How Do You Prove the Bessel Identity J-3/2(x)?

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The discussion centers on proving the Bessel identity for J-3/2(x), specifically J-3/2(x) = √(2/πx)(-cos(x)/x - sin(x)). The user, Alan, struggles with the gamma function's behavior, noting that Γ(n - 3/2 + 1) yields negative values for n=0 and positive for n=1, 2, 3. A solution is suggested to demonstrate that the right-hand side satisfies Bessel's differential equation and meets the necessary initial conditions, thereby confirming its identity as the required Bessel function.

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yungman
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I have been working on this for a few days and cannot prove this:

J-3/2 (x)=\sqrt{\frac{2}{\pi x}}[\frac{-cos(x)}{x} - sin(x) ]

Main reason is \Gamma(n-3/2+1) give a negative value for n=0 and possitive value for n=1,2,3... I cannot find a series representation of this gamma function.


Please advice me how to solve this problem. This is not a school homework.

thanks a million

Alan
 
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Maybe instead of the series representation for the Bessel function: show that your right-hand side satisfies Bessel's differential equation, and has the proper initial values, so that it therefore equals the Bessel function required.
 
yungman said:
Main reason is \Gamma(n-3/2+1) give a negative value for n=0 and possitive value for n=1,2,3... I cannot find a series representation of this gamma function.


Please advice me how to solve this problem. This is not a school homework.

thanks a million

Alan


For n > 0 we apply this formula
\Gamma (n+1) = n \Gamma (n)

but if n < 0 we apply this formula
\Gamma (n)=\frac{\Gamma (n+1)}{n}
 
matematikawan said:
For n > 0 we apply this formula
\Gamma (n+1) = n \Gamma (n)

but if n < 0 we apply this formula
\Gamma (n)=\frac{\Gamma (n+1)}{n}

I know this formula! This is embarassing! How can I over looked this and spent 3 days on this...Even joined two more math forums! I even plug in the numbers and hope this is not that simple! I use

\Gamma (-3/2)=\frac{\Gamma (-1/2)}{-3/2} all the time! Just never try with n in it!

Thanks a million...Even though you make me look really really bad!

Cheers.
Alan
 

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