• TimeSquirrel
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    011 months ago

    So math is like painting, you can just arbitrarily add a splash of color somewhere to change the mood…

    • Redjard
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      011 months ago

      This really seemed like a good simplification until you threw in that d’Alembert operator at the end

    • @InputZero@lemmy.ml
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      11 months ago

      Obviously the statement is obsurd. If you wanted to get super pedantic, if we let AI =√P*C and rearrange to get rid of the square root then we arrive Einstein’s full equation. E2 = (MC2 )2 + PC2. Where P is equal to the momentum of the object. Then AI is just a symbol for the energy stored in momentum. So they’re technically correct, which as we all know is the most important type of correct.

      • @Technus@lemmy.zip
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        011 months ago

        (MC^2 + C√P)^2 wouldn’t give you that result though, because you have to FOIL.

        Instead you’d get M^(2) C^4 + 2MC^(3)√P + PC^2

        And that’s not even the correct formula. It’s

        E^2 = (mc(2))2 + (pc)^2

        You can’t just naively apply a square root unless one of the terms is vanishing (momentum for a stationary mass, giving E = mc^2, or rest mass for a massless particle, giving E = pc = hf).

        The way to remember this is that it’s equivalent to the Pythagorean theorem, A^2 + B^2 = C^(2).

        So it in fact only makes sense if AI = 0.

        • @itslilith@lemmy.blahaj.zone
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          011 months ago

          In my experience, when E=mc² is written, physicists generally mean relativistic mass, making the formula extract, whereas m_0 is used for rest mass, as seen in the expansion E = m_0c² + m_0v²/2 + O(v⁴)

          • @Technus@lemmy.zip
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            011 months ago

            Where does that expansion come from? As far as I can tell, m0v^(2)/2 only gives you the kinetic energy of the object where v << c, in which case the difference between relativistic mass and rest mass is negligible?

            And where does the O(v^4) term come from?

  • Ayumu Tsukasa
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    011 months ago

    I love that’s it’s not “what are you on about?” it’s just a general what

  • @NauticalNoodle@lemmy.ml
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    011 months ago

    what is one unit of ‘Artificial(A)’ and also ‘Intelligence(I)’ mathemetically defined as?

    …besides e - mc^2

  • @Sam_Bass@lemmy.ml
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    011 months ago

    Wouldnt that be the same as using a multicore modeller computer, since AI is just semi randomizing code?

      • @qjkxbmwvz@startrek.website
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        011 months ago

        I can suggest an equation that has the potential to impact the future:

        H|ψ> = E|ψ> + AI

        Here, I have chosen the time-independent Schrödinger equation, to symbolize the fact that AI is the most important innovation of all time.

        This is all bullshit of course. Everyone knows that the AI term should be included in the Hamiltonian anyway 🙄

        • This doesn’t look right since you’ve written the equation for very slow movement (sub-relativistic) and including the AI term should increase all the velocities in your ensemble exponentially.

  • @CPMSP@midwest.social
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    010 months ago

    Ha! “Consultant / Technology Manager” – pretty sure he’s just working on the next buzzword buffet to justify his bloated comp package.

    Oh well, gives me an excuse to link this Weird Al song that makes far more sense.

    • Nemo's public admirer
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      10 months ago

      I’ve heard about an extended version of the equation:
      E2 = m2c4 + p2c2
      Or E = (m2c4 + p2c2)1/2

      If so, AI = (m2c4 + p2c2)1/2 - mc2

      I may have the capabilities to be a technology management consultant

  • @doctordevice@lemmy.ca
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    011 months ago

    As I said in another thread where this was posted, that original post has the distinctive voice of ChatGPT. Could be another similar model, but I’d bet money that was written by an LLM.

  • @toastal@lemmy.ml
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    011 months ago

    LinkedIn is just another Microsoft-owned account you should just delete for your sanity