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 $\pi$?
$\pi$ polynomials whose real zeros approximate $\pi$
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polynomials
approximation
pi
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4How did you obtain these polynomials in the first place? – 2011-12-28
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2Why not the Taylor series of $\cos(x/2)$? – 2011-12-29
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1With 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. – 2011-12-29
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0From 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. – 2011-12-29
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0I can't believe nobody has said $\ell(x)=x-\pi$... – 2013-01-24