Relationship between Bessel I sub 1/2 and sinh
From specialfunctionswiki
Theorem: The following formula holds: $$I_{\frac{1}{2}}(z)=\sqrt{\dfrac{2}{\pi z}}\sinh(z).$$
Proof: █
Theorem: The following formula holds: $$I_{\frac{1}{2}}(z)=\sqrt{\dfrac{2}{\pi z}}\sinh(z).$$
Proof: █
