In $x^{x^{x^{x^{…}}}}=2$, the answer is $x=sqrt{2}$. Why not also $-sqrt{2}$?












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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?






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  • 5




    A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
    – Matti P.
    Jan 4 at 9:07






  • 1




    By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
    – Michael Rozenberg
    Jan 4 at 9:07












  • @MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
    – Henry
    Jan 4 at 9:14










  • @Henry I said about domain only.
    – Michael Rozenberg
    Jan 4 at 9:15










  • The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
    – KM101
    Jan 4 at 9:15


















1














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







share|cite|improve this question









New contributor




b.ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 5




    A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
    – Matti P.
    Jan 4 at 9:07






  • 1




    By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
    – Michael Rozenberg
    Jan 4 at 9:07












  • @MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
    – Henry
    Jan 4 at 9:14










  • @Henry I said about domain only.
    – Michael Rozenberg
    Jan 4 at 9:15










  • The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
    – KM101
    Jan 4 at 9:15
















1












1








1


1





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







share|cite|improve this question









New contributor




b.ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






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b.ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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edited Jan 4 at 11:42









Blue

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asked Jan 4 at 9:02









b.benb.ben

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  • 5




    A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
    – Matti P.
    Jan 4 at 9:07






  • 1




    By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
    – Michael Rozenberg
    Jan 4 at 9:07












  • @MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
    – Henry
    Jan 4 at 9:14










  • @Henry I said about domain only.
    – Michael Rozenberg
    Jan 4 at 9:15










  • The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
    – KM101
    Jan 4 at 9:15
















  • 5




    A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
    – Matti P.
    Jan 4 at 9:07






  • 1




    By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
    – Michael Rozenberg
    Jan 4 at 9:07












  • @MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
    – Henry
    Jan 4 at 9:14










  • @Henry I said about domain only.
    – Michael Rozenberg
    Jan 4 at 9:15










  • The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
    – KM101
    Jan 4 at 9:15










5




5




A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
– Matti P.
Jan 4 at 9:07




A negative non-integer to a negative non-integer power is a little bit problematic (in real numbers) ...
– Matti P.
Jan 4 at 9:07




1




1




By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
– Michael Rozenberg
Jan 4 at 9:07






By the definition, if we wrote $x^x$ then the domain is $(0,+infty).$
– Michael Rozenberg
Jan 4 at 9:07














@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
– Henry
Jan 4 at 9:14




@MichaelRozenberg though if you want $x^{x^{x^{x^{...}}}}$ to converge then you need $e^{-e} leq x leq e^{1/e}$ roughly $0.066 lt x lt 1.44$
– Henry
Jan 4 at 9:14












@Henry I said about domain only.
– Michael Rozenberg
Jan 4 at 9:15




@Henry I said about domain only.
– Michael Rozenberg
Jan 4 at 9:15












The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
– KM101
Jan 4 at 9:15






The domain of $y = x^x$ is $x > 0$ if you’re dealing with only real numbers.
– KM101
Jan 4 at 9:15












2 Answers
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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?






share|cite|improve this answer





























    1














    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.






    share|cite|improve this answer























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      2 Answers
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      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?






      share|cite|improve this answer


























        1














        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?






        share|cite|improve this answer
























          1












          1








          1






          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?






          share|cite|improve this answer












          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?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 11:47









          J.G.J.G.

          23.3k22137




          23.3k22137























              1














              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.






              share|cite|improve this answer




























                1














                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.






                share|cite|improve this answer


























                  1












                  1








                  1






                  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.






                  share|cite|improve this answer














                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 4 at 11:59

























                  answered Jan 4 at 9:22









                  Todor MarkovTodor Markov

                  1,619410




                  1,619410






















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