Mixing
Nested powers of $sqrt 2$ has a solution different from its limit. What does this mean?
$begingroup$
The infinitely nested power expression below has a limit of $2$:
$$x=sqrt2^{sqrt2^{sqrt2^{...}}}$$
In finding this limit, we may use:
$$x=(sqrt2)^x$$
But this expression has two solutions, $2$ and $4$.
We know that $2$ is the right answer by evaluating some finite truncations, but this $4$ is bothering me. What does $4$ mean in this expression? Is it significant in some way?
limits power-towers
edited Jan 30 at 6:43
the_fox
2,90231538
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
$endgroup$
add a comment |
$begingroup$
The infinitely nested power expression below has a limit of $2$:
$$x=sqrt2^{sqrt2^{sqrt2^{...}}}$$
In finding this limit, we may use:
$$x=(sqrt2)^x$$
But this expression has two solutions, $2$ and $4$.
We know that $2$ is the right answer by evaluating some finite truncations, but this $4$ is bothering me. What does $4$ mean in this expression? Is it significant in some way?
limits power-towers
edited Jan 30 at 6:43
the_fox
2,90231538
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
$endgroup$
$begingroup$
To find the solution of $x=2$ you may use $x^2 + 8 = 6x$.This has two solutions, $2$ and $4$. Does this $4$ bother you?
$endgroup$
– Umberto P.
Jan 29 at 10:36
4
$begingroup$
It's not quite a duplicate, but you might be interested in my answer to this older question and the first comment on that answer.
$endgroup$
– Mees de Vries
Jan 29 at 10:37
1
$begingroup$
By writing $sqrt2^{sqrt2^{sqrt2^{cdots}}}$, do you mean $lim_{n to infty} underbrace{sqrt2^{sqrt2^{sqrt2^{cdots^{sqrt2}}}}}_{n text{ copies of } sqrt2}$?
$endgroup$
– L. F.
Jan 29 at 10:54
$begingroup$
@UmbertoP. It doesn’t, because I know in your case the 4 is introduced when you rewrote x=2 with the polynomial, and an additional root is included within the rearranged expression. Now can you tell me where did the 4 come from in my question?
$endgroup$
– Flying_Banana
Jan 29 at 11:46
add a comment |
$begingroup$
The infinitely nested power expression below has a limit of $2$:
$$x=sqrt2^{sqrt2^{sqrt2^{...}}}$$
In finding this limit, we may use:
$$x=(sqrt2)^x$$
But this expression has two solutions, $2$ and $4$.
We know that $2$ is the right answer by evaluating some finite truncations, but this $4$ is bothering me. What does $4$ mean in this expression? Is it significant in some way?
limits power-towers
edited Jan 30 at 6:43
the_fox
2,90231538
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
$endgroup$
The infinitely nested power expression below has a limit of $2$:
$$x=sqrt2^{sqrt2^{sqrt2^{...}}}$$
In finding this limit, we may use:
$$x=(sqrt2)^x$$
But this expression has two solutions, $2$ and $4$.
We know that $2$ is the right answer by evaluating some finite truncations, but this $4$ is bothering me. What does $4$ mean in this expression? Is it significant in some way?
limits power-towers
limits power-towers
edited Jan 30 at 6:43
the_fox
2,90231538
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
edited Jan 30 at 6:43
the_fox
2,90231538
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
edited Jan 30 at 6:43
the_fox
2,90231538
edited Jan 30 at 6:43
the_fox
2,90231538
edited Jan 30 at 6:43
the_fox
2,90231538
2,90231538
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
asked Jan 29 at 10:31
Flying_BananaFlying_Banana
1895
1895
$begingroup$
To find the solution of $x=2$ you may use $x^2 + 8 = 6x$.This has two solutions, $2$ and $4$. Does this $4$ bother you?
$endgroup$
– Umberto P.
Jan 29 at 10:36
4
$begingroup$
It's not quite a duplicate, but you might be interested in my answer to this older question and the first comment on that answer.
