So it's well-known that "i^i" is surprisingly a real number

RumChicken
RumChicken

So it's well-known that "i^i" is surprisingly a real number.

This got me wondering, what is "i to the ith root" (I'm not sure if I worded that correctly, pic related). Can anyone help me with this?

Attached: i-to-the-i.png (10 KB, 204x196)

DeathDog
DeathDog

insert 1/i into the following and solve

[math] \displaystyle
e^{ix} = cos(x) + i \cdot sin(x) \\
e^{i \frac{ \pi}{2}} = cos \left ( \frac{ \pi}{2} \right )
+ i \cdot sin \left ( \frac{ \pi}{2} \right ) = 0 + i \cdot 1 = i \\
i^i= ({e^{i \frac{ \pi}{2}}})^i = e^{i^2 \frac{ \pi}{2}} = e^{- \frac{ \pi}{2}}
= \frac{1}{e^{ \frac{\pi}{2}}} = \frac{1}{(e^ \pi)^ { \frac{1}{2}}}
= \frac{1}{ \sqrt{e^ \pi }} \approx 0.20788 \\
i^i = e^{ln(i^i)} = e^{i \cdot ln(i)} = e^{i \cdot ln(e^{i \frac{ \pi}{2}})}
= e^{i^2 \cdot \frac{ \pi}{2} \cdot ln(e)} = e^{- \frac{ \pi}{2}} \approx 0.20788
[/math]

AwesomeTucker
AwesomeTucker

[eqn]\sqrt[i]{i} = i^{\frac{1}{i}} = i^{-i} = \frac{1}{i^i}[/eqn]

You have a brain, use it.

PurpleCharger
PurpleCharger

What the fuck? this is nonsense. Why not just do the thing below?

GoogleCat
GoogleCat

Oh.

Yeah, that makes sense.

Thanks!

Illusionz
Illusionz

what's the fun in that

BlogWobbles
BlogWobbles

LOL OP BTFO
YOU ARE TARDED OP
GTFO OF THIS BOARD
YOU BRAINLET

Boy_vs_Girl
Boy_vs_Girl

one derives it from something known one derives it from the identity
don't really know what your issue is here

Deadlyinx
Deadlyinx

[eqn] \sqrt [i] { i } = i^{ \frac { 1 } { i } } = \ln \left( i^{ \frac { 1 } { i } }\right) = \frac { 1 } { i } \frac { i \pi } { 2 } = \frac { \pi } { 2 } [/eqn]

Booteefool
Booteefool

it's ugly
uh excuse me the complex log is not a function. please specify if you're using the principal branch :)

Techpill
Techpill

TIL pi/2 is about 4.81

TreeEater
TreeEater

easy there Cadet Capslock

LuckyDusty
LuckyDusty

Wait am I missing something? Why does 1/i = -i

MPmaster
MPmaster

1/i * i/i = i/-1 = -i

lostmypassword
lostmypassword

Well I'll be darned, ya learn something new every day

Skullbone
Skullbone

KILL YOURSELF NIGGER

LuckyDusty
LuckyDusty

crazy

TechHater
TechHater

it's ugly

Attached: 1521237121653.png (13 KB, 400x400)

eGremlin
eGremlin

crazy
why the ableism?

BlogWobbles
BlogWobbles

[math]
\begin{align*}
i^2&=-1 \\
-i^2&=1 \\
-i \cdot i&=1 \\
-i&= \dfrac{1}{i}
\end{align*}

[/math]

King_Martha
King_Martha

Well, the nth root of some number k can be expressed as k^(1/n). So the ith root of i is the same as i^(1/i). 1/i becomes -i if you multiply it by i/i, so therefore, i^1/i = i^-i. We're almost there now. Per Euler's Identity, i=e^(pi/2*i). -i = e^(3*pi/2), so i^-i = e^((pi/2*i)*3*pi/2*i) = e^(-3*pi^2/4), which is about .0006099.

WebTool
WebTool

the correct answer is about 4.81
try again

CouchChiller
CouchChiller

the correct answer is [math] \sqrt{e^ \pi} \approx 4.81[/math]

try again

Deadlyinx
Deadlyinx

If you want to bluff about expressing simple stuff in complex words then sociology is probably a better fit major for you. We as mathematicians must strive for simplicity.

TreeEater
TreeEater

there is nothing simplistic about mathematics.

Emberburn
Emberburn

This is sort of backwards thinking. For a number a, the number 1/a is defined as a number b such that a*b=1. now i*(-i)=-i^2=1, you do the same process but in reverse for some reason.

RavySnake
RavySnake

i = e^(i pi/2)
i^(-i) = e^(-i^2 pi/2) = e^(pi / 2)

you were on the right track

DeathDog
DeathDog

in reverse
It's an equation, not a sentence in English.
You're manipulating both sides of a scale,
not reciting a poem of Shakespeare.

AwesomeTucker
AwesomeTucker

No, you're verifying a statement "-i is the multiplicative inverse of i" and rather than solving an equation you can just show that it satisfies the criteria for being that. Your approach is exploratory however and shows how one would come up with the identity in the first place which could be useful for OP.

PurpleCharger
PurpleCharger

fuck off to /lit/

Fuzzy_Logic
Fuzzy_Logic

I'm just helping you get better at math, throwing everything in an equation blender is not the best solution every time.

Illusionz
Illusionz

rude

Sir_Gallonhead
Sir_Gallonhead

oof

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