
Re: Gamma Function  Bessel Function Identity
Posted:
Sep 7, 2012 10:16 AM


In article <9c1bfbe9826241b9944970820d92c3e8@CITESHT4.ad.uillinois.edu>, ksoileau <kmsoileau@gmail.com> wrote:
> I have observed and proved the following identity for all x\ne 0 : > $$ > (I_{k1}(x) +I_{k+1}(x) )I_{k}(x) > (I_{k1}(x)+I_{k+1}(x))I_k(x) > = \frac{4 k}{x \Gamma (1k) \Gamma (1+k)} > $$ > > Is this wellknown or trivially derived? Any comments will be appreciated. > Thanks, > Kerry M. Soileau >
Easily derived in Maple. So presumably it follows from the known identities for Bessel and Gamma function. Here it is...
First, we need the enhanced limits known to MultiSeries: >restart;with(MultiSeries):
Here is the claim to be proved: >claim:=(BesselI(k1,x)+BesselI(k+1,x))*BesselI(k,x)  (BesselI(k1,x)+BesselI(k+1,x))*BesselI(k,x) =(4*k)/(x*GAMMA(1k)*GAMMA(1+k));
claim := (BesselI(k1, x)+BesselI(k+1, x))*BesselI(k, x)(BesselI(k1, x)+BesselI(k+1, x))*BesselI(k, x) = 4*k/(x*GAMMA(k+1)*GAMMA(k+1))
Multiply by x so that the righthandside does not involve x: >claim*x;
x*((BesselI(k1, x)+BesselI(k+1, x))*BesselI(k, x)(BesselI(k1, x)+BesselI(k+1, x))*BesselI(k, x)) = 4*k/(GAMMA(k+1)*GAMMA(k+1))
Here is the lefthandside: >L:=lhs(%);
L := x*((BesselI(k1, x)+BesselI(k+1, x))*BesselI(k, x)(BesselI(k1, x)+BesselI(k+1, x))*BesselI(k, x))
Here is the righthandside: >R:=rhs(%%);
R := 4*k/(GAMMA(k+1)*GAMMA(k+1))
Since the righthandside does not involve x, we try to show that the lefthandside is also independent of x. Its derivative is zero: >simplify(diff(L,x));
0
What is the constant value of the lefthandside? It is undefined at x=0, but the limit there is: >simplify(limit(L,x=0)) assuming k::real;
4*sin(Pi*k)/Pi
On the other hand, the righthandside simplifies to the same thing: >expand(simplify(R));
4*sin(Pi*k)/Pi

