Bounds and asymptotic expansions for the radii of convexity and uniform convexity of normalized Bessel functions

Lemma 1. For any positive integer k and positive real $\nu > k$ , the Rayleigh sum in (1.7) has the convergent Laurent expansion (2.1) $$\eta_k(\nu) = \frac{1}{\nu^k} \sum_{n=0}^{\infty} \frac{\eta_n^{(k)}}{\nu^n},$$ where for any fixed non negative integer n, the coefficients $\eta_n^{(k)}$ can be evaluated by the recurrence relation $$\eta_n^{(k)} = -ka_n^{(k)} - \sum_{m=0}^n \sum_{i=1}^{k-1} a_m^{(i)} \eta_{n-m}^{(k-i)}$$ and $a_n^{(k)}$ is given by $$a_n^{(k)} = \frac{\left(-1\right)^k \left(2k+1\right)}{2^{2k}k!} \sum_{k_{k-1}=0}^n \dots \sum_{k_2=0}^{k_3} \sum_{k_1=0}^{k_2} \left(-1\right)^{k_n} \left(-2\right)^{k_2-k_1} \dots \left(-k\right)^{n-k_{k-1}}, \qquad n \in \mathbb{N}_0.$$

Lemma 2. For any positive integer k and positive real $\nu > k$ , the Rayleigh sum in (1.8) has the convergent Laurent expansion (2.2) $$\theta_k(\nu) = \frac{1}{\nu^k} \sum_{n=0}^{\infty} \frac{\theta_n^{(k)}}{\nu^n},$$ where for any fixed non negative integer n, the coefficients $\theta_n^{(k)}$ can be evaluated by the recurrence relation $$\theta_n^{(k)} = -kb_n^{(k)} - \sum_{m=0}^{n} \sum_{i=1}^{k-1} b_m^{(i)} \theta_{n-m}^{(k-i)}$$ and $b_n^{(k)}$ is given by $$b_n^{(k)} = \frac{(-1)^k (k+1)}{2^{2k} k!} \sum_{k_{k-1}=0}^n \dots \sum_{k_2=0}^{k_3} \sum_{k_1=0}^{k_2} (-1)^{k_n} (-2)^{k_2-k_1} \dots (-k)^{n-k_{k-1}}, \qquad n \in \mathbb{N}_0.$$ Furthermore, the next lemma provides the asymptotic form for the square of the radius of convexity of the normalized Bessel functions $g_{\nu}(z)$ . The proof of this lemma uses some results of [BPS14].

Lemma 3. For $\nu > -1$ the radius of convexity $r^c(g_{\nu})$ of the function $$z \mapsto g_{\nu}(z) = 2^{\nu} \Gamma(\nu + 1) z^{1-\nu} J_{\nu}(z)$$ has the asymptotic behavior $$(r^{c}(g_{\nu}))^{2} = \nu \left(c + \mathcal{O}\left(\frac{1}{\nu}\right)\right),$$ as $\nu \to \infty$ , where c is some positive constant. The next lemma provides the asymptotic form for the radius of convexity of the normalized Bessel functions $h_{\nu}(z)$ . In the proof we use analogous results of [BPS14, Theorem 1] for Dini function $e_{\nu}(z)$ .

