A better way to think about Taylor series

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  • Опубликовано: 18 окт 2024

Комментарии • 6

  • @algorithmicsimplicity
    @algorithmicsimplicity  2 месяца назад +6

    Trying out something a bit different with this video, instead of machine learning content I decided to make some more math-focused content for #SoMEpi.
    My next videos will be a series on Solomonoff induction, a mathematical model of the optimal learning algorithm and a core idea in the theory of machine learning. Despite its importance, almost no-one ever talks about it for some reason. Anyway, subscribe to the channel so you don't miss it!
    Also, while working on this video I was contacted by GiveInternet, a charity which aims to provide internet access for students in developing countries. And while they have not sponsored this video in any way, they seemed like a decent charity organization so I thought I'd share the link here cause why not ( giveinternet.org/AlgorithmicSimplicity ).

  • @jkzero
    @jkzero 2 месяца назад +10

    as physicist, I have used Taylor series since the beginning of my education but I had never seen this integral derivation of the Taylor series, thanks for such an illuminating approach

  • @chiichan3774
    @chiichan3774 2 месяца назад +5

    oh wow, that explanation of the error term clicked in my head for me! I have never thought of it that way; the reason why some functions cannot have a taylor polynomial is because the error term grows exponentially the more terms there are!

  • @Higgsinophysics
    @Higgsinophysics 2 месяца назад +4

    Clearly explained - thanks for sharing this awesome video - I never seen the integral derivation or heard about an infinitely differentiable function before, learned a lot :D

  • @NevinBR
    @NevinBR 2 месяца назад +3

    The statement of the fundamental theorem of calculus is quite subtle. A differentiable function may have a derivative which is not Riemann integrable. For example, Volterra’s function is differentiable everywhere and its derivative is bounded everywhere, but its derivative is not Riemann integrable.
    There are similar subtleties with the Lebesgue integral as well.
    The gauge integral has the property that the derivative of any differentiable function is gauge integrable, but most people have not heard of the gauge integral.

  • @Gekko-t4i
    @Gekko-t4i Месяц назад +1

    poggers