I'm in Limbo

Algebra Level 5

i i i . . . \LARGE { i }^{ { i }^{ i^{.^{.^{.}}} } }

For i = 1 i = \sqrt{-1} , the value of the infinitely nested exponent above is equal to A + B i A + Bi for real values A A and B B .

Find the value of A 2 + B 2 { A }^{ 2 }+{ B }^{ 2 } to 3 decimal places.


The answer is 0.322.

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3 solutions

i i i . . . = A + B i = i A + B i e π i 2 ( A + B i ) = e π B 2 e π A i 2 = A 2 + B 2 e tan 1 B A i i^{i^{i^{...}}} = A+ Bi = i^{A+Bi}\quad \Rightarrow e^{\frac{\pi i}{2}(A+Bi)} = e^{-\frac{\pi B}{2}} e^{\frac{\pi Ai}{2}} = \sqrt{A^2+B^2} e^{\tan^{-1} {\frac {B}{A}}i}

{ e π B 2 = A 2 + B 2 A 2 + B 2 = e π B . . . ( 1 ) π A 2 = tan 1 B A tan π A 2 = B A B = A tan π A 2 . . . ( 2 ) \begin{cases} e^{-\frac{\pi B}{2}} = \sqrt{A^2+B^2} & \Rightarrow A^2+B^2 = e^{-\pi B} & & ...(1) \\ \frac{\pi A}{2} = \tan^{-1} {\frac {B}{A}} & \Rightarrow \tan {\frac{\pi A}{2}} = \dfrac {B}{A} & \Rightarrow B = A \tan {\frac{\pi A}{2}} & ...(2) \end{cases}

Substituting Eqn. 2 in Eqn. 1:

A 2 + A 2 tan 2 π A 2 = e π A tan π A 2 A^2+A^2 \tan^2 {\frac{\pi A}{2}}= e^{-\pi A \tan {\frac{\pi A}{2}}}

Using numerical method, we find that A = 0.438282937 A=0.438282937

B = A tan π A 2 = 0.360592472 \Rightarrow B = A \tan {\frac{\pi A}{2}} = 0.360592472

A 2 + B 2 = 0.322118863 \Rightarrow A^2+B^2 = \boxed {0.322118863}

In response to the Master Challenger's question:

Thanks Otto Bretscher 's hint.

Infinite tetration of a complex number z z is related to Lambert W-function as follows:

z z z z . . . = h ( z ) = W ( ln z ) ln z z^{z^{z^{z^{^{.^{.^{.}}}}}}} = h(z) = - \frac {W(-\ln{z})} {\ln{z}}

h ( z ) h(z) converges iff e e x e 1 e ( 0.0659 ( z ) 1.4446 ) e^{-e} \le x \le e^{\frac{1}{e}} (0.0659\le \Re{(z)} \le 1.4446) . Since A = 0.438282937 A = 0.438282937 is within the limits, there is a solution for the problem.

Moderator note:

This solution is incomplete. It shows that if a solution exists, then it has to be of the form above. However, it doesn't show that such a solution must exist in the first place. We need to make an argument about limits, in order to be mathematically rigorous.

how did you got this A + B i e tan 1 B A i \sqrt { { A+Bi } } { e }^{ \tan ^{ -1 }{ \frac { B }{ A } i } } from this e π B 2 e π A i 2 { e }^{ \frac { -\pi B }{ 2 } }{ e }^{ \frac { \pi Ai }{ 2 } }

Vighnesh Raut - 6 years, 3 months ago

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No, it is from A + B i = A 2 + B 2 e tan 1 B A i A+Bi = \sqrt{A^2+B^2} e^{\tan^{-1}{\frac{B}{A}}i} .

Chew-Seong Cheong - 6 years, 1 month ago

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This formula works only for positive A A , though .

Otto Bretscher - 6 years, 1 month ago

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@Otto Bretscher It was positive. I made a mistake.