Lemma 4. For $\nu > -1$ the radius of convexity $r^c(h_{\nu})$ of the function $$z \mapsto h_{\nu}(z) = 2^{\nu} \Gamma(\nu + 1) z^{1 - \frac{\nu}{2}} J_{\nu}(\sqrt{z})$$ has the asymptotic behavior $$r^{c}\left(h_{\nu}\right) = \nu\left(d + \mathcal{O}\left(\frac{1}{\nu}\right)\right),$$ as $\nu \to \infty$ , where d is some positive constant. Before we state the main theorems of this paper, let us define the ordinary potential polynomials. Let $f(z) = 1 + \sum_{n=1}^{\infty} a_n z^n$ be a formal power series. Corresponding to f(z), for any complex number $\alpha$ , the ordinary potential polynomial $A_{\alpha,n}(a_1, a_2, ..., a_n)$ is defined by the generating function $$(f(z))^{\alpha} = \left(1 + \sum_{n=0}^{\infty} a_n z^n\right)^{\alpha} = \sum_{n=0}^{\infty} A_{\alpha,n}(a_1, \dots, a_n) z^n.$$ Thus, specifically $A_{\alpha,0} = 1$ , $A_{\alpha,1} = \alpha a_1$ and $A_{\alpha,2} = \alpha a_2 + {\alpha \choose 2} a_1^2$ . One can refer to [Ne13] for additional details about the ordinary potential polynomials. The next two theorems provide asymptotic expansions for the radius of convexity for two types of normalized Bessel functions of the first kind. The idea of the proofs of the next theorems is inspired from [BN21].

Theorem 1. Let $\eta_n^{(k)}$ denote the coefficients of the expansion in (2.1). Then, the square of the radius of convexity $r^c(g_{\nu})$ has the asymptotic expansion $$(r^{c}(g_{\nu}))^{2} \sim \nu \left(c + \sum_{n=1}^{\infty} \frac{\epsilon_{n}}{\nu^{n}}\right)$$ as $\nu \to \infty$ , where the coefficients $\epsilon_n$ can be determined by the recurrence relation $$(2.4) \qquad \frac{3}{4} \left( (-1)^{n+1}c + \sum_{m=0}^{n} (-1)^{n-m} \epsilon_{m+1} \right) + \sum_{k=0}^{n+1} \left( \sum_{m=1}^{\infty} A_{m+1,k} \left( \epsilon_1, \dots, \epsilon_k \right) \eta_{n-k+1}^{(m+1)} \right) = 0$$ and c satisfies (2.5) $$c = \frac{2}{3} - \frac{4}{3} \sum_{m=1}^{\infty} A_{m+1,0} \eta_0^{(m+1)}.$$ In particular, for n = 0 in equation (2.4) we arrive at (2.6) $$\epsilon_1 = \frac{\frac{3c}{4} - \sum_{m=1}^{\infty} c^{m+1} \eta_1^{(m+1)}}{\frac{3}{4} + \sum_{m=1}^{\infty} (m+1) c^m \eta_0^{(m+1)}}.$$ By using Mathematica, the coefficients, accurate up to $10^{-5}$ , in the above asymptotic expansion are (2.7) $$(r^c(g_\nu))^2 \sim \nu \left(0.535898 + \frac{0.335953}{\nu} + \dots\right).$$  <span id="page-4-0"></span>FIGURE 1. The image of the open disk $\mathbb{D}_r$ under the Bessel function $z \mapsto g_{\nu}(z)$ , where $r \sim 5.208...$ is the approximative value of the radius of convexity of $g_{\nu}(z)$ considering the first two terms of (2.7) for $\nu = 50$ . For large $\nu = 50$ , by using the first two terms in (2.7), we calculated the approximative value of the radius of convexity of $g_{\nu}(z)$ and plotted in Figure 1. <span id="page-4-1"></span>Remark 1. We would like to mention that we can write (2.4) explicitly to determine $\epsilon_n$ for n > 1. From the proof of Theorem 1 the expression $A_{k,n}(\epsilon_1,\ldots,\epsilon_n)$ is a potential polynomial given by the generating function $$\left(c + \sum_{n=1}^{\infty} \frac{\epsilon_n}{\nu^n}\right)^k = \sum_{n=0}^{\infty} \frac{A_{k,n}\left(\epsilon_1, \dots, \epsilon_n\right)}{\nu^n}.