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?

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?

2 months ago

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]

2 months ago

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

You have a brain, use it.

2 months ago

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

don't really know what your issue is here

2 months ago

[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]

2 months ago

it's ugly

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

2 months ago

[math]

\begin{align*}

i^2&=-1 \\

-i^2&=1 \\

-i \cdot i&=1 \\

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

\end{align*}

[/math]

2 months ago

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.

2 months ago

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.

2 months ago

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.

2 months ago

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.

2 months ago

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.

2 months ago

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

2 months ago

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