A race: i^i to 15dp, first answer...

OK, early morning quiz, first correct answer wins:

What is i^i to 15dp?

Gah! Why are you always on so early?

0.20787957635078_ or 0.20787957635077_ (I know, that's only 14) However, it is equal to all these things:

The ith root of 1/i
10 ^{(log(-1) * i/2)}
e ^{(ln(-1) * i/2)}

Winners prize goes to sql_lall. It's .20787957635077.

Giraffe, it's the insomnia :p Did you see Clare on Uni Challenge? A convincing win ;) Did it say when they were next on?

whats the i^i^i and i^i^i^i^i? first one gives the correct answer wins it all!

i^i^i = -i
i^i^i^i^i = i

Quote:

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Originally posted by sql_lall
The ith root of 1/i
10 ^{(log(-1) * i/2)}
e ^{(ln(-1) * i/2)}

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Edited parts hereinafter are in bold

From that could you say a^{(log_a(-1)* i/₂)} ?

Quote:

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Originally posted by Dreamlax
i^i^i = -i
i^i^i^i^i = i

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looks good. calculator or math?

Actually someone on the forum told me when I was looking for cool Math formulae... never forgotten it since!

that was me :D :D :D

I had a feeling it was you, but I didn't want to assume it in case someone else came and bit my head off for crediting someone else!

Cheers for those formlae though...

Quote:

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Originally posted by bugzpodder
whats the i^i^i and i^i^i^i^i? first one gives the correct answer wins it all!

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i^i^i=-1^(e^(-pi/2)/2) which is 1
i^i^i^i^i=e^(-pi*e^(-pi*sin(pi*e^(-pi/2)/2)/2)*sin(pi*cos(pi*e^(-pi/2)/2)/2)/2)*(-1)^(e^(-pi*sin(pi*e^(-pi/2)/2)/2)*cos(pi*cos(pi*e^(-pi/2)/2)/2)/2) which is -.388367808254

Yup. What do I win?

I have always wondered, what are the rules regarding orders of powers.

e.g. 3^3^3 = what??
is it 3^(3^3) = 3^27
or (3^3)^3 = 27^3 = 3^9 Blush
There is a huge difference, espescially with stuff like 2^2^2^2^2^2^2^2=??

3³³ I'm sure is equal to 3²⁷. Exponents are to be done first, and you can't find x (when x = c^ab) without first finding c, which requires you to find a^b beforehand. So I guess with exponents you have to read it backwards!

Quote:

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Originally posted by sql_lall

or (3^3)^3 = 9^3 = 3^6

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Just out of curiosity, did you mean to say (3^3)^3 = 27^3 = 3^9?

Quote:

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Originally posted by Dreamlax
3³³ I'm sure is equal to 3²⁷. Exponents are to be done first, and you can't find x (when x = c^ab) without first finding c, which requires you to find a^b beforehand. So I guess with exponents you have to read it backwards!

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Dreamlax is right.

but thats where the problem lies. i didn't give it a thought at first but now that i think about it,

in order for i^i^i^i^i=i

it would have to be ((((i^i)^i)^i)^i)=i^(i^4)=i

but apparently my TI 83+ calculator produces the wrong answer such as when entering

3^3^3 it gives me 27^3

and not 3^27

so apparently thats what happened to the i^i^i^i^i, i apologize for any cofusion i caused

Newer TI models do powers correctly.