$$ The above equation can be written as (2.8) $$c^k \left( 1 + \sum_{n=1}^{\infty} \frac{\epsilon_n}{c} \frac{1}{\nu^n} \right)^k = \sum_{n=0}^{\infty} A_{k,n}(\epsilon_1, \dots, \epsilon_n) \frac{1}{\nu^n}.$$ Moreover, in view of [Ne13, Appendix] we have (2.9) $$c^{k} \left( 1 + \sum_{n=1}^{\infty} \frac{\epsilon_{n}}{c} \frac{1}{\nu^{n}} \right)^{k} = c^{k} \sum_{n=0}^{\infty} \tilde{A}_{k,n} \left( \frac{\epsilon_{1}}{c}, \dots, \frac{\epsilon_{n}}{c} \right) \frac{1}{\nu^{n}},$$ where $$\tilde{A}_{k,n}\left(\frac{\epsilon_1}{c},\dots,\frac{\epsilon_n}{c}\right) = \sum \binom{k}{l} \frac{l!}{l_1! l_2! \dots l_n!} \epsilon_1^{l_1} \epsilon_2^{l_2} \dots \epsilon_n^{l_n} \frac{1}{c^l}$$ and the sum extends over all sequences $l_1, l_2, \ldots, l_n$ of non-negative integers such that $l_1 + 2l_2 + \ldots + nl_n = n$ and $l_1 + l_2 + \ldots + l_n = l$ . It is worth to note that for any value of l, $l_n = 0$ or 1. Particularly, for $n \in \mathbb{N}$ , $l_n = 1$ if and only if l = 1. In view of this observation, the above equation can be written as $$(2.10)$$ where each $l_i$ satisfies the above condition along with l > 1. From equations (2.8), (2.9) and (2.10) we obtain that $$(2.11) A_{k,n}(\epsilon_1,\ldots,\epsilon_n) = c^k \left( k \frac{\epsilon_n}{c} + \sum {k \choose l} \frac{l!}{l_1! \ldots l_n!} \epsilon_1^{l_1} \epsilon_2^{l_2} \ldots \epsilon_{n-1}^{l_{n-1}} \frac{1}{c^l} \right).$$ Moreover, we can rewrite equation (2.4) after separating the terms associated with $\epsilon_{n+1}$ as $$\frac{3}{4} \left( (-1)^{n+1} c + \sum_{m=0}^{n-1} (-1)^{n-m} \epsilon_{m+1} \right) + \frac{3}{4} \epsilon_{n+1} + \sum_{k=0}^{n} \left( \sum_{m=1}^{\infty} A_{m+1,k} \left( \epsilon_1, \dots, \epsilon_k \right) \eta_{n-k+1}^{(m+1)} \right) + \sum_{m=1}^{\infty} A_{m+1,n+1} \left( \epsilon_1, \dots, \epsilon_{n+1} \right) \eta_0^{(m+1)} = 0$$ Now, putting the value of $A_{m+1,n+1}(\epsilon_1,\ldots,\epsilon_{n+1})$ by using (2.11) in the above equation and solving for $\epsilon_{n+1}$ , we obtain that $$\epsilon_{n+1} \left( \frac{3}{4} + \sum_{m=1}^{\infty} \eta_0^{(m+1)}(m+1)c^m \right)$$ $$= -\sum_{m=1}^{\infty} c^{m+1} \eta_0^{(m+1)} \sum_{n=1}^{\infty} {m+1 \choose n} \frac{l!}{l_1! \dots l_{n+1}!} \epsilon_1^{l_1} \epsilon_2^{l_2} \dots \epsilon_n^{l_n} \frac{1}{c^l}$$ $$- \frac{3}{4} \left( (-1)^{n+1} c + \sum_{m=0}^{n-1} (-1)^{n-m} \epsilon_{m+1} \right) - \sum_{k=0}^{n} \left( \sum_{m=1}^{\infty} A_{m+1,k} \left( \epsilon_1, \dots, \epsilon_k \right) \eta_{n-k+1}^{(m+1)} \right),$$ where l > 1, $l_1 + 2l_2 + \ldots + nl_n = n$ and $l_1 + l_2 + \ldots + l_n = l$ . Notice that the condition l > 1 results in $l_{n+1} = 0$ . <span id="page-5-0"></span>Remark 2. We also note that from the proof of Theorem 1 we have that $A_{m+1,0} = c^{m+1}$ . Hence (2.5) can be written as (2.12) $$c = \frac{2}{3} - \frac{4}{3} \sum_{m=1}^{\infty} c^{m+1} \eta_0^{(m+1)}.