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 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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On My Favorite Number, 76923 (A Brief Survey of Cyclic Numbers)

(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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A silly little derivation of \( \zeta(2) \)

(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: \[ \int_0^1 \! \frac{\log(1 + x)}{x} \, dx = \int_0^1 \! \sum_{n=1}^{\infty} \frac{(-1)^nx^{n-1}}{n} \, dx \] ...
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100 is the only square that is the sum of 4 consecutive (positive) cubes.

The OEIS article on the number \( 100 \)  opens with an interesting factoid:  "\( 100 \) is the square of \( 10 \), and the smallest square that is the sum of four [positive] consecutive cubes: \( 1^3 + 2^3 + 3^3 + 4^3 = 100 \)." In fact, it is the only one. To see this, let's look at the equation \[ \begin{align*} y^2 &= x^3 + (x + 1)^3 + (x + 2)^3 + (x + 3)^3 \\ &= 4x^3 + 18x^2 + 42x + 36. \end{align*} \] Let \( X = 4x + 6, Y = 4y \); the above equation then reduces to \[ Y^2 = X^3 + 60X. \] Note that a positive integer solution in \( (x, y) \) will give a positive integer solution in \( (X, Y) \), though the converse is not true. Now, generally speaking, whenever one sees an equation of the form \( Y^2 = X^3 + aX + b \), one has an elliptic curve . Well, we have the extra condition \( 4a^3 + 27b^2 \ne 0 \) to get rid of problematic cases like \( y^2 = x^3 \), which are referred to as singular curves ; we'll see the logic of this later on. I will not go too...
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