Difference between revisions of "Relationship between Airy Bi and modified Bessel I"
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| − | + | ===Theorem=== | |
| − | + | The following formula holds: | |
$$\mathrm{Bi}(z)=\sqrt{\dfrac{z}{3}} \left( I_{\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}} \right) + I_{-\frac{1}{3}} \left( \frac{2}{3} x^{\frac{3}{2}} \right) \right),$$ | $$\mathrm{Bi}(z)=\sqrt{\dfrac{z}{3}} \left( I_{\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}} \right) + I_{-\frac{1}{3}} \left( \frac{2}{3} x^{\frac{3}{2}} \right) \right),$$ | ||
where $\mathrm{Bi}$ denotes the [[Airy Bi]] function and $I_{\nu}$ denotes the [[Modified Bessel I sub nu|modified Bessel $I$]]. | where $\mathrm{Bi}$ denotes the [[Airy Bi]] function and $I_{\nu}$ denotes the [[Modified Bessel I sub nu|modified Bessel $I$]]. | ||
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| − | + | ===Proof=== | |
| − | + | ||
| − | + | [[Category:Theorem]] | |
Revision as of 08:05, 5 June 2016
Theorem
The following formula holds: $$\mathrm{Bi}(z)=\sqrt{\dfrac{z}{3}} \left( I_{\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}} \right) + I_{-\frac{1}{3}} \left( \frac{2}{3} x^{\frac{3}{2}} \right) \right),$$ where $\mathrm{Bi}$ denotes the Airy Bi function and $I_{\nu}$ denotes the modified Bessel $I$.
