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Norman Wildberger's "rational trigonometry" has been viewed by some mathematicians as a clever new take on an ancient topic. Wildberger's "spread polynomials" $S_n$ are characterized by the identity $$ \sin^2(n\theta) = S_n(\sin^2\theta) $$ (except that Wildberger refuses to refer explicitly to the sine function in the definition and does it by other means). In one sense these are trivially equivalent to the Chebyshev polynomials $T_n$ characterized by $$ \cos(n\theta) = T_n(\cos\theta). $$ Wildberger notes that $1 - 2S_n(s) = T_n(1 - 2s)$.

In thinking about whether this polynomial sequence is even worth mentioning after Chebyshev polynomials have been treated, three questions come to mind:

  • Could it be that an essential difference justifying a separate treatment is the factorization of these polynomials? Wildberger factors the spread polynomials. Is there some important reason for doing that?

  • Is there a combinatorial interpretation of the coefficients? The coefficient of the $n$th-degree term in $S_n$ for $n=1,\dots,10$ is $n^2$ and the coefficient of the constant term is $4^{n-1}$. The first-degree term is $n/2$ times the constant term.

  • If there's no essential difference that justifies attending to the two sequences separately, the fact that the conventional way of viewing the sequence (Chebyshev polynomials) is chronologically first, doesn't mean it's necessarily better than Wildberger's way of viewing it (spread polynomials). Is it?

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what do you mean by "factorization"? Over integers? Then it is well known for Chebyshev polynomials. –  Fedor Petrov Nov 27 '10 at 19:32
    
OK, my statement about "not much written" on factoring Chebyshev polynomials may have been rash. I did mean over integers, but maybe over complex numbers to if that's of interest. So the first question above is whether differences in the way they factor might justify separate attention to the two sequences. –  Michael Hardy Nov 27 '10 at 20:29
    
Per your (seeming?) suggestion I've deleted that comment from the question. –  Michael Hardy Nov 27 '10 at 20:31
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Who are "some mathematicians" who view this as a clever new take on an ancient topic? It seems to me that geometric algebra (in the sense of Artin) subsumes all that stuff and it's been known for quite some time before Wildberger's book appeared. –  KConrad Nov 29 '10 at 5:09
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Oh, and of course, when working over arbitrary fields, you might want to assume $\sqrt 2$ to lie in the field for having squares, and $\sqrt 3$ for having equilateral triangles. The good news is that if you have, for instance, a proof for Napoleon's theorem over any field with $\sqrt 3$ in it, then it automatically also proves Napoleon's theorem with the equilateral triangles pointing inside instead of outside - because the field $\mathbb Q\left(X\right)\left(\sqrt 3\right)$ "won't notice" if you replace $\sqrt 3$ by $-\sqrt 3$ (this is a case of what John Horton Conway calls "extraversion"). –  darij grinberg Nov 29 '10 at 9:18
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2 Answers

Regarding which point of view is preferable, I don't see that there's much difference between them. The Chebyshev polynomials map $[-1,1]$ to $[-1,1]$, the spread polynomials map $[0,1]$ to $[0,1]$, and they are conjugate under a linear map between $[-1,1]$ and $[0,1]$, so all their properties translate easily between the two frameworks.

However, I'd vote for Chebyshev polynomials as being somewhat more fundamental, due to orthogonality. The spread polynomials aren't orthogonal with respect to any measure, because they are nonnegative everywhere. To get orthogonality, one must subtract $1/2$, after which they become orthogonal with respect to $dx/\sqrt{x(1-x)}$ on the interval $[0,1]$. By contrast, the Chebyshev polynomials are already orthogonal with respect to $dx/\sqrt{1-x^2}$ on $[-1,1]$, with no subtraction needed. This isn't a big deal, since it just amounts to subtracting $1/2$, but it's nice not to have to do the subtraction.

Overall, there's nothing sacred about using the domain $[-1,1]$ for Chebyshev polynomials. Of course it aligns beautifully with trigonometry, but Chebyshev polynomials are important in many other settings (such as approximation theory) in which $[-1,1]$ plays no special role, and they are simply rescaled to fit the interval of interest. From that perspective, $[0,1]$ is just as good a domain. On the other hand, I see no gain from making the range $[0,1]$ as well, and one has to undo it to recover orthogonality.

