What polynomial expansion is this? (sine) Reading through some articles on approximating the sine, I found this equation.
$\sin(x) / x = 1 - 0.16605 x^2 + 0.00761 x^4 + e(x)$
Those constants are different than what I'd expect from a Taylor series expansion (they should be 1/6 = 0.166667 and 1/120 = 0.0083333 respectively).  These constants also seem to give slightly better results than the Taylor series, so I'm really curious how they are derived.  Does anyone happen to know?
For the curious, the reference I got this from is Milton Abramowitz and Irene A. Stegun Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables published by Dover Press 1965.  That work cites "B. Carlson, M. Goldstein, Rational approximations of functions Los Alamos Scientific Laboratory, LA-1943, Los Alamos, NM, 1955."
 A: There are several ways to compute functions on an interval, including power series. A great book about this is "Approximations for Digital Computers" by Cecil Hastings Jr. The approximation you gave probably came from Chebyshev Polynomials.
A: Suppose that you want the best approximation of $\frac{\sin(x)} x$ by a quadratic polynomial in $x^2$ with a constant term equal to $1$.
You then need to minimize with respect to $a,b$
$$I=\int_0^{\frac \pi 2}\left(\frac{\sin(x)} x -(1+ax^2+b x^4)\right)^2$$ which is equivalent to a linear regression with an infinite number of data points.
This gives $$I=\frac{1}{160} \pi ^5 \left(a^2+2 b\right)+\frac{1}{448} \pi ^7 a b-2 a+\frac{\pi ^3
   a}{12}+\frac{\pi ^9 b^2}{4608}-\frac{3}{2} \left(\pi ^2-8\right) b+\text{Si}(\pi
   )-2 \text{Si}\left(\frac{\pi }{2}\right)+\frac{\pi }{2}-\frac{2}{\pi }$$ Then $$\frac{dI}{da}=\frac{\pi ^5 a}{80}+\frac{\pi ^7 b}{448}+\frac{\pi ^3}{12}-2=0$$
$$\frac{dI}{db}=\frac{\pi ^7 a}{448}+\frac{\pi ^9 b}{2304}+\frac{1}{80} \left(960-120 \pi ^2+\pi
   ^5\right)=0$$ Solving for $a,b$ gives $$a=-\frac{56 \left(-3240+300 \pi ^2+\pi ^5\right)}{3 \pi ^7}\approx -0.166259$$ $$b=\frac{336 \left(-5040+480 \pi ^2+\pi ^5\right)}{5 \pi ^9}\approx 0.00773195$$ 
This leads to $I\approx 6.19\times 10^{-9}$ while using Taylor coefficients ($a=-\frac 16$, $b=\frac 1 {120}$) would lead to $I\approx 1.01\times 10^{-6}$ which is almost $163$ times larger. The values given in the referenced bood would lead to $I\approx 1.41\times 10^{-8}$.
But the results depend on the range over which you integrate.
