morganrivers
commented
4 months ago Hi, well it seems you found my little problem set uploader. I worry if you are trying to beat GPT4 at handwriting recognition, you will not succeed, as GPT4 can actually understand the math being used and intuit illegible characters.
So, I'm not interested in an API interface, but I would love if you were to make your code more open source or at least offer a free version to provide more benefit to the scientific community?
As for handwritten latex, at the end of the semester I can provide you with about 100 handwritten pages of my problem sets and associated high quality latex conversions. I doubt this is of much interest of you as it's a pretty small dataset and it's just in my handwriting. I don't think anyone else is using this github repo.
That's a handwritten example for which I have properly latex'd solution.
Latex:
\section*{Exercise 4}
\begin{enumerate}
\item[(i)] I had to look at Wikihow for converting to polar and the square root idea, but did the rest myself.
\[
\int_{-\infty}^\infty e^{-ax^2} \, dx = \left(\int_{-\infty}^\infty e^{-ay^2} \, dy\right) \left(\int_{-\infty}^\infty e^{-az^2} \, dz\right)
\]
Note $y$ and $z$ are independent.
\[
= \sqrt{\int \int_{-\infty}^\infty e^{-a(y^2+z^2)} \, dy \, dz}
\]
The integral is over all area.
Set $y = r\cos\theta$, $z = r\sin\theta$, and $dA = dydz = rdrd\theta$.
Now, in polar coordinates
\[
= \sqrt{\int_0^{2\pi} \int_0^\infty e^{-ar^2} r \, dr \, d\theta} = \sqrt{2\pi} \sqrt{\int_0^\infty e^{-ar^2} r \, dr}
\]
Then I remembered, the derivative
\[
\frac{d}{dr} e^{-ar^2} = -2ar \, e^{-ar^2}
\]
Let $p = ar^2$,
\[
\frac{d}{dp} e^{-p} = \frac{d}{dr}e^{-ar^2} = e^{-ar^2}(-2ar) \rightarrow \frac{d}{dr} \left(\frac{-1}{2a}e^{-ar^2}\right) = e^{-ar^2}r
\]
Which means
\[
\int_0^\infty e^{-ar^2} r \, dr = -\frac{1}{2a} \left[e^{-ar^2}\right]_0^\infty = \frac{1}{2a}
\]
\boxed{
\int_{-\infty}^\infty e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}
}
YRQ66
Hello, we have the mathpix API interface, but we lack sufficient handwriting data. Perhaps we can collaborate