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Hypergeometric representations of Bessel functions

Related topics: Bessel functions, Confluent hypergeometric functions

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Jν ⁣(z)=(z2)ν0F1 ⁣(ν+1,z24)J_{\nu}\!\left(z\right) = {\left(\frac{z}{2}\right)}^{\nu} \,{}_0{\textbf F}_1\!\left(\nu + 1, -\frac{{z}^{2}}{4}\right)
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Iν ⁣(z)=(z2)ν0F1 ⁣(ν+1,z24)I_{\nu}\!\left(z\right) = {\left(\frac{z}{2}\right)}^{\nu} \,{}_0{\textbf F}_1\!\left(\nu + 1, \frac{{z}^{2}}{4}\right)
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Yν ⁣(z)=1sin ⁣(πν)(cos ⁣(πν)(z2)ν0F1 ⁣(ν+1,z24)(z2)ν0F1 ⁣(1ν,z24))Y_{\nu}\!\left(z\right) = \frac{1}{\sin\!\left(\pi \nu\right)} \left(\cos\!\left(\pi \nu\right) {\left(\frac{z}{2}\right)}^{\nu} \,{}_0{\textbf F}_1\!\left(\nu + 1, -\frac{{z}^{2}}{4}\right) - {\left(\frac{z}{2}\right)}^{-\nu} \,{}_0{\textbf F}_1\!\left(1 - \nu, -\frac{{z}^{2}}{4}\right)\right)
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Kν ⁣(z)=(2zπ)1/2ezU ⁣(ν+12,2ν+1,2z)K_{\nu}\!\left(z\right) = {\left(\frac{2 z}{\pi}\right)}^{-1 / 2} {e}^{-z} U^{*}\!\left(\nu + \frac{1}{2}, 2 \nu + 1, 2 z\right)
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Kν ⁣(z)=12πsin ⁣(πν)((z2)ν0F1 ⁣(1ν,z24)(z2)ν0F1 ⁣(1+ν,z24))K_{\nu}\!\left(z\right) = \frac{1}{2} \frac{\pi}{\sin\!\left(\pi \nu\right)} \left({\left(\frac{z}{2}\right)}^{-\nu} \,{}_0{\textbf F}_1\!\left(1 - \nu, \frac{{z}^{2}}{4}\right) - {\left(\frac{z}{2}\right)}^{\nu} \,{}_0{\textbf F}_1\!\left(1 + \nu, \frac{{z}^{2}}{4}\right)\right)
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Jν ⁣(z)=(z2)νeizΓ ⁣(ν+1)1F1 ⁣(ν+12,2ν+1,2iz)J_{\nu}\!\left(z\right) = {\left(\frac{z}{2}\right)}^{\nu} \frac{{e}^{-i z}}{\Gamma\!\left(\nu + 1\right)} \,{}_1F_1\!\left(\nu + \frac{1}{2}, 2 \nu + 1, 2 i z\right)
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Jν ⁣(z)=zν(2π)1/2((iz)1/2νeizU ⁣(ν+12,2ν+1,2iz)+(iz)1/2νeizU ⁣(ν+12,2ν+1,2iz))J_{\nu}\!\left(z\right) = \frac{{z}^{\nu}}{{\left(2 \pi\right)}^{1 / 2}} \left({\left(i z\right)}^{-1 / 2 - \nu} {e}^{i z} U^{*}\!\left(\nu + \frac{1}{2}, 2 \nu + 1, -2 i z\right) + {\left(-i z\right)}^{-1 / 2 - \nu} {e}^{-i z} U^{*}\!\left(\nu + \frac{1}{2}, 2 \nu + 1, 2 i z\right)\right)
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Jν ⁣(z)=(2πz)1/2(eiθU ⁣(ν+12,2ν+1,2iz)+eiθU ⁣(ν+12,2ν+1,2iz))   where θ=π(2ν+1)4zJ_{\nu}\!\left(z\right) = {\left(2 \pi z\right)}^{-1 / 2} \left({e}^{-i \theta} U^{*}\!\left(\nu + \frac{1}{2}, 2 \nu + 1, -2 i z\right) + {e}^{i \theta} U^{*}\!\left(\nu + \frac{1}{2}, 2 \nu + 1, 2 i z\right)\right)\; \text{ where } \theta = \frac{\pi \left(2 \nu + 1\right)}{4} - z

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2019-08-21 11:44:15.926409 UTC