Chew-Seong Cheong - 6 years, 1 month ago

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@Chew-Seong Cheong Can you please tell how did you find the value A A using numerical methods?

Vighnesh Raut - 6 years, 1 month ago

Are you sure that i i = i i^{ i ^ { \ldots}} = i ? If so, then you are saying that i i = i i ^ i = i , which is not true.

For complex exponentiation, the powers are identical doesn't mean that the bases are the same. For example, we have ( 1 ) 2 = 1 2 (-1) ^ 2 = 1 ^ 2 but 1 1 -1 \neq 1 .

Calvin Lin Staff - 6 years, 4 months ago

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Thanks for the comments. I was trying to fit in the answer. But now I got A = 1 A=1 and B = 0 B=0 . I wonder if it is right.

Chew-Seong Cheong - 6 years, 4 months ago

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It is not true that i i = 1 i ^ { i ^ { \ldots } } = 1 either. If so, then i 1 = 1 i ^ 1 = 1 , which is not true.

Calvin Lin Staff - 6 years, 4 months ago

FYI: The answer should be 0.322.

Calvin Lin Staff - 6 years, 4 months ago

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Sir,Could you please explain the solution?

Anandhu Raj - 6 years, 3 months ago

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Anandhu, I have provided an explanation. See my solution above.

Chew-Seong Cheong - 6 years, 3 months ago

I believe A A is positive. Luckily, your sign error does not affect the final answer ;)

Otto Bretscher - 6 years, 2 months ago

Can you prove that h ( z ) h(z) converges only when e e x e 1 e e^{-e} \le x \le e^{\frac{1}{e}} ?

Digvijay Singh - 6 years, 1 month ago

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https://en.wikipedia.org/wiki/Tetration#Formal_definition

https://math.stackexchange.com/questions/1089458/how-can-i-prove-the-convergence-of-a-power-tower

Damon Bernard - 2 years ago
Otto Bretscher
Apr 12, 2015

We will use the principal value of complex logarithms throughout, with the imaginary part being on the interval ( π , π ] (-\pi,\pi] . We need to solve the equation i z = z i^z=z , or, e ( ln i ) z = z e^{(\ln{i})z}=z , or, e π 2 i z = z e^{\frac{\pi}{2}iz}=z , or, z e π 2 i z = 1 ze^{-\frac{\pi}{2}iz}=1 . If we make the substitution w = π 2 i z w=-\frac{\pi}{2}iz , the equation becomes w e w = π i 2 we^w=-\frac{\pi{i}}{2} . Unfortunately, the equation w e w = x we^w=x cannot be solved in closed form (not even for real x x ) unless we use a special function, the Lambert W function in this case, defined by W ( x ) e W ( x ) = x W(x)e^{W(x)}=x . We find that w = W ( π i 2 ) w=W(-\frac{\pi{i}}{2}) and z = 2 i π W ( π i 2 ) 0.4383 + 0.3606 i z=\frac{2i}{\pi}W(-\frac{\pi{i}}{2})\approx0.4383+0.3606i . Thus the answer is 0.3221 \approx0.3221 .

Write a comment or ask a question...thank u sir

Ahnaf Arnob - 6 years, 2 months ago

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Comment or question about what? Give me a topic! ;)

Otto Bretscher - 6 years, 2 months ago

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Comment or question about what? ANYTHING! I need more questions, 2 ain't enough!

Pi Han Goh - 6 years, 1 month ago

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@Pi Han Goh I have a job and a life, Comrades! My wife is jealous of "Brilliant" already ;)

Otto Bretscher - 6 years, 1 month ago

Just wondering, can the value of A A and B B be written in a closed form ?

Vighnesh Raut - 3 years, 5 months ago

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No. \quad

Pi Han Goh - 3 years, 5 months ago

I had to use desmos to solve the pair of equations. I did the same way as @Chew-Seong Cheong . I am wondering is there any manual way to solve this pair of equation or do we have to take help of some graphing software.

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