Mixing
In $x^{x^{x^{x^{…}}}}=2$, the answer is $x=sqrt{2}$. Why not also $-sqrt{2}$?
$begingroup$
From this question
$$x^{x^{x^{x^{...}}}}=2$$
The answer is clearly $sqrt{2}$, but I'm curious why it's not also $-sqrt{2}$.
Am I missing something basic?
puzzle
edited Jan 4 at 11:42
Blue
47.8k870152
asked Jan 4 at 9:02
b.benb.ben
1113
$endgroup$
add a comment |
$begingroup$
From this question
$$x^{x^{x^{x^{...}}}}=2$$
The answer is clearly $sqrt{2}$, but I'm curious why it's not also $-sqrt{2}$.
Am I missing something basic?
puzzle
edited Jan 4 at 11:42
Blue
47.8k870152
asked Jan 4 at 9:02
b.benb.ben
1113
$endgroup$
5
$begingroup$
A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
$endgroup$
– Matti P.
Jan 4 at 9:07
1
$begingroup$
By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
$endgroup$
– Michael Rozenberg
Jan 4 at 9:07
$begingroup$
@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
$endgroup$
– Henry
Jan 4 at 9:14
$begingroup$
@Henry I said about domain only.
$endgroup$
– Michael Rozenberg
Jan 4 at 9:15
$begingroup$
The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
$endgroup$
– KM101
Jan 4 at 9:15
add a comment |
$begingroup$
From this question
$$x^{x^{x^{x^{...}}}}=2$$
The answer is clearly $sqrt{2}$, but I'm curious why it's not also $-sqrt{2}$.
Am I missing something basic?
puzzle
edited Jan 4 at 11:42
Blue
47.8k870152
asked Jan 4 at 9:02
b.benb.ben
1113
$endgroup$
From this question
$$x^{x^{x^{x^{...}}}}=2$$
The answer is clearly $sqrt{2}$, but I'm curious why it's not also $-sqrt{2}$.
Am I missing something basic?
puzzle
puzzle
edited Jan 4 at 11:42
Blue
47.8k870152
asked Jan 4 at 9:02
b.benb.ben
1113
edited Jan 4 at 11:42
Blue
47.8k870152
asked Jan 4 at 9:02
b.benb.ben
1113
edited Jan 4 at 11:42
Blue
47.8k870152
edited Jan 4 at 11:42
Blue
47.8k870152
edited Jan 4 at 11:42
Blue
47.8k870152
47.8k870152
asked Jan 4 at 9:02
b.benb.ben
1113
asked Jan 4 at 9:02
b.benb.ben
1113
asked Jan 4 at 9:02
b.benb.ben
1113
1113
5
$begingroup$
A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
$endgroup$
– Matti P.
Jan 4 at 9:07
1
$begingroup$
By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
$endgroup$
– Michael Rozenberg
Jan 4 at 9:07
$begingroup$
@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
$endgroup$
– Henry
Jan 4 at 9:14
$begingroup$
@Henry I said about domain only.
$endgroup$
– Michael Rozenberg
Jan 4 at 9:15
$begingroup$
The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
$endgroup$
– KM101
Jan 4 at 9:15
add a comment |
5
$begingroup$
A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
$endgroup$
– Matti P.
Jan 4 at 9:07
1
$begingroup$
By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
$endgroup$
– Michael Rozenberg
Jan 4 at 9:07
$begingroup$
@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
$endgroup$
– Henry
Jan 4 at 9:14
$begingroup$
@Henry I said about domain only.
$endgroup$
– Michael Rozenberg
Jan 4 at 9:15
$begingroup$
The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
$endgroup$
– KM101
Jan 4 at 9:15
5
5
$begingroup$
A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
$endgroup$
– Matti P.
Jan 4 at 9:07
$begingroup$
A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
$endgroup$
– Matti P.
Jan 4 at 9:07
1
1
$begingroup$
By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
$endgroup$
– Michael Rozenberg
Jan 4 at 9:07
$begingroup$
By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
$endgroup$
– Michael Rozenberg
Jan 4 at 9:07
$begingroup$
@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
$endgroup$
– Henry
Jan 4 at 9:14
$begingroup$
@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
$endgroup$
– Henry
Jan 4 at 9:14
$begingroup$
@Henry I said about domain only.
$endgroup$
– Michael Rozenberg
Jan 4 at 9:15
$begingroup$
@Henry I said about domain only.
