Welcome

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.

(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 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$$ is the Riemann zeta function. Consider the integral $I := \int_0^1 \frac{\log(1 - x + x^2)}{x(x - 1)} \, dx.$ We have, by using partial fractions and performing some other algebraic manipulations, \begin{align*} I &= -\int_0^1 \! \frac{\log(1 - x + x^2)}{x} \, dx - \int_0^1 \! \frac{\log(1 - x + x^2)}{1 - x} \, dx \\ &= -2\int_0^1 \! \frac{\log(1 - x + x^2)}{x} & (x \mapsto 1 - x ) \\ &= 2\left( \int_0^1 \! \frac{\log(1 + x)}{x} \, dx - \int_0^1 \! \frac{\log(1 + x^3)}{x} \, dx \right) \\ &= \frac{4}{3}\int_0^1 \frac{\log(1 + x)}{x} \, dx & (x \mapsto x^{1/3}). \end{align*} To evaluate this integral, we take the Maclaurin series: $\int_0^1 \! \frac{\log(1 + x)}{x} \, dx = \int_0^1 \! \sum_{n=1}^{\infty} \frac{(-1)^nx^{n-1}}{n} \, dx$ Since for
The OEIS article on the number $$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