1
$\begingroup$

Looking at the asymptotic expansion of the Bessel function of the first kind $J_0(x)$, the following bound seems an obvious conjecture

$$ \left | J_0(x) \right | \le \min \left (1, \sqrt{\frac{2}{\pi x}} \right) $$

I would also be interested in any bound of the form $J_0(x) \le A/ \sqrt{x +B}$ for some constants $A>0,\, B\ge0$. The bound looks tight when plotted on short segments and seems an obvious result. However I was a bit surprised when coming across this article.

Notes that the bounds are meant to hold for all real $x$, though $J_0(x)$ is even so one can restrict oneself to $x\ge0$.

$\endgroup$
1
$\begingroup$

This bound is known, and one citation given for it is G.N. Watson's book A Treatise On The Theory Of Bessel Functions, pp 205-8. Even accounting for the book's quite unfriendly writing style, it's not clear to me that the bound actually occurs therein.... Here is a complete proof.

Modulo a computation on the interval $[0,2]$, it suffices to show that $\frac{\pi x}2 J_0(x)^2 \le 1$ for $x\ge2$. At the top of page 206, equation (1), we have $$ J_0(x) = \sqrt{\frac2{\pi x}} \big( \cos(x-\tfrac\pi4) P(x,0) - \sin(x-\tfrac\pi4) Q(x,0) \big), $$ where $P$ and $Q$ are certain functions. Then on page 208, we take $p=2$ in equation (1) to get $$ 1 - \frac{1^23^2}{2!(8x)^2} \le P(x,0) \le 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^4}, $$ and we take $p=0$ in equation (2) to get $$ -\frac1{8x} \le Q(x,0) \le 0. $$ Therefore by Cauchy's inequality, \begin{align*} \frac{\pi x}2J_0(x)^2 &\le \big( \cos(x-\tfrac\pi4)^2 + \sin(x-\tfrac\pi4)^2 \big) \big( P(x,0)^2 + Q(x,0)^2 \big) \\ &\le 1 \cdot \bigg( \bigg( 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^4} \bigg)^2 + \frac1{(8x)^2} \bigg) \\ &\le \bigg( \bigg( 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^216^2} \bigg)^2 + \frac1{(8x)^2} \bigg) \\ &= 1-\frac{4517}{65536 x^2}+\frac{30702681}{17179869184 x^4}, \end{align*} where the final inequality used $x\ge2$. It's now easy to verify that this last bound is less than $1$ for $x\ge2$.

As for the interval $[0,2]$: for $x$ in this interval, the series $$ J_0(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!^2} \bigg(\frac x2 \bigg)^2 $$ satisfies the hypotheses of the alternating series test, and thus the series's truncations alternately form upper and lower bounds for $J_0(x)$ on this interval. The first two truncations give $1 \ge J_0(x) \ge 1-x^2/4 \ge 0$. The next truncation gives $$ J_0(x) \le 1-\frac{x^2}4 + \frac{x^4}{64} < \sqrt{\frac2{\pi x}}, $$ where we need to verify the last inequality to finish the proof. Taking square roots of both sides and subtracting, it suffices to show that $$ \sqrt[4]{\frac2{\pi x}} - \bigg( 1-\frac{x^2}8 \bigg) > 0 $$ on $[0,2]$. Standard calculus shows that the minimum of this function occurs at $x=(2/\pi)^{1/9}$ and has value $9/(2^{25/9} \pi ^{2/9})-1$; verifying that this minimum value is positive boils down to confirming that $\pi < 3^9/2^{25/2}$, which in turn (since $\pi<22/7$) follows from the inequality $3^{18}7^2 > 2^{27} 11^2$.

$\endgroup$
  • $\begingroup$ Thank you for the reference (which can be found here ). However I cannot but agree with your "quite unfriendly". I see on page 207 he writes "The remainder of the asymptotic expansion of $J_{0}(x)$ are numerically less than the first term neglected, [...] but are also of the same sign". How does this imply my bound? A naive interpretation of "remainder" results in a factor of two off. $\endgroup$ – lcv Sep 23 '15 at 0:22
  • $\begingroup$ I've seen your bound quoted as being from this section. One day I'll have to work through it myself to trust that it's in there.... $\endgroup$ – Greg Martin Sep 23 '15 at 3:33
  • $\begingroup$ Thanks for the update @Greg Martin .. $\endgroup$ – lcv Sep 23 '15 at 6:58
  • $\begingroup$ Thanks. It's almost a complete answer now. Quite astonishing how much work is needed (coming up with the $P$ and $Q$ in the first place) to prove an obvious bound. $\endgroup$ – lcv Sep 23 '15 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.