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Recently, I was watching Quanta Magazine's The Biggest Project in Modern Mathematics YouTube video. I was thrown off by the expansion of $(1-x)^{24}$ to $(1-24x+276x^2)$ at around timestamp 3:40. Is this an error? Or are they using some non-standard notation? Or completely glossing over a lot of details?

Of course, WolframAlpha confirms that the expansion of $(1-x)^{24}$ is not $(1-24x+276x^2)$, but something more unwieldy. $(1-x)^{24}$ and $(1-24x+276x^2)$ don't even have the same roots!

The video seems like it's targeted to a general audience, so if they are using some super non-standard notation, I would have hoped they had pointed it out.

I asked in the video comments, but no one answered, so I'm asking here!

joseville
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    They're just glossing over some details, to save writing lots of ellipses. If you just watch a bit more of the video it is clear enough. – Suzu Hirose Jun 09 '22 at 03:01
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    My main advice is that you avoid youtube videos if you want to make progress in math rather than be "entertained". – Peter Jun 10 '22 at 09:10

1 Answers1

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You're right (of course) that $(1-x)^{24}$ doesn't equal $1-24x+276x^2$. However, the expansion that's happening here is a Taylor series: taking the function $$x(1-x)^{24}(1-x^2)^{24}(1-x^3)^{24}\cdots$$ and expanding it as $x+a_2x^2+a_3x^3+\cdots$, where $a_2,a_3,\dots$ are coefficients (i.e. expanding around $0$ -- this information is useful for approximating the function for small $x$). This means that, to compute any particular $a_i$, you don't necessarily need the whole expansion of $(1-x)^{24}$, just the first few terms. This is presumably why it's expanded as $1-24x+276x^2$ in the video, although it would be clearer if it were written $1-24x+276x^2-\cdots$, to make it clear that there are more terms, and they just didn't want to take the space to write them down.

Carl Schildkraut
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    That answer definitely is not correct, you only have to watch a bit more of the video. It's not a Taylor series at all, and they're not approximating a function. – Suzu Hirose Jun 09 '22 at 02:56
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    It's quite categorically not a Taylor series. – Suzu Hirose Jun 09 '22 at 09:42
  • @SuzuHirose I've reread the answer and you're right. There's no sense in which this function is approximating anything around $0$. Rather, it's an infinite power series, which, when evaluated at $x = e^{2\pi i z}$ for complex numbers $z$, gives the values of the modular form. – Mathmo123 Jun 09 '22 at 13:06
  • There were plenty of errors in the video though. Just for example, 30 secs after this clip, the video states that it was Deligne that proved Ramanujan's conjecture. But that's not true.... the conjecture spoken about in the video, namely that the coefficients of the function are multiplicative, was solved much earlier by Mordell. – Mathmo123 Jun 09 '22 at 13:09