$$ By using the fact that the denominator of the terms in the series $\eta_0^{(k)}$ increases rapidly, the infinite series (2.12) can be reduced into a polynomial of a finite yet large degree in order to numerically approximate the value of c, which satisfies the series (2.12). For example the $20^{th}$ term of the series (2.7) has value less than $10^{-7}$ . We find out the value of c by solving the polynomial of degree 20 with the variable c, which is given in (2.7). Remark 3. By using the Euler-Rayleigh inequalities, Aktaş et al. [ABO18, Theorem 6] derived the bounds for the radius of convexity of normalized Bessel function $g_{\nu}(z)$ as follows $$(2.13) 4\sqrt{\frac{(\nu+1)^2(\nu+2)}{56\nu+137}} < (r^c(g_\nu))^2 < \frac{2(56\nu+137)(\nu+1)(\nu+3)}{208\nu^2+1172\nu+1693}.$$ The right-hand side of (2.13) can be expressed as $$\frac{2 \left(56 \nu+137\right) \left(\nu+1\right) \left(\nu+3\right)}{208 \nu^2+1172 \nu+1693}=\frac{7 \nu \left(1+\frac{137}{56 \nu}\right) \left(1+\frac{1}{\nu}\right) \left(1+\frac{3}{\nu}\right)}{13 \left(1+\frac{1172}{208 \nu}+\frac{1693}{208 \nu^2}\right)}=\nu \left(\frac{7}{13}+\frac{591}{1352}\frac{1}{\nu}+\mathcal{O}\left(\frac{1}{\nu^2}\right)\right).$$ Similarly, we can express the left-hand side of the inequality (2.13) as $$4\sqrt{\frac{(\nu+1)^2(\nu+2)}{56\nu+137}} = \nu\left(\sqrt{\frac{2}{7}} + \frac{87}{56\sqrt{14}}\frac{1}{\nu} + \mathcal{O}\left(\frac{1}{\nu^2}\right)\right).$$ In view of (2.3) and the above discussion, we can say that c lies in the interval $(\sqrt{2/7}, 7/13)$ . This interval could be further narrowed by taking higher order Euler-Rayleigh inequalities (see the proof of [ABO18, Theorem 6]). The value of c, which serves as the root of the polynomial mentioned in Remark 2 and lies within the precise interval, can be regarded as an approximate value for the constant c that fits into the asymptotic expansion of the square of $r^c(g_{\nu})$ . It is worth also to note that the accuracy of the value of c can be improved by narrowing the interval and using a higher degree polynomial. Remark 4. It is worth mentioning that besides using the recurrence relation (2.4) to calculate $\epsilon_1, \epsilon_2, \ldots$ , for large $\nu$ , we can find their bounds by using the Euler-Rayleigh inequalities. In view of relation (2.3) and inequality (2.13), which bounds the radius of convexity $r^c(g_{\nu})$ , for large $\nu$ we obtain a bound for $\epsilon_1$ as $$|\epsilon_1| < r$$ where (2.14) $$r = \frac{591}{1352} = \max\left(\left|\frac{591}{1352}\right|, \left|\frac{87}{56\sqrt{14}}\right|\right).$$ In a similar way we can find bounds for $\epsilon_2, \epsilon_3, \ldots$ , for large $\nu$ . For example it is clear from (2.14) that r = 0.4371302 and $$|\epsilon_1| < 0.4371302.$$ This bound is satisfied by the value we calculated by using the recurrence relation (2.4) which is given 0.335953 (see (2.7)). From the proof of the Lemma 3, we noted that the bounds of the radius of convexity converge, so we can find tighter bounds for $\epsilon_1$ by using other bounds for the radius of convexity (see also the proof of [ABO18, Theorem 6]). Note that we used modulus in (2.14) to emphasize that the coefficients could be negative for other Euler-Rayleigh inequalities. Before stating the next theorem, let us note that the approach used in both of the previous remarks is also applicable for the following theorem to approximate the constant d and finding bounds for the coefficients $\epsilon_1, \epsilon_2, \ldots$ in the asymptotic expansion of $r^c(h_\nu)$ .

