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

Posted 10 years ago
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 from
FourierSinSeries[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.
3 Replies
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 BY: Udo Krause
OMG, the first time for years to hold the Abramowitz/Stegun book in my hands again
In[193]:=
Sin[z Sin[x]] == 2 Sum[BesselJ[2 k + 1, z] Sin[(2 k + 1) x], {k, 0, Infinity}]
Out[193]= True
well, so go ahead with this, the even coefficients are as expected equal to zero
In[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, too
In[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 BY: Udo Krause
I should have typed:
Sin[z Sin] = 2 Sum[BesselJ[2 k + 1, z] Sin[(2 k + 1) x], {k, Infinity}]
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