$endgroup$
– Mees de Vries
Jan 29 at 10:37
1
$begingroup$
By writing $sqrt2^{sqrt2^{sqrt2^{cdots}}}$, do you mean $lim_{n to infty} underbrace{sqrt2^{sqrt2^{sqrt2^{cdots^{sqrt2}}}}}_{n text{ copies of } sqrt2}$?
$endgroup$
– L. F.
Jan 29 at 10:54
$begingroup$
@UmbertoP. It doesn’t, because I know in your case the 4 is introduced when you rewrote x=2 with the polynomial, and an additional root is included within the rearranged expression. Now can you tell me where did the 4 come from in my question?
$endgroup$
– Flying_Banana
Jan 29 at 11:46
add a comment |
$begingroup$
To find the solution of $x=2$ you may use $x^2 + 8 = 6x$.This has two solutions, $2$ and $4$. Does this $4$ bother you?
$endgroup$
– Umberto P.
Jan 29 at 10:36
4
$begingroup$
It's not quite a duplicate, but you might be interested in my answer to this older question and the first comment on that answer.
$endgroup$
– Mees de Vries
Jan 29 at 10:37
1
$begingroup$
By writing $sqrt2^{sqrt2^{sqrt2^{cdots}}}$, do you mean $lim_{n to infty} underbrace{sqrt2^{sqrt2^{sqrt2^{cdots^{sqrt2}}}}}_{n text{ copies of } sqrt2}$?
$endgroup$
– L. F.
Jan 29 at 10:54
$begingroup$
@UmbertoP. It doesn’t, because I know in your case the 4 is introduced when you rewrote x=2 with the polynomial, and an additional root is included within the rearranged expression. Now can you tell me where did the 4 come from in my question?
$endgroup$
– Flying_Banana
Jan 29 at 11:46
$begingroup$
To find the solution of $x=2$ you may use $x^2 + 8 = 6x$.This has two solutions, $2$ and $4$. Does this $4$ bother you?
$endgroup$
– Umberto P.
Jan 29 at 10:36
$begingroup$
To find the solution of $x=2$ you may use $x^2 + 8 = 6x$.This has two solutions, $2$ and $4$. Does this $4$ bother you?
$endgroup$
– Umberto P.
Jan 29 at 10:36
4
4
$begingroup$
It's not quite a duplicate, but you might be interested in my answer to this older question and the first comment on that answer.
$endgroup$
– Mees de Vries
Jan 29 at 10:37
$begingroup$
It's not quite a duplicate, but you might be interested in my answer to this older question and the first comment on that answer.
$endgroup$
– Mees de Vries
Jan 29 at 10:37
1
1
$begingroup$
By writing $sqrt2^{sqrt2^{sqrt2^{cdots}}}$, do you mean $lim_{n to infty} underbrace{sqrt2^{sqrt2^{sqrt2^{cdots^{sqrt2}}}}}_{n text{ copies of } sqrt2}$?
$endgroup$
– L. F.
Jan 29 at 10:54
$begingroup$
By writing $sqrt2^{sqrt2^{sqrt2^{cdots}}}$, do you mean $lim_{n to infty} underbrace{sqrt2^{sqrt2^{sqrt2^{cdots^{sqrt2}}}}}_{n text{ copies of } sqrt2}$?
$endgroup$
– L. F.
Jan 29 at 10:54
$begingroup$
@UmbertoP. It doesn’t, because I know in your case the 4 is introduced when you rewrote x=2 with the polynomial, and an additional root is included within the rearranged expression. Now can you tell me where did the 4 come from in my question?
$endgroup$
– Flying_Banana
Jan 29 at 11:46
$begingroup$
@UmbertoP. It doesn’t, because I know in your case the 4 is introduced when you rewrote x=2 with the polynomial, and an additional root is included within the rearranged expression. Now can you tell me where did the 4 come from in my question?
