The name is absent

74 Lectures on Scientific Subjects

illustrate this fact as simply as possible let us restrict at-
tention to the symmetric case, F = F(r), when, as we have
seen, the distribution function must also be symmetric,
∕=∕ω∙

If we consider two circles C_a and C_i of radius a, b with
a>⅛>0, concentric with the origin it is clear that the ratio
of the length of any chord of C_a to the length of the part
of the chord within C_i is at least a to b. It follows that the
amount of lead 2πffF(r, θ')rdrdθ within C_a is at least a∣b
times that within C_i, so that we must have

-[ F(r)rdr≥τ[ F(r)rdr (a>b),
aj_o           bj_o

if the drawing is to be possible without rectilinear erasures.
This conclusion may of course also be derived directly from
the fundamental equation (1) above, under the assumption
/(ʃ, φ) ≥0, the proof involving nothing more than an applica-
tion of the geometric fact about chords stated above.

APPENDED NOTE

To treat the general case we start from the equations

(1) H_m(r) =f_o^2πh_m(r sin u)e'^m,du (m =0, 1, 2, ∙ ∙ ∙)
where the H_m(r) are assumed to be continuous for r>0 and
vanish with r to the (m — l)st order. These equations may
be written

(2) H_m(r) =f* h_m(r sin u) (cos и — i sin u)e''^m''^ιudu

₌    —ɪɪ f f_lm(,-ι) (_r sin u')e^^il∙^m^^^,,du

Jo

d Γ f^2τ ,        .        '

— i-y                sin u')e'^if-^m~''^,udu ,

dτJ
where we define

(3)      h_m~∙'(s) = f°h_m(s)ds, h_mM(s) = f'h_m^(s)ds, ∙ ∙ ∙

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