The idea is to reduce the multiplier-$(-1)$ problem to the multiplier-$(+1)$ problem by passing to the second iterate $g := f \circ f$. Since $g'(p) = (f'(p))^2 = 1$, the fixed point $p$ of $f$ is also a fixed point of $g$ with unit multiplier, and the stability of $p$ under $f$ is equivalent to its stability under $g$. By the [Stability of a Nonhyperbolic Fixed Point with Multiplier One](/theorems/642), this stability is determined by the sign of $g'''(p)$ (provided $g''(p) = 0$). The main computation shows that $g''(p) = 0$ and $g'''(p) = 2 \, Sf(p)$, establishing the criterion. The proof proceeds in three claims: the first two compute $g''(p)$ and $g'''(p)$ via repeated [differentiation](/page/Derivative) of the chain rule, and the third identifies the result with the Schwarzian. **Step 1: Setup and reduction to the second iterate.** Define $g := f \circ f$, so $g(x) = f(f(x))$. Since $f(p) = p$, we have $g(p) = p$. By the chain rule, $g'(p) = f'(f(p)) \cdot f'(p) = f'(p)^2 = (-1)^2 = 1$. Thus $p$ is a fixed point of $g$ with multiplier $+1$. A point $x$ near $p$ converges to $p$ under iteration of $f$ if and only if both $x$ and $f(x)$ converge to $p$ under iteration of $g = f^2$. Thus the stability of $p$ for $f$ is equivalent to the stability of $p$ for $g$. By the [Stability of a Nonhyperbolic Fixed Point with Multiplier One](/theorems/642), this depends on $g''(p)$ and $g'''(p)$. [claim: Vanishing Of The Second Derivative] Under the hypotheses of the theorem, $g''(p) = 0$. [/claim] [proof] Differentiating $g(x) = f(f(x))$ twice using the chain rule: \begin{align*} g'(x) &= f'(f(x)) \, f'(x), \\ g''(x) &= f''(f(x)) \, (f'(x))^2 + f'(f(x)) \, f''(x). \end{align*} Evaluating at $x = p$ with $f(p) = p$ and $f'(p) = -1$: \begin{align*} g''(p) &= f''(p) \cdot (-1)^2 + (-1) \cdot f''(p) = f''(p) - f''(p) = 0. \end{align*} [/proof] Since $g'(p) = 1$ and $g''(p) = 0$, the [Stability of a Nonhyperbolic Fixed Point with Multiplier One](/theorems/642) (Cases 2 and 3) shows that stability is controlled by the sign of $g'''(p)$. [claim: Third Derivative Of The Second Iterate] Under the hypotheses of the theorem, \begin{align*} g'''(p) = -2f'''(p) - 3(f''(p))^2. \end{align*} [/claim] [proof] Differentiate $g''(x) = f''(f(x))(f'(x))^2 + f'(f(x))f''(x)$ once more using the product and chain rules: \begin{align*} g'''(x) &= f'''(f(x)) \, (f'(x))^3 + 2 f''(f(x)) \, f'(x) \, f''(x) \\ &\quad + f''(f(x)) \, f'(x) \, f''(x) + f'(f(x)) \, f'''(x). \end{align*} Collecting terms and evaluating at $x = p$ with $f(p) = p$ and $f'(p) = -1$: \begin{align*} g'''(p) &= f'''(p) \cdot (-1)^3 + 2 f''(p) \cdot (-1) \cdot f''(p) + f''(p) \cdot (-1) \cdot f''(p) + (-1) \cdot f'''(p) \\ &= -f'''(p) - 2(f''(p))^2 - (f''(p))^2 - f'''(p) \\ &= -2f'''(p) - 3(f''(p))^2. \end{align*} [/proof] [claim: Identification With The Schwarzian] \begin{align*} g'''(p) = 2 \, Sf(p). \end{align*} [/claim] [proof] Recall the definition $Sf(p) = \dfrac{f'''(p)}{f'(p)} - \dfrac{3}{2}\left(\dfrac{f''(p)}{f'(p)}\right)^2$. Substituting $f'(p) = -1$: \begin{align*} Sf(p) &= \frac{f'''(p)}{-1} - \frac{3}{2}\left(\frac{f''(p)}{-1}\right)^2 = -f'''(p) - \frac{3}{2}(f''(p))^2. \end{align*} Therefore \begin{align*} 2 \, Sf(p) = -2f'''(p) - 3(f''(p))^2 = g'''(p). \end{align*} [/proof] **Step 2: Conclusion.** Combining the three claims: $g'(p) = 1$, $g''(p) = 0$, and $g'''(p) = 2 \, Sf(p)$. By the [Stability of a Nonhyperbolic Fixed Point with Multiplier One](/theorems/642): - If $Sf(p) < 0$, then $g'''(p) < 0$, so $p$ is asymptotically stable for $g$ and hence for $f$. - If $Sf(p) > 0$, then $g'''(p) > 0$, so $p$ is unstable for $g$ and hence for $f$.