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InertialObservr's profile
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
@InertialObservr

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〈 Berger | Dillon 〉

@InertialObservr

PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …

DC → CA
youtube.com/c/InertialObse…
Joined August 2015

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    1. 〈 Berger | Dillon 〉‏ @InertialObservr 31 Mar 2019
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      Replying to @Corey_Yanofsky

      hahha do you have a feature request?

      1 reply 0 retweets 1 like
    2. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 31 Mar 2019
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      Replying to @InertialObservr

      more Lie groups and why they're important for physics SU(2) SO(3) ...E8?!!?!

      3 replies 0 retweets 2 likes
    3. 〈 Berger | Dillon 〉‏ @InertialObservr 31 Mar 2019
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      Replying to @Corey_Yanofsky

      yes yes yes!!! but I can't just jump right into those guys haha.. U(1) is just as important as those as it's isomorphic to SO(2)!

      1 reply 0 retweets 1 like
    4. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 31 Mar 2019
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      Replying to @InertialObservr

      so i mean if you already have a plan then that's the "more" i want i'm just an impatient little vampire girl

      1 reply 0 retweets 1 like
    5. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 31 Mar 2019
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      Replying to @Corey_Yanofsky @InertialObservr

      was originally looking for oliver but i went with claudia when her gif came uppic.twitter.com/nQ0LHTxOzX

      1 reply 0 retweets 1 like
    6. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 31 Mar 2019
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      Replying to @Corey_Yanofsky @InertialObservr

      (and i'm still noodling over your stupid distance-from-point-on-x-axis-to-point-on-unit-circle integral; not sure if you think of it that way but that's one of the things it is)

      2 replies 0 retweets 0 likes
    7. 〈 Berger | Dillon 〉‏ @InertialObservr 31 Mar 2019
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      Replying to @Corey_Yanofsky

      are you talking about the log one?

      1 reply 0 retweets 0 likes
    8. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 31 Mar 2019
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      Replying to @InertialObservr

      yep 1 - 2*a*cos(θ) + a^2 = sin^2(θ) + cos^2(θ) - 2*a*cos(θ) + a^2 = sin^2(θ) + [a - cos(θ)]^2 = the square of the distance from (a, 0) to (cos(θ), sin(θ)) maybe that has nothing to do with how to solve it but it's what immediately came to mind

      2 replies 0 retweets 1 like
    9. 〈 Berger | Dillon 〉‏ @InertialObservr 31 Mar 2019
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      Replying to @Corey_Yanofsky

      Neat! If you could solve it that way it'd be pretty cool

      1 reply 0 retweets 0 likes
    10. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 31 Mar 2019
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      Replying to @InertialObservr

      i feel like i'm supposed to take the derivative wrt a under the integral sign or maybe IBP; both can get rid of the log i've looked at the multiple angle formulae and i've tried the trick that solves the beta integral my trig-identity-fu is too weak

      1 reply 0 retweets 1 like
      〈 Berger | Dillon 〉‏ @InertialObservr 31 Mar 2019
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      Replying to @Corey_Yanofsky

      you're getting warmer..

      3:51 PM - 31 Mar 2019
      0 replies 0 retweets 0 likes

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