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Let's have the following polynomials
$x^4+105x^2-1134=0,$ $x^6+126x^4+10395x^2-115830=0,$ $3x^8+550x^6+45045x^4+3378375x^2-38288250=0$ The positive real zeros of these equations are good approximations of $\pi$. Does anyone know how to formulate the next polynomial so its real positive zeros give a better aproxmation of $\pi$?

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    I can't believe nobody has said $\ell(x)=x-\pi$...2013-01-24

2 Answers 2

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How do you measure the quality of these approximations? Here are the errors in the roots of your polynomials: $ \begin{eqnarray} \text{deg}&=&4; \qquad \rho=3.1419530007425911; \qquad\epsilon=3.6\times10^{-4}\\ \text{deg}&=&6;\qquad \rho=3.1415990271727633; \qquad \epsilon=6.4\times10^{-6}\\ \text{deg}&=&8; \qquad\rho=3.1415927638681944; \qquad \epsilon=1.1\times10^{-7}. \end{eqnarray} $ I would argue that these are fairly inefficient... you've used $24$ digits of coefficients to get only $7$ correct digits of $\pi$, for instance, in the degree-$8$ example. It would be more efficient to just use $1000000000x - 3141592653=0$!

There is a terrific compilation of approximations for $\pi$, including polynomial-root approximations, over at The Contest Center's Pi Competition. My favorite is $6x^6-4x^5+5x^4+2x^3-2x^2+3x-5083=0,$ which has just ten digits of coefficients and leads to the fourteen-digit approximation $ \rho=3.1415926535898031685143792;\qquad\epsilon=1.0\times10^{-14}. $ The best approximation of any kind on the page is $ \frac{\log\left((5!\times5336)^3 + 4! + 6!\right)}{\sqrt{163}}, $ which is correct to thirty decimal places.

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    More broadly, this leads to the concept of norms on polynomials, for instance the polynomial's _height_ (the largest of its coefficients in absolute value) or its $l_2$-norm (the square root of the sums of the squares of the coefficients). Of course, the OP's polynomials are fairly inefficient by either of these norms.2013-01-24
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From ancient relation, $\frac{\pi}{(\phi+1)}= \frac{6}{5}$ I had such approximation (not very good, but ancient) $25x^2 - 90x + 36$ error $3\times10^{-3}$

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    The solution is $\frac{9+3\sqrt{5}}{5}$, often written as $\frac{9}{5}+\sqrt{\frac{9}{5}}$, [by Ramanujan](https://en.wikipedia.org/wiki/Approximations_of_%CF%80#Miscellaneous_approximations).2016-04-16