Theorem 2. Let $\theta_n^{(k)}$ denote the coefficients of the expansion in (2.2). Then, the radius of convexity $r^c(h_\nu)$ , of the function $h_\nu(z)$ defined in (1.4), has the asymptotic expansion $$r^{c}(h_{\nu}) \sim \nu \left(d + \sum_{n=1}^{\infty} \frac{\epsilon_{n}}{\nu^{n}}\right)$$ as $\nu \to \infty$ , where the coefficients $\epsilon_n$ can be determined by the recurrence relation $$(2.15) \qquad \frac{3}{4} \left( (-1)^{n+1} d + \sum_{m=0}^{n} (-1)^{n-m} \epsilon_{m+1} \right) + \sum_{k=0}^{n+1} \left( \sum_{m=1}^{\infty} A_{m+1,k} \left( \epsilon_1, \dots, \epsilon_k \right) \theta_{n-k+1}^{(m+1)} \right) = 0$$ and d satisfies $$d = \frac{4}{3} \left( 1 - \sum_{m=1}^{\infty} A_{m+1,0} \theta_0^{(m+1)} \right).$$ In particular, for n = 0 in equation (2.15) we obtain that $$\epsilon_1 = \frac{\frac{3d}{4} - \sum_{m=1}^{\infty} d^{m+1} \theta_1^{(m+1)}}{\frac{3}{4} + \sum_{m=1}^{\infty} (m+1) d^m \theta_0^{(m+1)}}.$$ By using Mathematica, the coefficients, accurate up to $10^{-5}$ , in the above asymptotic expansion are (2.16) $$r^{c}(h_{\nu}) \sim \nu \left(1.17157 + \frac{0.858757}{\nu} + \ldots\right).$$ For large $\nu = 50$ , by using the first two terms in (2.16), we calculated the approximative value of the radius of convexity of $h_{\nu}(z)$ and we have shown this in Figure 2. 2.2. Bounds and asymptotic expansions for the radii of uniform convexity of normalized Bessel functions. Now, we focus on similar results as before related to uniform convexity. The next lemmas provide bounds for the radii of uniform convexity of normalized Bessel functions of the first kind. We use the so-called Euler-Rayleigh inequality technique [IM95] and properties of the Laguerre-Pólya class of entire functions to deduce these bounds.

Lemma 5. Let $\nu > -1$ . Then the radius of uniform convexity $r^{uc}(g_{\nu})$ of the function $$z \mapsto q_{\nu}(z) = 2^{\nu} \Gamma(\nu + 1) z^{1-\nu} J_{\nu}(z)$$ is the smallest positive root of the equation $$q'_{n}(z) + 2zq''_{n}(z) = 0$$  <span id="page-7-0"></span>FIGURE 2. The image of the open disk $\mathbb{D}_r$ under the Bessel function $z \mapsto h_{\nu}(z)$ , where $r \sim 59.437...$ is the approximative value of the radius of convexity of $h_{\nu}(z)$ considering the first two terms of (2.16) for $\nu = 50$ . and satisfies the following inequality $$\omega_k^{-\frac{1}{k}} < (r^{uc}(g_\nu))^2 < \frac{\omega_k}{\omega_{k+1}},$$ where $\omega_k$ is given by (3.31). In particular for k=1 we obtain that (2.17) $$2\sqrt{\frac{(\nu+1)}{15}} < r^{uc}(g_{\nu}) < 2\sqrt{\frac{\nu(\nu+1)}{3(4\nu-1)}}.$$

Lemma 6. Let $\nu > -1$ . Then the radius of uniform convexity $r^{uc}(h_{\nu})$ of the function $$z \mapsto h_{\nu}(z) = 2^{\nu} \Gamma(\nu + 1) z^{1 - \frac{\nu}{2}} J_{\nu}(\sqrt{z})$$ is the smallest positive root of the equation $$h'_{\nu}(z) + 2zh''_{\nu}(z) = 0$$ and satisfies the following inequality $$\sigma_k^{-\frac{1}{k}} < r^{uc}(h_\nu) < \frac{\sigma_k}{\sigma_{k+1}},$$ where $\sigma_k$ is given by (3.38). Furthermore, the next lemma provides the asymptotic form for the square of the radius of uniform convexity of the normalized Bessel functions $g_{\nu}(z)$ .

