## Archive for December, 2009

### Happy New Year

December 31, 2009

Let $x_1=1$ and $\displaystyle x_{n+1}=\frac{1}{\lfloor x_n \rfloor + 1 - \{ x_n \} }.$ Enjoy!

### Why Asians learn quicker II

December 29, 2009

I have managed to obtain the hardcopy of the SMH article referred to in my previous post and have scanned it. I now make the (main) missing part that didn’t appear online available for readers to view and discuss. (click to enlarge)