Consider a real number $a \ge 1$, and let $g(a)$ be the unique positive solution $x>0$ of $x^a - x^{a-1} - 1 = 0.$ We have $g(1) = 2$, $g(2) = {1+\sqrt{5}\over2}$ (the golden ratio), and $g$ is strictly decreasing, with limit $1$. As bounds we know $$4^{1\over 1+a} \le g(a) \le 2^{a^{-1/2}}.$$

Of interest is also $\operatorname{lng}(a) := \ln(g(a))$ (using the natural logarithm). So $\operatorname{lng}(1) = \ln(2)$, and $${\ln(4)\over 1+a} \le \operatorname{lng}(a) \le \ln(2) a^{-1/2};$$ $\operatorname{lng}(a)$ is the unique positive solution $x > 0$ of $\exp(-x) + \exp(-ax) = 1.$

Any information on $g(a)$ (or $\operatorname{lng}(a)$) would be helpful. We are especially interested in good bounds, especially in good lower bounds for $g(a)$ ($\operatorname{lng}(a)$).

Restricting parameter a to natural numbers is also natural here. For $a$ in $\{1,2,3,4\}$ there are solutions via radicals for $g(a)$, while for other values of a we are not aware of any. Information on this would also be interesting.

The background can be seen in the study of recurrences like $f(n) = f(n-1) + f(n-5)$, belonging to g(5). For that quantities like $g(a)$ need to be approximated well.

The natural way of computing $\operatorname{lng}(a)$ is to use Newton approximation, starting with the lower bound ${\ln(4)\over 1+a}$. To reach (close to) full precision with a floating point type needs in case of "long double" (an extended $80$ big floating point type) quite precisely $\ln(a)$ many iterations. Having a better lower bound $\operatorname{lb}(a)$ for $\operatorname{lng}(a)$, we could start the computation with $\operatorname{lb}(a)$ instead, using fewer iterations then.

I have been using implicitly $g(a)$ in my work for many years now, but never came across some systematic study of such a function.

ANSWER to comments and Answer: Thank you for your comments, all very useful. The general lower bound by Iosif Pinelis should be very helpful. Unfortunately I try now already for quite some time all possibilities to add a comment, but it all fails, only this edit-function works. So sorry for this misuse. Oliver

  • $\begingroup$ Try googling "trinomial equation", although to by-pass school algebra hits involving factoring trinomials, you'll want to include words such as "hypergeometric" and "series" and "Euler" and "nested radical" and "Bring radical" and "Lagrange inversion" and "Lambert function" and "Fibonacci Quarterly" (to get results likely related to the Fibonacci sequence) and "Galois" (to get results involving their Galois groups) and "Ramanujan" (this was a shot in the dark, but it works) and so on. $\endgroup$ – Dave L Renfro Jan 10 at 20:57
  • $\begingroup$ $lng(1)=\ln2$ not $0$. $\endgroup$ – T. Amdeberhan Jan 10 at 21:32
  • $\begingroup$ Using Lagrange inversion on can e.g. show that (with $\alpha=1-\frac{1}{a}$) \begin{align*} a\,g(a)&=\sum_{n=1}^\infty\frac{1}{n}{\alpha(n-1) \choose n-1}\mbox{ and }\\ (a-1)\,\operatorname{lng}(a)&=\sum_{n=1}^\infty \frac{1}{n} {\alpha n \choose n}\end{align*} But this is probably more of theoretical interest. $\endgroup$ – esg Jan 14 at 17:18

Here is a way to improve your lower bound (l.b.) and upper bound for large enough $a$. Let $x_a:=\text{lng}(a)$. Then for $x=x_a$ we have \begin{equation} 1-x+e^{-ax}<e^{-x}+e^{-ax}=1, \end{equation} whence $x>e^{-ax}$ and $ax e^{ax}>a=aye^{ay}$, where \begin{equation} y=y_a=\frac{L(a)}a \end{equation} and $L$ is the Lambert function, so that $L(a)e^{L(a)}=a$ and $L(a)>0$. So, $y_a=\frac{L(a)}a$ is a l.b. on $x_a$. One can also use the elementary l.b. \begin{equation} L_{\text{low}}(a):=\ln a - \ln\ln a + (\ln\ln a)/(2 \ln a) \end{equation} on $L(a)$ for $a>e$, found near the end of Section "Asymptotic expansions" at Lambert function, to we obtain an elementary l.b., say $\tilde y_a$, on $y_a$.

Similarly, for $x=x_a$, using the already proved inequality $x_a>y_a$, we have \begin{equation} \frac1{1+y_a}+e^{-ax}>\frac1{1+x}+e^{-ax}>e^{-x}+e^{-ax}=1, \end{equation} whence $e^{-ax}>1-\frac1{1+y_a}$ and
\begin{equation} x=x_a<z_a:=\frac1a\,\ln\Big(\frac1{y_a}+1\Big), \end{equation} so that $z_a$ is an upper bound on $x_a$. Moreover, replacing $y_a=\frac{L(a)}a$ in the above expression for $z_a$ by its lower bound $\frac{L_{\text{low}}(a)}a$, we obtain an elementary u.b., say $\tilde z_a$, on $z_a$.

Thus, \begin{equation} \tilde y_a<y_a<x_a<z_a<\tilde z_a. \end{equation}

For $a\to\infty$, we have $L(a)\sim \ln a$ and hence $y_a\sim \frac{\ln a}a$ and $z_a\sim y_a$, so that the l.b. $y_a$ and the u.b. $z_a$ on $x_a$ are both asymptotically optimal. Moreover, since $L(a)\sim L_{\text{low}}(a)$, the elementary l.b. $\tilde y_a$ and u.b. $\tilde z_a$ on $x_a$ are also asymptotically optimal.

Your l.b. and u.b. seem to be good for small values of $a$, whereas the l.b. and u.b. on $x_a$ obtained here are good for large enough $a$. Here is an illustration:

enter image description here

  • $\begingroup$ To the "Lambert" lower bound, I have now added an upper one, along with their elementary versions. All these bounds are asymptotically optimal for $a\to\infty$. $\endgroup$ – Iosif Pinelis Jan 11 at 4:09

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