this post was submitted on 05 Oct 2025
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[–] tetris11@lemmy.ml 30 points 1 week ago* (last edited 1 week ago) (1 children)

My high school teacher introduced this to us as a slow reveal over the course of weeks of what would be the proof of

e^iπ^ = -1

The happiest moment was when he brought in these two disparate field of mathematics, complex numbers and series expansions, and hit us with this magnificent revelation. Once he drew it up, he stood there shaking with excitement, beaming at us at how amazing this all was.

The class wasn't having it. We were teenagers. We understood it from a purely proof level, but did not get the implications. It was years years later that it all hit me how amazingly neat it was of the universe to unite these fields together like that and to unearth literally new tools we could use to explore further fields of maths.

Thankfully since then I've started dating Taylor Swift and reading the words of Samule Taylor Coleridge, whilst getting clothes fitted to size at my local clothes-maker guy to fit my enourmous expanding schwang.

[–] HowAbt2day 7 points 1 week ago (1 children)

Two questions for you my brother in god;

  1. what were the connections that were made in maths class that got the prof so excited?
  2. how long did you wait before removing the ever expanding schlong for your ever expanding sfinxter?
[–] tetris11@lemmy.ml 9 points 6 days ago* (last edited 6 days ago) (1 children)

For (1), we started with the Maclaurin series 1/x to get us familiar with the idea of differential expansions, and then we moved to Taylor to derive expansions of some common functions like cos and sin:

cos(x) = 1 - x^2^/2! + x^4^/4! - ...
sin(x) = x - x^3^/3! + x^5^/5! - ....

We now start with the definition of e^x^ Taylor expansion, and proceed to do some substitutions:

e^x^ = 1 + x + x^2^/2! + x^3^/3! + .... + x^n^/n!

We can then substitute in: x=iθ (remembering that i^2^ = -1) to get

e^iθ^ = 1 + iθ - θ^2^/2! - iθ^3^/3! + θ^4^/4! + iθ^5^/5! + ... etc...

If we group by real and complex, we can arrange the above as:

e^iθ^ = (1 - θ^2^/2! + θ^4^/4! + ... ) + i(θ - θ^3^/3! + θ^5^/5! + ... )

You should now realise that the left part resembles the expansion of cos(θ), and the right part resembles sin(θ). That is:

e^iθ^ = cos(θ) + i sin(θ)

Finally, we substitute in θ = π

e^iπ^ = cos(π) + i sin(π)

And we know that cos(π) = -1, and that sin(π) = 0, meaning that we end up with

e^iπ^ = -1 + i 0

or

e^iπ^ + 1 = 0

The teacher got excited because it is literally one of the most beautiful mathematical statements you can get, that connects five universal identities under a single statement: 0, 1, e, i, and π -- and does so using 3 different operators (times, power, plus).

For (2), I'm still waiting as I think it's currently holding the world together by sheer mass alone

[–] HowAbt2day 4 points 6 days ago

Thank you for your service for both 1) and 2)!