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5

We can find a 7th order polynomial that meets the requirement if we minimise the $\infty$-norm. For simplicity I construct an array of {x, 6 ArcTan[x]} over the required range and plug it into FindFit, no doubt there are better ways. {a, b} = {-1, 1}/Sqrt[3]; data = Table[{x, 6 ArcTan[x]}, {x, a, b, (b - a)/1000.}]; n = 7; expr = Sum[c[i] x^i, {i, 1, n, ...


6

Taylor polynomials of order n aren't necessarily the nth degree polynomials that optimally approximate a function on a given interval. We can use linear least squares to find the optimal polynomials for a fixed n. inn[f_, g_, x_] := Integrate[f g, {x, -1/Sqrt[3], 1/Sqrt[3]}] LeastSquarePolynomial[f_, x_, n_] := With[{pows = x^Range[0, n]}, pows . ...


8

From Weierstrass Approximation Theorem we know there is such a polynomial, moreover there are infinitely many polynomials satisfying given criterion. Therefore we would like to find those ones of the minimal order. Since we are supposed to exploit series approximations we define a polynomial of n - th order approximating 6 ArcTan[x] for x such that ...


4

Mathematica 10 provides new functionality dealing with curves, (see e.g. the Vector Analysis tutorial) like ArcLength, ArcCurvature and especially FrenetSerretSystem: FrenetSerretSystem[{ x1, ..., xn}, t] gives the generalized curvatures and Frenet-Serret basis for the parametric curve x[t] i.e. it returns {{ k1, ..., k(n-1)}, { e1, ..., en}}, where ki ...



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