# Why does the evaluation of this series fail?

The following series expression holds:

BesselJ[1,a x]BesselJ[1,b x]==a b (x^2)/4*Sum[(-1)^n*(a x/2)^(2n)*Hypergeometric2F1[-n,-1-n,2,b^2/a^2]/(n!Pochhammer[2,n]),{n,0,Infinity}]


Where BesselJ[1,x] is the Bessel function of the first kind of order 1. Hypergeometric2F1[a1,a2,b1,x] is Gauss hypergeometric function. Pochhammer[2,n] is the Pochhammer Symbol. This series expression can be found in e.g. Bateman Manuscript Project, Higher Transcendental Functions, Vol.2, and it holds without any constraints.

For this series expression, if the relatively small parameters are chosen, e.g. a=1, b=2.5, x=3, and just replace the sum index Infinite with 100, the product of Bessel functions gives 0.045857190997445 and the series gives 0.045857190997469. They coincide well. However, for a relatively large x, the product of Bessel functions must be decreasing, e.g. a=1, b=2.5, x=100, the product of Bessel functions gives 0.003338005199477126, but the series gives 6.410492282570008*10^131 which is apparently divergent. The reason of this divergence is unknown for me. It can't be better to increase the terms of the series expression or improve the precision of the Gauss hypergeometric function.

Could anyone help? Thanks!

• abx^2/4 is strange here and I think it doesn't affect your convergence.After putting values its a constant which does nothing to convergence or divergence of series. If I take it as a b (x^2)/4..for all value of x I get 0.045857190997445 Aug 25, 2013 at 9:00
• @Blackbird: Thanks! The coefficient a*b*(x^2)/4 is surely irrelevant. The decisive factor of the convergence is inside the sum[...]. " If I take it as a b (x^2)/4..for all value of x I get 0.045857190997445", this is incorrect please check whether the factor (ax/2)^(2n) is missed. Aug 26, 2013 at 4:20
• Hypergeometric2F1[-n,-1-n,2,b^2/a^2] has independent a,b but actually they are not independent in your supplied parameters, parameter is ax,bx it should be a x,b x.Formation of series is faulty, I suppose. Aug 26, 2013 at 6:24
• so did u get what I got ? I have tried the same already. Aug 26, 2013 at 13:06
• BesselJ[1,a x]BesselJ[1,b x]==a b (x^2)/4*Sum[(-1)^n*(a x/2)^(2n)*Hypergeometric2F1[-n,-1-n,2,b^2/a^2]/(n!Pochhammer[2,n]),{n,0,Infinity}]. Yes, this is the correct expression which is just evaluated by me. I thought the omitting space between the parameters would not lead to the misunderstanding of the expression. Now you can observe the above-mentioned phenomenon. The expression has no problem. The problem is the numerical evaluation of it. Aug 26, 2013 at 13:20

It's not obvious to me what exactly you did. When I do what you describe I do not get divergent values.

func = BesselJ[1, a x] BesselJ[1, b x];
ss = Series[func, {x, Infinity, 2}, Assumptions -> {a > 0, b > 0}]

(* Out= Cos[\[Pi]/4 + a x] Cos[\[Pi]/4 + b x] (
SeriesData[x,
DirectedInfinity, {2 a^Rational[-1, 2] b^Rational[-1, 2]/Pi}, 2, 6,
2]) + Cos[\[Pi]/4 + b x] (
SeriesData[x,
DirectedInfinity, {
Rational[-3, 4] a^Rational[-3, 2] b^Rational[-1, 2]/Pi}, 4, 6,
2]) Sin[\[Pi]/4 + a x] + Cos[\[Pi]/4 + a x] (
SeriesData[x,
DirectedInfinity, {
Rational[-3, 4] a^Rational[-1, 2] b^Rational[-3, 2]/Pi}, 4, 6,
2]) Sin[\[Pi]/4 + b x] + (
SeriesData[x,
DirectedInfinity, {
Rational[9, 32] a^Rational[-3, 2] b^Rational[-3, 2]/Pi}, 6, 8,
2]) Sin[\[Pi]/4 + a x] Sin[\[Pi]/4 + b x] *)


Numerical check:

{func, Normal[ss]} /. {a -> 1, b -> 2.5, x -> 100}

(* Out= {0.00333800519948, 0.00333795972052} *)

• Thank you Herr Lichtblau, but you haven't understand my series correctly. There are surely many different types of series expansion for BesselJ[1, a x] BesselJ[1, b x], but I'm just talking about the series written in the form of Gauss hypergeometric function. In fact I need a expression written in the form of power series of x. Aug 28, 2013 at 5:19
• Ah. There are two issues. One is that You simply did not take enough terms. Try 1000. The other is that you use machine arithmetic. I'm guessing this is suffering badly from cancellation error. In any case, use 5/2 instead of 2.5 for your b value. Aug 28, 2013 at 14:17