Bounding a modified Bessel function of the first kind.

Question

Let $I_0$ be the zeroth-order modified Bessel function of the first kind. We know that, asymptotically as $x\to \infty$, $I_0(x) \sim e^x/\sqrt{2\pi x}$. Does anybody have a reference for the maximum of $e^{-x} \sqrt{x} I_0(x)$ for $0\leq x\leq \infty$? (A plot shows that it's $0.4688...$, but of course that is no proof.)

Answer 1

By just considering the derivative of $\sqrt{x}\,I_0(x)\,e^{-x}$ and setting it equal to zero we have that the maximum is attained in the only point for which: $$f(x)=(1-2x)\,I_0(x)+2x\, I_1(x) = 0.\tag{1}$$ Such function is convex and decreasing over $(0,x_0=2.3555\ldots)$, concave and decreasing over $(x_0,+\infty)$. Since $f(0)=1$ and $f(x_0)<0$, we can just find the only zero $\xi$ of $f$ by applying the Newton's method with starting point $x=0$. After two iterations we get: $$\xi\geq \frac{1}{2}+\frac{I_1(1/2)}{I_0(1/2)},$$ after five iterations we get: $$\xi\approx 0.78997842\ldots $$ from which it follows that: $$\sqrt{x}\,I_0(x)\,e^{-x}\leq 0.4688223555.$$ From now on, this will be the Helfgott's constant.

Addendum. Since $I_n(z)=\frac{z}{2n}(I_{n-1}(z)-I_{n+1}(z))$ by the Briggs' formulas, we have the continued fraction representation: $$\frac{I_1}{I_0}(z)=\frac{1}{\frac{2}{z}+\frac{1}{\frac{4}{z}+\frac{1}{\frac{6}{z}+\ldots}}}\tag{2}$$ hence, due to $(1)$, our stationary point is just the solution of the continued fraction equation:

$$\frac{1}{\frac{2}{z}+\frac{1}{\frac{4}{z}+\frac{1}{\frac{6}{z}+\ldots}}}=1-\frac{1}{2z}.\tag{3}$$

Answer 2

Hint

$$\frac{d}{dx}\Big(e^{-x} \sqrt{x} I_0(x)\Big)=\frac{e^{-x} ((1-2 x) I_0(x)+2 x I_1(x))}{2 \sqrt{x}}$$ Then, appeared Jack D'Aurizio's answer to which nothing needs to be added !

Just to put some approximation beside Taylor and Newton, the function $$f(x)=(1-2 x) I_0(x)+2 x I_1(x)$$ is very well represented by a Pade approximant built at $x=0$. For example $$f(x) \simeq \frac{\frac{23 x^2}{144}-\frac{25 x}{18}+1}{\frac{19 x^2}{144}+\frac{11 x}{18}+1}$$ gives a solution $$x=\frac{4}{23} \left(25-\sqrt{418}\right) \simeq 0.792166$$ $$f(x) \simeq \frac{-\frac{103 x^2}{368}-\frac{24 x}{23}+1}{\frac{13 x^3}{184}+\frac{141 x^2}{368}+\frac{22 x}{23}+1}$$ gives a solution $$x=\frac{4}{103} \left(\sqrt{4673}-48\right) \simeq 0.790654$$ $$f(x) \simeq\frac{\frac{2675 x^2}{3708}-\frac{1702 x}{927}+1}{-\frac{1207 x^4}{59328}-\frac{197 x^3}{1854}-\frac{62 x^2}{309}+\frac{152 x}{927}+1}$$ gives a solution $$x=\frac{2 \left(1702-\sqrt{417079}\right)}{2675} \simeq 0.789670$$

Added later

In the above, the numerator of the Pade approximant was selected to be $2$ in order to only solve quadratic equations. What is interesting is that degree $1$ is selected instead, the approximate solutions are :$\frac{8}{11}$, $\frac{11}{14}$, $\frac{224}{283}$, $\frac{283}{358}$, $\frac{12888}{16313}$,$\frac{16313}{20650}$, $\frac{1321600}{1672963}$, $\frac{1672963}{2117734}$, $\frac{211773400}{268074929}$, $\frac{268074929}{339344618}$

Answer 3

Here's a crude but rigorous bound, while we wait for the question to be settled definitively.

