How can I plot a Chebyshev spiral?

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?

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PolarPlot[Evaluate@Table[n+ChebyshevT[n,t/Pi-1],{n,0,20,2}],{t,0,2Pi}] – Simon Woods Jul 18 '14 at 12:01
@eldo Thank you! I did not see this link. – Frank Jul 18 '14 at 12:14
@SimonWoods Thank you very much! That's all I needed as help. If you like, post as an answer with some short words to let future readers easily understand, I'd gladly upvote and accept it. – Frank Jul 18 '14 at 12:16

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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this so nice and the code so terse...FilledCurve – ubpdqn Jul 18 '14 at 12:51