i i i . . .
For i = − 1 , the value of the infinitely nested exponent above is equal to A + B i for real values A and B .
Find the value of A 2 + B 2 to 3 decimal places.
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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 A B i from this e 2 − π B e 2 π A i
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No, it is from A + B i = A 2 + B 2 e tan − 1 A B i .
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This formula works only for positive A , though .
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@Otto Bretscher – It was positive. I made a mistake.
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@Chew-Seong Cheong – Can you please tell how did you find the value A using numerical methods?
Are you sure that i i … = i ? If so, then you are saying that 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 but − 1 = 1 .
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Thanks for the comments. I was trying to fit in the answer. But now I got A = 1 and B = 0 . I wonder if it is right.
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It is not true that i i … = 1 either. If so, then i 1 = 1 , which is not true.
FYI: The answer should be 0.322.
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Sir,Could you please explain the solution?
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Anandhu, I have provided an explanation. See my solution above.
I believe A is positive. Luckily, your sign error does not affect the final answer ;)
Can you prove that h ( z ) converges only when e − e ≤ x ≤ e e 1 ?
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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
We will use the principal value of complex logarithms throughout, with the imaginary part being on the interval ( − π , π ] . We need to solve the equation i z = z , or, e ( ln i ) z = z , or, e 2 π i z = z , or, z e − 2 π i z = 1 . If we make the substitution w = − 2 π i z , the equation becomes w e w = − 2 π i . Unfortunately, the equation w e w = x cannot be solved in closed form (not even for real x ) unless we use a special function, the Lambert W function in this case, defined by W ( x ) e W ( x ) = x . We find that w = W ( − 2 π i ) and z = π 2 i W ( − 2 π i ) ≈ 0 . 4 3 8 3 + 0 . 3 6 0 6 i . Thus the answer is ≈ 0 . 3 2 2 1 .
Write a comment or ask a question...thank u sir
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Comment or question about what? Give me a topic! ;)
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Comment or question about what? ANYTHING! I need more questions, 2 ain't enough!
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@Pi Han Goh – I have a job and a life, Comrades! My wife is jealous of "Brilliant" already ;)
Just wondering, can the value of A and B be written in a closed form ?
@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.
I had to use desmos to solve the pair of equations. I did the same way asProblem Loading...
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i i i . . . = A + B i = i A + B i ⇒ e 2 π i ( A + B i ) = e − 2 π B e 2 π A i = A 2 + B 2 e tan − 1 A B i
⎩ ⎨ ⎧ e − 2 π B = A 2 + B 2 2 π A = tan − 1 A B ⇒ A 2 + B 2 = e − π B ⇒ tan 2 π A = A B ⇒ B = A tan 2 π A . . . ( 1 ) . . . ( 2 )
Substituting Eqn. 2 in Eqn. 1:
A 2 + A 2 tan 2 2 π A = e − π A tan 2 π A
Using numerical method, we find that A = 0 . 4 3 8 2 8 2 9 3 7
⇒ B = A tan 2 π A = 0 . 3 6 0 5 9 2 4 7 2
⇒ A 2 + B 2 = 0 . 3 2 2 1 1 8 8 6 3
In response to the Master Challenger's question:
Thanks Otto Bretscher 's hint.
Infinite tetration of a complex number z is related to Lambert W-function as follows:
z z z z . . . = h ( z ) = − ln z W ( − ln z )
h ( z ) converges iff e − e ≤ x ≤ e e 1 ( 0 . 0 6 5 9 ≤ ℜ ( z ) ≤ 1 . 4 4 4 6 ) . Since A = 0 . 4 3 8 2 8 2 9 3 7 is within the limits, there is a solution for the problem.