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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    $\begingroup$ How did you obtain these polynomials in the first place? $\endgroup$ – J. M. is a poor mathematician Dec 28 '11 at 23:36
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    $\begingroup$ Why not the Taylor series of $\cos(x/2)$? $\endgroup$ – Rahul Dec 29 '11 at 2:13
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    $\begingroup$ With Rahul's suggestion the degree 10 polynomial is $1-\frac{x^2}{2^2\cdot 2!}+\frac{x^4}{2^4\cdot 4!}-\frac{x^6}{2^6\cdot 6!}+\frac{x^8}{2^8\cdot 8!}-\frac{x^{10}}{2^{10}\cdot 10!}$, which has a zero at around $3.14159172406$, with error around $9.3\times 10^{-7}$. At least this gives a certain method of giving better and better approximations. $\endgroup$ – Jonas Meyer Dec 29 '11 at 2:40
  • $\begingroup$ From the equations (x+y)^n+(y-x)^n=z^n we create different series from the coefficients. In different arrangements of these series we formulate the coefficients of the presented equations. For the time being that is all I have to say because my work on this is not complete. $\endgroup$ – Vassilis Parassidis Dec 29 '11 at 4:01
  • $\begingroup$ I can't believe nobody has said $\ell(x)=x-\pi$... $\endgroup$ – JP McCarthy Jan 24 '13 at 16:16

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.

  • $\begingroup$ I quite like $86 x^4-610 x^3+1199 x^2-445 x+101$ and $16 x^6-32 x^5-61 x^4-53 x^3+162 x^2+147 x-65$ myself... $\endgroup$ – J. M. is a poor mathematician Dec 29 '11 at 2:00
  • $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Jan 24 '13 at 16:52

From ancient relation, $\frac{\pi}{(\phi+1)}= \frac{6}{5}$ I had such approximation (not very good, but ancient) $25x^2 - 90x + 36$ error $3\times10^{-3}$

  • $\begingroup$ The solution is $\frac{9+3\sqrt{5}}{5}$, often written as $\frac{9}{5}+\sqrt{\frac{9}{5}}$, by Ramanujan. $\endgroup$ – Jaume Oliver Lafont Apr 16 '16 at 6:34

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