The Chebyshev polynomials of the first kind are defined by the recurrence relation:
$$ T_n(x) = 2xT_{n-1}(x) - T_{n-2}(x)$$
By using these polynomials, you can multiply the frequency of a cosine by n via a polynomial, IE:
$$ T_n(\cos x) = \cos nx $$
This can be useful in additive synthesis.
I want to now do the opposite - instead of multiplying the frequency, I want to divide it. I want to know what $ T_{1/n}(x)$ is.
It's possible to determine $T_{1/2}(x)$ easily with a little bit of algebra:
$$T_2(x) = 2x^2 - 1$$ $$2x^2 = T_2(x) + 1$$ $$x^2 = {T_2(x) + 1 \over 2}$$ $$x = \sqrt{T_2(x) + 1 \over 2}$$
Where $T_1(x) = x$,
$$T_1(x) = \sqrt{T_2(x) + 1 \over 2}$$
or
$$T_{1/2}(x) = \sqrt{x + 1 \over 2}$$
So that's done. However, moving to $T_{1/3}(x)$ is not as easy.
$$ T_3(x) = 4x^3 - 3x $$ $$ 4x^3 = T_3(x) + 3x $$ $$ x^3 = {T_3(x) + 3x \over 4} $$ $$ x = \sqrt[3]{T_3(x) + 3x \over 4} $$
Since $x$ is the thing I'm trying to find, that $3x$ term underneath the cube root isn't helpful.
How can I simplify that last equation? Furthermore, how can I establish a recurrence relation for $T_{1/n}(x)$?