$endgroup$
– Flying_Banana
Jan 29 at 11:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is a common misconception that an expression like
$$sqrt2^{sqrt2^{sqrt2^{...}}}$$
denotes a well-defined number.
For more rigor, let us use
$$x=lim_{ntoinfty}a_n,text{ where }x_{n+1}=sqrt2^{x_n}.$$
Now if $x_0=2$, $x_n=2 forall n$ follows. Similarly, for $x_0=4$, $x_n=4$ follows. For other initial values, the sequence may converge to $2$, but may also diverge.
So the statement "has a limit of $2$" is dubious.
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_infty =;_{;_{;_{ ;_cdots} b}b}x_0 overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;)
$endgroup$
– Gottfried Helms
Jan 30 at 12:47
add a comment |
$begingroup$
If $sqrt{2}^{vdots}$ is well-defined, it must be as the limit of a sequence of the form $a_{n+1}=sqrt{2}^{a_n}$. The limit, if it exists, depends on $a_1$, but the most natural choice of $a_1$ is $sqrt{2}$; what else do we think is "at the top of the tower", which doesn't exist?
But we can prove by induction that if $a_1in[0,,2]$, then, for all $n$, $a_nle 2$, so $x:=limlimits_{ntoinfty}a_n$ is $x=2$, whereas $a_1=4$ implies $x=4$.
edited Jan 30 at 10:12
Did
248k23226466
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = sqrt2^x$? and expect only one answer, 2?
$endgroup$
– Flying_Banana
Jan 30 at 5:23
$begingroup$
@Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $vdots$ or down and right viz. $ddots$, I'll fix my notation.
$endgroup$
– J.G.
Jan 30 at 6:38
$begingroup$
This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $lim a_n=2$, for $a_1=4$, $lim a_n=4$, and, for every $a_1>4$, $a_ntoinfty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $ell$ the smallest of the two roots of $ell=c^ell$ and as "spurious" fixed point the largest of these two roots.
$endgroup$
– Did
Jan 30 at 7:32
$begingroup$
@did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead.
$endgroup$
– Gottfried Helms
Jan 30 at 9:45
$begingroup$
@GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment.
$endgroup$
– Did
Jan 30 at 10:10
|
show 3 more comments
$begingroup$
The problem here is that:
$$x = (sqrt{2})^x$$
is a necessary condition but not sufficient, that is, if $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ equals $x$ then it must hold that $x = (sqrt{2})^x$, but you don't know if $x$ can equal $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ in the first place.
One interesting property is that
$$x^{x^{x^{x^{cdots}}}}$$ for positive x converges only if $xin [e^{-e}, e^{1/e}]$ and converges to a value $yin[e^{-1}, e]$.
So if you are working with another expression like that and you get more than one solution, keep in mind that if it does not belong to that interval then it surely is not a solution of the expression.
https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0
Here is the proof of the property aforementioned and also, a more detailed analysis of the whole problem with nested power expressions.
edited Jan 30 at 6:42
the_fox
2,90231538
answered Jan 29 at 11:45
maxbpmaxbp
1467
$endgroup$
$begingroup$
Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $x^2=4 Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions?
$endgroup$
– Flying_Banana
Jan 30 at 5:31
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is a common misconception that an expression like
$$sqrt2^{sqrt2^{sqrt2^{...}}}$$
denotes a well-defined number.
For more rigor, let us use
$$x=lim_{ntoinfty}a_n,text{ where }x_{n+1}=sqrt2^{x_n}.$$
Now if $x_0=2$, $x_n=2 forall n$ follows. Similarly, for $x_0=4$, $x_n=4$ follows. For other initial values, the sequence may converge to $2$, but may also diverge.
So the statement "has a limit of $2$" is dubious.
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_infty =;_{;_{;_{ ;_cdots} b}b}x_0 overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;)
$endgroup$
– Gottfried Helms
Jan 30 at 12:47
add a comment |
$begingroup$
It is a common misconception that an expression like
$$sqrt2^{sqrt2^{sqrt2^{...}}}$$
denotes a well-defined number.
