The name is absent

Rectilinear Drawing          67

which we write in the form

f_o'⅛(r sin χ)(cos χ cos χ - sin χ sin χ')dχ=0.

On integrating by parts the first term evidently takes the
form

1 [^aτ

-   ∕2^w (r sin χ) sin χdχ,

^rJo

while the second is seen to be equal to
d f^aτ

-j_f I fz'^'' (r sɪn χ) sin χdχ.

Hence if we write

W = X*7?⁰ (r sin χ) sin χdχ,

we conclude that

1jγ-jγ≡o,

r

i.e., W = cτ. But since /F^z(0) =0 we must have c =0, and so
F≡0.

Employing now the same argument as in the case m = 1,
for the equation W = Owe find that f∙ι'∖s) =0 and so ∕_a(j^,)≡0.

Likewise we readily infer that gs(∕)≡O.

More generally, we may obtain the stated result for
m = 1, 2 ■ ∙ ∙ in succession, by writing at the mth stage

cos mχ =cos χ cos (tn — l)χ —sin χ sin (m-l')χ
sin mχ =sin χ cos (m — l)χ+ cos χ sin (m-l)χ

in the equations involving ∕_m(j) and g_m(j∙) respectively, and
thus reducing the question to one of the type considered at
the (tn —l)st stage.

In this manner the stated result is readily established.

We are now in a position to formulate preliminary neces-
sary and sufficient conditions for a continuous solution of the
general harmonic case of the mth order.

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