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First, let's make the substitution $x\to 1-x$, so the function $x-\frac12x^2$ becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the discriminant of even and odd polynomials are essentially squares. More precisely, $$\begin{aligned} \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2 \end{aligned} $$ and $$\begin{aligned} \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2x^2A'(x^2)\bigr)\\ &= \pm A(0)\text{Resultant}\bigl(A(x^2),2x^2A'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)^3\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\ \end{aligned} $$

Next, define three operators on the ring of rational functions: $$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad R(f)=\frac{1}{f}. $$ The question asksabout iterates of $T$, but note that $T=I\circ D$ and $D\circ I=-I\circ D$, so $$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$ So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$, so it's enough to look at the logarithmic derivative $D$. Finally, we note that if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function, then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even. From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares, up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.

First, let's make the substitution $x\to 1-x$, so the function $x-\frac12x^2$ becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the discriminant of even and odd polynomials are essentially squares. More precisely, $$\begin{aligned} \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2 \end{aligned} $$ and $$\begin{aligned} \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2x^2A'(x^2)\bigr)\\ &= \pm A(0)\text{Resultant}\bigl(A(x^2),2x^2A'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)^3\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\ \end{aligned} $$

Next, define three operators on the ring of rational functions: $$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad I(f)=\frac{1}{f}. $$ The question asks about iterates of $T$, but note that $T=I\circ D$ and $D\circ I=-D$, so $$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$ So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$, so it's enough to look at the logarithmic derivative $D$. Finally, we note that if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function, then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even. From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares, up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.

First, let's make the substitution $x\to 1-x$, so the function $x-\frac12x^2$ becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the discriminant of even and odd polynomials are essentially squares. More precisely, $$\begin{aligned} \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2 \end{aligned} $$ and $$\begin{aligned} \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2xA'(x^2)\bigr)\\ &= \pm A(0)\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)^2\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\ \end{aligned} $$

Next, define three operators on the ring of rational functions: $$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad R(f)=\frac{1}{f}. $$ The question asksabout iterates of $T$, but note that $T=I\circ D$ and $D\circ I=-I\circ D$, so $$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$ So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$, so it's enough to look at the logarithmic derivative $D$. Finally, we note that if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function, then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even. From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares, up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.

First, let's make the substitution $x\to 1-x$, so the function $x-\frac12x^2$ becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the discriminant of even and odd polynomials are essentially squares. More precisely, $$\begin{aligned} \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2 \end{aligned} $$ and $$\begin{aligned} \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2x^2A'(x^2)\bigr)\\ &= \pm A(0)\text{Resultant}\bigl(A(x^2),2x^2A'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)^3\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\ \end{aligned} $$

Next, define three operators on the ring of rational functions: $$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad R(f)=\frac{1}{f}. $$ The question asksabout iterates of $T$, but note that $T=I\circ D$ and $D\circ I=-I\circ D$, so $$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$ So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$, so it's enough to look at the logarithmic derivative $D$. Finally, we note that if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function, then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even. From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares, up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.

First, let's make the substitution $x\to x-1$, so the function $x-\frac12x^2$ becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the discriminant of even and odd polynomials are essentially squares. More precisely, $$\begin{aligned} \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2 \end{aligned} $$ and $$\begin{aligned} \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2xA'(x^2)\bigr)\\ &= \pm A(0)\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)^2\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\ \end{aligned} $$

Next, define three operators on the ring of rational functions: $$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad R(f)=\frac{1}{f}. $$ The question asksabout iterates of $T$, but note that $T=I\circ D$ and $D\circ I=-I\circ D$, so $$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$ So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$, so it's enough to look at the logarithmic derivative $D$. Finally, we note that if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function, then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even. From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares, up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.

First, let's make the substitution $x\to 1-x$, so the function $x-\frac12x^2$ becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the discriminant of even and odd polynomials are essentially squares. More precisely, $$\begin{aligned} \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2 \end{aligned} $$ and $$\begin{aligned} \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2xA'(x^2)\bigr)\\ &= \pm A(0)\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\ &= \pm 2^{\deg(A)} \cdot A(0)^2\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\ \end{aligned} $$

Next, define three operators on the ring of rational functions: $$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad R(f)=\frac{1}{f}. $$ The question asksabout iterates of $T$, but note that $T=I\circ D$ and $D\circ I=-I\circ D$, so $$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$ So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$, so it's enough to look at the logarithmic derivative $D$. Finally, we note that if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function, then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even. From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares, up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.