than "the assumption z exists leads to a fallacy", because it doesn't properly pinpoint what is going wrong here.

Answer 77

Sorry, I am having trouble in understanding the example ±x=2.

Answer 78

Well, if we said that \(g(x) = \pm x\), then we are assuming \(\pm x\) is a function, and it leads to many issues.

We could say find \(z\) where \(g(z) = c\). Then we find out \(g^{-1}(c) = z = |c|\). In the case where \(c=2\), then \(z=2\), in the case where \(c=-2\) then \(z=2\). We would then say \(2=g(2)=-2\). See the problem here?

Answer 79

Ah, by the way, when we draw conclusion that 2=4 from \(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}} = 2\) and \(z^{z^{z^{\cdot ^{\cdot^{\cdot}}}}}=4\), we do it by solving the two equations, getting both z's being equal to \(\sqrt{2}\). So, get left side of both equations are the same, so, the right side should also be the same, i.e. 2=4.

Answer 80

In your case, by solving your equation, you actually get two different answers.

Answer 81

Remember when you said \[
z^2=2
\]This gives 2 answers, doesn't it?
\[z=\pm \sqrt{2}
\]

Answer 82

Hmm... Yes.

Answer 83

When you said \[
z^4 = 4
\]You have 4 answers: \[
z^2=\pm 2 \implies z=\pm \sqrt{\pm 2}
\]

Answer 84

Which is just \(\pm \sqrt{2}\) and \(\pm \sqrt {2}i\).

Answer 85

\[
z^c-c=0 \implies z=c^{\frac{1}{c}}

\]
You do not Newton's method to do that.

Answer 86

@wio And then?

Answer 87

@eliassaab The problem is that \(1/c\) exponent is troublesome, especially if \(c=2^n\).

Answer 88

Newton's method makes it clear there will be many solutions, while that exponent ignores this fact.

Answer 89

@Callisto I would like to know the actual answer, but I only have many hypothesis.

Answer 90

My best answer is "you can't prove something is true by assuming it is true".
You started out assuming \(h(z)=4\), and then used algebra to show \(h^{-1}(4)=\sqrt{2}\).
The reality is that \(h(\sqrt{2})=2\neq 4\).

Answer 91

So, there must be something wrong when we solve it?

Answer 92

Yeah.

Answer 93

Consider \[\begin{array}{rcl}
\prod\limits_{k=0}^{\infty}x&=&c\\
x\prod\limits_{k=0}^{\infty}x&=&c\\
xc&=&c\\
x&=&1
\end{array}
\]Now: \[
\prod\limits_{k=0}^{\infty}x=5\implies x\prod\limits_{k=0}^{\infty}x=5\implies 5x=5\implies x=1
\]And:\[
\prod\limits_{k=0}^{\infty}x=6\implies x\prod\limits_{k=0}^{\infty}6=5\implies 6x=6\implies x=1
\]

Thus: \[
5=\prod\limits_{k=0}^{\infty}x=6
\]

Answer 94

The fallacy at play here isn't any different than yours, is it?

Answer 95

Doing it for sums: \[
\begin{array}{rcl}
\sum\limits_0^{\infty}x&=&c\\
x+\sum\limits_0^{\infty}x&=&c\\
x+c&=&c\\
x&=&0
\end{array}
\]

Answer 96

So you have a problem with dividing by \(c\) because \(c\) might be \(0\), but not taking both sides to the \(1/c\) power... when \(c\) might be \(0\)?

Answer 97

What If I just say \(c\neq 0\)? Does that change much?