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?

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?

12 months ago

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]

12 months ago

AwesomeTucker

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

You have a brain, use it.

12 months ago

PurpleCharger

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

12 months ago

GoogleCat

Oh.

Yeah, that makes sense.

Thanks!

12 months ago

Illusionz

what's the fun in that

12 months ago

BlogWobbles

LOL OP BTFO

YOU ARE TARDED OP

GTFO OF THIS BOARD

YOU BRAINLET

12 months ago

Boy_vs_Girl

one derives it from something known one derives it from the identity

don't really know what your issue is here

12 months ago

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]

12 months ago

Booteefool

it's ugly

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

12 months ago

Techpill

TIL pi/2 is about 4.81

12 months ago

TreeEater

easy there Cadet Capslock

12 months ago

LuckyDusty

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

12 months ago

MPmaster

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

12 months ago

lostmypassword

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

12 months ago

Skullbone

KILL YOURSELF NIGGER

12 months ago

LuckyDusty

crazy

12 months ago

TechHater

it's ugly

12 months ago

eGremlin

crazy

why the ableism?

12 months ago

BlogWobbles

[math]

\begin{align*}

i^2&=-1 \\

-i^2&=1 \\

-i \cdot i&=1 \\

-i&= \dfrac{1}{i}

\end{align*}

[/math]

12 months ago

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.

12 months ago

WebTool

the correct answer is about 4.81

try again

12 months ago

CouchChiller

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

try again

12 months ago

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.

12 months ago

TreeEater

there is nothing simplistic about mathematics.

12 months ago

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.

12 months ago

RavySnake

i = e^(i pi/2)

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

you were on the right track

12 months ago

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.

12 months ago

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.

12 months ago

PurpleCharger

fuck off to /lit/

12 months ago

Fuzzy_Logic

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

12 months ago

Illusionz

rude

12 months ago

Sir_Gallonhead

oof

12 months ago

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