Find zero of sum of 4 modified Bessel functions I am trying to find the (positive) root of the function
$f(x) = I_{-3/4}(x) + I_{3/4}(x) - I_{-1/4}(x) - I_{1/4}(x)$
where $I_\alpha(x)$ denotes the modified Bessel function of the first kind.
Mathematica tells me the numerical value of the root is $x=0.146287..$, but it is unable to find an analytical expression (and so am I).
As far as I can see, the usual identities, such as this one, don't help here. Still, it seems to be a rather special case, so maybe there is an analytical expression?
Since I am not that well versed in algebra, I would greatly appreciate help, including hints on how to make progress.
EDIT: Corrected typo pointed out by gammatester.
 A: This probably isn't the answer you're hoping for, but there isn't going to be an analytic expression for the root.  Bessel functions themselves (as in alone, not combined) have non-analytic roots that must be solved numerically, so combining these transcendental functions and looking for a root is surely only going to be solvable with further numerical methods.
If you're lucky, your expression might simplify to a result using fewer Bessel functions, but like you said, there don't appear to be any particularly applicable identities.  That being said, here are a couple representations of the BesselI functions that might show us a few things about your function:
$$\begin{align}
I_\alpha (x)&=  \Big(\frac{x}{2}\Big)^\alpha \ \sum^{\infty}_{k=0} \frac{(\frac{x}{2})^{2k}}{k! \ \Gamma(\alpha + k + 1)}
\end{align}$$
This would imply that your function looks like
$$\begin{align}
I_{3/4} (x) + 
I_{-3/4} (x) - 
I_{1/4} (x) - 
I_{-1/4} (x) &=  \Big(\frac{x}{2}\Big)^{3/4} \ \sum^{\infty}_{k=0} \frac{(\frac{x}{2})^{2k}}{k! \ \Gamma(3/4 + k + 1)} +  \Big(\frac{x}{2}\Big)^{-3/4} \ \sum^{\infty}_{k=0} \frac{(\frac{x}{2})^{2k}}{k! \ \Gamma(-3/4 + k + 1)} \\
& \ \ \ \ \ \ \ \  -  \Big(\frac{x}{2}\Big)^{1/4} \ \sum^{\infty}_{k=0} \frac{(\frac{x}{2})^{2k}}{k! \ \Gamma({1/4} + k + 1)} -  \Big(\frac{x}{2}\Big)^{-1/4} \ \sum^{\infty}_{k=0} \frac{(\frac{x}{2})^{2k}}{k! \ \Gamma({-1/4} + k + 1)} \\
f(x)&=  \sum^{\infty}_{k=0} \frac{(\frac{x}{2})^{2k}}{k!} \bigg[\frac{(\frac{x}{2})^{3/4}}{\Gamma (k+7/4)} + \frac{(\frac{x}{2})^{-3/4}}{\Gamma (k+1/4)} - \frac{(\frac{x}{2})^{1/4}}{\Gamma (k+5/4)} - \frac{(\frac{x}{2})^{-1/4}}{\Gamma (k+3/4)}\bigg]
\end{align}$$
If you can simplify what is in the brackets, then you might be able to simplify your expression to a different bessel-like function or two, but the solution will still involve numerically solving for the root.
An alternate representation is given by
$$\begin{align}
I_\alpha (x)&=  \frac{1}{\pi}\int^{\pi}_0 e^{x \cos\theta} \cos(\alpha \theta) \ d\theta - \frac{\sin(\alpha \pi)}{\pi} \int^{\infty}_{0} e^{-x \cosh(t) - \alpha t} \ dt
\end{align}$$
which you can plug your $\alpha=\{\pm 3/4, \pm 1/4\}$ values into to find
$$\begin{align}
f(x)&= \frac{1}{\pi}\int^{\pi}_0 e^{x \cos\theta} \bigg[ \cos(\frac{3}{4}\theta)+\cos(\frac{-3}{4}\theta)-\cos(\frac{1}{4}\theta)-\cos(\frac{-1}{4}\theta) \bigg]\ d\theta \\
& \ \ \ \ - \frac{\sin(\alpha \pi)}{\pi} \int^{\infty}_{0} e^{-x \cosh(t)} \bigg[e^{(3/4)t} + e^{(-3/4)t} -e^{(1/4)t} -e^{(-1/4)t} \bigg] \ dt \\
&= \frac{2}{\pi}\int^{\pi}_0 e^{x \cos\theta} \bigg[ \cos(\frac{3}{4}\theta)-\cos(\frac{1}{4}\theta)\bigg]\ d\theta \\
& \ \ \ \ - \int^{\infty}_{0} e^{-x \cosh(t)} \bigg[\frac{2}{\pi}\sin(\frac{3}{4}\pi) \sinh(\frac{3}{4}t) - \frac{2}{\pi}\sin(\frac{1}{4}\pi)\sinh(\frac{1}{4}t) \bigg] \ dt
\end{align}$$
but again, these integrals don't really simplify the expression into any sort of non-transcendental equation, so numeric methods are still going to be needed to compute the roots.