For more rigor, let us use
$$x=lim_{ntoinfty}a_n,text{ where }x_{n+1}=sqrt2^{x_n}.$$
Now if $x_0=2$, $x_n=2 forall n$ follows. Similarly, for $x_0=4$, $x_n=4$ follows. For other initial values, the sequence may converge to $2$, but may also diverge.
So the statement "has a limit of $2$" is dubious.
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_infty =;_{;_{;_{ ;_cdots} b}b}x_0 overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;)
$endgroup$
– Gottfried Helms
Jan 30 at 12:47
add a comment |
$begingroup$
It is a common misconception that an expression like
$$sqrt2^{sqrt2^{sqrt2^{...}}}$$
denotes a well-defined number.
For more rigor, let us use
$$x=lim_{ntoinfty}a_n,text{ where }x_{n+1}=sqrt2^{x_n}.$$
Now if $x_0=2$, $x_n=2 forall n$ follows. Similarly, for $x_0=4$, $x_n=4$ follows. For other initial values, the sequence may converge to $2$, but may also diverge.
So the statement "has a limit of $2$" is dubious.
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
$endgroup$
It is a common misconception that an expression like
$$sqrt2^{sqrt2^{sqrt2^{...}}}$$
denotes a well-defined number.
For more rigor, let us use
$$x=lim_{ntoinfty}a_n,text{ where }x_{n+1}=sqrt2^{x_n}.$$
Now if $x_0=2$, $x_n=2 forall n$ follows. Similarly, for $x_0=4$, $x_n=4$ follows. For other initial values, the sequence may converge to $2$, but may also diverge.
So the statement "has a limit of $2$" is dubious.
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
edited Jan 30 at 10:36
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
answered Jan 30 at 10:30
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_infty =;_{;_{;_{ ;_cdots} b}b}x_0 overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;)
$endgroup$
– Gottfried Helms
Jan 30 at 12:47
add a comment |
$begingroup$
I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_infty =;_{;_{;_{ ;_cdots} b}b}x_0 overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;)
$endgroup$
– Gottfried Helms
Jan 30 at 12:47
$begingroup$
I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_infty =;_{;_{;_{ ;_cdots} b}b}x_0 overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;)
$endgroup$
– Gottfried Helms
Jan 30 at 12:47
$begingroup$
I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_infty =;_{;_{;_{ ;_cdots} b}b}x_0 overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;)
$endgroup$
– Gottfried Helms
Jan 30 at 12:47
add a comment |
$begingroup$
If $sqrt{2}^{vdots}$ is well-defined, it must be as the limit of a sequence of the form $a_{n+1}=sqrt{2}^{a_n}$. The limit, if it exists, depends on $a_1$, but the most natural choice of $a_1$ is $sqrt{2}$; what else do we think is "at the top of the tower", which doesn't exist?
But we can prove by induction that if $a_1in[0,,2]$, then, for all $n$, $a_nle 2$, so $x:=limlimits_{ntoinfty}a_n$ is $x=2$, whereas $a_1=4$ implies $x=4$.
edited Jan 30 at 10:12
Did
248k23226466
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = sqrt2^x$? and expect only one answer, 2?
$endgroup$
– Flying_Banana
Jan 30 at 5:23
$begingroup$
@Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $vdots$ or down and right viz. $ddots$, I'll fix my notation.
$endgroup$
– J.G.
Jan 30 at 6:38
$begingroup$
This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $lim a_n=2$, for $a_1=4$, $lim a_n=4$, and, for every $a_1>4$, $a_ntoinfty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $ell$ the smallest of the two roots of $ell=c^ell$ and as "spurious" fixed point the largest of these two roots.
$endgroup$
– Did
Jan 30 at 7:32
$begingroup$
@did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead.
$endgroup$
– Gottfried Helms
Jan 30 at 9:45
$begingroup$
@GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment.