Lemma 7. The radius of uniform convexity $r^{uc}(g_{\nu})$ of the function $$z \to g_{\nu}(z) = 2^{\nu} \Gamma(\nu + 1) z^{1-\nu} J_{\nu}(z)$$ admits the asymptotic behavior $$(r^{uc}(g_{\nu}))^2 = \nu \left(c + \mathcal{O}\left(\frac{1}{\nu}\right)\right),$$ for $\nu \to \infty$ , where c is some positive constant. Similarly to the previous lemma, the next lemma provides the asymptotic form for the radius of uniform convexity of the normalized Bessel functions $h_{\nu}(z)$ .

Lemma 8. The radius of uniform convexity $r^{uc}(h_{\nu})$ of the function $$z \to h_{\nu}(z) = 2^{\nu} \Gamma(\nu + 1) z^{1 - \frac{\nu}{2}} J_{\nu}(\sqrt{z})$$ admits the asymptotic behavior $$r^{uc}\left(h_{\nu}\right) = \nu\left(d + \mathcal{O}\left(\frac{1}{\nu}\right)\right),$$ for $\nu \to \infty$ , where d is some positive constant. The next two theorems are the main results of this subsection. The idea of the proofs of the next theorems is also inspired from [BN21].

Theorem 3. Let $\eta_n^{(k)}$ denote the coefficients of the expansion in (2.1). Then, the square of the radius of uniform convexity $r^{uc}(g_{\nu})$ has the asymptotic expansion $$(r^{uc}(g_{\nu}))^{2} \sim \nu \left(\tilde{c} + \sum_{n=1}^{\infty} \frac{\varepsilon_{n}}{\nu^{n}}\right)$$ as $\nu \to \infty$ , where the coefficients $\varepsilon_n$ can be determined by the recurrence relation (2.19) $$\frac{3}{4} \left( (-1)^{n+1} \tilde{c} + \sum_{m=0}^{n} (-1)^{n-m} \varepsilon_{m+1} \right) + \sum_{k=0}^{n+1} \left( \sum_{m=1}^{\infty} A_{m+1,k} \left( \varepsilon_1, \dots, \varepsilon_k \right) \eta_{n-k+1}^{(m+1)} \right) = 0$$ and $\tilde{c}$ satisfies (2.20) $$\tilde{c} = \frac{1}{3} - \frac{4}{3} \sum_{m=1}^{\infty} A_{m+1,0} \eta_0^{(m+1)}.$$ In particular, for n = 0 in equation (2.19) we arrive at $$\varepsilon_1 = \frac{\frac{3\tilde{c}}{4} - \sum_{m=1}^{\infty} \tilde{c}^{m+1} \eta_1^{(m+1)}}{\frac{3}{4} + \sum_{m=1}^{\infty} (m+1) \tilde{c}^m \eta_0^{(m+1)}}.$$ By using Mathematica, the coefficients, accurate up to $10^{-5}$ , in the above asymptotic expansion are $$(2.21) (r^{uc}(g_{\nu}))^{2} \sim \nu \left(0.298438 + \frac{0.218612}{\nu} + \ldots\right).$$ For large $\nu = 50$ , by using the first two terms in (2.21), we calculated the approximative value of the radius of uniform convexity of $g_{\nu}(z)$ and plotted in Figure 3. Remark 5. By using a similar argument as in Remark 1, we can write (2.19) explicitly to determine $\varepsilon_n$ for n > 1 as $$\varepsilon_{n+1} \left( \frac{3}{4} + \sum_{m=1}^{\infty} \eta_0^{(m+1)}(m+1)\tilde{c}^m \right) \\ = -\sum_{m=1}^{\infty} \tilde{c}^{m+1} \eta_0^{(m+1)} \sum_{n=1}^{\infty} {m+1 \choose n} \frac{l!}{l_1! \dots l_{n+1}!