$endgroup$
– Michael Rozenberg
Jan 4 at 9:15
$begingroup$
The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
$endgroup$
– KM101
Jan 4 at 9:15
$begingroup$
The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
$endgroup$
– KM101
Jan 4 at 9:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The tower is $lim_{ntoinfty}u_n$ with $u_1=x,,u_{n+1}=x^{u_n}$. If $x=sqrt{2}$ this is an increasing sequence of positive numbers $<2$. If $x=-sqrt{2}$, on the other hand, it quickly goes off the rails. What is $(-sqrt{2})^{-sqrt{2}}$ supposed to be?
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
$endgroup$
add a comment |
$begingroup$
You have $x^{x^{x^{dots}}} = 2$. You can simply substitute in the exponent to conclude that $x^2 = 2$. But that's only one way. If, instead you have that $x^2=2$, it's not necessarily true that $x^{x^{x^{dots}}} = 2$. So solving $x^2=2$ lets us find candidate solutions for the original equation, but then we need to confirm that each candidate is in fact a solution.
First, let's check whether $x=sqrt{2}$ satisfy $x^{x^{x^{dots}}} = 2$. For this we need to know what $x^{x^{x^{dots}}} = 2$ actually is. We can define it as the limit of the sequence $x, x^x, x^{x^x}, x^{x^{x^x}}, dots$, and we can check that, if $x=sqrt{2}$, this sequence indeed converges to 2.
What about $x=-sqrt{2}$? Consider $(-sqrt{2})^{-sqrt{2}}$. This is not even defined in real numbers. See How do you compute negative numbers to fractional powers? So we need to go to complex numbers. Then $(-sqrt{2})^{-sqrt{2}}$ actually takes infinitely many values, which causes problems on its own. EDIT: We can consider the sequence of principal values and try to check if it converges, but that turns out to be more complex than I thought.
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
The tower is $lim_{ntoinfty}u_n$ with $u_1=x,,u_{n+1}=x^{u_n}$. If $x=sqrt{2}$ this is an increasing sequence of positive numbers $<2$. If $x=-sqrt{2}$, on the other hand, it quickly goes off the rails. What is $(-sqrt{2})^{-sqrt{2}}$ supposed to be?
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
$endgroup$
add a comment |
$begingroup$
The tower is $lim_{ntoinfty}u_n$ with $u_1=x,,u_{n+1}=x^{u_n}$. If $x=sqrt{2}$ this is an increasing sequence of positive numbers $<2$. If $x=-sqrt{2}$, on the other hand, it quickly goes off the rails. What is $(-sqrt{2})^{-sqrt{2}}$ supposed to be?
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
$endgroup$
add a comment |
$begingroup$
The tower is $lim_{ntoinfty}u_n$ with $u_1=x,,u_{n+1}=x^{u_n}$. If $x=sqrt{2}$ this is an increasing sequence of positive numbers $<2$. If $x=-sqrt{2}$, on the other hand, it quickly goes off the rails. What is $(-sqrt{2})^{-sqrt{2}}$ supposed to be?
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
$endgroup$
The tower is $lim_{ntoinfty}u_n$ with $u_1=x,,u_{n+1}=x^{u_n}$. If $x=sqrt{2}$ this is an increasing sequence of positive numbers $<2$. If $x=-sqrt{2}$, on the other hand, it quickly goes off the rails. What is $(-sqrt{2})^{-sqrt{2}}$ supposed to be?
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
answered Jan 4 at 11:47
J.G.J.G.
24.4k22539
24.4k22539
add a comment |
add a comment |
$begingroup$
You have $x^{x^{x^{dots}}} = 2$. You can simply substitute in the exponent to conclude that $x^2 = 2$. But that's only one way. If, instead you have that $x^2=2$, it's not necessarily true that $x^{x^{x^{dots}}} = 2$. So solving $x^2=2$ lets us find candidate solutions for the original equation, but then we need to confirm that each candidate is in fact a solution.
First, let's check whether $x=sqrt{2}$ satisfy $x^{x^{x^{dots}}} = 2$. For this we need to know what $x^{x^{x^{dots}}} = 2$ actually is. We can define it as the limit of the sequence $x, x^x, x^{x^x}, x^{x^{x^x}}, dots$, and we can check that, if $x=sqrt{2}$, this sequence indeed converges to 2.
What about $x=-sqrt{2}$? Consider $(-sqrt{2})^{-sqrt{2}}$. This is not even defined in real numbers. See How do you compute negative numbers to fractional powers? So we need to go to complex numbers. Then $(-sqrt{2})^{-sqrt{2}}$ actually takes infinitely many values, which causes problems on its own. EDIT: We can consider the sequence of principal values and try to check if it converges, but that turns out to be more complex than I thought.
