Consider a function from the Bessel family, for concreteness say $f(x) := J_0(x)$, depicted in blue below (the question can be asked for any order of the first or second kind):

enter image description here

I'm interested in the orange tangent curve to the function. That is, an analytic, decreasing, convex (possibly even completely monotonic) function $g(x)$ that always upper bounds $f(x)$ and meets the function (with a matching derivative) once for each oscillation (the intersection points converging to the local maxima of $f$ as $x$ grows).

We know from the asymptotics of the Bessel function that $g(x)$ asymptotically behaves like $\sqrt{2/(\pi x)}$. But is there an explicit exact expression known in the literature, in any form (in terms of special functions, as a series expansion, integral form, etc)?

In general, what are the known techniques to approach questions of this kind (deriving tangent curves to oscillating functions, or even proving that they exist)?

  • $\begingroup$ Certainly there are infinitely many such functions. Maybe one has add some simplifications constraints, like monicity, little oszillation or curvature.. $\endgroup$
    – hänsel
    Dec 22, 2017 at 9:03
  • $\begingroup$ For the Bessel function $J_0(x)$, see Lemma C.2 of this paper: math.ubc.ca/~gerg/papers/downloads/IPNR.pdf. They show that $|J_0(x)| \leq \min\{1, \sqrt{2/\pi |x|}\}$. Of course, this is not quite the same thing as the tangent curve, but perhaps it's good enough? $\endgroup$ Dec 22, 2017 at 9:55
  • $\begingroup$ @hansel: I'd ideally like the curve to be completely monotonic, or at least be convex. With that, I don't see how there can be infinitely many. $\endgroup$
    – MCH
    Dec 22, 2017 at 11:16
  • $\begingroup$ @Peter: I'm interested in the exact curve but in any nontrivial form (series, integral form, etc). $\endgroup$
    – MCH
    Dec 22, 2017 at 11:17
  • 6
    $\begingroup$ In your case of the Bessel function a good choice appear to be $|J_0(x)+iY_0(x)|$. This has to be proved, but the figure with Mathematica is spectacular. Perhaps this is how you get your figure. These are the Hankel functions. $\endgroup$
    – juan
    Dec 22, 2017 at 12:00

2 Answers 2


juan's suggestion of the Hankel function seems great. Here is $J_0(x)$ and $u(x):=|J_0(x)+iY_0(x)|$.
Of course $u(x) \ge J_0(x)$ everywhere and $u(x) = |J_0(x)|$ at all the zeros of $Y_0(x)$. For asymptotic properties of $u(x)$, from Maple I get asymptotics $$ u(x) = \sqrt{\frac{2}{\pi x}}+O(x^{-3/2})\qquad\text{as } x \to \infty $$ which explains the good fit in the picture of the OP.

For example, At the zero $x_0 = 19.64130970$ of $Y_0(x)$, we have $$ J_0(x_0) = u(x_0) = 0.180005\qquad\text{and}\qquad \sqrt{\frac{2}{\pi x_0}} = 0.180034 , $$ a very close fit.

  • $\begingroup$ Thank you. It can also be verified, from $|H_0(x)|=\sqrt{J_0(x)^2 + Y_0(x)^2}$, and directly taking the derivative, that the derivatives also match at every root of $Y_0(x)$. $\endgroup$
    – MCH
    Dec 22, 2017 at 15:07

To complete your answer, you could note note that a a consequence of Nicholson's integral identity, proved in Watson's book, section 13.73. is that if $x\geq n\geq 0$ then fact $$ \left|H_n(x)\right|\left(x^2-n^2\right)^{1/4} \text{ is an increasing function} $$ whereas if $n>\frac{1}{2}$ and $x>0$ $$ \left|H_n(x)\right|\sqrt{x} $$ is a decreasing function.


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