# A curious family of Chebyshev-like polynomials

Consider the family $$f_n(x)$$ of functions of $$x$$ for $$0\leq x\leq1$$, each indexed by a variable $$n \in \mathbb{N}$$, described by the following equation: $$f_n(x) = \sin^2\left(n \arcsin\left(\sqrt{x}\right)\right)$$ Evaluation in numerical algebra programs yields the following polynomial forms of each function $$f_n(x)$$ for the first few $$n$$:

1. $$f_1(x) = x$$
2. $$f_2(x) = 4x - 4x^2$$
3. $$f_3(x) = 16 x^3 - 24 x^2 + 9 x$$
4. $$f_4(x) = -64 x^4 + 128 x^3 - 80 x^2 + 16 x$$

and so on.

Is there any way to analytically derive the polynomial form of this family of functions in terms of $$n$$ and $$x$$ (i.e., $$f_n(x) = g(n,x)$$ where $$g$$ is a polynomial in $$x$$)? Or perhaps this family of polynomials is related to another "named" family of polynomials under an appropriate transformation?

COMMENT: The "Chebyshev-like" qualifier in the title comes from the fact that the Chebyshev polynomials $$T_n(x)$$ can be defined by a similar trigonometric identity: $$T_n(x) = \cos(n\arccos(x))$$.

Chebyshev polynomials of the second kind satisfy $$\,U_{n-1}(\cos \theta)\,\sin \theta =\sin(n\theta)\,$$.

With $$\,\theta=\arcsin\left(\sqrt{x}\right)\,$$ it follows that:

$$\sqrt{x} \; U_{n-1}\left(\sqrt{1-x}\right) = \sin\left(n \arcsin(\sqrt{x})\right) \;\;\iff\;\; f_n(x) = x \cdot U_{n-1}^2\left(\sqrt{1-x}\right)$$

Since $$\,U_n(x)\,$$ contains only powers of the same parity, the square root in $$\,\sqrt{1-x}\,$$ will either vanish or factor out then get squared again, so the end result is in fact a polynomial.

For example:

• $$n=3:$$

\begin{align} U_2(x) &= 4x^2-1 \\ f_3(x) &= x \cdot U_2^2\left(\sqrt{1-x}\right) \\ &= x \cdot \left(4\,(1-x)-1\right)^2 \\ &= x\,\left(-4x+3\right)^2 \\ &= 16 x^3 - 24 x^2 + 9 x \end{align}

• $$n=4:$$

\begin{align} U_3(x) &= 4x\,(2x^2-1) \\ f_4(x) &= x \cdot U_3^2\left(\sqrt{1-x}\right) \\ &= x \cdot 16(1-x) \, \left(2\,(1-x)-1\right)^2 \\ &= 16x(1-x)\,\left(-2x+1\right)^2 \\ &= -64 x^4 + 128 x^3 - 80 x^2 + 16 x \end{align}