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Using Mathematica, it is relatively straightforward to check that the zeroes of $$ f_n(z)\equiv \sqrt{z} J_{n+\frac{1}{2}}(\sqrt{z})+J_{n+\frac{3}{2}}(\sqrt{z}) $$ with $n=0,1,2,\ldots$ and $J_n$ a Bessel function, lie on the positive real axis, that is to say, if $f_n(z)=0$, then $z\in \mathbb{R}^+$. Is there a simple way to prove this? What about more general linear combinations?

Update 1

I have noted that the above linear combination can be written in a nicer form that might be easier to eventually prove the statement asked. Indeed

$$ f_n(z)\equiv F_n(\sqrt{z})\quad\text{with}\quad F_n(x)=\left(n+\frac{5}{2}\right)J_{n+\frac{3}{2}}(x)+x\,J^\prime_{n+\frac{3}{2}}(x)\,. $$

Update 2

This paper https://www.tandfonline.com/doi/epdf/10.1080/00036818608839633?needAccess=true appears to show exactly what I wanted in Corollary 2.3. A similar proof can be found in page 482 of Watson's treatise.

user12588
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  • I am not quite sure what you mean here. I tried using a Poisson-type argument for the derivative of the above quantity (such as the one presented in the solution to this question https://math.stackexchange.com/questions/4669585/complex-zeros-of-modified-bessel-functions-of-first-kind-with-order-zero) – user12588 Dec 23 '23 at 23:10
  • My comment didn’t address your actual question. My apologies, deleted now. – A rural reader Dec 23 '23 at 23:32
  • A discussion of the same question for the classical case of $J_0(z)$ can be found at https://mathoverflow.net/a/431579/46744. – Paul Enta Dec 24 '23 at 11:56
  • Why these functions in particular? Did you get them from some differential equation perhaps? Maybe starting there would help! – Radost Dec 25 '23 at 20:54

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