Largest root and asymptotics.

Question

I am considering the polynomial $$ p_c(x)=x^{c+1}-x^c-1. $$ Its largest root, denoted by $\lambda_c$, is contained in the open interval $(1,1+\frac{\ln c}{c})$.

Making the ansatz $\lambda_c=1+x_c$ for a zero sequence $(x_c)$, I want to express more explicitly what happens with $\lambda_c$ as $c\to\infty$.

Putting the ansatz into the polynomial, I get $$ (1+x_c)^cx_c=1 $$ which yields, when applying logarithm, $$ c\ln(1+x_c)+\ln(x_c)=0. $$ Using the approximation $\ln(1+x_c)=x_c+f(x_c)$ with $f(x_c)=O(x_c^2)$ as $c\to\infty$, this gives $$ cx_c+cf(x_c)+\ln(x_c)=0. $$ Exponentiating and multiplicating with the factor $c$ yields $$ cx_ce^{cx_c}=ce^{-cf(x_c)} $$ and using Lamberts W-Function, ones gets $$ x_c=c^{-1}W(ce^{-cf(x_c)}). $$ Using the large argument approximation $$ W(x)=\ln x-\ln(\ln(x))+o(1)\text{ as }x\to\infty, $$ gives me $$ x_c=\frac{\ln c}{c}-f(x_c)-\frac{\ln(\ln c)}{c}-\frac{\ln(1-\frac{cf(x_c)}{\ln c})}{c}+o\left(\frac{1}{c}\right)\text{ as }c\to\infty $$

My question is whether this can be also expressed as $$ x_c=\frac{\ln c}{c}+o\left(\frac{\ln c}{c}\right)\text{ as }c\to\infty? $$

For the summands $o\left(\frac{1}{c}\right)$ and $\frac{\ln(\ln c)}{c}$, I can see immediately that they are indeed of order $o\left(\frac{\ln c}{c}\right)$ as $c\to\infty$.

But for the summands $f(x_c)$ and $\frac{\ln(1-\frac{cf(x_c)}{\ln c})}{c}$ this is absolutely not clear to me.

Edit (due to the comments):

$x_c=o(1)$ and $x_c=O(\ln c/c)$. Both together imply $x_c^2=o(\ln c/c)$, as $c\to\infty$. Moreoever, $f(x_c)=O(x_c^2)$ and $x_c^2=o(\ln c/c)$ imply that $f(x_c)=o(\ln c/c)$ as $c\to\infty$.

Hence, it remains to clarify whether $\frac{\ln(1-\frac{cf(x_c)}{\ln c})}{c}=o(\ln c/c)$ as $c\to\infty$.

If $x_c=O(\ln c/c)$ then $x_c^2 =o(\ln c/c)$ and so $f(x_c)=O(x_c^2)=o(\ln c/c)$. — Marco, Sep 13 '18 at 16:38
I think the answer to you question is no. If we write $x_c = \frac{\ln c}{c}+g(c)$ and assume $g(c)=o\left(\frac{\ln c}{c}\right)$, by plugging this into $c\ln(1+x_c)+\ln(x_c)=0$ and using $\ln(1+x)=x+O(x^2)$ we get a contradiction. — leonbloy, Sep 13 '18 at 16:41
@Rhjg, since $x_c=o(1)$ and $x_c=O(\ln c/c)$ we should have $x_c^2=o(\ln c/c)$. In general, if $f=o(g)$ and $h=O(k)$, then $fh=o(gk)$ since $fh/gk=(f/g)(h/k)$ is a product of $f/g$ which goes to zero and $h/k$ which is bounded and the squeeze theorem shows that product should go to zero. Am I missing something here? — Marco, Sep 13 '18 at 17:22
Is it true that, since $f(x_c)=o(\ln c/c)$ as $c\to\infty$, we have that $cf(x_c)=o(\ln c)$ as $c\to\infty$? If yes, doesn't this imply that $\ln(1-\frac{cf(x_c)}{\ln c})=o(1)$ and hence $\frac{1}{c}\ln(1-\frac{cf(x_c)}{\ln c})=o(1/c)=o(\ln c/c)$ as $c\to\infty$? — Rhjg, Sep 13 '18 at 18:24
Yes, I saw this comment but did not understand it, this is why I wanted to show it in another way — Rhjg, Sep 14 '18 at 05:24

score 2 · Answer 1 · answered Sep 13 '18 at 18:28

2

The claim is equivalent to $$F(c)=\dfrac{\ln (1-cf(x_c)/\ln c)}{\ln c} \rightarrow 0,$$ as $c \rightarrow \infty$. Since $|f(x_c)|\leq Kx_c^2$ for some $K>0$ and $x_c<\ln c/c$, we have $$\ln(1-cf(x_c)/\ln c)\leq \ln (1+cKx_c^2/\ln c) \leq \ln (1+K \ln c/c) $$ and so $$F(c) \leq \dfrac{\ln (1+K \ln c/c)}{\ln c} \rightarrow 0,$$ since $\ln c /c \rightarrow 0$ as $c \rightarrow \infty$.

answered Sep 13 '18 at 18:28

Marco

2,733
6
15

Is $F(c)$ non-negative? In other words: Does your last inequality imply that indeed $F(c)\to 0$ as $c\to\infty$? – Rhjg Sep 13 '18 at 18:35
1

Yes, because $f(x)=\ln(1+x)-x<0$ for $x>0$ and so the numerator is the log of a number greater than 1 hence positive. – Marco Sep 13 '18 at 18:41
Thanks! So the whole thing is indeed true. Really great. My attempt was to use that I now know that $f(x_c)=o(\ln c/c)$. Since if I am not mistaken, this should imply that $cf(x_c)=o(\ln c)$. But that means that $\lim_{c\to\infty}\ln(1-\frac{cf(x_c)}{\ln c})=\ln(1-\lim_{c\to\infty}\frac{cf(x_c)}{\ln c})=\ln(1)=0$ and hence $\ln(1-\frac{cf(x_c)}{\ln c})=o(1)$ and $\frac{1}{c}\ln(1-\frac{cf(x_c)}{\ln c})=o(1/c)$ as $c\to\infty$. But each function in $o(1/c)$ for $c\to\infty$ is also $o(\ln c/c)$ as $c\to\infty$. - - Would you agree? – Rhjg Sep 13 '18 at 18:47
@Rhjg, Yes that is all right. – Marco Sep 13 '18 at 19:19

Largest root and asymptotics.

1 Answers1