Solve $i^i^i^ldots$ [duplicate]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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  • What is the value of $i^i^i^ldots$? [closed]

    1 answer



How to find $$i^i^i^ldots quad :quad i=sqrt-1$$
I'm able to find the solution for the finite powers
using
$$i=e^i(2kpi+fracpi2)quad:quad kinmathbbZ$$
$$i^i=e^-(2kpi+fracpi2)$$
$$i^i^i=e^-i(2pi k+fracpi2)=-i$$
$$i^i^i^i=e^(2pi k+fracpi2) $$
$$textand so on$$
but what should be the approach to solve for infitie powers$space$?




complex-numbers exponentiation tetration power-towers




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marked as duplicate by Math Lover, Simply Beautiful Art, lab bhattacharjee, Did, StubbornAtom Aug 12 at 4:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    Note that one should define exponentiation to be single-valued by choosing a branch, which would make this question clearer...
    – Simply Beautiful Art
    Aug 12 at 3:03






  • 2




    Your method of computing the iterates is not correct. Note that $(i^i)^i ne i^(i^i)$.
    – Shalop
    Aug 12 at 3:18











  • @Shalop What should be the correct method of computing the iterates?
    – Arpit Yadav
    Aug 12 at 3:21







  • 1




    Going by your rules, we say $i^z = e^big(2k+frac12big)ipi z$. Let us call this function $f(z)$. Now compute $f(i)$. Then compute $f(f(i))$. Then compute $f(f(f(i)))$. You will see that it is not the same as what you are doing.
    – Shalop
    Aug 12 at 3:44







  • 1




    See also: What is the order when doing $x^y^z$ and why?.
    – Simply Beautiful Art
    Aug 12 at 4:04














up vote
2
down vote

favorite
1













This question already has an answer here:



  • What is the value of $i^i^i^ldots$? [closed]

    1 answer



How to find $$i^i^i^ldots quad :quad i=sqrt-1$$
I'm able to find the solution for the finite powers
using
$$i=e^i(2kpi+fracpi2)quad:quad kinmathbbZ$$
$$i^i=e^-(2kpi+fracpi2)$$
$$i^i^i=e^-i(2pi k+fracpi2)=-i$$
$$i^i^i^i=e^(2pi k+fracpi2) $$
$$textand so on$$
but what should be the approach to solve for infitie powers$space$?




complex-numbers exponentiation tetration power-towers




share|cite|improve this question














marked as duplicate by Math Lover, Simply Beautiful Art, lab bhattacharjee, Did, StubbornAtom Aug 12 at 4:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    Note that one should define exponentiation to be single-valued by choosing a branch, which would make this question clearer...
    – Simply Beautiful Art
    Aug 12 at 3:03






  • 2




    Your method of computing the iterates is not correct. Note that $(i^i)^i ne i^(i^i)$.
    – Shalop
    Aug 12 at 3:18











  • @Shalop What should be the correct method of computing the iterates?
    – Arpit Yadav
    Aug 12 at 3:21







  • 1




    Going by your rules, we say $i^z = e^big(2k+frac12big)ipi z$. Let us call this function $f(z)$. Now compute $f(i)$. Then compute $f(f(i))$. Then compute $f(f(f(i)))$. You will see that it is not the same as what you are doing.
    – Shalop
    Aug 12 at 3:44







  • 1




    See also: What is the order when doing $x^y^z$ and why?.
    – Simply Beautiful Art
    Aug 12 at 4:04












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1






This question already has an answer here:



  • What is the value of $i^i^i^ldots$? [closed]

    1 answer



How to find $$i^i^i^ldots quad :quad i=sqrt-1$$
I'm able to find the solution for the finite powers
using
$$i=e^i(2kpi+fracpi2)quad:quad kinmathbbZ$$
$$i^i=e^-(2kpi+fracpi2)$$
$$i^i^i=e^-i(2pi k+fracpi2)=-i$$
$$i^i^i^i=e^(2pi k+fracpi2) $$
$$textand so on$$
but what should be the approach to solve for infitie powers$space$?




complex-numbers exponentiation tetration power-towers




share|cite|improve this question















This question already has an answer here:



  • What is the value of $i^i^i^ldots$? [closed]

    1 answer



How to find $$i^i^i^ldots quad :quad i=sqrt-1$$
I'm able to find the solution for the finite powers
using
$$i=e^i(2kpi+fracpi2)quad:quad kinmathbbZ$$
$$i^i=e^-(2kpi+fracpi2)$$
$$i^i^i=e^-i(2pi k+fracpi2)=-i$$
$$i^i^i^i=e^(2pi k+fracpi2) $$
$$textand so on$$
but what should be the approach to solve for infitie powers$space$?





