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

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











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













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

8,03421031




8,03421031







  • 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


-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier curve. Additionally it marks one side of the curve as inside t
閱讀完整內容

Carbon dioxide

圖片
"CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
閱讀完整內容