Comments added in edit:

As for the factorizations, this amounts to factoring $T_n(x)$ (for Chebyshev polynomials) or $T_n(x)+1$ (for spread polynomials - not quite, see comments below). Both are interesting, since both the roots and the extrema of the Chebyshev polynomials are important.

In fact, $T_{2n}(x)= T_2(T_n(x))$ and hence $T_{2n}(x)+1 = 2T_n(x)^2$, so factoring spread polynomials includes factoring Chebyshev polynomials as the even-index case. (In the odd-index case, $T_{2n+1}(x)+1 = (T_{n+1}(x)+T_n(x))^2/(x+1)$, but I'm not certain how to interpret this.)

So I'd say factoring spread polynomials is more general but slightly more obscure. Definitely both are interesting, though.

There are combinatorial interpretations of the Chebyshev polynomials involving weighted monomer-dimer configurations (although the conditions are a little odd: see http://www.math.hmc.edu/~benjamin/papers/CombTrig.pdf). The analogous idea doesn't work out as nicely for spread polynomials, but maybe some other approach is more appropriate. It's worth noting that the Chebyshev polynomials have somewhat simpler coefficients. For example, $T_8(x)=128x^8-256x^6+160x^4-32x^2+1$ while $S_8(x)=-16384x^8 + 65536x^7 - 106496x^6 + 90112x^5 - 42240x^4 + 10752x^3 - 1344x^2 + 64x$.

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OK, you've answered part of what I asked very well. I'm still wondering whether the factorizations hold some interest not available in the Chebyshev polynomials. And usually there's also a combinatorial interpreation. –  Michael Hardy Nov 28 '10 at 2:49
    
@ Henry: I don't know spread polynomials. But if it is true as stated above that $$ 1-2S_n(x)=T_n(1-2x)$$ then $$1+T_n(x)$$ does not coincide with spread polynomials except that $$ 2S_{2n+1}(x^2)=T_{4n+2}(x)+1=2T_{2n+1}(x)^2.$$ –  Johann Cigler Nov 28 '10 at 9:35
    
Oops, yeah: I chose the wrong extrema. In the odd case, $1-T_n(x) = 1+T_n(-x)$, so it doesn't really matter up to signs, but the even case seems to be genuinely different. Thanks! –  Henry Cohn Nov 28 '10 at 17:31
    
OK, still digesting stuff..... Maybe it's generally harder to notice patterns in the coefficients of the Chebyshev polynomials because the patterns are simpler. –  Michael Hardy Nov 29 '10 at 1:47
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Factorization of the spread polynomials can be reduced to the factorization of the Chebyshev polynomials by observing that $$ 1-T_{2n+2}(x)=2 U_n(x)^2 (1-x^2)$$ and $$ 1-T_{2n+3}(x)=(U_{n+1}(x)+U_n(x))^2 (1-x).$$ Edit: The following formulae show the connection between the factorization of spread polynomials and Chebyshev polynomials most clearly: $$ S_{2n}(x^2)= (1-x^2)U_{2n-1}(x)^2 $$ and $$ S_{2n+1}(x^2)= T_{2n+1}(x)^2.$$

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That's nice! I hadn't noticed that $S_n(x^2)$ had such a beautiful expression in terms of Chebyshev polynomials. This really makes me want a slick way to deal with both types of Chebyshev polynomials at once. I guess the point must be that the functions $\sqrt{1-x^2} U_n(x)$ are orthogonal with respect to the same measure as $T_n(x)$, namely $dx/\sqrt{1-x^2}$ on $[-1,1]$. Any idea what happens to the even-degree polynomials? If $(1-x^2)U_{2n-1}(x)^2$ and $T_{2n+1}(x)^2$ are spread polynomials (up to substituting $x^2$ for $x$), then what are $(1-x^2)U_{2n}(x)^2$ and $T_{2n}(x)^2$? –  Henry Cohn Nov 29 '10 at 23:23
    
Let $$ 2V_n(x)=1+T_n(1-2x).$$ Then $$ V_{2n}(x^2)= T_{2n}(x)^2$$ and $$ V_{2n+1}(x^2)= (1-x^2)U_{2n}(x)^2. $$ –  Johann Cigler Dec 1 '10 at 10:23
    
A further remark. These polynomials are characterized by $$\cos^2(n\theta) = V_n(\sin^2\theta)$$ –  Johann Cigler Dec 1 '10 at 13:13
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