The name is absent



38


The FT of function jn(r)∕ri can be easily found using following equality

^-2π∕f.r = ^(-07∙(2π∕r)yz,m(∕)^m(r),
l,m

(A.5)


which yields

⅛(r)
r3


= 4πY(-i)‰4∕) f j^-

, ,                            ⅜J ∙^∙

l,m'


Γ                 4π ʌ

J drY2,m(rWlm^ = --Y2,m(f), (A.6)

writing down explicitly

1 - 3cos20 4ττ ɔ , .

--Я3-- = -y(l-3∞*4)∙

The derivative terms can also be calculated using FT

V2<Mr) = F’1 [nv4(r)]] = -4π2<F^1 [∕2⅛(f)].             (A.7)

For practical numerical calculation, FT is implemented via discrete Fourier transform
(DFT). That is, for an array of
N uniformly sampled data hn = h(tn) (n = 0,1,2,..., N - 1),
if the sampling time-step is Δ, the frequency interval is ʌʃ = ɪ, i.e.,

I N                 N   1

f’-' (-2..... l∙ "∙' ∙-∙T.V∆'              (A'8)

The FT of ∕ι(r) is approximately calculated as

ʃv-l              N-I

dth(t)e-'f"' ≈ Y1 hke^2πif"tt A = Δ У hke-2nikn/N,            (A.9)

fc=0                     k=0

the DFT is then defined as

N-I

h^Yhke-2ikn/N,                         (A. 10)

Ic=O



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