} \varepsilon_1^{l_1} \varepsilon_2^{l_2} \dots \varepsilon_n^{l_n} \frac{1}{\tilde{c}^l} \\ - \frac{3}{4} \left( (-1)^{n+1} \tilde{c} + \sum_{m=0}^{n-1} (-1)^{n-m} \varepsilon_{m+1} \right) - \sum_{k=0}^{n} \left( \sum_{m=1}^{\infty} A_{m+1,k} \left( \varepsilon_1, \dots, \varepsilon_k \right) \eta_{n-k+1}^{(m+1)} \right)$$ Here, l > 1, $l_1 + 2l_2 + \ldots + nl_n = n$ and $l_1 + l_2 + \ldots + l_n = l$ . Notice that the condition l > 1 results in $l_{n+1} = 0$ . <span id="page-8-1"></span>Remark 6. Note that from the proof of Theorem 3, $A_{m+1,0} = \tilde{c}^{m+1}$ . Hence (2.20) can be written as (2.22) $$\tilde{c} = \frac{1}{3} - \frac{4}{3} \sum_{m=1}^{\infty} \tilde{c}^{m+1} \eta_0^{(m+1)}.$$ By using Lemma 1, we observed that $\eta_0^{m+1}$ decreases rapidly for large m, the infinite series (2.22) can be reduced into a polynomial of a finite yet large degree in order to numerically approximate the value of $\tilde{c}$ , which satisfies the series (2.22). For example, the $20^{th}$ term of the series (2.22) has value less than $10^{-7}$ . We find out the value of $\tilde{c}$ by solving the polynomial of degree 20 with the variable $\tilde{c}$ , which is given in (2.21).  <span id="page-9-0"></span>FIGURE 3. The image of the open disk $\mathbb{D}_r$ under the Bessel function $z \mapsto g_{\nu}(z)$ , where $r \sim 3.891...$ is the approximative value of the radius of uniform convexity of $q_{\nu}(z)$ considering the first two terms of (2.21) for $\nu = 50$ . Remark 7. From Lemma 5 we have that $$(2.23) \qquad \frac{4(\nu+1)}{15} < (r^{uc}(g_{\nu}))^2 < \frac{4\nu(\nu+1)}{3(4\nu-1)}.$$ The right-hand side of (2.23) can be expressed as $$\frac{4\nu(\nu+1)}{3(4\nu-1)} = \frac{\nu}{3}\left(1 + \frac{5}{4\nu} + \mathcal{O}\left(\frac{1}{\nu^2}\right)\right).$$ Similarly, we can express the left-hand side of the inequality (2.23) as $$\frac{4(\nu+1)}{15} = \frac{4\nu}{15} \left( 1 + \frac{1}{\nu} \right)$$ In view of (2.18) and the above discussion, we can say that $\tilde{c}$ lies in the interval $(\frac{4}{15}, \frac{1}{3})$ . This interval could be further narrowed by taking higher order Euler-Rayleigh inequalities (see the proof of Lemma 5). The value of $\tilde{c}$ , which serves as the root of the polynomial mentioned in Remark 6 and lies within the precise interval, can be regarded as an approximate value for the constant $\tilde{c}$ that fits into the asymptotic expansion of the square of $r^{uc}(g_{\nu})$ . It is noteworthy that the accuracy of the value of $\tilde{c}$ can be improved by narrowing the interval and using a higher degree polynomial. Remark 8. It is worth mentioning that besides using recurrence relation (2.19) to calculate $\varepsilon_1, \varepsilon_2, \ldots$ for large $\nu$ , we can also find their bounds by using the Euler-Rayleigh inequalities. In view of relation (2.18) and inequality (2.23), which bounds the radius of uniform convexity $r^{uc}(q_{\nu})$ , for large $\nu$ we obtain a bound for $\varepsilon_1$ as $$|\varepsilon_1| < r$$ , where $$(2.24) r = \frac{5}{12} = \max\left(\left|\frac{4}{15}\right|, \left|\frac{5}{12}\right|\right).