# How can Sin[z Sin[x]] be expanded ?

Posted 9 years ago
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 According to Abramowitz and Stegun eq. 9.1.43 Sin[z Sin] can be expanded as an infinite series as follows:Sin[z Sin[x]] = 2 Sum[BesselJ[k + 1, z] Sin[(2 k + 1) x], {k, Infinity}]How can I get the same result from Mathematica?A finite expansion can be obtained fromFourierSinSeries[Sin[z Sin[x]], x, 8]however, due to the reduction of the Bessel functions to small k in an awkward form.Also when applied to Frequency Modulation, we have e.g. Sin[x0 + z Sin[x1]] and then the FourierSinSeries does not seem appropiate.
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Posted 9 years ago
 Usually one has http://functions.wolfram.com/Bessel-TypeFunctions/BesselJ/11/Sum[BesselJ[k, z] Exp[I k t], {k, -Infinity, Infinity}] == Exp[I z Sin[t]]For real values of z and t one separates real and imaginary part and uses BesselJ[-k,z]=(-1)^k BesselJ[k,z] to see the relation you started with.
Posted 9 years ago
 OMG, the first time for years to hold the Abramowitz/Stegun book in my hands againIn[193]:= Sin[z Sin[x]] == 2 Sum[BesselJ[2 k + 1, z] Sin[(2 k + 1) x], {k, 0, Infinity}]Out[193]= Truewell, so go ahead with this, the even coefficients are as expected equal to zeroIn[223]:= Table[FourierSinCoefficient[Sin[z Sin[x]], x, k], {k, 0, 18, 2}]Out[223]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}and the odd coefficients fit into the relation as expected, tooIn[225]:= Table[FullSimplify[FourierSinCoefficient[Sin[z Sin[x]], x, k] - 2 BesselJ[k, z]], {k, 1, 19, 2}]Out[225]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}from a computational point of view you have it if you are willing to presume, that the relation valid for {1,3,5,...,17,19} is also valid for {21, 23, .... ,2k +1, ... }.
Posted 9 years ago
 I should have typed:Sin[z Sin] = 2 Sum[BesselJ[2 k + 1, z] Sin[(2 k + 1) x], {k, Infinity}]