/sci/ - Science & Math

>>9622490
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]

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

You have a brain, use it.

>>9622494
What the fuck? this is nonsense. Why not just do the thing below?

>>9622499
Oh.

Yeah, that makes sense.

Thanks!

>>9622502
what's the fun in that

>>9622499
LOL OP BTFO
YOU ARE TARDED OP
GTFO OF THIS BOARD
YOU BRAINLET

>>9622502
one derives it from something known one derives it from the identity
don't really know what your issue is here

>>9622490
[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]

>>9623265
it's ugly
>>9623386
uh excuse me the complex log is not a function. please specify if you're using the principal branch :)

>>9623386
TIL pi/2 is about 4.81

>>9623132
easy there Cadet Capslock

>>9622499
Wait am I missing something? Why does 1/i = -i

>>9624053
1/i * i/i = i/-1 = -i

>>9624186
Well I'll be darned, ya learn something new every day

>>9623942
KILL YOURSELF NIGGER

>>9624223
crazy

>>9623445
>it's ugly

>>9624329
>crazy
why the ableism?

>>9624053
[math]
\begin{align*}
i^2=-1 \\
-i^2=1 \\
-i \cdot i=1 \\
-i= \dfrac{1}{i}
\end{align*}
[/math]

>>9624053
[math]
\begin{align*}
i^2&=-1 \\
-i^2&=1 \\
-i \cdot i&=1 \\
-i&= \dfrac{1}{i}
\end{align*}

[/math]

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.

>>9625291
the correct answer is about 4.81
try again

>>9625291
the correct answer is [math] \sqrt{e^ \pi} \approx 4.81[/math]

try again

>>9624336
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.

>>9627158
there is nothing simplistic about mathematics.

>>9624972
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.

>>9625291
i = e^(i pi/2)
i^(-i) = e^(-i^2 pi/2) = e^(pi / 2)

you were on the right track

>>9627169
>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.

>>9627186
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.

>>9627191
fuck off to >>>/lit/

>>9627194
I'm just helping you get better at math, throwing everything in an equation blender is not the best solution every time.

>>9627199
rude

>>9624053
oof