Yes. Here's a very simple one.
If [math]d_\pi(n)[/math] is the [math]n^\textrm{th}[/math] decimal digit of [math]\pi[/math], where [math]d_\pi(0)[/math] is the integer part of [math]\pi[/math] (in decimal), then
[math]\displaystyle\pi=\sum_{k=0}^\infty \frac{d_\pi(k)}{10^k}.[/math]
If you don't get it, this expands to
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