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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: $T_n(x) = \cos(n\arccos(x))$.

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1 Answer 1

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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:

  • $n=3:$

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

  • $n=4:$

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

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