It may be covered by the articles referred to in the earlier answers, but if you integrate
$$\frac{(x-x^2)^{8k+4}}{1+x^2}$$ over the unit interval (for $k$ a non-negative integer), and
rewrite $(x-x^2)^{8k+4}$ as $x^{8k+4}(1+x^{2} -2x)^{4k+2}$, then rewrite
$2^{4k+2}x^{12k+6}$ as $2^{4k+2}(x^{12k+6} +1) -2^{4k+2}$, you can see that
you get a rational approximation to $2^{4k} \pi$ with an error less than
$4^{-(8k+4)}$. 4^{-(8k+4)}$, where the rational approximation is the integral of a polynomial
with integer coefficients over the unit interval. However the denominator is not usually so straightforward.
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It may be covered by the articles referred to in the earlier answers, but if you integrate
$$\frac{(x-x^2)^{8k+4}}{1+x^2}$$ over the unit interval (for $k$ a non-negative integer), and
rewrite |
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