Let: $$f(\Delta)=\sqrt{b \log ^n\left(\frac{E}{\Delta }\right)+4\frac{\Delta ^2}{E^2}},\ \ n=4,6$$ $$E,b,\Delta>0,\ \ \Delta \ll E$$

The function has a single minimum, but the explicit expression is difficult to find. Can an approximate location for the minimum of $f(\Delta)$ be found in terms of $E,b$?

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    $\begingroup$ Should we assume $b,E,\Delta \gt 0$? Have you considered dropping the square root? (It's monotonic.) $\endgroup$ – hardmath Jan 29 '14 at 12:41
  • $\begingroup$ @hardmath - yes. You can also assume $\Delta\ll E$, and I don't see how dropping the sqrt helps. $\endgroup$ – nbubis Jan 29 '14 at 12:42
  • $\begingroup$ Am I correct thinking that $b$ should be very samll ? $\endgroup$ – Claude Leibovici Jan 30 '14 at 10:45
  • $\begingroup$ @ClaudeLeibovici - yes, I think so. Let's assume $b \ll 1$. $\endgroup$ – nbubis Jan 30 '14 at 15:44

I'm not sure whether you don't understand that the square root can be dropped, or you don't agree that dropping it simplifies the calculation. The minimum of $f(\Delta)$ will be attained at the same $\Delta$ that minimizes $f^2$, so generally we expect working with $f^2$ to give a little easier work to find the same approximations to $\Delta$.

Here's my take. We substitute $x=\log(E/\Delta)$ into $f^2 = b \log^n \left(\frac{E}{\Delta}\right) + 4 \frac{\Delta^2}{E^2}$:

$$ g(x) = bx^n + 4e^{-2x} $$

The assumption that $\Delta \ll E$ means $x \gt 0$, so we can proceed with the usual setting of $g'(x)$ to zero to identify its minimum (critical point) on $(0,\infty)$.

$$ g'(x) = nb x^{n-1} - 8e^{-2x} $$

Thus $g'(x)=0$ occurs when:

$$ x^{n-1} e^{2x} = \frac{8}{nb} $$

While the left hand side doesn't have a convenient (functional) inverse, for $n=4,6$ it is monotone increasing on $(0,\infty)$. We don't have any information about the magnitude of $b$, but a narrow approximation of the minimizing $x$ (and thus of $\Delta$) can be extracted with a little work. First check to see if $x=1$ is too high or too low, ie. check if $e^2$ is above or below $\frac{8}{nb}$. It seems likely, given that $\Delta \ll E$, that you would expect $x \gt 1$, but this is easy to check.

I would then do some fixed-point iterations of:

$$ h(x) = \frac{\log{8} - \log{nb} - (n-1)\log(x)}{2} $$

starting from $x_0 = 1$, noting that $h(x)=x$ implies $x$ solves the critical point equation for $g'(x)=0$.

I understand that an (easily-computed) analytic approximation for the minimum value is sought, but let me at least point out that the exact location where the minimum occurs can be expressed in terms of the Lambert $W$ function. Taking the $n-1$ root of both sides of the critical point equation:

$$ x e^{\frac{2x}{n-1}} = \sqrt[n-1]{\frac{8}{nb}} $$

Thus $x = \left(\frac{n-1}{2}\right) W\left( \left( \frac{2}{n-1} \right) \sqrt[n-1]{\frac{8}{nb}} \right) $, and from $x = \log E - \log \Delta$ the exact minimizing value of $\Delta$ can be determined.

As we already see from the critical point equation, $x$ will be uniquely determined by $B = 8/nb$, so given that $n=4,6$, the assumption that $b \ll 1$ would say that $B \gg 1$. Of course the left-hand side $x^{n-1}e^{2x}$ is a rapidly increasing monotone function on $[0,\infty)$. One step of our proposed fixed-point iteration ($x_0 = 1$) gives $x_1 = \log(B)/2$, which is an overestimate of the actual root/critical point.

One idea would be to plug this first iteration back into the function, and get a better approximation $x_2 = h(\log(B)/2)$. It is well-known that if $|h'(x_k)| \le \alpha \lt 1$, then the iteration is a contraction map and converges at least linearly. To investigate from this point of view:

$$ h'(x) = \frac{1-n}{2x} $$

so that $|h'(x)| < 1$ in a neighborhood of root $x$ if and only if $x \gt (n-1)/2$.

Recall that the critical point equation allows us to verify easily if this is so:

$$ \text{root } x \gt (n-1)/2 \iff 4^{n-1} e^8 \lt \frac{8}{nb} = B $$

Therefore if this condition holds, e.g. if $b \lt 4^{4-n} e^{-8} n^{-1}$, then a sequence of analytic approximations that converge to the root $x$ is:

$$ \frac{n-1}{2}, h\left(\frac{n-1}{2}\right), h\left(h\left(\frac{n-1}{2}\right)\right), \ldots $$

If $b$ is not small enough, or if the sequence developed above doesn't converge rapidly enough for the (unstated) purpose, then perhaps an alternative approach based on the exact solution $x$ in terms of Lambert W (above) is more attractive. Numerical evaluation of Lambert W by Newton's method or by Halley's method (cubic convergence) turns out to be surprisingly elegant. For series and related iterative methods, see here and here.

  • $\begingroup$ if we're going to use fixed-point iterations, I could do that to the original equation as well, I was trying to see if I could come up with an approximate analytic form. $\endgroup$ – nbubis Jan 29 '14 at 19:00
  • $\begingroup$ @nbubis: Perhaps you could give me a specific $b$ and I can show how this gives an analytic approximation. $\endgroup$ – hardmath Jan 29 '14 at 20:27
  • $\begingroup$ I think we can assume $b \ll 1$. Does that help? $\endgroup$ – nbubis Jan 30 '14 at 15:44
  • $\begingroup$ Sure, as you can see from the rewriting of the problem, $E$ and $\Delta$ can be paired off into new variable $x$, so the real driver of the solution is $b$ (as well as $n$, but you specified $n=4,6$ as a couple of values to handle). $\endgroup$ – hardmath Jan 30 '14 at 16:02

Approximate the function $f$ by some low degree polynomial, whose coefficients will be functions of $E$ and $b$. Locate the minimum of this polynomial, and use it as an approximation of the minimum of $f$.

  • $\begingroup$ Tried that, it doesn't give very good results. I'd be glad to be proven otherwise. $\endgroup$ – nbubis Jan 29 '14 at 18:30

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