Let $x\geq 0$. By definition (and half-angle formulas), $$I_0(x) = \frac{1}{\pi} \int_0^\pi e^{x \cos \theta} d\theta = \frac{2 e^x}{\pi} \int_0^{\pi/2} e^{-2 x \sin^2 \alpha} d \alpha,$$ where we do the change of variables $\alpha = \theta/2$. Now, for $\alpha\in \lbrack 0, \pi/2\rbrack$, $\sin \alpha \geq 2 \alpha/\pi$. Hence, $$\int_0^{\pi/2} e^{-2 x \sin^2 \alpha} d \alpha \leq \int_0^{\pi/2} e^{- \frac{8 x}{\pi^2} \alpha^2} d\alpha < \frac{\pi}{\sqrt{8 x}} \int_0^\infty e^{-t^2} dt = \frac{\pi^{3/2}}{2^{5/2} \sqrt{x}}.$$ Hence, $$I_0(x) < \sqrt{\frac{\pi}{8}} \frac{e^x}{\sqrt{x}}$$ for all $x\geq 0$. Of course, $\sqrt{\pi/8} = 0.626657\dotsc$, so this is indeed crude in comparison with $0.468822\dotsc$.

What we really need is a bound better than $0.46882 e^x/\sqrt{x}$ for $x\geq 2$ (say), so that the problem gets reduced to that of maximizing the Bessel function on a compact interval (such as \lbrack 0,2\rbrack) where it can be estimated well by a truncated power series.

Answer 4

Here's a full answer, involving some (rigorous) computation towards the end.

Let $f(t) = e^{-t} I_0(t) \sqrt{t}$, as above. Let us start by showing that $f(t) < 0.45168$ for $t>8$. (This part of the lemma's proof was graciously provided by G. Kuperberg.)

First of all, $$f(t) = \frac{e^{-t}}{\pi} \sqrt{t} \int_0^\pi e^{t \cos \theta} d\theta = \frac{2}{\pi} \sqrt{t} \int_0^1 \frac{e^{-2 t s^2}}{\sqrt{1-s^2}} ds,$$ where we substitute $\cos \theta = 1 - 2 s^2$, and note that $d\theta = 2 ds/\sqrt{1-s^2}$. We break the integral on $s$ in two at $s=1/2$. The integrand from $0$ to $1/2$ is majorized by $(1.03+s^2/2) e^{-2 t s^2}$ (because $1/\sqrt{1-s^2} - s^2/2$ has positive derivative for $s\geq 0$, and $1/\sqrt{1-(1/2)^2} - (1/2)^2/2 = 1.0297\dotsc$); the integrand from $1/2$ to $1$ is majorized by $e^{-t/2}/\sqrt{1-s^2}$. The first integral can be extended from $0$ to $\infty$; the second one is exactly $e^{-t/2}\cdot \pi/3$. We conclude that $$\begin{aligned} f(t) &\leq \frac{2 \sqrt{t}}{\pi} \left(1.03\cdot \frac{\sqrt{\pi/2}}{ 2 \sqrt{t}} + \frac{1}{2} \frac{\sqrt{\pi/2}}{8 t^{3/2}}\right) + \frac{2 \sqrt{t} e^{-t/2}}{3}\\ &\leq \frac{1.03}{\sqrt{2 \pi}} + \frac{1}{8 \sqrt{2\pi} t} + \frac{2 \sqrt{t} e^{-t/2}}{3} \leq 0.45168\end{aligned}$$ for $t\geq 8$.

In the range $0\leq t\leq 8$, we will use the Taylor expansion around $t=0$. By (9.6.12) in Abramowitz-Stegun, $$\begin{aligned} I_0(t) = \sum_{n=0}^\infty \frac{(t^2/4)^2}{n!^2} \leq \sum_{n=0}^{15} \frac{(t^2/4)^n}{n!^2} + \frac{16^{16}}{16!^2} \cdot \sum_{m=0}^\infty \frac{1}{17^m} \leq \sum_{n=0}^{15} \frac{(t^2/4)^n}{n!^2} + 0.00000005.\end{aligned}$$

We now use the bisection method (with $25$ iterations), implemented via interval arithmetic (as described in, e.g., \S 5.2 of Tucker, Validated Numerics), to ascertain that the maximum of the series $\sum_{n=0}^{15} (t^2/4)^n/n!^2$ on the interval $\lbrack 0,8\rbrack$ lies between $0.46882234$ and $0.46882237$. End of proof!