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.
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.