Basically, how to show that $J_0(x) \leq 1 \ \forall x$.

I came across this in a tutorial assignment, and it was trivially proved using a result from a previous exercise, viz.

$$ J_0^2 + 2. J_1^2 + 2. J_3^2 + \dots = 1$$

giving, $$J_0 = \sqrt{1 - 2.(J_1^2 + J_2^2 + \dots )} \leq 1 \ , \ \because J_n^2 \gt 0$$

However, this solution appeared distinctly unsatisfactory for two reasons -

  1. This method is yet another example of what seems like a typical 'list all known identities and here goes nothing' procedure which seems typical of problems involving identities of Bessel functions. We had to do our tutorial problems out of order (or alternatively use results we had yet to prove) to complete them and were advised to memorize several of these identities.While there is nothing wrong with this per se, it seems like an incredibly ad hoc, hit-and-miss method of approaching problems. This seems especially true for an examination environment.

  2. We actually have some very 'nice' series representations of Bessel functions of the first kind. And it seems surprising that we cannot use that to bound the value of $J_0$, especially since the series often allows us to find interesting identities such as some involving derivatives of Bessel functions.

I tried making use of the series representation,

$$J_0(x) = \sum_{m=0}^{\infty} (\frac{x}{2})^{2m} \frac{(-1)^m}{(m!)^2}$$

It is clear that $J_0(0) = 0$, though the appearance of $0^0$ in the first term is disquieting. The series for $J_0(1)$ appears to converge absolutely and it is easily shown that $J_0(1), J_0(2) \lt 1$.

I tried to extend the above to obtain a general result of some kind, but drew a blank. The alternating series theorem shows convergence and is no help in obtaining the sum of a series. I thought about using the fact that factorials grow faster than exponentials, but was again unable to obtain anything definite. After trying a few other things, I am unsure of how to proceed further.

So, my questions are as follows -

  1. How to show that $J_0 \lt 1$ without using the above referenced identity. Preferably using the series.
  2. While evaluating $J_0(0)$ from the series, how do we overcome the apparent appearance of the indeterminate form $0^0$ in the first term?
  3. Any advice for approaching such problems.

1 Answer 1


$J_0(0)=1$, not $0$, for the same reason that $e^0 = \sum_{n=0}^\infty \frac{0^n}{n!} = 1$ not $0$, because the procedure of plugging in $x$ then the dummy variable is incorrect. But what you do have now is the following result; take $J_0(x)-1$ and integrate it:

$$\int_0^x J_0(z)-1dz = x\sum_{m=1}^\infty \frac{\left(-\frac{x^2}{4}\right)^m}{(2m+1)(m!)^2}$$

Consider $x>0$. For small enough $x$, we can guarantee that the $\frac{x^3}{12}$ term dominates the $\frac{x^5}{320}$ term, making the function negative. Suppose we have $x>4\sqrt{\frac{5}{3}}$, in which case the $x^5$ term is larger. But then

$$\frac{x^3}{12} + \frac{x^7}{16128} > \frac{x^5}{320}$$

will always hold in that region. One can prove via induction that every time a new positive term comes to dominate, it will always be less than the sum of all of the negative terms (up to the term just past it), thus the function is always negative for $x>0$.

One can make a similar argument via induction that the function is strictly positive for $x<0$. Thus the area accumulated towards $0$ is positive and away from $0$ negative, making $x=0$ a global maximum for $J_0(x)-1$, then we can use what we already know about $J_0(0)$ to conclude that

$$J_0(x) \leq 1$$

for all $x \in \mathbb{R}$.

  • 1
    $\begingroup$ This makes no sense because of the sign changes, and indeed the intermediate inequality you get is not true. Bessel J functions decay much slower than that. $\endgroup$
    – Ian
    Nov 5, 2019 at 10:54
  • $\begingroup$ @Ian you're absolutely right. I've updated the post with a completely different proof. $\endgroup$ Nov 5, 2019 at 14:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .