Difference between revisions of "Bessel at n+1/2 in terms of Bessel polynomial"

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==Theorem==
<strong>[[Bessel at n+1/2 in terms of Bessel polynomial|Theorem]]:</strong> The following formula holds:
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The following formula holds:
 
$$J_{n +\frac{1}{2}}(r) = (2\pi r)^{-\frac{1}{2}} \left[\dfrac{e^{ir}}{i^{n+1}} y_n \left( -\dfrac{1}{ir} \right) + i^{n+1}e^{-ir}y_n\left( \dfrac{1}{ir} \right) \right],$$
 
$$J_{n +\frac{1}{2}}(r) = (2\pi r)^{-\frac{1}{2}} \left[\dfrac{e^{ir}}{i^{n+1}} y_n \left( -\dfrac{1}{ir} \right) + i^{n+1}e^{-ir}y_n\left( \dfrac{1}{ir} \right) \right],$$
 
where $J_{n+\frac{1}{2}}$ denotes a [[Bessel function]] and $y_n$ denotes a [[Bessel polynomial]].
 
where $J_{n+\frac{1}{2}}$ denotes a [[Bessel function]] and $y_n$ denotes a [[Bessel polynomial]].
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<strong>Proof:</strong> █
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==Proof==
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==References==

Revision as of 19:56, 9 June 2016

Theorem

The following formula holds: $$J_{n +\frac{1}{2}}(r) = (2\pi r)^{-\frac{1}{2}} \left[\dfrac{e^{ir}}{i^{n+1}} y_n \left( -\dfrac{1}{ir} \right) + i^{n+1}e^{-ir}y_n\left( \dfrac{1}{ir} \right) \right],$$ where $J_{n+\frac{1}{2}}$ denotes a Bessel function and $y_n$ denotes a Bessel polynomial.

Proof

References