The Chebyshev polynomials T_n of the first kind are a certain set of orthogonal polynomials. They can be defined by T_n(cos(x))=cos(nx), the first of them are
- T_0(x) = 1
- T_1(x) = x
- T_2(x) = 2x^2 - 1
- T_3(x) = 4x^3 - 3x
- T_4(x) = 8x^4 - 8x^2 + 1
- T_5(x) = 16x^5 - 20x^3 + 5x
- T_6(x) = 32x^6 - 48x^4 + 18x^2 - 1
- T_7(x) = 64x^7 − 112x^5 + 56x^3 − 7x
A plot in the cartesian plane looks like this, seen on Wolfram Mathworld:
On the same page, I found a beautiful polar plot of those polynomials:
There they wrote, referring to "Graphica 1" by Michael Trott:
A beautiful plot can be obtained by plotting T_n(x) radially, increasing the radius for each value of n, and filling in the areas between the curves (Trott 1999, pp. 10 and 84).
A version without filling, posted by Paul Kemper:
I would like to understand the way that plot is generated. The filling doesn't matter to me, I just would like to know how I could plot such a polynomial radially. As the domain ranges from -1 to 1, I should take pi as a factor. Does anybody know a formula to display those functions in polar coordinates in such a way with increasing radius?
Answer
You can use PolarPlot to plot the curves. As noted in the question you need to map the polar angle onto the -1 to 1 domain of the polynomials. You should also note that only the even polynomials are plotted.
PolarPlot[Evaluate @ Table[n + ChebyshevT[n, t/Pi - 1], {n, 0, 40, 2}], {t, 0, 2 Pi}]
To get the filling effect you can used FilledCurve:
Graphics @ FilledCurve @ Cases[%, _Line, -1]
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