Chebyshev's series for square-root of x over $x\geq 0$.

Question

I am looking for the derivation of Chebyshev's series expansion for $\sqrt{x}$ over $x\in \mathbb{R}^{+}$. I am little confused on how to go from $[-1,1]$ range of Chebyshev's polynomial range to $\mathbb{R}^{+}$.

Answer 1

You need to map $x \in (0, \infty) $ to $y \in(-1, 1)$.

$\dfrac1{1+x} $ maps $(0, \infty)$ to $(0, 1) $ so $\dfrac{2}{1+x}-1 =\dfrac{1-x}{1+x} $ maps $(0, \infty)$ to $(-1, 1) $.

The inverse mapping for $y =\dfrac{1-x}{1+x} $ is $y+xy = 1-x $ or $1-y = x(1+y) $ or $x = \dfrac{1-y}{1+y} $.

Get the Chebychev series for $\sqrt{\dfrac{1-y}{1+y}} $ for $-1 < y < 1 $ and then, for any $x \in (0, \infty)$, get the series for $\dfrac{1-x}{1+x} $.