Use the product of the generating functions g(x, t) and g(x,−t) to show that 1 = [J0(x)]2+ 2∑n=1[Jn(x)]^2. (Hint: a sneaky use of the residue theorem will work wonders here.

For part (a), note that you also have

g(x,−t)=g(x,1/t)=∑n=0∞Jn(x)(1t)n.

You can then multiply this with the original series to get

[∑n=−∞∞Jn(x)t−n][∑n=−∞∞Jn(x)tn]=1.

By multiplying these two series together and matching coefficients, you should be able to prove the given identity.

Parts (b) and (c) follow from the identity in part (a). Note that if you have a set of quantities ai such that ai>0 for all i, then ai<∑jaj for all i.

Step-by-step explanation:

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