Hi,

Welcome to my little math blog.

This is where I will dump thoughts on recreational mathematics, fun puzzles, and other stuff every so often.

Thanks for stopping by! Hopefully you find something to enjoy here.

Hi,

Welcome to my little math blog.

This is where I will dump thoughts on recreational mathematics, fun puzzles, and other stuff every so often.

Thanks for stopping by! Hopefully you find something to enjoy here.

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(This post is a cleaned up and expanded version of this thread .) A cool fact I've seen shared around the internet a few times : The decimal expansion of \( 1/998001 \) starts with \[ \frac{1}{998001} = 0.000001002003\dots 996997999\dots \] That is, it begins with three-digit strings from \( 000 \) to \( 999 \), in order, except that it skips \(998\) for some reason. The first thing to observe is that \( 998001 = 999^2 \). Recall the formula for the infinite geometric series: \[ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}. \] If we differentiate both sides with respect to \( r\), we get \[ \sum_{n=1}^{\infty} nr^{n-1} = \frac{1}{(1 - r)^2}, \] and multiplying by \( r \) gives \[ \sum_{n=1}^{\infty} nr^n = \frac{r}{(1 - r)^2}. \] (This can also be obtained by some series manipulations.) Now, take \( r = 0.001 \). We have \[ \frac{0.001}{(1 - 0.001)^2} = \frac{1000}{998001} = 0.001 + 0.000002 + 0.000000003 + \dots \] From here, the appearance of the numbers from \( 001 \) to \( 997

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(This is a cleaned-up, somewhat revised/expanded version of my Twitter thread here .) Among math enthusiasts, the number \( 142857 \) is pretty cool. Move its leftmost digit to the right, and you get \( 428571 \), which is three times the original: \( 428571 = 142857 \times 3 \). Do this again, and you get \( 285714 \), which is two times the original: \( 285714 = 142857 \times 2 \). We can keep doing this until we return to \( 142857 \), as follows: \[ \begin{align*} 142857 &= 142857 \times 1 & 142857 \times 1 &= \color{red} 142857 \\ 428571 &= 142857 \times 3 & 142857 \times 2 &= 2857\color{red}14 \\ 285714 &= 142857 \times 2 & 142857 \times 3 &= 42857\color{red}1 \\ 857142 &= 142857 \times 6 & 142857 \times 4 &= 57\color{red}1428 \\ 571428 &= 142857 \times 4 & 142857 \times 5 &= 7\color{red}14285 \\ 714285 &= 142857 \times 5 & 142857 \times 6 &= 857\color{red}142 \end{align*} \] Numbers that give you consecutive

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(This is a cleaned-up and somewhat expanded version of this Twitter thread .) What follows is a silly little proof that \[ \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \] where \( \zeta \) is the Riemann zeta function. Consider the integral \[ I := \int_0^1 \frac{\log(1 - x + x^2)}{x(x - 1)} \, dx. \] We have, by using partial fractions and performing some other algebraic manipulations, \[ \begin{align*} I &= -\int_0^1 \! \frac{\log(1 - x + x^2)}{x} \, dx - \int_0^1 \! \frac{\log(1 - x + x^2)}{1 - x} \, dx \\ &= -2\int_0^1 \! \frac{\log(1 - x + x^2)}{x} & (x \mapsto 1 - x ) \\ &= 2\left( \int_0^1 \! \frac{\log(1 + x)}{x} \, dx - \int_0^1 \! \frac{\log(1 + x^3)}{x} \, dx \right) \\ &= \frac{4}{3}\int_0^1 \frac{\log(1 + x)}{x} \, dx & (x \mapsto x^{1/3}). \end{align*} \] To evaluate this integral, we take the Maclaurin series: \[ \int_0^1 \! \frac{\log(1 + x)}{x} \, dx = \int_0^1 \! \sum_{n=1}^{\infty} \frac{(-1)^nx^{n-1}}{n} \, dx \] Since for

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