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Addition Formulas for Bessel Functions

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Addition Formulas for Bessel Functions

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When trying to understand a complex formula, it helps to first ask what is being related before asking how they are related.

This post will look at addition theorems for Bessel functions. They related the values of Bessel functions at two points to the values at a third point.

Let ab, and c be lengths of the sides of a triangle and let θ be the angle between the sides of length a and b. The triangle could be degenerate, i.e. θ could be π. The addition theorems here give the value of Bessel functions at c in terms of values of Bessel functions evaluated at b andc.

The first addition theorem says that

J0(c) = ∑ Jm(aJm(b) exp(imθ)

where the sum is over all integer values of m.

This says that the value of the Bessel function J0 at c is determined by a sort of inner product of values of Bessel functions of all integer orders at a and b. It’s not exactly an inner product because it is not positive definite. (Sometimes you’ll see the formula above written with a factor λ multiplying ab, and c. This is because you can scale every side of a triangle by the same amount and not change the angles.)

Define the vectors xy, and z to be the values of all Bessel functions evaluated at ab, and crespectively. That is, for integer kxk = Jk(a) and similar for y and z. Also, define the vector w bywk = exp(ikθ). Then the first addition theorem says

z0 = ∑ xm ym wm.

This is a little unsatisfying because it relates the value of one particular Bessel function at c to the values of all Bessel functions at a and b. We’d like to relate all Bessel function values at c to the values at a and b. That is, we’d like to relate the whole vector z to the vectors x and y.

Define the vector v by vk = exp(inψ) where ψ is the angle between the sides of length b and c. Then the formula we’re looking for is

zn = v-n ∑ xn+m ym wm.

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Published at DZone with permission of John Cook, DZone MVB. See the original article here.

Opinions expressed by DZone contributors are their own.


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