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I encountered this when trying to solve a number theory problem. I have two variables $x,y$ related by

$$(\ln(x))^{y+1}=(\ln(xy))^y$$

and I want to know how big $y(x)$ is as $x\to\infty$. Ideally I want to know that $y(x)\sim\ln(x)$ or $y(x)\sim \sqrt{x}$ or whatever it is. But if that's not possible, a tight lower bound would suffice.

I've messed with this expression enough that I managed to convince myself it isn't possible to isolate either variable, except maybe using something like Lambert's W function, and even that I wasn't able to do. So the remaining alternative is to try to solve it as a problem about an implicit function. This sounds like something that is possible with standard calculus tools.

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  • $\begingroup$ using $ \log(x)+1\sim \log(x)$ and $\log( x \log(x))\sim \log(x)$ it seems that $y=\log(x)$ balances both sides in the big $x$ limit. my guess is actually that $y\sim \log(x)+1+o(1)$ but i haven't attempted that calcukations. $\endgroup$
    – asgeige
    Commented Nov 18, 2021 at 4:42
  • $\begingroup$ @asgeige. Just a bit more. $\endgroup$ Commented Nov 18, 2021 at 10:50

2 Answers 2

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I will assume that $y>1$ and $x >e$. Note that your equation is equivalent to $$ \log x = \left( {1 + \frac{{\log y}}{{\log x}}} \right)^y . $$ From this, $$ \log x = \left( {1 + \frac{{\log y}}{{\log x}}} \right)^y \le e^{y\frac{{\log y}}{{\log x}}} , $$ i.e., $\log x\log \log x \le y\log y$. This implies $y \ge \log x$. Therefore, $$ \log x = \left( {1 + \frac{{\log y}}{{\log x}}} \right)^y \ge \left( {1 + \frac{{\log \log x}}{{\log x}}} \right)^y , $$ i.e., $$ \frac{{\log \log x}}{{\log \left( {1 + \frac{{\log \log x}}{{\log x}}} \right)}} \ge y. $$ The left-hand side is $\log x +\mathcal{O}(\log \log x)$. Thus, we can also see that $y(x)=\log x +\mathcal{O}(\log \log x)$ as $x \to +\infty$.

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To make things easier to handle, start letting $xy=z$ and solve for $y$. In the resulting expression, reset $z=xy$ to face the implicit function $$F(x,y)=y \log \left(\frac{\log (x y)}{\log (x)}\right)-\log (\log (x))$$ Finding its zero does not make any problem using Newton method since, using the implicit function theorem $$\frac {dy}{dx}=\frac{(y+1) \log (x y)-y \log (x)}{x \log (x) \left(1+\log (x y) \log \left(\frac{\log (x y)}{\log (x)}\right)\right)}$$

Now, for the computation, let $\color{red}{x=e^k}$. This makes the equation to be $$\color{blue}{y \log \left(1+\frac{\log (y)}{k}\right)-\log (k)=0}$$ Some results $$\left( \begin{array}{cc} k & y \\ 1 & 1.00000\\ 2 & 2.14477 \\ 3 & 3.28784 \\ 4 & 4.39968 \\ 5 & 5.49064 \\ 6 & 6.56722 \\ 7 & 7.63335 \\ 8 & 8.69156 \\ 9 & 9.74355 \\ 10 & 10.7905\\ 20 & 21.1086 \\ 30 & 31.3002 \\ 40 & 41.4378 \\ 50 & 51.5453 \\ 60 & 61.6335 \\ 70 & 71.7083 \\ 80 & 81.7732 \\ 90 & 91.8306 \\ 100 & 101.882 \\ 200 & 202.221 \\ 300 & 302.421 \\ 400 & 402.562 \\ 500 & 502.672 \\ 600 & 602.762 \\ 700 & 702.838 \\ 800 & 802.904 \\ 900 & 902.962 \\ 1000 & 1003.01 \end{array} \right)$$ which do not need much comments.

Notice that $e^{1000}=1.97\times 10^{434}$.

In fact making one single iteration of Newton method with $y_0=k$, we have $$\color{red}{y_1=k+\frac{(k+\log (k)) \left(\log (k)-k \log \left(1+\frac{\log (k)}{k}\right)\right)}{(k+\log (k)) \log \left(1+\frac{\log (k)}{k}\right)+1}}$$

For $k=10$, this simple expression gives $y=10.7984$ and, for $k=1000$, it gives $y=1003.01$.

Expanded as series for large values of $k$ $$y_1=k+\frac{\log ^2(k)}{2 (\log (k)+1)}\Bigg[1-\frac{(\log (k)-2) \log (k)}{6 k (\log (k)+1)} +O\left(\frac{1}{k^2}\right)\Bigg]$$

Looking at this last result, there is tight upper bound $$y < y_*=k+\frac 12 \log(k)$$ starting Newton iteration with this starting point, we should converge without any overshoot of the solution since $f(y_*)\,f''(y_*) >0$ (Darboux theorem).

If, if the "blue" formula, we make inside the first logarithm $y=y_*$, we end with tight bounds $$ k+\frac 12 \log(k)-\frac 12 <y < k+\frac 12 \log(k)$$

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