This question already has an answer here:



  • What is the value of $i^i^i^ldots$? [closed]

    1 answer





complex-numbers exponentiation tetration power-towers





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 12 at 3:07









Simply Beautiful Art

49.4k572172




49.4k572172










asked Aug 12 at 1:02









Arpit Yadav

331215




331215




marked as duplicate by Math Lover, Simply Beautiful Art, lab bhattacharjee, Did, StubbornAtom Aug 12 at 4:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Math Lover, Simply Beautiful Art, lab bhattacharjee, Did, StubbornAtom Aug 12 at 4:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 2




    Note that one should define exponentiation to be single-valued by choosing a branch, which would make this question clearer...
    – Simply Beautiful Art
    Aug 12 at 3:03






  • 2




    Your method of computing the iterates is not correct. Note that $(i^i)^i ne i^(i^i)$.
    – Shalop
    Aug 12 at 3:18











  • @Shalop What should be the correct method of computing the iterates?
    – Arpit Yadav
    Aug 12 at 3:21







  • 1




    Going by your rules, we say $i^z = e^big(2k+frac12big)ipi z$. Let us call this function $f(z)$. Now compute $f(i)$. Then compute $f(f(i))$. Then compute $f(f(f(i)))$. You will see that it is not the same as what you are doing.
    – Shalop
    Aug 12 at 3:44







  • 1




    See also: What is the order when doing $x^y^z$ and why?.
    – Simply Beautiful Art
    Aug 12 at 4:04












  • 2




    Note that one should define exponentiation to be single-valued by choosing a branch, which would make this question clearer...
    – Simply Beautiful Art
    Aug 12 at 3:03






  • 2




    Your method of computing the iterates is not correct. Note that $(i^i)^i ne i^(i^i)$.
    – Shalop
    Aug 12 at 3:18











  • @Shalop What should be the correct method of computing the iterates?
    – Arpit Yadav
    Aug 12 at 3:21







  • 1




    Going by your rules, we say $i^z = e^big(2k+frac12big)ipi z$. Let us call this function $f(z)$. Now compute $f(i)$. Then compute $f(f(i))$. Then compute $f(f(f(i)))$. You will see that it is not the same as what you are doing.
    – Shalop
    Aug 12 at 3:44







  • 1




    See also: What is the order when doing $x^y^z$ and why?.
    – Simply Beautiful Art
    Aug 12 at 4:04







2




2




Note that one should define exponentiation to be single-valued by choosing a branch, which would make this question clearer...
– Simply Beautiful Art
Aug 12 at 3:03




Note that one should define exponentiation to be single-valued by choosing a branch, which would make this question clearer...
– Simply Beautiful Art
Aug 12 at 3:03




2




2




Your method of computing the iterates is not correct. Note that $(i^i)^i ne i^(i^i)$.
– Shalop
Aug 12 at 3:18





Your method of computing the iterates is not correct. Note that $(i^i)^i ne i^(i^i)$.
– Shalop
Aug 12 at 3:18













@Shalop What should be the correct method of computing the iterates?
– Arpit Yadav
Aug 12 at 3:21





@Shalop What should be the correct method of computing the iterates?
– Arpit Yadav
Aug 12 at 3:21





1




1




Going by your rules, we say $i^z = e^big(2k+frac12big)ipi z$. Let us call this function $f(z)$. Now compute $f(i)$. Then compute $f(f(i))$. Then compute $f(f(f(i)))$. You will see that it is not the same as what you are doing.
– Shalop
Aug 12 at 3:44





Going by your rules, we say $i^z = e^big(2k+frac12big)ipi z$. Let us call this function $f(z)$. Now compute $f(i)$. Then compute $f(f(i))$. Then compute $f(f(f(i)))$. You will see that it is not the same as what you are doing.
– Shalop
Aug 12 at 3:44





1




1




See also: What is the order when doing $x^y^z$ and why?.
– Simply Beautiful Art
Aug 12 at 4:04




See also: What is the order when doing $x^y^z$ and why?.
– Simply Beautiful Art
Aug 12 at 4:04










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










$X = i^i^i^cdot$



So $X = i^X$.



So $log (X) = X log (i)$.



So $log (X) over X = log (i)$.



So $X = i W(-i)$, where $W$ is the Lambert $W$ or PolyLog function.



Mathematica evaluates this to $X = 0.44 + 0.36 i$.