$endgroup$
– Did
Jan 30 at 10:10
|
show 3 more comments
$begingroup$
If $sqrt{2}^{vdots}$ is well-defined, it must be as the limit of a sequence of the form $a_{n+1}=sqrt{2}^{a_n}$. The limit, if it exists, depends on $a_1$, but the most natural choice of $a_1$ is $sqrt{2}$; what else do we think is "at the top of the tower", which doesn't exist?
But we can prove by induction that if $a_1in[0,,2]$, then, for all $n$, $a_nle 2$, so $x:=limlimits_{ntoinfty}a_n$ is $x=2$, whereas $a_1=4$ implies $x=4$.
edited Jan 30 at 10:12
Did
248k23226466
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = sqrt2^x$? and expect only one answer, 2?
$endgroup$
– Flying_Banana
Jan 30 at 5:23
$begingroup$
@Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $vdots$ or down and right viz. $ddots$, I'll fix my notation.
$endgroup$
– J.G.
Jan 30 at 6:38
$begingroup$
This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $lim a_n=2$, for $a_1=4$, $lim a_n=4$, and, for every $a_1>4$, $a_ntoinfty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $ell$ the smallest of the two roots of $ell=c^ell$ and as "spurious" fixed point the largest of these two roots.
$endgroup$
– Did
Jan 30 at 7:32
$begingroup$
@did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead.
$endgroup$
– Gottfried Helms
Jan 30 at 9:45
$begingroup$
@GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment.
$endgroup$
– Did
Jan 30 at 10:10
|
show 3 more comments
$begingroup$
If $sqrt{2}^{vdots}$ is well-defined, it must be as the limit of a sequence of the form $a_{n+1}=sqrt{2}^{a_n}$. The limit, if it exists, depends on $a_1$, but the most natural choice of $a_1$ is $sqrt{2}$; what else do we think is "at the top of the tower", which doesn't exist?
But we can prove by induction that if $a_1in[0,,2]$, then, for all $n$, $a_nle 2$, so $x:=limlimits_{ntoinfty}a_n$ is $x=2$, whereas $a_1=4$ implies $x=4$.
edited Jan 30 at 10:12
Did
248k23226466
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
$endgroup$
If $sqrt{2}^{vdots}$ is well-defined, it must be as the limit of a sequence of the form $a_{n+1}=sqrt{2}^{a_n}$. The limit, if it exists, depends on $a_1$, but the most natural choice of $a_1$ is $sqrt{2}$; what else do we think is "at the top of the tower", which doesn't exist?
But we can prove by induction that if $a_1in[0,,2]$, then, for all $n$, $a_nle 2$, so $x:=limlimits_{ntoinfty}a_n$ is $x=2$, whereas $a_1=4$ implies $x=4$.
edited Jan 30 at 10:12
Did
248k23226466
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
edited Jan 30 at 10:12
Did
248k23226466
edited Jan 30 at 10:12
Did
248k23226466
edited Jan 30 at 10:12
Did
248k23226466
248k23226466
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
answered Jan 29 at 12:04
J.G.J.G.
32.5k23250
32.5k23250
$begingroup$
Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = sqrt2^x$? and expect only one answer, 2?
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– Flying_Banana
Jan 30 at 5:23
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@Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $vdots$ or down and right viz. $ddots$, I'll fix my notation.
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– J.G.
Jan 30 at 6:38
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This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $lim a_n=2$, for $a_1=4$, $lim a_n=4$, and, for every $a_1>4$, $a_ntoinfty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $ell$ the smallest of the two roots of $ell=c^ell$ and as "spurious" fixed point the largest of these two roots.
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– Did
Jan 30 at 7:32
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@did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead.
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– Gottfried Helms
Jan 30 at 9:45
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@GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment.
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– Did
Jan 30 at 10:10
|
show 3 more comments
$begingroup$
Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = sqrt2^x$? and expect only one answer, 2?
$endgroup$
– Flying_Banana
Jan 30 at 5:23
$begingroup$
@Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $vdots$ or down and right viz. $ddots$, I'll fix my notation.
$endgroup$
– J.G.
