Fantastic Imaginary unit!

From Euler's identity : \[e^{i\theta}=\cos(\theta)+i\sin(\theta)\] For \(\theta=\frac{1}{2}\pi\) \[e^{i\frac{1}{2}\pi}=i\] Let \(a=e^{\frac{1}{2}\pi}\) \[\Rightarrow a^i=i \Rightarrow a^{a^{a^{a^{...}}}}=i\] Now a question which I'm unable to solve and it is very interesting one! (see tetration if you aren't understanding ni{^n i}) limnni=?\lim_{n \to \infty}{^n i}=?

Note by Zakir Husain
3 months, 2 weeks ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

I have plotted the graph of ni{^ni} on the complex plain using python: The limnni\lim_{n \to \infty}{^ni} is converging in the darker middle region.

960i0.43828...+0.36059...i{^{960}i}\approx 0.43828...+0.36059...i

Zakir Husain - 3 months, 2 weeks ago

Log in to reply

Let the limit be LL. Does iL=L?i^L=L?

Jeff Giff - 3 months, 2 weeks ago

Log in to reply

Yes

Justin Travers - 3 months, 1 week ago

Log in to reply

BTW, I don’t really understand ni,n^i, although I do know tetration. Could you explain it to me? Thanks :)

Jeff Giff - 3 months, 2 weeks ago

Log in to reply

ni=iiii...{^ni}=i^{i^{i^{i^{.^{.^{.}}}}}} where there are nn isi's

Sahar Bano - 3 months, 2 weeks ago

Log in to reply

@Sahar Bano Well that I know... :) but iii^i I do not understand :D ii times itself ii times?

Jeff Giff - 3 months, 2 weeks ago

Log in to reply

@Jeff Giff See how we evaluate iii^i:

From Euler's identity (It comes every place you talk about ii): eiθ=cos(θ)+isin(θ)e^{i\theta}=\cos(\theta)+i\sin(\theta) At θ=12π\theta = \frac{1}{2}\pi ei12π=ie^{i\frac{1}{2}\pi}=i Raising both sides to ii (ei12π)i=ii(e^{i\frac{1}{2}\pi})^i=i^i ii=(ei12π)i=ei12πi=e12πi^i=(e^{i\frac{1}{2}\pi})^i=e^{\red{i}\frac{1}{2}\pi \red{i}}=e^{-\frac{1}{2}\pi}

Zakir Husain - 3 months, 2 weeks ago

Log in to reply

@Zakir Husain Oic. Thank you :)

Jeff Giff - 3 months, 2 weeks ago

Log in to reply

Log in to reply

Sorry but I can't be any help here since I don't know Calculus that well (I have just started the Fundamentals Course)

Kumudesh Ghosh - 3 months, 2 weeks ago

Log in to reply

I don't know tetration , but it seems you must calculate exponents from top to botom . Tower of "a" you equal to "i" is right if at the end of tower is and "i" . To calculate the tower you must consider where braquets are . See tetration explanation.

jordi curto - 3 months, 2 weeks ago

Log in to reply

Tetration of a complex number to a infinite height can be done using the Lambert W function. The formula is W(-log(z))/(-log(z)). Here log(i)=iπ/2 and the formula gives the value that Zakir Husain has calculated.

Justin Travers - 3 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...