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
$endgroup$
add a comment |
$begingroup$
You have $x^{x^{x^{dots}}} = 2$. You can simply substitute in the exponent to conclude that $x^2 = 2$. But that's only one way. If, instead you have that $x^2=2$, it's not necessarily true that $x^{x^{x^{dots}}} = 2$. So solving $x^2=2$ lets us find candidate solutions for the original equation, but then we need to confirm that each candidate is in fact a solution.
First, let's check whether $x=sqrt{2}$ satisfy $x^{x^{x^{dots}}} = 2$. For this we need to know what $x^{x^{x^{dots}}} = 2$ actually is. We can define it as the limit of the sequence $x, x^x, x^{x^x}, x^{x^{x^x}}, dots$, and we can check that, if $x=sqrt{2}$, this sequence indeed converges to 2.
What about $x=-sqrt{2}$? Consider $(-sqrt{2})^{-sqrt{2}}$. This is not even defined in real numbers. See How do you compute negative numbers to fractional powers? So we need to go to complex numbers. Then $(-sqrt{2})^{-sqrt{2}}$ actually takes infinitely many values, which causes problems on its own. EDIT: We can consider the sequence of principal values and try to check if it converges, but that turns out to be more complex than I thought.
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
$endgroup$
add a comment |
$begingroup$
You have $x^{x^{x^{dots}}} = 2$. You can simply substitute in the exponent to conclude that $x^2 = 2$. But that's only one way. If, instead you have that $x^2=2$, it's not necessarily true that $x^{x^{x^{dots}}} = 2$. So solving $x^2=2$ lets us find candidate solutions for the original equation, but then we need to confirm that each candidate is in fact a solution.
First, let's check whether $x=sqrt{2}$ satisfy $x^{x^{x^{dots}}} = 2$. For this we need to know what $x^{x^{x^{dots}}} = 2$ actually is. We can define it as the limit of the sequence $x, x^x, x^{x^x}, x^{x^{x^x}}, dots$, and we can check that, if $x=sqrt{2}$, this sequence indeed converges to 2.
What about $x=-sqrt{2}$? Consider $(-sqrt{2})^{-sqrt{2}}$. This is not even defined in real numbers. See How do you compute negative numbers to fractional powers? So we need to go to complex numbers. Then $(-sqrt{2})^{-sqrt{2}}$ actually takes infinitely many values, which causes problems on its own. EDIT: We can consider the sequence of principal values and try to check if it converges, but that turns out to be more complex than I thought.
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
$endgroup$
You have $x^{x^{x^{dots}}} = 2$. You can simply substitute in the exponent to conclude that $x^2 = 2$. But that's only one way. If, instead you have that $x^2=2$, it's not necessarily true that $x^{x^{x^{dots}}} = 2$. So solving $x^2=2$ lets us find candidate solutions for the original equation, but then we need to confirm that each candidate is in fact a solution.
First, let's check whether $x=sqrt{2}$ satisfy $x^{x^{x^{dots}}} = 2$. For this we need to know what $x^{x^{x^{dots}}} = 2$ actually is. We can define it as the limit of the sequence $x, x^x, x^{x^x}, x^{x^{x^x}}, dots$, and we can check that, if $x=sqrt{2}$, this sequence indeed converges to 2.
What about $x=-sqrt{2}$? Consider $(-sqrt{2})^{-sqrt{2}}$. This is not even defined in real numbers. See How do you compute negative numbers to fractional powers? So we need to go to complex numbers. Then $(-sqrt{2})^{-sqrt{2}}$ actually takes infinitely many values, which causes problems on its own. EDIT: We can consider the sequence of principal values and try to check if it converges, but that turns out to be more complex than I thought.
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
edited Jan 4 at 11:59
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
answered Jan 4 at 9:22
Todor MarkovTodor Markov
1,819410
1,819410
add a comment |
add a comment |
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5
$begingroup$
A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
$endgroup$
– Matti P.
Jan 4 at 9:07
1
$begingroup$
By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
$endgroup$
– Michael Rozenberg
Jan 4 at 9:07
$begingroup$
@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
$endgroup$
– Henry
Jan 4 at 9:14
$begingroup$
@Henry I said about domain only.
$endgroup$
– Michael Rozenberg
Jan 4 at 9:15
$begingroup$
The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
$endgroup$
– KM101
Jan 4 at 9:15