Jan 30 at 6:38
$begingroup$
This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $lim a_n=2$, for $a_1=4$, $lim a_n=4$, and, for every $a_1>4$, $a_ntoinfty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $ell$ the smallest of the two roots of $ell=c^ell$ and as "spurious" fixed point the largest of these two roots.
$endgroup$
– Did
Jan 30 at 7:32
$begingroup$
@did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead.
$endgroup$
– Gottfried Helms
Jan 30 at 9:45
$begingroup$
@GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment.
$endgroup$
– Did
Jan 30 at 10:10
$begingroup$
Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = sqrt2^x$? and expect only one answer, 2?
$endgroup$
– Flying_Banana
Jan 30 at 5:23
$begingroup$
Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = sqrt2^x$? and expect only one answer, 2?
$endgroup$
– Flying_Banana
Jan 30 at 5:23
$begingroup$
@Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $vdots$ or down and right viz. $ddots$, I'll fix my notation.
$endgroup$
– J.G.
Jan 30 at 6:38
$begingroup$
@Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $vdots$ or down and right viz. $ddots$, I'll fix my notation.
$endgroup$
– J.G.
Jan 30 at 6:38
$begingroup$
This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $lim a_n=2$, for $a_1=4$, $lim a_n=4$, and, for every $a_1>4$, $a_ntoinfty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $ell$ the smallest of the two roots of $ell=c^ell$ and as "spurious" fixed point the largest of these two roots.
$endgroup$
– Did
Jan 30 at 7:32
$begingroup$
This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $lim a_n=2$, for $a_1=4$, $lim a_n=4$, and, for every $a_1>4$, $a_ntoinfty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $ell$ the smallest of the two roots of $ell=c^ell$ and as "spurious" fixed point the largest of these two roots.
$endgroup$
– Did
Jan 30 at 7:32
$begingroup$
@did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead.
$endgroup$
– Gottfried Helms
Jan 30 at 9:45
$begingroup$
@did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead.
$endgroup$
– Gottfried Helms
Jan 30 at 9:45
$begingroup$
@GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment.
$endgroup$
– Did
Jan 30 at 10:10
$begingroup$
@GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment.
$endgroup$
– Did
Jan 30 at 10:10
|
show 3 more comments
$begingroup$
The problem here is that:
$$x = (sqrt{2})^x$$
is a necessary condition but not sufficient, that is, if $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ equals $x$ then it must hold that $x = (sqrt{2})^x$, but you don't know if $x$ can equal $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ in the first place.
One interesting property is that
$$x^{x^{x^{x^{cdots}}}}$$ for positive x converges only if $xin [e^{-e}, e^{1/e}]$ and converges to a value $yin[e^{-1}, e]$.
So if you are working with another expression like that and you get more than one solution, keep in mind that if it does not belong to that interval then it surely is not a solution of the expression.
https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0
Here is the proof of the property aforementioned and also, a more detailed analysis of the whole problem with nested power expressions.
edited Jan 30 at 6:42
the_fox
2,90231538
answered Jan 29 at 11:45
maxbpmaxbp
1467
$endgroup$
$begingroup$
Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $x^2=4 Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions?
$endgroup$
– Flying_Banana
Jan 30 at 5:31
add a comment |
$begingroup$
The problem here is that:
$$x = (sqrt{2})^x$$
is a necessary condition but not sufficient, that is, if $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ equals $x$ then it must hold that $x = (sqrt{2})^x$, but you don't know if $x$ can equal $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ in the first place.
One interesting property is that
$$x^{x^{x^{x^{cdots}}}}$$ for positive x converges only if $xin [e^{-e}, e^{1/e}]$ and converges to a value $yin[e^{-1}, e]$.
So if you are working with another expression like that and you get more than one solution, keep in mind that if it does not belong to that interval then it surely is not a solution of the expression.
https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0
Here is the proof of the property aforementioned and also, a more detailed analysis of the whole problem with nested power expressions.
edited Jan 30 at 6:42
the_fox
2,90231538
answered Jan 29 at 11:45
maxbpmaxbp
1467
$endgroup$
$begingroup$
Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $x^2=4 Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions?