Here's a graph on the imaginary plane of 200 successive exponentiations with the solution as a red dot:



enter image description here






share|cite|improve this answer


















  • 4




    This is a fixed point but proving that it converges is more work.
    – parsiad
    Aug 12 at 1:18






  • 4




    @parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence!
    – David G. Stork
    Aug 12 at 1:19






  • 1




    @DavidG.Stork Does series converge to any particular value?
    – Arpit Yadav
    Aug 12 at 1:21










  • This is purely formal, obviously.
    – Did
    Aug 12 at 3:51

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










$X = i^i^i^cdot$



So $X = i^X$.



So $log (X) = X log (i)$.



So $log (X) over X = log (i)$.



So $X = i W(-i)$, where $W$ is the Lambert $W$ or PolyLog function.



Mathematica evaluates this to $X = 0.44 + 0.36 i$.



Here's a graph on the imaginary plane of 200 successive exponentiations with the solution as a red dot:



enter image description here






share|cite|improve this answer


















  • 4




    This is a fixed point but proving that it converges is more work.
    – parsiad
    Aug 12 at 1:18






  • 4




    @parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence!
    – David G. Stork
    Aug 12 at 1:19






  • 1




    @DavidG.Stork Does series converge to any particular value?
    – Arpit Yadav
    Aug 12 at 1:21










  • This is purely formal, obviously.
    – Did
    Aug 12 at 3:51














up vote
3
down vote



accepted










$X = i^i^i^cdot$



So $X = i^X$.



So $log (X) = X log (i)$.



So $log (X) over X = log (i)$.



So $X = i W(-i)$, where $W$ is the Lambert $W$ or PolyLog function.



Mathematica evaluates this to $X = 0.44 + 0.36 i$.



Here's a graph on the imaginary plane of 200 successive exponentiations with the solution as a red dot:



enter image description here






share|cite|improve this answer


















  • 4




    This is a fixed point but proving that it converges is more work.
    – parsiad
    Aug 12 at 1:18






  • 4




    @parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence!
    – David G. Stork
    Aug 12 at 1:19






  • 1




    @DavidG.Stork Does series converge to any particular value?
    – Arpit Yadav
    Aug 12 at 1:21










  • This is purely formal, obviously.
    – Did
    Aug 12 at 3:51












up vote
3
down vote



accepted







up vote
3
down vote



accepted






$X = i^i^i^cdot$



So $X = i^X$.



So $log (X) = X log (i)$.



So $log (X) over X = log (i)$.



So $X = i W(-i)$, where $W$ is the Lambert $W$ or PolyLog function.



Mathematica evaluates this to $X = 0.44 + 0.36 i$.



Here's a graph on the imaginary plane of 200 successive exponentiations with the solution as a red dot:



enter image description here






share|cite|improve this answer














$X = i^i^i^cdot$



So $X = i^X$.



So $log (X) = X log (i)$.



So $log (X) over X = log (i)$.



So $X = i W(-i)$, where $W$ is the Lambert $W$ or PolyLog function.



Mathematica evaluates this to $X = 0.44 + 0.36 i$.



Here's a graph on the imaginary plane of 200 successive exponentiations with the solution as a red dot:



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 12 at 1:49

























answered Aug 12 at 1:16









David G. Stork

8,05521131




8,05521131







  • 4




    This is a fixed point but proving that it converges is more work.
    – parsiad
    Aug 12 at 1:18






  • 4




    @parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence!
    – David G. Stork
    Aug 12 at 1:19






  • 1




    @DavidG.Stork Does series converge to any particular value?
    – Arpit Yadav
    Aug 12 at 1:21










  • This is purely formal, obviously.
    – Did
    Aug 12 at 3:51












  • 4




    This is a fixed point but proving that it converges is more work.
    – parsiad
    Aug 12 at 1:18






  • 4




    @parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence!
    – David G. Stork
    Aug 12 at 1:19






  • 1




    @DavidG.Stork Does series converge to any particular value?
    – Arpit Yadav
    Aug 12 at 1:21










  • This is purely formal, obviously.
    – Did
    Aug 12 at 3:51







4




4




This is a fixed point but proving that it converges is more work.
– parsiad
Aug 12 at 1:18




This is a fixed point but proving that it converges is more work.
– parsiad
Aug 12 at 1:18




4




4




@parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence!
– David G. Stork
Aug 12 at 1:19




@parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence!
– David G. Stork
Aug 12 at 1:19




1




1




@DavidG.Stork Does series converge to any particular value?
– Arpit Yadav
Aug 12 at 1:21




@DavidG.Stork Does series converge to any particular value?
– Arpit Yadav
Aug 12 at 1:21












This is purely formal, obviously.
– Did
Aug 12 at 3:51




This is purely formal, obviously.
– Did
Aug 12 at 3:51


-

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