Difference between revisions of "Relationship between Airy Ai and modified Bessel K"
From specialfunctionswiki
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| − | + | ==Theorem== | |
The following formula holds: | The following formula holds: | ||
$$\mathrm{Ai}(z)=\dfrac{1}{\pi} \sqrt{\dfrac{z}{3}} \mathrm{K}_{\frac{1}{3}} \left( \dfrac{2}{3} x^{\frac{3}{2}} \right),$$ | $$\mathrm{Ai}(z)=\dfrac{1}{\pi} \sqrt{\dfrac{z}{3}} \mathrm{K}_{\frac{1}{3}} \left( \dfrac{2}{3} x^{\frac{3}{2}} \right),$$ | ||
where $\mathrm{Ai}$ is the [[Airy Ai]] function and $K_{\nu}$ denotes the [[Modified Bessel K sub nu|modified Bessel $K$]]. | where $\mathrm{Ai}$ is the [[Airy Ai]] function and $K_{\nu}$ denotes the [[Modified Bessel K sub nu|modified Bessel $K$]]. | ||
| − | === | + | |
| + | ==Proof== | ||
| + | |||
| + | ==References== | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 23:07, 9 June 2016
Theorem
The following formula holds: $$\mathrm{Ai}(z)=\dfrac{1}{\pi} \sqrt{\dfrac{z}{3}} \mathrm{K}_{\frac{1}{3}} \left( \dfrac{2}{3} x^{\frac{3}{2}} \right),$$ where $\mathrm{Ai}$ is the Airy Ai function and $K_{\nu}$ denotes the modified Bessel $K$.