$endgroup$
– Flying_Banana
Jan 30 at 5:31
add a comment |
$begingroup$
The problem here is that:
$$x = (sqrt{2})^x$$
is a necessary condition but not sufficient, that is, if $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ equals $x$ then it must hold that $x = (sqrt{2})^x$, but you don't know if $x$ can equal $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ in the first place.
One interesting property is that
$$x^{x^{x^{x^{cdots}}}}$$ for positive x converges only if $xin [e^{-e}, e^{1/e}]$ and converges to a value $yin[e^{-1}, e]$.
So if you are working with another expression like that and you get more than one solution, keep in mind that if it does not belong to that interval then it surely is not a solution of the expression.
https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0
Here is the proof of the property aforementioned and also, a more detailed analysis of the whole problem with nested power expressions.
edited Jan 30 at 6:42
the_fox
2,90231538
answered Jan 29 at 11:45
maxbpmaxbp
1467
$endgroup$
The problem here is that:
$$x = (sqrt{2})^x$$
is a necessary condition but not sufficient, that is, if $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ equals $x$ then it must hold that $x = (sqrt{2})^x$, but you don't know if $x$ can equal $sqrt{2}^{sqrt{2}^{sqrt{2}^{cdots}}}$ in the first place.
One interesting property is that
$$x^{x^{x^{x^{cdots}}}}$$ for positive x converges only if $xin [e^{-e}, e^{1/e}]$ and converges to a value $yin[e^{-1}, e]$.
So if you are working with another expression like that and you get more than one solution, keep in mind that if it does not belong to that interval then it surely is not a solution of the expression.
https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0
Here is the proof of the property aforementioned and also, a more detailed analysis of the whole problem with nested power expressions.
edited Jan 30 at 6:42
the_fox
2,90231538
answered Jan 29 at 11:45
maxbpmaxbp
1467
edited Jan 30 at 6:42
the_fox
2,90231538
edited Jan 30 at 6:42
the_fox
2,90231538
edited Jan 30 at 6:42
the_fox
2,90231538
2,90231538
answered Jan 29 at 11:45
maxbpmaxbp
1467
answered Jan 29 at 11:45
maxbpmaxbp
1467
answered Jan 29 at 11:45
maxbpmaxbp
1467
1467
$begingroup$
Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $x^2=4 Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions?
$endgroup$
– Flying_Banana
Jan 30 at 5:31
add a comment |
$begingroup$
Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $x^2=4 Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions?
$endgroup$
– Flying_Banana
Jan 30 at 5:31
$begingroup$
Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $x^2=4 Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions?
$endgroup$
– Flying_Banana
Jan 30 at 5:31
$begingroup$
Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $x^2=4 Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions?
$endgroup$
– Flying_Banana
Jan 30 at 5:31
add a comment |
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$begingroup$
To find the solution of $x=2$ you may use $x^2 + 8 = 6x$.This has two solutions, $2$ and $4$. Does this $4$ bother you?
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– Umberto P.
Jan 29 at 10:36
4
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It's not quite a duplicate, but you might be interested in my answer to this older question and the first comment on that answer.
$endgroup$
– Mees de Vries
Jan 29 at 10:37
1
$begingroup$
By writing $sqrt2^{sqrt2^{sqrt2^{cdots}}}$, do you mean $lim_{n to infty} underbrace{sqrt2^{sqrt2^{sqrt2^{cdots^{sqrt2}}}}}_{n text{ copies of } sqrt2}$?
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– L. F.
Jan 29 at 10:54
$begingroup$
@UmbertoP. It doesn’t, because I know in your case the 4 is introduced when you rewrote x=2 with the polynomial, and an additional root is included within the rearranged expression. Now can you tell me where did the 4 come from in my question?
$endgroup$
– Flying_Banana
Jan 29 at 11:46