Bessel function difference Hello I'm looking for estimates for the difference
$I_{\nu+1}(x)-I_{\nu}(x)$
where $I_{\nu}$ is the modified Bessel functions of the first kind of order, $\nu\in \mathbb{N}$ is a positive integer and $x>0$ a positive real number.
Thank you very much!
 A: Let $D_\nu(x):=I_{\nu+1}(x)-I_\nu(x)$
$\underline{\text{Positive }\nu}$
For small $x$ or large $\nu$ we have the asymptotic form
$$
I_\nu(x)\sim\frac{1}{\Gamma(\nu+1)}\left(\frac{x}{2}\right)^\nu \qquad ;\qquad x\ll\sqrt{\nu+1}
$$
Simplifying, we find
$$
D_\nu(x)\sim\frac{1}{\Gamma(\nu+2)}\left(\frac{x}{2}\right)^\nu\left[\frac{x}{2}-(\nu+1) \right] \qquad ;\qquad x\ll\sqrt{\nu+1}
$$
For which the leading order is
$$
D_\nu(x)\sim-\frac{1}{\Gamma(\nu+1)}\left(\frac{x}{2}\right)^\nu \qquad ;\qquad x\ll\sqrt{\nu+1} \tag{1}
$$
Which is also valid for negative non-integer $\nu$. The case $\nu=0$ requires a little care
$$
D_0(x) \sim \frac{x}{2}-1  \qquad ;\qquad x \to 0
$$
Here is a plot of the ratio exact/leading order for a few $\nu$ values:

For large $x$ and fixed $\nu$ we have the asymptotic form
$$
I_\nu(x)\sim\frac{e^x}{\sqrt{2\pi x}} \sum\limits_{k=0}^\infty \left(\frac{-1}{x}\right)^k a_k(\nu) \qquad ;\qquad x \to \infty 
$$
Where the coefficients $a_k(\nu)$ are defined here. At leading order, we find
$$
D_\nu(x) \sim -\frac{e^x}{\sqrt{8\pi x^3}}(1+2\nu) \qquad ;\qquad x \to \infty \tag{2}
$$
Which is also valid for negative $\nu \neq -1/2$. Here is a plot of the ratio exact/leading order for a few $\nu$ values:

$\underline{\text{Exact result at }\nu=-1/2}$
Looking at the recurrence relations for the modified Bessel functions $K$, we find the exact result
$$
D_{-1/2}(x)=-\frac{2}{\pi}K_{1/2}(x) \tag{3}
$$
$\underline{\text{Negative }\nu}$
For $\nu<0$ we have the connection formula
$$
I_{-\nu}(x)=I_\nu(x)+\frac{2}{\pi} \sin(\nu \pi) K_\nu(x) \tag{4}
$$
From which we find the 'reflection formula' for any integer $\nu$
$$
D_{-\nu}(x)=-D_{\nu-1}(x) \tag{5}
$$
Eq (5) allows us to extend (1) to negative integers
$$
D_{-\nu}(x) \sim \frac{1}{\Gamma(\nu)}\left(\frac{x}{2} \right)^{\nu-1} \qquad ; \qquad x \to 0 \quad , \quad \nu =2,3,4 \dots\tag{6}
$$
As an aside, we could try to get the asymptotic behavior directly from the differential equation for $D_\nu$, with standard local analysis. From Bessel's equation for $I_\nu$ and $I_{\nu+1}$ we find the inhomogeneous equation
$$
x^2D_\nu''+xD_\nu'-(x^2+\nu^2)D_\nu=(2\nu+1)I_{\nu+1}
$$
Where $I_{\nu+1}$ is known. Not really necessary here as we already know the asymptotic forms of $D_\nu$.
