Every even degree polynomial is eventually symmetric

If $A(y)>0$, such that $$ a\big(A(y)\big)^{2n}+b\big(A(y)\big)^{2n-1}+\cdots+q=y, \tag{1} $$ then $A(y)\to\infty$, as $yas\to\infty$, and hence $\big(A(y)\big)^{2n}\gg \big(A(y)\big)^{2n-1}$. So from $(1)$ we obtain $A(y)\approx\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}$. In particular, there exists a function $e_+(y)$, such that $$ A(y)=\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y), \quad |e_+(y)|\ll y^\frac{1}{2n}. $$ To get a better estimate for $e_+$, we have $$ a\bigg(\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y)\bigg)^{2n}+b\bigg(\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_+(y)\bigg)^{2n-1}+\cdots+q=y, $$ or $$ \bigg(y+2na\Big(\frac{y}{a}\Big)^{\frac{2n-1}{2n}}e_+(y)+\frac{2n(2n-1)}{2}a\Big(\frac{y}{a}\Big)^{\frac{2n-2}{2n}}e_+^2(y)+\cdots\bigg)+\bigg(b\Big(\frac{y}{a}\Big)^{\frac{2n-1}{2n}}+b\Big(\frac{y}{a}\Big)^{\frac{2n-2}{2n}}e_+(y)+\cdots\bigg)\\+\cdots+q=y. $$ The assumption $|e_+(y)|\ll y^\frac{1}{2n}$ implies now that $$ e_+(y)=-\frac{b}{2an}+d_+(y) $$ where $\lim_{y\to\infty} d_+(y)=0$.

Repeating this argument for $B(y)<0$, with $$ a\big(B(y)\big)^{2n}+b\big(B(y)\big)^{2n-1}+\cdots+q=y, $$ we obtain that $$ B(y)=-\Big(\frac{y}{a}\Big)^{\frac{1}{2n}}+e_-(y), \quad |e_+(y)|\ll y^\frac{1}{2n}. $$ and similarly we obtain that $$ e_-(y)=-\frac{b}{2an}+d_-(y) $$ where $\lim_{y\to\infty} d_-(y)=0$.

Hence $$ A(y)+B(y)=-\frac{b}{na}+d_-(y)+d_+(y)\to -\frac{b}{na}. $$

It is clear that $A(y), B(y) \sim \pm (y/a)^{1/2n}$, and the problem is to determine the lower order term in their expansion. Wlog suppose $B(y) < 0 < A(y)$. By the mean value theorem, $$(A(y) - (y/a)^{1/2n}) \cdot F'(\xi) \sim y - F((y/a)^{1/2n}) \,,$$ for some $\xi \sim (y/a)^{1/2n}$. The LHS is $\sim 2an \cdot (y/a)^{(2n-1)/2n}$ times what we are looking for, and the RHS is $b(y/a)^{(2n-1)/2n} (1 + o(1))$. We conclude that $$A(y) - (y/a)^{1/2n} \sim \frac{b}{2an} \,.$$ Replacing $F(x)$ by $F(-x)$, we obtain $$-B(y) - (y/a)^{1/2n} \sim \frac{-b}{2an} \,.$$ The conclusion follows.

Let's replace $A(y), B(y) $ by $A, B$ to simplify typing and let $A>0>B$ and we write $C=-B$ so that $C>0$. Then we have $$y=aA^{2n}+bA^{2n-1}+\dots$$ and $$y=aC^{2n}-bC^{2n-1}+\dots$$ From these equations we get $$y/A^{2n}\to a, y/C^{2n}\to a$$ so that $A/C\to 1$.

Subtracting these equations we get $$a(A^{2n}-C^{2n})+b(A^{2n-1}+C^{2n-1})+\dots=0$$ or $$a(A-C) (A^{2n-1}+A^{2n-2}C+\dots+C^{2n-1})+b(A^{2n-1}+C^{2n-1})+\dots =0$$ Dividing the above equation by $C^{2n-1}$ we get $$a(A-C)\{1+(A/C)+(A/C)^2+\dots+(A/C)^{2n-1}\} +b\{1+(A/C)^{2n-1}\} + \text{ (terms tending to zero)} =0$$ Letting $y\to \infty $ in above equation we get $$2na\lim_{y\to\infty} (A-C) +2b=0$$ or $$A-C\to-\frac{b} {na} $$ as $y\to\infty $.

One has to observe that when we divide by $C^{2n-1}$ the terms like $A^{r} / C^{2n-1}$ tend to $0$ for $r<2n-1$ because we can write it as $(A/C) ^r(C^r/C^{2n-1})$.

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The argument above is entirely elementary and simple. We just have to understand that $A, C$ are functions of $y$ which are strictly increasing as $y\to\infty $ and $$A\to\infty, C\to\infty, A/C\to 1$$ as $y\to\infty $. Further they satisfy the relation $F(A) =y=F(-C) $.

In general most of the algebraic limits do not involve anything more algebraic manipulation.