The uniform second-order Taylor remainder of $z\mapsto (1-z{)}^{-1\mathrm{/}2}$ on $[0,z_0]$

The composite form of $f$. Define $g:(-\mathrm{\infty },1)\to (0,\mathrm{\infty })$ by $g(z):=1-z$ and $h:(0,\mathrm{\infty })\to (0,\mathrm{\infty })$ by $h(y):=y^{-1\mathrm{/}2}$. For every $z<1$ the base $1-z$ is positive, so $f=h\circ g$; and by [L1] the rational power $h(y)=y^{-1\mathrm{/}2}$ equals the real power $\exp (-\frac{1}{2}\log y)$, so the derivative statement of [L2] applies to $h$.

The inner derivative. The constant function $z\mapsto 1$ has derivative $0$ and the identity $p_1$ has $p_1^{\mathrm{\prime }}\equiv 1$ by [L3]; the difference rule of Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$ therefore gives $g^{\mathrm{\prime }}(z)=-1$ for every $z\in (-\mathrm{\infty },1)$, and each point of $(-\mathrm{\infty },1)$ is a limit point of that interval.

The outer derivative. By [L2] with $\alpha =-\frac{1}{2}$, the function $y\mapsto y^{-1\mathrm{/}2}$ is differentiable on $(0,\mathrm{\infty })$ with derivative $(y^{-1\mathrm{/}2}{)}^{\mathrm{\prime }}=-\frac{1}{2}{y}^{-3\mathrm{/}2}$; by step 1.1 this is $h^{\mathrm{\prime }}$.

First derivative of $f$. Fix $z\in (-\mathrm{\infty },1)$. Then $g(z)=1-z>0$ is a limit point of $(0,\mathrm{\infty })$, so the chain rule [L3] applied to $f=h\circ g$ at $z$ gives $f^{\mathrm{\prime }}(z)=h^{\mathrm{\prime }}(g(z))g^{\mathrm{\prime }}(z)=-\frac{1}{2}(1-z{)}^{-3\mathrm{/}2}\cdot (-1)=\frac{1}{2}(1-z{)}^{-3\mathrm{/}2}\mathrm{.}$

Second derivative of $f$. The function $z\mapsto \frac{1}{2}(1-z{)}^{-3\mathrm{/}2}$ is the scalar multiple by $\frac{1}{2}$ of the composite of $g$ with $y\mapsto y^{-3\mathrm{/}2}$; the rational power $y^{-3\mathrm{/}2}$ agrees with the real power by [L1], so [L2] with $\alpha =-\frac{3}{2}$ gives it derivative $-\frac{3}{2}{y}^{-5\mathrm{/}2}$ on $(0,\mathrm{\infty })$, and the scalar rule and the chain rule [L3] give $f^{\mathrm{\prime }\mathrm{\prime }}(z)=\frac{1}{2}(-\frac{3}{2}(1-z{)}^{-5\mathrm{/}2})(-1)=\frac{3}{4}(1-z{)}^{-5\mathrm{/}2}$ for every $z\in (-\mathrm{\infty },1)$, which also covers the claim that $f$ is twice differentiable there.

Uniform bound on $[0,z_0]$. Let $z\in [0,z_0]$. By [L4], $0<1-z_0\le 1-z$, so $(1-z_0{)}^{5\mathrm{/}2}\le (1-z{)}^{5\mathrm{/}2}$ and, taking reciprocals of positive elements, $(1-z{)}^{-5\mathrm{/}2}\le (1-z_0{)}^{-5\mathrm{/}2}$. Multiplying by $\frac{3}{4}>0$ and using $f^{\mathrm{\prime }\mathrm{\prime }}>0$ by [L1] to read $\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }=f^{\mathrm{\prime }\mathrm{\prime }}$ ([L6]) gives $\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}(z)\mathrm{\mid }=f^{\mathrm{\prime }\mathrm{\prime }}(z)\le \frac{3}{4}(1-z_0{)}^{-5\mathrm{/}2}$ for every $z\in [0,z_0]$.

The Taylor remainder. Fix $z\in [0,z_0]$. Since $[0,z]\subseteq (-\mathrm{\infty },1)$, the function $f$ has derivatives through order $2$ on the closed interval between $0$ and $z$, continuous at the endpoints; by step 5.1 the second derivative is bounded there by $M:=\frac{3}{4}(1-z_0{)}^{-5\mathrm{/}2}$. Applying [L5] with $n=1$, $a=0$ and this $z$ gives, for $R(z):=R_{1,0}f(z)=f(z)-f(0)-f^{\mathrm{\prime }}(0)z$, $\mathrm{\mid }R(z)\mathrm{\mid }\le \frac{M}{2!}{z}^2=\frac{3}{8}(1-z_0{)}^{-5\mathrm{/}2}{z}^2=K(z_0)z^2\mathrm{.}$ Here $f(0)=1^{-1\mathrm{/}2}=1$ by [L1] and $f^{\mathrm{\prime }}(0)=\frac{1}{2}$ by step 3.1, so the displayed expansion of $f$ is $f(z)=1+\frac{z}{2}+R(z)$.

The quantitative limit. For $z\in [0,z_0]$, steps 6.1 and 3.1 give $\mathrm{\mid }f(z)-1\mathrm{\mid }=\mid \frac{z}{2}+R(z)\mid \le \frac{z}{2}+K(z_0)z^2=z(\frac{1}{2}+K(z_0)z)\le z(\frac{1}{2}+K(z_0)z_0)$ by [L6] and $K(z_0)>0$, $z\le z_0$, which is the first displayed assertion of (b); if in addition $0\le z\le \eta \mathrm{/}(\frac{1}{2}+K(z_0)z_0)$ for an $\eta >0$, then $\mathrm{\mid }f(z)-1\mathrm{\mid }\le \eta$, which is the second displayed assertion; and the bound is not asserted for $z$ outside $[0,z_0]$.

Failure of a single remainder constant on $[0,1)$. For every integer $N\ge 2$ put $z_N=1-N^{-2}\in (0,1)$. By [L1], $f(z_N)=N$, so the exact remainder $r_N:=f(z_N)-1-z_N\mathrm{/}2=N-1-z_N\mathrm{/}2$ satisfies $r_N>N-3\mathrm{/}2$. Given $B\ge 0$, choose $N>\max \{2,B+3\mathrm{/}2\}$ by [L7]; then $\mathrm{\mid }{r}_N\mathrm{\mid }>B\ge B{z}_N^2$. Thus no finite $B\ge 0$ bounds $\mathrm{\mid }f(z)-1-z\mathrm{/}2\mathrm{\mid }$ by $B{z}^2$ on all of $[0,1)$.