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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 26 Aug 2019
      • Report Tweet
      • Report NetzDG Violation

      A straight line (y=x) can be written as a sum over sinusoids, known as a Fourier Series.pic.twitter.com/Obii1qd62T

      42 replies 416 retweets 1,750 likes
      Show this thread
    2. Alex Kontorovich‏ @AlexKontorovich 26 Aug 2019
      • Report Tweet
      • Report NetzDG Violation
      Replying to @InertialObservr @DynamicsSIAM

      No, you only get y=x in (-pi,pi), as you’ve drawn, and then extended periodically from there (since every function in the series is 2pi periodic). The dual of the line is the line, so you’d need an integral to recover the actual line. Presumably you know this, so why fib?...

      1 reply 0 retweets 8 likes
    3. 〈 Berger | Dillon 〉‏ @InertialObservr 26 Aug 2019
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      Replying to @AlexKontorovich @DynamicsSIAM

      that's right, it converges to y=x in (-π,π) for the Fourier series (as opposed to the transform). Given that it does converge to a line on (-π,π), i wouldn't say i fibbed.. This is twitter,not an academic conference.The purpose of my tweets is to get across the basic idea in..

      2 replies 0 retweets 37 likes
    4. Claire  🏳️‍🌈 🏳️‍⚧️ 🧙🏻‍♀️‏ @oe1cxw 27 Aug 2019
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      Replying to @InertialObservr @AlexKontorovich @DynamicsSIAM

      But ... isn't the "basic idea" that one can use a Fourier series to approximate any periodic signal? So why start with a "straight line", an inherently aperiodic object? Also, sorry for the nitpick, but it doesn't converge to y=x in (-π,π). The integral error converges to zero.

      2 replies 0 retweets 0 likes
      〈 Berger | Dillon 〉‏ @InertialObservr 28 Aug 2019
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      Replying to @oe1cxw @AlexKontorovich @DynamicsSIAM

      lines are precisely what sawtooth waves are..

      6:24 PM - 28 Aug 2019
      1 reply 0 retweets 0 likes
        1. New conversation
        1. Claire  🏳️‍🌈 🏳️‍⚧️ 🧙🏻‍♀️‏ @oe1cxw 29 Aug 2019
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          Replying to @InertialObservr @AlexKontorovich @DynamicsSIAM

          Respectfully no. They're line segments. A sawtooth wave is piecewise linear, not linear.

          1 reply 0 retweets 0 likes
        2. 〈 Berger | Dillon 〉‏ @InertialObservr 29 Aug 2019
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          Replying to @oe1cxw @AlexKontorovich @DynamicsSIAM

          You can see from the boundary conditions that this gif precisely corresponds to a sawtooth wave with period 2 pi

          1 reply 0 retweets 0 likes
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