$$ In a similar way we can find bounds for $\varepsilon_2, \varepsilon_3, \ldots$ , for large $\nu$ . For example it is clear from (2.24) that r = 0.41666 and $$|\varepsilon_1| < 0.41666.$$ This bound is satisfied by the value we calculated by using the recurrence relation (2.19) which is given 0.218612 (see (2.21)). From the proof of the Lemma 7, we noted that the bounds of the radius of uniform convexity converge, so we can find tighter bounds for $\varepsilon_1$ by using other bounds for the radius of uniform convexity (see also the proof of Lemma 5). Note that we used modulus in (2.24) to emphasize that the coefficients could be negative for other Euler-Rayleigh inequalities. Before stating the next theorem, let us note that the approach used in both of the previous remarks is also applicable for the following theorem to approximate the constant $\tilde{d}$ and finding bounds for the coefficients $\varepsilon_1, \varepsilon_2, \ldots$ in the asymptotic expansion of $r^{uc}(h_{\nu})$ .

Theorem 4. Let $\theta_n^{(k)}$ denote the coefficients of the expansion in (2.2). Then, the radius of uniform convexity $r^{uc}(h_{\nu})$ has the asymptotic expansion $$r^{uc}(h_{\nu}) \sim \nu \left( \tilde{d} + \sum_{n=1}^{\infty} \frac{\varepsilon_n}{\nu^n} \right)$$ as $\nu \to \infty$ , where the coefficients $\varepsilon_n$ can be determined by the recurrence relation (2.25) $$\frac{3}{4} \left( (-1)^{n+1} \tilde{d} + \sum_{m=0}^{n} (-1)^{n-m} \varepsilon_{m+1} \right) + \sum_{k=0}^{n+1} \left( \sum_{m=1}^{\infty} A_{m+1,k} \left( \varepsilon_1, \dots, \varepsilon_k \right) \theta_{n-k+1}^{(m+1)} \right) = 0$$ and $\tilde{d}$ satisfies $$\tilde{d} = \frac{2}{3} - \frac{4}{3} \sum_{m=1}^{\infty} A_{m+1,0} \theta_0^{(m+1)}.$$ In particular, for n = 0 in equation (2.25) we obtain that $$\varepsilon_1 = \frac{\frac{3\tilde{d}}{4} - \sum_{m=1}^{\infty} \tilde{d}^{m+1} \theta_1^{(m+1)}}{\frac{3}{4} + \sum_{m=1}^{\infty} (m+1) \tilde{d}^m \theta_0^{(m+1)}}.$$ By using Mathematica, the coefficients, accurate up to $10^{-5}$ , in the above asymptotic expansion are (2.26) $$r^{c}(h_{\nu}) \sim \nu \left(0.627719 + \frac{0.478612}{\nu} + \ldots\right).$$ <span id="page-10-0"></span>For large $\nu = 50$ , by using the first two terms in (2.26), we calculated the approximative value of the radius of uniform convexity of $h_{\nu}(z)$ and plotted in Figure 4.